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\begin_body
\begin_layout Standard
COMSW4281 Lecture Notes
\end_layout
\begin_layout Standard
Professor Anargyros Papageorgiou
\end_layout
\begin_layout Standard
Spring 2007
\end_layout
\begin_layout Section*
Lecture 1 (17 January)
\end_layout
\begin_layout Subsection*
Course Information
\end_layout
\begin_layout Standard
Course email: cs4281@columbia.edu (to contact TA, submit homeworks)
\newline
Professor
email: ap@cs
\newline
Textbook: Nielsen & Chuang
\newline
Courseworks
\newline
Grading: Homework 30%,
Midterm 30%, Final 40%
\newline
Matlab recommended for programming
\newline
Quizzes might be
given, averaged into midterm/final grades
\newline
Office Hours: Tues 1-2, Wed 12-1
\end_layout
\begin_layout Subsection*
Why Quantum Computing
\end_layout
\begin_layout Standard
1.
Moore's Law - Density of transistors increases every 18 months, which means
within 10-20 years quantum effects will become a barrier
\end_layout
\begin_layout Standard
2.
Quantum Algorithms - Factoring, Search Algorithms
\newline
\newline
Best Classical Factoring
Algorithm: Number Field
\newline
Performance: log
\begin_inset Formula $2^{C(\log N)^{1/2}}\left(\log\log N\right)^{2/3}$
\end_inset
\newline
Quantum Algorithm Discovered 1994 by Peter Shor
\newline
Performance:
\begin_inset Formula $\left(\log N\right)^{3}$
\end_inset
w/ probability O(1) (or
\begin_inset Formula $>\frac{1}{2}$
\end_inset
)
\newline
In general, quantum algorithms do not give results with certainty, but
if probablity is greater than one half, they can be repeated to get sufficient
certainty
\newline
\newline
Discrete Logarithm Problem: given numbers
\begin_inset Formula $a$
\end_inset
and
\begin_inset Formula $b=a^{s}$
\end_inset
, find s.
\newline
Quantum solution: 1 query
\begin_inset Formula $O\left(\left(\log s\right)^{2}\right)$
\end_inset
\newline
When looking at quantum performance, have to look at the number of queries
as well as number of operations.
\newline
\newline
Search Algorithm
\newline
Given binary string length
N (for some huge N), find position of subtring
\newline
Classical Lower Bound O(N),
even for randomized algorithms, no success guarantee
\newline
Quantum Algorithm O(
\begin_inset Formula $\sqrt{N}$
\end_inset
)
\newline
Quantum Algorithm only has polynomial advantage over classical, not exponential
as in factoring
\newline
Grover's Algorithm
\newline
\newline
Boolean mean, Numerical integration -
related to search, same polynomial speedup
\end_layout
\begin_layout Standard
3.
Quantum cryptography for key distribution
\newline
Can replace PKI, which assumes
that factoring is hard.
Private key distribution over a quantum channel can't be eavesdropped upon
without detection.
Earliest successes and applications of quantum computing are in this area.
\end_layout
\begin_layout Standard
4.
Study of Quantum Mechanics as a model of computation
\newline
Church-Turing Thesis
- any algorithmic computation can be simulated (efficiently) by a turing
machine (which is probabalistic).
\newline
(efficiently) - part of strong version
of thesis, weaker one leaves it out
\newline
(Thesis) - means conjecture or belief,
not a provable theorem.
\newline
(which is probalistic) - fix for thesis added in 1970s.
Solovay Strassev came up with a randomized algorithm for primality testing
which can give arbitrary levels of certainty at vastly improved efficiency
over deterministic algorithm
\newline
\newline
Deutsch 1985
\newline
Came up with quantum model of computati
on in order to produce more solid version of CT thesis.
Conjectured a Universal Quantum Computer that can simulate any physical
process.
It isn't known if this is really possible.
(His paper is in courseworks, first 3-4 pages are philosophical and accessible,
rest is more technical)
\newline
\newline
5.
Study of Entanglement.
2 qubit system that can't be seperated...
EPR state...
Quantum Teleportation...
More later in the course.
\end_layout
\begin_layout Subsection*
Quantum Systems
\end_layout
\begin_layout Standard
Size of quantum system is
\begin_inset Formula $C^{N}$
\end_inset
where N is number of particles, C is number of coordinates, set of complex
numbers used to represent a state.
\end_layout
\begin_layout Standard
Quantum states can be expressed as vectors
\newline
\begin_inset Formula $x=\left(\begin{array}{c}
x_{1}\\
x_{1}\\
\vdots\\
x_{n}\end{array}\right)\in\mathbb{C}^{N}$
\end_inset
\end_layout
\begin_layout Standard
Dirac notation is used in many cases instead of standard linear algebra
notation.
Dirac
\begin_inset Quotes eld
\end_inset
captures the transpose of a matrix
\begin_inset Quotes erd
\end_inset
(
\begin_inset Formula $X^{H}$
\end_inset
- hermetian transpose of X)
\end_layout
\begin_layout Standard
For quantum states,
\begin_inset Formula $\left\Vert x\right\Vert =1$
\end_inset
.
(
\begin_inset Formula $\left\Vert x\right\Vert $
\end_inset
- euclidean norm of X, square root of sum of squares)
\end_layout
\begin_layout Standard
States are transformed by unitary matrices
\begin_inset Formula $U_{x}$
\end_inset
.
Unitary means preserves length, that
\begin_inset Formula $UU^{H}=I$
\end_inset
, or
\begin_inset Formula $U^{-1}=U^{H}$
\end_inset
.
Examples of unitary matrices: identity matrix, any rotation matrix, and
the Hausholder matrix
\newline
(
\begin_inset Formula $P=I-2UU^{H}$
\end_inset
,
\begin_inset Formula $\left\Vert U\right\Vert =1$
\end_inset
).
Hausholder matrix has something to do with mirrors/reflection.
\end_layout
\begin_layout Subsection*
Quantum Computing Nutshell
\end_layout
\begin_layout Standard
\begin_inset Formula $X_{0}$
\end_inset
- initial states
\newline
\begin_inset Formula $Y=V_{T}=U_{3}U_{2}U_{1}X_{0}$
\end_inset
- result is unitary matrices applied to input
\end_layout
\begin_layout Standard
Complexity is
\begin_inset Formula $n\cdot T$
\end_inset
, where
\begin_inset Formula $T$
\end_inset
is number of matrices, n=log N=number of qubits, and N=length of quantum
register
\end_layout
\begin_layout Section*
Lecture 2 (22 January)
\end_layout
\begin_layout Subsection*
Quantum Computation
\end_layout
\begin_layout Standard
Start with
\begin_inset Formula $X_{0}$
\end_inset
, initial quantum state which is a vector.
Apply unitary operations,
\begin_inset Formula $U_{1}\ldots U_{t}$
\end_inset
(Unitary meaning
\begin_inset Formula $U_{j}^{H}U_{j}=I$
\end_inset
).
Cost can be measured in terms of number of qubits n, and number of operations
t.
\end_layout
\begin_layout Subsection*
Qubits
\end_layout
\begin_layout Standard
Classical computation uses bits with states 0 and 1
\newline
Quantum computation uses
qubits whose states are superpositions of states
\begin_inset Formula $\left|0\right\rangle $
\end_inset
and
\begin_inset Formula $\left|1\right\rangle $
\end_inset
.
(pronounced
\begin_inset Formula $\texttt{"}$
\end_inset
ket zero
\begin_inset Formula $\texttt{"}$
\end_inset
and
\begin_inset Formula $\texttt{"}$
\end_inset
ket one.
\begin_inset Formula $\texttt{"}$
\end_inset
)
\end_layout
\begin_layout Standard
Superposition states of qubits occur in a variety of physical systems.
Simplest example is an electron that can be in a ground state 0 and excited
state 1.
You can shine light to put electron in excited state and reduce light to
put it into ground state, or you can increase and decrease light repeatedly
to put electron into superposition state.
Other examples of superposition states occur in polarizations of photons,
and intermediate angle spins of electrons traveling through magnetic fields
\end_layout
\begin_layout Standard
Mathematical description of qubits.
Formally, a qubit is an element of a two dimensional hilbert space
\begin_inset Formula $\mathcal{H}$
\end_inset
.
\end_layout
\begin_layout Standard
A hilbert space is ___.
(couldn't make out word)
\end_layout
\begin_layout Standard
Qubit states can be represented as two complex numbers a and b, in a vector
like
\newline
\begin_inset Formula $\left(\begin{array}{c}
a\\
b\end{array}\right)\in\mathbb{C}^{2}$
\end_inset
\newline
\begin_inset Formula $\left|0\right\rangle =\left(\begin{array}{c}
1\\
0\end{array}\right)$
\end_inset
,
\begin_inset Formula $\left|1\right\rangle =\left(\begin{array}{c}
0\\
1\end{array}\right)$
\end_inset
\newline
Euclidean norms are 1 (
\begin_inset Formula $\left\Vert \left|0\right\rangle \right\Vert =1$
\end_inset
,
\begin_inset Formula $\left\Vert \left|1\right\rangle \right\Vert =1$
\end_inset
)
\newline
\begin_inset Formula $\left|0\right\rangle $
\end_inset
and
\begin_inset Formula $\left|1\right\rangle $
\end_inset
are orthonormal (
\begin_inset Formula $\left|0\right\rangle \cdot\left|1\right\rangle =0$
\end_inset
)
\end_layout
\begin_layout Subsubsection*
Dirac Notation
\end_layout
\begin_layout Standard
\begin_inset Formula $\left|x\right\rangle $
\end_inset
called
\begin_inset Quotes eld
\end_inset
ket
\begin_inset Quotes erd
\end_inset
, denotes coordinate vector
\begin_inset Formula $\left(\begin{array}{c}
x_{1}\\
x_{2}\\
\vdots\\
x_{n}\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left\langle x\right|$
\end_inset
called
\begin_inset Quotes eld
\end_inset
bra
\begin_inset Quotes erd
\end_inset
denotes
\begin_inset Formula $\left(\begin{array}{cccc}
\overline{x_{1}} & \overline{x_{2}} & \cdots & \overline{x_{n}}\end{array}\right)$
\end_inset
\newline
(where
\begin_inset Formula $\bar{n}$
\end_inset
is complex conjugate of
\begin_inset Formula $n$
\end_inset
)
\end_layout
\begin_layout Standard
\begin_inset Formula $\left\langle x|y\right\rangle $
\end_inset
called
\begin_inset Quotes eld
\end_inset
braket
\begin_inset Quotes erd
\end_inset
denotes inner product, magnitude of projection of y on to x.
\newline
\begin_inset Formula $\left\langle x|y\right\rangle =\left\langle x\right|\cdot\left|y\right\rangle =\left(\begin{array}{cccc}
\overline{x_{1}} & \overline{x_{2}} & \cdots & \overline{x_{n}}\end{array}\right)\left(\begin{array}{c}
y_{1}\\
y_{2}\\
\vdots\\
y_{n}\end{array}\right)=\sum_{j=1}^{n}\bar{x_{j}}y_{j}$
\end_inset
\newline
\begin_inset Formula $\left\langle x|x\right\rangle =\left\Vert \left|x\right\rangle \right\Vert ^{2}$
\end_inset
\end_layout
\begin_layout Standard
A qubit is a unitary vector in hilbert space.
\newline
\begin_inset Formula $\left|y\right\rangle =a\left|0\right\rangle +b\left|1\right\rangle $
\end_inset
where
\begin_inset Formula $a,b\in\mathbb{C}$
\end_inset
and
\begin_inset Formula $\left|a\right|^{2}+\left|b\right|^{2}=1$
\end_inset
\end_layout
\begin_layout Standard
A qubit is a linear combination of ket 0 and ket 1.
\newline
\begin_inset Formula $\left|y\right\rangle =a\left(\begin{array}{c}
1\\
0\end{array}\right)+b\left(\begin{array}{c}
0\\
1\end{array}\right)=\left(\begin{array}{c}
a\\
b\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
Showing qubit y is unitary.
\newline
\begin_inset Formula $\left\Vert \left|y\right\rangle \right\Vert ^{2}=\left(\begin{array}{cc}
\bar{a} & \bar{b}\end{array}\right)\left(\begin{array}{c}
a\\
b\end{array}\right)=\bar{a}a+\bar{b}b=\left|a\right|^{2}+\left|b\right|^{2}=1$
\end_inset
\end_layout
\begin_layout Standard
Showing qubit y is unitary in dirac notation.
\begin_inset Formula \begin{eqnarray*}
\left\Vert \left|y\right\rangle \right\Vert ^{2} & = & \left\langle y|y\right\rangle =\left\langle y\right|\cdot\left|y\right\rangle \\
& = & \left(\bar{a}\left\langle 0\right|+\bar{b}\left\langle 1\right|\right)\cdot\left(a\left|0\right\rangle +b\left|1\right\rangle \right)\\
& = & \bar{a}a\left\langle 0|0\right\rangle +\bar{a}b\left\langle 0|1\right\rangle +\bar{b}a\left\langle 1|0\right\rangle +\bar{b}b\left\langle 1|1\right\rangle \\
& = & \bar{a}a+\bar{b}b=\left|a\right|^{2}+\left|b\right|^{2}=1\end{eqnarray*}
\end_inset
One feature of dirac notation demonstrated here is that you can treat matrix
multiplication as regular multiplication
\end_layout
\begin_layout Subsection*
Bloch Sphere representation of Qubits
\end_layout
\begin_layout Standard
Bloch sphere lets you represent qubits in a picture
\newline
\begin_inset Formula $\left|\psi\right\rangle =\alpha\left|0\right\rangle +\beta\left|1\right\rangle $
\end_inset
\newline
Substitute
\begin_inset Formula $\alpha=\left|\alpha\right|e^{i\varphi_{\alpha}}$
\end_inset
,
\begin_inset Formula $\beta=\left|\beta\right|e^{i\varphi_{\beta}}$
\end_inset
\newline
\begin_inset Formula $\left|\psi\right\rangle =\left|\alpha\right|e^{i\varphi_{\alpha}}\left|0\right\rangle +\left|\beta\right|e^{i\varphi_{\beta}}\left|1\right\rangle $
\end_inset
\newline
\begin_inset Formula $\left|\psi\right\rangle =e^{i\varphi_{\alpha}}\left(\left|\alpha\right|\left|0\right\rangle +\left|\beta\right|e^{i\left(\varphi_{\beta}-\varphi_{\alpha}\right)}\left|1\right\rangle \right)$
\end_inset
\newline
Substitute
\begin_inset Formula $\varphi=\varphi_{\beta}-\varphi_{\alpha}$
\end_inset
,
\begin_inset Formula $\gamma=\varphi_{\alpha}$
\end_inset
for
\begin_inset Formula $\varphi\in\left[0,2\pi\right]$
\end_inset
\newline
\begin_inset Formula $\left|\psi\right\rangle =e^{i\gamma}\left(\left|\alpha\right|\left|0\right\rangle +\left|\beta\right|e^{i\varphi}\left|1\right\rangle \right)$
\end_inset
\newline
Substitute
\begin_inset Formula $\left|\alpha\right|=\cos\vartheta$
\end_inset
,
\begin_inset Formula $\left|\beta\right|=\sin\vartheta$
\end_inset
, for
\begin_inset Formula $\vartheta\in\left[0,\frac{\pi}{2}\right]$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left|\psi\right\rangle =e^{i\gamma}\left(\cos\vartheta\left|0\right\rangle +\sin\vartheta e^{i\varphi}\left|1\right\rangle \right)$
\end_inset
\newline
Substitute
\begin_inset Formula $\left|\alpha\right|=\cos\frac{\vartheta}{2}$
\end_inset
,
\begin_inset Formula $\left|\beta\right|=\sin\frac{\vartheta}{2}$
\end_inset
, for
\begin_inset Formula $\vartheta\in\left[0,\pi\right]$
\end_inset
\newline
\begin_inset Formula $\left|\psi\right\rangle =e^{i\gamma}\left(\cos\frac{\vartheta}{2}\left|0\right\rangle +\sin\frac{\vartheta}{2}e^{i\varphi}\left|1\right\rangle \right)$
\end_inset
\newline
Any qubit can be expressed this way, in terms of
\begin_inset Formula $\gamma$
\end_inset
,
\begin_inset Formula $\vartheta$
\end_inset
, and
\begin_inset Formula $\varphi$
\end_inset
.
And if you disregard
\begin_inset Formula $\gamma$
\end_inset
, (global phase which it turns out to be impossible to physically measure
anyway), you can take angles
\begin_inset Formula $\vartheta$
\end_inset
and
\begin_inset Formula $\varphi$
\end_inset
and use them to plot points on a unit sphere.
\end_layout
\begin_layout Standard
(Drawing bloch spheres
\newline
- z axis up, y axis right, x axis out
\newline
-
\begin_inset Formula $\vartheta$
\end_inset
= angle between point and z axis (latitude, but starting at north pole
and extending down)
\newline
-
\begin_inset Formula $\varphi$
\end_inset
= angle between projection of point at z=0 and x axis (longitude)
\end_layout
\begin_layout Standard
Textbook Figure 1.3, page 15)
\end_layout
\begin_layout Subsection*
Measuring Qubits
\end_layout
\begin_layout Standard
Attempting to measure a qubit in a superposition state,
\begin_inset Formula $\left|y\right\rangle =a\left|0\right\rangle +b\left|1\right\rangle $
\end_inset
, will give you state
\begin_inset Formula $\left|0\right\rangle $
\end_inset
with probability
\begin_inset Formula $\left|a\right|^{2}$
\end_inset
, and
\begin_inset Formula $\left|1\right\rangle $
\end_inset
with probability
\begin_inset Formula $\left|b\right|^{2}$
\end_inset
.
After measuring the state, the qubit collapses, it ceases to be a superposition
, and it changes to whatever state the measurement gave, either
\begin_inset Formula $\left|0\right\rangle $
\end_inset
or
\begin_inset Formula $\left|1\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
Collapse is a non-unitary transformation.
It can be considered a projection and renormalization.
\end_layout
\begin_layout Standard
Using standard linear algebra notation, you can express measurement of some
state
\begin_inset Formula $u$
\end_inset
(where
\begin_inset Formula $u\in\mathbb{C}^{N}$
\end_inset
) from a superposition state
\begin_inset Formula $x$
\end_inset
as being a projection of
\begin_inset Formula $x$
\end_inset
onto
\begin_inset Formula $u$
\end_inset
, followed by the normalization.
To compute the result of the projection, you can multiply
\begin_inset Formula $x$
\end_inset
by a matrix
\begin_inset Formula $M$
\end_inset
where
\begin_inset Formula $M=u\cdot u^{H}$
\end_inset
.
\begin_inset Formula $M$
\end_inset
is called a projection matrix and
\begin_inset Formula $Mx$
\end_inset
will be the result of projecting
\begin_inset Formula $x$
\end_inset
onto
\begin_inset Formula $u$
\end_inset
.
You don't actually need to form the matrix, however, since
\begin_inset Formula $Mx=uu^{H}x$
\end_inset
, and
\begin_inset Formula $u^{H}x$
\end_inset
can be computed as a simple dot product.
\end_layout
\begin_layout Standard
Using dirac notation, and taking a measurement of
\begin_inset Formula $\left|0\right\rangle $
\end_inset
as an example:
\begin_inset Formula \begin{eqnarray*}
M\left|x\right\rangle & = & \left|0\right\rangle \left\langle 0\right|\left|x\right\rangle \\
& = & \left|0\right\rangle \left\langle 0\right|\left(a\left|0\right\rangle +b\left|1\right\rangle \right)\\
& = & a\left|0\right\rangle \left\langle 0\right|\left|0\right\rangle +b\left|0\right\rangle \left\langle 0\right|\left|1\right\rangle \\
& = & a\left|0\right\rangle \end{eqnarray*}
\end_inset
After normalizing, final state is
\begin_inset Formula $\frac{a}{\left|a\right|}\left|0\right\rangle $
\end_inset
.
Same process can be used for a measurement of
\begin_inset Formula $\left|1\right\rangle $
\end_inset
to express final state as
\begin_inset Formula $\frac{b}{\left|b\right|}\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Section*
Lecture 3 (24 January)
\end_layout
\begin_layout Subsection*
Course Information
\end_layout
\begin_layout Standard
Midterm - Wednesday, 7 March
\newline
Midterm Review - Monday before
\end_layout
\begin_layout Subsection*
Lecture 2 Review
\end_layout
\begin_layout Standard
\begin_inset Formula $\left|y\right\rangle =a\left|0\right\rangle +b\left|1\right\rangle =e^{i\gamma}\left(\cos\frac{\vartheta}{2}\left|0\right\rangle +e^{i\varphi}\sin\frac{\vartheta}{2}\left|0\right\rangle \right)$
\end_inset
\newline
Collapse probability
\newline
\begin_inset Formula $\left|0\right\rangle $
\end_inset
,
\begin_inset Formula $p=\left|a\right|^{2}$
\end_inset
\newline
\begin_inset Formula $\left|1\right\rangle $
\end_inset
,
\begin_inset Formula $p=\left|b\right|^{2}$
\end_inset
\end_layout
\begin_layout Subsection*
Alternate Basis
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|+\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|0\right\rangle +\left|1\right\rangle \right)=\cos\left(\frac{\pi}{4}\right)\left|0\right\rangle +e^{i0}\sin\left(\frac{\pi}{4}\right)\left|1\right\rangle \\
\left|-\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|0\right\rangle -\left|1\right\rangle \right)=\cos\left(\frac{\pi}{4}\right)\left|0\right\rangle +e^{i\pi}\sin\left(\frac{\pi}{4}\right)\left|1\right\rangle \end{eqnarray*}
\end_inset
Bloch sphere representations
\newline
\begin_inset Formula $\left|+\right\rangle $
\end_inset
,
\begin_inset Formula $\vartheta=\pi/2$
\end_inset
,
\begin_inset Formula $\varphi=0$
\end_inset
, on equator at front of sphere
\newline
\begin_inset Formula $\left|-\right\rangle $
\end_inset
,
\begin_inset Formula $\vartheta=\pi/2$
\end_inset
,
\begin_inset Formula $\varphi=\pi$
\end_inset
, on equator at back of sphere
\end_layout
\begin_layout Standard
Proving orthogonality of
\begin_inset Formula $\left|+\right\rangle $
\end_inset
and
\begin_inset Formula $\left|-\right\rangle $
\end_inset
:
\begin_inset Formula \begin{eqnarray*}
\left\langle +|-\right\rangle & = & \left(\frac{1}{\sqrt{2}}\left(\left\langle 0\right|+\left\langle 1\right|\right)\right)\cdot\left(\frac{1}{\sqrt{2}}\left(\left|0\right\rangle -\left|1\right\rangle \right)\right)\\
& = & \frac{1}{2}\left(\left\langle 0|0\right\rangle -\left\langle 0|1\right\rangle +\left\langle 1|0\right\rangle -\left\langle 1|1\right\rangle \right)\\
& = & \frac{1}{2}\left(1-0+0-1\right)=0\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Proving
\begin_inset Formula $\left|+\right\rangle $
\end_inset
is unit vector (proof is similar for
\begin_inset Formula $\left|-\right\rangle $
\end_inset
):
\begin_inset Formula \begin{eqnarray*}
\left\langle +|+\right\rangle & = & \left(\frac{1}{\sqrt{2}}\left(\left\langle 0\right|+\left\langle 1\right|\right)\right)\cdot\left(\frac{1}{\sqrt{2}}\left(\left|0\right\rangle +\left|1\right\rangle \right)\right)\\
& = & \frac{1}{2}\left(\left\langle 0|0\right\rangle +\left\langle 0|1\right\rangle +\left\langle 1|0\right\rangle +\left\langle 1|1\right\rangle \right)\\
& = & \frac{1}{2}\left(1+0+0+1\right)=1\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Equivalences, change of base
\begin_inset Formula \begin{eqnarray*}
\left|0\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|+\right\rangle +\left|-\right\rangle \right)\\
\left|1\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|+\right\rangle -\left|-\right\rangle \right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Hadamard Gate
\end_layout
\begin_layout Standard
Hadamard matrix maps
\begin_inset Formula $\left|0\right\rangle $
\end_inset
to
\begin_inset Formula $\left|+\right\rangle $
\end_inset
,
\begin_inset Formula $\left|1\right\rangle $
\end_inset
to
\begin_inset Formula $\left|-\right\rangle $
\end_inset
\begin_inset Formula \[
H=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & 1\\
1 & -1\end{array}\right)\]
\end_inset
You can see it by looking at columns.
In general when looking at a transformation matrix, the first column shows
what first basis vector (
\begin_inset Formula $\left|0\right\rangle $
\end_inset
) is transformed to, second column shows second basis vector (
\begin_inset Formula $\left|1\right\rangle $
\end_inset
), and so on.
\end_layout
\begin_layout Standard
Hadamard mapping for general state
\begin_inset Formula $\left|y\right\rangle =a\left|0\right\rangle +b\left|1\right\rangle $
\end_inset
\begin_inset Formula \begin{eqnarray*}
H\left|y\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left(a+b\right)\left|0\right\rangle +\left(a-b\right)\left|1\right\rangle \right)\\
& = & \frac{1}{\sqrt{2}}\left(a\left(\left|0\right\rangle +\left|1\right\rangle \right)+b\left(\left|0\right\rangle -\left|1\right\rangle \right)\right)\\
& = & a\left|+\right\rangle +b\left|-\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Hadamard transform is unitary and is it's own inverse.
\end_layout
\begin_layout Standard
Example:
\begin_inset Formula \begin{eqnarray*}
H\left|0\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|0\right\rangle +\left|1\right\rangle \right)\\
HH\left|0\right\rangle & = & \frac{1}{\sqrt{2}}\left(H\left|0\right\rangle +H\left|1\right\rangle \right)=\frac{1}{2}\left(\left|0\right\rangle +\left|1\right\rangle +\left|0\right\rangle -\left|1\right\rangle \right)=\left|0\right\rangle \end{eqnarray*}
\end_inset
The two half-
\begin_inset Formula $\left|0\right\rangle $
\end_inset
kets adding up is called constructive interference.
The two half-
\begin_inset Formula $\left|1\right\rangle $
\end_inset
kets cancelling out is called destructive interference.
\end_layout
\begin_layout Subsection*
Other Gates
\end_layout
\begin_layout Standard
Pauli Matrix:
\begin_inset Formula \[
X=\left(\begin{array}{cc}
0 & 1\\
1 & 0\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
X is analogous to NOT gate:
\begin_inset Formula \begin{eqnarray*}
X\left|0\right\rangle & = & \left|1\right\rangle \\
X\left|1\right\rangle & = & \left|0\right\rangle \\
X\left|y\right\rangle & = & a\left|1\right\rangle +b\left|0\right\rangle \end{eqnarray*}
\end_inset
in Dirac notation:
\begin_inset Formula \begin{eqnarray*}
X & = & \left|1\right\rangle \left\langle 0\right|+\left|0\right\rangle \left\langle 1\right|\\
& = & \left(\begin{array}{c}
0\\
1\end{array}\right)\left(\begin{array}{cc}
1 & 0\end{array}\right)+\left(\begin{array}{c}
1\\
0\end{array}\right)\left(\begin{array}{cc}
0 & 1\end{array}\right)\\
& = & \left(\begin{array}{cc}
0 & 0\\
1 & 0\end{array}\right)+\left(\begin{array}{cc}
0 & 1\\
0 & 0\end{array}\right)=\left(\begin{array}{cc}
0 & 1\\
1 & 0\end{array}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
In general, when you have a unitary operation U and you know how it transforms
basis states:
\newline
\begin_inset Formula $U\left|0\right\rangle =\left|s\right\rangle $
\end_inset
,
\begin_inset Formula $U\left|1\right\rangle =\left|t\right\rangle $
\end_inset
, you can express U as:
\begin_inset Formula \[
U=\left|s\right\rangle \left\langle 0\right|+\left|t\right\rangle \left\langle 1\right|\]
\end_inset
Verification:
\begin_inset Formula \begin{eqnarray*}
U\left|0\right\rangle & = & \left|s\right\rangle \left\langle 0|0\right\rangle +\left|t\right\rangle \left\langle 1|0\right\rangle =\left|s\right\rangle \\
U\left|1\right\rangle & = & \left|s\right\rangle \left\langle 0|1\right\rangle +\left|t\right\rangle \left\langle 1|1\right\rangle =\left|t\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Linear Algebra Review
\end_layout
\begin_layout Standard
Given U is unitary matrix,
\begin_inset Formula $\left\langle y|y\right\rangle =1$
\end_inset
,
\begin_inset Formula $U\left|y\right\rangle =\lambda\left|y\right\rangle $
\end_inset
.
Eigenvalues
\begin_inset Formula $\lambda\in\mathbb{C}$
\end_inset
can be expressed as
\begin_inset Formula $\lambda=e^{it}$
\end_inset
,
\begin_inset Formula $t\in\mathbb{R}$
\end_inset
.
\end_layout
\begin_layout Standard
Show unitary matrix has eigenvalues of unit length:
\begin_inset Formula \begin{eqnarray*}
\left\langle y\right|U^{H} & = & \bar{\lambda}\left\langle y\right|\\
\left\langle y|U^{H}U|y\right\rangle & = & \bar{\lambda}\lambda\left\langle y|y\right\rangle \\
1 & = & \left|\lambda\right|^{2}\end{eqnarray*}
\end_inset
Because they have unit length, eigenvalues can be written in the form
\begin_inset Formula $\lambda=e^{it}.$
\end_inset
\end_layout
\begin_layout Standard
Show a hermitian matrix
\begin_inset Formula $A^{H}=A$
\end_inset
has eigenvalues that are real numbers:
\begin_inset Formula \begin{eqnarray*}
A\left|y\right\rangle & = & \lambda\left|y\right\rangle \\
\left\langle y\right|A\left|y\right\rangle & = & \lambda\left\langle y|y\right\rangle =\lambda\\
\left\langle y\right|A^{H} & = & \bar{\lambda}\left\langle y\right|\\
\left\langle y\right|A^{H}\left|y\right\rangle & = & \bar{\lambda}\left\langle y|y\right\rangle =\bar{\lambda}\end{eqnarray*}
\end_inset
\begin_inset Formula $A=A^{H}$
\end_inset
therefore
\begin_inset Formula $\lambda=\bar{\lambda}$
\end_inset
therefore
\begin_inset Formula $\lambda\in\mathbb{R}$
\end_inset
\end_layout
\begin_layout Subsubsection*
Spectral Theorem
\end_layout
\begin_layout Standard
Unitary matrix U can be diagnalized as
\begin_inset Formula $U=V\Lambda V^{H}$
\end_inset
where
\begin_inset Formula \[
\Lambda=\left(\begin{array}{cccc}
\lambda_{1} & 0 & 0 & 0\\
0 & \lambda_{1} & 0 & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & 0 & \lambda_{n}\end{array}\right)\]
\end_inset
and
\begin_inset Formula $V=\left(\begin{array}{cccc}
Y_{1} & Y_{2} & \cdots & Y_{n}\end{array}\right)$
\end_inset
, columns
\begin_inset Formula $Y_{i}$
\end_inset
are eigenvectors.
\end_layout
\begin_layout Standard
The spectral theorem applies more generally to any Normal matrix.
Normal matrices commute with their transpose,
\begin_inset Formula $UU^{H}=U^{H}U$
\end_inset
.
\end_layout
\begin_layout Standard
(Matrix types
\newline
Normal -
\begin_inset Formula $U^{H}U=UU^{H}$
\end_inset
\newline
Unitary -
\begin_inset Formula $U^{H}U=UU^{H}=I$
\end_inset
, type of normal matrix
\newline
Hermetian -
\begin_inset Formula $U^{H}=U$
\end_inset
, type of normal matrix)
\end_layout
\begin_layout Subsection*
Pauli Matrices
\end_layout
\begin_layout Standard
\begin_inset Formula \[
X=\left(\begin{array}{cc}
0 & 1\\
1 & 0\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Eigenvalues are
\begin_inset Formula $\lambda_{1}=1$
\end_inset
,
\begin_inset Formula $\lambda_{2}=-1$
\end_inset
.
Eigenvectors:
\begin_inset Formula \begin{eqnarray*}
\left|+\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|0\right\rangle +\left|1\right\rangle \right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1\\
1\end{array}\right)\\
\left|-\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|0\right\rangle -\left|1\right\rangle \right)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1\\
-1\end{array}\right)\end{eqnarray*}
\end_inset
Called X matrix because on Bloch Sphere, eigenvectors are aligned with X
axis.
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
Y & = & \left(\begin{array}{cc}
0 & -i\\
i & 0\end{array}\right)\\
Y\left|0\right\rangle & = & i\left|1\right\rangle \\
Y\left|1\right\rangle & = & -i\left|0\right\rangle \\
Y\left|y\right\rangle & = & ia\left|1\right\rangle -ib\left|0\right\rangle \end{eqnarray*}
\end_inset
Eigenvectors are
\end_layout
\begin_layout Standard
\begin_inset Formula $|\otimes\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1\\
i\end{array}\right)=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle +i\left|1\right\rangle \right)$
\end_inset
\begin_inset Formula $ $
\end_inset
,
\begin_inset Formula $\vartheta=\frac{\pi}{2}$
\end_inset
,
\begin_inset Formula $\varphi=\frac{\pi}{2}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $|\oplus\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1\\
-i\end{array}\right)=\frac{1}{\sqrt{2}}\left(\left|0\right\rangle -i\left|1\right\rangle \right)$
\end_inset
,
\begin_inset Formula $\vartheta=\frac{\pi}{2}$
\end_inset
,
\begin_inset Formula $\varphi=\frac{3\pi}{2}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
Z & = & \left(\begin{array}{cc}
1 & 0\\
0 & -1\end{array}\right)\\
Z\left|y\right\rangle & = & a\left|0\right\rangle -b\left|1\right\rangle =a\left|0\right\rangle -e^{i\pi}b\left|1\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Section*
Lecture 4 (29 Janary)
\end_layout
\begin_layout Subsection*
Lecture 3 Review
\end_layout
\begin_layout Standard
Hadamard matrix
\begin_inset Formula \[
H=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & 1\\
1 & -1\end{array}\right)\]
\end_inset
Pauli Matrices
\begin_inset Formula \begin{eqnarray*}
X & = & \left(\begin{array}{cc}
0 & 1\\
1 & 0\end{array}\right)\\
Y & = & \left(\begin{array}{cc}
0 & -i\\
i & 0\end{array}\right)\\
Z & = & \left(\begin{array}{cc}
1 & 0\\
0 & -1\end{array}\right)=\left(\begin{array}{cc}
1 & 0\\
0 & e^{i\pi}\end{array}\right)\end{eqnarray*}
\end_inset
First column of each matrix tells you where
\begin_inset Formula $\left|0\right\rangle $
\end_inset
maps, second column tells you where
\begin_inset Formula $\left|1\right\rangle $
\end_inset
maps.
\end_layout
\begin_layout Subsection*
New Gates
\end_layout
\begin_layout Standard
Phase gate
\begin_inset Formula \begin{eqnarray*}
S & = & \left(\begin{array}{cc}
1 & 0\\
0 & i\end{array}\right)=\left(\begin{array}{cc}
1 & 0\\
0 & e^{i\frac{\pi}{2}}\end{array}\right)\\
S^{2} & = & X\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Pi over 8 gate
\begin_inset Formula \[
T=\left(\begin{array}{cc}
1 & 0\\
0 & e^{i\frac{\pi}{4}}\end{array}\right)=e^{i\frac{\pi}{8}}\left(\begin{array}{cc}
e^{-i\frac{\pi}{8}} & 0\\
0 & e^{i\frac{\pi}{8}}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Homework tip: We are allowed to cite eigenvalues from class.
We can also guess and verify other eigenvalues without going through characteri
stic equations.
Also, for diagonal matrices, eigenvalues can be read off the diagonal and
a eigenvectors are
\begin_inset Formula $\left|0\right\rangle $
\end_inset
and
\begin_inset Formula $\left|1\right\rangle $
\end_inset
.
\end_layout
\begin_layout Subsection*
Multiple Qubit Systems
\end_layout
\begin_layout Standard
Starting with 2 qubits.
2 qubits mean 4 base states:
\begin_inset Formula $\left|00\right\rangle $
\end_inset
,
\begin_inset Formula $\left|01\right\rangle $
\end_inset
,
\begin_inset Formula $\left|10\right\rangle $
\end_inset
,
\begin_inset Formula $\left|11\right\rangle $
\end_inset
or, equivalently
\begin_inset Formula $\left|0\right\rangle \left|0\right\rangle $
\end_inset
,
\begin_inset Formula $\left|0\right\rangle \left|1\right\rangle $
\end_inset
,
\begin_inset Formula $\left|1\right\rangle \left|0\right\rangle $
\end_inset
,
\begin_inset Formula $\left|1\right\rangle \left|1\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
An arbitrary state is a superposition
\begin_inset Formula \[
\left|y\right\rangle =a_{00}\left|00\right\rangle +a_{01}\left|01\right\rangle +a_{10}\left|10\right\rangle +a_{11}\left|11\right\rangle \]
\end_inset
constrained by
\begin_inset Formula $\sum_{j,k}\left|a_{jk}\right|^{2}=1$
\end_inset
,
\begin_inset Formula $a_{jk}\in\mathbb{C}$
\end_inset
\end_layout
\begin_layout Standard
Measurement of any outcome j occurs with probability
\begin_inset Formula $\left|a_{j}\right|^{2}$
\end_inset
.
Following measurement
\begin_inset Formula $\left|y\right\rangle $
\end_inset
collapses to
\begin_inset Formula $\left|j\right\rangle $
\end_inset
.
It is also possible to make partial measurements, measuring some qubits
but not others, taking sums as you would expect.
(The probabilty of making the partial measurement is just the sum of probabilit
ies of the full measurements containing the partial measurement.
Following measurement, base states that aren't compatible with the measurement
have their coefficients set to 0 and the rest are renormalized.)
\end_layout
\begin_layout Standard
Combining qubit states
\begin_inset Formula \begin{eqnarray*}
\left|y_{1}\right\rangle & = & a_{1}\left|0\right\rangle +b_{1}\left|1\right\rangle \\
\left|y_{2}\right\rangle & = & a_{2}\left|0\right\rangle +b_{2}\left|1\right\rangle \end{eqnarray*}
\end_inset
You can combine two independent qubit states to get a state describing the
probability of each joint outcome
\begin_inset Formula \begin{eqnarray*}
\left|y_{1}y_{2}\right\rangle & = & \left|y_{1}\right\rangle \left|y_{2}\right\rangle =\left(a_{1}\left|0\right\rangle +b_{1}\left|1\right\rangle \right)\left(a_{2}\left|0\right\rangle +b_{2}\left|1\right\rangle \right)\\
& = & a_{1}\left|0\right\rangle a_{2}\left|0\right\rangle +a_{1}\left|0\right\rangle b_{2}\left|1\right\rangle +b_{1}\left|1\right\rangle a_{2}\left|0\right\rangle +b_{1}\left|1\right\rangle b_{2}\left|1\right\rangle \\
& = & a_{1}a_{2}\left|00\right\rangle +a_{1}b_{2}\left|01\right\rangle +b_{1}a_{2}\left|10\right\rangle +b_{1}b_{2}\left|11\right\rangle \end{eqnarray*}
\end_inset
(Note: all the products in the above equations are really tensor products,
which are defined below.
