Physics:CHSH game

An illustration of the CHSH game.

The CHSH game, named after the CHSH inequality discovered by John Clauser, Michael Horne, Abner Shimony, and Richard Holt, 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 phenomenon, at least in a "local" fashion. The CHSH inequality can be seen as a restatement of Bell's theorem.

Definition

In the CHSH game, there are two cooperating players, Alice and Bob, and a referee, Charlie. These agents will be abbreviated [math]\displaystyle{ A, B, C }[/math] respectively. At the start of the game, Charlie chooses bits [math]\displaystyle{ x,y \in \{0,1\} }[/math] uniformly at random, and then sends [math]\displaystyle{ x }[/math] to Alice and [math]\displaystyle{ y }[/math] to Bob. Alice and Bob must then each respond to Charlie with bits [math]\displaystyle{ a,b \in \{0,1\} }[/math] respectively. Now, once Alice and Bob send their responses back to Charlie, Charlie tests if [math]\displaystyle{ a\oplus b = x \land y }[/math]. If this equality holds, then Alice and Bob win, and if not then they lose.

It is also required that Alice and Bob's responses can only depend on the bits they see: so Alice's response [math]\displaystyle{ a }[/math] depends only on [math]\displaystyle{ x }[/math], and similarly for Bob. This means that Alice and Bob are forbidden from directly communicating with each another about the values of the bits sent to them by Charlie. However, Alice and Bob are allowed to decide on a common strategy before the game begins.

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%.

Optimal classical strategy

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 [math]\displaystyle{ x,y }[/math]). By a deterministic strategy, we mean a pair of functions [math]\displaystyle{ f_A, f_B: \{0,1\} \mapsto \{0,1\} }[/math], where [math]\displaystyle{ f_A }[/math] is a function determining Alice's response as a function of the message she receives from Charlie, and [math]\displaystyle{ f_B }[/math] 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 [math]\displaystyle{ \{0,1\} \mapsto \{0,1\} }[/math]). It can be verified that for each of those 8 strategies there is always at least one out of the four possible input pairs [math]\displaystyle{ (x, y) \in \{0,1\}^2 }[/math] 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 [math]\displaystyle{ x=y=1 }[/math], so using this strategy their win probability is exactly 75%.

Now, consider the case of randomized classical strategies, where Alice and Bob have the ability to make some random coin flips and then determine their output as a function of both Charlie's message and the outcomes of their coin flips. Such a strategy can be equivalently 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.

Optimal quantum strategy

Now, imagine that Alice and Bob each possess one qubit of the following 2-qubit entangled state: [math]\displaystyle{ \Phi = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) }[/math]. This state is commonly referred to as the EPR Pair, and can be equivalently written as [math]\displaystyle{ \Phi = \frac{1}{\sqrt{2}}(|++\rangle + |--\rangle) }[/math]. Alice and Bob will use this entangled pair in their strategy as described below. The optimality of this strategy follows from Tsirelson's inequality.

Alice's strategy

When Alice receives her bit [math]\displaystyle{ x }[/math] from Charlie, if [math]\displaystyle{ x=0 }[/math] she will measure her qubit in the basis [math]\displaystyle{ |0\rangle, |1\rangle }[/math], and then respond with 0 if the measurement outcome is [math]\displaystyle{ |0\rangle }[/math], and 1 if it is [math]\displaystyle{ |1\rangle }[/math].

Otherwise, if [math]\displaystyle{ x=1 }[/math] she will measure her qubit in the basis [math]\displaystyle{ |+\rangle, |-\rangle }[/math], and respond with 0 if the measurement outcome is [math]\displaystyle{ |+\rangle }[/math], and 1 if it is [math]\displaystyle{ |-\rangle }[/math].

Bob's strategy

When Bob receives his bit [math]\displaystyle{ y }[/math] from Charlie, if [math]\displaystyle{ y=0 }[/math] he will measure his qubit in the basis [math]\displaystyle{ |a_0\rangle, |a_1\rangle }[/math] where [math]\displaystyle{ |a_0\rangle = \left(\cos\frac{\pi}{8}\right)|0\rangle + \left(\sin\frac{\pi}{8}\right)|1\rangle }[/math], and [math]\displaystyle{ |a_1\rangle = \left(-\sin\frac{\pi}{8}\right)|0\rangle + \left(\cos\frac{\pi}{8}\right)|1\rangle }[/math]. He then responds with 0 if the result is [math]\displaystyle{ |a_0\rangle }[/math], and 1 if it is [math]\displaystyle{ |a_1\rangle }[/math].

