Computing CHSH violation bound

[facenian]

It seems that the upper bound of the CHSH inequality is ##2\sqrt{2}##
How is it analytically derived?

 

[stevendaryl]

It seems that the upper bound of the CHSH inequality is ##2\sqrt{2}##
How is it analytically derived?

That's not an upper bound of the CHSH inequality. The upper bound is 2. The fact that actual correlations are [itex]2\sqrt{2}[/itex] shows that the inequality is violated by QM.

 

[facenian]

That's not an upper bound of the CHSH inequality. The upper bound is 2. The fact that actual correlations are [itex]2\sqrt{2}[/itex] shows that the inequality is violated by QM.

I'm sorry I did not express it correctly. I was talking about the quantum mechanical bound not the hidden variable bound.
How is ##2\sqrt{2}## analytically calculated?

 

[stevendaryl]

I don't know how to prove it for an arbitrary QM system, but in the particular case of anti-correlated spin-1/2 particles, I can prove it.

In that case, we have:

[itex]E(a,b) = - cos(b-a)[/itex]

So the quantity of interest is:
[itex]C(a,b,a',b') = E(a,b) + E(a,b') + E(a', b') - E(a',b)[/itex]
[itex]= -cos(b-a) -cos(b'-a) -cos(b'-a') + cos(b-a') [/itex]

If we let [itex]\alpha = b-a, \beta = b'-a, \gamma = b'-a'[/itex], then we have:

[itex]C(\alpha, \beta, \gamma) = -cos(\alpha) - cos(\beta) -cos(\gamma) + cos(\alpha + \gamma - \beta)[/itex]

For a minimum or maximum, the partial derivatives with respect to [itex]\alpha, \beta, \gamma[/itex] must all be zero. This implies:

1. [itex]sin(\alpha) - sin(\alpha + \gamma - \beta) = 0[/itex]
   
2. [itex]sin(\beta) + sin(\alpha + \gamma - \beta) = 0[/itex]
3. [itex]sin(\gamma) - sin(\alpha + \gamma - \beta) = 0[/itex]

We can use some trigonometry:

• If [itex]sin(A) = sin(B)[/itex], then either [itex]A=B[/itex] or [itex]B = \pi - A[/itex]
• If [itex]sin(A) = -sin(B)[/itex], then either [itex]A=-B[/itex] or [itex]B = \pi + A[/itex]

So we have 8 possibilities:

1. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
2. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi -(\alpha + \gamma - \beta)[/itex]
3. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
4. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi - (\alpha + \gamma - \beta)[/itex]
5. [itex]\alpha = \pi - (\alpha + \gamma - \beta)[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
6. [itex]\alpha = \pi - (\alpha + \gamma - \beta)[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi - (\alpha + \gamma - \beta)[/itex]
7. [itex]\alpha = \pi -(\alpha + \gamma - \beta)[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
8. [itex]\alpha = \pi - (\alpha + \gamma - \beta)[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi - (\alpha + \gamma - \beta)[/itex]

These can be solved to yield the possibilities:

1. [itex]\alpha = \beta = \gamma = 0[/itex]
2. Impossible
3. Impossible
4. [itex]\alpha = 0, \beta = \gamma = \pi[/itex]
5. Impossible
6. [itex]\alpha = \gamma = 0, \beta = -\pi[/itex]
7. [itex]\alpha = \beta = \pi, \gamma = 0[/itex]
8. [itex]\alpha = \gamma = \frac{3\pi}{4}, \beta = \frac{5\pi}{4}[/itex]

To find out whether these are minima or maxima, we plug them back into the expression for [itex]C(\alpha, \beta, \gamma)[/itex]

1. [itex]C(\alpha, \beta, \gamma) = -1-1-1+1 = -2[/itex]
2. Impossible
3. Impossible
4. [itex]C(\alpha, \beta, \gamma) = -1+1+1+1 = +2[/itex]
5. Impossible
6. [itex]C(\alpha, \beta, \gamma) = -1+1-1-1 = -2[/itex]
7. [itex]C(\alpha, \beta, \gamma) = +1+1-1+1 = +2[/itex]
8. [itex]C(\alpha, \beta, \gamma) =+\sqrt{2}/2 +\sqrt{2}/2 +\sqrt{2}/2+\sqrt{2}/2 = 2\sqrt{2}[/itex]

So, [itex]2\sqrt{2}[/itex] is the biggest value that you can get with anti-correlated spin-1/2 particles. It's not at all obvious to me why that is the best you can possibly do with any QM system.

 

[facenian]

Excellent! Thank you very much.
There is one more question however, since 2 and -2 occur it seems that ##-2\sqrt{2}## should also appear.

 

[stevendaryl]

Excellent! Thank you very much.
There is one more question however, since 2 and -2 occur it seems that ##-2\sqrt{2}## should also appear.

Now, that makes me think I must have done something wrong. Let me recheck it.

 

[stevendaryl]

Just trying some values, I found that there is a minimum at [itex]\alpha = \gamma = \frac{\pi}{4}, \beta = \frac{7\pi}{4}[/itex], and that gives:

[itex]C(\alpha, \beta, \gamma) = -2 \sqrt{2}[/itex]

So, it looks I should have used a more general condition:

• If [itex]sin(A) = sin(B)[/itex], then either [itex]B = A + 2n\pi[/itex] or [itex]B = (2n+1)\pi - A[/itex]
• If [itex]sin(A) = -sin(B)[/itex], then either [itex]B = A + (2n+1)\pi[/itex] or [itex]B = 2n\pi - A[/itex]

 

[facenian]

Just trying some values, I found that there is a minimum at [itex]\alpha = \gamma = \frac{\pi}{4}, \beta = \frac{7\pi}{4}[/itex], and that gives:

[itex]C(\alpha, \beta, \gamma) = -2 \sqrt{2}[/itex]

So, it looks I should have used a more general condition:

• If [itex]sin(A) = sin(B)[/itex], then either [itex]B = A + 2n\pi[/itex] or [itex]B = (2n+1)\pi - A[/itex]
• If [itex]sin(A) = -sin(B)[/itex], then either [itex]B = A + (2n+1)\pi[/itex] or [itex]B = 2n\pi - A[/itex]

Yes, and it is only a minor detail. I seems reasonable to accept only solutions in the range ##[0,2\pi]## so maybe instead of ##B=\pi-A## putting ##B=\pi\pm A## will suffice