In previous lecture notes, products inside dirac expressions implied standard
matrix multiplication, which makes no sense here.)
\end_layout
\begin_layout Standard
You cannot generally break a multiple qubit state up into separate independent
states.
Example: EPR state
\begin_inset Formula $\frac{1}{\sqrt{2}}\left(\left|00\right\rangle +\left|11\right\rangle \right)$
\end_inset
.
You can see visually that there are no values for
\begin_inset Formula $a_{1}$
\end_inset
,
\begin_inset Formula $b_{1}$
\end_inset
,
\begin_inset Formula $a_{2}$
\end_inset
,
\begin_inset Formula $b_{2}$
\end_inset
that will let you write the EPR state in above form.
\end_layout
\begin_layout Standard
For an n-qubit system, each basis state can be expressed as a bitstring
\begin_inset Formula $\left|j_{n-1}j_{n-2}\cdots j_{0}\right\rangle $
\end_inset
for
\begin_inset Formula $j_{m}\in\left\{ 0,1\right\} $
\end_inset
.
Or, for convenience, it can just be written as a normal number
\begin_inset Formula $\left|j\right\rangle $
\end_inset
where
\begin_inset Formula $j=\sum_{m=0}^{n-1}2^{m}j_{m}$
\end_inset
\end_layout
\begin_layout Standard
Now that we have a notion for multiple qubit states, have to answer 3 questions:
\newline
Q1: What is the coordinate representation of a state
\begin_inset Formula $\left|j\right\rangle $
\end_inset
?
\newline
Q2: How can you decompose and recompose states? For example, how do you
compute
\begin_inset Formula $\left|y\right\rangle =\left|y_{1}\right\rangle \left|y_{2}\right\rangle $
\end_inset
where
\begin_inset Formula $\left|y_{1}\right\rangle $
\end_inset
and
\begin_inset Formula $\left|y_{2}\right\rangle $
\end_inset
are multiple qubit states.
\newline
Q3: How can you compose operations.
Given
\begin_inset Formula $\left|y_{1}\right\rangle $
\end_inset
and
\begin_inset Formula $\left|y_{2}\right\rangle $
\end_inset
which are multiple qubit states and unitary operators
\begin_inset Formula $U_{1}$
\end_inset
and
\begin_inset Formula $U_{2}$
\end_inset
which apply to
\begin_inset Formula $\left|y_{1}\right\rangle $
\end_inset
and
\begin_inset Formula $\left|y_{2}\right\rangle $
\end_inset
, respectively, how can you find a U matrix that satisfies
\begin_inset Formula $U\left|y\right\rangle =U_{1}\left|y_{1}\right\rangle U_{2}\left|y_{2}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Once you answer these questions you can start looking at algorithms.
\end_layout
\begin_layout Subsection*
Tensor Products
\end_layout
\begin_layout Standard
Tensor products are a new concept.
\end_layout
\begin_layout Standard
Given two matrices, A which has size
\begin_inset Formula $m\times n$
\end_inset
, and B which has size
\begin_inset Formula $p\times q$
\end_inset
, the tensor product
\begin_inset Formula $A\otimes B$
\end_inset
will be a huge matrix of size
\begin_inset Formula $mp\times nq$
\end_inset
:
\begin_inset Formula \[
A\otimes B=\left(\begin{array}{cccc}
a_{11}B & a_{12}B & \cdots & a_{1n}B\\
a_{21}B & a_{22}B & \text{ }\cdots & a_{2n}B\\
\vdots & \vdots & \ddots\\
a_{m1}B & a_{m2}B & & a_{mn}B\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Example:
\begin_inset Formula \begin{eqnarray*}
\left(\begin{array}{ccc}
1 & 2 & 3\\
4 & 5 & 6\end{array}\right))\otimes\left(\begin{array}{rr}
10 & 0\\
0 & 20\end{array}\right) & = & \begin{array}{ccc}
1\left(\begin{array}{rr}
10 & 0\\
0 & 20\end{array}\right) & 2\left(\begin{array}{rr}
10 & 0\\
0 & 20\end{array}\right) & 3\left(\begin{array}{rr}
10 & 0\\
0 & 20\end{array}\right)\\
4\left(\begin{array}{rr}
10 & 0\\
0 & 20\end{array}\right) & 5\left(\begin{array}{rr}
10 & 0\\
0 & 20\end{array}\right) & 6\left(\begin{array}{rr}
10 & 0\\
0 & 20\end{array}\right)\end{array}\\
& = & \left(\begin{array}{rrrrrr}
10 & 0 & 20 & 0 & 30 & 0\\
0 & 20 & 0 & 40 & 0 & 60\\
40 & 0 & 50 & 0 & 60 & 0\\
0 & 80 & 0 & 100 & 0 & 120\end{array}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Example: (using column vectors):
\newline
\begin_inset Formula $X=\left(\begin{array}{c}
1\\
2\\
3\end{array}\right)$
\end_inset
,
\begin_inset Formula $Y=\left(\begin{array}{c}
10\\
100\end{array}\right)$
\end_inset
\newline
\begin_inset Formula $X\otimes Y=\left(\begin{array}{c}
10\\
100\\
20\\
200\\
30\\
300\end{array}\right)$
\end_inset
,
\begin_inset Formula $Y\otimes X=\left(\begin{array}{c}
10\\
20\\
30\\
100\\
200\\
300\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
Example: (using qubits)
\newline
\begin_inset Formula $X=\left(\begin{array}{c}
1\\
0\end{array}\right)=\left|0\right\rangle $
\end_inset
,
\begin_inset Formula $Y=\left(\begin{array}{c}
1\\
0\end{array}\right)=\left|0\right\rangle $
\end_inset
\newline
\begin_inset Formula $X\otimes Y=\left|0\right\rangle \otimes\left|0\right\rangle =\left|0\right\rangle \left|0\right\rangle =\left|00\right\rangle =\left(\begin{array}{c}
1\\
0\\
0\\
0\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
Similarly,
\begin_inset Formula $\left|11\right\rangle =\left(\begin{array}{c}
0\\
0\\
0\\
1\end{array}\right)$
\end_inset
\newline
And more generally,
\begin_inset Formula $\left|j\right\rangle =\left|j_{1}j_{0}\right\rangle $
\end_inset
will be a column vector which is all zeros except for single
\begin_inset Formula $1$
\end_inset
entry at position
\begin_inset Formula $j+1$
\end_inset
(see
\begin_inset Quotes eld
\end_inset
Multiple Qubit Systems
\begin_inset Quotes erd
\end_inset
above,
\begin_inset Formula $j=\sum_{m=0}^{n-1}2^{m}j_{m}$
\end_inset
).
\end_layout
\begin_layout Standard
Example: (using qubits again):
\begin_inset Formula \begin{eqnarray*}
H\left|0\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|0\right\rangle +\left|1\right\rangle \right)\\
\left(H\left|0\right\rangle \right)\otimes\left(H\left|0\right\rangle \right) & = & \left(\frac{1}{\sqrt{2}}\left(\left(\begin{array}{c}
1\\
0\end{array}\right)+\left(\begin{array}{c}
0\\
1\end{array}\right)\right)\right)\otimes\left(\frac{1}{\sqrt{2}}\left(\left(\begin{array}{c}
1\\
0\end{array}\right)+\left(\begin{array}{c}
0\\
1\end{array}\right)\right)\right)\\
& = & \frac{1}{\sqrt{2}}\left(\begin{array}{c}
1\\
1\end{array}\right)\otimes\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1\\
1\end{array}\right)=\frac{1}{2}\left(\begin{array}{c}
1\\
1\\
1\\
1\end{array}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Repeat example using dirac notation:
\begin_inset Formula \begin{eqnarray*}
\left(H\left|0\right\rangle \right))\otimes(\left(H\left|0\right\rangle \right)) & = & \frac{1}{2}\left(\left|0\right\rangle +\left|1\right\rangle \right)\left(\left|0\right\rangle +\left|1\right\rangle \right)\\
& = & \frac{1}{2}\left(|00\rangle+\left|01\right\rangle +\left|10\right\rangle +\left|11\right\rangle \right)\\
& = & \frac{1}{2}\left(\left(\begin{array}{c}
1\\
0\\
0\\
0\end{array}\right)+\left(\begin{array}{c}
0\\
1\\
0\\
0\end{array}\right)+\left(\begin{array}{c}
0\\
0\\
1\\
0\end{array}\right)+\left(\begin{array}{c}
0\\
0\\
0\\
1\end{array}\right)\right)=\frac{1}{2}\left(\begin{array}{c}
1\\
1\\
1\\
1\end{array}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Associative property of tensor product:
\begin_inset Formula \[
\left|x\right\rangle \otimes\left|y\right\rangle \otimes\left|z\right\rangle =\left|x\right\rangle \otimes\left|yz\right\rangle =\left|xy\right\rangle \otimes\left|z\right\rangle =\left|xyz\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Notation for tensor exponentiation:
\begin_inset Formula \[
\left|\psi\right\rangle ^{\otimes k}=\left|\psi\right\rangle \otimes\left|\psi\right\rangle \cdots\otimes\left|\psi\right\rangle \]
\end_inset
\end_layout
\begin_layout Subsection*
Powers of
\begin_inset Formula $H\left|0\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left(H\left|0\right\rangle \right)^{\otimes k}=\frac{1}{2^{k/2}}\left(\begin{array}{c}
1\\
1\\
\vdots\\
1\end{array}\right)\]
\end_inset
where length of column vector is
\begin_inset Formula $2^{k}$
\end_inset
.
\end_layout
\begin_layout Standard
The formula is obvious for
\begin_inset Formula $k=1$
\end_inset
.
Proof for rest of cases is by induction, (base case
\begin_inset Formula $k=2$
\end_inset
was shown in previous section.) Inductive case is:
\begin_inset Formula \begin{eqnarray*}
\left(\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\right)^{\otimes k} & = & \left(\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\right)^{\otimes k-1}\otimes\left(\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\right)\\
& = & \frac{1}{2^{(k-1)/2}}\left(\begin{array}{c}
1\\
1\\
\vdots\\
1\end{array}\right)\otimes\frac{1}{\sqrt{2}}\left(\begin{array}{c}
1\\
1\end{array}\right)=\frac{1}{2^{k/2}}\left(\begin{array}{c}
1\\
1\\
\vdots\\
1\end{array}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Tensor Products Continued
\end_layout
\begin_layout Standard
[DUNNO] I'm not exactly sure what the following equations are supposed to
show.
I think I was lost at the time I was taking these notes.
\begin_inset Formula \[
\left|j\right\rangle =\left|j_{n-1}j_{n-2}\cdots j_{0}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Now make an arbitary split at qubit k.
\begin_inset Formula \begin{eqnarray*}
\left|j\right\rangle & = & \left|j_{n-1}\cdots j_{k}\right\rangle \left|j_{k-1}\cdots j_{0}\right\rangle \\
& = & \left|j^{(1)}\right\rangle \left|j^{(2)}\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
j & = & \sum_{m=0}^{n-1}2^{m}j_{m}\\
j^{(1)} & = & \sum_{m=k}^{n-1}2^{m-k}j_{m}\\
j^{(2)} & = & \sum_{m=0}^{k-1}2^{m}j_{m}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Known n=1, n=2.
\begin_inset Formula \begin{eqnarray*}
\left|j\right\rangle & = & \left|j^{(1)}\right\rangle \left|j^{(2)}\right\rangle \\
& = & \left(\begin{array}{c}
0\\
\vdots\\
1\\
\vdots\\
0\end{array}\right)\otimes\left(\begin{array}{c}
0\\
\vdots\\
1\\
\vdots\\
0\end{array}\right)=\left(\begin{array}{c}
0\\
\vdots\\
\vdots\\
1\\
\vdots\\
\vdots\\
0\end{array}\right)\end{eqnarray*}
\end_inset
\begin_inset Formula $\left|j^{(1)}\right\rangle $
\end_inset
has 1 at position
\begin_inset Formula $j^{(1)}+1$
\end_inset
\newline
\begin_inset Formula $\left|j^{(2)}\right\rangle $
\end_inset
has 1 at position
\begin_inset Formula $j^{(2)}+1$
\end_inset
\newline
\begin_inset Formula $\left|j\right\rangle $
\end_inset
has 1 at position
\begin_inset Formula $j+1=j^{(1)}\cdot2^{k}+j^{(2)}+1$
\end_inset
\end_layout
\begin_layout Subsection*
New Course Information
\end_layout
\begin_layout Standard
The class now has a TA: James Li (sp?)
\end_layout
\begin_layout Section*
Lecture 5 (31 January)
\end_layout
\begin_layout Subsection*
Lecture 4 Review
\end_layout
\begin_layout Standard
In n-dimensional qubit systems, basis vectors are
\begin_inset Formula $\left(\begin{array}{c}
0\\
\vdots\\
1\\
\vdots\\
0\end{array}\right)=\left|j\right\rangle $
\end_inset
with the 1 at position
\begin_inset Formula $j+1$
\end_inset
.
\end_layout
\begin_layout Standard
Any state
\begin_inset Formula $\left|j\right\rangle $
\end_inset
is a linear combination of j vectors.
\end_layout
\begin_layout Standard
Any outcome
\begin_inset Formula $\left|j\right\rangle $
\end_inset
has probability
\begin_inset Formula $\left|c_{j}\right|^{2}$
\end_inset
for
\begin_inset Formula $\left|y\right\rangle =\sum_{j=0}^{2^{n-1}}c_{j}\left|j\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
First homework assignment is out today, 3 problems, due 14 Feb.
\end_layout
\begin_layout Standard
Showed last time that
\begin_inset Formula \[
\left(H\left|0\right\rangle \right)^{\otimes k}=\left(\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\right)^{\otimes k}=\frac{1}{2^{k/2}}\left(\begin{array}{c}
1\\
\vdots\\
1\end{array}\right)\]
\end_inset
producing column vector of
\begin_inset Formula $2^{k}$
\end_inset
rows.
Will be generalizing today for inputs other than
\begin_inset Formula $\left|0\right\rangle $
\end_inset
.
\end_layout
\begin_layout Subsection*
Properties of Tensor Products
\end_layout
\begin_layout Standard
1.
Associative property (see lecture 4)
\newline
2.
Distributing scalar multiple over tensor product
\newline
\begin_inset Formula $a\left(\left|x\right\rangle \otimes\left|y\right\rangle \right)=\left(a\left|x\right\rangle \right)\otimes\left|y\right\rangle =\left|x\right\rangle \otimes\left(a\left|y\right\rangle \right)$
\end_inset
\newline
3.
Distributing tensor product over addition
\newline
\begin_inset Formula $\left|y\right\rangle \otimes\left(\left|x_{1}\right\rangle +\left|x_{2}\right\rangle \right)=\left|y\right\rangle \otimes\left|x_{1}\right\rangle +\left|0y\right\rangle \otimes\left|x_{2}\right\rangle $
\end_inset
\newline
4.
Applying tensor products of operations individually to vectors.
\newline
\begin_inset Formula $\left(A\otimes B\right)\left(\left|x\right\rangle \otimes\left|y\right\rangle \right)=\left(A\left|x\right\rangle \right)\otimes\left(B\left|y\right\rangle \right)$
\end_inset
\end_layout
\begin_layout Standard
Example:
\begin_inset Formula \begin{eqnarray*}
\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)^{\otimes k} & = & H\left|0\right\rangle \otimes\cdots\otimes H\left|0\right\rangle \\
& = & \left(H\otimes\cdots\otimes H\right)\left(\left|0\right\rangle \otimes\cdots\otimes\left|0\right\rangle \right)\\
& = & H^{\otimes k}\left|0\right\rangle ^{\otimes k}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Example:
\begin_inset Formula \[
\left(X\otimes S\right)\left|00\right\rangle =X\left|0\right\rangle \otimes S\left|0\right\rangle \]
\end_inset
(X is NOT gate from lecture 3, S is phase gate from lecture 4)
\newline
Lets you apply
transformations to individual inputs instead of applying huge combination
operations with big matrices.
\end_layout
\begin_layout Standard
Example:
\begin_inset Formula \[
\left(A\otimes B\right)\left(\sum_{j}a_{j}\left|x_{j}\right\rangle \left|y_{j}\right\rangle \right)=\sum_{j}a_{j}\left(A\left|x_{j}\right\rangle \right)\otimes\left(B\left|y_{j}\right\rangle \right)\]
\end_inset
6.
Distributing Hermitian transpose over tensor product
\newline
\begin_inset Formula $\left(A\otimes B\right)^{H}=A^{H}\otimes B^{H}$
\end_inset
\newline
7.
Applying tensor product of operations to other operations.
\newline
\begin_inset Formula $\left(A\otimes B\right)\left(C\otimes D\right)=AC\otimes BD$
\end_inset
\newline
Proof can be stated in terms of property 4, just break down C and D into
individual column vectors
\begin_inset Formula $\left|x\right\rangle $
\end_inset
and
\begin_inset Formula $\left|y\right\rangle $
\end_inset
.
This is a useful technique in general, substituting vectors to figure out
how matrices work.
\newline
8.
If A and B are unitary operations, then
\begin_inset Formula $A\otimes B$
\end_inset
is unitary.
\newline
Proof:
\begin_inset Formula $A^{H}A=I$
\end_inset
,
\begin_inset Formula $B^{H}B=I$
\end_inset
\newline
\begin_inset Formula $\left(A\otimes B\right)^{H}\left(A\otimes B\right)=\left(A^{H}\otimes B^{H}\right)\left(A\otimes B\right)=\left(A^{H}A\right)\otimes\left(B^{H}B\right)=I\otimes I=I$
\end_inset
\end_layout
\begin_layout Subsection*
Powers of Hadamard Gate
\end_layout
\begin_layout Standard
Result of applying hadamard transforms to each qubit in some base (non-superposi
tion) state
\begin_inset Formula $j$
\end_inset
, made of
\begin_inset Formula $k$
\end_inset
qubits.
\begin_inset Formula \begin{eqnarray*}
H^{\otimes k}\left|j_{k-1}\cdots j_{0}\right\rangle & = & H\left|j_{k-1}\right\rangle \cdots H\left|j_{0}\right\rangle \\
& = & \frac{\left|0\right\rangle +\left(-1\right)^{j_{k-1}}\left|1\right\rangle }{\sqrt{2}}\otimes\cdots\otimes\frac{\left|0\right\rangle +\left(-1\right)^{j_{0}}\left|1\right\rangle }{\sqrt{2}}\\
& = & \bigotimes_{s=k-1}^{0}\frac{\left|0\right\rangle +\left(-1\right)^{j_{s}}\left|1\right\rangle }{\sqrt{2}}\\
& = & \frac{1}{2^{k/2}}\sum_{m=0}^{2^{k-1}}\left(-1\right)^{j\cdot m}\left|m\right\rangle \end{eqnarray*}
\end_inset
Where
\begin_inset Formula $j\cdot m=j_{k-1}m_{k-1}+j_{k-2}m_{k-2}+\cdots+j_{0}m_{0}$
\end_inset
.
\begin_inset Formula $j_{i},m_{i}\in\left\{ 0,1\right\} $
\end_inset
are digits of
\begin_inset Formula $j$
\end_inset
and
\begin_inset Formula $m$
\end_inset
expressed as binary strings.
\end_layout
\begin_layout Standard
(Observe that for some single qubit state,
\begin_inset Formula $j_{s}$
\end_inset
,
\begin_inset Formula $H|j_{s}\rangle=\frac{\left|0\right\rangle +\left(-1\right)^{j_{s}}\left|1\right\rangle }{\sqrt{2}}$
\end_inset
to see the intuition in the last step.
At
\begin_inset Formula $k=2$
\end_inset
,
\begin_inset Formula \begin{eqnarray*}
& & \frac{\left|0\right\rangle +\left(-1\right)^{j_{1}}\left|1\right\rangle }{\sqrt{2}}\otimes\frac{\left|0\right\rangle +\left(-1\right)^{j_{0}}\left|1\right\rangle }{\sqrt{2}}\\
& & =\left|00\right\rangle +\left(-1\right)^{j_{0}}\left|01\right\rangle +\left(-1\right)^{j_{1}}\left|10\right\rangle +\left(-1\right)^{j_{0}+j_{1}}\left|11\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
At
\begin_inset Formula $k=3$
\end_inset
,
\begin_inset Formula \begin{eqnarray*}
& & \frac{\left|0\right\rangle +\left(-1\right)^{j_{2}}\left|1\right\rangle }{\sqrt{2}}\otimes\frac{\left|0\right\rangle +\left(-1\right)^{j_{1}}\left|1\right\rangle }{\sqrt{2}}\otimes\frac{\left|0\right\rangle +\left(-1\right)^{j_{0}}\left|1\right\rangle }{\sqrt{2}}\\
& & =\left|000\right\rangle +\left(-1\right)^{j_{0}}\left|001\right\rangle +\left(-1\right)^{j_{1}}\left|010\right\rangle +\left(-1\right)^{j_{0}+j}\left|011\right\rangle \\
& & +\left(-1\right)^{j_{2}}\left|100\right\rangle +\left(-1\right)^{j_{0}+j_{2}}\left|101\right\rangle +\left(-1\right)^{j_{1}+j_{2}}\left|110\right\rangle +\left(-1\right)^{j_{0}+j_{1}+j_{2}}\left|111\right\rangle \end{eqnarray*}
\end_inset
You can see that
\begin_inset Formula $\left(-1\right)^{j_{i}}$
\end_inset
coefficients follow the
\begin_inset Formula $\left|1\right\rangle $
\end_inset
's and get included in the expanded terms where
\begin_inset Formula $m_{i}$
\end_inset
is 1 instead of 0.)
\end_layout
\begin_layout Standard
For two qubit system, inputs
\begin_inset Formula $\left|+\right\rangle $
\end_inset
,
\begin_inset Formula $\left|+\right\rangle $
\end_inset
:
\begin_inset Formula \[
\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\otimes\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\rightarrow\begin{array}{c}
\left|00\right\rangle \\
\left|01\right\rangle \\
\left|10\right\rangle \\
\left|11\right\rangle \end{array}\]
\end_inset
For
\begin_inset Formula $n+1$
\end_inset
qubit system, inputs:
\begin_inset Formula $\left|+\right\rangle $
\end_inset
,
\begin_inset Formula $\left|y\right\rangle $
\end_inset
:
\begin_inset Formula $\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\left|y\right\rangle =\frac{\left|0y\right\rangle +\left|1y\right\rangle }{\sqrt{2}}$
\end_inset
(has
\begin_inset Formula $n+1$
\end_inset
qubits if
\begin_inset Formula $\left|y\right\rangle $
\end_inset
has
\begin_inset Formula $n$
\end_inset
qubits)
\end_layout
\begin_layout Standard
Solution is then made up of terms like
\begin_inset Formula $\left|m_{k-1}m_{k-2}\cdots m_{0}\right\rangle $
\end_inset
\newline
if
\begin_inset Formula $m_{k-2}=1$
\end_inset
then term has
\begin_inset Formula $\left|1\right\rangle $
\end_inset
at second cubit, but may be + or -
\newline
if
\begin_inset Formula $j_{k-2}=1$
\end_inset
then it's
\begin_inset Formula $-\left|1\right\rangle $
\end_inset
else if
\begin_inset Formula $j_{k-2}=0$
\end_inset
then
\begin_inset Formula $\left|1\right\rangle $
\end_inset
\newline
if
\begin_inset Formula $m_{k-2}=0$
\end_inset
or
\begin_inset Formula $j_{k-2}=0$
\end_inset
then no problem with + or - since we have 0 at location k-2
\end_layout
\begin_layout Standard
Summary:
\begin_inset Formula $\left(-1\right)^{j_{k-2}m_{k-2}}\left|?\mbox{(0 or 1)}???\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Repeat for all qubits to get
\begin_inset Formula $\left(-1\right)^{j_{k-1}m_{k-1}+\cdots+j_{0}m_{0}}\left|m_{k-1}\cdots m_{0}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Upshot:
\begin_inset Formula $H^{\otimes k}\left|j\right\rangle =\frac{1}{2^{k/2}}\sum_{m=0}^{2^{k-1}}\left(-1\right)^{m\cdot j}\left|m\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Homework hint: This is half of work needed to solve one of the problems.
It tells you columns of the matrix.
Homework asks you to find the whole matrix.
\end_layout
\begin_layout Subsection*
Properties of Tensor Products (continued)
\end_layout
\begin_layout Standard
9.
If you have two states
\begin_inset Formula $\left|x\right\rangle $
\end_inset
and
\begin_inset Formula $\left|y\right\rangle $
\end_inset
which you can decompose with same dimensions.
\newline
\begin_inset Formula $\left|y\right\rangle =\left|y_{1}\right\rangle \left|y_{2}\right\rangle $
\end_inset
\newline
\begin_inset Formula $\left|x\right\rangle =\left|x_{1}\right\rangle \left|x_{2}\right\rangle $
\end_inset
\newline
Then
\begin_inset Formula $\left\langle x|y\right\rangle =\left\langle x_{1}x_{2}|y_{1}y_{2}\right\rangle =\left\langle x_{1}|y_{1}\right\rangle \left\langle x_{2}|y_{2}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Example:
\newline
\begin_inset Formula $\left|d\right\rangle =|011000\rangle$
\end_inset
\newline
\begin_inset Formula $\left|k\right\rangle =|011010\rangle$
\end_inset
\newline
You can tell
\begin_inset Formula $\left\langle d|k\right\rangle =0$
\end_inset
based solely on the fact that the fifth qubit's inner product (
\begin_inset Formula $\left\langle 0|1\right\rangle $
\end_inset
) is zero.
\end_layout
\begin_layout Standard
Example:
\newline
\begin_inset Formula $\left\langle z_{1}\right|\left\langle z_{2}\right|X\otimes Y\left|\psi_{1}\right\rangle \left|\psi_{2}\right\rangle =\left\langle z_{1}\right|X\left|\psi_{1}\right\rangle \otimes\left\langle z_{2}\right|Y\left|\psi_{2}\right\rangle $
\end_inset
\end_layout
\begin_layout Subsection*
Completeness Relation
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|j\right\rangle \left\langle i\right|=\left(\begin{array}{c}
0\\
\vdots\\
1\\
\vdots\\
0\end{array}\right)\left(\begin{array}{ccccc}
0 & \cdots & 1 & \cdots & 0\end{array}\right)=\left(\begin{array}{ccccc}
0 & \cdots & \cdots & \cdots & 0\\
\vdots & \ddots & & .\cdot & \vdots\\
\vdots & & 1 & & \vdots\\
\vdots & .\cdot & & \ddots & \vdots\\
0 & \cdots & \cdots & \cdots & 0\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left|j\right\rangle $
\end_inset
is column vector with 1 at position j+1
\newline
\begin_inset Formula $\left\langle i\right|$
\end_inset
is row vector with 1 at position i+i
\newline
\begin_inset Formula $\left|j\right\rangle \left\langle i\right|$
\end_inset
is projection matrix with 1 at row j+1, col i+i
\end_layout
\begin_layout Standard
Completeness relation
\begin_inset Formula \[
\sum_{i=o}^{2^{n-1}}\left|i\right\rangle \left\langle i\right|=I\]
\end_inset
\end_layout
\begin_layout Standard
Next lecture will show completeness relation holds on any orthonormal basis,
not just
\begin_inset Formula $i$
\end_inset
.
\end_layout
\begin_layout Section*
Lecture 6 (5 February)
\end_layout
\begin_layout Subsection*
Lecture 5 Review
\end_layout
\begin_layout Standard
1.
\begin_inset Formula $H^{\otimes k}\left|j\right\rangle =\frac{1}{2^{k/2}}\sum_{m=0}^{2^{k-1}}\left(-1\right)^{j\cdot m}\left|m\right\rangle $
\end_inset
\newline
2.
\begin_inset Formula $\left\langle x_{1}\right|\otimes\left\langle x_{2}\right|\cdot\left|y_{1}\right\rangle \otimes\left|y_{2}\right\rangle =\left\langle x_{1}|y_{1}\right\rangle \otimes\left\langle x_{2}|y_{2}\right\rangle $
\end_inset
\newline
Example:
\newline
\begin_inset Formula $\left\langle 01\right|X\otimes A\left|01\right\rangle =\left\langle 0\right|X\left|0\right\rangle \otimes\left\langle 1\right||A\left|1\right\rangle $
\end_inset
\newline
3.
\begin_inset Formula $I=\sum_{j=0}^{2^{n-1}}\left|j\right\rangle \left\langle j\right|$
\end_inset
\end_layout
\begin_layout Subsection*
Completeness Relation
\end_layout
\begin_layout Standard
A proof of the completeness relation (3 above) was shown last lecture based
on adding up projections to get the diagonal matrix.
This is another proof:
\end_layout
\begin_layout Standard
An arbitrary state is a linear combination of basis states:
\begin_inset Formula \[
\left|y\right\rangle =\sum_{j}c_{j}\left|x_{j}\right\rangle \]
\end_inset
Each constant is just:
\begin_inset Formula \[
c_{j}=\left\langle x_{j}|y\right\rangle \in\mathbb{C}\]
\end_inset
Substituting
\begin_inset Formula $c_{j}$
\end_inset
above gives:
\begin_inset Formula \[
\left|y\right\rangle =\sum_{j}\left|x_{j}\right\rangle \left\langle x_{j}|y\right\rangle \]
\end_inset
In general
\begin_inset Formula \[
\left|y\right\rangle =A\left|y\right\rangle \rightarrow A=I\]
\end_inset
So
\begin_inset Formula \[
\sum_{j}\left|x_{j}\right\rangle \left\langle x_{j}\right|=I\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $A=AI=A\sum_{j}\left|x_{j}\right\rangle \left\langle x_{j}\right|=\sum_{j}\left(A\left|x_{j}\right\rangle \right)\left\langle x_{j}\right|$
\end_inset
\end_layout
\begin_layout Subsection*
More Linear Algebra
\end_layout
\begin_layout Standard
An
\begin_inset Formula $n\times n$
\end_inset
matrix has
\begin_inset Formula $n$
\end_inset
eigenvalues (
\begin_inset Formula $\lambda_{i}\in\mathbb{C}$
\end_inset
) and orthonormal eigenvectors (
\begin_inset Formula $\left|x_{j}\right\rangle $
\end_inset
\begin_inset Formula $j=\left\{ 1\ldots n\right\} $
\end_inset
) iff A is normal (
\begin_inset Formula $A^{H}A=AA^{H}$
\end_inset
).
\end_layout
\begin_layout Standard
\begin_inset Formula $V=\left(\begin{array}{ccc}
\left|x_{1}\right\rangle & \cdots & \left|x_{n}\right\rangle \end{array}\right)$
\end_inset
, eigenvector matrix,
\begin_inset Formula $V^{H}V=I$
\end_inset
\newline
\begin_inset Formula $\Lambda=\left(\begin{array}{cccc}
\lambda_{1} & 0 & 0 & 0\\
0 & \lambda_{2} & 0 & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & 0 & \lambda_{n}\end{array}\right)$
\end_inset
, eigenvalue matrix
\newline
\begin_inset Formula \begin{eqnarray*}
A & = & V\Lambda V^{H}=\left(\begin{array}{ccc}
\lambda_{1}\left|x_{1}\right\rangle & \ldots & \lambda_{n}\left|x_{n}\right\rangle \end{array}\right)\left(\begin{array}{c}
\left\langle x_{1}\right|\\
\vdots\\
\left\langle x_{n}\right|\end{array}\right)\\
& = & \sum_{j=1}^{n}\lambda_{j}\left|x_{j}\right\rangle \left\langle x_{j}\right|\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Eigenvalues and Tensor Products:
\end_layout
\begin_layout Standard
If
\begin_inset Formula $A$
\end_inset
eigenvalues are
\begin_inset Formula $\lambda_{j}$
\end_inset
, eigenvectors are
\begin_inset Formula $\left|x_{j}\right\rangle $
\end_inset
, for j=1..n,
\newline
and
\begin_inset Formula $B$
\end_inset
eigenvalues are
\begin_inset Formula $\lambda_{k}$
\end_inset
, eigenvectors are
\begin_inset Formula $\left|x_{k}\right\rangle $
\end_inset
, for k=1..n
\newline
then
\begin_inset Formula $A\otimes B$
\end_inset
eigenvalues are
\begin_inset Formula $\lambda_{j}$
\end_inset
\begin_inset Formula $\lambda_{k}$
\end_inset
, eigenvectors are
\begin_inset Formula $\left|x_{j}\right\rangle $
\end_inset
\begin_inset Formula $\left|x_{k}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
For any matrix A, you can compute
\begin_inset Formula $A^{2}$
\end_inset
,
\begin_inset Formula $A^{3}$
\end_inset
,
\begin_inset Formula $A^{4}$
\end_inset
\newline
If A is nonsingular, you can compute
\begin_inset Formula $A^{-1}$
\end_inset
,
\begin_inset Formula $A^{-2}$
\end_inset
\newline
What about a general f: D
\begin_inset Formula $\rightarrow$
\end_inset
C, like sin(A),
\begin_inset Formula $e^{A}$
\end_inset
, f(A)
\newline
If A is normal and f(
\begin_inset Formula $\lambda_{j}$
\end_inset
) is well defined for all eigenvalues
\begin_inset Formula \begin{eqnarray*}
f(A) & = & \sum_{j}f\left(\lambda_{j}\right)\left|x_{j}\right\rangle \left\langle x_{j}\right|\\
& = & V\left(\begin{array}{cccc}
f\left(\lambda_{1}\right) & 0 & 0 & 0\\
0 & f\left(\lambda_{2}\right) & 0 & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & 0 & f\left(\lambda_{n}\right)\end{array}\right)V^{H}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Example:
\newline
\begin_inset Formula $f(z)=z^{k}$
\end_inset
\newline
\begin_inset Formula $f(A)=A^{k}=(V\Lambda V^{H})(V\Lambda V^{H})\cdots(V\Lambda V^{H})$
\end_inset
[multiplied k times]
\newline
\begin_inset Formula $=V\Lambda^{k}V^{H}$
\end_inset
\end_layout
\begin_layout Standard
If
\begin_inset Formula $A\in\mathbb{R}^{n,n}$
\end_inset
and
\begin_inset Formula $A=A^{T}$
\end_inset
then
\begin_inset Formula $e^{iA}$
\end_inset
is unitary.
Proof:
\begin_inset Formula \begin{eqnarray*}
\left(e^{iA}\right)^{H}\left(e^{iA}\right) & = & \left(\sum_{j}e^{i\lambda_{j}}\left|x_{j}\right\rangle \left\langle x_{j}\right|\right)^{H}\left(\sum_{j}e^{i\lambda_{j}}\left|x_{j}\right\rangle \left\langle x_{j}\right|\right)\\
& = & \left(\sum_{j}e^{-i\lambda_{j}}\left|x_{j}\right\rangle \left\langle x_{j}\right|\right)\left(\sum_{j}e^{\text{i$\lambda$}_{j}}\left|x_{j}\right\rangle \left\langle x_{j}\right|\right)\\
& = & \sum_{j,k}e^{-i\lambda_{j}}e^{i\lambda_{k}}\left|x_{j}\right\rangle \left\langle x_{j}\right|\left|x_{k}\right\rangle \left\langle x_{k}\right|\\
& = & \sum_{j}1\left|x_{j}\right\rangle \left\langle x_{j}\right|=I\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Controlled Gates
\end_layout
\begin_layout Standard
Controlled gates have a control input and a normal input.
Control output is always the same as the control input.
If control input is
\begin_inset Formula $\left|0\right\rangle $
\end_inset
, normal output is the same as the normal input.
If control input
\begin_inset Formula $\left|1\right\rangle $
\end_inset
, normal output is some function of the normal input, where the function
depends on the type of controlled gate.
\end_layout
\begin_layout Subsubsection*
Controlled Not Gate
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|00\right\rangle & \rightarrow & \left|00\right\rangle \\
\left|01\right\rangle & \rightarrow & \left|01\right\rangle \\
\left|10\right\rangle & \rightarrow & \left|11\right\rangle \\
\left|11\right\rangle & \rightarrow & \left|10\right\rangle \end{eqnarray*}
\end_inset
\begin_inset Formula \[
Q_{\text{CNOT}}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle \left|i\oplus j\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
Q_{\text{CNOT}}=\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\end{array}\right)=\left(\begin{array}{cc}
I & 0\\
0 & X\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Each column of matrix is derived directly by writing CNOT mappings listed
above for
\begin_inset Formula $\left|00\right\rangle $
\end_inset
,
\begin_inset Formula $\left|01\right\rangle $
\end_inset
,
\begin_inset Formula $\left|10\right\rangle $
\end_inset
, and
\begin_inset Formula $\left|11\right\rangle $
\end_inset
inputs.
Matrix can be shown to be unitary by multipling with its conjugate transpose
and getting the identity matrix.
\end_layout
\begin_layout Standard
CNOT for Hadamard inputs:
\begin_inset Formula \begin{eqnarray*}
\left|+\right\rangle \left|+\right\rangle & \rightarrow & \left|+\right\rangle \left|+\right\rangle \\
\left|-\right\rangle \left|+\right\rangle & \rightarrow & \left|-\right\rangle \left|+\right\rangle \\
\left|+\right\rangle \left|-\right\rangle & \rightarrow & \left|-\right\rangle \left|-\right\rangle \\
\left|-\right\rangle \left|-\right\rangle & \rightarrow & \left|+\right\rangle \left|-\right\rangle \end{eqnarray*}
\end_inset
When dealing with inputs that aren't 0 and 1, calling the same input the
\begin_inset Quotes eld
\end_inset
control input
\begin_inset Quotes erd
\end_inset
doesn't neccessarily make sense.
When superposition states are fed to the CNOT, as above, the state of the
second qubit controls a NOT operation on the first.
