\documentclass{article}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem{proposition}{Proposition}
\newcommand{\Pb}{{\bf P}}
\newcommand{\trace}{{\rm tr}}
\title{Lectures on Quantum Information}
\author{William G. Faris\\
University of Arizona}
\begin{document}
\maketitle
\tableofcontents
\section{Introduction}
These lectures are an elementary introduction to entangled
states and to quantum information. They begin with a brief
account of quantum mechanics,
focusing on quantum events and their probabilities.
The first topic is a review of the theory of square complex
matrices. Then quantum events are realized
mathematically as orthogonal projection matrices.
The first example is the system consisting of a
single spin 1/2 particle.
This system is formulated in terms of 2 by 2 matrices.
Each non-trivial event
corresponds to whether the spin
of the particle is in a certain direction.
Next there is a brief introduction to the logic of quantum events.
The main difference between quantum mechanics and probability is
that in quantum mechanics there are events that are not
compatible.
The next example is the main subject of the lectures. This is the
system of two spin 1/2 particles. This example is formulated
in terms of 4 by 4 matrices. For each of the two
particles there is an event associated with each given direction.
Two events associated with two different particles are compatible.
It turns out that there is a state of the
system where these events are very far from independent.
The lectures conclude with a
discussion of the significance of this example for our
understanding of nature. The dependence between the
spins of the two particles presents the simplest and
most striking example of an entangled quantum state. Such examples
point to a holistic nature of quantum mechanics that has
startling consequences. This is made precise in Bell's
theorem, which says that the dependence has no explanation
in terms of conventional probability. There is also some mention of quantum
cryptography as a development in the relatively new field of
quantum information.
\section{Matrices}
\subsection{Algebra of square matrices}
An $n$ by $n$ complex matrix is a square array $A$ of complex
numbers with $n$ rows and $n$ columns. The entry in the $i$th row
and $j$th column is denoted $A_{ij}$. Matrices admit several
algebraic operations.
Addition and subtraction. The {\em sum} or {\em difference} of two
$n$ by $n$ matrices $A$, $B$ is denoted $A\pm B$ and is defined by
\begin{equation}
(A \pm B)_{ij} = A_{ij} \pm B_{ij}.
\end{equation}
Multiplication. The {\em product} of two $n$ by $n$ matrices $A$,
$B$ is denoted $AB$ and is defined by
\begin{equation}
(A B)_{ij} = \sum_{k=1}^n A_{ik} B_{kj}.
\end{equation}
For each $n$ there is an $n$ by $n$ {\em zero matrix} $O$ that has
all entries zero. There is another $n$ by $n$ {\em identity}
matrix $I$ that has the diagonal entries $I_{ii} = 1$, but all
off-diagonal entries $I_{ij} = 0$ for $i \neq j$.
The algebra of matrices includes such addition identities as
$A+B= B +A$, $A+0 = A$, $A-A=0$. Two matrices are said to {\em
commute} if $AB=BA$. Typical matrices do not commute. However
there are useful multiplication identities such as $A(B\pm C) =
AB \pm AC$, $(B\pm C)A = BA \pm CA$, $A0 = 0$, $0A = 0$, $AI =
A$, $IA = A$.
A {\em diagonal} $n$ by $n$ matrix is a matrix with zeros
everywhere except on the main diagonal. Thus a diagonal matrix is
specified by a list of $n$ complex numbers. Thus one can think of
a complex matrix as a generalization of the concept of a list of
complex numbers.
\subsection{The adjoint of a matrix}
The {\em adjoint} of an $n$ by $n$ matrix $A$ is the matrix $A^*$
defined by
\begin{equation}
(A^*)_{ij} = \overline{A_{ji}}.
\end{equation}
That is, one reflects the matrix over the diagonal and takes the
complex conjugate of each entry.
The adjoint satisfies algebraic properties like $A^{**}=A$,
$(A\pm B)^* = A^* \pm B^*$ and $(AB)^* = B^* A^*$. It can be
thought of as a generalization of the concept of complex
conjugation.
A matrix $A$ is said to be {\em self-adjoint} if $A = A^*$. The
diagonal entries of a self-adjoint matrix are real. Thus a
self-adjoint matrix is a generalization of the concept of a list
of $n$ real numbers.
