CHSH inequality

The following is based on page 37 of Bell's Speakable and Unspeakable,[4] the main change being to use the symbol ‘E’ instead of ‘P’ for the expected value of the quantum correlation. This avoids any suggestion that the quantum correlation is itself a probability.

We start with the standard assumption of independence of the two sides, enabling us to obtain the joint probabilities of pairs of outcomes by multiplying the separate probabilities, for any selected value of the "hidden variable" λ. λ is assumed to be drawn from a fixed distribution of possible states of the source, the probability of the source being in the state λ for any particular trial being given by the density function ρ(λ), the integral of which over the complete hidden variable space is 1. We thus assume we can write:

$${\displaystyle E(a,b)=\int {\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda )\rho (\lambda )d\lambda }$$

where A and B are the outcomes. Since the possible values of A and B are −1, 0 and +1, it follows that:

${\displaystyle \left|{\underline {A}}\right|\leq 1\quad \left|{\underline {B}}\right|\leq 1}$

(4)

Then, if a, a′, b and b′ are alternative settings for the detectors,

${\displaystyle {\begin{aligned}&E(a,b)-E\left(a,b'\right)\\={}&\int \left[{\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda )-{\underline {A}}(a,\lambda ){\underline {B}}\left(b',\lambda \right)\right]\rho (\lambda )d\lambda \\={}&\int \left[{\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda )-{\underline {A}}(a,\lambda ){\underline {B}}\left(b',\lambda \right)\pm {\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda ){\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\mp {\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda ){\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\right]\rho (\lambda )d\lambda \\={}&\int {\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda )\left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\right]\rho (\lambda )d\lambda -\int {\underline {A}}(a,\lambda ){\underline {B}}\left(b',\lambda \right)\left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}(b,\lambda )\right]\rho (\lambda )d\lambda \end{aligned}}}$

Taking absolute values of both sides, and applying the triangle inequality to the right-hand side, we obtain

${\displaystyle \left|E(a,b)-E\left(a,b'\right)\right|\leq \left|\int {\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda )\left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\right]\rho (\lambda )d\lambda \right|+\left|\int {\underline {A}}(a,\lambda ){\underline {B}}\left(b',\lambda \right)\left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}(b,\lambda )\right]\rho (\lambda )d\lambda \right|}$

We use the fact that ${\displaystyle \left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\right]\rho (\lambda )}$ and ${\displaystyle \left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}(b,\lambda )\right]\rho (\lambda )}$ are both non-negative to rewrite the right-hand side of this as

$${\displaystyle \int \left|{\underline {A}}(a,\lambda ){\underline {B}}(b,\lambda )\right|\left|\left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\right]\rho (\lambda )d\lambda \right|+\int \left|{\underline {A}}(a,\lambda ){\underline {B}}(b',\lambda )\right|\left|\left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}(b,\lambda )\right]\rho (\lambda )d\lambda \right|}$$

By (4), this must be less than or equal to

$${\displaystyle \int \left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\right]\rho (\lambda )d\lambda +\int \left[1\pm {\underline {A}}\left(a',\lambda \right){\underline {B}}(b,\lambda )\right]\rho (\lambda )d\lambda }$$

which, using the fact that the integral of ρ(λ) is 1, is equal to

$${\displaystyle 2\pm \left[\int {\underline {A}}\left(a',\lambda \right){\underline {B}}\left(b',\lambda \right)\rho (\lambda )d\lambda +\int {\underline {A}}\left(a',\lambda \right){\underline {B}}(b,\lambda )\rho (\lambda )d\lambda \right]}$$

which is equal to ${\displaystyle 2\pm \left[E\left(a',b'\right)+E\left(a',b\right)\right]}$.

Putting this together with the left-hand side, we have:

$${\displaystyle \left|E(a,b)-E\left(a,b'\right)\right|\;\leq 2\;\pm \left[E\left(a',b'\right)+E\left(a',b\right)\right]}$$

which means that the left-hand side is less than or equal to both ${\displaystyle 2+\left[E\left(a',b'\right)+E\left(a',b\right)\right]}$ and ${\displaystyle 2-\left[E\left(a',b'\right)+E\left(a',b\right)\right]}$. That is:

$${\displaystyle \left|E(a,b)-E\left(a,b'\right)\right|\;\leq \;2-\left|E\left(a',b'\right)+E\left(a',b\right)\right|}$$

from which we obtain

$${\displaystyle 2\;\geq \;\left|E(a,b)-E\left(a,b'\right)\right|+\left|E\left(a',b'\right)+E\left(a',b\right)\right|\;\geq \;\left|E(a,b)-E\left(a,b'\right)+E\left(a',b'\right)+E\left(a',b\right)\right|}$$

(by the triangle inequality again), which is the CHSH inequality.

