\documentclass[11pt,fleqn,twoside]{article}
\usepackage{cs191}
\begin{document}
\def\BC{\C}
\begin{center}
{\bf Two Qubit Entanglement and Bell Inequalities.}
\end{center}
\section{Two qubits:}
\iffalse
***
partial measurement
product states, entangled states
cnot gates
how to prepare bell states
bell inequalities
quantum teleportation...
***
\fi
Now let us examine a system of two qubits. Consider the two electrons
in two hydrogen atoms, each regarded as a $2$-state quantum system:
\begin{figure}[!hbp]
\includegraphics*[bb=78 419 152 493,scale=0.65]{lec2images}\quad
\includegraphics*[bb=78 419 152 493,scale=0.65]{lec2images}
\end{figure}
Since each electron can be in either of the ground or excited state,
classically the two electrons are in one of four states -- 00, 01,
10, or 11 -- and represent 2 bits of classical information.
By the superposition principle, the quantum state of the two electrons
can be any linear combination of these four classical states:
$$
\ket{\psi} = \alpha_{00}\ket{00} + \alpha_{01}\ket{01} + \alpha_{10}\ket{10} +\alpha_{11}\ket{11} \enspace ,
$$
where $\alpha_{ij} \elt \C$, $\sum_{ij}{\abs{\alpha_{ij}}^2} = 1$.
Again, this is just Dirac notation for the unit vector in $\C^4$:
\[
\left(
\begin{matrix}
\alpha_{00}\\
\alpha_{01}\\
\alpha_{10}\\
\alpha_{11}
\end{matrix}
\right)
\]
\noindent
{\bf Measurement:}
Measuring $\ket{\psi}$ now reveals two bits of information. The
probability that the outcome of the measurement is the two bit string
$x \in \{0,1\}^2$ is $\abs{\alpha_x}^2$. Moreover, following the
measurement the state of the two qubits is $\ket{x}$. i.e. if
the first bit of $x$ is $j$ and the second bit $k$, then following
the measurement, the state of the first qubit is $\ket{j}$ and
the state of the second is $\ket{k}$.
An interesting question comes up here: what if we measure just the
first qubit? What is the probability that the outcome is $0$?
This is simple. It is exactly the same as it would have been if we
had measured both qubits:
$\prob{\mbox{1st bit }= 0} = \prob{00} + \prob{01} = {|\alpha_{00}|\,}^2 +
{|\alpha_{01}|\,}^2$.
Ok, but how does this partial measurement disturb the state of the system?
The answer is obtained by an elegant generalization of our previous
rule for obtaining the new state after a measurement. The new superposition
is obtained by crossing out all those terms of $\ket{\psi}$ that are
inconstent with the outcome of the measurement (i.e. those whose
first bit is $1$). Of course, the sum of the squared amplitudes is
no longer $1$, so we must renormalize to obtain a unit vector:
$$\ket{\phi}_{new} = \frac{\alpha_{00}\ket{00} + \alpha_{01}\ket{01}}{\sqrt{
{|\alpha_{00}|\,}^2 + {|\alpha_{01}|\,}^2}}$$.
{\bf Entanglement}
Suppose the first qubit is in the state
$3/5 \ket{0} + 4/5 \ket{1}$ and
the second qubit is in the state
$1/\sqrt{2} \ket{0} - 1/\sqrt{2} \ket{1}$,
then the joint state of the two qubits is
$(3/5 \ket{0} + 4/5 \ket{1})(1/\sqrt{2} \ket{0} - 1/\sqrt{2} \ket{1})
= 3/5\sqrt{2} \ket{00} - 3/5\sqrt{2} \ket{01} +
4/5\sqrt{2} \ket{10} - 4/5\sqrt{2} \ket{11} $
But there are states such as
$\ket{\Phi^+} = \tfrac{1}{\sqrt{2}} \left( \ket{00} + \ket{11} \right)$
which cannot be decomposed in this way as a state of the
first qubit and that of the second qubit. Can you see why?
Such a state is called an entangled state.
