CHSH game in detail

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CHSH (John Clauser, Michael Horne, Abner Shimony, and Richard Holt) game is another view on the Bell’s inequalities, showing that under the hood our world is not classical. The game is played by Alice and Bob, and proceeds as follows:

  • a referee generates two independent uniformly chosen random bits (x and y, also called “questions”) and sends x bit to Alice and y bit to Bob
  • After receiving the questions, Alice and Bob send their answers back to the referee; an answer is one bit too (a for Alice and b for Bob)
  • Alice and Bob win if x \cdot y = a \oplus b, where \oplus stands for xor operation.

The game is cooperative – Alice and Bob does not play against each other, their goal is to get the highest winning probability together. Before the game starts, Alice and Bob can communicate and decide on the strategy they will use; after the game started, Alice and Bob are not allowed to communicate anymore; Alice doesn’t know what question the referee sent to Bob, and Bob doesn’t know what question the referee sent to Alice.

In the classical world, Alice and Bob have 16 possible deterministic strategies – both Alice and Bob have two possible answers for each of two possible questions, combining them we get 2^2 \cdot 2^2 = 16 possible strategies. Since the probability of x \cdot y = 0 is 75%, it can be easily seen that there exists 75%-winning strategy – Alice and Bob simply ignore the questions and answer identically – either by bit 0 both or by bit 1 both; it seems natural that this is the best strategy, and it is indeed so, as can be formally shown by checking all 16 possible strategies.

In the quantum world, Alice and Bob have additional degree of freedom – before the game starts, they share an entangled bipartite state, for example two-qubit EPR state:

|\Psi_{AB}\rangle = \frac{1}{\sqrt{2}}(|0\rangle_A\otimes|0\rangle_B + |1\rangle_A\otimes|1\rangle_B)

Now, Alice and Bob choose the measurement bases according to the questions they receive from the referee, and answer by measuring their qubits in the chosen bases.

Let us denote the basis vectors as \nu_i(\theta):

|\nu_0(\theta)\rangle = \cos \theta |0\rangle + \sin \theta |1\rangle
|\nu_1(\theta)\rangle = \sin \theta |0\rangle - \cos \theta |1\rangle

Alice uses \theta_{A0} on receiving the question 0, and \theta_{A1} on receiving the question 1; Bob uses \theta_{B0} on receiving the question 0, and \theta_{B1} on receiving the question 1.

If the referee sends questions x = 0, y = 0, Alice and Bob win if they answer identically a = 0, b = 0 or a = 1, b = 1. The correspondent probability of winning (given x = 0, y = 0) is:

P(win|x=0,y=0) =
|\langle_A\nu_0(\theta_{A0})|\otimes\langle_B\nu_0(\theta_{B0})|\Psi_{AB}\rangle|^2 + |\langle_A\nu_1(\theta_{A0})|\otimes\langle_B\nu_1(\theta_{B0})|\Psi_{AB}\rangle|^2

Substituting |\nu_i(\theta)\rangle we get
P(win|x=0,y=0) =
|(\cos\theta_{A0} \langle_A 0| + \sin\theta_{A0} \langle_A 1|) \otimes (\cos\theta_{B0} \langle_B 0| + \sin\theta_{B0} \langle_B 1|)|\Psi_{AB}\rangle|^2 +
|(\sin\theta_{A0} \langle_A 0| - \cos\theta_{A0} \langle_A 1|) \otimes (\sin\theta_{B0} \langle_B 0| - \cos\theta_{B0} \langle_B 1|)|\Psi_{AB}\rangle|^2

Substituting |\Psi_{AB}\rangle we get
P(win|x=0,y=0) =
\frac{1}{2}|\cos\theta_{A0} \cos\theta_{B0} + \sin\theta_{A0} \sin\theta_{B0}|^2 + \frac{1}{2}|\sin\theta_{A0} \sin\theta_{B0} + \cos\theta_{A0} \cos\theta_{B0}|^2

Finally,
P(win|x=0,y=0) = \cos^2 (\theta_{A0} - \theta_{B0})

The cases x = 0, y = 1 and x = 1, y = 0 are similar to x = 0, y = 0: Alice and Bob win if they answer identically:

P(win|x=0,y=1) =
|\langle_A\nu_0(\theta_{A0})|\otimes\langle_B\nu_0(\theta_{B1})|\Psi_{AB}\rangle|^2 + |\langle_A\nu_1(\theta_{A0})|\otimes\langle_B\nu_1(\theta_{B1})|\Psi_{AB}\rangle|^2 =
|(\cos\theta_{A0} \langle_A 0| + \sin\theta_{A0} \langle_A 1|) \otimes (\cos\theta_{B1} \langle_B 0| + \sin\theta_{B1} \langle_B 1|)|\Psi_{AB}\rangle|^2 +
|(\sin\theta_{A0} \langle_A 0| - \cos\theta_{A0} \langle_A 1|) \otimes (\sin\theta_{B1} \langle_B 0| - \cos\theta_{B1} \langle_B 1|)|\Psi_{AB}\rangle|^2