In their 1974 paper,[11] Clauser and Horne show that the CHSH inequality can be derived from the CH74 one. As they tell us, in a two-channel experiment the CH74 single-channel test is still applicable and provides four sets of inequalities governing the probabilities p of coincidences.

Working from the inhomogeneous version of the inequality, we can write:

$${\displaystyle -1\;\leq \;p_{jk}(a,b)-p_{jk}(a,b')+p_{jk}(a',b)+p_{jk}(a',b')-p_{jk}(a')-p_{jk}(b)\;\leq \;0}$$

where j and k are each '+' or '−', indicating which detectors are being considered.

To obtain the CHSH test statistic S (2), all that is needed is to multiply the inequalities for which j is different from k by −1 and add these to the inequalities for which j and k are the same.

There exists an optimal configuration of the measurement bases a, a', b, b' for a given ${\displaystyle \rho }$ that yields S_max with at least one free parameter.[14][15]

The projective measurement that yields either +1 or −1 for two orthogonal states ${\displaystyle |\alpha \rangle ,|\alpha ^{\perp }\rangle }$ respectively, can be expressed by an operator ${\displaystyle \mathrm {A} =|\alpha \rangle \langle \alpha |-|\alpha ^{\perp }\rangle \langle \alpha ^{\perp }|}$. The choice of this measurement basis can be parametrized by a real unit vector ${\displaystyle {\boldsymbol {a}}\in \mathbb {R} ^{3},|{\boldsymbol {a}}|=1}$ and the Pauli vector ${\displaystyle {\boldsymbol {\sigma }}}$ by expressing ${\displaystyle \mathrm {A} ={\boldsymbol {a}}\cdot {\boldsymbol {\sigma }}}$. Then, the expected correlation in bases a, b is

$${\displaystyle E(a,b)=\operatorname {Tr} [\rho ({\boldsymbol {a}}\cdot {\boldsymbol {\sigma }})\otimes ({\boldsymbol {b}}\cdot {\boldsymbol {\sigma }})]={\boldsymbol {a}}^{\text{T}}T_{\rho }{\boldsymbol {b}}.}$$

The numerical values of the basis vectors, when found, can be directly translated to the configuration of the projective measurements.[15]

The optimal set of bases for the state ${\displaystyle \rho }$ is found by taking the two greatest eigenvalues ${\displaystyle \lambda _{1,2}}$ and the corresponding eigenvectors ${\displaystyle {\boldsymbol {e}}_{1,2}}$ of ${\displaystyle U_{\rho }}$, and finding the auxiliary unit vectors

$${\displaystyle {\begin{aligned}{\boldsymbol {c}}&={\boldsymbol {e}}_{1}\cos \varphi +{\boldsymbol {e}}_{2}\sin \varphi ,\\{\boldsymbol {c}}'&={\boldsymbol {e}}_{1}\sin \varphi -{\boldsymbol {e}}_{2}\cos \varphi ,\end{aligned}}}$$

where ${\displaystyle \varphi }$ is a free parameter. We also calculate the acute angle

$${\displaystyle \theta =\operatorname {arctan} {\sqrt {\frac {\lambda _{2}+\lambda _{1}\tan ^{2}{\varphi }}{\lambda _{1}+\lambda _{2}\tan ^{2}{\varphi }}}}}$$

to obtain the bases that maximize Eq. 2,

$${\displaystyle {\begin{aligned}{\boldsymbol {a}}&=T_{\rho }{\boldsymbol {c}}'/|T_{\rho }{\boldsymbol {c}}'|,\\{\boldsymbol {a}}'&=T_{\rho }{\boldsymbol {c}}/|T_{\rho }{\boldsymbol {c}}|,\\{\boldsymbol {b}}&={\boldsymbol {c}}\cos \theta +{\boldsymbol {c}}'\sin \theta ,\\{\boldsymbol {b}}'&={\boldsymbol {c}}\cos \theta -{\boldsymbol {c}}'\sin \theta .\end{aligned}}}$$

In entanglement-based quantum key distribution, there is another measurement basis used to communicate the secret key (${\displaystyle {\boldsymbol {a}}_{0}}$ assuming Alice uses the side A). The bases ${\displaystyle {\boldsymbol {a}}_{0},{\boldsymbol {b}}}$ then need to minimize the quantum bit error rate Q, which is the probability of obtaining different measurement outcomes (+1 on one particle and −1 on the other).[13] The corresponding bases are[15]

$${\displaystyle {\begin{aligned}{\boldsymbol {a}}_{0}&=T_{\rho }{\boldsymbol {e}}_{1}/|T_{\rho }{\boldsymbol {e}}_{1}|,\\{\boldsymbol {b}}&={\boldsymbol {e}}_{1}.\end{aligned}}}$$