If the first (resp. second) qubit of $\ket{\Phi^+}$ is measured then
the outcome is $0$ with probability $1/2$ and $1$ with
probability $1/2$. However if the outcome is $0$, then a
measurement of the second qubit results in $0$ with certainty.
Furthermore this is true even if both qubits are measured in
a rotated basis $\ket{v},\ket{v^{\perp}}$, where
$\ket{v}=\alpha\ket{0}+\beta\ket{1}$ and
$\ket{v^{\perp}}=-\beta\ket{0}+\alpha\ket{1}$.
{\bf Claim:}
$\ket{\Phi^+} = \tfrac{1}{\sqrt{2}} \left( \ket{00} + \ket{11} \right) \\
= \tfrac{1}{\sqrt{2}}\left(\ket{vv} + \ket{v^{\perp}v^{\perp}}\right)$.
{\bf Proof:}
Then
$\frac{1}{\sqrt{2}}(\ket{vv}+\ket{v^{\perp}v^{\perp}}) \\
= \frac{1}{\sqrt{2}}(\alpha^2\ket{00}+\alpha\beta\ket{01}+\alpha\beta\ket{10}+\beta^2\ket{11})
+
\frac{1}{\sqrt{2}}(\beta^2\ket{00}-\alpha\beta\ket{01}-\alpha\beta\ket{10}+\alpha^2\ket{11}) \\
= \frac{1}{\sqrt{2}}(\alpha^2+\beta^2)(\ket{00}+\ket{11}) \\
= \frac{1}{\sqrt{2}}(\ket{00}+\ket{11})$
\iffalse
$\ket{v} = cos\theta \ket{0} + sin\theta \ket{1}$.
To see this just observe that
$\ket{0} = cos\theta \ket{v} - sin\theta \ket{v^{\perp}}$
and $\ket{1} = sin\theta \ket{v} + cos\theta \ket{v^{\perp}}$.
Substituting, we get that:
$\ket{\Phi^+} = \tfrac{1}{\sqrt{2}} \left( \ket{00} + \ket{11} \right)
= \tfrac{1}{\sqrt{2}} \left( (cos\theta \ket{v} - sin\theta \ket{v^{\perp}})
(cos\theta \ket{v} - sin\theta \ket{v^{\perp}}) +
(sin\theta \ket{v} + cos\theta \ket{v^{\perp}})(sin\theta \ket{v} + cos\theta \ket{v^{\perp}}) \right )= \tfrac{1}{\sqrt{2}}\left(\ket{vv} + \ket{v^{\perp}v^{\perp}}\right)$.
\fi
\subsection{Two Qubit Gates}
Let us now consider how a system of two qubits evolves in time.
Recall that the third axiom of quantum physics states that the evolution of
a quantum system is necessarily unitary. Intuitively, a unitary transformation
is a rigid body rotation (or reflection) of the Hilbert space, thus
resulting in a transformation of the state vector that doesn't change
its length.
Let us consider what this means for the evolution of a two qubit system.
A unitary transformation on the Hilbert space ${\cal C}^4$ is
specified by a $4x4$ matrix $U$ that satisfies the condition
$UU^{\dagger} = U^{\dagger}U = I$. The four columns of $U$ specify
the four orthonormal vectors
$\ket{v_{00}}$, $\ket{v_{01}}$, $\ket{v_{10}}$ and $\ket{v_{11}}$
that the basis states
$\ket{00}$, $\ket{01}$, $\ket{10}$ and $\ket{11}$
are mapped to by $U$.
A very basic two qubit gate is the controlled-not gate or the CNOT:
\begin{itemize}
\item Controlled Not (CNOT).
\begin{displaymath}
\mathrm{CNOT } = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right)
\end{displaymath}
The first bit of a CNOT gate is the ``control bit;'' the second is the
``target bit.'' The control bit never changes, while the target bit flips if
and only if the control bit is $1$.