A matrix $A^*A$ is automatically self-adjoint. It is the analog
of the absolute value squared of a complex number. The diagonal
entries of such a matrix are real and positive.
\subsection{The trace of a matrix}
The {\em trace} of an $n$ by $n$ matrix $A$ is the complex number
defined by
\begin{equation}
\trace(A) = \sum_{i=1}^n A_{ii}.
\end{equation}
Thus the trace is the sum of the diagonal entries.
The trace satisfies algebraic properties that include $\trace(A
\pm B) = \trace(A) \pm \trace(B)$, $\trace(AB) = \trace(BA)$, and
$\trace(A^*) = \overline{\trace(A)}$. The trace of a self-adjoint
matrix is a real number. Furthermore, it is easy to see that
$\trace(A^*A) \geq 0$.
\subsection{Projection matrices}
A matrix $P$ is said to be a {\em projection} if $P^2 = P$. The
trace of a projection is a natural number $0,1,2, \ldots, n$.
This number is called the {\em rank} or {\it dimension} of the
projection.
A matrix $E$ is said to be an {\em orthogonal projection} if $E =
E^*$ and $E^2 = E$, that is, if $E$ is a self-adjoint projection.
An orthogonal projection that is diagonal has just the numbers
one and zero on the diagonal. An orthogonal projection is a
generalization of the concept of a list of $n$ numbers, each of
which is either one or zero. In general, one can think of an
orthogonal projection as assuming the value one with a certain
multiplicity, which is the rank $r = \trace(E)$ of the orthogonal
projection. Corresponding, it can assume the value zero with
multiplicity $n-r$.
\subsection{Problems}
\begin{enumerate}
\item Which of the following are correct trace identities?
(a) $\trace(AB) = \trace(A) \trace(B)$. (b) $\trace(AA^*) \geq 0$.
(c) $\trace(ABC) = \trace(CAB)$. (d) $\trace(ABC) = \trace(CBA)$.
\item Define the Hilbert-Schmidt norm $\|A\|_2$ of a matrix to be given by
$\|A\|_2^2 = \trace(A^*A)$. Prove that $|\trace(A^*B)| \leq
\|A\|_2 \|B\|_2$.
\item If $E$ is an orthogonal projection, then when is
$I-E$ an orthogonal projection? If $E$ and $F$ are orthogonal
projections, then when is $EF$ an orthogonal projection?
\item If $E$ and $F$ are orthogonal projections, express
$\trace(EF)$ as the square of the Hilbert-Schmidt norm of a
certain matrix.
\item Find all 2 by 2 complex matrices $C$ such that
$C^* = C$ and $C^2 = I$. Such a matrix is called a
{\em complex reflection} if also $\trace(C) = 0$. Find all complex
reflections. Show that the set of all complex reflections
form a sphere.
\item Find all 2 by 2 complex matrices $E$ that satisfy the
equations $E^* = E$ and $E^2 = E$ for an orthogonal projection.
Show that the orthogonal projections with $\trace(E) = 1$ form
a sphere. Hint: Let $E = \frac{1}{2} (I + C)$.
\end{enumerate}
\section{Quantum mechanics}
\subsection{Quantum events and quantum states}
Quantum mechanics for an $n$ level system is formulated in terms
of $n$ by $n$ matrices. A {\em quantum event} is an orthogonal
projection $E$. That is, it is an $n$ by $n$ matrix with $E= E^*$
and $E^2 = E$.
A {\em quantum state} is an orthogonal projection $R$ with
$\trace(R) = 1$.
\subsection{Quantum probability}
\begin{theorem}
If the probability of a quantum event $E$ in the quantum state
$R$ is defined by
\begin{equation}
\Pb_R[E] = \trace(ER),
\end{equation}
then $0 \leq \Pb_R[E] \leq 1$.
\end{theorem}\
Proof: Observe that $\Pb_R(E) = \trace(ER) = \trace(ER^2) =
\trace(RER)$. Furthermore, $RER = (RE)(ER) = (ER)^*(ER)$. It
follows that $0 \leq \Pb_R(E) $. Thus the probability of an
arbitrary event is positive. The orthogonal projection $I-E$
plays the role of the complementary event. Since $\trace(ER) +
\trace((I-E)R) = \trace(R) =1$, it follows that $\Pb_R(E)+
\Pb_R(I-E) = 1$. Thus the probability of the event and its
complement sum to one. From this we see that $0 \leq \Pb_R(E)
\leq 1$.