\end_layout
\begin_layout Standard
Formally:
\begin_inset Formula \[
Q_{\text{CNOT}}H\left|i\right\rangle H\left|j\right\rangle =H\left|i\oplus j\right\rangle H\left|j\right\rangle \]
\end_inset
for
\begin_inset Formula $i,j\in0,1$
\end_inset
\end_layout
\begin_layout Standard
Proof:
\begin_inset Formula \begin{eqnarray*}
Q_{\text{CNOT}}\left(H|i\rangle H|j\rangle\right) & = & Q_{\text{CNOT}}\left(\left(\frac{\left|0\right\rangle +\left(-1\right)^{i}\left|1\right\rangle }{\sqrt{2}}\right)\left(\frac{\left|0\right\rangle +\left(-1\right)^{j}\left|1\right\rangle }{\sqrt{2}}\right)\right)\\
& = & Q_{\text{CNOT}}\left(\frac{1}{2}\left(\left|00\right\rangle +\left(-1\right)^{j}\left|01\right\rangle +\left(-1\right)^{i}\left|10\right\rangle +\left(-1\right)^{i+j}\left|11\right\rangle \right)\right)\\
& = & \frac{1}{2}\left(\left|00\right\rangle +\left(-1\right)^{j}\left|01\right\rangle +\left(-1\right)^{i}\left|11\right\rangle +\left(-1\right)^{i+j}\left|10\right\rangle \right)\\
& = & \frac{1}{2}\left(\left|00\right\rangle +\left(-1\right)^{j}\left|01\right\rangle +\left(-1\right)^{i+j}\left|10\right\rangle +\left(-1\right)^{i}\left|11\right\rangle \right)\\
& = & \frac{1}{2}\left(\left|00\right\rangle +\left(-1\right)^{j}\left|01\right\rangle +\left(-1\right)^{i+j}\left|10\right\rangle +\left(-1\right)^{i+j+j}\left|11\right\rangle \right)\\
& = & \left(\frac{\left|0\right\rangle +\left(-1\right)^{i+j}\left|1\right\rangle }{\sqrt{2}}\right)\left(\frac{\left|0\right\rangle +\left(-1\right)^{j}\left|1\right\rangle }{\sqrt{2}}\right)\\
& = & H\left|i\oplus j\right\rangle H\left|j\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Example circuit:
\begin_inset Formula \[
\left(H\otimes H\right)\left(Q_{\text{CNOT}}\right)\left(H\otimes H\right)\left|i\right\rangle \left|j\right\rangle \]
\end_inset
\begin_inset Formula \[
\left|i\right\rangle \left|j\right\rangle \xrightarrow{H\otimes H}H\left|i\right\rangle H\left|j\right\rangle \xrightarrow{Q_{\text{CNOT}}}H\left|i\oplus j\right\rangle H\left|j\right\rangle \xrightarrow{H\otimes H}H^{\otimes2}\left|i\oplus j\right\rangle H^{\otimes2}\left|j\right\rangle =\left|i\oplus j\right\rangle \left|j\right\rangle \]
\end_inset
\end_layout
\begin_layout Subsubsection*
Controlled-U Gate
\end_layout
\begin_layout Standard
Notation looks like
\begin_inset Formula $\left|i\right\rangle \left|j\right\rangle \xrightarrow{Q_{C-U}}\left|i\right\rangle U^{i}\left|j\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|00\right\rangle & \rightarrow & \left|00\right\rangle \\
\left|01\right\rangle & \rightarrow & \left|01\right\rangle \\
\left|10\right\rangle & \rightarrow & \left|1\right\rangle Q\left|0\right\rangle \\
\left|11\right\rangle & \rightarrow & \left|1\right\rangle Q\left|1\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
Q_{C-U}=\left(\begin{array}{cc}
I & 0\\
0 & U\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Section*
Lecture 7 (7 February)
\end_layout
\begin_layout Subsection*
Lecture 6 Review
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }(\left|A\right\rangle \left|B\right\rangle )=Q_{CNOT}\left|i\right\rangle \left|j\right\rangle \]
\end_inset
If
\begin_inset Formula $i$
\end_inset
and
\begin_inset Formula $j$
\end_inset
are defined on the computational basis, output is
\begin_inset Formula $\left|i\right\rangle \left|i\oplus j\right\rangle $
\end_inset
, but output in terms of some other set of basis states will be expressed
differently.
\end_layout
\begin_layout Subsection*
Controlled Z Gate
\end_layout
\begin_layout Standard
\begin_inset Formula \[
Z=\left(\begin{array}{rr}
1 & 0\\
0 & -1\end{array}\right)\]
\end_inset
\begin_inset Formula \[
Q_{Z}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle Z^{i}\left|j\right\rangle \]
\end_inset
\begin_inset Formula \[
\begin{array}{l}
Z\left|0\right\rangle =\left|0\right\rangle \\
Z\left|1\right\rangle =-\left|1\right\rangle \end{array}\Rightarrow Z\left|j\right\rangle =\left(-1\right)^{j}\left|j\right\rangle \]
\end_inset
\begin_inset Formula \[
Q_{Z}\left|i\right\rangle \left|j\right\rangle =\left(-1\right)^{ij}\left|i\right\rangle \left|j\right\rangle \]
\end_inset
The gate with the control reversed,
\begin_inset Formula $Z^{j}\left|i\right\rangle \left|j\right\rangle $
\end_inset
, going through the steps above, has the exact same definition,
\begin_inset Formula $\left(-1\right)^{ij}\left|i\right\rangle \left|j\right\rangle $
\end_inset
.
So for the controlled Z gate, it is reasonable to think of either input
as being the controlling input on the computational basis.
\end_layout
\begin_layout Subsection*
Swap Gate
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }\left(Q_{CNOT}\right)\left({Q'}_{CNOT}\right)\left(Q_{CNOT}\right)\left|i\right\rangle \left|j\right\rangle \]
\end_inset
where
\begin_inset Formula $Q_{CNOT}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle X^{i}\left|j\right\rangle $
\end_inset
and
\begin_inset Formula ${Q'}_{CNOT}\left|i\right\rangle \left|j\right\rangle =X^{j}\left|i\right\rangle \left|j\right\rangle $
\end_inset
)
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|i\right\rangle \left|j\right\rangle ^{\underrightarrow{Q_{CNOT}}}\left|i\right\rangle \left|i\oplus j\right\rangle {}^{\underrightarrow{{Q'}_{CNOT}}}\left|i\oplus i\oplus j=j\right\rangle \left|i\oplus j\right\rangle {}^{\underrightarrow{Q_{CNOT}}}\left|j\right\rangle \left|i\right\rangle \]
\end_inset
\end_layout
\begin_layout Subsubsection*
For arbitary states:
\begin_inset Formula \begin{eqnarray*}
\left|x\right\rangle \left|y\right\rangle & = & \left(a\left|0\right\rangle +b\left|1\right\rangle \right)\left(c\left|0\right\rangle +d\left|1\right\rangle \right)\\
& = & ac\left|00\right\rangle +ad\left|01\right\rangle +bc\left|10\right\rangle +bd\left|11\right\rangle \\
Swap\left|x\right\rangle \left|y\right\rangle & = & ac\left|00\right\rangle +ad\left|10\right\rangle +bc\left|01\right\rangle +bd\left|11\right\rangle \\
& = & a\left(c\left|0\right\rangle +d\left|1\right\rangle \right)\left|0\right\rangle +b\left(c\left|0\right\rangle +d\left|1\right\rangle \right)\left|1\right\rangle \\
& = & \left(a\left|0\right\rangle +b\left|1\right\rangle \right)\left(c\left|0\right\rangle +d\left|1\right\rangle \right)\\
& = & \left|y\right\rangle \left|x\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
1-Qubit Gates as Rotations
\end_layout
\begin_layout Standard
(See also section 4.2 of the textbook)
\end_layout
\begin_layout Standard
\begin_inset Formula \[
X=\left(\begin{array}{rr}
0 & 1\\
1 & 0\end{array}\right),\quad Y=\left(\begin{array}{rr}
0 & -i\\
i & 0\end{array}\right),\quad Z=\left(\begin{array}{rr}
1 & 0\\
0 & -1\end{array}\right)\]
\end_inset
Implementation of all 1-qubit gates can be seen as counterclockwise rotations
by an angle
\begin_inset Formula $\vartheta$
\end_inset
, of points plotted on the Bloch sphere around the X, Y, and Z axes.
\begin_inset Formula \begin{eqnarray*}
e^{-i\vartheta X/2} & = & R_{X}\left(\vartheta\right)=\cos\left(\frac{\vartheta}{2}\right)I-i\sin\left(\frac{\vartheta}{2}\right)X=\left(\begin{array}{rr}
\cos\left(\frac{\vartheta}{2}\right) & -i\sin\left(\frac{\vartheta}{2}\right)\\
-i\sin\left(\frac{\vartheta}{2}\right) & \cos\left(\frac{\vartheta}{2}\right)\end{array}\right)\\
e^{-i\vartheta Y/2} & = & R_{Y}\left(\vartheta\right)=\cos\left(\frac{\vartheta}{2}\right)I-i\sin\left(\frac{\vartheta}{2}\right)Y=\left(\begin{array}{rr}
\cos\left(\frac{\vartheta}{2}\right) & -\sin\left(\frac{\vartheta}{2}\right)\\
\sin\left(\frac{\vartheta}{2}\right) & \cos\left(\frac{\vartheta}{2}\right)\end{array}\right)\\
e^{-i\vartheta Z/2} & = & R_{Z}\left(\vartheta\right)=\cos\left(\frac{\vartheta}{2}\right)I-i\sin\left(\frac{\vartheta}{2}\right)Z=\left(\begin{array}{lr}
e^{-i\vartheta/2} & 0\\
0 & e^{i\vartheta/2}\end{array}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Proof of the equations above will be done in two parts.
First, by deriving the matrices from rotations around the axes, and then
by showing that the exponential forms are equivalent.
\end_layout
\begin_layout Standard
One general note to make here is that any matrix of the form
\begin_inset Formula $\left(\begin{array}{rr}
e^{ia} & 0\\
0 & e^{ib}\end{array}\right)$
\end_inset
can be expressed as a rotation around the Z axis by some angle.
Just observe:
\begin_inset Formula \[
\left(\begin{array}{rr}
e^{ia} & 0\\
0 & e^{ib}\end{array}\right)=e^{i\frac{a+b}{2}}\left(\begin{array}{rr}
e^{i\frac{a-b}{2}} & 0\\
0 & e^{i\frac{b-a}{2}}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Rotation around X axis
\end_layout
\begin_layout Standard
Take an arbitrary qubit
\begin_inset Formula $\left|y\right\rangle =\cos\frac{\vartheta_{1}}{2}\left|0\right\rangle +e^{i\varphi}\sin\frac{\vartheta_{1}}{2}\left|1\right\rangle $
\end_inset
.
Trying to find an expression for this qubit after a rotation about the
X axis is messy.
But for the case where
\begin_inset Formula $\varphi=\frac{\pi}{2}$
\end_inset
, finding the rotation is easy, and using that case is sufficient for finding
the general rotation matrix.
So let
\begin_inset Formula \[
\left|\widetilde{y}\right\rangle =\cos\frac{\vartheta_{1}}{2}\left|0\right\rangle +e^{i\pi/2}\sin\frac{\vartheta_{1}}{2}\left|1\right\rangle =\cos\frac{\vartheta_{1}}{2}\left|0\right\rangle +i\sin\frac{\vartheta_{1}}{2}\left|1\right\rangle \]
\end_inset
This is a projection of
\begin_inset Formula $\left|y\right\rangle $
\end_inset
onto the YZ plane (x=0).
Rotating this point
\begin_inset Formula $\vartheta$
\end_inset
radians counterclockwise around the X axis is the same thing as subtracting
\begin_inset Formula $\vartheta$
\end_inset
from
\begin_inset Formula $\vartheta_{1}$
\end_inset
.
The rotated point is
\begin_inset Formula \[
\left|\widetilde{y}'\right\rangle =\cos\frac{\vartheta_{1}-\vartheta}{2}\left|0\right\rangle +i\sin\frac{\vartheta_{1}-\vartheta}{2}\left|1\right\rangle \]
\end_inset
In terms of a rotation matrix, the mapping from
\begin_inset Formula $\left|\widetilde{y}\right\rangle $
\end_inset
to
\begin_inset Formula $\left|\widetilde{y}'\right\rangle $
\end_inset
looks like:
\begin_inset Formula \[
\left(\begin{array}{rr}
a & b\\
c & d\end{array}\right)\left(\begin{array}{r}
\cos\frac{\vartheta_{1}}{2}\\
i\sin\frac{\vartheta_{1}}{2}\end{array}\right)=\left(\begin{array}{r}
\cos\frac{\vartheta_{1}-\vartheta}{2}\\
i\sin\frac{\vartheta_{1}-\vartheta}{2}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Using the angle sum identies:
\begin_inset Formula \begin{eqnarray*}
\sin\left(x+y\right) & = & \sin x\cos y+\cos x\sin y\\
\cos\left(x+y\right) & = & \cos x\cos y-\sin x\sin y\end{eqnarray*}
\end_inset
on the right hand side, and matrix multiplication on the left hand side
gives:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left(\begin{array}{r}
a\cos\frac{\vartheta_{1}}{2}+ib\sin\frac{\vartheta_{1}}{2}\\
c\cos\frac{\vartheta_{1}}{2}+id\sin\frac{\vartheta_{1}}{2}\end{array}\right)=\left(\begin{array}{r}
\cos\frac{\vartheta_{1}}{2}\cos\frac{\vartheta}{2}+\sin\frac{\vartheta_{1}}{2}\sin\frac{\vartheta}{2}\\
i\sin\frac{\vartheta_{1}}{2}\cos\frac{\vartheta}{2}-i\cos\frac{\vartheta_{1}}{2}\sin\frac{\vartheta}{2}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Comparing the two sides of this equation gives you the following solution:
\begin_inset Formula \begin{eqnarray*}
a & = & \cos\frac{\vartheta}{2}\\
b & = & -\sin\frac{\vartheta}{2}\\
c & = & -i\sin\frac{\vartheta}{2}\\
d & = & \cos\frac{\vartheta}{2}\end{eqnarray*}
\end_inset
So
\begin_inset Formula \[
R_{X}\left(\vartheta\right)=\left(\begin{array}{rr}
a & b\\
c & d\end{array}\right)=\left(\begin{array}{rr}
\cos\left(\frac{\vartheta}{2}\right) & -i\sin\left(\frac{\vartheta}{2}\right)\\
-i\sin\left(\frac{\vartheta}{2}\right) & \cos\left(\frac{\vartheta}{2}\right)\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Rotation around Z axis
\end_layout
\begin_layout Standard
Again, start with an arbitrary qubit,
\begin_inset Formula $\left|y\right\rangle =\cos\frac{\vartheta}{2}\left|0\right\rangle +e^{i\varphi}\sin\frac{\vartheta}{2}\left|1\right\rangle $
\end_inset
.
Taking
\begin_inset Formula $\vartheta=\frac{\pi}{2}$
\end_inset
projects
\begin_inset Formula $\left|y\right\rangle $
\end_inset
onto the XY plane (on the equator at z=0).
\begin_inset Formula \[
\left|\widetilde{y}\right\rangle =\frac{1}{\sqrt{2}}\left|0\right\rangle +e^{i\varphi}\frac{1}{\sqrt{2}}\left|1\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Rotating that point by
\begin_inset Formula $\vartheta$
\end_inset
radians counterclockwise gives:
\begin_inset Formula \[
\left|\widetilde{y}'\right\rangle =\frac{1}{\sqrt{2}}\left|0\right\rangle +e^{i(\varphi+\vartheta)}\frac{1}{\sqrt{2}}\left|1\right\rangle \]
\end_inset
Expressed as a rotation matrix:
\begin_inset Formula \[
\left(\begin{array}{rr}
a & b\\
c & d\end{array}\right)\left(\begin{array}{r}
\frac{1}{\sqrt{2}}\\
e^{i\varphi}\frac{1}{\sqrt{2}}\end{array}\right)=\left(\begin{array}{r}
\frac{1}{\sqrt{2}}\\
e^{i(\varphi+\vartheta)}\frac{1}{\sqrt{2}}\end{array}\right)\]
\end_inset
Simplifying:
\begin_inset Formula \[
\left(\begin{array}{r}
a+be^{i\varphi}\\
c+de^{i\varphi}\end{array}\right)=\left(\begin{array}{r}
1\\
e^{i\varphi_{1}}e^{i\vartheta}\end{array}\right)\]
\end_inset
\begin_inset Formula \[
R_{Z}\left(\vartheta\right)=\left(\begin{array}{rr}
a & b\\
c & d\end{array}\right)=\left(\begin{array}{rr}
1 & 0\\
0 & e^{i\vartheta}\end{array}\right)=e^{i\vartheta/2}\left(\begin{array}{lr}
e^{-i\vartheta/2} & 0\\
0 & e^{i\vartheta/2}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Rotation around Y axis
\end_layout
\begin_layout Standard
This was left as an exercise and not covered in the lecture.
But, taking a qubit on the XZ plane so
\begin_inset Formula $\varphi=0$
\end_inset
gives:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|\widetilde{y}\right\rangle =\cos\frac{\vartheta_{1}}{2}\left|0\right\rangle +\sin\frac{\vartheta_{1}}{2}\left|1\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Rotating that point by
\begin_inset Formula $\vartheta$
\end_inset
radians counterclockwise gives:
\begin_inset Formula \[
\left|\widetilde{y}'\right\rangle =\cos\frac{\vartheta_{1}+\vartheta}{2}\left|0\right\rangle +\sin\frac{\vartheta_{1}+\vartheta}{2}\left|1\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Expressed using a rotation matrix this is:
\begin_inset Formula \[
\left(\begin{array}{rr}
a & b\\
c & d\end{array}\right)\left(\begin{array}{r}
\cos\frac{\vartheta_{1}}{2}\\
\sin\frac{\vartheta_{1}}{2}\end{array}\right)=\left(\begin{array}{r}
\cos\frac{\vartheta_{1}+\vartheta}{2}\\
\sin\frac{\vartheta_{1}+\vartheta}{2}\end{array}\right)\]
\end_inset
Which becomes:
\begin_inset Formula \[
\left(\begin{array}{r}
a\cos\frac{\vartheta_{1}}{2}+b\sin\frac{\vartheta_{1}}{2}\\
c\cos\frac{\vartheta_{1}}{2}+d\sin\frac{\vartheta_{1}}{2}\end{array}\right)=\left(\begin{array}{r}
\cos\frac{\vartheta_{1}}{2}\cos\frac{\vartheta}{2}-\sin\frac{\vartheta_{1}}{2}\sin\frac{\vartheta}{2}\\
\sin\frac{\vartheta_{1}}{2}\cos\frac{\vartheta}{2}+\cos\frac{\vartheta_{1}}{2}\sin\frac{\vartheta}{2}\end{array}\right)\]
\end_inset
So
\begin_inset Formula \[
R_{Y}\left(\vartheta\right)=\left(\begin{array}{rr}
a & b\\
c & d\end{array}\right)=\left(\begin{array}{rr}
\cos\frac{\vartheta}{2} & -\sin\frac{\vartheta}{2}\\
\sin\frac{\vartheta}{2} & \cos\frac{\vartheta}{2}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Subsection*
Rotations as Exponentials of Pauli Matrices
\end_layout
\begin_layout Standard
Theorem 1 (exercise 4.2): If
\begin_inset Formula $A^{2}=I$
\end_inset
then
\begin_inset Formula $e^{ixA}=\cos xI-i\sin x$
\end_inset
A
\end_layout
\begin_layout Standard
Proof (using Taylor series expansions from calculus):
\begin_inset Formula \begin{eqnarray*}
e^{-ixA} & = & \sum_{k=0}^{\infty}\frac{i^{k}x^{k}A^{k}}{k!}=\sum_{k\textrm{ odd}}^{\infty}\frac{i^{k}x^{k}A^{k}}{k!}+\sum_{k\textrm{ even}}^{\infty}\frac{i^{k}x^{k}A^{k}}{k!}\\
e^{-ixA} & = & \sum_{k=0}^{\infty}\frac{i^{\left(2k+1\right)}x^{\left(2k+1\right)}A^{\left(2k+1\right)}}{\left(2k+1\right)!}+\sum_{k=0}^{\infty}\frac{i^{\left(2k\right)}x^{\left(2k\right)}A^{\left(2k\right)}}{\left(2k\right)!}\\
& = & iA\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}x^{\left(2k+1\right)}}{\left(2k+1\right)!}+I\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}x^{\left(2k\right)}}{\left(2k\right)!}\\
& = & iA\sin x+I\cos x\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Section*
Lecture 8 (12 February)
\end_layout
\begin_layout Subsection*
Z-Y Decomposition
\end_layout
\begin_layout Standard
Theorem 2: Any 1-qubit gate can be decomposed as
\begin_inset Formula \[
U=e^{i\alpha}R_{Z}\left(\beta\right)R_{Y}\left(\gamma\right)R_{Z}\left(\delta\right)\]
\end_inset
(This is theorem 4.1 in textbook)
\end_layout
\begin_layout Standard
Proof:
\end_layout
\begin_layout Standard
Start by expressing U as:
\begin_inset Formula \[
U=\left(\begin{array}{cc}
X_{11}e^{i\varphi_{11}} & X_{12}e^{i\varphi_{12}}\\
X_{21}e^{i\varphi_{21}} & X_{22}e^{i\varphi_{22}}\end{array}\right)\]
\end_inset
Because U is orthonormal, vector norm of rows and columns is 1, so:
\begin_inset Formula \begin{eqnarray*}
X_{11}^{2}+X_{12}^{2} & = & 1\\
X_{21}^{2}+X_{22}^{2} & = & 1\\
X_{11}^{2}+X_{21}^{2} & = & 1\\
X_{12}^{2}+X_{22}^{2} & = & 1\end{eqnarray*}
\end_inset
Begin decomposing U by pulling out Z rotations (note that in general, post-multi
plying with a diagonal matrix gives you multiples of the original columns,
pre-multiplying with diagonal matrix gives you multiples of the original
rows):
\begin_inset Formula \begin{eqnarray*}
U & = & \left(\begin{array}{cc}
X_{11} & X_{12}\\
X_{21}e^{i\left(\varphi_{21}-\varphi_{11}\right)} & X_{22}e^{i\left(\varphi_{22}{-\varphi}_{12}\right)}\end{array}\right)\left(\begin{array}{cc}
e^{i\varphi_{11}} & 0\\
0 & e^{i\varphi_{12}}\end{array}\right)\\
& = & \left(\begin{array}{cc}
1 & 0\\
0 & e^{i\left(\varphi_{21}-\varphi_{11}\right)}\end{array}\right)\left(\begin{array}{cl}
X_{11} & X_{12}\\
X_{21} & X_{22}e^{i\left(\varphi_{22}{-\varphi}_{12}-\left(\varphi_{21}-\varphi_{11}\right)\right)}\end{array}\right)\left(\begin{array}{cc}
e^{i\varphi_{11}} & 0\\
0 & e^{i\varphi_{12}}\end{array}\right)\end{eqnarray*}
\end_inset
The two outer matrices are Z rotations, and middle matrix can also be expressed
as rotations, but different kinds of rotations, depending on the values
inside.
\end_layout
\begin_layout Standard
Case 1:
\begin_inset Formula $X_{11}=0\quad\Longrightarrow\quad X_{12}^{2}=1\quad\Longrightarrow\quad X_{22}=0\quad\Longrightarrow\quad X_{21}^{2}=1$
\end_inset
\end_layout
\begin_layout Standard
Case 1.1:
\begin_inset Formula $X_{12}$
\end_inset
and
\begin_inset Formula $X_{21}$
\end_inset
have the same sign.
In this case, the matrix is the product of Y and Z rotations:
\begin_inset Formula \[
\pm\left(\begin{array}{rr}
0 & 1\\
1 & 0\end{array}\right)=\pm\left(\begin{array}{rr}
0 & -1\\
1 & 0\end{array}\right)\left(\begin{array}{rr}
1 & 0\\
0 & -1\end{array}\right)=\pm R_{Y}\left(\pi\right)Z\]
\end_inset
\end_layout
\begin_layout Standard
Case 1.2:
\begin_inset Formula $X_{12}$
\end_inset
and
\begin_inset Formula $X_{21}$
\end_inset
have different signs.
In this case, the matrix is a Y rotation:
\begin_inset Formula \[
\pm\left(\begin{array}{rr}
0 & -1\\
1 & 0\end{array}\right)=\pm R_{Y}\left(\pi\right)\]
\end_inset
\end_layout
\begin_layout Standard
Case 2:
\begin_inset Formula $X_{12}=0\quad\Longrightarrow\quad X_{22}^{2}=1\quad\Longrightarrow\quad X_{21}=0\quad\Longrightarrow\quad X_{11}^{2}=1$
\end_inset
\end_layout
\begin_layout Standard
Case 2.1:
\begin_inset Formula $X_{11}$
\end_inset
and
\begin_inset Formula $X_{22}$
\end_inset
have the same sign.
In this case the matrix is just identity:
\begin_inset Formula \[
\pm\left(\begin{array}{rr}
1 & 0\\
0 & 1\end{array}\right)=\pm I\]
\end_inset
\end_layout
\begin_layout Standard
Case 2.2:
\begin_inset Formula $X_{11}$
\end_inset
and
\begin_inset Formula $X_{22}$
\end_inset
have different signs.
In this case the matrix is Z:
\begin_inset Formula \[
\pm\left(\begin{array}{rr}
1 & 0\\
0 & -1\end{array}\right)=\pm Z\]
\end_inset
\end_layout
\begin_layout Standard
Case 3: All
\begin_inset Formula $X_{ij}\neq0$
\end_inset
.
Letting
\begin_inset Formula ${\varphi=\varphi}_{22}{-\varphi}_{12}-\left(\varphi_{21}-\varphi_{11}\right)$
\end_inset
, because the matrix is unitary we know:
\begin_inset Formula \begin{eqnarray*}
\left(\begin{array}{cc}
X_{11} & X_{21}\end{array}\right)\left(\begin{array}{c}
X_{12}\\
X_{22}e^{i\varphi}\end{array}\right) & = & 0\\
X_{11}X_{12}+X_{21}X_{22}e^{i\varphi} & = & 0\end{eqnarray*}
\end_inset
In order for that to be true, the expression
\begin_inset Formula $e^{i\varphi}=\cos\varphi+i\sin\varphi$
\end_inset
cannot have an imaginary component, so
\begin_inset Formula $\sin\varphi=0$
\end_inset
and
\begin_inset Formula $\varphi=k\pi$
\end_inset
.
\end_layout
\begin_layout Standard
[DUNNO: The fact that the matrix is unitary and all entries are real is
enough to make it rotation about Y?]
\end_layout
\begin_layout Standard
Theorem 3:
\begin_inset Formula $U$
\end_inset
can be decomposed as
\begin_inset Formula $U=e^{i\alpha}AXBXC$
\end_inset
where
\begin_inset Formula $ABC=I$
\end_inset
.
Book has recipe for
\begin_inset Formula $ABC$
\end_inset
which serves as proof.
(Corollary 4.2).
In summary,
\begin_inset Formula \begin{eqnarray*}
A & = & R_{z}R_{Y}\\
B & = & R_{Y}R_{Z}\\
C & = & R_{Z}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Controlled U, C(U), From 1-Qubit U
\end_layout
\begin_layout Standard
Use the Theorem 3 decomposition above.
Start with just the phase shift,
\begin_inset Formula $e^{i\alpha}$
\end_inset
.
A controlled circuit for the phase shift
\begin_inset Formula $C\left(e^{i\alpha}I\right)$
\end_inset
should have the following behavior:
\end_layout
\begin_layout Standard
\begin_inset Tabular
|
\begin_inset Text
\begin_layout Standard
Input
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
Output
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle \left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle \left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle \left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle e^{i\alpha}\left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Standard
or
\end_layout
\begin_layout Standard
\begin_inset Tabular
|
\begin_inset Text
\begin_layout Standard
Input
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
Output
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle \left|0\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle \left|0\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle \left|1\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle \left|1\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle \left|0\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle e^{i\alpha}\left|0\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle \left|1\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle e^{i\alpha}\left|1\right\rangle $
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Standard
The outputs from this controlled gate can be produced by a simple unitary
transformation applied to the first qubit:
\begin_inset Formula \[
\left(\begin{array}{cc}
1 & 0\\
0 & e^{i\alpha}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }\left(I\otimes\left(\begin{array}{cc}
e^{i\alpha} & 0\\
0 & e^{i\alpha}\end{array}\right)^{\#1}\right)=\left(\left(\begin{array}{cc}
1 & 0\\
0 & e^{i\alpha}\end{array}\right)\otimes I\right)\]
\end_inset
\end_layout
\begin_layout Standard
(My weird circuit notation would look a lot better in picture form, but
it just shows the gates applied to each qubit (from top to bottom) written
as tensor products.
\begin_inset Formula $I$
\end_inset
means no gate operating on a qubit.
Variables or matrices stand for real gates.
Controlled gates are written with numeric superscripts indicating which
qubit they are controlled by.
For example,
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $X^{\#1}$
\end_inset
is
\begin_inset Formula $C_{NOT}$
\end_inset
controlled by first qubit.)
\end_layout
\begin_layout Standard
Using the circuit above, you can write the circuit for a general
\begin_inset Formula $C\left(U\right)$
\end_inset
where
\begin_inset Formula $U=e^{i\alpha}AXBXC$
\end_inset
and
\begin_inset Formula $ABC=I$
\end_inset
:
\begin_inset Formula \[
\textrm{Circuit: }\left(\left(\begin{array}{cc}
1 & 0\\
0 & e^{i\alpha}\end{array}\right)\otimes A\right)\left({I\otimes X}^{\#1}\right)\left(I\otimes B\right)\left({I\otimes X}^{\#1}\right)\left(I\otimes C\right)\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Tabular
|
\begin_inset Text
\begin_layout Standard
Input
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
Output
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle \left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle ABC\left|j\right\rangle =\left|0\right\rangle I\left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle \left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $e^{i\alpha}\left|1\right\rangle AXBXC\left|j\right\rangle =\left|1\right\rangle U\left|j\right\rangle $
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Standard
The circuit works because when
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $\left|i\right\rangle $
\end_inset
is 0, the
\begin_inset Formula $C_{NOT}$
\end_inset
gates have no effect and the output for the second qubit is
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\begin_inset Formula $ABC$
\end_inset
, exactly the same as the input because
\begin_inset Formula $ABC=I$
\end_inset
\end_layout
\begin_layout Subsection*
Controlled U,
\begin_inset Formula $C^{n}\left(U\right)$
\end_inset
, with multiple control inputs
\end_layout
\begin_layout Standard
Using
\begin_inset Formula $n$
\end_inset
control qubits to control unitary operation of
\begin_inset Formula $k$
\end_inset
qubits:
\begin_inset Formula \begin{eqnarray*}
C^{n}\left(U\right)\left|x\right\rangle \left|y\right\rangle & = & C^{n}\left(U\right)\left|x_{0}\ldots x_{n-1}\right\rangle \left|y_{0}\ldots y_{k-1}\right\rangle \\
& = & \left|x_{0}\ldots x_{n-1}\right\rangle U^{x_{0}\cdot x_{1}\ldots{\cdot x}_{n-1}}\left|y_{0}\ldots y_{k-1}\right\rangle \\
& = & \left|x\right\rangle U^{x_{0}\cdot x_{1}\ldots{\cdot x}_{n-1}}\left|y\right\rangle \end{eqnarray*}
\end_inset
\begin_inset Formula $U$
\end_inset
is controlled by product of bits of
\begin_inset Formula $\left|x\right\rangle $
\end_inset
, it is only applied when every bit is 1.
\end_layout
\begin_layout Standard
In matrix form,
\begin_inset Formula \[
C^{n}\left(U\right)=\left(\begin{array}{cc}
I & 0\\
0 & U\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Size of
\begin_inset Formula $C^{n}\left(U\right)$
\end_inset
is
\begin_inset Formula $2^{n+k}$
\end_inset
, size of
\begin_inset Formula $I$
\end_inset
is
\begin_inset Formula $\left(2^{n}-1\right)\left(2^{k}\right)$
\end_inset
, size of
\begin_inset Formula $U$
\end_inset
is
\begin_inset Formula $2^{k}$
\end_inset
.
Each column of the matrix is the output of a basis state.
The columns from
\begin_inset Formula $\left|0\cdots0\right\rangle \left|0\cdots0\right\rangle $
\end_inset
to
\begin_inset Formula $\left|1\cdots10\right\rangle \left|1\cdots1\right\rangle $
\end_inset
for the first part of the matrix (the bulk of it) just give identify.
Then, the last columns from
\begin_inset Formula $\left|1\cdots1\right\rangle \left|0\cdots0\right\rangle $
\end_inset
to
\begin_inset Formula $\left|1\cdots1\right\rangle \left|1\cdots1\right\rangle $
\end_inset
apply
\begin_inset Formula $U$
\end_inset
to the last
\begin_inset Formula $k$
\end_inset
qubits of the input.
\end_layout
\begin_layout Subsection*
Toffolli Gate
\end_layout
\begin_layout Standard
An example of
\begin_inset Formula $C^{n}\left(U\right)$
\end_inset
, with
\begin_inset Formula $n=2$
\end_inset
and
\begin_inset Formula $U=X$
\end_inset
.
\begin_inset Formula \[
X^{x\cdot y}\left|x\right\rangle \left|y\right\rangle \left|z\right\rangle =\left|x\right\rangle \left|y\right\rangle \left|\left(xy\right)\oplus z\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
NAND gate can be implemented in terms of Toffoli:
\begin_inset Formula \[
X^{x\cdot y}\left|x\right\rangle \left|y\right\rangle \left|1\right\rangle =\left|x\right\rangle \left|y\right\rangle \left|\left(xy\right)\oplus1\right\rangle =\left|x\right\rangle \left|y\right\rangle \left|\overline{xy}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
(
\begin_inset Formula $\overline{xy}$
\end_inset
is logical inverse of xy)
\end_layout
\begin_layout Standard
Fanout can be implemented with:
\begin_inset Formula \[
X^{1\cdot x}\left|1\right\rangle \left|x\right\rangle \left|0\right\rangle =\left|1\right\rangle \left|x\right\rangle \left|\left(1\cdot x\right)\oplus0\right\rangle =\left|1\right\rangle \left|x\right\rangle \left|x\right\rangle \]
\end_inset
\end_layout
\begin_layout Section*
Lecture 9 (14 February)
\end_layout
\begin_layout Subsection*
Implementing Controlled 1-Qubit Gate
\begin_inset Formula $C^{2}\left(U\right)$
\end_inset
with
\begin_inset Formula $V^{2}=U$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }\left({I\otimes I\otimes U}^{\#1,\#2}\right)=\left(I\otimes I\otimes V^{\#1}\right)\left(I\otimes X^{\#1}\otimes I\right)\left(I\otimes I\otimes{V^{H}}^{\#2}\right)\left(I\otimes X^{\#1}\otimes I\right)\left(I\otimes I\otimes V^{\#2}\right)\]
\end_inset
\end_layout
\begin_layout Standard
(Figure 4.8 in book)
\end_layout
\begin_layout Standard
Evaluated:
\end_layout
\begin_layout Standard
\begin_inset Tabular
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{2}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{3}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{2}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $V^{x_{2}}\left|x_{3}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|{x_{1}\oplus x}_{2}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $V^{x_{2}}\left|x_{3}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|{x_{1}\oplus x}_{2}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula ${V^{H}}^{{x_{1}\oplus x}_{2}}{V^{x_{2}}}\left|x_{3}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{2}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula ${V^{H}}^{{x_{1}\oplus x}_{2}}{V^{x_{2}}}\left|x_{3}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|x_{2}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula ${V^{x_{1}}}{V^{H}}^{{x_{1}\oplus x}_{2}}{V^{x_{2}}}\left|x_{3}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Standard
Evaluate third qubit output case by case
\end_layout
\begin_layout Standard
Case 1: If
\begin_inset Formula $X_{1}\oplus X_{2}=1$
\end_inset
then
\begin_inset Formula ${V^{x_{1}}}V^{H}{V^{x_{2}}}\left|x_{3}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case 1.1: If
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $X_{1}=0$
\end_inset
then
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\begin_inset Formula $X_{2}=1$
\end_inset
and
\begin_inset Formula $V^{H}V\left|x_{3}\right\rangle =\left|x_{3}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case 1.2: If
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $X_{2}=0$
\end_inset
then
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\begin_inset Formula $X_{1}=1$
\end_inset
and
\begin_inset Formula ${VV}^{H}\left|x_{3}\right\rangle =\left|x_{3}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case 2: If
\begin_inset Formula $X_{1}\oplus X_{2}=0$
\end_inset
then
\begin_inset Formula ${V^{x_{1}}}{V^{x_{2}}}\left|x_{3}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case 2.1: If
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $X_{1}=X_{2}=0$
\end_inset
then
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\begin_inset Formula $\left|x_{3}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case 2.2: If
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $X_{1}=X_{2}=1$
\end_inset
then
\family default
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\begin_inset Formula $VV\left|x_{3}\right\rangle =U\left|x_{3}\right\rangle $
\end_inset
\end_layout
\begin_layout Subsubsection*
Toffoli Gate
\end_layout
\begin_layout Standard
Using
\begin_inset Formula $V=\frac{1-i}{2}\left(I+iX\right)$
\end_inset
above gives the tofolli gate.
Verify:
\begin_inset Formula \begin{eqnarray*}
V^{2} & = & \frac{\left(1-2i-1\right)^{2}}{4}\left(I^{2}+2iX-X^{2}\right)\\
& = & \frac{-2i}{4}2iX=X\end{eqnarray*}
\end_inset
\begin_inset Formula $V$
\end_inset
can also be rewritten:
\begin_inset Formula \begin{eqnarray*}
V & = & \frac{1-i}{2}\left(I+iX\right)\\
& = & \frac{1-i}{\sqrt{2}}\left(\frac{\sqrt{2}}{2}I+\frac{\sqrt{2}}{2}iX\right)\\
& = & \frac{1-i}{\sqrt{2}}e^{i\frac{\pi}{4}X}=\frac{1-i}{\sqrt{2}}R_{X}\left(-\frac{\pi}{2}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
No Cloning Theorem
\end_layout
\begin_layout Standard
(Box 12.1, Page 532 in book)
\end_layout
\begin_layout Standard
Is there an operator
\begin_inset Formula $U$
\end_inset
that satisfies
\begin_inset Formula $U\left(\left|\psi\right\rangle \left|s\right\rangle \right)=\left|\psi\right\rangle \left|\psi\right\rangle $
\end_inset
for arbitrary states
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
with some standard input
\begin_inset Formula $\left|s\right\rangle $
\end_inset
?