P(win|x=0,y=1) = \cos^2 (\theta_{A0} - \theta_{B1})

P(win|x=1,y=0) =
|\langle_A\nu_0(\theta_{A1})|\otimes\langle_B\nu_0(\theta_{B0})|\Psi_{AB}\rangle|^2 + |\langle_A\nu_1(\theta_{A1})|\otimes\langle_B\nu_1(\theta_{B0})|\Psi_{AB}\rangle|^2=
|(\cos\theta_{A1} \langle_A 0| + \sin\theta_{A1} \langle_A 1|) \otimes (\cos\theta_{B0} \langle_B 0| + \sin\theta_{B0} \langle_B 1|)|\Psi_{AB}\rangle|^2 +
|(\sin\theta_{A1} \langle_A 0| - \cos\theta_{A1} \langle_A 1|) \otimes (\sin\theta_{B0} \langle_B 0| - \cos\theta_{B0} \langle_B 1|)|\Psi_{AB}\rangle|^2

P(win|x=1,y=0) =\cos^2 (\theta_{A1} - \theta_{B0})

In the last case x = 1, y = 1 Alice and Bob win if they answer differently, so:

P(win|x=1,y=1) =
|\langle_A\nu_0(\theta_{A1})|\otimes\langle_B\nu_1(\theta_{B1})|\Psi_{AB}\rangle|^2 + |\langle_A\nu_1(\theta_{A1})|\otimes\langle_B\nu_0(\theta_{B1})|\Psi_{AB}\rangle|^2=

|(\cos\theta_{A1} \langle_A 0| + \sin\theta_{A1} \langle_A 1|) \otimes (\sin\theta_{B1} \langle_B 0| - \cos\theta_{B1} \langle_B 1|)|\Psi_{AB}\rangle|^2 +
|(\sin\theta_{A1} \langle_A 0| - \cos\theta_{A1} \langle_A 1|) \otimes (\cos\theta_{B1} \langle_B 0| + \sin\theta_{B1} \langle_B 1|)|\Psi_{AB}\rangle|^2=

\frac{1}{2}|\cos\theta_{A1} \sin\theta_{B1} - \sin\theta_{A1} \cos\theta_{B1}|^2 + \frac{1}{2}|\sin\theta_{A1} \cos\theta_{B1} - \cos\theta_{A1} \sin\theta_{B1}|^2

P(win|x=1,y=1) = \sin^2 (\theta_{A1} - \theta_{B1})

Putting it all together:

P(win) = \frac{1}{4}(P(win|x=0,y=0) + P(win|x=0,y=1)+ P(win|x=1,y=0)+ P(win|x=1,y=1)) =
\frac{1}{4}(\cos^2 (\theta_{A0} - \theta_{B0})+\cos^2 (\theta_{A0} - \theta_{B1})+\cos^2 (\theta_{A1} - \theta_{B0})+\sin^2 (\theta_{A1} - \theta_{B1}))

if Alice and Bob choose \{\theta_{A0}=0, \theta_{A1}=\frac{\pi}{4}, \theta_{B0}=\frac{\pi}{8}, \theta_{B1}=-\frac{\pi}{8}\}, then

P(win) = \frac{3}{4} \cos^2 \frac{\pi}{8} + \frac{1}{4} \sin^2 \frac{3\pi}{8}= \cos^2 \frac{\pi}{8} = \frac{1+cos \frac{\pi}{4}}{2}= \frac{1}{2} + \frac{1}{2\sqrt{2}}\approx 0.85355

and we obtain the winning probability above 85%, which is significantly better than classical 75%.

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Notes on EPR Paradox, Entanglement and Bell’s Inequality

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1. Elements of reality (aka hidden variables) vs uncertainty.

fig1Suppose we have a spin-1/2 particle (ex electron) in the state |n+\rangle (fig. 1). Quantum Mechanics (QM) states that if we measure the spin along z direction we get |z+\rangle with probability cos^2(\theta/2) and |z-\rangle with probability sin^2(\theta/2). There is no way to tell which result will be obtained; the uncertainty of measurement of non-commuting observables is a fundamental law of nature.

Einstein, Podolsky and Rosen (EPR) say – no, there is nothing fundamental here. The particle has elements of reality which determine the measurement result in a unique deterministic way; the particle is either of type (z+,n+) (with probability cos^2(\theta/2)), or of type (z-,n+) (with probability sin^2(\theta/2)), but not of both types at the same time. Statistically the results of spin measurements exactly reproduce the QM predictions.

2. Entanglement.

QM rules out the “elements of reality” because generally they cannot be measured simultaneously; if we have a particle in state |n+\rangle and measure spin along z direction the original state |n+\rangle “collapses” either to |z+\rangle or to |z-\rangle and does not “exist” anymore.