The CNOT gate is usually drawn as follows, with the control bit on top and the
target bit on the bottom:
\begin{center}
\begin{picture}(150,40)
% Two horizontal lines for the qubits
\put (0,30){\line(1,0){150}}
\put (0,10){\line(1,0){150}}
% The CNOT gate
\put (75,30){\circle*{5}}
\put (75,10){\circle{5}}
\put (75,10){\line(0,1){20}}
\end{picture}
\end{center}
\end{itemize}
Though the CNOT gate looks very simple, any unitary transformation
on two qubits can be closely approximated by a sequence of CNOT gates
and single qubit gates. This brings us to an important point. What happens
to the quantum state of two qubits when we apply a single qubit gate
to one of them, say the first? Let's do an example. Suppose we apply
a Hadamard gate to the superposition:
$\ket{\psi} = 1/2\ket{00} -i/\sqrt{2} \ket{01} + 1/\sqrt{2}\ket{11}$.
Then this maps the first qubit as follows:
$\ket{0} \rightarrow 1/\sqrt{2}\ket{0} + 1/\sqrt{2}\ket{1}$, and \\
$\ket{1} \rightarrow 1/\sqrt{2}\ket{0} - 1/\sqrt{2}\ket{1}$.\\
So $\ket{\psi} \rightarrow
1/2\sqrt{2}\ket{00} + 1/2\sqrt{2}\ket{01} -i/2 \ket{00} +i/2 \ket{01} + 1/2\ket{10} - 1/2\ket{11}$ \\
$= (1/2\sqrt{2} - i/2)\ket{00} + (1/2\sqrt{2} + i/2)\ket{01}
+ 1/2\ket{10} - 1/2\ket{11}$.
{\bf Bell states:}
We can generate the Bell states
$\ket{\Phi^+} = \tfrac{1}{\sqrt{2}} \left( \ket{00} + \ket{11} \right)$
with the following simple qauntum circuit consisting of
a Hadamard and CNOT gate:
\begin{center}
\begin{picture}(150,40)
\put (0,30){\line(1,0){45}}
\put (0,10){\line(1,0){150}}
% The Hadamard Gate
\put (45,27){\framebox{H}}
\put (59,30){\line(1,0){89}}
% The CNOT gate
\put (100,30){\circle*{5}}
\put (100,10){\circle{5}}
\put (100,10){\line(0,1){20}}
\end{picture}
\end{center}
The first qubit is passed through a Hadamard gate and then both qubits are
entangled by a CNOT gate.
If the input to the system is $|0\> \otimes |0\>$, then the Hadamard gate
changes the state to $$\tfrac{1}{\sqrt{2}}(|0\>+|1\>) \otimes |0\> = \tfrac{1}{\
sqrt{2}} \tket{00} + \tfrac{1}{\sqrt{2}} \tket{10} \enspace ,$$ and after
the CNOT gate the state becomes $\frac{1}{\sqrt{2}}(|00\>+|11\>)$, the Bell
state $|\Phi^+\>$.
The state $\ket{\Phi^+} = \tfrac{1}{\sqrt{2}} \left( \ket{00} + \ket{11} \right)$
is one of four Bell basis states:
\bea
\ket{\Phi^\pm} &=& \tfrac{1}{\sqrt{2}} \left( \ket{00} \pm \ket{11} \right) \\
\ket{\Psi^\pm} &=& \tfrac{1}{\sqrt{2}} \left( \ket{01} \pm \ket{10} \right) \enspace .
\eea
These are maximally entangled states on two qubits. Show how to generate
all these states by a simple quantum circuit, and verify that the
four Bell states form an orthonormal basis.
\iffalse
In fact, one can verify that the four possible inputs
produce the four Bell states:
\begin{align*}
|00\> & \mapsto \frac{1}{\sqrt{2}}(|00\>+|11\>) = |\Phi^+\>; &
|01\> & \mapsto \frac{1}{\sqrt{2}}(|01\>+|10\>) = |\Psi^+\>; \\
|10\> & \mapsto \frac{1}{\sqrt{2}}(|00\>-|11\>) = |\Phi^-\>; &
|11\> & \mapsto \frac{1}{\sqrt{2}}(|01\>-|10\>) = |\Psi^-\>. \\
\end{align*}
\subsection{Creating Bell States:}
Here is a simple quantum circuit for creating the Bell state
$\ket{\Phi^+}$:
Figure of hadamard + CNOT.