\subsection{Problems}
\begin{enumerate}
\item A quantum state is an
orthogonal projection of rank one. Define the distance $d(R,S)$
between two quantum states to be given in terms of the
Hilbert-Schmidt norm by $d(R,S) = (1/\sqrt{2}) \|R-S\|_2$. What
is the maximum distance between two quantum states?
\item Say that $R$ and $S$ are quantum states with $RS = 0$. Let
$G = R+S$, so that $G$ is an orthogonal projection with rank 2.
Let $C$ be a self-adjoint matrix satisfying $CG = GC = C$ and $C^2 = G$ and $\trace(C) = 0$.
Let $E = (1/2) (G + C)$. Show that $E$ is also a quantum state and
that $E$ satisfies $EG = GE = E$. Such a state is
called a {\em superposition} of $R$ and $S$. The superpositions of
$R$ and $S$ form a sphere, with $R$ at the north pole (corresponding to
$C = R-S$) and with $S$ at
the south pole (corresponding to $C = S-R$). The latitude circles are
parametrized by $p = \trace(RER)$ and $q = \trace(SES)$ with
$p + q = 1$. We may write $E = GEG = RER + SES + RES + SER$, or
$E = pR + qS + RES + SER$. The last two terms are called the {\em interference terms}.
\end{enumerate}
\section{One quantum bit}
\subsection{One spin 1/2 particle}
In classical computation one {\em bit} is the information
contained in a choice of 0 or 1. In quantum computation one {\em
qubit} is the information contained in a choice of a quantum
state in a two-level system. Sometimes this system is referred to
as a single spin 1/2 particle.
Let $(x,y,z)$ be a triple of real numbers with $x^2 + y^2 + z^2 =
1$. For each such unit vector there is a orthogonal projection
\begin{equation}
E(x,y,z) = \frac{1}{2} \left[ \matrix{ 1+z & x-iy \cr x+iy & 1-z } \right].
\end{equation}
\begin{proposition} The matrix $E(x,y,z)$ satisfies the
equations $E = E^*$ and $E^2 = E$ for a projection matrix.
Furthermore, it has rank one.
\end{proposition}
Thus for every unit vector $(x,y,z)$ there is a corresponding
quantum state for the single spin $1/2$ particle given by
$E(x,y,z)$.
The quantum events are also given in this way. The only
non-trivial quantum events are given by the projections
$E(x',y',z')$. Thus for every unit vector $(x',y',z')$ there is a
corresponding quantum event given by $E(x',y',z')$. Of course
there are also the trivial quantum events $0$ and $I$.
For each direction $(x',y',z')$ there is a spin variable that can
either be along this direction or opposite to this direction. The
interpretation of the quantum event $E(x',y',z')$ is that the
spin variable is along this direction. Sometimes one thinks of the
spin variable for a given direction as having the value $1/2$ when
the spin is along the direction and the value $-1/2$ when the spin
is opposite to the direction.
\begin{proposition} The quantum events associated with unit
vectors in opposite directions satisfy
\begin{equation}
E(-x',-y',-z') + E(x',y',z') = I
\end{equation}
and so are the negations of each other. All pairs of quantum
events for this system determined by unit vectors that are not in
the same or opposite directions are incompatible.
\end{proposition}
\begin{theorem} Consider a system consisting of a single spin 1/2
particle. If the state is determined by the unit vector $(x,y,z)$,
then the probability that the spin is in the direction of
the unit vector $(x',y',z')$
is
\begin{equation}
\Pb_{(x,y,z)} [ E(x',y',z') ] = \frac{1}{2} (1 + x x' + y y' + z z')
= \frac{1}{2} ( 1 + \cos(\theta) ) = \cos^2(\theta/2).
\end{equation}
\end{theorem}
Proof. Multiply the matrices. Take the trace of the result.
Remark. The presence of the $\theta/2$ in the last formula
is the reason for the terminology spin 1/2.
\subsection{Problems}
\begin{enumerate}
\item Recall the definition of distance between quantum
states in a previous problem. What is the distance between the two
quantum states $E(x,y,z)$ and $E(x',y',z')$?