The CHSH polynomial S needs to be maximized as well, which together with the bases above creates the constraint ${\displaystyle \varphi =\pi /4}$.[15]

We first establish that any deterministic classical strategy has success probability at most 75% (where the probability is taken over Charlie's uniformly random choice of ${\displaystyle x,y}$). By a deterministic strategy, we mean a pair of functions ${\displaystyle f_{A},f_{B}:\{0,1\}\mapsto \{0,1\}}$, where ${\displaystyle f_{A}}$ is a function determining Alice's response as a function of the message she receives from Charlie, and ${\displaystyle f_{B}}$ is a function determining Bob's response based on what he receives. To prove that any deterministic strategy fails at least 25% of the time, we can simply consider all possible pairs of strategies for Alice and Bob, of which there are at most 8 (for each party, there are 4 functions ${\displaystyle \{0,1\}\mapsto \{0,1\}}$). It can be verified that for each of those 8 strategies there is always at least one out of the four possible input pairs ${\displaystyle (x,y)\in \{0,1\}^{2}}$ which makes the strategy fail. For example, in the strategy where both players always answer 0, we have that Alice and Bob win in all cases except for when ${\displaystyle x=y=1}$, so using this strategy their win probability is exactly 75%.

Now, consider the case of randomized classical strategies, where Alice and Bob have access to correlated random numbers. They can be produced by jointly flipping a coin several times before the game has started and Alice and Bob are still allowed to communicate. The output they give at each round is then a function of both Charlie's message and the outcome of the corresponding coin flip. Such a strategy can be viewed as a probability distribution over deterministic strategies, and thus its success probability is a weighted sum over the success probabilities of the deterministic strategies. But since every deterministic strategy has a success probability of at most 75%, this weighted sum cannot exceed 75% either.

Now, imagine that Alice and Bob share the two-qubit entangled state: ${\textstyle |\Phi \rangle ={\frac {1}{\sqrt {2}}}(|00\rangle +|11\rangle )}$, commonly referred to as an EPR pair. Alice and Bob will use this entangled pair in their strategy as described below. The optimality of this strategy then follows from Tsirelson's bound.

Upon receiving the bit ${\displaystyle x}$ from Charlie, Alice will measure her qubit in the basis ${\displaystyle \{|0\rangle ,|1\rangle \}}$ or in the basis ${\displaystyle \{|+\rangle ,|-\rangle \}}$, conditionally on whether ${\displaystyle x=0}$ or ${\displaystyle x=1}$, respectively. She will then label the two possible outputs resulting from each measurement choice as ${\displaystyle a=0}$ if the first state in the measurement basis is observed, and ${\displaystyle a=1}$ otherwise.

Bob also uses the bit ${\displaystyle y}$ received from Charlie to decide which measurement to perform: if ${\displaystyle y=0}$ he measures in the basis ${\displaystyle \{|a_{0}\rangle ,|a_{1}\rangle \}}$, while if ${\displaystyle y=1}$ he measures in the basis ${\displaystyle \{|b_{0}\rangle ,|b_{1}\rangle \}}$, where

$${\displaystyle {\begin{aligned}&|a_{0}\rangle =\cos \theta \,|0\rangle +\sin \theta \,|1\rangle ,\qquad |a_{1}\rangle =-\sin \theta \,|0\rangle +\cos \theta \,|1\rangle ,\\&|b_{0}\rangle =\cos \theta \,|0\rangle -\sin \theta \,|1\rangle ,\qquad |b_{1}\rangle =\sin \theta \,|0\rangle +\cos \theta \,|1\rangle ,\end{aligned}}}$$

with ${\displaystyle \theta =\pi /8}$.

To analyze the success probability, it suffices to analyze the probability that they output a winning value pair on each of the four possible inputs ${\displaystyle (x,y)}$, and then take the average. We analyze the case where ${\displaystyle x=y=0}$ here: In this case the winning response pairs are ${\displaystyle a=b=0}$ and ${\displaystyle a=b=1}$. On input ${\displaystyle x=y=0}$, we know that Alice will measure in the basis ${\displaystyle |0\rangle ,|1\rangle }$, and Bob will measure in the basis ${\displaystyle |a_{0}\rangle ,|a_{1}\rangle }$. Then the probability that they both output 0 is the same as the probability that their measurements yield ${\displaystyle |0\rangle ,|a_{0}\rangle }$ respectively, so precisely ${\textstyle |(\langle 0|\otimes \langle a_{0}|)|\Phi \rangle |^{2}={\frac {1}{2}}\cos ^{2}\left({\frac {\pi }{8}}\right)}$. Similarly, the probability that they both output 1 is exactly ${\textstyle |(\langle 1|\otimes \langle a_{1}|)|\Phi \rangle |^{2}={\frac {1}{2}}\cos ^{2}\left({\frac {\pi }{8}}\right)}$. So the probability that either of these successful outcomes happens is ${\textstyle \cos ^{2}\left({\frac {\pi }{8}}\right)}$.