Notice how the action of the CNOT gate is not so much copying
as our classical intuition would suggest, but rather to entangle.
What state does the circuit output on input $11$? What about ...?
Can you conclude that orthonormal?
\fi
\iffalse
BOX
%\begin{box}
{\bf Tensor products:}
How do we glue two $2$-dimensional Hilbert spaces (corresponding to
the two qubits above) to get a $4$-dimensional Hilbert space? This
is done by taking a tensor product of the two spaces.
More generally, suppose we have two quantum systems - a $k$-state
system with associated $k$-dimensional Hilbert space $V$ with
basis $\ket{0}, \ldots , \ket{k-1}$ and
a $l$-state system with associated $l$-dimensional Hilbert space $W$
with basis $\ket{0}, \ldots , \ket{l-1}$.
What is resulting Hilbert space obtained by gluing these two
Hilbert spaces together?
We can answer this question as follows: there are $kl$ distinguishable states
of the composite system --- one for each choice of basis state
$\ket{i}$ of the first system and basis state $\ket{j}$ of the second
system. We denote the resulting of dimension $kl$ Hilbert space by
$V \otimes W$ (pronounced ``$V$ tensor $W$''). The orthonormal basis for this
new Hilbert space is given by:
\begin{displaymath}
\{|i\> \otimes |j\> : 0 \leq i \leq k-1, 0 \leq j \leq l-1\},
\end{displaymath}
So a typical element of $V \otimes W$ will
be of the form $\sum_{ij} \alpha_{ij} (|i\> \otimes |j\>)$.
For example, consider $X = \BC^2 \otimes \BC^2$. $V$ is a Hilbert space of
dimension 4, so $X \cong \BC^4$. So we can write $|00\>$ alternatively as
$|0\> \otimes |0\>$. More generally, for $n$ qubits we have
$\BC^2 \otimes \cdots$ ($n$ times) $\otimes \cdots \BC^2 \cong \BC^{2^n}$.
A typical element of this space is of the form
\begin{displaymath}
\sum_{x \in \{0,1\}^n} \alpha_x|x\>.
\end{displaymath}
Surprisingly, there are states $\ket{\psi} = \alpha_{00}\ket{00} +
\alpha_{01}\ket{01} + \alpha_{10}\ket{10} +\alpha_{11}\ket{11}$ for
which we cannot specify the individual states of each of the two
qubits. Here is an example: The state $\ket{\psi} =
\frac{1}{\sqrt{2}}(\ket{00}+\ket{11})$ cannot be decomposed as
$\ket{\psi_1} \otimes \ket{\psi_2}$ for any choice of $\alpha_1$,
$\beta_1$, $\alpha_2$, $\beta_2$. This is because all the four complex
numbers must be non-zero to achieve non-zero amplitudes for $\ket{00}$
and $\ket{11}$, but then the amplitudes of $\ket{01}$ and $\ket{10}$
are also non-zero. We say that such states are entangled states -
they cannot be decomposed as a product of states of the two
individual qubits. The particular state in this example is one of
the famous Bell states --- it is a maximally entangled state of two
qubits.
If we were to pick the amplitudes of a two qubit system at random,
would it typically be an entangled state? We can answer this question
by counting parameters. A single qubit state can be specified by
two real parameters - since there are two complex numbers (four real
numbers), but the overall phase doesn't matter, and we have the constraint
that the norm is $1$. So the unentangled states can be specified by
four real numbers (two for each qubit). On the other hand the state of
a two qubit system requires six real parameters to describe - since
we now have four complex numbers (eight real numbers), but still with
two constraints which reduces this the count to six. Thus putting two
qubits together into a system leads to many more possible configurations
than keeping them separate. We will see that this phenomenon becomes
even more extreme as the size of the subsystems increases, and as the
number of subsystems increase.