\end_layout
\begin_layout Standard
Assuming there is such an operator, the following will be true for two states
\begin_inset Formula $\left|\psi_{1}\right\rangle $
\end_inset
and
\begin_inset Formula $\left|\psi_{2}\right\rangle $
\end_inset
\begin_inset Formula \begin{eqnarray*}
U\left(\left|\psi_{1}\right\rangle \left|s\right\rangle \right) & = & \left|\psi_{1}\right\rangle \left|\psi_{1}\right\rangle \\
U\left(\left|\psi_{2}\right\rangle \left|s\right\rangle \right) & = & \left|\psi_{2}\right\rangle \left|\psi_{2}\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Inner product of above equations, left hand side:
\begin_inset Formula \[
\left(U\left(\left|\psi_{1}\right\rangle \left|s\right\rangle \right)\right)^{H}\left(U\left(\left|\psi_{2}\right\rangle \left|s\right\rangle \right)\right)\]
\end_inset
\begin_inset Formula \begin{eqnarray*}
& = & \left(\left|\psi_{1}\right\rangle \left|s\right\rangle \right)^{H}U^{H}U\left(\left|\psi_{2}\right\rangle \left|s\right\rangle \right)\\
& = & \left(\left\langle \psi_{1}\right|\left\langle s\right|\right)\left(\left|\psi_{2}\right\rangle \left|s\right\rangle \right)\\
& = & \left\langle \psi_{1}\mid\psi_{2}\right\rangle \left\langle s\mid s\right\rangle \\
& = & \left\langle \psi_{1}\mid\psi_{2}\right\rangle \end{eqnarray*}
\end_inset
Right hand side:
\begin_inset Formula \[
\left(\left|\psi_{1}\right\rangle \left|\psi_{1}\right\rangle \right)^{H}\left(\left|\psi_{2}\right\rangle \left|\psi_{2}\right\rangle \right)\]
\end_inset
\begin_inset Formula \begin{eqnarray*}
& = & \left(\left\langle \psi_{1}\right|\left\langle \psi_{1}\right|\right)\left(\left|\psi_{2}\right\rangle \left|\psi_{2}\right\rangle \right)\\
& = & \left\langle \psi_{1}\mid\psi_{2}\right\rangle \left\langle \psi_{1}\mid\psi_{2}\right\rangle \\
& = & \left\langle \psi_{1}\mid\psi_{2}\right\rangle ^{2}\end{eqnarray*}
\end_inset
Both sides together:
\begin_inset Formula \[
\left\langle \psi_{1}\mid\psi_{2}\right\rangle =\left\langle \psi_{1}\mid\psi_{2}\right\rangle ^{2}\]
\end_inset
which can only be true in two cases
\end_layout
\begin_layout Standard
Case 1:
\begin_inset Formula $\left\langle \psi_{1}\mid\psi_{2}\right\rangle =0$
\end_inset
means
\begin_inset Formula $\left|\psi_{1}\right\rangle \perp\left|\psi_{2}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case 2:
\begin_inset Formula $\left\langle \psi_{1}\mid\psi_{2}\right\rangle =1$
\end_inset
means
\begin_inset Formula $\left|\psi_{1}\right\rangle =\left|\psi_{2}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
So a cloning operator
\begin_inset Formula $U$
\end_inset
will can work for a single state, or for two states that are orthogonal,
there is no
\begin_inset Formula $U$
\end_inset
that can clone states generally.
\end_layout
\begin_layout Subsection*
Implementing Controlled n-Qubit Gates
\begin_inset Formula $C^{n}\left(U\right)$
\end_inset
(n>2)
\end_layout
\begin_layout Standard
Start off with simple examples and build in complexity
\end_layout
\begin_layout Subsubsection*
U=X, k=1, n=3
\end_layout
\begin_layout Standard
This just means taking a normal Toffili gate
\begin_inset Formula \[
\textrm{Circuit: }\left({I\otimes I\otimes X}^{\#1,\#2}\right)\left|x_{1}\right\rangle \left|x_{2}\right\rangle \left|x_{3}\right\rangle \]
\end_inset
and extending it to get an additional control line:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }\left({I\otimes I\otimes I\otimes I\otimes X}^{\#3,\#4}\right)\left({I\otimes I\otimes X}^{\#1,\#2}\otimes I\otimes I\right)\left|x_{1}\right\rangle \left|x_{2}\right\rangle \left|0\right\rangle \left|x_{3}\right\rangle \left|x_{4}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Output of circuit will be
\begin_inset Formula $\left|x_{1}\right\rangle \left|x_{2}\right\rangle \left|x_{1}x_{2}\right\rangle \left|x_{3}\right\rangle \left|\left(x_{1}x_{2}x_{3}\right)\oplus x_{4}\right\rangle $
\end_inset
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\color none
,
\family default
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\shape default
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\emph default
\bar default
\noun default
and the last qubit has the output we are looking for.
\end_layout
\begin_layout Subsubsection*
U=any, k=1, n=any
\end_layout
\begin_layout Standard
In the general case, if you want to control a 1 qubit
\begin_inset Formula $U$
\end_inset
with
\begin_inset Formula $n$
\end_inset
inputs, you need to have
\begin_inset Formula $n-1$
\end_inset
\begin_inset Formula $\left|0\right\rangle $
\end_inset
inputs as well.
Add
\begin_inset Formula $n-1$
\end_inset
toffoli gates, taking the product of the first two qubit lines to the first
\family roman
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\shape up
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\emph off
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\noun off
\color none
\begin_inset Formula $\left|0\right\rangle $
\end_inset
input, the product of the second two lines (#3 and #4) onto the second
\begin_inset Formula $\left|0\right\rangle $
\end_inset
input, and so on.
Halfway down, you reach the first
\begin_inset Formula $\left|0\right\rangle $
\end_inset
lines, but you keep taking products in the same pattern, and at the end,
the last
\begin_inset Formula $\left|0\right\rangle $
\end_inset
line will have the product of the first
\begin_inset Formula $n$
\end_inset
qubits.
Below that, the unitary operation
\begin_inset Formula $U$
\end_inset
can be placed it's own line and can it be controlled by the last
\begin_inset Formula $\left|0\right\rangle $
\end_inset
line, right above it, which holds the product of all the control inputs.
An additional
\begin_inset Formula $n-1$
\end_inset
toffoli gates can be placed after the unitary operation, in the same pattern
as before, to make the
\begin_inset Formula $\left|0\right\rangle $
\end_inset
lines have
\begin_inset Formula $\left|0\right\rangle $
\end_inset
outputs.
\end_layout
\begin_layout Subsubsection*
U=
\begin_inset Formula $U^{\otimes k}$
\end_inset
, k=any, n=1
\end_layout
\begin_layout Standard
\begin_inset Formula \[
Q_{U}\left|c\right\rangle \left|t_{1}t_{2}\cdots t_{k}\right\rangle =\left|c\right\rangle U^{C}\left|t_{1}\right\rangle U^{C}\left|t_{2}\right\rangle \cdots U^{C}\left|t_{k}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
A special case is
\begin_inset Formula $U=X$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
Q_{X}\left|c\right\rangle \left|t_{1}t_{2}\cdots t_{k}\right\rangle & = & \left|c\right\rangle \left|{c\oplus t}_{1}\right\rangle \left|{c\oplus t}_{2}\right\rangle \cdots\left|{c\oplus t}_{k}\right\rangle \end{eqnarray*}
\end_inset
which can be drawn with a single control node,
\begin_inset Formula $k$
\end_inset
\begin_inset Formula $X$
\end_inset
nodes (
\begin_inset Formula $\oplus$
\end_inset
) for each affected qubit, and a line connecting all the nodes.
\end_layout
\begin_layout Standard
In matrix form,
\begin_inset Formula \[
C\left(X^{\otimes k}\right)=\left(\begin{array}{cc}
I & 0\\
0 & X^{\otimes k}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Size of
\begin_inset Formula $C\left(X^{\otimes k}\right)$
\end_inset
is
\begin_inset Formula $2^{k+1}$
\end_inset
, size of
\begin_inset Formula $I$
\end_inset
,
\begin_inset Formula $X^{\otimes k}$
\end_inset
and the 0 matrices is
\begin_inset Formula $2^{k}$
\end_inset
.
Each column of the matrix is the output of a basis state.
The columns from
\begin_inset Formula $\left|0\right\rangle \left|0\cdots0\right\rangle $
\end_inset
to
\begin_inset Formula $\left|0\right\rangle \left|1\cdots1\right\rangle $
\end_inset
for the first half of the matrix just give identify.
Then, the last columns from
\begin_inset Formula $\left|1\right\rangle \left|0\cdots0\right\rangle $
\end_inset
to
\begin_inset Formula $\left|1\right\rangle \left|1\cdots1\right\rangle $
\end_inset
apply
\begin_inset Formula $X^{\otimes k}$
\end_inset
to the last
\begin_inset Formula $k$
\end_inset
qubits of the input.
\begin_inset Formula $X^{\otimes k}$
\end_inset
looks like a reflected identify matrix, with 1s going from the bottom left
corner to the top right, and 0s everywhere else.
\end_layout
\begin_layout Standard
Above can be generalized,
\begin_inset Formula \begin{eqnarray*}
Q_{U}\left|c\right\rangle \left|t_{1}t_{2}\cdots t_{k}\right\rangle & = & \left|c\right\rangle U^{C}\left|t_{1}\right\rangle U^{C}\left|t_{2}\right\rangle \cdots U^{C}\left|t_{k}\right\rangle \\
& = & \left|c\right\rangle \left(U^{C}\right)^{\otimes k}\left|t_{1}\cdots t_{k}\right\rangle \end{eqnarray*}
\end_inset
And
\begin_inset Formula \[
Q_{U}=\left(\begin{array}{cc}
I & 0\\
0 & U^{\otimes k}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Subsection*
2 Level Matrix Gate
\end_layout
\begin_layout Standard
Acts non-linearly on at most 2 components of a vector, example:
\begin_inset Formula \[
\left(U\right)_{dxd}\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}\\
\vdots\\
x_{d}\end{array}\right)=\left(\begin{array}{c}
{x'}_{1}\\
{x'}_{2}\\
x_{3}\\
\vdots\\
x_{d}\end{array}\right)\textrm{ or }=\left(\begin{array}{c}
{x'}_{1}\\
x_{2}\\
{x'}_{3}\\
\vdots\\
x_{d}\end{array}\right)\textrm{ or }=\left(\begin{array}{c}
x_{1}\\
{x'}_{2}\\
{x'}_{3}\\
\vdots\\
x_{d}\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
Mentioned: QR Decomposition, Hausholder matrix
\end_layout
\begin_layout Section*
Lecture 10 (19 February)
\end_layout
\begin_layout Standard
[DUNNO: Was late to class, no idea what this is]
\end_layout
\begin_layout Standard
Implementation Topic
\end_layout
\begin_layout Standard
\begin_inset Formula $C^{2}\left(U\right)$
\end_inset
,
\begin_inset Formula $C^{n}\left(U\right)$
\end_inset
\end_layout
\begin_layout Standard
2-Level Gates /Matrices
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left(\begin{array}{ccc}
\frac{\alpha_{1}}{x} & \frac{\alpha_{2}}{x} & 0\\
\frac{\alpha_{2}}{x} & \frac{\alpha_{1}}{x} & 0\\
0 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}
\alpha_{1} & \beta_{1} & \gamma_{1}\\
\alpha_{2} & \beta_{2} & \gamma_{2}\\
\alpha_{3} & \beta_{3} & \gamma_{3}\end{array}\right) & = & \left(\begin{array}{ccc}
\alpha_{1} & \beta_{1} & \gamma_{1}\\
0 & \beta_{2} & \gamma_{2}\\
\alpha_{3} & \beta_{3} & \gamma_{3}\end{array}\right)\end{eqnarray*}
\end_inset
\begin_inset Formula \[
x=\sqrt{\left|\alpha_{1}\right|^{2}+\left|\alpha_{2}\right|^{2}}\]
\end_inset
\begin_inset Formula \[
\left(\begin{array}{cc}
\frac{\alpha_{1}}{x} & \frac{{-\alpha}_{2}}{x}\end{array}\right)\left(\begin{array}{c}
\frac{{-\alpha}_{2}}{x}\\
\frac{\alpha_{1}}{x}\end{array}\right)=0\]
\end_inset
\end_layout
\begin_layout Standard
First matrix is
\begin_inset Formula $U_{1}$
\end_inset
, second matrix is
\begin_inset Formula $U$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left(\begin{array}{ccc}
{\bar{\alpha}'}_{1} & 0 & \bar{\gamma}_{1}\\
0 & 1 & 0\\
{-\alpha}_{3} & 0 & {\gamma'}_{3}\end{array}\right)\left(\begin{array}{ccc}
{\alpha''}_{1} & {\beta''}_{1} & {\gamma''}_{1}\\
0 & {\beta''}_{2} & {\gamma''}_{2}\\
0 & {\beta''}_{3} & {\gamma''}_{3}\end{array}\right)=\left(\begin{array}{ccc}
{\alpha''}_{1} & 0 & 0\\
0 & {\beta''}_{2} & {\gamma''}_{2}\\
0 & {\beta''}_{3} & {\gamma''}_{3}\end{array}\right)\]
\end_inset
Repeat as submatrix
\begin_inset Formula $\left|{\alpha''}_{1}\right|=1$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
U_{3}U_{2}U_{1}U=\left(\begin{array}{ccc}
{\alpha''}_{1} & 0 & 0\\
0 & {\beta''}_{2} & 0\\
0 & 0 & {\gamma''}_{3}\end{array}\right)D\]
\end_inset
\begin_inset Formula $D$
\end_inset
is some diagonal matrix holding relative phases.
\end_layout
\begin_layout Standard
\begin_inset Formula \[
U=U_{1}^{-1}U_{2}^{-1}U_{3}^{-1}D=U_{1}^{H}U_{2}^{H}U_{3}^{H}D\]
\end_inset
\end_layout
\begin_layout Subsection*
Measurements
\end_layout
\begin_layout Standard
(Book 2.2.3) Measurements are made with collections of measurement operators,
\begin_inset Formula $\left\{ M_{j}\right\} $
\end_inset
, where operators are Hermitian matrices, not necessarily unitary.
There is one measurement operator
\begin_inset Formula $M_{j}$
\end_inset
for each possible outcome,
\begin_inset Formula $j$
\end_inset
.
The measurements satisfy the completeness equation:
\begin_inset Formula \[
\sum_{j=0}^{k-1}M_{j}^{H}M_{j}=I\]
\end_inset
Given a state
\begin_inset Formula $\left|\psi\right\rangle ,$
\end_inset
an outcome
\begin_inset Formula $j$
\end_inset
occurs with probability
\begin_inset Formula \[
p\left(j\right)=\left\langle \psi\right|M_{j}^{H}M_{j}\left|\psi\right\rangle =\left\Vert M_{j}\left|\psi\right\rangle \right\Vert ^{2}\]
\end_inset
causing the state to collapse to
\begin_inset Formula $\frac{M_{j}\left|\psi\right\rangle }{\sqrt{p\left(j\right)}}$
\end_inset
.
\end_layout
\begin_layout Standard
The probabilities of all possible outcomes
\begin_inset Formula $j$
\end_inset
sum to one.
Proof
\begin_inset Formula \begin{eqnarray*}
1 & = & \left\langle \psi\mid\psi\right\rangle =\left\langle \psi\right|I\left|\psi\right\rangle =\left\langle \psi\right|\sum_{j}M_{j}^{H}M_{j}\left|\psi\right\rangle \\
& = & \sum_{j}\left\langle \psi\right|M_{j}^{H}M_{j}\left|\psi\right\rangle =\sum_{j}p\left(j\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Projective Measurements
\end_layout
\begin_layout Standard
Projective measurements are measurements on the computational basis.
\end_layout
\begin_layout Standard
Example 1:
\end_layout
\begin_layout Standard
\begin_inset Formula $M_{j}=\left|j\right\rangle \left\langle j\right|=\left(\begin{array}{ccccc}
& & 0\\
& & \vdots\\
0 & \ldots & 1 & \ldots & 0\\
& & \vdots\\
& & 0\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $M_{j}$
\end_inset
is zero matrix with single one row and column
\begin_inset Formula $j+1$
\end_inset
.
\end_layout
\begin_layout Standard
Observe
\begin_inset Formula $M_{j}^{H}=M_{j}$
\end_inset
(matrix is Hermitian) and
\begin_inset Formula $M_{j}^{H}M_{j}=M_{j}$
\end_inset
(because it's a projection matrix and
\begin_inset Formula $\left|j\right\rangle \left\langle j\mid j\right\rangle \left\langle j\right|=\left|j\right\rangle \left\langle j\right|$
\end_inset
).
\end_layout
\begin_layout Standard
\begin_inset Formula \[
p\left(j\right)=\left\Vert M_{j}\left|\psi\right\rangle \right\Vert ^{2}=\left\Vert \left|j\right\rangle \left\langle j\mid\psi\right\rangle \right\Vert ^{2}=\left|\left\langle j\mid\psi\right\rangle \right|^{2}\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
M_{j}\left|\psi\right\rangle =\left|j\right\rangle \left\langle j\mid\psi\right\rangle =\left\langle j\mid\psi\right\rangle \left|j\right\rangle \]
\end_inset
\begin_inset Formula \[
\left|\psi\right\rangle =\sum_{j}\left|j\right\rangle \left\langle j\mid\psi\right\rangle =\sum_{j}c_{j}\left|j\right\rangle \]
\end_inset
\begin_inset Formula \[
\sum_{j=0}^{2^{n}-1}M_{j}^{H}M_{j}=\sum_{j}M_{j}=\sum_{j}=\left|j\right\rangle \left\langle j\right|=I\]
\end_inset
\end_layout
\begin_layout Standard
Example 2:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|\psi\right\rangle & = & a\left|0\right\rangle +b\left|1\right\rangle \\
M_{1} & = & \left|+\right\rangle \left\langle +\right|\\
M_{2} & = & \left|-\right\rangle \left\langle -\right|\\
\left|+\right\rangle & = & \frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\\
\left|-\right\rangle & = & \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\\
M_{j}^{H} & = & M_{j}\\
M_{j}^{H}M_{j} & = & M_{j}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Completeness relation
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|+\right\rangle \left\langle +\right|+\left|-\right\rangle \left\langle -\right|=\frac{1}{2}\left(\begin{array}{rr}
1 & 1\\
1 & 1\end{array}\right)+\frac{1}{2}\left(\begin{array}{rr}
1 & -1\\
-1 & 1\end{array}\right)=\left(\begin{array}{rr}
1 & 0\\
0 & 1\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
M_{1}\left|\psi\right\rangle =\left|+\right\rangle \left\langle +\right|\left|\psi\right\rangle =\cdots=\left(\frac{a}{\sqrt{2}}+\frac{b}{\sqrt{2}}\right)\left|+\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
M_{2}\left|\psi\right\rangle =\left|-\right\rangle \left\langle -\right|\left|\psi\right\rangle =\frac{1}{\sqrt{2}}\left(a-b\right)\left|-\right\rangle \]
\end_inset
\begin_inset Formula \[
p\left(1\right)=\left\Vert M_{1}\left|\psi\right\rangle \right\Vert ^{2}=\frac{\left|a+b\right|^{2}}{2}\]
\end_inset
\begin_inset Formula \[
p\left(2\right)=\left\Vert M_{2}\left|\psi\right\rangle \right\Vert ^{2}=\frac{\left|a-b\right|^{2}}{2}\]
\end_inset
\end_layout
\begin_layout Standard
Example 3
\end_layout
\begin_layout Standard
Measuring some qubits and not others.
Say measuring
\begin_inset Formula $k$
\end_inset
qubits in a system of
\begin_inset Formula $k+n$
\end_inset
qubits.
Then
\begin_inset Formula $M_{j}=\left|j\right\rangle \left\langle j\right|\otimes I$
\end_inset
where
\begin_inset Formula $\left|j\right\rangle \left\langle j\right|$
\end_inset
is a size
\begin_inset Formula $2^{k}$
\end_inset
matrix and
\begin_inset Formula $I$
\end_inset
is size
\begin_inset Formula $2^{n}$
\end_inset
.
\end_layout
\begin_layout Standard
Properties
\end_layout
\begin_layout Standard
\begin_inset Formula $M_{j}^{H}=M_{j}$
\end_inset
\end_layout
\begin_layout Standard
Means matrix is Hermetian and therefore that it has real eigenvalues.
In other words it's observable because eigenvalues are what you observe.
\begin_inset Formula \[
M_{j}^{H}M_{j}=\left(\left|j\right\rangle \left\langle j\right|\otimes I\right)\left(\left|j\right\rangle \left\langle j\right|\otimes I\right)=\left|j\right\rangle \left\langle j\mid j\right\rangle \left\langle j\right|\otimes II=\left|j\right\rangle \left\langle j\right|\otimes I=M_{j}\]
\end_inset
\begin_inset Formula \begin{eqnarray*}
M_{j}\left|\psi\right\rangle & = & M_{j}\left|\psi_{1}\right\rangle \left|\psi_{2}\right\rangle \\
& = & \left(\left|j\right\rangle \left\langle j\right|\otimes I\right)\left|\psi_{1}\right\rangle \left|\psi_{2}\right\rangle \\
& = & \left|j\right\rangle \left\langle j\mid\psi_{1}\right\rangle \otimes I\left|\psi_{2}\right\rangle \\
& = & \left\langle j\mid\psi_{1}\right\rangle \left(\left|j\right\rangle \left|\psi_{2}\right\rangle \right)\end{eqnarray*}
\end_inset
\begin_inset Formula \begin{eqnarray*}
p\left(j\right) & = & \left\Vert M_{j}\left|\psi\right\rangle \right\Vert ^{2}\\
& = & \left\Vert \left\langle j\mid\psi_{1}\right\rangle \left(\left|j\right\rangle \left|\psi_{2}\right\rangle \right)\right\Vert ^{2}\\
& = & \left|\left\langle j\mid\psi_{1}\right\rangle \right|^{2}\left\Vert \left|j\right\rangle \left|\psi_{2}\right\rangle \right\Vert ^{2}\\
& = & \left|\left\langle j\mid\psi_{1}\right\rangle \right|^{2}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Reason for last step is that
\begin_inset Formula $\left|j\right\rangle $
\end_inset
and
\begin_inset Formula $\left|\psi_{2}\right\rangle $
\end_inset
are both normal:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left\Vert \left|j\right\rangle \left|\psi_{2}\right\rangle \right\Vert ^{2} & = & \left(\left|j\right\rangle \left|\psi_{2}\right\rangle \right)^{H}\left(\left|j\right\rangle \left|\psi_{2}\right\rangle \right)\\
& = & \left(\left\langle j\right|\left\langle \psi_{2}\right|\right)\left(\left|j\right\rangle \left|\psi_{2}\right\rangle \right)\\
& = & \left\langle j\mid j\right\rangle \left\langle \psi_{2}\mid\psi_{2}\right\rangle \\
& = & 1\cdot1=1\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Section*
Lecture 11 (21 February)
\end_layout
\begin_layout Subsection*
Distinguishing states with certainty
\end_layout
\begin_layout Standard
Can we distinguish between two states
\begin_inset Formula $\left|\psi_{1}\right\rangle $
\end_inset
and
\begin_inset Formula $\left|\psi_{2}\right\rangle $
\end_inset
with certainty (with probability 1)? (Similar proof page 87, box 2.3)
\end_layout
\begin_layout Standard
Assume it is possible, then there are measurement operators
\begin_inset Formula $M_{1},M_{2},\ldots,M_{k}$
\end_inset
so that
\begin_inset Formula \[
I=\sum_{j}M_{j}^{H}M_{j}\]
\end_inset
Given states
\begin_inset Formula $\left|\psi_{1}\right\rangle $
\end_inset
and
\begin_inset Formula $\left|\psi_{2}\right\rangle $
\end_inset
, then probabilities of outcomes
\begin_inset Formula $1$
\end_inset
and
\begin_inset Formula $2$
\end_inset
(respectively) should be:
\begin_inset Formula \begin{eqnarray*}
p_{\psi_{1}}\left(1\right) & = & \left\langle \psi_{1}\right|M_{1}^{H}M_{1}\left|\psi_{1}\right\rangle =1\\
p_{\psi_{2}}\left(2\right) & = & \left\langle \psi_{2}\right|M_{2}^{H}M_{2}\left|\psi_{2}\right\rangle =1\end{eqnarray*}
\end_inset
This implies that
\begin_inset Formula $M_{1}\left|\psi_{1}\right\rangle $
\end_inset
and
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\begin_inset Formula $M_{1}\left|\psi_{2}\right\rangle $
\end_inset
are unit vectors.
Using the completeness equation, it also implies
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\begin_inset Formula $M_{1}\left|\psi_{2}\right\rangle =0$
\end_inset
and
\begin_inset Formula $M_{2}\left|\psi_{1}\right\rangle =0$
\end_inset
.
\end_layout
\begin_layout Standard
(Completeness equation
\begin_inset Formula \[
M_{1}^{H}M_{1}+M_{2}^{H}M_{2}=I\]
\end_inset
\end_layout
\begin_layout Standard
leads to
\begin_inset Formula \[
1=\left\langle \psi_{1}\right|I\left|\psi_{1}\right\rangle =\left\langle \psi_{1}\right|M_{1}^{H}M_{1}\left|\psi_{1}\right\rangle +\left\langle \psi_{1}\right|M_{2}^{H}M_{2}\left|\psi_{1}\right\rangle \]
\end_inset
\begin_inset Formula \[
1=\left\langle \psi_{2}\right|I\left|\psi_{2}\right\rangle =\left\langle \psi_{2}\right|M_{1}^{H}M_{1}\left|\psi_{2}\right\rangle +\left\langle \psi_{2}\right|M_{2}^{H}M_{2}\left|\psi_{2}\right\rangle \]
\end_inset
and because
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\begin_inset Formula $\left\langle \psi_{1}\right|M_{2}^{H}M_{2}\left|\psi_{1}\right\rangle =1$
\end_inset
and
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\begin_inset Formula $\left\langle \psi_{2}\right|M_{2}^{H}M_{2}\left|\psi_{2}\right\rangle =1$
\end_inset
the other terms must be 0.)
\end_layout
\begin_layout Standard
State
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\begin_inset Formula $\left|\psi_{1}\right\rangle $
\end_inset
can be written in using
\begin_inset Formula $\left|\psi_{2}\right\rangle $
\end_inset
and another vector
\begin_inset Formula $\left|Z\right\rangle $
\end_inset
as a basis, where
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\begin_inset Formula $\left\Vert \left|Z\right\rangle \right\Vert =1$
\end_inset
,
\begin_inset Formula $\left|Z\right\rangle \perp\left|\psi_{2}\right\rangle $
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|\psi_{1}\right\rangle =\alpha\left|\psi_{2}\right\rangle +\beta\left|Z\right\rangle \]
\end_inset
\begin_inset Formula \[
M_{1}\left|\psi_{1}\right\rangle =\alpha M_{1}\left|\psi_{2}\right\rangle +\beta M_{1}\left|Z\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Assume
\begin_inset Formula $\left|\psi_{1}\right\rangle $
\end_inset
and
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\begin_inset Formula $\left|\psi_{2}\right\rangle $
\end_inset
are not orthogonal,
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then
\begin_inset Formula $\left\langle \psi_{1}\mid\psi_{2}\right\rangle \neq0$
\end_inset
and
\begin_inset Formula $\alpha\neq$
\end_inset
0, so
\begin_inset Formula $\left|\beta\right|<1$
\end_inset
.
Then below, you have a contradiction, proving they must be orthogonal.
\begin_inset Formula \[
1=\left\Vert M_{1}\left|\psi_{1}\right\rangle \right\Vert ^{2}=\left|\beta\right|^{2}\left\Vert M_{1}\left|Z\right\rangle \right\Vert ^{2}\leq\left|\beta\right|^{2}<1\]
\end_inset
Note that
\begin_inset Formula $\left\Vert M_{1}\left|Z\right\rangle \right\Vert ^{2}$
\end_inset
just means
\color inherit
\begin_inset Formula $\left\langle Z\right|M_{1}^{H}M_{1}\left|Z\right\rangle $
\end_inset
.
\end_layout
\begin_layout Subsection*
Some Properties of Operators and Completeness
\end_layout
\begin_layout Standard
Given:
\begin_inset Formula \[
\left|\psi_{1}\right\rangle \perp\left|\psi_{2}\right\rangle \]
\end_inset
\begin_inset Formula \[
M_{1}=\left|\psi_{1}\right\rangle \left\langle \psi_{1}\right|\]
\end_inset
\begin_inset Formula \[
M_{2}=\left|\psi_{2}\right\rangle \left\langle \psi_{2}\right|\]
\end_inset
\begin_inset Formula $M_{1}$
\end_inset
and
\begin_inset Formula $M_{2}$
\end_inset
are symmetric non-negative definate, which means that for all
\begin_inset Formula $\left|x\right\rangle $
\end_inset
,
\begin_inset Formula $\left\langle x\right|M\left|x\right\rangle \geq0$
\end_inset
.
This can be verified as follows:
\begin_inset Formula \[
\left\langle x\right|M\left|x\right\rangle =\left\langle x\right|M^{H}M\left|x\right\rangle =\left\Vert M\left|x\right\rangle \right\Vert ^{2}\geq0\]
\end_inset
\end_layout
\begin_layout Standard
Symmetric non-negative definate matrices have non-negative eigenvalues.
\end_layout
\begin_layout Standard
Define:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
M=I-M_{1}-M_{2}\]
\end_inset
\end_layout
\begin_layout Standard
Observe it's symmetric.
This means eigenvalues are real.
Check that it is positive semi-definite, that for all
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
,
\begin_inset Formula \[
\left\langle \psi\right|M\left|\psi\right\rangle =\left\langle \psi\mid\psi\right\rangle -\left\langle \psi\right|M_{1}\left|\psi\right\rangle -\left\langle \psi\right|M_{2}\left|\psi\right\rangle \geq0\]
\end_inset
\begin_inset Formula \[
M_{1}\left|\psi\right\rangle =\left|\psi_{1}\right\rangle \left\langle \psi_{1}\mid\psi\right\rangle \]
\end_inset
\begin_inset Formula \[
M_{2}\left|\psi\right\rangle =\left|\psi_{2}\right\rangle \left\langle \psi_{2}\mid\psi\right\rangle \]
\end_inset
\begin_inset Formula \[
\left\langle \psi\right|M_{1}\left|\psi\right\rangle =\left|\left\langle \psi_{1}\mid\psi\right\rangle \right|^{2}\]
\end_inset
\begin_inset Formula \[
\left\langle \psi\right|M_{2}\left|\psi\right\rangle =\left|\left\langle \psi_{2}\mid\psi\right\rangle \right|^{2}\]
\end_inset
\begin_inset Formula \[
\left\langle \psi\right|M\left|\psi\right\rangle =1-\left|\left\langle \psi_{1}\mid\psi\right\rangle \right|^{2}-\left|\left\langle \psi_{2}\mid\psi\right\rangle \right|^{2}\overset{?}{=}0\]
\end_inset
\end_layout
\begin_layout Standard
Need to find prove that above is
\begin_inset Formula $\geq0$
\end_inset
.
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
can be expressed as
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|\psi\right\rangle =\left|\psi_{1}\right\rangle \left\langle \psi_{1}\mid\psi\right\rangle +\left|\psi_{2}\right\rangle \left\langle \psi_{2}\mid\psi\right\rangle +\left|z\right\rangle \]
\end_inset
For some
\begin_inset Formula $\left|z\right\rangle \perp\left|\psi_{1}\right\rangle ,\left|\psi_{2}\right\rangle $
\end_inset
.
Taking inner product with
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
gives:
\begin_inset Formula \[
1=\left\langle \psi\mid\psi\right\rangle =\left|\left\langle \psi_{1}\mid\psi\right\rangle \right|^{2}+\left|\left\langle \psi_{2}\mid\psi\right\rangle \right|^{2}+\left\Vert \left|z\right\rangle \right\Vert ^{2}\geq0\]
\end_inset
Rearranging,
\begin_inset Formula \[
1-\left|\left\langle \psi_{1}\mid\psi\right\rangle \right|^{2}-\left|\left\langle \psi_{2}\mid\psi\right\rangle \right|^{2}=\left\Vert \left|z\right\rangle \right\Vert ^{2}=\left\langle \psi\right|M\left|\psi\right\rangle \geq0\]
\end_inset
\end_layout
\begin_layout Standard
Which tells us
\begin_inset Formula $M$
\end_inset
is symmetric non-negative definite.
This means we can do spectral decomposition:
\begin_inset Formula \begin{eqnarray*}
M & = & V\left(\begin{array}{cccc}
\lambda_{1} & 0 & 0 & 0\\
0 & \lambda_{2} & 0 & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & 0 & \lambda_{4}\end{array}\right)V^{H}\\
& = & V\left(\begin{array}{cccc}
\sqrt{\lambda_{1}} & 0 & 0 & 0\\
0 & \sqrt{\lambda_{2}} & 0 & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & 0 & \sqrt{\lambda_{4}}\end{array}\right)V^{H}V\left(\begin{array}{cccc}
\sqrt{\lambda_{1}} & 0 & 0 & 0\\
0 & \sqrt{\lambda_{2}} & 0 & 0\\
0 & 0 & \ddots & 0\\
0 & 0 & 0 & \sqrt{\lambda_{4}}\end{array}\right)V^{H}\end{eqnarray*}
\end_inset
since
\begin_inset Formula $V^{H}V=I$
\end_inset
.
Now, define
\begin_inset Formula $M_{3}=\sqrt{M}$
\end_inset
\end_layout
\begin_layout Standard
Results of measuring state
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\begin_inset Formula $\left|\psi_{1}\right\rangle $
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Tabular
|
\begin_inset Text
\begin_layout Standard
Outcome
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
Probability
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
1
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $p_{\psi_{1}}\left(1\right)=1=\left\langle \psi_{1}\right|M_{1}\left|\psi_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
2
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $p_{\psi_{1}}\left(2\right)=0=\left\langle \psi_{1}\right|M_{2}\left|\psi_{1}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
3
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $p_{\psi_{1}}\left(3\right)=0$
\end_inset
(see below)
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
p_{\psi_{1}}\left(3\right)=0 & = & \left\langle \psi_{1}\right|M_{3}^{H}M_{3}\left|\psi_{1}\right\rangle \\
& = & \left\langle \psi_{1}\right|\left(I-M_{1}-M_{2}\right)\left|\psi_{1}\right\rangle \\
& = & \left\langle \psi_{1}\mid\psi_{1}\right\rangle -\left\langle \psi_{1}\right|M_{1}\left|\psi_{1}\right\rangle -\left\langle \psi_{1}\right|M_{2}\left|\psi_{1}\right\rangle \\
& = & 1-1-0\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
For
\begin_inset Formula $\left|\psi_{2}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Tabular
|
\begin_inset Text
\begin_layout Standard
Outcome
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
Probability
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
1
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $p_{\psi_{2}}\left(1\right)=0$
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
2
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $p_{\psi_{2}}\left(2\right)=1$
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
3
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $p_{\psi_{2}}\left(3\right)=0$
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Standard
(Review of matrix types:
\end_layout
\begin_layout Standard
Normal -
\begin_inset Formula $A^{H}A=AA^{H}$
\end_inset
, matrix commutes with it's transpose
\end_layout
\begin_layout Standard
Unitary -
\begin_inset Formula $A^{H}A=I$
\end_inset
or
\begin_inset Formula $A^{H}=A^{-1}$
\end_inset
, type of normal matrix, eigenvalues all have absolute value of 1.
\end_layout
\begin_layout Standard
Hermetian -
\begin_inset Formula $A=A^{H}$
\end_inset
, type of normal matrix, eigenvalues are real
\end_layout
\begin_layout Standard
Positive Definite - hermitian and
\begin_inset Formula $x^{H}Ax>0$
\end_inset
for all
\begin_inset Formula $x$
\end_inset
.
(if not hermetian, some values of
\begin_inset Formula $x^{H}Ax$
\end_inset
would be complex)
\end_layout
\begin_layout Standard
Projection matrix - hermetian and
\begin_inset Formula $A^{2}=A$
\end_inset
, is positive semi-definite, eigenvalues are all 0 or all 1)
\end_layout
\begin_layout Subsection*
EPR States / Bell States
\end_layout
\begin_layout Standard
(page 25)
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit:}\left(I\otimes X^{\#1}\right)\left(H\otimes I\right)\]
\end_inset
\begin_inset Formula \[
\left|0\right\rangle \left|0\right\rangle ^{\underrightarrow{\left(H\otimes I\right)}}\frac{\left|00\right\rangle +\left|10\right\rangle }{\sqrt{2}}{}^{\underrightarrow{\left(I\otimes X^{\#1}\right)}}\frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}=\left|\beta_{00}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|i\right\rangle \left|j\right\rangle ^{\underrightarrow{\left(H\otimes I\right)}}\frac{\left|0\right\rangle +\left(-1\right)^{i}\left|1\right\rangle }{\sqrt{2}}\left|j\right\rangle {}^{\underrightarrow{\left(I\otimes X^{\#1}\right)}}\frac{\left|0j\right\rangle +\left(-1\right)^{i}\left|1\bar{j}\right\rangle }{\sqrt{2}}=\left|\beta_{ij}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|\beta_{00}\right\rangle & = & \frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}\\
\left|\beta_{01}\right\rangle & = & \frac{\left|01\right\rangle +\left|10\right\rangle }{\sqrt{2}}\\
\left|\beta_{10}\right\rangle & = & \frac{\left|00\right\rangle -\left|11\right\rangle }{\sqrt{2}}\\
\left|\beta_{11}\right\rangle & = & \frac{\left|01\right\rangle -\left|10\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
Property 1: if you measure first qubit, you know second qubit.
\end_layout
\begin_layout Standard
Property 2: Bell states are entangled, and therefore can't be written as
the product of two entangled qubits.
\begin_inset Formula \[
\left(a_{1}\left|0\right\rangle +b_{1}\left|1\right\rangle \right)\left(a_{2}\left|0\right\rangle +b_{2}\left|1\right\rangle \right)\]
\end_inset
\begin_inset Formula \[
a_{1}a_{2}\left|00\right\rangle +a_{1}b_{2}\left|01\right\rangle +b_{1}a_{2}\left|10\right\rangle +b_{1}b_{2}\left|11\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
There are no values of
\begin_inset Formula $a_{1}$
\end_inset
,
\begin_inset Formula $b_{1}$
\end_inset
,
\begin_inset Formula $a_{2}$
\end_inset
,
\begin_inset Formula $b_{2}$
\end_inset
that can give one of the bell states.
\end_layout
\begin_layout Standard
Property 3: Bell states are pairwise orthogonal.