In the case of the 3 other possible input pairs, essentially identical analysis shows that Alice and Bob will have the same win probability of ${\displaystyle \cos ^{2}\left({\frac {\pi }{8}}\right)}$, so overall the average win probability for a randomly chosen input is ${\textstyle \cos ^{2}\left({\frac {\pi }{8}}\right)}$. Since ${\textstyle \cos ^{2}\left({\frac {\pi }{8}}\right)\approx 85\%}$, this is strictly better than what was possible in the classical case.

An arbitrary quantum strategy for the CHSH game can be modeled as a triple ${\displaystyle {\mathcal {S}}=\left(|\psi \rangle ,(A_{0},A_{1}),(B_{0},B_{1})\right)}$ where

• ${\displaystyle |\psi \rangle \in \mathbb {C} ^{d}\otimes \mathbb {C} ^{d}}$ is a bipartite state for some ${\displaystyle d}$,
• ${\displaystyle A_{0}}$ and ${\displaystyle A_{1}}$ are Alice's observables each corresponding to receiving ${\displaystyle x\in \{0,1\}}$ from the referee, and
• ${\displaystyle B_{0}}$ and ${\displaystyle B_{1}}$ are Bob's observables each corresponding to receiving ${\displaystyle y\in \{0,1\}}$ from the referee.

The optimal quantum strategy described above can be recast in this notation as follows: ${\displaystyle |\psi \rangle \in \mathbb {C} ^{2}\otimes \mathbb {C} ^{2}}$ is the EPR pair ${\textstyle |\psi \rangle ={\frac {1}{\sqrt {2}}}(|00\rangle +|11\rangle )}$, the observable ${\displaystyle A_{0}=Z}$ (corresponding to Alice measuring in the ${\displaystyle \{|0\rangle ,|1\rangle \}}$ basis), the observable ${\displaystyle A_{1}=X}$ (corresponding to Alice measuring in the ${\displaystyle \{|+\rangle ,|-\rangle \}}$ basis), where ${\displaystyle X}$ and ${\displaystyle Z}$ are Pauli matrices. The observables ${\textstyle B_{0}={\frac {1}{\sqrt {2}}}(X+Z)}$ and ${\textstyle B_{1}={\frac {1}{\sqrt {2}}}(Z-X)}$ (corresponding to each of Bob's choice of basis to measure in). We will denote the success probability of a strategy ${\displaystyle {\mathcal {S}}}$ in the CHSH game by ${\displaystyle \omega _{\text{CHSH}}^{*}({\mathcal {S}})}$, and we define the bias of the strategy ${\displaystyle {\mathcal {S}}}$ as ${\displaystyle \beta _{\text{CHSH}}^{*}({\mathcal {S}}):=2\omega _{\text{CHSH}}^{*}({\mathcal {S}})-1}$, which is the difference between the winning and losing probabilities of ${\displaystyle {\mathcal {S}}}$.

In particular, we have

$${\displaystyle \beta _{\text{CHSH}}^{*}({\mathcal {S}})={\frac {1}{4}}\sum _{x,y\in \{0,1\}}(-1)^{x\wedge y}\cdot \langle \psi |A_{x}\otimes B_{y}|\psi \rangle .}$$

The bias of the quantum strategy described above is ${\textstyle {\frac {1}{\sqrt {2}}}}$.

Tsirelson's inequality, discovered by Boris Tsirelson in 1980,[16] states that for any quantum strategy ${\displaystyle {\mathcal {S}}}$ for the CHSH game, the bias ${\textstyle \beta _{\text{CHSH}}^{*}({\mathcal {S}})\leq {\frac {1}{\sqrt {2}}}}$. Equivalently, it states that success probability

$${\displaystyle \omega _{\text{CHSH}}^{*}({\mathcal {S}})\leq \cos ^{2}\left({\frac {\pi }{8}}\right)={\frac {1}{2}}+{\frac {1}{2{\sqrt {2}}}}}$$

for any quantum strategy ${\displaystyle {\mathcal {S}}}$ for the CHSH game. In particular, this implies the optimality of the quantum strategy described above for the CHSH game.