\subsection{The Significance of Tensor Products}
Classically, if we put together a subsystem
that stores $k$ bits of information with
one that stores $l$ bits of information, the total capacity of the
composite system is $k+l$ bits.
From this viewpoint, the situation with quantum systems is extremely
paradoxical. We need $k$ complex numbers to describe the state of
a k-level quantum system. Now consider a system that consists of
a k-level subsystem and an l-level subsystem. To describe the composite
system we need $kl$ complex numbers. One might wonder where nature
finds the extra storage space when we put these two subsystems together.
An extreme case of this phenomenon occurs when we consider an $n$ qubit
quantum system. The Hilbert space associated with this system is the
n-fold tensor product of ${\C}^2 \equiv {\C}^{2^n}$. Thus nature
must ``remember'' of $2^n$ complex numbers to keep track of the state of
an $n$ qubit system. For modest values of $n$ of a few hundred, $2^n$ is
larger than estimates on the number of elementary particles in the Universe.
This is the fundamental property of quantum systems that is used in
quantum information processing.
Finally, note that when we actually a measure an $n$-qubit quantum state, we
see only an $n$-bit string - so we can recover from the system only $n$, rather
than $2^n$, bits of information.
\section{Unitary Operators and Quantum Gates}
\iffalse
\subsection{Unitary Operators}
The final postulate of quantum physics states that the evolution of
a quantum system is necessarily unitary. Intuitively, a unitary transformation
is a rigid body rotation (or reflection) of the Hilbert space, thus
resulting in a transformation of the state vector that doesn't change
its length. How do we physically
effect such a transformation on a quantum system? To explain this we must
first introduce the notion of the Hamiltonian acting on a system; you will
have to wait for three to four lectures before we get to those concepts.
Suppose we have a $k$-state quantum system. Then a unitary transformation
over the space is a linear transformation that can be specified by a
$k \times k$ matrix $U$ with complex entries that satisfies
$U^{-1} = U^\dagger$. For example, for an op erator on $\C^2$,
$$
U = \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)
\Rightarrow U^\dagger =
\left(\begin{smallmatrix}\bar{a}&\bar{c}\\\bar{b}&\bar{d}\end{smallmatrix}
\right) \enspace .
$$
It is easily verified that the composition of two unitary transformations
is also unitary (Proof: $U, V$ unitary, then
$(U V)^\dagger = V^\dagger U^\dagger = V^{-1} U^{-1} = (UV)^{-1}$).
Some properies of a unitary transformation $U$:
\begin{itemize}
\item The rows of $U$ form an orthonormal basis.
\item The columns of $U$ form an orthonormal basis.
\item $U$ preserves inner products, i.e. the inner product between
$\ket{u}$ and $\ket{v}$ is the same as the inner product between
$U\ket{u}$ and $U\ket{v}$.
%$(\vec{v},\vec{w}) = (U\vec{v},U\vec{w})$.
The latter quantity can be written as
$\bra{v}U^\dagger U\ket{w} = \braket{v}{w}$. Therefore,
$U$ preserves norms and angles (up to sign).
\item The eigenvalues of $U$ are all of the form $e^{i\theta}$ (since $U$ is
length-preserving, i.e., $(\vec{v},\vec{v}) = (U\vec{v},U\vec{v})$).
\item $U$ can be diagonalized into the form
\begin{displaymath}
\left(
\begin{array}{cccc}
e^{i\theta_1} & 0 & \cdots & 0 \\
0 & \ddots & \ddots & 0 \\
\vdots & \ddots & \ddots & \vdots \\
0 & \cdots & 0 & e^{i\theta_k}
\end{array}
\right)
\end{displaymath}
\end{itemize}
\subsection{Quantum Gates}
We give some examples of simple unitary transforms, or ``quantum gates.''
Some quantum gates with one qubit:
\begin{itemize}
\item Hadamard Gate. Can be viewed as a reflection around $\pi/8$, or a
rotation around $\pi/4$ followed by a reflection.
\begin{displaymath}
H = \frac{1}{\sqrt{2}}\left(
\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}
\right)
\end{displaymath}
The Hadamard Gate is one of the most important gates. Note that $H^\dagger = H$
-- since $H$ is real and symmetric -- and $H^2 = I$.