Proof:
\begin_inset Formula \begin{eqnarray*}
\left\langle \beta_{i_{1}j_{1}}\mid\beta_{i_{2}j_{1}}\right\rangle & = & \left(\frac{\left\langle 0j_{1}\right|+\left(-1\right)^{i_{1}}\left\langle 1\bar{j_{1}}\right|}{\sqrt{2}}\right)\left(\frac{\left|0j_{2}\right\rangle +\left(-1\right)^{i_{2}}\left|1\bar{j_{2}}\right\rangle }{\sqrt{2}}\right)\\
& = & \frac{1}{2}\left(\left\langle 0j_{1}\mid0j_{2}\right\rangle +0+0+\left(-1\right)^{i_{1}+i_{2}}\left\langle 1\bar{j_{1}}\mid1\bar{j_{2}}\right\rangle \right)\\
& = & \frac{1}{2}\left(\left\langle j_{1}\mid j_{2}\right\rangle +\left(-1\right)^{i_{1}+i_{2}}\left\langle \bar{j_{1}}\mid\bar{j_{2}}\right\rangle \right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Case 1:
\begin_inset Formula $j_{1}\neq j_{2}$
\end_inset
then
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\begin_inset Formula $\left\langle \beta_{i_{1}j_{1}}\mid\beta_{i_{2}j_{1}}\right\rangle =0$
\end_inset
\end_layout
\begin_layout Standard
Case 2:
\begin_inset Formula $j_{1}=j_{2}$
\end_inset
\end_layout
\begin_layout Standard
Case 2.1:
\begin_inset Formula $i_{1}\neq i_{2}$
\end_inset
then
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\begin_inset Formula $\left\langle \beta_{i_{1}j_{1}}\mid\beta_{i_{2}j_{1}}\right\rangle =0$
\end_inset
(because exponent
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\begin_inset Formula $i_{1}+i_{2}$
\end_inset
is odd)
\end_layout
\begin_layout Standard
Case 2.2:
\begin_inset Formula $i_{1}=i_{2}$
\end_inset
then
\begin_inset Formula $\left\langle \beta_{i_{1}j_{1}}\mid\beta_{i_{2}j_{1}}\right\rangle =1$
\end_inset
\end_layout
\begin_layout Standard
So inner product is zero unless
\begin_inset Formula $i_{1}=i_{2}$
\end_inset
and
\begin_inset Formula $j_{1}=j_{2}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left\langle \beta_{i_{1}j_{1}}\mid\beta_{i_{2}j_{1}}\right\rangle =\left(\frac{\left\langle 0j_{2}\right|+\left(-1\right)^{i_{2}}\left\langle 1\bar{j_{2}}\right|}{\sqrt{2}}\right)\left(\frac{\left|0j_{2}\right\rangle +\left(-1\right)^{i_{2}}\left|1\bar{j_{2}}\right\rangle }{\sqrt{2}}\right)\]
\end_inset
\end_layout
\begin_layout Subsection*
Quantum Teleportation
\end_layout
\begin_layout Standard
[Book 1.3.7 p27]
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit:}\left(I\otimes I\otimes Z^{M\#1}X^{M\#2}\right)\left(H\otimes I\otimes I\right)\left(I\otimes X^{\#1}\otimes I\right)\left|\psi\right\rangle \left|\beta_{00}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|\psi\right\rangle \left|\beta_{00}\right\rangle & = & \left|\psi\right\rangle \frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}\\
& ^{\underrightarrow{\left(I\otimes X^{\#1}\otimes I\right)}} & a\left|0\right\rangle \frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}+b\left|1\right\rangle \frac{\left|10\right\rangle +\left|01\right\rangle }{\sqrt{2}}\\
& ^{\underrightarrow{\left(H\otimes I\otimes I\right)}} & \frac{a}{2}\left(\left|0\right\rangle +\left|1\right\rangle \right)\left(\left|00\right\rangle +\left|11\right\rangle \right)+\frac{b}{2}\left(\left|0\right\rangle -\left|1\right\rangle \right)\left(\left|10\right\rangle +\left|01\right\rangle \right)\\
& = & \frac{1}{2}\left|00\right\rangle \left(a\left|0\right\rangle +b\left|1\right\rangle \right)+\frac{1}{2}\left|01\right\rangle \left(a\left|1\right\rangle +b\left|0\right\rangle \right)\\
& & +\frac{1}{2}\left|10\right\rangle \left(a\left|0\right\rangle -b\left|1\right\rangle \right)+\frac{1}{2}\left|11\right\rangle \left(a\left|1\right\rangle -b\left|0\right\rangle \right)\\
& = & \frac{1}{2}\left(\left|00\right\rangle \left|\psi\right\rangle +\left|01\right\rangle X\left|\psi\right\rangle +\left|10\right\rangle Z\left|\psi\right\rangle +\left|10\right\rangle XZ\left|\psi\right\rangle \right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
By measuring first two qubits, you can know what
\begin_inset Formula $X$
\end_inset
and
\begin_inset Formula $Z$
\end_inset
filters to apply the third qubit in to make it equivalent original input
value
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\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
.
This means that if you have two entangled qubits (of the bell state
\begin_inset Formula $\beta_{00}$
\end_inset
) in seperate locations, you can use them to transmit an arbitrary qubit
over a classical channel.
\end_layout
\begin_layout Section*
Lecture 12 (26 February)
\end_layout
\begin_layout Subsection*
Lecture 11 Review
\end_layout
\begin_layout Standard
EPR States:
\begin_inset Formula $\beta_{ij}=\left|0j\right\rangle +-\left(-1\right)^{i}1\left|1\bar{j}\right\rangle $
\end_inset
\end_layout
\begin_layout Subsection*
Superdense Coding
\end_layout
\begin_layout Standard
[Book 2.3, p97]
\end_layout
\begin_layout Standard
Just like in teleportation, Alice and Bob each have 1 qubit of a 2-qubit
entangled state,
\begin_inset Formula $\beta_{00}$
\end_inset
.
Alice wants to send 2 classical bits of information by sending Bob a single
quantum bit.
She can do this by sending Bob her half of the entangled qubit after she
applies one of four operations to it, depending on the classical bits she
wants to send.
The operations are shown below, and result in Bell states which Bob can
distinguish to determine the original classical bits.
\end_layout
\begin_layout Standard
\begin_inset Tabular
|
\begin_inset Text
\begin_layout Standard
Bits
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
Qubit
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
00
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|\beta_{00}\right\rangle =\frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}\underrightarrow{I\otimes I}\frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}=\left|\beta_{00}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
01
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|\beta_{00}\right\rangle =\frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}\underrightarrow{Z\otimes I}\frac{\left|00\right\rangle -\left|11\right\rangle }{\sqrt{2}}=\left|\beta_{10}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
10
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|\beta_{00}\right\rangle =\frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}\underrightarrow{X\otimes I}\frac{\left|10\right\rangle +\left|01\right\rangle }{\sqrt{2}}=\left|\beta_{01}\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
11
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|\beta_{00}\right\rangle =\frac{\left|00\right\rangle +\left|11\right\rangle }{\sqrt{2}}\underrightarrow{iY\otimes I}\frac{\left|01\right\rangle -\left|10\right\rangle }{\sqrt{2}}=\left|\beta_{11}\right\rangle $
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Subsection*
Quantum Queries
\end_layout
\begin_layout Standard
Mechanism to insert data into a quantum computer.
\end_layout
\begin_layout Standard
Assume you have a boolean function
\begin_inset Formula $f=\left\{ 0,1\right\} \rightarrow\left\{ 0,1\right\} $
\end_inset
, then this is a corresponding unitary operation
\begin_inset Formula $U_{f}$
\end_inset
which is defined on the basis states as:
\begin_inset Formula $U_{f}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
Proving
\begin_inset Formula $U_{f}$
\end_inset
is unitary:
\end_layout
\begin_layout Standard
\begin_inset Formula $\left|i\right\rangle \left|j\right\rangle \underrightarrow{U_{f}}\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle \underrightarrow{U_{f}}\left|i\right\rangle \left|j\oplus f\left(i\right)\oplus f\left(i\right)\right\rangle =\left|i\right\rangle \left|j\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left(U_{f}\right)^{2}=I$
\end_inset
therefore
\begin_inset Formula $U_{f}=U_{f}^{-1}$
\end_inset
and
\begin_inset Formula $U_{f}$
\end_inset
is unitary.
\end_layout
\begin_layout Subsubsection*
Quantum Parallelism
\end_layout
\begin_layout Standard
Inputing a superposition state to
\begin_inset Formula $U_{f}$
\end_inset
computes a result of multiple inputs to the function
\begin_inset Formula $f$
\end_inset
in just a single operation.
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }U_{f}\left(H\otimes I\right)\left|0\right\rangle \left|0\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|00\right\rangle & \underrightarrow{H\otimes I} & \frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\left|0\right\rangle \\
& \underrightarrow{U_{f}} & \frac{\left|0\right\rangle }{\sqrt{2}}\left|0\oplus f\left(0\right)\right\rangle +\frac{\left|1\right\rangle }{\sqrt{2}}\left|0\oplus f\left(1\right)\right\rangle \\
& = & \frac{\left|0f\left(0\right)\right\rangle +\left|1f\left(1\right)\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Single output state contains both values of function
\begin_inset Formula $f$
\end_inset
.
\end_layout
\begin_layout Subsection*
General Quantum Queries
\end_layout
\begin_layout Standard
Assume you have a boolean function
\begin_inset Formula $f=\left\{ 0,\ldots,2^{n}-1\right\} \rightarrow\left\{ 0,1\right\} $
\end_inset
, then there is a corresponding unitary operation
\begin_inset Formula $U_{f}$
\end_inset
which is defined on the basis states as:
\begin_inset Formula $U_{f}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }U_{f}\left(H^{\otimes n}\otimes I\right)\left|0\right\rangle ^{\otimes n}\left|0\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|0\ldots0\right\rangle \left|0\right\rangle & \underrightarrow{H^{\otimes n}\otimes I} & H^{\otimes n}\left|0\ldots0\right\rangle \left|0\right\rangle \\
& = & \left(H\left|0\right\rangle \otimes\cdots\otimes H\left|0\right\rangle \right)\left|0\right\rangle \\
& = & \frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left|j\right\rangle \left|0\right\rangle \\
& \underrightarrow{U_{f}} & \frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left|j\right\rangle \left|f\left(j\right)\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Quantum parallelism is not enough to exploit power of quantum computing,
because even though the final state is a combination of all possible outputs
of the function
\begin_inset Formula $f$
\end_inset
, when measurement occurs it will only give a single, random output.
Need to find ways to combine the values to be able to efficiently measure
some global property of the function.
\end_layout
\begin_layout Subsubsection*
Matrix Representation of
\begin_inset Formula $U_{f}$
\end_inset
\end_layout
\begin_layout Standard
When
\begin_inset Formula $f$
\end_inset
is boolean function of 1 bit:
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}=\left(\begin{array}{cc}
X^{f\left(0\right)}\\
& X^{f\left(1\right)}\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $i=0$
\end_inset
,
\begin_inset Formula $f\left(0\right)=0$
\end_inset
,
\begin_inset Formula $\left|0j\right\rangle \rightarrow\left|0j\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $i=0$
\end_inset
,
\begin_inset Formula $f\left(0\right)=1$
\end_inset
,
\begin_inset Formula $\left|0j\right\rangle \rightarrow\left|0\bar{j}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $i=1$
\end_inset
,
\begin_inset Formula $f\left(1\right)=0$
\end_inset
,
\begin_inset Formula $\left|1j\right\rangle \rightarrow\left|1j\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $i=1$
\end_inset
,
\begin_inset Formula $f\left(1\right)=1$
\end_inset
,
\begin_inset Formula $\left|1j\right\rangle \rightarrow\left|1\bar{j}\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
When
\begin_inset Formula $f$
\end_inset
is boolean function of
\begin_inset Formula $m$
\end_inset
bits:
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}=\left(\begin{array}{cccc}
X^{f\left(0\right)}\\
& X^{f\left(1\right)}\\
& & \ddots\\
& & & X^{f\left(2^{m}-1\right)}\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
When
\begin_inset Formula $f$
\end_inset
is function of
\begin_inset Formula $m$
\end_inset
bits returning
\begin_inset Formula $n$
\end_inset
bits:
\end_layout
\begin_layout Standard
\begin_inset Formula $f=\left\{ 0,\ldots,2^{m}-1\right\} \rightarrow\left\{ 0,\ldots,2^{n}-1\right\} $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}\left|i_{1}\ldots i_{n}\right\rangle \left|j_{1}\ldots j_{n}\right\rangle =U_{f}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
where
\begin_inset Formula $\oplus$
\end_inset
is addition mod
\begin_inset Formula $n$
\end_inset
.
\begin_inset Formula $U_{f}$
\end_inset
is a permutation matrix (which makes it unitary) with a single 1 in each
column.
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}$
\end_inset
is made up of shifting matrixes, which work like:
\begin_inset Formula $A_{p}\left|j\right\rangle =\left|p\oplus j\right\rangle $
\end_inset
for
\begin_inset Formula $j,p=0,\ldots,2^{n}-1$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $A_{p}=\left(\begin{array}{cccccccc}
& & & & 1\\
& & & & & 1\\
& & & & & & 1\\
& & & & & & & 1\\
1\\
& 1\\
& & 1\\
& & & 1\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
Matrix
\begin_inset Formula $A_{p}$
\end_inset
has 1 in first column at position
\begin_inset Formula $p+1$
\end_inset
, then one row down in each column after that.
Some special cases of shifting matrices are
\begin_inset Formula $A_{0}=I$
\end_inset
and when
\begin_inset Formula $n=1$
\end_inset
,
\begin_inset Formula $A_{1}=X$
\end_inset
.
But, back to
\begin_inset Formula $U_{f}:$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle =\left|i\right\rangle A_{f\left(i\right)}\left|j\right\rangle =\left(I\otimes A_{f\left(i\right)}\right)U_{f}\left|i\right\rangle \left|j\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}=\left(\begin{array}{cccc}
A_{f\left(0\right)}\\
& A_{f\left(1\right)}\\
& & \ddots\\
& & & A_{f\left(2^{m}-1\right)}\end{array}\right)$
\end_inset
\end_layout
\begin_layout Section*
Lecture 13 (28 February)
\end_layout
\begin_layout Subsection*
Lecture 12 Review
\end_layout
\begin_layout Standard
\begin_inset Formula $U_{f}\left|i\right\rangle \left|j\right\rangle =\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Homework Hint: When
\begin_inset Formula $U_{f}^{2}\neq I$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
U_{f}U_{f}\left|i\right\rangle \left|j\right\rangle & = & U_{f}\left|i\right\rangle \left|j\oplus f\left(i\right)\right\rangle \\
& = & \left|i\right\rangle \left|j\oplus f\left(i\right)\oplus f\left(i\right)\right\rangle \\
& = & \left|i\right\rangle \left|j\oplus2f\left(i\right)\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Deutsch's Algorithm
\end_layout
\begin_layout Standard
Given
\begin_inset Formula $f:\left\{ 0,1\right\} \rightarrow\left\{ 0,1\right\} $
\end_inset
compute
\begin_inset Formula $f\left(0\right)\oplus f\left(1\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }\left[M\otimes I\right]\left(H\otimes I\right)U_{f}\left(H\otimes H\right)\left|0\right\rangle \left|1\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|0\right\rangle \left|1\right\rangle & \underrightarrow{H\otimes H} & \left(\frac{\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}\right)\left(\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\right)\\
& = & \frac{1}{2}\left(\left|00\right\rangle -\left|01\right\rangle +\left|10\right\rangle -\left|11\right\rangle \right)\\
& \underrightarrow{U_{f}} & \frac{1}{2}\left(\left|0f\left(0\right)\right\rangle -\left|0\overline{f\left(0\right)}\right\rangle +\left|1f\left(1\right)\right\rangle -\left|1\overline{f\left(1\right)}\right\rangle \right)\\
& = & \frac{1}{2}\left(\left|0\right\rangle \left(\left|f\left(0\right)\right\rangle -\left|\overline{f\left(0\right)}\right\rangle \right)+\left|1\right\rangle \left(\left|f\left(1\right)\right\rangle -\left|\overline{f\left(1\right)}\right\rangle \right)\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $f\left(0\right)=0$
\end_inset
,
\begin_inset Formula $\left|f\left(0\right)\right\rangle -\left|\overline{f\left(0\right)}\right\rangle =\left|0\right\rangle -\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $f\left(0\right)=1$
\end_inset
,
\begin_inset Formula $\left|f\left(0\right)\right\rangle -\left|\overline{f\left(0\right)}\right\rangle =\left|1\right\rangle -\left|0\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Therefore
\begin_inset Formula $\left|f\left(0\right)\right\rangle -\left|\overline{f\left(0\right)}\right\rangle =\left(-1\right)^{f\left(0\right)}\left(\left|0\right\rangle -\left|1\right\rangle \right)$
\end_inset
\end_layout
\begin_layout Standard
Continuing\SpecialChar \ldots{}
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
& = & \frac{1}{2}\left(\left(-1\right)^{f\left(0\right)}\left|0\right\rangle \left(\left|0\right\rangle -\left|1\right\rangle \right)+\left(-1\right)^{f\left(1\right)}\left|1\right\rangle \left(\left|0\right\rangle -\left|1\right\rangle \right)\right)\\
& = & \frac{1}{\sqrt{2}}\left(\left(-1\right)^{f\left(0\right)}\left|0\right\rangle +\left(-1\right)^{f\left(1\right)}\left|1\right\rangle \right)\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\\
& = & \frac{1}{\sqrt{2}}\left(-1\right)^{f\left(0\right)}\left(\left|0\right\rangle +\left(-1\right)^{f\left(1\right)-f\left(0\right)}\left|1\right\rangle \right)\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $f\left(0\right)=f\left(1\right)$
\end_inset
, then
\begin_inset Formula $\left(\left|0\right\rangle +\left(-1\right)^{f\left(1\right)-f\left(0\right)}\left|1\right\rangle \right)=\left|0\right\rangle +\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $f\left(0\right)\neq f\left(1\right)$
\end_inset
, then
\begin_inset Formula $\left(\left|0\right\rangle +\left(-1\right)^{f\left(1\right)-f\left(0\right)}\left|1\right\rangle \right)=\left|0\right\rangle -\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Continuing\SpecialChar \ldots{}
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
& \underrightarrow{H\otimes I} & \left\{ \begin{array}{cc}
\left(-1\right)^{f\left(0\right)}\left|0\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}} & \textrm{ if }f\left(0\right)=f\left(1\right)\\
\left(-1\right)^{f\left(0\right)}\left|1\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}} & \textrm{ if }f\left(0\right)\neq f\left(1\right)\end{array}\right.\\
& = & \pm\left|f\left(0\right)\oplus f\left(1\right)\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Deutsch-Jonsa Algorithm
\end_layout
\begin_layout Standard
Given
\begin_inset Formula $f:\left\{ 0,\ldots,2^{n}-1\right\} \rightarrow\left\{ 0,1\right\} $
\end_inset
\end_layout
\begin_layout Standard
where function
\begin_inset Formula $f$
\end_inset
is either balanced or constant.
Constant means
\begin_inset Formula $f\left(j\right)=f\left(0\right)\forall j$
\end_inset
.
Balanced means if
\begin_inset Formula $A_{o}=\left\{ j:f\left(j\right)=0\right\} $
\end_inset
and
\begin_inset Formula $A_{1}=\left\{ j:f\left(j\right)=1\right\} $
\end_inset
, then
\begin_inset Formula $\left|A_{0}\right|=\left|A_{1}\right|=2^{n-1}$
\end_inset
.
The algorithm determines whether a function is constant or balanced, assuming
it won't be any thing else.
\end_layout
\begin_layout Subsubsection*
Classical Solution
\end_layout
\begin_layout Standard
Requires
\begin_inset Formula $2^{n-1}+1$
\end_inset
evaluations worst case.
Algorithm is to loop through the inputs, checking to see if the function
ever returns two different values.
If it does return two different values, the function is not constant and
the loop can be terminated.
After the halfway point, if only one value has been returned, the function
is not balanced and can be labeled constant.
\end_layout
\begin_layout Subsubsection*
Quantum Solution
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }\left[M\otimes I\right]\left(H\otimes I\right)U_{f}\left(H^{\otimes n}\otimes H\right)\left|0\right\rangle ^{\otimes n}\left|1\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|0\right\rangle ^{\otimes n}\left|1\right\rangle & \underrightarrow{H^{\otimes n}\otimes H} & \left(\frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left|j\right\rangle \right)\left(\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\right)\\
& = & \frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left(\frac{\left|j0\right\rangle -\left|j1\right\rangle }{\sqrt{2}}\right)\\
& \underrightarrow{U_{f}} & \frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left(\frac{\left|jf\left(j\right)\right\rangle -\left|j\overline{f\left(j\right)}\right\rangle }{\sqrt{2}}\right)\\
& = & \frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left|j\right\rangle \left(\frac{\left|f\left(j\right)\right\rangle -\left|\overline{f\left(j\right)}\right\rangle }{\sqrt{2}}\right)\\
& = & \frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left|j\right\rangle \left(-1\right)^{f\left(j\right)}\left(\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\right)\\
& \underrightarrow{H^{\otimes n}\otimes I} & \frac{1}{2^{n/2}}\sum_{j=0}^{2^{n}-1}\left(\frac{1}{2^{n/2}}\sum_{k=0}^{2^{n}-1}\left(-1\right)^{j\cdot k}\left|k\right\rangle \right)\left(-1\right)^{f\left(j\right)}\left(\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\right)\\
& = & \sum_{k}a_{k}\left|k\right\rangle \left(\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Setting
\begin_inset Formula $a_{k}=\frac{1}{2^{n}}\sum_{j}\left(-1\right)^{j\cdot k}\left(-1\right)^{f\left(j\right)}$
\end_inset
to represent the
\begin_inset Quotes eld
\end_inset
amplitude
\begin_inset Quotes erd
\end_inset
of each
\begin_inset Formula $k$
\end_inset
.
\end_layout
\begin_layout Standard
Looking at
\begin_inset Formula $a_{0}=\frac{1}{2^{n}}\sum_{j}\left(-1\right)^{f\left(j\right)}$
\end_inset
:
\end_layout
\begin_layout Standard
If function
\begin_inset Formula $f$
\end_inset
is constant then
\begin_inset Formula $a_{0}=\pm1$
\end_inset
which means
\begin_inset Formula $a_{k}=0$
\end_inset
for all
\begin_inset Formula $k\neq0$
\end_inset
.
This is because of completeness, if coefficient of one basis state 1, all
others must be zero.
If function
\begin_inset Formula $f$
\end_inset
is balanced then
\begin_inset Formula $a_{0}=0.$
\end_inset
\end_layout
\begin_layout Standard
Homework: What do we need to do to detect 3/4 balanced instead of 1/2 balanced.
\end_layout
\begin_layout Subsubsection*
Randomized Classical Algorithm
\end_layout
\begin_layout Standard
Generate
\begin_inset Formula $k$
\end_inset
random input to function
\begin_inset Formula $f$
\end_inset
, if function evaluated at any two of the inputs is different, the function
will be labeled balanced, otherwise it's considered constant.
Algorithm can fail, outputting constant when the function is actually balanced.
\end_layout
\begin_layout Standard
\begin_inset Formula $A_{0}=\left\{ j:f\left(j\right)=0\right\} $
\end_inset
,
\begin_inset Formula $A_{1}=\left\{ j:f\left(j\right)=1\right\} $
\end_inset
\end_layout
\begin_layout Standard
Balanced when
\begin_inset Formula $\left|A_{0}\right|=\left|A_{1}\right|$
\end_inset
\end_layout
\begin_layout Standard
Probability of picking 1 sample which is 0:
\begin_inset Formula $p\left(1,0\right)=\frac{1}{2}$
\end_inset
\end_layout
\begin_layout Standard
Probability of picking
\begin_inset Formula $k$
\end_inset
samples which are 0:
\begin_inset Formula $p\left(k,0\right)=\frac{1}{2^{k}}$
\end_inset
\end_layout
\begin_layout Standard
Probability of picking 1 sample which is 1:
\begin_inset Formula $p\left(1,1\right)=\frac{1}{2}$
\end_inset
\end_layout
\begin_layout Standard
Probability of picking
\begin_inset Formula $k$
\end_inset
samples which are 1:
\begin_inset Formula $p\left(k,1\right)=\frac{1}{2^{k}}$
\end_inset
\end_layout
\begin_layout Standard
Probability of failure given
\begin_inset Formula $f$
\end_inset
is balanced:
\begin_inset Formula $p\left(k,0\right)+p\left(k,1\right)=\frac{2}{2^{k}}$
\end_inset
\end_layout
\begin_layout Standard
Set
\begin_inset Formula $p>\frac{2}{2^{k}}$
\end_inset
to succeed with arbitrary probability.
\end_layout
\begin_layout Section*
Lectures 14/15 (5/7 March)
\end_layout
\begin_layout Subsection*
Midterm and Midterm Review
\end_layout
\begin_layout Section*
Lecture 16 (19 March)
\end_layout
\begin_layout Standard
[Book Chapter 5]
\end_layout
\begin_layout Subsection*
Discrete Fourier Transform
\end_layout
\begin_layout Standard
Maps vectors
\begin_inset Formula $x_{0},\ldots,x_{N-1}\rightarrow y_{0},\ldots,y_{N-1}$
\end_inset
like:
\begin_inset Formula \[
y_{k}=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}x_{j}e^{2\pi ikj/N}\]
\end_inset
Cost
\begin_inset Formula $O\left(N^{2}\right)$
\end_inset
\end_layout
\begin_layout Standard
1965 -- Cooley-Tukey Fast Fourier Transform (FFT) cost
\begin_inset Formula $O\left(N\log N\right)$
\end_inset
\end_layout
\begin_layout Subsection*
Quantum Fourier Transform
\end_layout
\begin_layout Standard
Input state
\begin_inset Formula $\left|j\right\rangle $
\end_inset
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
for
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\color inherit
\begin_inset Formula $j=0,\ldots,N-1$
\end_inset
where
\begin_inset Formula $N=2^{n}$
\end_inset
(
\begin_inset Formula $N$
\end_inset
is power of 2):
\begin_inset Formula \[
F\left|j\right\rangle =\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}e^{2\pi ijk/N}\left|k\right\rangle \]
\end_inset
\end_layout
\begin_layout Subsubsection*
Equivalence to Discrete Transform
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
F\left(\sum_{j=0}^{N-1}x_{j}\left|j\right\rangle \right) & = & \sum_{j=0}^{N-1}x_{j}F\left|j\right\rangle \\
& = & \sum_{j=0}^{N-1}x_{j}\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}e^{2\pi ijk/N}\left|k\right\rangle \\
& = & \sum_{k=0}^{N-1}\left(\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}x_{j}e^{2\pi ijk/N}\right)\left|k\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsubsection*
Applied to
\begin_inset Formula $\left|0\right\rangle $
\end_inset
\begin_inset Formula \[
F\left|0\right\rangle =\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}e^{2\pi i0k/N}\left|k\right\rangle =\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\left|k\right\rangle =H^{\bigotimes n}\left|0\right\rangle ^{\bigotimes n}\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Matrix representation
\end_layout
\begin_layout Standard
\begin_inset Formula $F\left|j\right\rangle =\frac{1}{\sqrt{N}}\left(\begin{array}{c}
\vdots\\
e^{2\pi ijk/N}\\
e^{2\pi ij\left(k+1\right)/N}\\
\vdots\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $F=\frac{1}{\sqrt{N}}\left(\begin{array}{cccc}
1 & 1 & \cdots & 1\\
1 & e^{2\pi i1/N} & & e^{2\pi i\left(N-1\right)/N}\\
\vdots & \vdots & \ddots & \vdots\\
1 & e^{2\pi i\left(N-1\right)/N} & & e^{2\pi i\left(N-1\right)^{2}/N}\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $F=\frac{1}{\sqrt{N}}\left(e^{2\pi ijk/N}\right)_{k,j=0,\ldots,N-1}$
\end_inset
\end_layout
\begin_layout Subsubsection*
Proof F is Unitary
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
F^{H} & = & \frac{1}{\sqrt{N}}\left(e^{-2\pi ijk/N}\right)_{j,k=0,\ldots,N-1}\\
F^{H}F & = & \frac{1}{N}\left(\sum_{k=0}^{N-1}e^{-2\pi ipk/N}e^{2\pi ikq/N}\right)_{p,q=0,\ldots,N-1}\\
& = & \frac{1}{N}\left(\sum_{k=0}^{N-1}e^{2\pi ik\left(q-p\right)/N}\right)_{p,q}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
If
\begin_inset Formula $p=q$
\end_inset
,
\begin_inset Formula $\left(F^{H}F\right)_{p,q}=1$
\end_inset
\end_layout
\begin_layout Standard
If
\begin_inset Formula $p\neq q$
\end_inset
,
\begin_inset Formula \begin{eqnarray*}
\left(F^{H}F\right)_{p,q} & = & \sum_{k=0}^{N-1}e^{2\pi ik\left(q-p\right)/N}\\
& = & \frac{e^{2\pi i\left(q-p\right)\frac{N-1}{N}}e^{2\pi i\left(q-p\right)\frac{1}{N}}-1}{e^{2\pi i\left(q-p\right)\frac{1}{N}}-1}\\
& = & \frac{e^{2\pi i\left(q-p\right)}-1}{e^{2\pi i\left(q-p\right)\frac{1}{N}}-1}\\
& = & 1-1=0\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Therefore,
\begin_inset Formula $F^{H}F=I$
\end_inset
, and
\begin_inset Formula $F$
\end_inset
is unitary.
\end_layout
\begin_layout Standard
(Formula for sum of a geometric progression used above:
\begin_inset Formula $\sum_{k=0}^{n}r^{k}=\frac{r^{n+1}-1}{r-1}$
\end_inset
)
\end_layout
\begin_layout Subsubsection*
Tensor Product Representation
\end_layout
\begin_layout Standard
Notation:
\end_layout
\begin_layout Standard
\begin_inset Formula $j=j_{1}j_{2}\ldots j_{n}=j_{1}2^{n-1}+j_{2}2^{n-2}+\ldots+j_{n}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\frac{j}{2^{n}}=0.j_{1}j_{2}\ldots j_{n}=j_{1}\frac{1}{2}+j_{2}\frac{1}{2^{2}}+j_{n}\frac{1}{2^{n}}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\frac{j}{2^{\ell}}=j_{1}\ldots j_{n-\ell}.j_{n-\ell+1}\ldots j_{n}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $e^{2\pi i\frac{j}{2^{\ell}}}=e^{2\pi i\left(j_{1}\ldots j_{n-\ell}.j_{n-\ell+1}\ldots j_{n}\right)}$
\end_inset
\end_layout
\begin_layout Standard
Lemma:
\begin_inset Formula \[
F\left|j\right\rangle =F\left|j_{1}\ldots j_{n}\right\rangle =\frac{\left|0\right\rangle +e^{2\pi i0.j_{n}}\left|1\right\rangle }{\sqrt{2}}\otimes\frac{\left|0\right\rangle +e^{2\pi i0.j_{n-1}j_{n}}\left|1\right\rangle }{\sqrt{2}}\otimes\cdots\otimes\frac{\left|0\right\rangle +e^{2\pi i0.j_{1}j_{2}\ldots j_{n}}\left|1\right\rangle }{\sqrt{2}}\]
\end_inset
\end_layout
\begin_layout Standard
Proof, start with:
\begin_inset Formula \[
F\left|j\right\rangle =\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}e^{2\pi ijk/N}\left|k\right\rangle \]
\end_inset
Decompose
\begin_inset Formula $k=\left(k_{1}\ldots k_{n}\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
F\left|j\right\rangle & = & \frac{1}{\sqrt{N}}\sum_{k_{1}=0}^{1}\cdots\sum_{k_{n}=0}^{1}e^{2\pi ij0.k_{1}\ldots k_{n}}\left|k_{1}\right\rangle \ldots\left|k_{n}\right\rangle \\
& = & \frac{1}{\sqrt{N}}\sum_{k_{1}=0}^{1}\cdots\sum_{k_{n}=0}^{1}e^{2\pi ij\left(k_{1}\frac{1}{2}+k_{2}\frac{1}{2^{2}}+\cdots+k_{n}\frac{1}{2^{n}}\right)}\left|k_{1}\right\rangle \ldots\left|k_{n}\right\rangle \\
& = & \frac{1}{\sqrt{N}}\sum_{k_{1}=0}^{1}\cdots\sum_{k_{n}=0}^{1}\bigotimes_{\ell=1}^{n}e^{2\pi ijk_{\ell}\frac{1}{2^{\ell}}}\left|k_{\ell}\right\rangle \\
& = & \frac{1}{\sqrt{N}}\bigotimes_{\ell=1}^{n}\sum_{k_{\ell}=0}^{1}e^{2\pi ijk_{\ell}\frac{1}{2^{\ell}}}\left|k_{\ell}\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Simple example illustrating last step:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
& & \sum_{k_{1}=0}^{1}\sum_{k_{2}=0}^{1}e^{2\pi ijk_{1}\frac{1}{2}/N}e^{2\pi ijk_{2}\frac{1}{4}}\left|k_{1}\right\rangle \left|k_{2}\right\rangle \\
& & =\left(\sum_{k_{1}=0}^{1}e^{2\pi ijk_{1}\frac{1}{2}}\left|k_{1}\right\rangle \right)\left(\sum_{k_{2}=0}^{1}e^{2\pi ijk_{2}\frac{1}{4}}\left|k_{2}\right\rangle \right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Continuing:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
F\left|j\right\rangle & = & \bigotimes_{\ell=1}^{n}\left(\frac{\left|0\right\rangle +e^{2\pi ij/2^{\ell}}\left|1\right\rangle }{\sqrt{2}}\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Dividing
\begin_inset Formula $j$
\end_inset
by
\begin_inset Formula $2^{\ell}$
\end_inset
is equivalent to moving the decimal point in the binary representation
of
\begin_inset Formula $j$
\end_inset
by
\begin_inset Formula $\ell$
\end_inset
places to the left.
Additionally, after the shift, any digits to the left of the dot can be
discarded.
This is because the function
\begin_inset Formula $e^{xi}$
\end_inset
has a period of
\begin_inset Formula $2\pi$
\end_inset
, so the only the fractional part of
\begin_inset Formula $\frac{j}{2^{\ell}}$
\end_inset
matters, and the whole number part can be set to
\begin_inset Formula $0.$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \[
F\left|j\right\rangle =F\left|j_{1}\ldots j_{n}\right\rangle =\frac{\left|0\right\rangle +e^{2\pi i0.j_{n}}\left|1\right\rangle }{\sqrt{2}}\otimes\frac{\left|0\right\rangle +e^{2\pi i0.j_{n-1}j_{n}}\left|1\right\rangle }{\sqrt{2}}\otimes\cdots\otimes\frac{\left|0\right\rangle +e^{2\pi i0.j_{1}j_{2}\ldots j_{n}}\left|1\right\rangle }{\sqrt{2}}\]
\end_inset
\end_layout
\begin_layout Standard
At
\begin_inset Formula $\ell=1$
\end_inset
, term is:
\begin_inset Formula $\left(\frac{\left|0\right\rangle +e^{2\pi i0.j_{n}}\left|1\right\rangle }{\sqrt{2}}\right)$
\end_inset
\end_layout
\begin_layout Standard
At
\begin_inset Formula $\ell=2$
\end_inset
, term is:
\begin_inset Formula $\left(\frac{\left|0\right\rangle +e^{2\pi i0.j_{n-1}j_{n}}\left|1\right\rangle }{\sqrt{2}}\right)$
\end_inset
\end_layout
\begin_layout Standard
At
\begin_inset Formula $\ell=3$
\end_inset
, term is:
\begin_inset Formula $\left(\frac{\left|0\right\rangle +e^{2\pi i0.j_{n-2}j_{n-1}j_{n}}\left|1\right\rangle }{\sqrt{2}}\right)$
\end_inset
\end_layout
\begin_layout Standard
Tip: Remember this lemma for final
\end_layout
\begin_layout Subsection*
Fourier Transform as Circuit
\end_layout
\begin_layout Standard
Define
\begin_inset Formula $R_{k}=\left(\begin{array}{cc}
1 & 0\\
0 & e^{2\pi i/2^{k}}\end{array}\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $R_{0}=\left(\begin{array}{cc}
1 & 0\\
0 & 1\end{array}\right)=I$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $R_{1}=\left(\begin{array}{cc}
1 & 0\\
0 & -1\end{array}\right)=Z$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $R_{2}=\left(\begin{array}{cc}
1 & 0\\
0 & i\end{array}\right)=S$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $R_{3}=\left(\begin{array}{cc}
1 & 0\\
0 & e^{\pi i/4}\end{array}\right)=T$
\end_inset
\end_layout
\begin_layout Standard
When evaluating cost of implementing fourier transform, assume implementing
each of one these
\begin_inset Formula $R$
\end_inset
gates has unit cost.
The assumption may not necessarily be true in an actual implementation.
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit:}\bigotimes_{\ell=1}^{n}\left(R_{n-\ell+1}^{\# n}\ldots R_{2}^{\#\ell+1}H\right)\left|j_{\ell}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
First Qubit:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\left|j_{1}\right\rangle \xrightarrow{H}\frac{\left|0\right\rangle +\left(-1\right)^{j_{1}}\left|1\right\rangle }{\sqrt{2}}=\frac{\left|0\right\rangle +e^{2i\pi0.j_{1}}\left|1\right\rangle }{\sqrt{2}}\]
\end_inset
\end_layout
\begin_layout Standard
Trick used:
\begin_inset Formula $e^{2\pi i0.j_{1}}=\left\{ \begin{array}{cc}
1 & j_{1}=0\\
e^{2\pi i/2} & j_{1}=1\end{array}\right.=\left(-1\right)^{j_{1}}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
& \xrightarrow[w/j_{2}]{R_{2}} & \frac{R_{2}^{j_{2}}\left|0\right\rangle +e^{2i\pi0.j_{1}}R_{2}^{j_{2}}\left|1\right\rangle }{\sqrt{2}}\\
& = & \frac{\left|0\right\rangle +e^{2i\pi0.j_{1}}e^{2\pi ij_{2}/2^{2}}\left|1\right\rangle }{\sqrt{2}}\\
& = & \frac{\left|0\right\rangle +e^{2i\pi0.j_{1}j_{2}}\left|1\right\rangle }{\sqrt{2}}\\
& \xrightarrow[w/j_{3}]{R_{3}} & \frac{R_{3}^{j_{3}}\left|0\right\rangle +e^{2i\pi0.j_{1}}R_{3}^{j_{3}}\left|1\right\rangle }{\sqrt{2}}\\
& = & \frac{\left|0\right\rangle +e^{2i\pi0.j_{1}j_{2}j_{3}}\left|1\right\rangle }{\sqrt{2}}\\
& \vdots\\
& \xrightarrow[w/j_{n}]{R_{n}} & \frac{\left|0\right\rangle +e^{2i\pi0.j_{1}j_{2}\ldots j_{n}}\left|1\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Second qubit:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|j_{2}\right\rangle & \xrightarrow{H} & \frac{\left|0\right\rangle +e^{2i\pi0.j_{2}}\left|1\right\rangle }{\sqrt{2}}\\
& \xrightarrow[w/j_{3}]{R_{2}} & \frac{\left|0\right\rangle +e^{2i\pi0.j_{2}}e^{2\pi ij_{3}/2^{2}}\left|1\right\rangle }{\sqrt{2}}\\
& = & \frac{\left|0\right\rangle +e^{2i\pi0.j_{2}j_{3}}\left|1\right\rangle }{\sqrt{2}}\\
& \xrightarrow[w/j_{4}]{R_{3}} & \frac{\left|0\right\rangle +e^{2i\pi0.j_{2}j_{3}}e^{2\pi ij_{4}/2^{3}}\left|1\right\rangle }{\sqrt{2}}\\
& = & \frac{\left|0\right\rangle +e^{2i\pi0.j_{2}j_{3}j_{4}}\left|1\right\rangle }{\sqrt{2}}\\
& \vdots\\
& \xrightarrow[w/j_{n}]{R_{n-1}} & \frac{\left|0\right\rangle +e^{2i\pi0.j_{2}\ldots j_{n}}\left|1\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Last qubit:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|j_{n}\right\rangle & \xrightarrow{H} & \frac{\left|0\right\rangle +e^{2i\pi0.j_{n}}\left|1\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Circuit output is
\begin_inset Formula \[
\frac{\left|0\right\rangle +e^{2i\pi0.j_{1}\ldots j_{n}}\left|1\right\rangle }{\sqrt{2}}\otimes\cdots\otimes\frac{\left|0\right\rangle +e^{2i\pi0.j_{n}}\left|1\right\rangle }{\sqrt{2}}\]
\end_inset
\end_layout
\begin_layout Standard
which is the Lemma 1 expression for the Fourier transform with qubits in
reverse order.
To correct this,
\begin_inset Formula $\left\lfloor \frac{n}{2}\right\rfloor $
\end_inset
swap gates can be used to swap the top and bottom bits, second to top and
second to bottom bits, and so on.