\end{enumerate}
\section{Quantum logic}
\subsection{Compatible quantum events}
Quantum events $E, F$ are said to be {\em compatible} if they
commute, that is, if $EF= FE$. In that case $EF$ is also a
quantum event.
\subsection{Negation, conjunction, disjunction}
The {\em negation} or {\em complement} of a quantum event is the
quantum event
\begin{equation}
\lnot E = I - E.
\end{equation}
The {\em conjunction} of two compatible quantum
events $E,F$ is defined to be
\begin{equation}
E \land F = EF.
\end{equation}
The conjunction is not defined for events that are not compatible.
The peculiarity of quantum mechanics (compared to probability
theory) is that conjunction is defined only for events that are
compatible.
The {\em disjunction} of two compatible quantum events $E,F$ is
defined to be
\begin{equation}
E \lor F = \lnot( \lnot E \land \lnot F).
\end{equation}
The disjunction is not defined for events that are not compatible.
Note: The disjunction of two compatible events may be written in
matrix algebra rather than in logic. It is then
\begin{equation}
E \lor F = I - (I-E)(I-F) = E + F - EF.
\end{equation}
The probability of the complement of a quantum event is
\begin{equation}
\Pb_R[\lnot E] = 1 - \Pb_R[E].
\end{equation}
The probability of the disjunction of two compatible quantum
events is
\begin{equation}
\Pb_R[E \lor F] = \Pb_R[E] + \Pb_R[F] - \Pb_R[E \land F].
\end{equation}
Two compatible quantum events
are said to be {\em exclusive}
if $E \land F = 0$. For exclusive events we have the law $\Pb_R[
E \lor F] = \Pb_R[E] + \Pb_R[F]$.
Two compatible quantum events
are said to be {\em independent}
if $\Pb_R[ E \land F] = \Pb_R[E] \Pb_R[F]$. Sometimes the size
of the {\em correlation} $4(\Pb(E \land F] - \Pb[E] \Pb[F])$ is
taken as a measure of the lack of independence. This correlation
is always between $-1$ and $1$.
\subsection{Problems}
\begin{enumerate}
\item Prove the following {\em uncertainty principle}:
\begin{equation}
P_S[E] + P_S[F] \leq 1 + 2 \sqrt{\trace(EF)}.
\end{equation}
Hint: Expand $(E+F-1)^2$. This principle implies that certain
events that are not compatible are nevertheless almost exclusive,
in a certain sense. The most famous example is when $E$ is the
event that a particle has its position in a certain small
interval and $F$ is the event that the particle has its momentum
in a certain small interval. Then $\trace(EF)$ is a physical
constant times the product of the lengths of the two intervals,
divided by $2\pi$. If the product of the lengths is sufficiently
small, then, for every state, it is impossible that the position
probability and the momentum probability both be near one.
\item Some authors have proposed to define the conjunction of
two events $E,F$ that are not compatible. For instance, it may be
shown that $(EFE)^n$ converges to a matrix $G$ as $n \to
\infty$. Show that $G$ is an orthogonal projection.
\end{enumerate}
\section{Two quantum bits}
\subsection{Two spin 1/2 particles}
Now we look at a 4-level system consisting of two spin 1/2
particles. One belongs to Alice and one belongs to Bob. There are
now two sets of spin matrices. For Alice we have the Kronecker
product of the matrices $E(x,y,z)$ and $I$, that is, $E^A(x,y,z)$
given by
\begin{equation}
\frac{1}{2} \left[ \matrix{ 1+z \left[\matrix{ 1 & 0 \cr 0 & 1
\cr} \right] & x-iy \left[\matrix{ 1 & 0 \cr 0 & 1 \cr} \right]
\cr x+iy \left[\matrix{ 1 & 0 \cr 0 & 1 \cr} \right] & 1- z
\left[\matrix{ 1 & 0 \cr 0 & 1 \cr} \right] \cr} \right] =
\frac{1}{2} \left[ \matrix{ 1+z & 0 & x-iy & 0 \cr 0 & 1+ z & 0
& x-iy \cr x+iy & 0 & 1-z & 0 \cr 0 & x+iy & 0 & 1-z \cr} \right].
