CHSH inequality

\[\newcommand{\ol}{\overline} \newcommand{\sqrtt}{\frac{1}{\sqrt 2}} \newcommand{\tinv}[1]{\frac{1}{#1}} \newcommand{\win}{\text{win}} \newcommand{\bb}{\mathbb}\]

In physics, the Clauser–Horne–Shimony–Holt (CHSH) inequality can be used in the proof of Bell’s theorem, which states that certain consequences of entanglement in quantum mechanics cannot be reproduced by local hidden-variable theories.

Classical states

As illustrated in Fig. 1, let us consider two persons, Alice and Bob, who have each two independent observables $A,\bar{A}$ and $B,\bar{B}$ with possibles outcomes $\pm 1$.

We define the variable

\[\begin{equation} \label{eq:S} S = \langle AB \rangle + \langle \bar{A}B \rangle + \langle A\bar{B} \rangle + \langle \bar{A}\bar{B} \rangle. \end{equation}\]

The usual form of the CHSH inequality is

\[\begin{equation} \label{eq:CHSH_ineq} |S| \leq 2. \end{equation}\]

Expliciting all possible combinations of Alice’ and Bob’s observables in Table 1, it is easy to verify \eqref{eq:CHSH_ineq} in the classical case. Indeed,

\[\begin{align} \begin{split} S &= \big| \braket{AB} + \braket{A\bar{B}} + \braket{\bar{A}B} - \braket{\bar{A}\bar{B}} \big| \\ &= \big| \braket{A(B+\bar{B}) + \bar{A}(B-\bar{B})} \big| \\ &= 2 \end{split} \end{align}\]

if the only outcomes are $\pm 1$ and cannot be in between.

Table 1: All possible combinations of Alice and Bob measurements in the classical case.

\(A\)	\(\bar{A}\)	\(B\)	\(\bar{B}\)	\(A(B+\bar{B})\)	\(\bar{A}(B-\bar{B})\)
-1	-1	-1	-1	2	0
-1	-1	-1	1	0	2
-1	-1	1	-1	0	-2
-1	-1	1	1	-2	0
-1	1	-1	-1	2	0
-1	1	-1	1	0	-2
-1	1	1	-1	0	2
-1	1	1	1	-2	0
1	-1	-1	-1	-2	0
1	-1	-1	1	0	2
1	-1	1	-1	0	-2
1	-1	1	1	2	0
1	1	-1	-1	-2	0
1	1	-1	1	0	-2
1	1	1	-1	0	2
1	1	1	1	2	0

Quantum states

In the more general quantum case, illustrated in Fig. 2, we write

\(\begin{equation} \ket{\psi} = a \ket{00} + b \ket{01} + c \ket{10} + d \ket{11} = \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \end{equation}\) any general 2-qbit quantum state, with $|a|^2+|b|^2+|c|^2+|d|^2 =1$. Now, in contrast to the classical case, $A,\bar{A},B,\bar{B}$ are all bases for measuring the same state $\ket{\psi}$ shared by Alice and Bob. They are given by

\[\begin{align} \begin{split} A &= Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \\ \bar{A} &= X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \\ B &= \sqrtt (Z+X) = \sqrtt \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \\ \bar{B} &= \sqrtt (Z-X) = \sqrtt \begin{bmatrix} 1 & -1 \\ -1 & -1 \end{bmatrix} \end{split} \end{align}\]

We remind the definition in \eqref{eq:S} of the variable

\[\begin{equation*} S = \langle AB \rangle + \langle \bar{A}B \rangle + \langle A\bar{B} \rangle + \langle \bar{A}\bar{B} \rangle. \end{equation*}\]

In quantum algebra, the statistical expectation of the output of the observables $AB$ can be computed as

\[\begin{align} \label{eq:AB} \begin{split} \braket{AB} &= \braket{\psi | A \otimes B | \psi} \\ &= \begin{bmatrix} a & b & c & d \end{bmatrix} \sqrtt \begin{bmatrix} 1 & 1 & & \\ 1 & -1 & & \\ & & -1 & -1 \\ & & -1 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \\ &= \sqrtt \begin{bmatrix} a & b & c & d \end{bmatrix} \begin{bmatrix} a+b \\ a-b \\ -c-d \\ -c + d \end{bmatrix} \\ &= \sqrtt (a^2-b^2-c^2+d^2). \end{split} \end{align}\]

And, we similarly obtain

\[\begin{align} \braket{\bar{A}B} &= \sqrtt (2ad+2bc) \label{eq:A_B} \\ \braket{A\bar{B}} &= \sqrtt (a^2-b^2-c^2+d^2) \label{eq:AB_} \\ \braket{\bar{A}\bar{B}} &= \sqrtt (-2ad-2bc). \label{eq:A_B_} \end{align}\]

Injecting \eqref{eq:AB}-\eqref{eq:A_B_} into \eqref{eq:S} yields

\[\begin{align} \label{eq:S_quantum} \begin{split} S &= \frac{2}{\sqrt 2} (a^2-b^2-c^2+d^2+2ad+2bc) \\ &= \frac{2}{\sqrt 2} [(a+d)^2-(b-c)^2]. \end{split} \end{align}\]

From \eqref{eq:S_quantum}, we can distinguish different situations. Firstly, the simplest case $a=1$, which is a classical state, gives $S=\frac{2}{\sqrt 2}$ and satisfies the CHSH inequality \eqref{eq:CHSH_ineq}. However, choosing $a=d=\sqrtt$ gives the entangled state

\[\begin{equation} \ket{\psi} = \ket{\Phi^+} = \sqrtt (\ket{00} + \ket{11}) \end{equation}\]

and $S=2\sqrt{2} > 2$ and breaks the CHSH inequality!