Otherwise, if [math]\displaystyle{ y=1 }[/math], he will measure his qubit in the basis [math]\displaystyle{ |b_0\rangle, |b_1\rangle }[/math] where [math]\displaystyle{ |b_0\rangle = \left(\cos\frac{\pi}{8}\right)|0\rangle - \left(\sin\frac{\pi}{8}\right)|1\rangle }[/math], and [math]\displaystyle{ |b_1\rangle = (\sin\frac{\pi}{8})|0\rangle + \left(\cos\frac{\pi}{8}\right)|1\rangle }[/math]. In this case, he responds with 0 if the result is [math]\displaystyle{ |b_0\rangle }[/math], and 1 if it is [math]\displaystyle{ |b_1\rangle }[/math].

Analysis of success probability

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 [math]\displaystyle{ (x,y) }[/math], and then take the average. We analyze the case where [math]\displaystyle{ x=y=0 }[/math] here: In this case the winning response pairs are [math]\displaystyle{ a = b = 0 }[/math] and [math]\displaystyle{ a=b=1 }[/math]. On input [math]\displaystyle{ x=y=0 }[/math], we know that Alice will measure in the basis [math]\displaystyle{ |0\rangle, |1\rangle }[/math], and Bob will measure in the basis [math]\displaystyle{ |a_0\rangle, |a_1\rangle }[/math]. Then the probability that they both output 0 is the same as the probability that their measurements yield [math]\displaystyle{ |0\rangle, |a_0\rangle }[/math] respectively, so precisely [math]\displaystyle{ |(\langle 0|\otimes \langle a_0|) |\Phi \rangle|^2 = \frac{1}{2}\cos^2\left(\frac{\pi}{8}\right) }[/math]. Similarly, the probability that they both output 1 is exactly [math]\displaystyle{ |(\langle 1|\otimes \langle a_1|) |\Phi \rangle|^2 = \frac{1}{2}\cos^2\left(\frac{\pi}{8}\right) }[/math]. So the probability that either of these successful outcomes happens is [math]\displaystyle{ \cos^2\left(\frac{\pi}{8}\right) }[/math].

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 [math]\displaystyle{ \cos^2\left(\frac{\pi}{8}\right) }[/math], so overall the average win probability for a randomly chosen input is [math]\displaystyle{ \cos^2\left(\frac{\pi}{8}\right) }[/math]. Since [math]\displaystyle{ \cos^2\left(\frac{\pi}{8}\right) \approx 85% }[/math], this is strictly better than what was possible in the classical case.

Modeling general quantum strategies

An arbitrary quantum strategy for the CHSH game can be modeled as a triple [math]\displaystyle{ \mathcal{S} = \left(|\psi\rangle, (A_{0}, A_1), (B_0, B_1)\right) }[/math] where

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

The optimal quantum strategy described above can be recast in this notation as follows: [math]\displaystyle{ |\psi\rangle \in\mathbb{C}^2\otimes\mathbb{C}^2 }[/math] is the EPR pair [math]\displaystyle{ |\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) }[/math], the observable [math]\displaystyle{ A_0 = Z }[/math] (corresponding to Alice measuring in the [math]\displaystyle{ \{|0\rangle, |1\rangle\} }[/math] basis), the observable [math]\displaystyle{ A_1 = X }[/math] (corresponding to Alice measuring in the [math]\displaystyle{ \{|+\rangle, |-\rangle\} }[/math] basis), where [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Z }[/math] are Pauli matrices. The observables [math]\displaystyle{ B_{0} = \frac{(X+Z)}{\sqrt{2}} }[/math] and [math]\displaystyle{ B_{1}=\frac{(Z-X)}{\sqrt{2}} }[/math] (corresponding to each of Bob's choice of basis to measure in). We will denote the success probability of a strategy [math]\displaystyle{ \mathcal{S} }[/math] in the CHSH game by [math]\displaystyle{ \omega^*_{\text{CHSH}}(\mathcal{S}) }[/math], and we define the bias of the strategy [math]\displaystyle{ \mathcal{S} }[/math] as [math]\displaystyle{ \beta^*_{\text{CHSH}}(\mathcal{S}) := 2\omega^*_{\text{CHSH}}(\mathcal{S}) - 1 }[/math], which is the difference between the winning and losing probabilities of [math]\displaystyle{ \mathcal{S} }[/math].