EPR say – no, this does not seem to be true because we can measure a particle’s spin without acting on the particle. Suppose we have the singlet state of two spin-1/2 particles A and B:

\psi = \frac{1}{\sqrt{2}}(|+\rangle_A\otimes|-\rangle_B - |-\rangle_A\otimes|+\rangle_B)

this state is fully invariant under rotations, so it can be written as

\psi = \frac{1}{\sqrt{2}}(|z+\rangle_A\otimes|z-\rangle_B - |z-\rangle_A\otimes|z+\rangle_B)

or

\psi = \frac{1}{\sqrt{2}}(|n+\rangle_A\otimes|n-\rangle_B - |n-\rangle_A\otimes|n+\rangle_B)

We can create an experimental setup such that a spin measurement on the A particle does not affect the B particle in any way, and if we measure spin of the particle A along some direction then spin of the particle B is always opposite, so in the end we measure spin of the particle B without disturbing it.

N.B. Prior to the measurement the particles A and B had no (pure) states of their own, only the common 2-particle singlet state. The measurement does not really “collapse” a state of the particle B because it had no (pure) state before the measurement but instead “create” a (pure) state of the particle B.

3. EPR paradox.

fig2The challenge of EPR is to reproduce all statistical QM predictions using the classical concept of “elements of reality”. As an example let us consider the same singlet state and measure spins of the particles A and B along two different directions z_1 and z_2 (fig.2). QM predicts that the probability of obtaining |z_1+\rangle for the particle A and |z_2+\rangle for the particle B is equal to P(++)=\frac{1}{2}sin^2(\theta/2); the other possibilities are P(+-)=\frac{1}{2}cos^2(\theta/2), P(-+)=\frac{1}{2}cos^2(\theta/2), P(--)=\frac{1}{2}sin^2(\theta/2), all add up to 1.

EPR say: in this experiment we have following statistical mixture of pairs of particles:

  1. a fraction \frac{1}{2}cos^2(\theta/2) of pairs having the particle A of type (z_1+,z_2+) and the particle B of type (z_1-,z_2-)
  2. a fraction \frac{1}{2}sin^2(\theta/2) of pairs having the particle A of type (z_1+,z_2-) and the particle B of type (z_1-,z_2+)
  3. a fraction \frac{1}{2}sin^2(\theta/2) of pairs having the particle A of type (z_1-,z_2+) and the particle B of type (z_1+,z_2-)
  4. a fraction \frac{1}{2}cos^2(\theta/2) of pairs having the particle A of type (z_1-,z_2-) and the particle B of type (z_1+,z_2+)

It can be seen that the mixture reproduces the QM predictions whether we measure the spins along the same direction or along different directions. For example, if we measure the spins of both particles along z_1 then the probability of obtaining + for the particle A and - for the particle B is equal to the sum of fractions (1) and (2), giving \frac{1}{2}cos^2(\theta/2) + \frac{1}{2}sin^2(\theta/2)=\frac{1}{2} as one should expect; if we measure spin of the particle A along z_1 and spin of the particle B along z_2, then the probability of obtaining + for the particle A and - for the particle B is equal to the fraction (1), giving \frac{1}{2}cos^2(\theta/2), again consistent with QM prediction.

It turns out that it is not easy to propose an experiment disproving the EPR paradox.

4. Bell’s inequality.

fig3Let us take the same 2-particle singlet state, but now measure spins along 3 directions (fig.3). Using the EPR’s “elements of reality” framework, the statistical mixture consists of 8 fractions of pairs of the particles A and B:

fraction particle A particle B
P_1 (z_1+,z_2+,z_3+) (z_1-,z_2-,z_3-)
P_2 (z_1+,z_2+,z_3-) (z_1-,z_2-,z_3+)
P_3 (z_1+,z_2-,z_3+) (z_1-,z_2+,z_3-)
P_4 (z_1+,z_2-,z_3-) (z_1-,z_2+,z_3+)
P_5 (z_1-,z_2+,z_3+) (z_1+,z_2-,z_3-)
P_6 (z_1-,z_2+,z_3-) (z_1+,z_2-,z_3+)
P_7 (z_1-,z_2-,z_3+) (z_1+,z_2+,z_3-)
P_8 (z_1-,z_2-,z_3-) (z_1+,z_2+,z_3+)

We choose 2 directions z_i and z_j and measure spin of the particle A along z_i and spin of the particle B along z_j.
Let us denote the probability of obtaining (+) for spin of the particle A and (+) for spin of the particle B as P_{ij}. We have from the table above:

  • P_{12}=P_3+P_4
  • P_{13}=P_2+P_4
  • P_{32}=P_3+P_7

Using little algebra

P_{13}+P_{32}=P_2+P_4+P_3+P_7=P_{12}+P_2+P_7 \ge P_{12},

or finally

P_{12} \le P_{13} + P_{32}

But this inequality is easily violated in QM. Let us consider a planar configuration of the axis z_i and choose \theta_{13}=\theta_{32}=\theta (fig.3), then the inequality becomes

\frac{1}{2}sin^2(\theta) \le sin^2(\theta/2)

which is wrong for any \theta < \pi/2

The result can be formulated as the Bell’s theorem:


No physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics.