# I Computing CHSH violation bound

1. Dec 21, 2017

### facenian

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

2. Dec 21, 2017

### stevendaryl

Staff Emeritus
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.

3. Dec 22, 2017

### facenian

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?

4. Dec 23, 2017

### stevendaryl

Staff Emeritus
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.

5. Dec 23, 2017

### 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.

6. Dec 24, 2017

### stevendaryl

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

7. Dec 24, 2017

### stevendaryl

Staff Emeritus
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 $B = 2n\pi - A$

8. Dec 24, 2017

### facenian

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