In particular, we have

[math]\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. }[/math]

(1)

Note that the bias of the quantum strategy described above is [math]\displaystyle{ \frac{1}{\sqrt{2}} }[/math].

Tsirelson's inequality

Tsirelson's inequality, discovered by Boris Tsirelson in 1980, states that for any quantum strategy [math]\displaystyle{ \mathcal{S} }[/math] for the CHSH game, the bias [math]\displaystyle{ \beta^*_{\text{CHSH}}(\mathcal{S}) \leq \frac{1}{\sqrt{2}} }[/math]. Equivalently, it states that success probability [math]\displaystyle{ \omega^*_{\text{CHSH}}(\mathcal{S}) \leq \cos^2\left(\frac{\pi}{8}\right) = \frac{1}{2} + \frac{1}{2\sqrt{2}} }[/math] for any quantum strategy [math]\displaystyle{ \mathcal{S} }[/math] for the CHSH game. In particular, this implies the optimality of the quantum strategy described above for the CHSH game.

Proof of Tsirelson's inequality

Let [math]\displaystyle{ \mathcal{S} = \left(|\psi\rangle, (A_0, A_1), (B_0, B_1)\right) }[/math] be a general quantum strategy for the CHSH game. From Equation (1), it suffices to show that [math]\displaystyle{ \langle \psi | A_0\otimes B_0 + A_1\otimes B_0 + A_0\otimes B_1 - A_1\otimes B_1 | \psi \rangle \leq 2\sqrt{2}. }[/math] We can rewrite this expression as

[math]\displaystyle{ \|A_0\otimes C_0 + A_1\otimes C_1 \| \leq 2 }[/math]

(2)

where [math]\displaystyle{ C_0 := (B_0 + B_1)/\sqrt{2} }[/math] and [math]\displaystyle{ C_1 := (B_0 - B_1)/\sqrt{2} }[/math], and [math]\displaystyle{ \|\cdot\| }[/math] denotes the operator norm. This follows from the fact that for any Hermitian operator [math]\displaystyle{ H }[/math], we have [math]\displaystyle{ \|H\| = \sup_{|\psi\rangle} \langle \psi | H | \psi \rangle }[/math]. We have [math]\displaystyle{ \begin{align} \|(A_0\otimes C_0 + A_1\otimes C_1)^2\| &= \|A_0^2 \otimes C_0^2 + A_1^2\otimes C_1^2 + A_0A_1\otimes C_0C_1 + A_1A_0\otimes C_1C_0\| \\ & = \|2\mathbb{I}\otimes\mathbb{I} + A_0A_1 \otimes C_0C_1 + A_1A_0\otimes C_1C_0\| \\ & \leq 2 + \|A_0A_1\otimes C_0C_1\|+ \|A_1A_0 \otimes C_1C_0\| \\ & = 2 + 2\|A_0A_1\|\cdot\|C_0C_1\| \end{align} }[/math] where the second equality follows from the fact that [math]\displaystyle{ A_0^2\otimes C_0^2 = \mathbb{I}\otimes \left(\mathbb{I}+\frac{1}{2}(B_0B_1 + B_1B_0)\right) }[/math]and [math]\displaystyle{ A_1^2\otimes C_1^2 = \mathbb{I}\otimes \left(\mathbb{I}-\frac{1}{2}(B_0B_1 + B_1B_0)\right) }[/math], the inequality follows from the triangle inequality, and the final equality from the facts that [math]\displaystyle{ \|A\otimes B\| = \|A\|\cdot\|B\| }[/math] and [math]\displaystyle{ \|AB\| = \|BA\| }[/math]. Finally, [math]\displaystyle{ \|A_0A_1\| \leq \|A_0\|\cdot\|A_1\| }[/math], and as [math]\displaystyle{ A_0, A_1 }[/math] are binary observables, this is at most 1. We also have [math]\displaystyle{ \|C_0C_1\| = \left\|\frac{1}{2}(B_0^2 - B_1^2 - B_0B_1 + B_1B_0)\right\| = \frac{1}{2}\|B_1B_0 - B_0B_1\| \leq 1 }[/math] where we used the fact that [math]\displaystyle{ B_y^2 = \mathbb{I} }[/math] as [math]\displaystyle{ B_y }[/math] is a binary observable. In particular, we have that [math]\displaystyle{ \begin{align} \|(A_0\otimes C_0 + A_1\otimes C_1)^2\| \leq 4. \end{align} }[/math] Equation (2) now follows from the fact that [math]\displaystyle{ \|H^2\| = \|H\|^2 }[/math] for any operator [math]\displaystyle{ H }[/math].

CHSH rigidity

Tsirelsen's inequality establishes that the maximum success probability of any quantum strategy is [math]\displaystyle{ \cos^2\left(\frac{\pi}{8}\right) }[/math], 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. More formally, we have the following result:

Theorem (Exact CHSH rigidity). Let [math]\displaystyle{ \mathcal{S} = \left(|\psi\rangle, (A_0, A_1), (B_0, B_1)\right) }[/math] be a quantum strategy for the CHSH game where [math]\displaystyle{ |\psi\rangle \in \mathcal{A}\otimes\mathcal{B} }[/math] such that [math]\displaystyle{ \omega_{\text{CHSH}}(\mathcal{S}) = \cos^2\left(\frac{\pi}{8}\right) }[/math]. Then there exist isometries [math]\displaystyle{ V : \mathcal{A}\to\mathcal{A}_1\otimes\mathcal{A}_2 }[/math] and [math]\displaystyle{ W : \mathcal{B}\to\mathcal{B}_1\otimes\mathcal{B}_2 }[/math] where [math]\displaystyle{ \mathcal{A}_1,\mathcal{B}_1 }[/math] are isomorphic to [math]\displaystyle{ \mathbb{C}^2 }[/math] such that letting [math]\displaystyle{ |\theta\rangle = (V\otimes W)|\psi\rangle }[/math] we have [math]\displaystyle{ |\theta\rangle = |\Phi\rangle_{\mathcal{A}_1, \mathcal{B}_1} \otimes |\phi\rangle_{\mathcal{A}_2,\mathcal{B}_2} }[/math] where [math]\displaystyle{ |\Phi\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right) }[/math] denotes the EPR pair and [math]\displaystyle{ |\phi\rangle_{\mathcal{A}_2,\mathcal{B}_2} }[/math] denotes some pure state , and [math]\displaystyle{ \begin{align} (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{align} }[/math]

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 [math]\displaystyle{ V, W }[/math]) for Alice and Bob such that their shared state [math]\displaystyle{ |\psi\rangle }[/math] factors into the tensor of an EPR pair [math]\displaystyle{ |\Phi\rangle }[/math] and an additional auxiliary state [math]\displaystyle{ |\phi\rangle }[/math]. Furthermore, Alice and Bob's observables [math]\displaystyle{ (A_0, A_1) }[/math] and [math]\displaystyle{ (B_0, B_1) }[/math] behave, up to unitary transformations, like the [math]\displaystyle{ Z }[/math] and [math]\displaystyle{ X }[/math] observables on their respective qubits from the EPR pair. An approximate or quantitative version of CHSH rigidity was obtained by McKague, et al. who proved that if you have a quantum strategy [math]\displaystyle{ \mathcal{S} }[/math] such that [math]\displaystyle{ \omega_{\text{CHSH}}(\mathcal{S}) = \cos^2\left(\frac{\pi}{8}\right) - \epsilon }[/math] for some [math]\displaystyle{ \epsilon \gt 0 }[/math], then there exist isometries under which the strategy [math]\displaystyle{ \mathcal{S} }[/math] is [math]\displaystyle{ O(\sqrt{\epsilon}) }[/math]-close to the canonical quantum strategy. Representation-theoretic proofs of approximate rigity are also known.

Applications

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, certifiable randomness expansion, and device-independent cryptography.

See also

• Bell's theorem
• CHSH inequality
• Quantum entanglement
• Quantum nonlocality

External links

•  Henry Yuen, The Complexity of Entanglement, lecture notes for CSC2429/MAT1751 (Fall 2020) at the University of Toronto