\end{equation}
This is just two copies of the projection matrix. The first copy
belongs with the first and third rows and columns, while the
second copy belongs with the second and fourth rows and columns.
For Bob we have the Kronecker product of $I$ and $E(x,y,z)$,
that is $E^B(x,y,z)$ given by
\begin{equation}
\frac{1}{2} \left[ \matrix{
1 \left[\matrix{ 1+z & x-iy \cr x+iy & 1-z \cr} \right] &
0 \left[\matrix{ 1+z & x-iy \cr x+iy & 1-z \cr} \right] \cr
0 \left[\matrix{ 1+z & x-iy \cr x+iy & 1-z \cr} \right] &
1 \left[\matrix{ 1+z & x-iy \cr x+iy & 1-z \cr} \right] \cr
} \right]
=
\frac{1}{2} \left[ \matrix{ 1+z & x-iy & 0 & 0 \cr
x+iy & 1-z & 0 & 0 \cr
0 & 0 & 1+z & x-iy \cr 0 & 0 & x+iy & 1-z \cr} \right].
\end{equation}
This is again two copies of the projection matrix. The first copy
belongs with the first and second rows and columns, while the
second copy belongs with the third and fourth rows and columns.
Each of these matrices has rank two.
\begin{proposition} The orthogonal projection
operators $E^A(x,y,z)$ and $E^B(x',y',z')$ commute with each other.
\end{proposition}
The orthogonal projection $E^A(x,y,z)E^B(x',y',z')$ has rank one
and thus defines a state. This is a state in which the A spin and
the $B$ spin are independent. The $A$ spin in the $(x,y,z)$
direction is $1/2$ and the $B$ spin in the $(x',y',z')$ direction
is also $1/2$.
In this system there are other interesting orthogonal projections
that define quantum events and quantum states. The orthogonal
projection
\begin{equation}
E_0 = \frac{1}{2} \left[ \matrix{ 0 & 0 & 0 & 0 \cr 0 & 1 & -1
& 0 \cr 0 & -1 & 1 & 0 \cr 0 & 0 & 0 & 0 \cr} \right]
\end{equation}
corresponds to the event of total spin 0, while the complementary
orthogonal projection
\begin{equation}
E_1 = \frac{1}{2} \left[ \matrix{ 2 & 0 & 0 & 0 \cr 0 & 1 & 1
& 0 \cr 0 & 1 & 1 & 0 \cr 0 & 0 & 0 & 2 \cr} \right]
\end{equation}
corresponds to the event of total spin 1. These projections have
rank one and three respectively. The projection matrix $R_0 = E_0$
may thus be regarded also as a quantum state and is called the
{\em singlet state}.
The singlet state has the property that for an arbitrary direction the
probability that both particles have spin in this direction is zero. On the
other hand, the probability that one or the other of the
particles has spin in this direction is one.
\begin{theorem} The singlet state $R_0$ is the unique state such
that for each direction the conjunction $E^A(x,y,z) E^B(x,y,z)$
has probability zero.
\end{theorem}
\begin{theorem}
Consider a system of two spin 1/2 particles in the singlet
state. The probability that the spin of particle $A$ is in the
$(x,y,z)$ direction is
\begin{equation}
\Pb_0[E^A(x,y,z)] = \trace(E^A(x,y,z) R_0) = \frac{1}{2}.
\end{equation}
Similarly, the probability that the spin of particle $B$ is in the
$(x',y',z')$ direction is
\begin{equation}
\Pb_0[E^B(x',y',z')] = \trace(E^B(x',y',z') R_0) = \frac{1}{2}.
\end{equation}
\end{theorem}
\begin{theorem}
Consider a system of two spin 1/2 particles in the singlet
state. The probability that the spin of particle $A$ is in the
$(x,y,z)$ direction and the spin of particle $B$ is
in the $(x',y',z')$ direction is
\begin{eqnarray}
\Pb_0[E^A(x,y,z)E^B(x',y',z')] &=&
\trace(E^A(x,y,z) E^B(x',y',z')R_0) \nonumber \\
= \frac{1}{4} (1 - xx' - yy' - z z')
&=& \frac{1}{4} (1- \cos(\theta)) = \frac{1}{2} \sin^2(\theta/2).
