The CHSH game

Figure 1: Playing the CHSH game

Consider the following game (see Figure 1):

Alice and Bob are in separate rooms with no way to communicate with each other. A referee in each room randomly flips a coin which the players see. Each player responds by saying "+1" or "-1".

For Alice and Bob to win the game, their responses have to be the same if and only if at least one of the coins show heads. They can agree on a strategy beforehand.

The game is repeated many times. What classical strategy maximizes their probability for winning and what is that probability? Have a go at answering it, then check below.

The best strategy is for both players to ignore the coin and just say "+1" every time.[1] They will win 75% of the time on average, losing only when both coins are tails. Thus we can say that any classical strategy wins ≤ 75% of the time. In this context, that's called a Bell inequality.

Figure 2: Playing the CHSH game (quantum strategy)

Figure 3: Entangled particle pair

Now let's try a specific quantum strategy (see Figure 2). For each game, Alice and Bob share a single entangled particle pair (see Figure 3).

Alice and Bob agree beforehand to each measure their particle at an angle that depends on the coin flips. On heads, Alice's measurement angle will be 0°, on tails 45°. On heads, Bob's measurement angle will be 22.5°, on tails -22.5°. They each measure their particle and report the "+1" or "-1" result.[2]

In three of the cases, at least one of the coins will be heads and the difference in Alice and Bob's measurement angles is 22.5°. QM predicts that they will measure the same result cos²(22.5°) ≈ 85% of the time.[3] In the double-tails case, the difference in Alice and Bob's measurement angles is 67.5°. QM predicts that they will measure a different result sin²(67.5°) ≈ 85% of the time. Thus Alice and Bob win the game on average 85% of the time.

So the quantum strategy is better than the best possible classical strategy, violating the Bell inequality. That's Bell's Theorem (one of many variants). So at least one of the classical assumptions is incorrect (hidden variables, locality, or statistical independence).

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As an alternative presentation of the CHSH inequality, consider the following two-qubit observable composed of the various single-qubit observables that Alice and Bob can measure.

O = A_HB_H + A_HB_T + A_TB_H – A_TB_T

Each term specifies the product of a particular pair of measurements that Alice and Bob perform. For example, suppose when Alice's coin was heads, she measured -1 (with the 0° angle measurement) and when Bob's coin was tails, he measured +1 (with the -22.5° measurement). Thus the product A_HB_T would be -1 * + 1 = -1. Since the measured values are always either +1 or -1, a product term must also be either +1 or -1.

Note that it's not possible to make the four measurements jointly because they don't commute. However let's assume that each of the measurement pairs do have definite values even if they are not measured. This assumption is called realism (or hidden variables). Let's further assume that no communication occurs between Alice and Bob's particle. This assumption is called locality.

Now consider that observable O can be factored as:

O = (A_H + A_T)B_H + (A_H - A_T)B_T

Note that one of the parenthesized quantities must be equal to zero. That's because Alice's two possible measurement results (±1) would either add to 0 (if different) or subtract to 0 (if equal). So the maximum possible value for observable O is 2. This equation (and inequality) can be written as:

|E(O)| = |E(A_HB_H) + E(A_HB_T) + E(A_TB_H) – E(A_TB_T)| ≤ 2

where E specifies the expectation value. The expectation value for observable O can be empirically determined by running the experiment many times (using a new entangled particle pair and coin pair each time) and then summing the expectation values for each term.

The formula for the expectation value is:

E(ab) = (+1)cos²(θ) + (-1)sin²(θ) = cos(2*θ)

where θ is the relative angle between Alice and Bob’s measurements. So:

|E(O)| = |E(22.5°) + E(22.5°) + E(22.5°) – E(67.5°)|

           = |1/√2 + 1/√2 + 1/√2 – -1/√2| = 2√2 ≈ 2.828 ≰ 2

thus violating the CHSH inequality.[4]

As noted with the CHSH game, the Bell inequality depends on three assumptions. The first is locality (or no spooky action at a distance), which means that Alice and Bob have no way of communicating and thus potentially influencing the results. The second is hidden variables (or realism or counterfactual definiteness or non-contextuality), which means that there are definite values for the spin angles even if measurements are not performed at those angles. The third is statistical independence (or free-will or no conspiracy), which means that the coin flips are random and independent.

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References:

Introduction to Quantum Information Science Lecture Notes, Lecture 13: Hidden Variables, Bell’s Inequality, Lecture 14: Nonlocal Games - Scott Aaronson

Quantum Randomness - American Scientist, Scott Aaronson

Bell's Inequality: The weirdest theorem in the world | Nobel Prize 2022 - Qiskit (video)

Bell's Theorem - Brilliant

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[1] Enumeration of all possible deterministic strategies for the CHSH game:

  ==================================
 | Strategy | Coin| Say | Win | Win |
 |          | A B | A B |     |  %  |
  ==================================
 |          | H H | + + | Yes |     |
 | Heads:+1 | H T | + + | Yes | 75% |
 | Tails:+1 | T H | + + | Yes |     |
 |          | T T | + + | No  |     |
  ----------+-----+-----+-----+-----
 |          | H H | - - | Yes |     |
 | Heads:-1 | H T | - - | Yes | 75% |
 | Tails:-1 | T H | - - | Yes |     |
 |          | T T | - - | No  |     |
  ----------+-----+-----+-----+-----
 |          | H H | + + | Yes |     |
 | Heads:+1 | H T | + - | No  | 25% |
 | Tails:-1 | T H | - + | No  |     |
 |          | T T | - - | No  |     |
  ----------+-----+-----+-----+-----
 |          | H H | - - | Yes |     |
 | Heads:-1 | H T | - + | No  | 25% |
 | Tails:+1 | T H | + - | No  |     |
 |          | T T | + + | No  |     |
  ================================== 

[2] Figure 4 shows the quantum circuit diagram that creates the Bell state and rotates Alice and Bob's axes for the A_HB_H measurement combination.