Tsirelsen's inequality establishes that the maximum success probability of any quantum strategy is ${\textstyle \cos ^{2}\left({\frac {\pi }{8}}\right)}$, and we saw that this maximum success probability is achieved by the quantum strategy described above. In fact, any quantum strategy that achieves this maximum success probability must be isomorphic (in a precise sense) to the canonical quantum strategy described above; this property is called the rigidity of the CHSH game, first attributed to Summers and Werner.[17] More formally, we have the following result:

Theorem (Exact CHSH rigidity) — Let ${\displaystyle {\mathcal {S}}=\left(|\psi \rangle ,(A_{0},A_{1}),(B_{0},B_{1})\right)}$ be a quantum strategy for the CHSH game where ${\displaystyle |\psi \rangle \in {\mathcal {A}}\otimes {\mathcal {B}}}$ such that ${\textstyle \omega _{\text{CHSH}}({\mathcal {S}})=\cos ^{2}\left({\frac {\pi }{8}}\right)}$. Then there exist isometries ${\displaystyle V:{\mathcal {A}}\to {\mathcal {A}}_{1}\otimes {\mathcal {A}}_{2}}$ and ${\displaystyle W:{\mathcal {B}}\to {\mathcal {B}}_{1}\otimes {\mathcal {B}}_{2}}$ where ${\displaystyle {\mathcal {A}}_{1},{\mathcal {B}}_{1}}$ are isomorphic to ${\displaystyle \mathbb {C} ^{2}}$ such that letting ${\displaystyle |\theta \rangle =(V\otimes W)|\psi \rangle }$ we have

$${\displaystyle |\theta \rangle =|\Phi \rangle _{{\mathcal {A}}_{1},{\mathcal {B}}_{1}}\otimes |\phi \rangle _{{\mathcal {A}}_{2},{\mathcal {B}}_{2}}}$$

where ${\textstyle |\Phi \rangle ={\frac {1}{\sqrt {2}}}\left(|00\rangle +|11\rangle \right)}$ denotes the EPR pair and ${\displaystyle |\phi \rangle _{{\mathcal {A}}_{2},{\mathcal {B}}_{2}}}$ denotes some pure state, and

$${\displaystyle {\begin{aligned}(V\otimes W)A_{0}|\psi \rangle =Z_{{\mathcal {A}}_{1}}|\theta \rangle ,&\qquad (V\otimes W)B_{0}|\psi \rangle =Z_{{\mathcal {B}}_{1}}|\theta \rangle ,\\(V\otimes W)A_{1}|\psi \rangle =X_{{\mathcal {A}}_{1}}|\theta \rangle ,&\qquad (V\otimes W)B_{1}|\psi \rangle =Z_{{\mathcal {B}}_{1}}|\theta \rangle .\end{aligned}}}$$

Informally, the above theorem states that given an arbitrary optimal strategy for the CHSH game, there exists a local change-of-basis (given by the isometries ${\displaystyle V,W}$) for Alice and Bob such that their shared state ${\displaystyle |\psi \rangle }$ factors into the tensor of an EPR pair ${\displaystyle |\Phi \rangle }$ and an additional auxiliary state ${\displaystyle |\phi \rangle }$. Furthermore, Alice and Bob's observables ${\displaystyle (A_{0},A_{1})}$ and ${\displaystyle (B_{0},B_{1})}$ behave, up to unitary transformations, like the ${\displaystyle Z}$ and ${\displaystyle X}$ observables on their respective qubits from the EPR pair. An approximate or quantitative version of CHSH rigidity was obtained by McKague, et al.[18] who proved that if you have a quantum strategy ${\displaystyle {\mathcal {S}}}$ such that ${\textstyle \omega _{\text{CHSH}}({\mathcal {S}})=\cos ^{2}\left({\frac {\pi }{8}}\right)-\epsilon }$ for some ${\displaystyle \epsilon >0}$, then there exist isometries under which the strategy ${\displaystyle {\mathcal {S}}}$ is ${\displaystyle O({\sqrt {\epsilon }})}$-close to the canonical quantum strategy. Representation-theoretic proofs of approximate rigidity are also known.[19]

Note that the CHSH game can be viewed as a test for quantum entanglement and quantum measurements, and that the rigidity of the CHSH game lets us test for a specific entanglement as well as specific quantum measurements. This in turn can be leveraged to test or even verify entire quantum computations—in particular, the rigidity of CHSH games has been harnessed to construct protocols for verifiable quantum delegation,[20][21] certifiable randomness expansion,[22] and device-independent cryptography.[23]