Each swap gate is made of 3 CNOT gates.
\end_layout
\begin_layout Standard
Cost: Each qubit requires
\begin_inset Formula $O\left(n\right)$
\end_inset
gates to transform, total cost is
\begin_inset Formula $O\left(n^{2}\right)$
\end_inset
for transforming all qubits and
\begin_inset Formula $O\left(n\right)$
\end_inset
for swaps, which is
\begin_inset Formula $O\left(n^{2}\right)$
\end_inset
total.
Best classical cost is
\begin_inset Formula $O\left(N\log N\right)=O\left(2^{n}n\right)$
\end_inset
, so quantum implementation represents an exponential speedup.
\end_layout
\begin_layout Standard
Next Lecture: Phase Estimation
\end_layout
\begin_layout Standard
Homework: Implement
\emph on
reverse
\emph default
fourier transform, finding algorithm and cost.
\end_layout
\begin_layout Section*
Lecture 17 (21 March)
\end_layout
\begin_layout Subsection*
Lecture 16 Review
\end_layout
\begin_layout Standard
\begin_inset Formula $F\left|j\right\rangle =\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}e^{2\pi ijk/N}\left|k\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $F\left|j_{1}\ldots j_{n}\right\rangle =\left(\frac{\left|0\right\rangle +e^{2\pi i0.j_{n}}\left|k\right\rangle }{\sqrt{0}}\right)\otimes\cdots\otimes\left(\frac{\left|0\right\rangle +e^{2\pi i0.j_{1}\ldots j_{n}}\left|k\right\rangle }{\sqrt{0}}\right)$
\end_inset
\end_layout
\begin_layout Standard
Cost: quantum implementation
\begin_inset Formula $O\left(n^{2}\right)$
\end_inset
beats classical
\begin_inset Formula $O\left(2^{n}n\right)=O\left(N\log N\right)$
\end_inset
\end_layout
\begin_layout Standard
Hint: Finding
\begin_inset Formula $F^{H}$
\end_inset
, problem 1 next homework.
Two approaches.
One is to look at the circuit and determine meaning of the conjugate transpose
as an operator.
Other is to look at the definition of
\begin_inset Formula $F^{H}$
\end_inset
which differs only by a minus sign.
\begin_inset Formula \[
F^{H}\left|j\right\rangle =\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}e^{-2\pi ijk/N}\left|k\right\rangle \]
\end_inset
Cost should be the same.
\end_layout
\begin_layout Subsection*
Phase Estimation
\end_layout
\begin_layout Subsubsection*
Overview
\end_layout
\begin_layout Standard
Heart of many quantum algorithms.
Related to solution to Schrodinger's equation, important for quantum simulation.
\end_layout
\begin_layout Standard
Problem: Given unitary matrix
\begin_inset Formula $U$
\end_inset
, size
\begin_inset Formula $N\times N$
\end_inset
where
\begin_inset Formula $N=2^{k}$
\end_inset
(using
\begin_inset Formula $k$
\end_inset
instead of
\begin_inset Formula $n$
\end_inset
as in previous lecture because
\begin_inset Formula $n$
\end_inset
is used for something else here).
Also given
\begin_inset Formula $\left|u\right\rangle $
\end_inset
, eigenvector of
\begin_inset Formula $U$
\end_inset
, so
\end_layout
\begin_layout Standard
\begin_inset Formula \[
U\left|u\right\rangle =\lambda\left|u\right\rangle =e^{2\pi i\varphi}\left|u\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
for
\begin_inset Formula $\varphi\in\left[0,1\right]$
\end_inset
.
Goal is to find approximation of
\begin_inset Formula $\varphi$
\end_inset
with accuracy
\begin_inset Formula $2^{-n}$
\end_inset
.
Algorithm is covered in this lecture, the correctness in shown next lecture.
\begin_inset Formula $\varphi$
\end_inset
can be represented as:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\varphi=0.\varphi_{1}\varphi_{2}\ldots\varphi_{n}\varphi_{n+1}\]
\end_inset
\end_layout
\begin_layout Standard
If only
\begin_inset Formula $n$
\end_inset
digits are given, precision is lost but bounded by a maximum error.
Maximum error can be computed by assuming every digit after the
\begin_inset Formula $n$
\end_inset
th is 1 when it should be zero:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\sum_{j=n+1}^{\infty}\frac{1}{2^{j}}=\frac{1}{2^{n+1}}\sum_{j=0}^{\infty}\frac{1}{2^{j}}=\frac{1}{2^{n+1}}\left(\frac{1}{1-\frac{1}{2}}\right)=2^{^{-n}}\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Givens
\end_layout
\begin_layout Standard
1.
Given
\begin_inset Formula $\left|u\right\rangle $
\end_inset
, eigenvector as superposition state
\begin_inset Formula $k$
\end_inset
qubits long.
\end_layout
\begin_layout Standard
2.
Given controlled operators
\begin_inset Formula $U^{2^{j}}$
\end_inset
for
\begin_inset Formula $j=0,2,\ldots$
\end_inset
implemented as black boxes.
\end_layout
\begin_layout Standard
[See also paper: Quantum Algorithms Revisited]
\end_layout
\begin_layout Subsubsection*
Algorithm
\end_layout
\begin_layout Standard
To see understand the algorithm, it helps to look at how a
\begin_inset Formula $U^{2^{j}}$
\end_inset
gate controlled by an
\begin_inset Formula $H\left|0\right\rangle $
\end_inset
qubit acts:
\begin_inset Formula \begin{eqnarray*}
\frac{\left|0\right\rangle \left|u\right\rangle +\left|1\right\rangle \left|u\right\rangle }{\sqrt{2}} & \xrightarrow{C\left(U^{2^{j}}\right)} & \frac{\left|0\right\rangle \left|u\right\rangle +\left|1\right\rangle U^{2^{j}}\left|u\right\rangle }{\sqrt{2}}\\
& = & \frac{\left|0\right\rangle \left|u\right\rangle +\left|1\right\rangle e^{2\pi i\varphi2^{j}}\left|u\right\rangle }{\sqrt{2}}\\
& = & \frac{\left|0\right\rangle +\left|1\right\rangle e^{2\pi i\varphi2^{j}}}{\sqrt{2}}\left|u\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Using different values of
\begin_inset Formula $j$
\end_inset
results in qubits that can be expressed in terms of different portions
of the bitwise representation of
\begin_inset Formula $\varphi$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula $U^{2^{t-1}}\rightarrow\left|0\right\rangle +e^{2\pi i\varphi2^{t-1}}\left|1\right\rangle =\left|0\right\rangle +e^{2\pi i0.\varphi_{t}\varphi_{t+1}\ldots}\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U^{2^{t-2}}\rightarrow\left|0\right\rangle +e^{2\pi i\varphi2^{t-2}}\left|1\right\rangle =\left|0\right\rangle +e^{2\pi i0.\varphi_{t-1}\varphi_{t}\varphi_{t+1}\ldots}\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\vdots$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U^{2^{1}}\rightarrow\left|0\right\rangle +e^{2\pi i\varphi2}\left|1\right\rangle =\left|0\right\rangle +e^{2\pi i0.\varphi_{2}\ldots\varphi_{t}\varphi_{t+1}\ldots}\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $U^{2^{0}}\rightarrow\left|0\right\rangle +e^{2\pi i\varphi}\left|1\right\rangle =\left|0\right\rangle +e^{2\pi i0.\varphi_{1}\ldots\varphi_{t}\varphi_{t+1}\ldots}\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
As can be seen above,
\begin_inset Formula $U^{2^{j}}$
\end_inset
essentially means shift
\begin_inset Formula $\varphi$
\end_inset
by
\begin_inset Formula $j$
\end_inset
bits to the left.
The whole number portion of the resulting numbers is discarded because
the exponential function is periodic.
\end_layout
\begin_layout Standard
The states shown above look like the states that would result from
\begin_inset Formula $F\left|\varphi_{1}\ldots\varphi_{t}\right\rangle $
\end_inset
, where
\begin_inset Formula $F$
\end_inset
is the quantum fourier transform.
\end_layout
\begin_layout Subsubsection*
Circuit
\end_layout
\begin_layout Standard
Circuit is made of two registers, top register is
\begin_inset Formula $t$
\end_inset
\begin_inset Formula $\left|0\right\rangle $
\end_inset
qubits, bottom register is
\begin_inset Formula $\left|u\right\rangle $
\end_inset
(which is
\begin_inset Formula $k$
\end_inset
qubits long).
Hadamard gates are applied to each
\begin_inset Formula $\left|0\right\rangle $
\end_inset
qubit in the first register, and output of those is used to control a sequence
of
\begin_inset Formula $U^{2^{j}}$
\end_inset
gates on the second register.
The first gate on the second register,
\begin_inset Formula $U^{2^{0}}$
\end_inset
, is controlled by the
\family roman
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\shape up
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\emph off
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\noun off
\color none
\begin_inset Formula $H\left|0\right\rangle $
\end_inset
output on the first qubit, then there is a
\begin_inset Formula $U^{2^{1}}$
\end_inset
gate controlled by
\begin_inset Formula $H\left|0\right\rangle $
\end_inset
from the second qubit, followed by a
\begin_inset Formula $U^{2^{2}}$
\end_inset
gate controlled by the third qubit, and so on.
[Textbook figure 5.2 page 222]
\end_layout
\begin_layout Standard
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
After these gates, an inverse fourier transform is applied to the first
register, and measurement of that register on the computational basis yields
the binary digits of the representation of
\begin_inset Formula $\varphi$
\end_inset
.
\end_layout
\begin_layout Standard
We need to show that before the inverse fourier transform is applied, the
top register has value
\begin_inset Formula $\frac{1}{2^{t/2}}\sum_{k=0}^{2^{t}-1}e^{2\pi ik\varphi}\left|k\right\rangle $
\end_inset
.
This can be proved with induction.
Start with the last two qubits:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
& & \frac{1}{2}\left(\left|0\right\rangle +e^{2\pi i2\varphi}\left|1\right\rangle \right)\left(\left|0\right\rangle +e^{2\pi i\varphi}\left|1\right\rangle \right)\\
& & =\left|00\right\rangle +e^{2\pi i\varphi}\left|01\right\rangle +e^{2\pi i2\varphi}\left|10\right\rangle +e^{2\pi i3\varphi}\left|11\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Then the inductive step is:
\begin_inset Formula \begin{eqnarray*}
& & \frac{1}{\sqrt{2}}\left(\left|0\right\rangle +e^{2\pi i2^{t-1}\varphi}\left|1\right\rangle \right)\left(\frac{1}{2^{\left(t-1\right)/2}}\sum_{k=0}^{2^{t-1}-1}e^{2\pi ik\varphi}\left|k\right\rangle \right)\\
& & =\frac{1}{2^{t/2}}\sum_{k=0}^{2^{t-1}-1}\left(e^{2\pi ik\varphi}\left|0k\right\rangle +e^{2\pi i\left(k+2^{t-1}\right)\varphi}\left|1k\right\rangle \right)\\
& & =\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}e^{2\pi ij\varphi}\left|j\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsubsection*
Circuit and Expression
\end_layout
\begin_layout Standard
A second interpretation of phase estimation can be seen by looking at the
overall circuit diagram [Textbook figure 5.3 page 223].
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|0\right\rangle ^{\otimes t}\left|u\right\rangle & \xrightarrow{H^{\otimes t}\otimes I} & \frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle \left|u\right\rangle \\
& \xrightarrow{U^{j}} & \frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle U^{j}\left|u\right\rangle \\
& = & \frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle e^{2\pi ij\varphi}\left|u\right\rangle \\
& \xrightarrow{F^{H}} & \sum_{\ell=0}^{2^{t}-1}g\left(\ell,\varphi\right)\left|\ell\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
We aren't solving for the coefficients of possible output basis states right
now, we just refer to them here as
\begin_inset Formula $g\left(\ell,\varphi\right)$
\end_inset
or
\begin_inset Formula $\alpha_{\ell}$
\end_inset
.
(The next lecture solves for
\begin_inset Formula $\alpha_{\ell}$
\end_inset
).
Now when
\begin_inset Formula $\varphi$
\end_inset
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
can be expressed as
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\color inherit
\begin_inset Formula $0.\varphi_{1}\ldots\varphi_{t}$
\end_inset
exactly, there is unique
\begin_inset Formula $\ell$
\end_inset
so that
\begin_inset Formula \[
\left|a_{\ell}\right|=\left|g\left(\ell_{1}\varphi\right)\right|=1\]
\end_inset
\end_layout
\begin_layout Standard
and all the other
\begin_inset Formula $\alpha_{\ell}$
\end_inset
values are 0.
The algorithm succeeds in this case, but it also succeeds more generally,
and this is shown in the next lecture.
\end_layout
\begin_layout Standard
(Above depends on the fact that the sequence of controlled-U operations
in the circuit transform a basis state
\begin_inset Formula $\left|j\right\rangle \left|u\right\rangle $
\end_inset
to
\begin_inset Formula $\left|j\right\rangle U^{j}\left|u\right\rangle .$
\end_inset
This is exercise 5.7 in the book, and can be seen from the fact that if
\begin_inset Formula $j$
\end_inset
has a
\begin_inset Formula $t$
\end_inset
bit representation:
\begin_inset Formula \begin{eqnarray*}
\left|j\right\rangle U^{j}\left|u\right\rangle & = & \left|j\right\rangle U^{j_{t}2^{o}+j_{t-1}2^{t}+\cdots+j_{1}2^{t-1}}\left|u\right\rangle \\
& = & \left|j\right\rangle U^{2^{o}j_{t}}\times U^{2^{1}j_{t-1}}\times\cdots\times U^{2^{t-1}j_{t}}\left|u\right\rangle \end{eqnarray*}
\end_inset
)
\end_layout
\begin_layout Standard
Cost of circuit in gates is
\begin_inset Formula $t$
\end_inset
H gates,
\begin_inset Formula $t$
\end_inset
controlled
\begin_inset Formula $U^{2^{j}}$
\end_inset
gates (assuming the exponents don't affect cost), and an
\begin_inset Formula $F^{H}$
\end_inset
gate which has cost
\begin_inset Formula $t^{2}$
\end_inset
.
\begin_inset Formula $O\left(t+t+t^{2}\right)=O\left(t^{2}\right)$
\end_inset
.
Cost in qubits is
\begin_inset Formula $t+k$
\end_inset
,
\begin_inset Formula $O\left(t+k\right)$
\end_inset
\end_layout
\begin_layout Standard
Application: If you have real matrix,
\begin_inset Formula $A$
\end_inset
, so that
\begin_inset Formula $A=A^{T}$
\end_inset
,
\begin_inset Formula $e^{iA}$
\end_inset
is unitary,
\begin_inset Formula $ih\frac{g\psi\left(x,t\right)}{gt}=H\psi\left(x,t\right)$
\end_inset
where
\begin_inset Formula $h$
\end_inset
is Planck's constant,
\begin_inset Formula $H$
\end_inset
is Hamiltonian.
\end_layout
\begin_layout Section*
Lecture 18 (26 March)
\end_layout
\begin_layout Standard
Missed Class, filling in blanks from Textbook section 5.2.1.
\end_layout
\begin_layout Standard
Output of the first register of the phase estimation circuit before inverse
fourier transform is:
\begin_inset Formula \begin{eqnarray*}
& & \frac{1}{2^{t/2}}\left(\left|0\right\rangle +e^{2\pi i2^{t-1}\varphi}\left|1\right\rangle \right)\left(\left|0\right\rangle +e^{2\pi i2^{t-2}\varphi}\left|1\right\rangle \right)\ldots\left(\left|0\right\rangle +e^{2\pi i2^{0}\varphi}\left|1\right\rangle \right)\\
& & =\frac{1}{2^{t/2}}\sum_{k=0}^{2^{t}-1}e^{2\pi i\varphi k}\left|k\right\rangle \end{eqnarray*}
\end_inset
If
\begin_inset Formula $\varphi=0.\varphi_{1}\ldots\varphi_{t}$
\end_inset
exactly, applying inverse fourier transform to this state gives state
\begin_inset Formula $\left|\varphi_{1}\ldots\varphi_{t}\right\rangle $
\end_inset
.
When
\begin_inset Formula $\varphi$
\end_inset
cannot be represented with
\begin_inset Formula $t$
\end_inset
bits, the analysis below applies.
\end_layout
\begin_layout Standard
Let
\begin_inset Formula $b$
\end_inset
be integer in the range 0 to
\begin_inset Formula $2^{t}-1$
\end_inset
such that
\begin_inset Formula $b/2^{t}=0.b_{1}\ldots b_{t}$
\end_inset
is the best
\begin_inset Formula $t$
\end_inset
bit approximation to
\begin_inset Formula $\varphi$
\end_inset
which is less than
\begin_inset Formula $\varphi$
\end_inset
.
Error is
\begin_inset Formula $\delta\equiv\varphi-b/2^{t}$
\end_inset
and
\begin_inset Formula $0\le\delta\le2^{t-1}$
\end_inset
.
Applying the inverse fourier transform to the first register gives:
\begin_inset Formula \begin{eqnarray*}
& & \frac{1}{2^{t}}\sum_{\ell=0}^{2^{t}-1}\sum_{k=0}^{2^{t}-1}e^{-2\pi ik\ell/2^{t}}e^{2\pi i\varphi k}\left|\ell\right\rangle \\
& & =\sum_{\ell=0}^{2^{t}-1}\frac{1}{2^{t}}\sum_{k=0}^{2^{t}-1}\left(e^{2\pi i\left(\varphi-\ell/2^{t}\right)}\right)^{k}\left|\ell\right\rangle \\
& & =\sum_{\ell=0}^{2^{t}-1}\alpha_{\ell-b}\left|\ell\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Let
\begin_inset Formula $\alpha_{\ell}$
\end_inset
be the amplitude of
\begin_inset Formula $\left|b+\ell\right\rangle $
\end_inset
(taking addition inside the state and subtraction in the subscript of
\begin_inset Formula $\alpha$
\end_inset
to be modulo
\begin_inset Formula $2^{t}$
\end_inset
):
\begin_inset Formula \begin{eqnarray*}
\alpha_{\ell} & \equiv & \frac{1}{2^{t}}\sum_{k=0}^{2^{t}-1}\left(e^{2\pi i\left(\varphi-\left(b+\ell\right)/2^{t}\right)}\right)^{k}\\
& = & \frac{1}{2^{t}}\frac{1-e^{2\pi i\left(2^{t}\varphi-\left(b+\ell\right)\right)}}{1-e^{2\pi i\left(\varphi-\left(b+\ell\right)/2^{t}\right)}}\\
& = & \frac{1}{2^{t}}\frac{1-e^{2\pi i\left(2^{t}\delta-\ell\right)}}{1-e^{2\pi i\left(\delta-\ell/2^{t}\right)}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
The second step follows from the formula for sum of a geometric series,
the third from substituting
\begin_inset Formula $\delta=\varphi-b/2^{t}$
\end_inset
.
\end_layout
\begin_layout Standard
Introduce new variables.
Take the output of the final measurement to be
\begin_inset Formula $m$
\end_inset
, and chose an error tolerance,
\begin_inset Formula $e$
\end_inset
which is a positive integer such that if
\begin_inset Formula $\left|m-b\right|>e$
\end_inset
, the algorithm is considered to have failed.
The probability of that failure condition is:
\begin_inset Formula \[
p\left(\left|m-b\right|>e\right)=\sum_{\ell=-2^{t-1}+1}^{-\left(e+1\right)}\left|\alpha_{\ell}\right|^{2}+\sum_{\ell=e+1}^{2^{t-1}}\left|\alpha_{\ell}\right|^{2}\]
\end_inset
For any real
\begin_inset Formula $\theta$
\end_inset
,
\begin_inset Formula $\left|1-e^{i\theta}\right|\le2$
\end_inset
, so
\begin_inset Formula \[
\left|\alpha_{\ell}\right|\le\frac{1}{2^{t}\left|1-e^{2\pi i\left(\delta-\ell/2^{t}\right)}\right|}\]
\end_inset
Whenever
\begin_inset Formula $-\pi\le\theta\le\pi$
\end_inset
, then
\begin_inset Formula $\left|1-e^{i\theta}\right|\ge2\left|\theta\right|/\pi$
\end_inset
.
And when
\begin_inset Formula $-2^{t-1}<\ell\le2^{t-1}$
\end_inset
, then
\begin_inset Formula $-\pi\le2\pi\left(\delta-\ell/2^{t}\right)\le\pi$
\end_inset
, therefore:
\begin_inset Formula \[
\left|\alpha_{\ell}\right|\le\frac{1}{2^{t}\cdot2\left(\delta-\ell/2^{t}\right)}=\frac{1}{-\frac{1}{2}\left(\ell-2^{t}\delta\right)}\]
\end_inset
And
\begin_inset Formula \begin{eqnarray*}
p\left(\left|m-b\right|>e\right) & \le & \frac{1}{4}\left(\sum_{\ell=-2^{t-1}+1}^{-\left(e+1\right)}\frac{1}{\left(\ell-2^{t}\delta\right)^{2}}+\sum_{\ell=e+1}^{2^{t-1}}\frac{1}{\left(\ell-2^{t}\delta\right)^{2}}\right)\\
& \le & \frac{1}{4}\left(\sum_{\ell=-2^{t-1}+1}^{-\left(e+1\right)}\frac{1}{\ell^{2}}+\sum_{\ell=e+1}^{2^{t-1}}\frac{1}{\left(\ell-1\right)^{2}}\right)\\
& \le & \frac{1}{2}\sum_{\ell=e}^{2^{t-1}+1}\frac{1}{\ell^{2}}\\
& \le & \frac{1}{2}\int_{e-1}^{2^{t-1}-1}d\ell\frac{1}{\ell^{2}}\\
& = & \frac{1}{2\left(e-1\right)}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Second step follows because
\begin_inset Formula $0\le2^{t}\delta\le1$
\end_inset
.
\end_layout
\begin_layout Standard
When approximating
\begin_inset Formula $\varphi$
\end_inset
to an accuracy of
\begin_inset Formula $2^{-n}$
\end_inset
,
\begin_inset Formula $e=2^{t-n}-1$
\end_inset
.
When using
\begin_inset Formula $t=n+p$
\end_inset
qubits in phase estimation, then the probability of success is
\begin_inset Formula $1-\frac{1}{2\left(2^{p}-2\right)}$
\end_inset
.
Let
\begin_inset Formula $\epsilon$
\end_inset
be the probability of failure, then you can find minimum
\begin_inset Formula $t$
\end_inset
that won't exceed that failure rate:
\begin_inset Formula \begin{eqnarray*}
\epsilon & = & \frac{1}{2\left(2^{p}-2\right)}\\
2^{p} & = & \frac{1}{2\epsilon}+2\\
p & = & \log_{2}\left(\frac{1}{2\epsilon}+2\right)\end{eqnarray*}
\end_inset
So for success probability of at least
\begin_inset Formula $1-\epsilon$
\end_inset
, choose
\begin_inset Formula $t=n+\left\lceil \log_{2}\left(2+\frac{1}{2\epsilon}\right)\right\rceil $
\end_inset
:
\end_layout
\begin_layout Section*
Lecture 19 (28 March)
\end_layout
\begin_layout Subsection*
Homework Hint
\end_layout
\begin_layout Standard
Homework 4, Problem 2: Convolution Theorem and Fourier Transform
\end_layout
\begin_layout Standard
Given coefficients of basis states
\begin_inset Formula $\alpha_{0}$
\end_inset
,
\begin_inset Formula $\alpha_{N-1}$
\end_inset
and
\begin_inset Formula $\beta_{0}$
\end_inset
,
\begin_inset Formula $\beta_{N-1}$
\end_inset
which tranform into
\begin_inset Formula $\gamma_{0}$
\end_inset
,
\begin_inset Formula $\gamma_{N-1}$
\end_inset
and
\begin_inset Formula $\delta_{0}$
\end_inset
,
\begin_inset Formula $\delta_{N-1}$
\end_inset
\end_layout
\begin_layout Standard
Discrete FT:
\begin_inset Formula \[
\gamma_{j}=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\alpha_{k}e^{2\pi ijk/N},\quad\delta_{j}=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\beta_{k}e^{2\pi ijk/N}\]
\end_inset
Inverse FT:
\begin_inset Formula \[
\alpha_{j}=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\gamma_{k}e^{-2\pi ijk/N},\quad\beta_{j}=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\delta_{k}e^{-2\pi ijk/N}\]
\end_inset
Convolution:
\begin_inset Formula \[
\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}\sum_{\ell=0}^{N-1}\alpha_{\ell}\beta_{j-\ell}\left|j\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
Use FT formulas to substitute
\begin_inset Formula $\alpha_{\ell}$
\end_inset
and
\begin_inset Formula $\beta_{j-\ell}$
\end_inset
above:
\begin_inset Formula \[
\beta_{j-\ell}=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\delta_{k}e^{-2\pi i\left(j-\ell\right)k/N}\]
\end_inset
\end_layout
\begin_layout Standard
Result is messy, you end up with four sums, but it simplifies.
\end_layout
\begin_layout Subsection*
Performance Analysis of Phase Estimation
\end_layout
\begin_layout Standard
We can compute bounded probability of failure and use that to determine
how many qubits to use in top register to get desired accuracy.
Last lecture proved:
\begin_inset Formula \[
Pr\left\{ \left|m-b\right|>e=2^{t-n}-1\right\} \le\frac{1}{2\left(e-1\right)}<\epsilon\]
\end_inset
where
\begin_inset Formula $\epsilon$
\end_inset
is highest allowed probability of failure,
\begin_inset Formula $m$
\end_inset
is the measurement of the first register on the computational basis,
\begin_inset Formula $b$
\end_inset
is the representation of
\begin_inset Formula $\varphi$
\end_inset
expressed as a measurement,
\begin_inset Formula $e$
\end_inset
is our error tolerance, expressed as the highest allowed absolute difference
between
\begin_inset Formula $m$
\end_inset
and
\begin_inset Formula $b$
\end_inset
.
\begin_inset Formula $t$
\end_inset
is the number of qubits in the top register and
\begin_inset Formula $n$
\end_inset
is the desired accuracy in bits, which is just an alternate expression
of error tolerance.
\end_layout
\begin_layout Standard
This expression of probability of failure and accuracy is unwieldy and not
what we originally set out to determine, which was finding:
\begin_inset Formula \[
Pr\left\{ \left|\varphi-\hat{\varphi}\right|\leq2^{-n}\right\} \]
\end_inset
where
\begin_inset Formula $\hat{\varphi}=\frac{m}{2^{t}}$
\end_inset
.
To find this, we switch to finding probability of failure instead of probabilit
y of success because that it is easier to bound that from above.
\begin_inset Formula \[
Pr\left\{ \left|\varphi-\hat{\varphi}\right|>2^{-n}\right\} =Pr\left\{ \left|\varphi-\frac{b}{2^{t}}+\frac{b}{2^{t}}-\hat{\varphi}\right|>2^{-n}\right\} \]
\end_inset
Using the triangle inequality (
\begin_inset Formula $\left|a+b\right|\le\left|a\right|+\left|b\right|$
\end_inset
) :
\begin_inset Formula \begin{eqnarray*}
& \le & Pr\left\{ \left|\varphi-\frac{b}{2^{t}}\right|+\left|\frac{b}{2^{t}}-\hat{\varphi}\right|>2^{-n}\right\} \\
& \le & Pr\left\{ 2^{-t}+\left|\frac{b}{2^{t}}-\hat{\varphi}\right|>2^{-n}\right\} \\
& = & Pr\left\{ \left|\frac{b}{2^{t}}-\hat{\varphi}\right|>2^{-n}-2^{-t}\right\} \\
& = & Pr\left\{ \left|b-m\right|>2^{t-n}-1\right\} \\
& = & Pr\left\{ \left|b-m\right|>e\right\} \\
& \le & \frac{1}{2\left(e-1\right)}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
P.E.
w/ approx Eigenvector
\end_layout
\begin_layout Standard
Phase estimation algorithm requires an eigenvector of the matrix
\begin_inset Formula $U$
\end_inset
, but it is also possible to use approximations of an eigenvector and still
get meaningful results.
[Paper: Abrams + Lloyd]
\end_layout
\begin_layout Standard
Given
\begin_inset Formula $\left|u\right\rangle $
\end_inset
, an approximate eigenvector which can be expressed in terms of real eigenvector
s
\begin_inset Formula $\left|u_{k}\right\rangle $
\end_inset
as:
\begin_inset Formula \[
\left|u\right\rangle =\sum_{k=0}^{N-1}d_{k}\left|u_{k}\right\rangle \]
\end_inset
We would like to estimate the phase
\begin_inset Formula $\varphi_{0}$
\end_inset
corresponding to
\begin_inset Formula $\lambda_{0}=e^{2\pi i\varphi_{0}}\left|u_{0}\right\rangle $
\end_inset
.
The of the P.E.
on this input is:
\begin_inset Formula \begin{eqnarray*}
\left|0\right\rangle ^{\otimes t}\left|u\right\rangle & \xrightarrow{H^{\otimes t}\otimes I} & \frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle \left|u\right\rangle \\
& = & \frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle \sum_{k=0}^{2^{t}-1}d_{k}\left|u_{k}\right\rangle \\
& = & \sum_{k=0}^{2^{t}-1}d_{k}\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle \left|u_{k}\right\rangle \\
& \xrightarrow{U^{j}} & \sum_{k=0}^{2^{t}-1}d_{k}\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle U^{j}\left|u_{k}\right\rangle \\
& = & \sum_{k=0}^{2^{t}-1}d_{k}\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}e^{2\pi i\varphi_{k}j}\left|j\right\rangle \left|u_{k}\right\rangle \\
& \xrightarrow{F^{H}\otimes I} & \sum_{k=0}^{2^{t}-1}d_{k}\sum_{j=0}^{2^{t}-1}g\left(\varphi_{k},j\right)\left|j\right\rangle \left|u_{k}\right\rangle \end{eqnarray*}
\end_inset
\begin_inset Formula $g\left(\varphi_{k},j\right)$
\end_inset
in the previous lecture was
\begin_inset Formula $\alpha_{j}$
\end_inset
, the amplitude of output state
\begin_inset Formula $j$
\end_inset
in the top register.
The last lecture showed that
\begin_inset Formula $\alpha_{j}$
\end_inset
amplitudes were bounded from above, and that if
\begin_inset Formula $j$
\end_inset
was far from
\begin_inset Formula $b$
\end_inset
, then
\begin_inset Formula $\alpha_{j}$
\end_inset
would be low.
\end_layout
\begin_layout Standard
Next, measure top register to find
\begin_inset Formula $Pr\left\{ \left|m-b_{o}\right|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|y\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $U\left|y\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|0\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|2\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|2\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|4\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|3\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|1\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|4\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|3\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|5\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|5\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|6\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|6\right\rangle $
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|7\right\rangle $
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left|7\right\rangle $
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Section*
Lecture 20 (2 April)
\end_layout
\begin_layout Subsection*
Lecture 19 Review
\end_layout
\begin_layout Standard
Item 1: With initial state
\begin_inset Formula $\left|u\right\rangle $
\end_inset
, eigenvector for phase estimation satisfies
\begin_inset Formula $\varphi-\hat{\varphi}\le2^{-n}$
\end_inset
where
\begin_inset Formula $n$
\end_inset
is the number of bits of desired accuracy, and
\begin_inset Formula $\hat{\varphi}=\frac{m}{2^{t}}$
\end_inset
is the measured value approximating
\begin_inset Formula $\varphi$
\end_inset
.
This condition has to hold with probability
\begin_inset Formula $\ge\left(1-\epsilon\right)$
\end_inset
,
\begin_inset Formula $\epsilon$
\end_inset
being the allowed probability of failure.
\end_layout
\begin_layout Standard
Item 2: If the initial state is some arbitrary initial vector,
\begin_inset Formula $\left|\tilde{u}\right\rangle $
\end_inset
, instead of an eigenvector, the output will still satisfy
\begin_inset Formula $\varphi-\hat{\varphi}\le2^{-n}$
\end_inset
with probability
\begin_inset Formula $\ge\left|\left\langle u|\tilde{u}\right\rangle \right|^{2}\left(1-\epsilon\right)$
\end_inset
.
[There is another proof of this fact in 2 lines in a paper online about
constructing initial states]
\end_layout
\begin_layout Standard
Example: Take unitary matrix
\begin_inset Formula $U$
\end_inset
, which has two phases
\begin_inset Formula $\varphi_{1}$
\end_inset
,
\begin_inset Formula $\varphi_{2}$
\end_inset
and two eigenvectors
\begin_inset Formula $u_{1}$
\end_inset
,
\begin_inset Formula $u_{2}$
\end_inset
.
The eigenvalues are related to the phases like
\begin_inset Formula $\lambda_{1}=e^{2\pi i\varphi_{1}}$
\end_inset
,
\begin_inset Formula $\lambda_{2}=e^{2\pi i\varphi_{2}}$
\end_inset
.
If the initial state is
\begin_inset Formula $\left|\tilde{u}\right\rangle =\frac{1}{2}\left|u_{1}\right\rangle +\frac{\sqrt{3}}{2}\left|u_{2}\right\rangle $
\end_inset
, the phase estimation algorithm will give close approximation of
\begin_inset Formula $\varphi_{1}$
\end_inset
with probability
\begin_inset Formula $\left(\frac{1}{2}\right)^{2}\left(1-\epsilon\right)$
\end_inset
, and a close approximation of
\begin_inset Formula $\varphi_{2}$
\end_inset
with probability
\begin_inset Formula $\left(\frac{\sqrt{3}}{2}\right)^{2}\left(1-\epsilon\right)$
\end_inset
.
\end_layout
\begin_layout Subsection*
Order Finding
\end_layout
\begin_layout Standard
Given
\begin_inset Formula $x$
\end_inset
,
\begin_inset Formula $N$
\end_inset
:
\begin_inset Formula $x1$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
2b & = & 1\left(\textrm{mod }4\right)\\
2b-1 & = & 4\ell\\
\textrm{odd} & = & \textrm{even}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $x$
\end_inset
has multiplicative inverse
\begin_inset Formula $x^{-1}\left(\textrm{mod }N\right)$
\end_inset
iff
\begin_inset Formula $\gcd\left(x,N\right)=1$
\end_inset
.
\end_layout
\begin_layout Subsubsection*
Showing U is Unitarry
\end_layout
\begin_layout Standard
To show that the mapping
\begin_inset Formula $xy\left(\textrm{mod }N\right)$
\end_inset
is 1:1 for different values of
\begin_inset Formula $y$
\end_inset
, we need to show no two values of
\begin_inset Formula $y$
\end_inset
give the same output.
\end_layout
\begin_layout Standard
Take
\begin_inset Formula $z=xy_{1}\left(\textrm{mod }N\right)$
\end_inset
and
\begin_inset Formula $z=xy_{2}\left(\textrm{mod }N\right)$
\end_inset
.
The order finding problem assumes
\begin_inset Formula $x$
\end_inset
and
\begin_inset Formula $N$
\end_inset
are coprime, so we don't have to worry about multiplicative inverses not
existing and:
\begin_inset Formula \begin{eqnarray*}
x^{-1}xy_{1}\left(\textrm{mod }N\right) & = & x^{-1}xy_{2}\left(\textrm{mod }N\right)\\
\left(1+\ell N\right)y_{1}\left(\textrm{mod }N\right) & = & \left(1+\ell N\right)y_{2}\left(\textrm{mod }N\right)\\
y_{1}\left(\textrm{mod }N\right) & = & y_{2}\left(\textrm{mod }N\right)\\
y_{1} & = & y_{2}\end{eqnarray*}
\end_inset
Therefore
\begin_inset Formula $P$
\end_inset
is 1:1
\begin_inset Formula $\Rightarrow$
\end_inset
\begin_inset Formula $U$
\end_inset
is 1:1
\begin_inset Formula $\Rightarrow$
\end_inset
\begin_inset Formula $U$
\end_inset
is unitary .
\end_layout
\begin_layout Standard
If
\begin_inset Formula $P$
\end_inset
is a permutation matrix then
\begin_inset Formula $P^{-1}=P^{T}$
\end_inset
\end_layout
\begin_layout Subsubsection*
Eigenvector of U
\end_layout
\begin_layout Standard
Definition: For
\begin_inset Formula $S=0,\ldots,r-1$
\end_inset
define
\begin_inset Formula $\left|u_{s}\right\rangle =\frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}e^{-2\pi isk/r}\left|x^{k}\left(\textrm{mod }N\right)\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Then
\begin_inset Formula \begin{eqnarray*}
U\left|u_{s}\right\rangle & = & \frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}e^{-2\pi isk/r}U\left|x^{k}\left(\textrm{mod }N\right)\right\rangle \\
& = & \frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}e^{-2\pi isk/r}\left|x^{k+1}\left(\textrm{mod }N\right)\right\rangle \\
& = & \frac{1}{\sqrt{r}}\sum_{k=1}^{r}e^{-2\pi is\left(k-1\right)/r}\left|x^{k}\left(\textrm{mod }N\right)\right\rangle \\
& = & \frac{1}{\sqrt{r}}e^{2\pi is/r}\sum_{k=1}^{r}e^{-2\pi isk/r}\left|x^{k}\left(\textrm{mod }N\right)\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $k=r$
\end_inset
, then
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $e^{-2\pi isr/r}=1$
\end_inset
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\color inherit
,
\begin_inset Formula $\left|x^{r}\left(\textrm{mod }N\right)\right\rangle =\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $k=0$
\end_inset
, then
\begin_inset Formula $e^{-2\pi is0/r}=1$
\end_inset
,
\begin_inset Formula $\left|x^{0}\left(\textrm{mod }N\right)\right\rangle =\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
This means we can sum from
\begin_inset Formula $0$
\end_inset
to
\begin_inset Formula $r-1$
\end_inset
instead of
\begin_inset Formula $1$
\end_inset
to
\begin_inset Formula $r$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
U\left|u_{s}\right\rangle & = & \frac{1}{\sqrt{r}}e^{2\pi is/r}\sum_{k=0}^{r-1}e^{-2\pi isk/r}\left|x^{k}\left(\textrm{mod }N\right)\right\rangle \\
& = & e^{-2\pi is/r}\left|u_{s}\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsubsection*
Phase Estimation for Order Finding
\end_layout
\begin_layout Standard
The phase of matrix
\begin_inset Formula $U$
\end_inset
given eigenvector
\begin_inset Formula $\left|u_{s}\right\rangle $
\end_inset
will be
\begin_inset Formula $\frac{s}{r}$
\end_inset
and the output of phase estimation
\begin_inset Formula $\hat{\varphi}$
\end_inset
, will approximate this.
\end_layout
\begin_layout Standard
Possible problems with using this approach to find
\begin_inset Formula $r$
\end_inset
:
\end_layout
\begin_layout Standard
1.
We do not know how to construct initial state
\begin_inset Formula $\left|u_{s}\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
2.
We do not know how to get
\begin_inset Formula $r$
\end_inset
from
\begin_inset Formula $\varphi=\frac{s}{r}$
\end_inset
.
\end_layout
\begin_layout Standard
3.
We do not know how to compute
\begin_inset Formula $U^{j}$
\end_inset
.