\item Rotation Gate. This rotates the plane by $\theta$.
\begin{displaymath}
U = \left(
\begin{array}{cc}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{array}
\right)
\end{displaymath}
\item NOT Gate. This flips a bit from 0 to 1 and vice versa.
\begin{displaymath}
NOT = \left(\begin{matrix}0&1\\1&0\end{matrix}\right)
\end{displaymath}
\item Phase Flip.
\begin{displaymath}
Z = \left(
\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right)
\end{displaymath}
The phase flip is a NOT gate acting in the
$\ket{+} = \tfrac{1}{\sqrt{2}}(\ket{0}+\ket{1}), \ket{-} =
\tfrac{1}{\sqrt{2}}(\ket{0}-\ket{1})$ basis.
Indeed, $Z\ket{+}=\ket{-}$ and $Z\ket{-} = \ket{+}$.
\end{itemize}
And a two-qubit quantum gate:
\begin{itemize}
\item Controlled Not (CNOT).
\begin{displaymath}
\mathrm{CNOT } = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right)
\end{displaymath}
The first bit of a CNOT gate is the ``control bit;'' the second is the
``target bit.'' The control bit never changes, while the target bit flips if
and only if the control bit is $1$.
The CNOT gate is usually drawn as follows, with the control bit on top and the
target bit on the bottom:
\begin{center}
\begin{picture}(150,40)
% Two horizontal lines for the qubits
\put (0,30){\line(1,0){150}}
\put (0,10){\line(1,0){150}}
% The CNOT gate
\put (75,30){\circle*{5}}
\put (75,10){\circle{5}}
\put (75,10){\line(0,1){20}}
\end{picture}
\end{center}
\end{itemize}
\fi
\subsection{Tensor product of operators}
Suppose we have two quantum systems: a $k$-state system with associated
Hilbert space $V$ and a $l$-state system with associated Hilbert space
$W$. Suppose we apply a unitary transformation $A$ to the first system
and $B$ to the second system. What is the resulting transformation
on the combined system $V \otimes W$? To figure this out, let us first see how
the combined transformation acts on basis states of $V \otimes W$.
Consider a basis state $\ket{i} \otimes \{ket{j}$ where $0\leq i \leq k-1$
and $0\leq j\leq l-1$. Since $A$ is only acting on $V$ and $B$ only on
$W$, this state is transformed to $A\ket{i} \otimes B\ket{j}$.
Suppose $\ket{v}$ and $\ket{w}$ are unentangled states on $\C^m$ and
$\C^n$, respectively. The state of the combined system is
$\ket{v}\tensor \ket{w}$ on $\C^{m n}$. If the unitary operator $A$
is applied to the first subsystem, and $B$ to the second subsystem,
the combined state becomes $A\ket{v}\tensor B\ket{w}$.
In general, the two subsystems will be entangled with each other, so
the combined state is not a tensor-product state. We can still apply
$A$ to the first subsystem and $B$ to the second subsystem. This
gives the operator $A\tensor B$ on the combined system, defined on
entangled states by linearly extending its action on unentangled
states.
(For example, $(A \tensor B) (\ket{0} \tensor \ket{0}) = A \ket{0}
\tensor B \ket{0}$. $(A \tensor B) (\ket{1} \tensor \ket{1}) = A
\ket{1} \tensor B \ket{1}$. Therefore, we define $(A \tensor B)
(\tfrac{1}{\sqrt{2}} \ket{00} + \tfrac{1}{\sqrt{2}} \ket{11})$ to be
$\tfrac{1}{\sqrt{2}} (A\tensor B) \ket{00} + \tfrac{1}{\sqrt{2}}
(A\tensor B) \ket{11} = \tfrac{1}{\sqrt{2}} \left( A\ket{0}\tensor
B\ket{0} + A\ket{1}\tensor B\ket{1} \right)$.)
