In physics, the 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. Experimental verification of the inequality being violated is seen as confirmation that nature cannot be described by such theories. CHSH stands for John Clauser, Michael Horne, Abner Shimony, and Richard Holt, who described it in a much-cited paper published in 1969. They derived the CHSH inequality, which, as with John Stewart Bell's original inequality,[1] is a constraint—on the statistical occurrence of "coincidences" in a Bell test—which is necessarily true if an underlying local hidden-variable theory exists. In practice, the inequality is routinely violated by modern experiments in quantum mechanics.[2]
The usual form of the CHSH inequality iswhere
a
a'
A
b
b'
B
E(a,b)
A(a) x B(b)
A,B
The mathematical formalism of quantum mechanics predicts that the value of
S
2\sqrt{2}
Many Bell tests conducted subsequent to Alain Aspect's second experiment in 1982 have used the CHSH inequality, estimating the terms using (3) and assuming fair sampling. Some dramatic violations of the inequality have been reported.[6]
In practice most actual experiments have used light rather than the electrons that Bell originally had in mind. The property of interest is, in the best known experiments, the polarisation direction, though other properties can be used. The diagram shows a typical optical experiment. Coincidences (simultaneous detections) are recorded, the results being categorised as '++', '+-', '-+' or '--' and corresponding counts accumulated.
Four separate subexperiments are conducted, corresponding to the four terms
E(a,b)
For each selected value of
a,b
\left\{N++,N--,N+-,N-+\right\}
E(a,b)
Once all the 's have been estimated, an experimental estimate of S (Eq.) can be found. If it is numerically greater than 2 it has infringed the CHSH inequality and the experiment is declared to have supported the quantum mechanics prediction and ruled out all local hidden-variable theories.
The CHSH paper lists many preconditions (or "reasonable and/or presumable assumptions") to derive the simplified theorem and formula. For example, for the method to be valid, it has to be assumed that the detected pairs are a fair sample of those emitted. In actual experiments, detectors are never 100% efficient, so that only a sample of the emitted pairs are detected. A subtle, related requirement is that the hidden variables do not influence or determine detection probability in a way that would lead to different samples at each arm of the experiment.
The CHSH inequality has been violated with photon pairs, beryllium ion pairs, ytterbium ion pairs, rubidium atom pairs, whole rubidium-atom cloud pairs, nitrogen vacancies in diamonds, and Josephson phase qubits.[7]
The original 1969 derivation will not be given here since it is not easy to follow and involves the assumption that the outcomes are all +1 or -1, never zero. Bell's 1971 derivation is more general. He effectively assumes the "Objective Local Theory" later used by Clauser and Horne. It is assumed that any hidden variables associated with the detectors themselves are independent on the two sides and can be averaged out from the start. Another derivation of interest is given in Clauser and Horne's 1974 paper, in which they start from the CH74 inequality.
The following is based on page 37 of Bell's Speakable and Unspeakable,[3] 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:where A and B are the outcomes. Since the possible values of A and B are -1, 0 and +1, it follows that:
Then, if a, a′, b and b′ are alternative settings for the detectors,
\begin{align} &E(a,b)-E\left(a,b'\right)\\ ={}&\int\left[\underline{A}(a,λ)\underline{B}(b,λ)-\underline{A}(a,λ)\underline{B}\left(b',λ\right)\right]\rho(λ)dλ\\ ={}&\int\left[\underline{A}(a,λ)\underline{B}(b,λ)-\underline{A}(a,λ)\underline{B}\left(b',λ\right)\pm\underline{A}(a,λ)\underline{B}(b,λ)\underline{A}\left(a',λ\right)\underline{B}\left(b',λ\right)\mp\underline{A}(a,λ)\underline{B}(b,λ)\underline{A}\left(a',λ\right)\underline{B}\left(b',λ\right)\right]\rho(λ)dλ\\ ={}&\int\underline{A}(a,λ)\underline{B}(b,λ)\left[1\pm\underline{A}\left(a',λ\right)\underline{B}\left(b',λ\right)\right]\rho(λ)dλ-\int\underline{A}(a,λ)\underline{B}\left(b',λ\right)\left[1\pm\underline{A}\left(a',λ\right)\underline{B}(b,λ)\right]\rho(λ)dλ \end{align}
Taking absolute values of both sides, and applying the triangle inequality to the right-hand side, we obtain
\left|E(a,b)-E\left(a,b'\right)\right|\leq\left|\int\underline{A}(a,λ)\underline{B}(b,λ)\left[1\pm\underline{A}\left(a',λ\right)\underline{B}\left(b',λ\right)\right]\rho(λ)dλ\right|+\left|\int\underline{A}(a,λ)\underline{B}\left(b',λ\right)\left[1\pm\underline{A}\left(a',λ\right)\underline{B}(b,λ)\right]\rho(λ)dλ\right|
We use the fact that
\left[1\pm\underline{A}\left(a',λ\right)\underline{B}\left(b',λ\right)\right]\rho(λ)
\left[1\pm\underline{A}\left(a',λ\right)\underline{B}(b,λ)\right]\rho(λ)
By, this must be less than or equal towhich, using the fact that the integral of is 1, is equal towhich is equal to
2\pm\left[E\left(a',b'\right)+E\left(a',b\right)\right]
Putting this together with the left-hand side, we have:which means that the left-hand side is less than or equal to both
2+\left[E\left(a',b'\right)+E\left(a',b\right)\right]
2-\left[E\left(a',b'\right)+E\left(a',b\right)\right]
In their 1974 paper, 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:where j and k are each '+' or '-', indicating which detectors are being considered.