Figure 4: Quantum circuit

The state vector is:

\(\begin{aligned}\hspace{20 mm}|\Psi\rangle = 0.653|00\rangle - 0.271|01\rangle + 0.271|10\rangle + 0.653|11\rangle\end{aligned}\)

The results "+1" and "-1" map to the \(|0\rangle\) and \(|1\rangle\) basis states respectively. The terms that represent Alice and Bob getting the same results are the \(|00\rangle\) and \(|11\rangle\) terms. Squaring the amplitudes for those terms and adding them gives 85% which is the expected win percentage for the A_HB_H measurement combination.

To view the state vectors for each measurement combination, click on the circuit diagram link in Figure 4 and use the following rotations (which will produce the associated state vector):

A_HB_H =  0° |  22.5° = RY(0 * pi/8 * 2) | RY( 1 * pi/8 * 2)

\(\begin{aligned}\hspace{10 mm}|\Psi\rangle = 0.653|00\rangle - 0.271|01\rangle + 0.271|10\rangle + 0.653|11\rangle\end{aligned}\)

A_HB_T =  0° | -22.5° = RY(0 * pi/8 * 2) | RY(-1 * pi/8 * 2)

\(\begin{aligned}\hspace{10 mm}|\Psi\rangle = 0.653|00\rangle + 0.271|01\rangle - 0.271|10\rangle + 0.653|11\rangle\end{aligned}\)

A_TB_H = 45° |  22.5° = RY(2 * pi/8 * 2) | RY( 1 * pi/8 * 2)

\(\begin{aligned}\hspace{10 mm}|\Psi\rangle = 0.653|00\rangle + 0.271|01\rangle - 0.271|10\rangle + 0.653|11\rangle\end{aligned}\)

A_TB_T = 45° | -22.5° = RY(2 * pi/8 * 2) | RY(-1 * pi/8 * 2)

\(\begin{aligned}\hspace{10 mm}|\Psi\rangle = 0.271|00\rangle + 0.653|01\rangle - 0.653|10\rangle + 0.271|11\rangle\end{aligned}\)

Note that on the Bloch sphere representation, orthogonal vectors are antiparallel. So the RY multiplier is π/8 * 2 (45°) rather than just π/8 (22.5°) because RY will internally divide the angle by 2.

[3] The measurement bases for each angle are:

0° \(\begin{aligned}\hspace{14 mm}\{ |0\rangle, |1\rangle \}\end{aligned}\)

45° \(\begin{aligned}\hspace{12 mm}\{ |\frac{\pi}{4}\rangle, |\frac{3\pi}{4}\rangle \}\end{aligned}\)

22.5° \(\begin{aligned}\hspace{9 mm}\{ |\frac{\pi}{8}\rangle, |\frac{5\pi}{8}\rangle \}\end{aligned}\)

-22.5° \(\begin{aligned}\hspace{8 mm}\{ |-\frac{\pi}{8}\rangle, |\frac{3\pi}{8}\rangle \}\end{aligned}\)

where

\(\begin{aligned}\hspace{20 mm}|\frac{\pi}{4}\rangle = cos(\frac{\pi}{4})|0\rangle + sin(\frac{\pi}{4})|1\rangle, |3\frac{\pi}{4}\rangle = cos(3\frac{\pi}{4})|0\rangle + sin(3\frac{\pi}{4})|1\rangle\end{aligned}\)

\(\begin{aligned}\hspace{20 mm}|\frac{\pi}{8}\rangle = cos(\frac{\pi}{8})|0\rangle + sin(\frac{\pi}{8})|1\rangle, |5\frac{\pi}{8}\rangle = cos(5\frac{\pi}{8})|0\rangle + sin(5\frac{\pi}{8})|1\rangle\end{aligned}\)

\(\begin{aligned}\hspace{20 mm}|-\frac{\pi}{8}\rangle = cos(-\frac{\pi}{8})|0\rangle + sin(-\frac{\pi}{8})|1\rangle, |\frac{3\pi}{8}\rangle = cos(\frac{3\pi}{8})|0\rangle + sin(\frac{3\pi}{8})|1\rangle\end{aligned}\)

and (see calculator)

 \(\begin{aligned}\hspace{20 mm}cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} = 0.7071..., sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} = 0.7071... \end{aligned}\)

 \(\begin{aligned}\hspace{20 mm}cos(\frac{\pi}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{2} = 0.9238..., sin(\frac{\pi}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2} = 0.3826... \end{aligned}\)

[4] The result 2√2 is also the Tsirelson bound, which is the upper limit of the Bell violation. The disparity between the classical and quantum predictions for detector angle differences is shown in Figure 5.

Figure 5: Classical and quantum predictions