Let $\ket{e_1},\ldots,\ket{e_m}$ be a basis for the first subsystem,
and write $A = \sum_{i,j=1}^{m} a_{ij} \ketbra{e_i}{e_j}$ (the
$i$,$j$th element of $A$ is $a_{ij}$). Let
$\ket{f_1},\ldots,\ket{f_n}$ be a basis for the second subsystem, and
write $B = \sum_{k,l=1}^{n} b_{kl} \ketbra{f_k}{f_l}$. Then a basis
for the combined system is $\ket{e_i}\tensor \ket{f_j}$, for
$i=1,\ldots,m$ and $j=1,\ldots,n$. The operator $A \tensor B$ is \bea
A \tensor B &=& \left( \sum_{ij} a_{ij} \ketbra{e_i}{e_j} \right) \tensor \left(
\sum_{kl} b_{kl} \ketbra{f_k}{f_l} \right) \\
&=& \sum_{ijkl} a_{ij} b_{kl} \ketbra{e_i}{e_j} \tensor \ketbra{f_k}{f_l} \\
&=& \sum_{ijkl} a_{ij} b_{kl}
(\ket{e_i}\tensor\ket{f_k})(\bra{e_j}\tensor\bra{f_l}) \enspace .
\eea Therefore the $(i,k),(j,l)$th element of $A \tensor B$ is $a_{ij}
b_{kl}$. If we order the basis $\ket{e_i}\tensor \ket{f_j}$
lexicographically, then the matrix for $A \tensor B$ is
$$
\left(\begin{matrix}
a_{11} B & a_{12} B & \cdots \\
a_{21} B & a_{22} B & \cdots \\
\vdots & \vdots & \ddots
\end{matrix}\right) \enspace ;
$$
in the $i,j$th subblock, we multiply $a_{ij}$ by the matrix for $B$.
\fi
\subsection{EPR Paradox:}
Everyone has heard Einstein's famous quote ``God does not play dice''.
It is lifted from Einstein's 1926 letter to Max Born where he
expressed his dissatisfaction with quantum physics by writing:
"Quantum mechanics is certainly imposing.
But an inner voice tells me that it is not
yet the real thing. The theory says a lot,
but does not really bring us any closer to
the secret of the Old One. I, at any rate,
am convinced that He does not throw dice."
Even to the end of his life he held on to the view that quantum physics
is just an incomplete theory and that some day we would learn
a more complete and satisfactory theory that describes nature.
For example, consider coin-flipping. We can model coin-flipping as a
random process giving heads $50\%$ of the time, and tails $50\%$ of
the time. This model is perfectly predictive, but incomplete. If we
knew the initial conditions of the coin with perfect accuracy
(position, momentum), then we could solve Newton's equations to
determine the eventual outcome of the coin flip with certainty.
Einstein sharpened this line of reasoning in a paper he wrote
with Podolsky and Rosen in 1935, where they introduced the
famous Bell states. Recall that
for Bell state $\frac{1}{\sqrt{2}}(\ket{00}+\ket{11})$, when you
measure first qubit, the second qubit is determined. However, if two
qubits are far apart, then the second qubit must have had a determined
state in some time interval before measurement, since the speed of
light is finite. Moreover this holds in any basis. This appears
analogous to the coin flipping example. EPR therefore suggested that
there is a more complete theory where ``God does not throw dice.''
Until his death in 1955, Einstein tried to formulate a more complete
"local hidden variable theory" that would describe the predictions
of quantum mechanics, but without resorting to probabilistic outcomes.
But in 1964, almost three decades after the EPR paper, John Bell
showed that properties of Bell (EPR) states were not merely fodder
for a philosophical discussion, but had verifiable consequences:
local hidden variables are not the answer.
How does one rule out every possible hidden variable theory?
Here's how: we will consider an extravagant framework within
which every possible hidden variable theory. And then we will
show that there is a particular quantum mechanical experiment
using Bell states, whose results cannot be duplicated by any
theory in this framework. The framework is this:
when the Bell state is created, the two particles
each make up a (infinitely long!) list of all possible experiments that they
might be subjected to, and decide how they will behave under each such
experiment. When the two particles separate and can no longer
communicate, they consult their respective lists to coordinate their
actions.
