How Do I Calculate the Score of a Bell's Game

Question

I am reading "Quantum Chance" by Nicolas Gisin. I would like to know how to calculate the Bell Game outcome given in the following section of the book.

BACKGROUND

The Bell Game begins with a pair boxes. Alice and Bob each have a box with a joy stick that can move either left or right. This leads to four combinations of joystick positions(Alice/Bob): L/L, R/L, L/R, R/R. At the front of each box is a screen that displays either 0 or 1 every time they move their joystick. A hypothetical move looks like the following. Alice moves her joystick to the left and sees the number 1. Bob moves his joystick to the left and sees the number 0. They record these outcomes and later come back to compare notes.

Scoring works as the follow: For every combination of Left/Left, Right/Left, Left/Right they get one point if the numbers they recorded were the same. For Right/Right, they get one point if the numbers they recorded are different.

Example: Alice: Left, 0 | Bob: Left, 0...they get one point. Alice: Left, 1 | Bob: Right, 0... they do not get one point. Alice: Right, 1 | Bob: Right, 0... they get one point.

The total score is a sum over the fraction of points/total for each category of joystick position, ie L/L, R/L, L/R, and R/R. Because there are four categories, the score is tyically written as a fraction of 4, ie 3/4. A perfect score in three categories with a 0 score in the fourth would be 1 + 1 + 1 + 0 = 3...3/4.

It can be shown that if the world were limited to non-quantum effects, the maximum score one could achieve is a 3/4.

Bell's games are meant to demonstrate how entanglement allows Alice and Bob to achieve a score over 3. They achieve this by using a pair of entangled qubits.

Assume we entangle a pair of qubits in a [00] + [11] state. This means that if Alice measures a qubit to be 1, the other qubit must be 1. If she measures 0, the other qubit must be 0.

We attach the joysticks to a measuring device and perform the following experiment:

First we create a large number of entangled qubit pairs. For each pair, we give one qubit to Alice and the corresponding qubit to Bob.

Second, Alice and Bob measure the qubit by moving the joystick either left or right.

The measurement, as illustrated in the picture below, is based on the polarization of the qubit. When Alice measures the qubit, the box generates a 0 if the qubit is polarized parallel to that direction, or a 1 if it is anti-parallel to that direction. The direction of Bob's device is slightly rotated compared to Alice's measurements.

When Alice makes a measurement on her qubit, she generates either 1 or 0. Because of entanglement, Bob's qubit now shares the same state as Alice's. By making a measurement similar to Alice's measurement, he has a high chance of measuring the same value (either a 1 or a 0). Note, the measurement is similar - but not the same - because Bob's device was rotated. This means that there is still a chance that Bob measures a different value than Alice.

With the arrangement on the right side of the photo, Alice and Bob should achieve a score of ~3.41.

But I do not see how the author arrived at the value of ~3.41. Can someone explain how I would go about calculating this value?

Answer 1

This "Bell Game" is just the famous CHSH game. It was rather unfortunate that Gisin didn't use the regular name, as it makes it much harder to search information about it. It is explained here, for example, or here, or here.

There is no easy way to calculate the quantum "score", you need to use quantum mechanics. Let me first rephrase the game a bit, though, so that the answer makes sense. The game is played between Alice and Bob and a referee. The referee sends a question to Alice and Bob (the position of the joystick, in Gisin's exposition, L or R), and they send back an answer to the referee, either 0 or 1. Alice and Bob win the game if they manage to send the same answer whenever the questions are L/L, L/R, or R/L, and different answers whenever the questions are R/R. The "score" of the game is the probability that Alice and Bob win the game, assuming that each possible question is sent with the same probability.

Now without quantum effects the maximal probability that they can win the game is 3/4, and using entanglement the maximal probability is $(2+\sqrt2)/4$, which gets you the "score" of $\approx 3.41$ if you multiply by 4.

To calculate that, you need to use Born's rule to calculate the probability that Alice and Bob get answers $a$ and $b$ (either 0 or 1) when making measurements $x$ and $y$ (either L or R). This is given by $$ p(a,b|x,y) = |\langle A^a_x| \otimes \langle B^b_y| |\psi \rangle |^2, $$where $|\psi\rangle$ is the quantum state they are using, $(|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle)/\sqrt2$, and $\langle A^a_x|$ and $\langle B^b_y|$ describe their measurements.

Gisin stated that the quantum probability is obtained when Alice makes a measurement along the directions horizontal and vertical, and Bob along diagonal and antidiagonal. Quantum mechanics then implies that $$\langle A^0_L| = \langle 0|, \quad \langle A^1_L| = \langle 1|, $$ $$\langle A^0_R| = \frac1{\sqrt2}(\langle 0|+\langle 1|), \quad \langle A^1_R| = \frac1{\sqrt2}(\langle 0|-\langle 1|), $$ $$\langle B^0_L| = \cos\left(\frac\pi8\right) \langle 0|+\sin\left(\frac\pi8\right) \langle 1|, \quad \langle B^1_L| = \sin\left(\frac\pi8\right) \langle 0|-\cos\left(\frac\pi8\right) \langle 1|, $$ $$\langle B^0_R| = \cos\left(-\frac\pi8\right) \langle 0|+\sin\left(-\frac\pi8\right) \langle 1|, \quad \langle B^1_R| = \sin\left(-\frac\pi8\right) \langle 0|-\cos\left(-\frac\pi8\right) \langle 1|, $$ and the rest is just calculation.