To obtain the CHSH test statistic S, 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.
\rho
Let us introduce a 3×3 real matrix
T\rho
tij=\operatorname{Tr}[\rho ⋅ (\sigmai ⊗ \sigmaj)]
\sigma1,\sigma2,\sigma3
U\rho=
T | |
T | |
\rho |
T\rho
λ1\geqλ2\geqλ3
There exists an optimal configuration of the measurement bases a, a', b, b for a given
\rho
The projective measurement that yields either +1 or −1 for two orthogonal states
|\alpha\rangle,|\alpha\perp\rangle
\Alpha=|\alpha\rangle\langle\alpha|-|\alpha\perp\rangle\langle\alpha\perp|
\boldsymbol{a}\inR3,|\boldsymbol{a}|=1
\boldsymbol{\sigma}
\Alpha=\boldsymbol{a} ⋅ \boldsymbol{\sigma}
The optimal set of bases for the state
\rho
λ1,2
\boldsymbol{e}1,2
U\rho
\varphi
In entanglement-based quantum key distribution, there is another measurement basis used to communicate the secret key (
\boldsymbol{a}0
\boldsymbol{a}0,\boldsymbol{b}
\varphi=\pi/4
The CHSH game is a thought experiment involving two parties separated at a great distance (far enough to preclude classical communication at the speed of light), each of whom has access to one half of an entangled two-qubit pair. Analysis of this game shows that no classical local hidden-variable theory can explain the correlations that can result from entanglement. Since this game is indeed physically realizable, this gives strong evidence that classical physics is fundamentally incapable of explaining certain quantum phenomena, at least in a "local" fashion.
In the CHSH game, there are two cooperating players, Alice and Bob, and a referee, Charlie. These agents will be abbreviated
A,B,C
x,y\in\{0,1\}
x
y
a,b\in\{0,1\}
a ⊕ b=x\landy
It is also required that Alice and Bob's responses can only depend on the bits they see: so Alice's response
a
x
In the following sections, it is shown that if Alice and Bob use only classical strategies involving their local information (and potentially some random coin tosses), it is impossible for them to win with a probability higher than 75%. However, if Alice and Bob are allowed to share a single entangled qubit pair, then there exists a strategy which allows Alice and Bob to succeed with a probability of ~85%.
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
x,y
fA,fB:\{0,1\}\mapsto\{0,1\}
fA
fB
\{0,1\}\mapsto\{0,1\}
(x,y)\in\{0,1\}2
x=y=1
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: , 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
x
\{|0\rangle,|1\rangle\}
\{|+\rangle,|-\rangle\}
x=0
x=1
a=0
a=1
Bob also uses the bit
y
y=0
\{|a0\rangle,|a1\rangle\}
y=1
\{|b0\rangle,|b1\rangle\}
\theta=\pi/8
The following table shows how the game is played. The states are arranged in the order that puts each state between the two most similar. They could correspond, for example, to photons polarized at angles of 0°, 22.5°, 45°, ... 180° (with 180° and 0° being the same state).
State | 1\rangle ! | b_1\rangle ! | +\rangle ! | a_0\rangle ! | 0\rangle ! | b_0\rangle ! | -\rangle ! | a_1\rangle ! | 1\rangle | |||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
At what value of x does Alice test for the state? | 0 | 1 | 0 | 1 | 0 | |||||||||||||
At what value of y does Bob test for the state? | 1 | 0 | 1 | 0 | ||||||||||||||
What does Alice send if she finds the state? | 1 | 0 | 0 | 1 | 1 | |||||||||||||
What does Bob send if she finds the state? | 1 | 0 | 0 | 1 |
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
(x,y)
x=y=0
a=b=0
a=b=1
x=y=0
|0\rangle,|1\rangle
|a0\rangle,|a1\rangle
|0\rangle,|a0\rangle
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
| ||||
\cos |
\right)
An arbitrary quantum strategy for the CHSH game can be modeled as a triple
l{S}=\left(|\psi\rangle,(A0,A1),(B0,B1)\right)
|\psi\rangle\inCd ⊗ Cd
d
A0
A1
x\in\{0,1\}
B0
B1
y\in\{0,1\}
|\psi\rangle\inC2 ⊗ C2
A0=Z
\{|0\rangle,|1\rangle\}
A1=X
\{|+\rangle,|-\rangle\}
X
Z
l{S}
* | |
\omega | |
CHSH |
(l{S})
l{S}
* | |
\beta | |
CHSH |
(l{S}):=
* | |
2\omega | |
CHSH |
(l{S})-1
l{S}
In particular, we have The bias of the quantum strategy described above is .
See also: Tsirelson's bound.
Tsirelson's inequality, discovered by Boris Tsirelson in 1980,[8] states that for any quantum strategy
l{S}
l{S}
Tsirelson's inequality establishes that the maximum success probability of any quantum strategy is , 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.[9] More formally, we have the following result:
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
V,W
|\psi\rangle
|\Phi\rangle
|\phi\rangle
(A0,A1)
(B0,B1)
Z
X
l{S}
\epsilon>0
l{S}
O(\sqrt{\epsilon})
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,[12] [13] certifiable randomness expansion,[14] and device-independent cryptography.[15]