\section{Bell's Thought Experiment}
Bell considered the following experiment: the two particles in a Bell
pair move in opposite directions to two distant apparatus. A decision
about which of two experiments is to be performed at each apparatus is
made randomly at the last moment, so that speed of light
considerations rule out information about the choice at one apparatus
being transmitted to the other.
How correlated can the outcomes on the two
experiments be? It can be shown that any theory in the
classical hidden variable framework above gives a correlation of
at most $0.75$ whereas the quantum experiments described below
give a correlation of about $0.8$. Therefore the predictions of
quantum mechanics are not consistent with any local hidden variable
theory. We now describe the experiment in more detail.
The two experimenters A and B (for Alice and Bob) each receive a
random bit $r_A$ and $r_B$ respectively. Each also receives one half
of a Bell state, and makes a suitable measurement described below
based on the received random bit. Call the outcomes of the measurements
$a$ and $b$ respectively. We are interested in the achievable correlation
between the two quantities $r_A \times r_B$ and $a + b (mod 2)$.
We will show that for the particular quantum measurements
described below $P[r_A \times r_B = a + b (mod 2)] \approx .8$.
What would a classical hidden variable theory predict for this setting?
Now, when the Bell state was created, the two particles could share an
arbitrary amount of information. But by the time the random bits
$r_A$ and $r_B$ are generated, the two particles are too far apart to
exchange information. Thus in any experiment, the outcome can only be
a function of the previously shared information and one of the random
bits. It can be shown that in this setting the best correlation is
achieved by always letting the outcomes of the two experiments be
$a = 0$ and $b = 0$ (see homework exercise). This gives
$P[r_A \times r_B = a + b (mod 2)] \leq .75$. This experiment therefore
distinguishes between the predictions of quantum physics and those of
any arbitrary local hidden variable theory. It has now been performed
in several different ways, and the results are consistent with quantum
physics and inconsistent with any classical hidden variable theory.
Here is the protocol:
\begin{itemize}
\item if $X_A = 0$, then Alice measures in the $-\pi/16$ basis.
\item if $X_A = 1$, then Alice measures in the $3\pi/16$ basis.
\item if $X_B = 0$, then Bob measures in the $\pi/16$ basis.
\item if $X_B = 1$, then Bob measures in the $-3\pi/16$ basis.
\end{itemize}
Now an easy calculation shows that in each of the four cases
$X_A = X_B = 0$, etc, the success probability is $cos^2 \pi/8$.
This is because in the three cases where $x_A \cdot x_B = 0$,
Alice and Bob measure in bases that differ by $/pi/8$. In
the last case they measure in bases that differ by $3\pi/8$,
but in this case they must output different bits.
We still have to prove that Bell state
$\frac{1}{\sqrt{2}}(\ket{00}+\ket{11}) =
\frac{1}{\sqrt{2}}(\ket{v_Av_A}+\ket{v_A^{\perp}v_A^{\perp}})$ Let
$\ket{v_A}=\alpha\ket{0}+\beta\ket{1}$ and
$\ket{v_A^{\perp}}=-\beta\ket{0}+\alpha\ket{1}$. Then
$\frac{1}{\sqrt{2}}(\ket{v_Av_A}+\ket{v_A^{\perp}v_A^{\perp}}) =
\frac{1}{\sqrt{2}}(\alpha^2\ket{00}+\alpha\beta\ket{01}+\alpha\beta\ket{10}+\beta^2\ket{11})
+
\frac{1}{\sqrt{2}}(\beta^2\ket{00}-\alpha\beta\ket{01}-\alpha\beta\ket{10}+\alpha^2\ket{11})
= \frac{1}{\sqrt{2}}(\alpha^2+\beta^2)(\ket{00}+\ket{11}) =
\frac{1}{\sqrt{2}}(\ket{00}+\ket{11})$
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