\end_layout
\begin_layout Subsubsection*
Initial State for Phase Estimation
\end_layout
\begin_layout Standard
Regarding problem 1, it turns out there is a trivial way to construct a
suitable approximate initial state.
Take the combination of all eigenvectors:
\begin_inset Formula \begin{eqnarray*}
\frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\left|u_{s}\right\rangle & = & \frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\frac{1}{\sqrt{r}}\sum_{k=0}^{r-1}e^{-2\pi isk/r}\left|x^{k}\left(\textrm{mod }N\right)\right\rangle \\
& = & \frac{1}{r}\sum_{k=0}^{r-1}\sum_{s=0}^{r-1}e^{-2\pi isk/r}\left|x^{k}\left(\textrm{mod }N\right)\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $k=0$
\end_inset
, this simplifies to
\begin_inset Formula $\left|1\right\rangle $
\end_inset
\end_layout
\begin_layout Standard
Case
\begin_inset Formula $k>0$
\end_inset
, the coefficients for each output state are given by geometric series:
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\frac{e^{-2\pi ik/r\cdot\left(r-1+1\right)}-1}{e^{-2\pi ik/r}-1}=\frac{1-1}{e^{-2\pi ik/r}-1}=0\]
\end_inset
\end_layout
\begin_layout Standard
(Geometric series
\begin_inset Formula \begin{eqnarray*}
\sum_{k=0}^{n}r^{k} & = & r^{0}+r^{1}+\cdots+r^{n}\\
\left(1-r\right)\sum_{k=0}^{n}r^{k} & = & \left(r^{0}+r^{1}+\cdots+r^{n}\right)-\left(r^{1}+r^{2}+\cdots+r^{n+1}\right)\\
& = & r^{0}-r^{n+1}\\
\sum_{k=0}^{n}r^{k} & = & \frac{1-r^{n+1}}{1-r}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
So,
\begin_inset Formula \begin{eqnarray*}
\frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\left|u_{s}\right\rangle & = & \frac{1}{r}r\left|x^{0}\left(\textrm{mod }N\right)\right\rangle =\left|1\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
The state
\begin_inset Formula $\left|1\right\rangle $
\end_inset
is equal to a combination of all the eigenvectors of
\begin_inset Formula $U$
\end_inset
, and also happens to be extremely easy to implement, making it a good initial
input for phase estimation.
Note that the state
\begin_inset Formula $\left|1\right\rangle $
\end_inset
is
\begin_inset Formula $L$
\end_inset
qubits long, expressed as
\begin_inset Formula $\left|0\ldots01\right\rangle $
\end_inset
in binary.
\end_layout
\begin_layout Standard
Showing P.E.
with this initial state (following circuit diagram 5.3 again, page 223):
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|0\right\rangle ^{\otimes t}\left|1\right\rangle & = & \left|0\right\rangle ^{\otimes t}\frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\left|u_{s}\right\rangle \\
& = & \frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\left|0\right\rangle ^{\otimes t}\left|u_{s}\right\rangle \\
& \xrightarrow{H^{\otimes t}\otimes I} & \frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle \left|u_{s}\right\rangle \\
& \xrightarrow{U^{j}} & \frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle U^{j}\left|u_{s}\right\rangle \\
& = & \frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}\left|j\right\rangle e^{2\pi isj/r}\left|u_{s}\right\rangle \\
& = & \sum_{s=0}^{r-1}\frac{1}{\sqrt{r}}\left(\frac{1}{2^{t/2}}\sum_{j=0}^{2^{t}-1}e^{2\pi isj/r}\left|j\right\rangle \right)\left|u_{s}\right\rangle =\left|\psi\right\rangle \\
& \xrightarrow{F^{H}\otimes I} & \sum_{s=0}^{r-1}\frac{1}{\sqrt{r}}\left(\sum_{\ell=0}^{2^{t}-1}g\left(\varphi_{s},\ell\right)\left|\ell\right\rangle \right)\left|u_{s}\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\varphi_{s}$
\end_inset
is just
\begin_inset Formula $\frac{s}{r}$
\end_inset
.
If
\begin_inset Formula $\varphi_{s}2^{t}$
\end_inset
is an exact whole number, then
\begin_inset Formula $g\left(\varphi_{s},\ell\right)$
\end_inset
is
\begin_inset Formula $1$
\end_inset
when
\begin_inset Formula $\ell=\varphi_{s}2^{t}$
\end_inset
and
\begin_inset Formula $0$
\end_inset
otherwise.
\end_layout
\begin_layout Standard
In the general case, assuming initial state is
\begin_inset Formula $\left|u_{s}\right\rangle $
\end_inset
, phase estimation produces an approximation with
\begin_inset Formula $n$
\end_inset
bits accuracy satisfying
\begin_inset Formula $\left|\varphi-\hat{\varphi}\right|\le2^{-n}$
\end_inset
with probability
\begin_inset Formula $\left(1-\epsilon\right)$
\end_inset
.
Since we are starting with state
\begin_inset Formula $\left|1\right\rangle $
\end_inset
instead of
\begin_inset Formula $\left|u_{s}\right\rangle $
\end_inset
, the success probability is
\begin_inset Formula $\ge\left(\frac{1}{\sqrt{r}}\right)^{2}\left(1-\epsilon\right)$
\end_inset
\end_layout
\begin_layout Standard
This success probability, which is the probability of
\begin_inset Formula $\hat{\varphi}$
\end_inset
being close to
\begin_inset Formula $\frac{s}{r}$
\end_inset
for some specific value of
\begin_inset Formula $s$
\end_inset
, is very small.
In reality though, we don't care which value of
\begin_inset Formula $s$
\end_inset
(which eigenvector) phase estimation returns the phase for, because we
only care about the ratio
\begin_inset Formula $\frac{s}{r}$
\end_inset
which we can recover from any
\begin_inset Formula $s$
\end_inset
.
The success probability is high enough to allow this.
\end_layout
\begin_layout Section*
Lecture 21 (4 April)
\end_layout
\begin_layout Subsection*
Lecture 20 Review
\end_layout
\begin_layout Standard
Phase estimation for order finding
\end_layout
\begin_layout Standard
Initial state
\begin_inset Formula $\left|1\right\rangle =\frac{1}{\sqrt{r}}\sum_{s=0}^{r-1}\left|u_{s}\right\rangle $
\end_inset
(
\begin_inset Formula $L$
\end_inset
qubits long)
\end_layout
\begin_layout Standard
U matrix given by:
\begin_inset Formula \[
U\left|y\right\rangle =\left\{ \begin{array}{ll}
\left|xy\textrm{ mod }N\right\rangle & y=0,1,2,\ldots,N-1\\
\left|y\right\rangle & y=N,N+1,\ldots,2^{L}-1\end{array}\right.\]
\end_inset
where
\begin_inset Formula $L=\left\lceil \log_{2}N\right\rceil $
\end_inset
.
\end_layout
\begin_layout Standard
Accuracy needed is
\begin_inset Formula $n=2L+1$
\end_inset
.
\begin_inset Formula \[
\left|\varphi_{s}-\hat{\varphi}_{s}\right|=\left|\frac{s}{r}-\hat{\varphi}_{s}\right|\le2^{-\left(2L+1\right)}\]
\end_inset
\begin_inset Formula \[
\lambda_{s}=e^{2\pi is/r}\]
\end_inset
Success probability is
\begin_inset Formula $\frac{1}{r}\left(1-\epsilon\right)$
\end_inset
for each
\emph on
individual
\emph default
\begin_inset Formula $s$
\end_inset
using
\begin_inset Formula $t=2L+1+\left\lceil \log_{2}\left(\frac{1}{2\epsilon}+2\right)\right\rceil $
\end_inset
\end_layout
\begin_layout Standard
Recovering denominator of
\begin_inset Formula $\frac{s}{r}$
\end_inset
is impossible for individual numerator because success probability is too
low.
But overall, if we don't care about individual values of
\begin_inset Formula $s$
\end_inset
, then we can get a high enough probability ratio.
\end_layout
\begin_layout Subsection*
Deriving order from estimated phase
\end_layout
\begin_layout Standard
Question 2 unanswered from last time: How to get
\begin_inset Formula $r$
\end_inset
from
\begin_inset Formula $\hat{\varphi}_{s}$
\end_inset
, assuing phase estimation suceeded, or
\begin_inset Formula \[
\left|\frac{s}{r}-\hat{\varphi}_{s}\right|\le2^{-\left(2L+1\right)}\]
\end_inset
for some
\begin_inset Formula $s$
\end_inset
.
\end_layout
\begin_layout Standard
Theorem: If
\begin_inset Formula $\left|\frac{s}{r}-\hat{\varphi}_{s}\right|\le\frac{1}{2r^{2}}$
\end_inset
, then
\begin_inset Formula $\frac{s}{r}$
\end_inset
is a convergent of a continued fraction for
\begin_inset Formula $\varphi$
\end_inset
, and can be computed from
\begin_inset Formula $\varphi$
\end_inset
in
\begin_inset Formula $O\left(L^{3}\right)$
\end_inset
operations, using continued fraction algorithm.
[Theorem 5.1, and A.4.16 p637]
\end_layout
\begin_layout Standard
It is not hard to satisfy condition needed to apply this theorem:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
L=\left\lceil \log_{2}N\right\rceil & \ge & \log_{2}N\\
2L+1 & \ge & 2\log_{2}N+1=2\log_{2}N+\log_{2}2=\log_{2}\left(2N^{2}\right)\\
-\left(2L+1\right) & \le & -\log_{2}\left(2N^{2}\right)\\
2^{-\left(2L+1\right)} & \le & 2^{-\log_{2}\left(2N^{2}\right)}=2^{\log_{2}\left(\frac{1}{2N^{2}}\right)}=\frac{1}{2N^{2}}\le\frac{1}{2r^{2}}\\
\left|\varphi-\hat{\varphi}_{s}\right| & \le & \frac{1}{2r^{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsubsection*
Continued Fractions
\end_layout
\begin_layout Standard
Any real number
\begin_inset Formula $x$
\end_inset
be represented as a sequence of integers
\begin_inset Formula $\left[a_{0},a_{1},\ldots a_{n}\right]$
\end_inset
which are terms in a continued fraction:
\begin_inset Formula \[
x=a_{o}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{\cdots+\frac{1}{a_{n-1}+\frac{1}{a_{n}}}}}}\]
\end_inset
Example:
\begin_inset Formula \begin{eqnarray*}
\frac{43}{18} & = & 2+\frac{7}{18}\\
& = & 2+\frac{1}{\frac{18}{7}}\\
& = & 2+\frac{1}{2+\frac{4}{7}}\\
& = & 2+\frac{1}{2+\frac{1}{1+\frac{3}{4}}}\\
& = & 2+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{3}}}}\end{eqnarray*}
\end_inset
The sequence of
\begin_inset Formula $a_{i}$
\end_inset
values can continue forever when
\begin_inset Formula $x$
\end_inset
is an arbitrary real, but will end when
\begin_inset Formula $x$
\end_inset
is a rational number because the sequence of remainders will be strictly
decreasing, and the procedure for determining
\begin_inset Formula $a_{i}$
\end_inset
terminates in
\begin_inset Formula $\log\left(\textrm{numerator or denominator}\right)$
\end_inset
.
In the case of order finding, it converges in
\begin_inset Formula $O\left(t\right)$
\end_inset
or
\begin_inset Formula $O\left(L\right)$
\end_inset
steps.
\end_layout
\begin_layout Standard
Reason: The procedure divides when remainder is
\begin_inset Formula $\ge2$
\end_inset
and stops when remainer is
\begin_inset Formula $1$
\end_inset
or
\begin_inset Formula $0$
\end_inset
.
Since are dividing repeatedly by numbers which are
\begin_inset Formula $\ge2$
\end_inset
, it only takes
\begin_inset Formula $\log N$
\end_inset
iterations before reaching
\begin_inset Formula $0$
\end_inset
or
\begin_inset Formula $1$
\end_inset
.
\end_layout
\begin_layout Standard
Continued fraction cost per step is
\begin_inset Formula $t^{2}$
\end_inset
for the division of a
\begin_inset Formula $t$
\end_inset
bit number.
This is
\begin_inset Formula $O\left(L^{2}\right)$
\end_inset
.
Total cost of the algorithm is the cost per steps times number of steps,
or
\begin_inset Formula $O\left(L^{3}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
[More information on Continued Fraction Algorithm in Appendix p635-6, theorem
A.4.15).
\end_layout
\begin_layout Subsubsection*
Continued Fractions as Simple Fractions
\end_layout
\begin_layout Standard
Given
\begin_inset Formula $\left[a_{0},a_{1},\ldots,a_{N}\right]$
\end_inset
then
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none
\begin_inset Formula $\left[a_{0},a_{1},\ldots,a_{n}\right]=\frac{P_{n}}{Q_{n}}$
\end_inset
for
\family default
\series default
\shape default
\size default
\emph default
\bar default
\noun default
\color inherit
\begin_inset Formula $n
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left[a_{0}\right]$
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $a_{0}=\frac{a_{0}}{1}=\frac{p_{0}}{q_{0}}$
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left[a_{0},a_{1}\right]$
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $a_{0}+\frac{1}{a_{1}}=\frac{a_{0}a_{1}+1}{a_{1}}=\frac{p_{1}}{q_{1}}$
\end_inset
\end_layout
\end_inset
|
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $\left[a_{0},a_{1},a_{2}\right]$
\end_inset
\end_layout
\end_inset
|
\begin_inset Text
\begin_layout Standard
\begin_inset Formula $a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}}}=\frac{a_{0}\left(a_{1}+\frac{1}{a_{2}}\right)+1}{a_{1}+\frac{1}{a_{2}}}=\frac{p_{1}+\frac{p_{0}}{a_{2}}}{\frac{a_{1}a_{2}+1}{a_{2}}}=\frac{p_{1}a_{2}+p_{0}}{a_{2}q_{1}+q_{0}}=\frac{p_{2}}{q_{2}}$
\end_inset
\end_layout
\end_inset
|
\end_inset
\end_layout
\begin_layout Subsubsection*
Continued Fraction Examples
\end_layout
\begin_layout Standard
Ex 1:
\begin_inset Formula $\frac{9}{15}=0+\frac{1}{\frac{15}{9}}$
\end_inset
, continue running with
\begin_inset Formula $\frac{15}{9}$
\end_inset
\end_layout
\begin_layout Standard
Ex 2:
\begin_inset Formula $0.333=\frac{333}{1000}=0+\frac{1}{\frac{1000}{333}}=0+\frac{1}{3+\frac{1}{333}}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $p_{0}=a_{0}=0$
\end_inset
,
\begin_inset Formula $q_{0}=1$
\end_inset
,
\begin_inset Formula $\frac{p_{0}}{q_{0}}=0$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $p_{1}=a_{0}a_{1}+1=1$
\end_inset
,
\begin_inset Formula $q_{1}=a_{1}=3$
\end_inset
,
\begin_inset Formula $\frac{p_{0}}{q_{0}}=\frac{1}{3}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $p_{2}=a_{2}p_{1}+p_{0}=333\cdot1+0=333$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $q_{2}=a_{2}q_{1}+q_{0}=333\cdot3+1=1000$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\frac{p_{2}}{q_{2}}=\frac{333}{1000}$
\end_inset
\end_layout
\begin_layout Standard
If we would have started over with
\begin_inset Formula $\frac{1000}{333}=3+\frac{1}{333}$
\end_inset
, then:
\end_layout
\begin_layout Standard
\begin_inset Formula $p_{0}=a_{0}=3$
\end_inset
,
\begin_inset Formula $q_{0}=1$
\end_inset
,
\begin_inset Formula $\frac{p_{0}}{q_{0}}=\frac{3}{1}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $p_{1}=333\cdot3+1=1000$
\end_inset
,
\begin_inset Formula $q_{1}=333\cdot1=333$
\end_inset
,
\begin_inset Formula $\frac{p_{1}}{q_{1}}=\frac{1000}{333}$
\end_inset
\end_layout
\begin_layout Subsubsection*
Order Finding Algorithm
\end_layout
\begin_layout Standard
Algorithm after phase estimation is classical.
Get
\begin_inset Formula $\hat{\varphi}=\left[a_{0},\ldots,a_{m}\right]$
\end_inset
determine
\begin_inset Formula $\frac{p_{n}}{q_{n}}$
\end_inset
, for
\begin_inset Formula $n=0,\ldots,m$
\end_inset
.
For each
\begin_inset Formula $q_{n}$
\end_inset
, check if
\begin_inset Formula $x^{q_{n}}\overset{?}{=}1\left(\textrm{mod }N\right)$
\end_inset
.
If the equation is satisfied, then
\begin_inset Formula $r=q_{n}$
\end_inset
.
\end_layout
\begin_layout Standard
The algorithm may fail in two seperate cases:
\end_layout
\begin_layout Standard
1.
If phase estimation fails
\end_layout
\begin_layout Standard
2.
If
\begin_inset Formula $\gcd\left(s,r\right)>1$
\end_inset
.
If this is the case, recovering the denominator of
\begin_inset Formula $\frac{s}{r}$
\end_inset
will give
\begin_inset Formula $r$
\end_inset
divided by the gcd, instead of just
\begin_inset Formula $r$
\end_inset
.
\begin_inset Formula $\frac{s}{r}=\frac{s'}{r'}$
\end_inset
,
\begin_inset Formula $r'0$
\end_inset
.
\end_layout
\begin_layout Standard
Related to order finding.
\end_layout
\begin_layout Standard
Definitions:
\end_layout
\begin_layout Standard
\begin_inset Formula $N$
\end_inset
- composite number
\end_layout
\begin_layout Standard
\begin_inset Formula $x$
\end_inset
,
\begin_inset Formula $N$
\end_inset
- coprime,
\begin_inset Formula $11-\frac{1}{2^{2}}=\frac{3}{4}$
\end_inset
\end_layout
\begin_layout Standard
even if
\begin_inset Formula $n=1$
\end_inset
prob
\begin_inset Formula $>\frac{1}{2}$
\end_inset
\end_layout
\begin_layout Standard
If you can verify solution, even probabilities
\begin_inset Formula $<\frac{1}{2}$
\end_inset
are ok, just repeat.
As long as probability isn't exponentially tiny, verification is all you
need to get away with tiny probabilities.
\end_layout
\begin_layout Subsection*
Reduction of factoring to order finding
\end_layout
\begin_layout Standard
High level summary, steps later
\end_layout
\begin_layout Standard
Choose random
\begin_inset Formula $x$
\end_inset
and find
\begin_inset Formula $r$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula $x^{r}=1\left(\textrm{mod }N\right)$
\end_inset
\end_layout
\begin_layout Standard
if
\begin_inset Formula $r$
\end_inset
is even
\begin_inset Formula $=\left(x^{r/2}\right)^{2}$
\end_inset
and you can use theorem 2
\end_layout
\begin_layout Standard
return
\begin_inset Formula $\gcd\left(x^{r/2}-1,N\right)$
\end_inset
or
\begin_inset Formula $\gcd\left(x^{r/2}+1,N\right)$
\end_inset
\end_layout
\begin_layout Standard
Wednesday: details, grover's algorithm
\end_layout
\begin_layout Section*
Lecture 23 (11 April)
\end_layout
\begin_layout Standard
Theorem 1:
\begin_inset Formula $x^{2}=1\left(\textrm{mod }N\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left(x-1\right)\left(x+1\right)=\ell N$
\end_inset
\begin_inset Formula $\gcd\left(x-1,N\right)$
\end_inset
or
\begin_inset Formula $\gcd\left(x+1,N\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $x\neq\pm\left(\textrm{mod }N\right)$
\end_inset
\end_layout
\begin_layout Standard
Ex 1:
\begin_inset Formula $N=15$
\end_inset
,
\begin_inset Formula $x=4$
\end_inset
,
\begin_inset Formula $x^{2}=16=1\left(\textrm{mod }15\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left(x-1\right)\left(x+1\right)=15$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\gcd\left(3,15\right)\times\gcd\left(5,15\right)=\ell N$
\end_inset
\end_layout
\begin_layout Standard
Both gcd's are factors
\end_layout
\begin_layout Standard
Ex 2:
\begin_inset Formula $N=12$
\end_inset
,
\begin_inset Formula $x=7$
\end_inset
,
\begin_inset Formula $x^{2}=49=1\left(\textrm{mod }12\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left(x-1\right)\left(x+1\right)=6\cdot8=48=4\cdot12$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\ell=4$
\end_inset
,
\begin_inset Formula $N=12$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\gcd\left(6,12\right)=6$
\end_inset
,
\begin_inset Formula $\gcd\left(8,12\right)=4$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $6\cdot4\ne12$
\end_inset
, both gcd's are not factors
\end_layout
\begin_layout Standard
Therefore take one gcd, get another factor by division.Look at pages 15-16
of schor's papers.
\end_layout
\begin_layout Subsection*
Reduction to order finding
\end_layout
\begin_layout Standard
Choose
\begin_inset Formula $x\in\left\{ 1,\ldots,N-1\right\} $
\end_inset
\end_layout
\begin_layout Standard
Find order r,
\begin_inset Formula $x^{r}=1\left(\textrm{mod }N\right)$
\end_inset
, use theorem
\end_layout
\begin_layout Standard
If r is even
\begin_inset Formula $\left(x^{r/2}-1\right)\left(x^{r/2}+1\right)=\ell N$
\end_inset
\end_layout
\begin_layout Standard
then
\begin_inset Formula $\gcd\left(x^{2}-1,N\right)$
\end_inset
or
\begin_inset Formula $\gcd\left(x^{2}+1,N\right)$
\end_inset
\end_layout
\begin_layout Standard
there are constraints, we cover them later
\end_layout
\begin_layout Standard
assuming r is even, assuming not a trivial solution (as indicated by theorem).
In even case, you know
\begin_inset Formula $x^{r/2}\ne1\left(\textrm{mod }N\right)$
\end_inset
because
\begin_inset Formula $r$
\end_inset
is order, order is smallest
\begin_inset Formula $r$
\end_inset
.
\end_layout
\begin_layout Standard
Theorem: If
\begin_inset Formula $N$
\end_inset
is odd and composite and factors as
\begin_inset Formula $N=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\ldots p_{n}^{\alpha_{n}}$
\end_inset
\begin_inset Formula $p_{n}$
\end_inset
primes and pick one
\begin_inset Formula $X=\left\{ 1,\ldots,N-1\right\} $
\end_inset
uniformly at random.
\begin_inset Formula $Pr\left\{ \textrm{r even and }x^{r/2}\neq-1\left(\textrm{mod }N\right)\right\} \ge1-\frac{1}{2^{n}}$
\end_inset
\end_layout
\begin_layout Subsection*
Shor Factoring Algorithm
\end_layout
\begin_layout Standard
Input: Composite number N
\end_layout
\begin_layout Standard
Output: Factor of N
\end_layout
\begin_layout Standard
Runtime:
\begin_inset Formula $O\left(L^{3}\right)=O\left(\left(\log N\right)^{3}\right)$
\end_inset
\end_layout
\begin_layout Standard
Probability success
\begin_inset Formula $\ge\frac{3}{4}$
\end_inset
\end_layout
\begin_layout Standard
Steps:
\end_layout
\begin_layout Standard
1.
If
\begin_inset Formula $N$
\end_inset
even output 2
\end_layout
\begin_layout Standard
2.
Determine if
\begin_inset Formula $N=a^{b}$
\end_inset
, for some
\begin_inset Formula $a\ge1$
\end_inset
,
\begin_inset Formula $b\ge2$
\end_inset
.
If so, determine
\begin_inset Formula $n$
\end_inset
and stop.
\end_layout
\begin_layout Standard
3.
Uniformly chose
\begin_inset Formula $x\in\left\{ 1,\ldots,N-1\right\} $
\end_inset
.
If
\begin_inset Formula $\gcd\left(x,N\right)>1m$
\end_inset
return
\begin_inset Formula $\gcd\left(x,N\right)$
\end_inset
.
\end_layout
\begin_layout Standard
4.
Only quantum step.
Use order finding algorithm, obtain
\begin_inset Formula $r$
\end_inset
order of
\begin_inset Formula $x$
\end_inset
mod
\begin_inset Formula $N$
\end_inset
.
\end_layout
\begin_layout Standard
5.
If
\begin_inset Formula $r$
\end_inset
is even and
\begin_inset Formula $x^{r/2}\ne-1\left(\textrm{mod }N\right)$
\end_inset
, then return
\begin_inset Formula $\gcd\left(x^{r/2}-1,N\right)$
\end_inset
or
\begin_inset Formula $\gcd\left(x^{r/2}+1,N\right)$
\end_inset
doesn't matter which.
Otherwise algorithm fails.
\end_layout
\begin_layout Standard
Remarks
\end_layout
\begin_layout Standard
Step 1: Step 1 is easy
\end_layout
\begin_layout Standard
Step 2: How to check
\begin_inset Formula $N=a^{b}$
\end_inset
for
\begin_inset Formula $a\ge3$
\end_inset
,
\begin_inset Formula $b\ge2$
\end_inset
, we know
\begin_inset Formula $b\le\log N\le L$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $b=\left\{ 2,3,4,\ldots,L\right\} $
\end_inset
\end_layout
\begin_layout Standard
Check if
\begin_inset Formula $a^{k}=N$
\end_inset
for each
\begin_inset Formula $b$
\end_inset
and integer
\begin_inset Formula $a$
\end_inset
.
Book uses algorithm to find
\begin_inset Formula $a$
\end_inset
, but you can use binary search.
Bisection isn't so bad because only need sign information, don't need to
compute whole powers.
Cost of bisection in
\begin_inset Formula $\log N=L$
\end_inset
steps, error
\begin_inset Formula $\le\frac{1}{2N}$
\end_inset
, cost per step is cost of repeated querying
\begin_inset Formula $O\left(\log\log N\right)$
\end_inset
.
Total cost
\begin_inset Formula $O\left(\log^{2}N\log\log N\right)$
\end_inset
\end_layout
\begin_layout Standard
Alternative: Newton's method
\end_layout
\begin_layout Standard
\begin_inset Formula $x_{i+1}=x_{i}-\frac{f\left(x\right)}{f'\left(x\right)}=x_{i}-\frac{x_{i}^{-b}-N}{bx_{i}^{b-1}}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left|x_{i}-N^{1/b}\right|=O\left(\left|x_{i}-N^{1/b}\right|^{2}\right)$
\end_inset
\end_layout
\begin_layout Standard
Quadratic convergence, taking as few steps as needed.
\end_layout
\begin_layout Standard
After steps 1 and 2 know
\begin_inset Formula $x$
\end_inset
is odd and has more than one prime factor.
\end_layout
\begin_layout Standard
Combine w/ theorem 2 to see that prob
\begin_inset Formula $\ge\frac{3}{4}=1-\frac{1}{2^{2}}$
\end_inset
.
2 is number of factors (
\begin_inset Formula $2>1$
\end_inset
).
\end_layout
\begin_layout Standard
Step 3: Use euclid algorithm with cost
\begin_inset Formula $O\left(L^{3}\right)=O\left(\log^{3}N\right)$
\end_inset
to get
\begin_inset Formula $\gcd\left(x,N\right)$
\end_inset
.
if
\begin_inset Formula $>1$
\end_inset
stop, else continue knowing
\begin_inset Formula $x$
\end_inset
,
\begin_inset Formula $N$
\end_inset
are coprime.
\end_layout
\begin_layout Standard
Step 4:
\begin_inset Formula $x$
\end_inset
,
\begin_inset Formula $N$
\end_inset
coprime, so order finding.
Cost
\begin_inset Formula $O\left(L^{3}\right)$
\end_inset
or
\begin_inset Formula $O\left(L^{4}\right)$
\end_inset
, depending which way you do, not significant.
Gives
\begin_inset Formula $r$
\end_inset
, so
\begin_inset Formula $x^{r}=1\left(\textrm{mod }N\right)$
\end_inset
.
\end_layout
\begin_layout Standard
Theorem 2:
\begin_inset Formula $Pr\left\{ \textrm{r even and }x^{r/2}\ne-1\left(\textrm{mod }N\right)\right\} \ge1-\frac{1}{2^{2}}=\frac{3}{4}$
\end_inset
.
\end_layout
\begin_layout Standard
Step 5: Check r even
\end_layout
\begin_layout Standard
\begin_inset Formula $x^{r/2}\ne-1\left(\textrm{mod }N\right)$
\end_inset
?
\end_layout
\begin_layout Standard
If so,
\begin_inset Formula $\gcd\left(x^{r/2}-1,N\right)$
\end_inset
or
\begin_inset Formula $\gcd\left(x^{r/2}+1,N\right)$
\end_inset
\end_layout
\begin_layout Standard
Example:
\end_layout
\begin_layout Standard
\begin_inset Formula $N=a\left(13\times7\right)$
\end_inset
\end_layout
\begin_layout Standard
1.
Not even
\end_layout
\begin_layout Standard
2.
Not
\begin_inset Formula $N\ne a^{b}$
\end_inset
\end_layout
\begin_layout Standard
3.
Say
\begin_inset Formula $x=4$
\end_inset
,
\begin_inset Formula $\gcd\left(4,9\right)=1$
\end_inset
coprime
\end_layout
\begin_layout Standard
4.
Order of
\begin_inset Formula $x=4\left(\textrm{mod }91\right)=1$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $r=6$
\end_inset
,
\begin_inset Formula $4^{6}=2^{12}=4096=1\left(\textrm{mod }91\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $4096=45\cdot91+1$
\end_inset
\end_layout
\begin_layout Standard
5.
\begin_inset Formula $r=6$
\end_inset
even
\end_layout
\begin_layout Standard
\begin_inset Formula $x^{r/2}=4^{3}=2^{6}=64=64\left(\textrm{mod }91\right)\ne-1\left(\textrm{mod }91\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\gcd\left(63,91\right)=\gcd\left(63,28\right)=\gcd\left(28,7\right)=7$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\gcd\left(65,91\right)=\gcd\left(65,26\right)=\gcd\left(26,13\right)=13$
\end_inset
\end_layout
\begin_layout Standard
Both of them are not factors, but they are both divisors.
\end_layout
\begin_layout Subsection*
Search, counting algorithms
\end_layout
\begin_layout Standard
Grover's algorithms
\end_layout
\begin_layout Standard
Boolean mean, Brascyd et al (amplitude, amplification, estimations)
\end_layout
\begin_layout Standard
Applications many problems science / engineering, integration, approximation,
path integrations, differential equations (Schrodinger eqn).
\end_layout
\begin_layout Section*
Lecture 24 (16 April)
\end_layout
\begin_layout Subsection*
Searching / Counting Algorithm
\end_layout
\begin_layout Standard
Given a function
\begin_inset Formula \[
f:\left\{ 0,1,\ldots,N-1\right\} \rightarrow\left\{ 0,1\right\} \]
\end_inset
as in Deutsch Josza, find which inputs produce an output.
\end_layout
\begin_layout Standard
\begin_inset Formula $N=2^{n}$
\end_inset
,
\begin_inset Formula $N$
\end_inset
is huge
\end_layout
\begin_layout Standard
We use queries, or oracle calls
\begin_inset Formula \[
Q_{f}\left|x\right\rangle \left|y\right\rangle =\left|x\right\rangle \left|y\oplus f\left(x\right)\right\rangle \]
\end_inset
\begin_inset Formula $\left|x\right\rangle $
\end_inset
is
\begin_inset Formula $n$
\end_inset
qubits,
\begin_inset Formula $\left|y\right\rangle $
\end_inset
is 1 qubit.
\end_layout
\begin_layout Standard
Using
\begin_inset Formula $H\left|1\right\rangle $
\end_inset
as an input:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
Q_{f}\left|x\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}} & = & \frac{1}{\sqrt{2}}\left(\left|x\right\rangle \left|0\oplus f\left(x\right)\right\rangle -\left|x\right\rangle \left|1\oplus f\left(x\right)\right\rangle \right)\\
& = & \frac{1}{\sqrt{2}}\left(\left|x\right\rangle \left|f\left(x\right)\right\rangle -\left|x\right\rangle \left|\overline{f\left(x\right)}\right\rangle \right)\end{eqnarray*}
\end_inset
if
\begin_inset Formula $f\left(x\right)=0$
\end_inset
then
\begin_inset Formula $\left|x\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}$
\end_inset
.
\end_layout
\begin_layout Standard
if
\begin_inset Formula $f\left(x\right)=1$
\end_inset
then
\begin_inset Formula $\left|x\right\rangle \frac{\left|1\right\rangle -\left|0\right\rangle }{\sqrt{2}}$
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
Q_{f}\left|x\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}} & = & \left(-1\right)^{f\left(x\right)}\left|x\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
With this input, the
\begin_inset Formula $\left|y\right\rangle $
\end_inset
qubit is unchanged, there is just an overall phase shift.
This is more convenient for analysis than the general
\begin_inset Formula $Q_{f}\left|x\right\rangle \left|y\right\rangle $
\end_inset
function.
\end_layout
\begin_layout Subsection*
Search
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
M_{f} & = & \left\{ x:f\left(x\right)=1\right\} \\
M & = & \left|M_{f}\right|\ge1\end{eqnarray*}
\end_inset
Problem is to find elements of
\begin_inset Formula $M_{f}$
\end_inset
.
This is a reverse lookup problem, like searching an unordered database.
Unlike the quantum solution for the factoring problem, the quantum solution
for this problem is not exponentially faster, just polylog.
Another difference is that this quantum solution beats the upper bound
of the equivalent classical solution, whereas the quantum factoring algorithm
only beats known classical algorithms.
\end_layout
\begin_layout Standard
Related Problem: Boolean Mean
\end_layout
\begin_layout Standard
\begin_inset Formula \[
S\left(f\right)=\frac{1}{N}\sum_{x=0}^{N-1}f\left(x\right)=\frac{M}{N}\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Classsical Search Algorithms
\end_layout
\begin_layout Standard
1.
Deterministic Algorithm.
Lower bound
\begin_inset Formula $O\left(N\right)$
\end_inset
, because you may have to evaluate
\begin_inset Formula $N$
\end_inset
times before search succeeds.
\end_layout
\begin_layout Standard
2.
Randomized Algorithms.
(See paper Beals et al).
Lower bound is also
\begin_inset Formula $O\left(N\right)$
\end_inset
.
No proof, but basic idea follows.
\end_layout
\begin_layout Standard
Algorithm:
\end_layout
\begin_layout Standard
Choose
\begin_inset Formula $x$
\end_inset
uniformly at random with replacement.
\end_layout
\begin_layout Standard
If
\begin_inset Formula $f\left(x\right)=1$
\end_inset
stop with success.
\end_layout
\begin_layout Standard
If
\begin_inset Formula $f\left(x\right)=0$
\end_inset
fail.
\end_layout
\begin_layout Standard
Repeat
\begin_inset Formula $k$
\end_inset
times.
\end_layout
\begin_layout Standard
Probability of failure first time is
\begin_inset Formula $1-\frac{M}{N}$
\end_inset
.
If
\begin_inset Formula $M=1$
\end_inset
,
\begin_inset Formula $1-\frac{1}{N}>C$
\end_inset
.
\end_layout
\begin_layout Standard
Probability to fail in
\begin_inset Formula $k$
\end_inset
trials is
\begin_inset Formula $\left(1-\frac{M}{N}\right)^{k}\le\delta$
\end_inset
.
Set desired tolerance to
\begin_inset Formula $\delta$
\end_inset
.
\begin_inset Formula \[
k\log\left(1-\frac{M}{N}\right)=\log\delta\]
\end_inset
\begin_inset Formula \begin{eqnarray*}
k & = & \frac{\log d}{\log\left(1-\frac{M}{N}\right)}\approx\frac{\log d}{-\frac{M}{N}}=\frac{N}{M}\log\frac{1}{d}\end{eqnarray*}
\end_inset
Approximation holds when
\begin_inset Formula $M\ll N$
\end_inset
.
In this case, algorithm is
\begin_inset Formula $O\left(N\right)$
\end_inset
, in general case algorithm is
\begin_inset Formula $O\left(\frac{N}{M}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
General hint:
\begin_inset Formula $\log\left(1+x\right)\approx x$
\end_inset
when
\begin_inset Formula $x$
\end_inset
is tiny.
This is used frequently in complexity analysis, and is based on the power
series expansion.
\end_layout
\begin_layout Standard
When algorithm is run without replacement, probability of failure is
\begin_inset Formula \[
\frac{\left(\begin{array}{c}
N-M\\
k\end{array}\right)\left(\begin{array}{c}
M\\
0\end{array}\right)}{\left(\begin{array}{c}
N\\
k\end{array}\right)}=\frac{\left(\begin{array}{c}
N-1\\
k\end{array}\right)}{\left(\begin{array}{c}
N\\
k\end{array}\right)}=\frac{\left(N-1\right)!}{k!\left(N-k-1\right)!}=\frac{N-k}{N}=1-\frac{k}{N}\]
\end_inset
this is bounded by a constant unless
\begin_inset Formula $k$
\end_inset
is
\begin_inset Formula $O\left(N\right)$
\end_inset
\end_layout
\begin_layout Subsubsection*
Repeating
\begin_inset Formula \[
\left(1-\frac{k}{N}\right)^{S}\le\delta\]
\end_inset
\begin_inset Formula \[
S=\frac{N}{k}\log\frac{1}{\delta}\Rightarrow S_{k}=N\log\frac{1}{\delta}=O\left(N\right)\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Quantum Search Algorithm
\end_layout
\begin_layout Standard
Grover's Algorithm,
\begin_inset Formula $O\left(\sqrt{\frac{N}{M}}\right)$
\end_inset
counting the number of queries of
\begin_inset Formula $Q_{f}$
\end_inset
.
\end_layout
\begin_layout Standard
Algorithm: Given some initial state
\begin_inset Formula $\left|\psi_{0}\right\rangle $
\end_inset
.
Apply operator then query, then operator, repeating as neccesary.
\begin_inset Formula \[
\left|\psi_{1}\right\rangle =Q_{F}U_{T}Q_{F}U_{T-1}\ldots U_{2}Q_{F}U_{1}\left|\psi_{0}\right\rangle \]
\end_inset
\end_layout
\begin_layout Standard
using
\begin_inset Formula $T$
\end_inset
queries.
\end_layout
\begin_layout Subsection*
Boolean Mean
\end_layout
\begin_layout Standard
\begin_inset Formula \[
S\left(f\right)=\frac{1}{N}\sum_{x=0}^{N-1}f\left(x\right)=\frac{M}{N}\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Classical Algorithms
\end_layout
\begin_layout Standard
1.
Deterministic Algorithm.
Find lower bound on number of evaluations given error tolerance
\begin_inset Formula $\epsilon$
\end_inset
.
To do this we need to ensure
\begin_inset Formula \[
\max_{f}\left|S\left(f\right)-\hat{S}\left(f\right)\right|\le\epsilon\]
\end_inset
For
\begin_inset Formula $k$
\end_inset
evaluations, assume
\begin_inset Formula $f\left(x\right)=0$
\end_inset
to get lower bound, we then know that
\begin_inset Formula $0\le S\left(f\right)\le\frac{M-k}{N}$
\end_inset
.