\end{eqnarray}
\end{theorem}
A state in which the event $E_1$ has probability one is called a
{\em triplet state}. An example of a triplet state is a state
$E^A(x,y,z) E^B(x,y,z)$ in which the spins are aligned in a fixed
direction $(x,y,z)$. Each individual spin in this direction has
the alue $1/2$, so the total spin in this direction is 1.
\subsection{Problems}
\begin{enumerate}
\item Let $0 \leq p \leq 1$ and $0 \leq q \leq 1$
and $p+q = 1$. Let $\chi$ be arbitrary. Show that the state
\begin{equation}
R = \left[ \matrix{ p & 0 & 0 & \sqrt{pq}e^{i\chi} \cr 0 & 0 & 0
& 0 \cr 0 & 0 & 0 & 0 \cr \sqrt{pq}e^{-i\chi} & 0 & 0 & q \cr}
\right].
\end{equation}
is a triplet state. (This is a superposition of the triplet state
where both spins are up and the triplet state where both spins are
down.)
\item Find the probabilities of $E(x,y,z)$ and $E(x',y',z')$ and
of $E(x,y,z)E(x',y',z')$ in the superposition triplet state $R$.
\item Look at $E^{A}(0,0,1)$ and $E^{B}(0,0,1)$ in the superposition
triplet state $R$. When are these independent? That is, when is
the probability of the conjunction equal to the product
of the probabilities?
\end{enumerate}
\section{Entangled states}
\subsection{Quantum mechanics as a physical theory}
The quantum theory has several astonishing characteristics:
\begin{itemize}
\item It explains all of physics and chemistry on the molecular,
atomic, and subatomic scales.
\item It involves linear algebra and nothing else.
\item It gives a holistic description of nature in which
an arbitrary collection of particles can display correlated
behavior.
\item Its interpretation is confused and controversial.
The most common interpretation says (roughly) that nothing is
real unless it is measured.
\end{itemize}
The holistic nature of quantum theory was first brought out in a
famous paper of Einstein, Rosen, and Podolsky. They proposed a
thought experiment that illustrated this feature. Later David
Bohm proposed a variant of this experiment involving spin that
could actually be performed. Subsequently, John Bell gave a
profound analysis of this spin experiment.
\subsection{The spin experiment}
A spin 1/2 particle is a quantum system with a particularly
simple structure. For each direction in space, there is a {\em
spin variable} that can have two possible values, say $+1/2$ and
$-1/2$. For each such direction, one can set up an experimental
apparatus that measures this variable. The values obtained are
random and are predicted by quantum theory. If one takes the
opposite direction in space, then spin $-1/2$ and $+1/2$ in that
opposite direction are regarded as the same as spin $+1/2$ and
$-1/2$ in the original direction.
The Bohm experiment involves two experimenters, Alice and Bob.
They are in distant locations. One of the experimenters prepares
a system consisting of two spin 1/2 particles prepared in the
{\em singlet state}. This is a special quantum mechanical state
in which the particles have a certain correlated behavior. Then
one of the particles is sent to the other experimenter. So Alice
has a particle, Bob has a particle, Alice and Bob are far apart,
and each is ready to do an experiment.
The singlet state has the following properties. For each particle
and for each direction in space, the probability of the spin
variable for that direction having the value $+1/2$ is 1/2, and
the probability of the spin variable for that direction having the
value $-1/2$ is also 1/2. However the values for the two particles
are correlated in a special way. Say that the angle between the
axis chosen for the Alice particle and the Bob particle is
$\theta$. Then the probability that Alice and Bob get the same
answer (both $+1/2$ or both $-1/2$) is $\sin^2(\theta/2)$. It
follows that the probability that Alice and Bob get different
answers is $\cos^2(\theta/2)$.
More specifically, the probability that Alice and Bob get
particular values when the angle between their measurement
directions is $\theta$ is given by the following table:
\begin{eqnarray}
P[+1/2,+1/2] & = & (1/2)\sin^2(\theta/2) \nonumber \\
P[+1/2,-1/2] & = & (1/2)\cos^2(\theta/2) \nonumber \\
P[-1/2,+1/2] & = & (1/2)\cos^2(\theta/2) \nonumber \\
P[-1/2,-1/2] & = & (1/2)\sin^2(\theta/2) .
