I 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 2\sqrt{2} 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 2\sqrt{2} 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:

E(a,b) = - cos(b-a)

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

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

C(\alpha, \beta, \gamma) = -cos(\alpha) - cos(\beta) -cos(\gamma) + cos(\alpha + \gamma - \beta)

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

1. sin(\alpha) - sin(\alpha + \gamma - \beta) = 0
   
2. sin(\beta) + sin(\alpha + \gamma - \beta) = 0
3. sin(\gamma) - sin(\alpha + \gamma - \beta) = 0

We can use some trigonometry:

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

So we have 8 possibilities:

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

These can be solved to yield the possibilities:

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

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

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

So, 2\sqrt{2} 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 \alpha = \gamma = \frac{\pi}{4}, \beta = \frac{7\pi}{4}, and that gives:

C(\alpha, \beta, \gamma) = -2 \sqrt{2}

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

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

 

[facenian]

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

C(\alpha, \beta, \gamma) = -2 \sqrt{2}

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

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

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