Take an estimate at the middle of this range to get least possible error
in worst case:
\begin_inset Formula \[
\hat{S}\left(f\right)=\frac{0+\left(1-\frac{k}{N}\right)}{2}\]
\end_inset
In this case
\begin_inset Formula \[
\max_{f}\left|S\left(f\right)-\hat{S}\left(f\right)\right|^{2}=\frac{1}{2}\left(1-\frac{k}{N}\right)\]
\end_inset
\begin_inset Formula \begin{eqnarray*}
\frac{1}{2}\left(1-\frac{k}{N}\right) & < & \epsilon\\
N-k & \le & 2N\epsilon\\
k & \ge & N\left(1-2\epsilon\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Section*
Lecture 25 (18 April)
\end_layout
\begin_layout Subsection*
Searching + Counting
\end_layout
\begin_layout Standard
Classical Algorithms, Deterministic and Random
\begin_inset Formula $O\left(N\right)$
\end_inset
,
\begin_inset Formula $1\le M\ll N$
\end_inset
.
\end_layout
\begin_layout Standard
Quantum Algorithm
\begin_inset Formula $O\left(\sqrt{\frac{N}{M}}\right)$
\end_inset
\end_layout
\begin_layout Subsection*
Counting / Boolean Mean
\end_layout
\begin_layout Standard
Classical deterministic algorithm:
\begin_inset Formula $k\ge N\left(1-2\epsilon\right)$
\end_inset
\end_layout
\begin_layout Standard
Classical randomized algorithm: Choose
\begin_inset Formula $k$
\end_inset
inputs at random computing
\begin_inset Formula \[
\hat{S}\left(f\right)=\frac{1}{k}\sum_{i=1}^{k}f\left(x_{i}\right)\]
\end_inset
\begin_inset Formula \[
k:x_{i}\in\left\{ 0,\ldots,N-1\right\} \]
\end_inset
This is a Monte Carlo algorithm.
Instead of finding the lower bound error, find the expected error
\begin_inset Formula \[
\left(E_{x_{1}\ldots x_{k}}\left[S\left(f\right)-\hat{S}\left(f\right)\right]^{2}\right)^{1/2}\le\frac{1}{\sqrt{k}}<\epsilon\]
\end_inset
\begin_inset Formula \[
k=\min\left(\frac{1}{\epsilon^{2}},N\right)\]
\end_inset
Chebyshev inequality
\begin_inset Formula \begin{eqnarray*}
Pr\left\{ \left|S\left(f\right)-\hat{S}\left(f\right)\right|>\tau\right\} & \le & \frac{E\left(\left(S\left(f\right)-\hat{S}\left(f\right)\right)^{2}\right)}{\tau^{2}}\\
\tau & = & 2\epsilon\\
Pr\left\{ \left|S\left(f\right)-\hat{S}\left(f\right)\right|>2\epsilon\right\} & \le & \frac{E\left(\left(S\left(f\right)-\hat{S}\left(f\right)\right)^{2}\right)}{4\epsilon^{2}}\end{eqnarray*}
\end_inset
\begin_inset Formula \[
k=\min\left(\frac{1}{\epsilon},N\right)\]
\end_inset
\end_layout
\begin_layout Subsubsection*
Quantum Algorithm
\end_layout
\begin_layout Standard
Provides provable polynomial speedup over classical randomized algorithm.
This is unlike factoring where there is no known lower bound and we are
beating known classical algorithms without any proof that there isn't an
unknown classical algorithm which could be faster.
\end_layout
\begin_layout Subsection*
Grover's Search
\end_layout
\begin_layout Standard
1.
Apply oracle
\begin_inset Formula $Q_{f}$
\end_inset
(from Deutsch-Jozsa) or the nice oracle
\begin_inset Formula $O_{f}$
\end_inset
(from previous lecture),
\begin_inset Formula $O_{f}\left|x\right\rangle =\left(-1\right)^{f\left(x\right)}\left|x\right\rangle $
\end_inset
.
\end_layout
\begin_layout Standard
2.
Apply
\begin_inset Formula $H^{\otimes n}$
\end_inset
\end_layout
\begin_layout Standard
3.
Apply phase shift
\begin_inset Formula $Q_{0}$
\end_inset
which is a reflaction about
\begin_inset Formula $\left|0\right\rangle $
\end_inset
\begin_inset Formula \begin{eqnarray*}
Q_{0} & = & 2\left|0\right\rangle \left\langle 0\right|-I\\
Q_{0}\left|0\right\rangle & = & 2\left|0\right\rangle \left\langle 0|0\right\rangle -I\left|0\right\rangle =\left|0\right\rangle \\
Q_{0}\left|y\right\rangle & = & 2\left|0\right\rangle \left\langle 0|y\right\rangle -I\left|y\right\rangle =-\left|y\right\rangle ,\: y\ne0\end{eqnarray*}
\end_inset
so for any basis state
\begin_inset Formula $\left|y\right\rangle $
\end_inset
\begin_inset Formula \[
Q_{0}\left|y\right\rangle =\left(-1\right)^{\delta_{0}y}\left|y\right\rangle \:\forall y=0,\ldots,N-1\]
\end_inset
\end_layout
\begin_layout Standard
4.
Apply
\begin_inset Formula $H^{\otimes n}$
\end_inset
\end_layout
\begin_layout Standard
All 4 operations can be combined with one operator
\begin_inset Formula \begin{eqnarray*}
G & = & H^{\otimes n}\left(2\left|0\right\rangle \left\langle 0\right|-I\right)H^{\otimes n}O_{f}\\
& = & \left(2H^{\otimes n}\left|0\right\rangle \left\langle 0\right|H^{\otimes n}-H^{\otimes n}H^{\otimes n}\right)O_{f}\\
& = & \left(2\left|\psi\right\rangle \left\langle \psi\right|-I\right)O_{f}\end{eqnarray*}
\end_inset
where
\begin_inset Formula \[
\left|\psi\right\rangle =\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}\left|j\right\rangle \]
\end_inset
\begin_inset Formula $2\left|\psi\right\rangle \left\langle \psi\right|-I$
\end_inset
acts as a reflection about
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
because for
\begin_inset Formula $\left|z\right\rangle \perp\left|\psi\right\rangle $
\end_inset
\begin_inset Formula \begin{eqnarray*}
\left(2\left|\psi\right\rangle \left\langle \psi\right|-I\right)\left|\psi\right\rangle & = & \left|\psi\right\rangle \\
\left(2\left|\psi\right\rangle \left\langle \psi\right|-I\right)\left|z\right\rangle & = & -\left|z\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Analysis
\end_layout
\begin_layout Standard
\begin_inset Formula \[
M_{f}=\left\{ x:f\left(x\right)=1\right\} ,\:1\le M=\left|M_{f}\right|\le N-1\]
\end_inset
\end_layout
\begin_layout Standard
Define two states
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
and
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
so
\begin_inset Formula \begin{eqnarray*}
\left|\alpha\right\rangle & = & \frac{1}{\sqrt{N-M}}\sum_{x\notin M_{f}}\left|x\right\rangle \\
\left|\beta\right\rangle & = & \frac{1}{\sqrt{M}}\sum_{x\in M_{f}}\left|x\right\rangle \end{eqnarray*}
\end_inset
\begin_inset Formula $\left\Vert \left|\alpha\right\rangle \right\Vert =\left\Vert \left|b\right\rangle \right\Vert =1$
\end_inset
and
\begin_inset Formula $\left\langle a|b\right\rangle =0$
\end_inset
\begin_inset Formula \begin{eqnarray*}
\left|\psi\right\rangle & = & \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}\left|j\right\rangle \\
& = & \frac{1}{\sqrt{N}}\left(\sqrt{N-M}\left|\alpha\right\rangle +\sqrt{M}\left|\beta\right\rangle \right)\\
& = & \frac{\sqrt{N-M}}{\sqrt{N}}\left|\alpha\right\rangle +\frac{\sqrt{M}}{\sqrt{N}}\left|\beta\right\rangle \end{eqnarray*}
\end_inset
Idea of algorithm is to boost change initial state
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
boosting the amplitude of
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
so the state that is measured will likely be in
\begin_inset Formula $M_{f}$
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
\left|\psi\right\rangle & = & \left\langle \alpha|\psi\right\rangle \left|\alpha\right\rangle +\left\langle \beta|\psi\right\rangle \left|\beta\right\rangle \\
\left\langle \alpha|\psi\right\rangle & = & \frac{1}{\sqrt{N-M}}\frac{1}{\sqrt{N}}\sum_{x\notin M_{f}}\sum_{y=0}^{N-1}\left\langle x|y\right\rangle \\
& = & \frac{1}{\sqrt{N-M}}\frac{1}{\sqrt{N}}\sum_{x\notin M_{f}}1\\
& = & \frac{1}{\sqrt{N-M}}\frac{1}{\sqrt{N}}\left(N-M\right)\\
& = & \sqrt{\frac{N-M}{N}}\end{eqnarray*}
\end_inset
Similarly for
\begin_inset Formula $\left|\beta\right\rangle .$
\end_inset
Because
\begin_inset Formula $\left\Vert \left|\psi\right\rangle \right\Vert ^{2}=\frac{N-M}{N}\left\Vert \left|\alpha\right\rangle \right\Vert ^{2}+\frac{M}{N}\left\Vert \left|\beta\right\rangle \right\Vert ^{2}=1$
\end_inset
,
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
can be written as
\begin_inset Formula \[
\left|\psi\right\rangle =\cos\frac{\vartheta}{2}\left|\alpha\right\rangle +\sin\frac{\vartheta}{2}\left|\beta\right\rangle \]
\end_inset
where
\begin_inset Formula \begin{eqnarray*}
\cos\frac{\vartheta}{2} & = & \sqrt{\frac{N-M}{N}}\\
\sin\frac{\vartheta}{2} & = & \sqrt{\frac{M}{N}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsubsection*
Powers of G
\end_layout
\begin_layout Standard
Want to determine
\begin_inset Formula $G\left|\psi\right\rangle ,G^{2}\left|\psi\right\rangle ,G^{3}\left|\psi\right\rangle ,\ldots,G^{k}\left|\psi\right\rangle $
\end_inset
\begin_inset Formula \begin{eqnarray*}
G & = & \left(2\left|\psi\right\rangle \left\langle \psi\right|-I\right)O_{f}\\
O_{f}\left|\alpha\right\rangle & = & \frac{1}{\sqrt{N-M}}\sum_{x\notin M_{f}}O_{f}\left|x\right\rangle \\
& = & \frac{1}{\sqrt{N-M}}\sum_{x\notin M_{f}}\left(-1\right)^{f\left(x\right)=1}\left|x\right\rangle \\
& = & \left|\alpha\right\rangle \\
O_{f}\left|\beta\right\rangle & = & \frac{1}{\sqrt{M}}\sum_{x\in M_{f}}O_{f}\left|x\right\rangle \\
& = & \frac{1}{\sqrt{M}}\sum_{x\in M_{f}}\left(-1\right)^{f\left(x\right)=1}\left|x\right\rangle \\
& = & -\left|\beta\right\rangle \end{eqnarray*}
\end_inset
So when operator
\begin_inset Formula $O_{f}$
\end_inset
is applied to
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
, it is unchanged.
When the operator is applied to
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
it is reflected.
Given this, we can see
\begin_inset Formula $G$
\end_inset
applied to a state creates a rotation of that state in the
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
,
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
plane.
Previously we showed expressed
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
in terms of an angle
\begin_inset Formula $\frac{\vartheta}{2}$
\end_inset
\begin_inset Formula \[
\left|\psi\right\rangle =\cos\frac{\vartheta}{2}\left|\alpha\right\rangle +\sin\frac{\vartheta}{2}\left|\beta\right\rangle \]
\end_inset
allowing the state
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
to be represented graphically as a 2 dimensional vector in the
\begin_inset Formula $\left|\alpha\right\rangle ,\left|\beta\right\rangle $
\end_inset
plane oriented
\begin_inset Formula $\frac{\vartheta}{2}$
\end_inset
degrees above the
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\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
axis.
Applying
\begin_inset Formula $O_{f}$
\end_inset
to this state negates
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
component reflecting it about the
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
axis resulting in a vector
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\begin_inset Formula $\frac{\vartheta}{2}$
\end_inset
degrees below the
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\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
axis.
Applying
\begin_inset Formula $2\left|\psi\right\rangle \left\langle \psi\right|-I$
\end_inset
is the same as a reflection about
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
.
This results in a state oriented
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\begin_inset Formula $\frac{\vartheta}{2}+\left(\frac{\vartheta}{2}--\frac{\vartheta}{2}\right)=\frac{3\vartheta}{2}$
\end_inset
degrees above the
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\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
axis.
(
\begin_inset Formula $\frac{\vartheta}{2}$
\end_inset
being the angle of the
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\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
state relative to
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\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
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,
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\begin_inset Formula $\left(\frac{\vartheta}{2}--\frac{\vartheta}{2}\right)$
\end_inset
being the angular distance between
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\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
and
\begin_inset Formula $O_{f}\left|\psi\right\rangle $
\end_inset
).
The result of the two reflections is
\begin_inset Formula \[
G\left|\psi\right\rangle =\cos\frac{3\vartheta}{2}\left|\alpha\right\rangle +\sin\frac{3\vartheta}{2}\left|\beta\right\rangle \]
\end_inset
which taken together are equivalent to a rotation by
\begin_inset Formula $2\vartheta$
\end_inset
.
Applying the
\begin_inset Formula $G$
\end_inset
operator repeatedly is then equivalent to
\begin_inset Formula \[
G^{k}\left|\psi\right\rangle =\cos\left(\frac{2k+1}{2}\vartheta\right)\left|\alpha\right\rangle +\sin\left(\frac{2k+1}{2}\vartheta\right)\left|\beta\right\rangle \]
\end_inset
when a good value for
\begin_inset Formula $k$
\end_inset
is chosen,
\begin_inset Formula $\sin\left(\frac{2k+1}{2}\vartheta\right)$
\end_inset
will be close to 1 and measuring
\begin_inset Formula $G^{k}\left|\psi\right\rangle $
\end_inset
will yield a state from the
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\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
superposition with high probability.
Recall that any state in
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\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
is a solution to the search problem.
More formally, for a measurement
\begin_inset Formula $\left|x\right\rangle $
\end_inset
on the computational basis
\begin_inset Formula \begin{eqnarray*}
Pr\left\{ x\in M_{f}\right\} & = & \sin^{2}\left(\frac{2k+1}{2}\vartheta\right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Section*
Lecture 26 (23 April)
\end_layout
\begin_layout Subsection*
Grover's Review
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
G & = & H^{\otimes n}\left(2\left|0\right\rangle \left\langle 0\right|-I\right)H^{\otimes n}O_{f}\\
& = & \left(2\left|\psi\right\rangle \left\langle \psi\right|-I\right)O_{f}\\
O_{f}\left|x\right\rangle & = & \left(-1\right)^{f\left(x\right)}\left|x\right\rangle \\
\left|\alpha\right\rangle & = & \frac{1}{\sqrt{N-M}}\sum_{x\notin M_{f}}\left|x\right\rangle \\
\left|\beta\right\rangle & = & \frac{1}{\sqrt{M}}\sum_{x\in M_{f}}\left|x\right\rangle \\
\left|\psi\right\rangle & = & \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}\left|j\right\rangle =H^{\otimes n}\left|0\right\rangle ^{\otimes n}\\
G^{k}\left|\psi\right\rangle & = & \cos\left(\frac{2k+1}{2}\vartheta\right)\left|\alpha\right\rangle +\sin\left(\frac{2k+1}{2}\vartheta\right)\left|\beta\right\rangle \\
\sin\left(\frac{\vartheta}{2}\right) & = & \sqrt{\frac{M}{N}}\\
\cos\left(\frac{\vartheta}{2}\right) & = & \sqrt{\frac{M-N}{N}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Grover's Analyis (continued)
\end_layout
\begin_layout Standard
Need to find value of
\begin_inset Formula $k$
\end_inset
that gives
\begin_inset Formula $G^{k}\left|\psi\right\rangle $
\end_inset
close to
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
.
Start with success probability of algorithm assuming
\begin_inset Formula $k$
\end_inset
is known.
For
\begin_inset Formula $x\in M_{f}$
\end_inset
\begin_inset Formula \begin{eqnarray*}
\left\langle x\right|G^{k}\left|\psi\right\rangle & = & \left|\sin\left(\frac{2k+1}{2}\vartheta\right)\left\langle x|\beta\right\rangle \right|^{2}\\
& = & \left|\sin\left(\frac{2k+1}{2}\vartheta\right)\frac{1}{\sqrt{M}}\right|^{2}\\
& = & \sin^{2}\left(\frac{2k+1}{2}\vartheta\right)\frac{1}{M}\\
Pr\left\{ \textrm{measuring }x\in M_{f}\right\} & = & M\sin^{2}\left(\frac{2k+1}{2}\vartheta\right)\frac{1}{M}\\
& = & \sin^{2}\left(\frac{2k+1}{2}\vartheta\right)\frac{1}{M}\end{eqnarray*}
\end_inset
Success probability is therefore
\begin_inset Formula $\sin^{2}\left(\frac{2k+1}{2}\vartheta\right)$
\end_inset
given a
\begin_inset Formula $k$
\end_inset
.
\begin_inset Formula $\frac{2k+1}{2}\vartheta$
\end_inset
should be close to
\begin_inset Formula $\frac{\pi}{2}$
\end_inset
, so
\begin_inset Formula \[
\frac{\pi}{2}-\frac{\vartheta}{2}\le\frac{2k+1}{2}\vartheta<\frac{\pi}{2}+\frac{\vartheta}{2}\]
\end_inset
If
\begin_inset Formula $\frac{M}{N}<\frac{1}{2}$
\end_inset
then
\begin_inset Formula $\sqrt{\frac{M}{N}}=\sin\frac{\vartheta}{2}<\sin\frac{\pi}{4}=\sqrt{\frac{1}{2}}$
\end_inset
and
\begin_inset Formula $0\le\vartheta\le\frac{\pi}{2}$
\end_inset
\begin_inset Formula \[
\pi-\vartheta\le\left(2k+1\right)\vartheta<\pi+\vartheta\]
\end_inset
\begin_inset Formula \[
\frac{\pi}{\vartheta}-1\le2k+1<\frac{\pi}{\vartheta}+1\]
\end_inset
\begin_inset Formula \[
\frac{\pi}{2\vartheta}-1\le k<\frac{\pi}{2\vartheta}\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\sqrt{\frac{M}{N}}=\sin\frac{\vartheta}{2}\approx\frac{\vartheta}{2}$
\end_inset
, a valid approximation when
\begin_inset Formula $\vartheta$
\end_inset
is tiny.
In this case
\begin_inset Formula $\varphi\approx2\sqrt{\frac{M}{N}}$
\end_inset
, and
\begin_inset Formula $k=\frac{\pi}{2\vartheta}\approx\frac{\pi}{4}\sqrt{\frac{N}{M}}$
\end_inset
, so a good
\begin_inset Formula $k$
\end_inset
can be set
\begin_inset Formula $k=\left\lceil \frac{\pi}{4}\sqrt{\frac{N}{M}}\right\rceil $
\end_inset
, making the complexity of the algorithm in terms of oracle calls
\begin_inset Formula $O\left(N\right)$
\end_inset
.
\end_layout
\begin_layout Standard
When
\begin_inset Formula $M$
\end_inset
is larger, you need a more precise analysis without the approximation.
In this case use
\begin_inset Formula \begin{eqnarray*}
\sin\frac{\vartheta}{2} & = & \sqrt{\frac{M}{N}}\\
\cos\frac{\vartheta}{2} & = & \sqrt{\frac{N-M}{N}}\end{eqnarray*}
\end_inset
Then use a trig identity
\begin_inset Formula \begin{eqnarray*}
\sin\vartheta & = & 2\sin\frac{\vartheta}{2}\cos\frac{\vartheta}{2}=2\sqrt{\frac{M}{N}}\sqrt{\frac{N-M}{N}}\\
\varphi & = & \arcsin\left(2\sqrt{\frac{M\left(N-M\right)}{N^{2}}}\right)\\
k & = & \frac{\pi}{2\vartheta}\end{eqnarray*}
\end_inset
To see what happens when
\begin_inset Formula $\frac{M}{N}\ge\frac{1}{2}$
\end_inset
look at the relation
\begin_inset Formula \[
\sin^{2}\vartheta=\frac{4M\left(N-M\right)}{N^{2}}=4\frac{M}{N}\left(1-\frac{M}{N}\right)\]
\end_inset
and note that
\begin_inset Formula $\vartheta$
\end_inset
gets smaller as
\begin_inset Formula $M$
\end_inset
increases from
\begin_inset Formula $\frac{N}{2}$
\end_inset
to
\begin_inset Formula $N$
\end_inset
.
Because
\begin_inset Formula $\vartheta$
\end_inset
gets smaller,
\begin_inset Formula $k$
\end_inset
gets larger, and algorithm gets slower to the point where it is not better
than a random classical algorithm.
\end_layout
\begin_layout Subsection*
Grover's Algorithm Circuit
\end_layout
\begin_layout Standard
\begin_inset Formula \[
\textrm{Circuit: }\left(G^{k}\right)\left(H^{\otimes n}\otimes I\right)\left(\left|0\right\rangle ^{\otimes n}\otimes\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\right)\]
\end_inset
(Diagram Figure 6.1, Page 251)
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|0\right\rangle ^{\otimes t}\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}} & \xrightarrow{H^{\otimes n}\otimes I} & \left|\psi\right\rangle \frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\\
& \xrightarrow{G^{k}} & \left(\cos\left(\frac{2k+1}{2}\vartheta\right)\left|\alpha\right\rangle +\sin\left(\frac{2k+1}{2}\vartheta\right)\left|\beta\right\rangle \right)\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsection*
Estimating M
\end_layout
\begin_layout Standard
Grovers assumes
\begin_inset Formula $M$
\end_inset
is known.
If
\begin_inset Formula $M$
\end_inset
is unknown, there are two approaches.
\end_layout
\begin_layout Standard
1.
Brasard, et al.
Apply
\begin_inset Formula $G^{j}$
\end_inset
for random
\begin_inset Formula $j$
\end_inset
, and show expected number of steps is
\begin_inset Formula $O\left(\sqrt{N/M}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
2.
Find
\begin_inset Formula $M$
\end_inset
using
\begin_inset Formula $\sin\frac{\vartheta}{2}=\sqrt{\frac{M}{N}}$
\end_inset
and estimating
\begin_inset Formula $\vartheta$
\end_inset
.
Since this also gives you the boolean mean it lets you
\begin_inset Quotes eld
\end_inset
hit 2 birds for the price of one.
\begin_inset Quotes erd
\end_inset
We cover this second approach.
\begin_inset Formula \begin{eqnarray*}
G & = & \left(2\left|\psi\right\rangle \left\langle \psi\right|-I\right)O_{f}\\
G\left|\alpha\right\rangle & = & \cos\left(\vartheta\right)\left|\alpha\right\rangle +\sin\left(\vartheta\right)\left|\beta\right\rangle \\
G\left|\beta\right\rangle & = & \cos\left(\vartheta+\frac{\pi}{2}\right)\left|\alpha\right\rangle +\sin\left(\varphi+\frac{\pi}{2}\right)\left|\beta\right\rangle \\
& = & -\sin\left(\vartheta\right)\left|\alpha\right\rangle +\cos\left(\vartheta\right)\left|\beta\right\rangle \end{eqnarray*}
\end_inset
The effect of
\begin_inset Formula $G$
\end_inset
on the
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
vector above is determined by looking at reflections and rotations on the
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
,
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
plane (Figure 6.3, page 253).
Applying
\begin_inset Formula $O_{f}$
\end_inset
to
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\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
does not change anything because as shown earlier,
\begin_inset Formula $O_{f}$
\end_inset
is a reflection about the
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
axis.
Applying
\begin_inset Formula $\left(2\left|\psi\right\rangle \left\langle \psi\right|-I\right)$
\end_inset
to
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
reflects the state about
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
.
Since
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
is
\begin_inset Formula $\frac{\vartheta}{2}$
\end_inset
degrees above
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
, reflecting
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
about it is equivalent to rotating the state by
\begin_inset Formula $2\cdot\frac{\vartheta}{2}=\vartheta$
\end_inset
degrees.
The resulting vector has magnitudes of
\begin_inset Formula $\cos\left(\vartheta\right)$
\end_inset
in the
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
direction and
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\begin_inset Formula $\sin\left(\vartheta\right)$
\end_inset
in the
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
direction.
\begin_inset Formula $G\left|\beta\right\rangle $
\end_inset
is
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derived similarly.
The relation above can be used to express
\begin_inset Formula $G$
\end_inset
as a transformation on the
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\begin_inset Formula $\left|\alpha\right\rangle ,$
\end_inset
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
basis:
\begin_inset Formula \[
G=\left(\begin{array}{cc}
\cos\vartheta & -\sin\vartheta\\
\sin\vartheta & \cos\vartheta\end{array}\right)\]
\end_inset
\end_layout
\begin_layout Standard
The eigenvalues of
\begin_inset Formula $G$
\end_inset
are given by
\begin_inset Formula \begin{eqnarray*}
0 & = & \left|\begin{array}{cc}
\cos\vartheta-\lambda & -\sin\vartheta\\
\sin\vartheta & \cos\vartheta-\lambda\end{array}\right|\\
& = & \left(\cos\left(\vartheta\right)-\lambda\right)^{2}+\sin^{2}\left(\vartheta\right)\\
\left(\cos\left(\vartheta\right)-\lambda\right)^{2} & = & -\sin^{2}\left(\vartheta\right)\\
\cos\left(\vartheta\right)-\lambda & = & \pm i\sin\left(\vartheta\right)\\
\lambda & = & \cos\left(\vartheta\right)\pm i\sin\left(\vartheta\right)\\
& = & e^{\pm i\vartheta}\end{eqnarray*}
\end_inset
Phase estimation is performed on
\begin_inset Formula $G$
\end_inset
to find
\begin_inset Formula $\vartheta$
\end_inset
, which can in turn be used to find
\begin_inset Formula $M$
\end_inset
and
\begin_inset Formula $k$
\end_inset
.
\end_layout
\begin_layout Section*
Lecture 27 (25 April)
\end_layout
\begin_layout Subsection*
Review: Grover's Algorithm
\end_layout
\begin_layout Standard
\begin_inset Formula $G^{k}$
\end_inset
,
\begin_inset Formula $k=\frac{\pi}{2\vartheta}$
\end_inset
,
\begin_inset Formula $\sin\frac{\vartheta}{2}=\sqrt{\frac{M}{N}}$
\end_inset
\end_layout
\begin_layout Standard
When
\begin_inset Formula $M$
\end_inset
is small, the following approximation for
\begin_inset Formula $k$
\end_inset
is valid:
\begin_inset Formula $k\approx\left\lceil \frac{\pi}{4}\sqrt{\frac{M}{N}}\right\rceil $
\end_inset
\end_layout
\begin_layout Standard
Otherwise
\begin_inset Formula $k$
\end_inset
can be found using
\begin_inset Formula $\vartheta=\arcsin\left(2\frac{M\left(N-M\right)}{N^{2}}\right)$
\end_inset
.
\end_layout
\begin_layout Standard
If
\begin_inset Formula $M=1$
\end_inset
,
\begin_inset Formula $k=O\left(\sqrt{N}\right)$
\end_inset
\end_layout
\begin_layout Standard
If
\begin_inset Formula $M$
\end_inset
is unknown, can use estimate of
\begin_inset Formula $\vartheta$
\end_inset
to find it.
Estimating
\begin_inset Formula $\vartheta$
\end_inset
happens with phase estimation on
\begin_inset Formula $G$
\end_inset
, which expressed as a transformation on the
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
,
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
basis looks like
\begin_inset Formula \[
G=\left(\begin{array}{cc}
\cos\vartheta & -\sin\vartheta\\
\sin\vartheta & \cos\vartheta\end{array}\right)\]
\end_inset
and has eigenvalues,
\begin_inset Formula $\lambda_{\pm}=e^{\pm i\vartheta}$
\end_inset
.
\end_layout
\begin_layout Subsection*
Phase Estimation for
\begin_inset Formula $\vartheta$
\end_inset
\end_layout
\begin_layout Standard
Phase estimation requires us to generate an initial state which approximates
one or more eigenvectors of
\begin_inset Formula $G$
\end_inset
.
\end_layout
\begin_layout Subsubsection*
Finding Eigenvectors of G
\end_layout
\begin_layout Standard
Denote unknown eigenvector as
\begin_inset Formula $\left(x\left|\alpha\right\rangle +y\left|\beta\right\rangle \right)$
\end_inset
so:
\begin_inset Formula \[
G\left(x\left|\alpha\right\rangle +y\left|\beta\right\rangle \right)=e^{\pm i\vartheta}\left(x\left|\alpha\right\rangle +y\left|\beta\right\rangle \right)\]
\end_inset
Expanding the left side gives:
\begin_inset Formula \begin{eqnarray*}
xG\left|\alpha\right\rangle +yG\left|\beta\right\rangle & = & x\left(\cos\vartheta\left|\alpha\right\rangle +\sin\vartheta\left|\beta\right\rangle \right)+y\left(-\sin\vartheta\left|\alpha\right\rangle +\cos\vartheta\left|\beta\right\rangle \right)\end{eqnarray*}
\end_inset
Which is true when
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\begin_inset Formula $x\cos\vartheta-y\sin\vartheta=xe^{\pm i\vartheta}$
\end_inset
and
\begin_inset Formula $x\sin\vartheta+y\cos\vartheta=ye^{\pm i\vartheta}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
x\cos\vartheta-y\sin\vartheta & = & xe^{\pm ig}\\
x\cos\vartheta-y\sin\vartheta & = & x\left(\cos\vartheta\pm i\sin\vartheta\right)\\
-y & = & \pm ix\textrm{ when }\sin\vartheta\ne0,\frac{\vartheta}{2}\ne k\frac{\pi}{2}\end{eqnarray*}
\end_inset
Which gives eigenvector of the form
\begin_inset Formula $x\left|\alpha\right\rangle \pm ix\left|\beta\right\rangle $
\end_inset
.
Normalized, the eigenvectors are
\begin_inset Formula \begin{eqnarray*}
\left|\psi_{+}\right\rangle & = & \frac{\left|\alpha\right\rangle +i\left|\beta\right\rangle }{\sqrt{2}}\\
\left|\psi_{-}\right\rangle & = & \frac{\left|\alpha\right\rangle -i\left|\beta\right\rangle }{\sqrt{2}}\end{eqnarray*}
\end_inset
\begin_inset Formula $\left|\alpha\right\rangle $
\end_inset
and
\begin_inset Formula $\left|\beta\right\rangle $
\end_inset
can be rewritten as
\begin_inset Formula \begin{eqnarray*}
\left|\alpha\right\rangle & = & \frac{1}{\sqrt{2}}\left(\left|\psi_{+}\right\rangle +\left|\psi_{-}\right\rangle \right)\\
\left|\beta\right\rangle & = & \frac{i}{\sqrt{2}}\left(\left|\psi_{+}\right\rangle -\left|\psi_{-}\right\rangle \right)\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Subsubsection*
Generating Initial State
\end_layout
\begin_layout Standard
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
can be used as an initial state because it is a combination of eigenvectors:
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\left|\psi\right\rangle & = & \frac{1}{\sqrt{N}}\sum_{x}\left|x\right\rangle \\
& = & \cos\left(\frac{\vartheta}{2}\right)\left|\alpha\right\rangle +\sin\left(\frac{\vartheta}{2}\right)\left|\beta\right\rangle \\
& = & \cos\left(\frac{\vartheta}{2}\right)\frac{1}{\sqrt{2}}\left(\left|\psi_{+}\right\rangle +\left|\psi_{-}\right\rangle \right)+\sin\left(\frac{\vartheta}{2}\right)\frac{i}{\sqrt{2}}\left(\left|\psi_{+}\right\rangle -\left|\psi_{-}\right\rangle \right)\\
& = & \frac{1}{\sqrt{2}}\left(\cos\left(\frac{\vartheta}{2}\right)+i\sin\left(\frac{\vartheta}{2}\right)\right)\left|\psi_{+}\right\rangle +\frac{1}{\sqrt{2}}\left(\cos\left(\frac{\vartheta}{2}\right)-i\sin\left(\frac{\vartheta}{2}\right)\right)\left|\psi_{-}\right\rangle \\
& = & \frac{1}{\sqrt{2}}e^{i\frac{\vartheta}{2}}\left|\psi_{+}\right\rangle +\frac{1}{\sqrt{2}}e^{-i\frac{\vartheta}{2}}\left|\psi_{-}\right\rangle \end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Note: Coefficients to
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\begin_inset Formula $\left|\psi_{+}\right\rangle $
\end_inset
and
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\begin_inset Formula $\left|\psi_{-}\right\rangle $
\end_inset
above are not important.
Any state which was a combination of the two eigenvectors would work for
finding the boolean mean.
\end_layout
\begin_layout Subsubsection*
Phase Estimation Results
\end_layout
\begin_layout Standard
The result of phase estimation with
\begin_inset Formula $G$
\end_inset
and initial state
\begin_inset Formula $\left|\psi\right\rangle $
\end_inset
can be used to find boolean mean
\begin_inset Formula \[
S\left(f\right)=\frac{1}{N}\sum_{x}f\left(x\right)=\frac{M}{N}=\sin^{2}\left(\frac{\vartheta}{2}\right)\]
\end_inset
Phase estimation gives
\begin_inset Formula $\left|\varphi-\hat{\varphi}\right|\le2^{\eta_{0}}$
\end_inset
,
\begin_inset Formula $\hat{\varphi}=\frac{d}{2^{t}}$
\end_inset
with probability
\begin_inset Formula $\left(1-\epsilon\right)$
\end_inset
using
\begin_inset Formula $t=\eta_{0}+\left\lceil \log\left(2+\frac{1}{2\epsilon}\right)\right\rceil $
\end_inset
qubits in the top register.
Phase estimation will approximate
\begin_inset Formula $\lambda_{+}$
\end_inset
with probability
\begin_inset Formula \[
\left(1-\epsilon\right)\left|\frac{1}{\sqrt{2}}e^{i\frac{\vartheta}{2}}\right|^{2}\]
\end_inset
and approximate
\begin_inset Formula $\lambda_{-}$
\end_inset
with probability
\begin_inset Formula \[
\left(1-\epsilon\right)\left|\frac{1}{\sqrt{2}}e^{-i\frac{\vartheta}{2}}\right|^{2}\]
\end_inset
Eigenvalues again are
\end_layout
\begin_layout Standard
\begin_inset Formula \begin{eqnarray*}
\lambda_{+} & = & e^{i\vartheta}=e^{2\pi i\vartheta/\left(2\pi\right)}\\
\lambda_{-} & = & e^{i\left(2\pi-\vartheta\right)}=e^{-2\pi i\left(2\pi-\vartheta\right)/\left(2\pi\right)}\\
\varphi_{+} & = & \frac{\vartheta}{2\pi}\\
\varphi_{-} & = & \frac{2\pi-\vartheta}{2\pi}\end{eqnarray*}
\end_inset
\end_layout
\begin_layout Standard
Error bounds look like
\begin_inset Formula \begin{eqnarray*}
\left|\varphi_{\pm}-\hat{\varphi}\right| & \le & 2^{-\eta_{0}}\\
\left|\pi\varphi_{\pm}-\pi\hat{\varphi}\right| & \le & \frac{\pi}{2^{\eta_{0}}}\\
\pi\varphi_{+} & = & \frac{\vartheta}{2}\\
\pi\varphi_{-} & = & \pi-\frac{\vartheta}{2}\\
\left|\frac{\vartheta}{2}-\pi\hat{\varphi}\right| & \le & \frac{\pi}{2^{\eta_{0}}}\textrm{ w/prob }\frac{1}{2}\left(1-\epsilon\right)\\
\left|\pi-\frac{\vartheta}{2}-\pi\hat{\varphi}\right| & \le & \frac{\pi}{2^{\eta_{0}}}\textrm{ w/prob }\frac{1}{2}\left(1-\epsilon\right)\end{eqnarray*}
\end_inset
But we aren't using phase estimation to compute phase, we are using it to
compute
\begin_inset Formula $M$
\end_inset
which is related to the sin, so we need a different bound:
\begin_inset Formula \begin{eqnarray*}
\left|\sin^{2}\left(\frac{\vartheta}{2}\right)-\sin^{2}\left(\frac{\pi j}{2^{t}}\right)\right| & \le & \frac{\pi}{2^{\eta_{0}}}\sqrt{S\left(f\right)\left(1-S\left(f\right)\right)}+\frac{\pi^{2}}{2^{2\eta_{0}}}\end{eqnarray*}
\end_inset
(from paper BHMT lemma 7, page 15)
\end_layout
\begin_layout Standard
\begin_inset Formula $j$
\end_inset
is measurement in computational basis
\end_layout
\begin_layout Standard
since
\begin_inset Formula $\sin^{2}\left(\frac{\varphi}{2}\right)=\sin^{2}\left(\pi-\frac{\varphi}{2}\right)=S\left(f\right)$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $\left.\begin{array}{c}
\sin^{2}\left(\frac{\vartheta}{2}\right)\approx\frac{\vartheta}{2}\\
\sin^{2}\left(\frac{\pi j}{2}\right)\approx\frac{\pi j}{2}\end{array}\right\} \rightarrow\left(\frac{\vartheta}{2}\right)^{2}-\left(\frac{\pi i}{2}\right)^{2}=\left(\frac{\vartheta}{2}+\frac{\pi i}{2}\right)\left(\frac{\vartheta}{2}-\frac{\pi i}{2}\right)$
\end_inset
\end_layout
\begin_layout Subsubsection*
Phase Estimation Circuit
\end_layout
\begin_layout Standard
The circuit for using phase estimation to compute the boolean mean is the
same as the circuit for normal phase estimation.
(Figure 6.7 page 262)
\end_layout
\begin_layout Standard
The output of the circuit in the bottom register will be either
\begin_inset Formula $\left|\psi_{+}\right\rangle $
\end_inset
or
\begin_inset Formula $\left|\psi_{-}\right\rangle $
\end_inset
.
The top register needs just enough bits to be able to distinguish
\begin_inset Formula $\frac{1}{N}$
\end_inset
from
\begin_inset Formula $0$
\end_inset
.
\end_layout
\begin_layout Standard
In general, there is no efficient way to make
\begin_inset Formula $G^{2^{j}}$
\end_inset
gates, just have to apply
\begin_inset Formula $G$
\end_inset
repeatedly, so the number of quesries is
\begin_inset Formula $\tau=2^{t-1}=\Theta\left(2^{t}\right)=\Theta\left(2^{\eta_{0}}\right)$
\end_inset
.
Error is
\begin_inset Formula $O\left(\frac{1}{\tau}\right)$
\end_inset
.
\begin_inset Formula $t=\eta_{0}+\left\lceil log_{2}\left(2+\frac{1}{2\epsilon}\right)\right\rceil $
\end_inset
.
This is the best possible, see paper [Nayak+Wu], algorithm is optimal.
\end_layout
\begin_layout Section*
Lecture 28 (30 April)
\end_layout
\begin_layout Standard
Review for Final
\end_layout
\end_body
\end_document