\end{eqnarray}
Notice that if Alice and Bob happen to choose the same direction,
so $\theta = 0$, then they always get opposite answers. If they
choose opposite directions, so $\theta = \pi$, then they also
always get the same answer. So it is reasonable to say that for
any common choice of axis the spins are in opposite directions.
The total spin is zero.
If, on the other hand, Alice and Bob choose directions that are
at right angles, so that $\theta = \pi/2$, then the chance that
they get the same answer is $1/2$.
It will specially important for us to have the answer when the
angle is $2\pi/3$. In this case the probability that they get the
same answer is $3/4$, and the probability that they get
different answers is $1/4$. This is again saying that spins tend
to be in opposite directions, but in a probabilistic sense.
In probability, we may say that two events $U$ and $V$ are
probabilistically equivalent if $\Pb[U] = \Pb[V] = \Pb[U \land
V]$. If $U$ and $V$ are probabilistically equivalent, then for
every event $W$ we have $\Pb[U \land W] = \Pb[V \land W]$.
Pick three directions in space in a plane that are at angle
$2\pi/3$ with each other. Let $E^A$, $F^A$, and $G^A$ be the
events that the spin of the Alice particle is $+1/2$ in the first,
second, or third of these directions. Let $E^B$, $F^B$,and $G^B$
be the events that the spin of the Bob particle is $+1/2$ in the
same directions. Then for the first direction $\Pb[ \lnot E^A] =
\frac{1}{2}$, $\Pb[ E^B] = \frac{1}{2} $, and $\Pb[ \lnot E^A
\land E^B] = \frac{1}{2}$. Thus $\lnot E^A$ and $E^B$ are
probabilistically equivalent. Similarly, for the second direction,
$\lnot F^A$ and $F^B$ are probabilistically equvalent. And
finally, for the third direction $\lnot G^A$ and $G^B$ are
probabilistically equivalent.
In summary, these equations imply that whenever the two directions
for the two particles are the same, the event that the $A$
particle has spin $-1/2$ is probabilistically equivalent to the
event that the $B$ particle has spin $+1/2$.
\begin{theorem} Consider the system of two spin 1/2
particles in the singlet state. Suppose that there
were a probability assignment obeying the usual
probability laws that agrees with the probability predictions
of quantum mechanics for this system.
Then we would have the inequality
\begin{equation}
\Pb[ E^A \land F^B] + \Pb[ F^A \land G^B] + \Pb[ G^A \land E^B] \leq 1.
\end{equation}
\end{theorem}
The hypothesis of the theorem implies in particular that events
involving two different directions for one particle can be
combined in accordance with the laws of a fixed probability
assignment. Of course quantum mechanics regards these events as
incompatible, but the probability assignment is supposed to be an
extension of quantum mechanics. This theorem is called Bell's
first theorem, and the inequalities are Bell's inequalities.
Here is the proof of Bell's first
theorem. Suppose that probabilities associated with the same
particle were meaningful. Since $E^B$ is equivalent to $\lnot
E^A$, $F^B$ is equivalent to $\lnot F^A$, and $G^B$ is equivalent
to $\lnot G^A$, we would have the equations
\begin{eqnarray}
\Pb[E^A \land F^B] &=& \Pb[ E^A \land \lnot F^A] \nonumber \\
\Pb[F^A \land G^B] &=& \Pb[ F^A \land \lnot G^A] \nonumber \\
\Pb[G^A \land E^B] &=& \Pb[ G^A \land \lnot E^A] .
\end{eqnarray}
However from elementary probability
\begin{equation}
\Pb[ E^A \land \lnot F^A] + \Pb[ F^A \land \lnot G^A] +
\Pb[ G^A \land \lnot E^A] \leq 1.
\end{equation}
\begin{corollary} There is no probability assignment obeying the usual
probability laws that agrees with the probability predictions
of quantum mechanics for this system.
\end{corollary}
Proof: According to quantum mechanics,
\begin{eqnarray}
\Pb[ E^A \land F^B] &=& \frac{3}{8} \nonumber \\
\Pb[ F^A \land G^B] &=& \frac{3}{8} \nonumber \\
\Pb[ G^A \land E^B] &=& \frac{3}{8} .
\end{eqnarray}
This contradicts Bell's inequalities. Thus for
quantum mechanics there are no joint
probabilities for spin in different directions for the same
particle.
Since Bell's inequalities contradict the prediction of quantum
mechanics, it follows that quantum mechanics must violate the
hypothesis of the theorem. Events involving measurements in two
different directions for the same particle cannot be combined in
accordance with the laws of a fixed probability assignment.
Making one measurement interferes in an essential way with the
possibility of making the other measurement. There is much more
to be said about the implications of this theorem. An excellent
source is the appendix to the book {\it The Infamous Boundary},
by David Wick.
\subsection{Quantum cryptography}
It is tempting to put quantum entanglement to practical use. There
are several ways to proceed. Here is one example. Say that Alice
and Bob are located far apart. They can communicate openly.
However they desire to generate a key to encode secret
information. This key must have the property that a third party
Eve cannot obtain it without being detected.
Here is a first attempt. Alice and Bob order pairs of spin 1/2
particles in the singlet state from a supplier. They decide on a
certain direction $a$ in space in which to perform spin
measurements. With probability $1/2$, Alice measures a spin $+1/2$
in this direction, Bob measures a spin $-1/2$ in this direction.
With probability $1/2$, Alice measures a spin $-1/2$ in this
direction, Bob measures a spin $+1/2$ in this direction. Thus for
each pair of particles Alice and Bob get opposite results. It is
as if they each could look at the flip of a single coin. They
record the results, and this gives each of them the key to the
code in which they will communicate.
This first attempt is not as secure as might be desired, since Eve
might also somehow manage to make spin measurements in the $a$
direction.
The second attempt is more subtle. Alice and Bob each randomly
choose to use one of two perpendicular directions $a$ and $b$ for
their spin measurements. When Alice and Bob both happen to choose
$a$, then they get opposite results as before. When they both
happen to choose $b$, the same thing happens. When they choose
different directions, then their results are completely
independent of each other, as if they flipped two coins. After
all this experimentation takes place, Alice and Bob communicate
openly and tell what directions they chose. They then use the
cases when they happened to choose the same direction to
establish their key.
Now Eve cannot detect the code without being detected. Say, for
instance, that Eve made measurements in the $a$ direction. These
would perhaps not be detected when Alice and Bob themselves made
measurements in the $a$ direction. However when Alice and Bob both
make measurements in the $b$ direction, they will notice that
they no longer get results that are exact opposites. They can
establish this by open communication about the results of a
certain fraction of their experiments. The measurements Eve made
in the $a$ direction have betrayed themselves. This is because a
measurement in the $a$ direction in the entangled state precludes
measurements by Alice and Bob in the $b$ direction in that state.
That is, this measurements by Alice or Bob in the $b$ direction no
longer give the results predicted by the entangled state, and this
is something they would eventually notice. Of course all this
requires a detailed analysis to prove. Such a proof is provided
by quantum information theory.
\subsection{Problems}
\begin{enumerate}
\item Prove that if $U$ and $V$ are probabilistically equivalent,
then $\Pb[U \land W] = \Pb[V \land W]$.
\item Prove the probability inequality
$\Pb[ E \land \lnot F] + \Pb[ F \land \lnot G] + \Pb[ G \land
\lnot E] \leq 1$.
\item In probability two events are said to be {\em independent}
if the probability that they both occur is the product of the
individual probabilities. Consider the system of two spin 1/2
particles in the singlet state. For which relative orientations
$\theta$ are the events that the A particle has spin $+1/2$ and
that the B particle has spin $+1/2$ independent?
\item For each direction in space, there is a triplet state of
the two spin 1/2 particle system in which both particles are
aligned in this direction. Let $\theta_a$ be the angle between the
measured spin component of the A particle and the direction for
the triplet state. Let $\theta_b$ be the angle between the
measured spin component of the B particle and the direction for
the triplet state. Then the probability that the A measurement
and the B measurement both give $+1/2$ is $\cos^2(\theta_a/2)
\cos^2(\theta_b/2)$. Use this to work out the probabilities for
all four outcomes. Show that these probabilities sum to one. For
which parameter values are the events for the A and B particle
independent? Suppose the direction of this triplet state is known.
Does the result of a measurement on the $A$ particle give any
additional useful information about the result of a measurement
on the B particle?
\end{enumerate}
\end{document}