# What is Tsirelson's bound for the original bell's inequality

As its well known, quantum correlations break Bell's inequalities only to a certain limit called the Tsirelson's bound. The bound was written for the CHSH inequality.

My question is what is the Tsirelson bound for the original case that Bell dealt with in his article, and how it is proven.

The folklore says the upper bound is 3/2 but I don't remember a reference. I am going to reproduce some old notes of mine demonstrating that bound.

First, Bell's original inequality is eqn (15) of :

$$1+P(\vec{b},\vec{c})\ge|P(\vec{a},\vec{b})-P(\vec{a},\vec{c})|,$$

where $a,b,c$ are unit vectors. The associated quantum prediction is obtained by using eqn (3) which states that

$$P(\vec{x},\vec{y})=-\vec{x}\cdot\vec{y}.$$

So the quantum version of the inequality can be written

$$\underbrace{|\vec{a}\cdot\vec{b}-\vec{a}\cdot\vec{c}|+\vec{b}\cdot\vec{c}}_R}\le 1$$

Your question would then be how badly $R$ can violate that upper bound of 1. By a suitable choice of basis, we can always write

\begin{align} \vec{c}&=(\cos\varphi,\sin\varphi,0),\\ \vec{b}&=(\cos\varphi,-\sin\varphi,0),\\ \vec{a}&=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta), \end{align}

with $\varphi,\phi\in[0,2\pi]$ and $\theta\in[0,\pi]$.

Then

\begin{align} \vec{a}\cdot\vec{b}&=\cos(\phi+\varphi)\sin\theta,\\ \vec{a}\cdot\vec{c}&=\cos(\phi-\varphi)\sin\theta,\\ \vec{b}\cdot\vec{c}&=\cos^2\varphi-\sin^2\varphi, \end{align}

and

$$R=2|\sin\varphi\sin\phi\sin\theta|+1-2\sin^2\varphi.$$

This is a second-order polynomial in $|\sin\varphi|$ whose maximum is reached for $|\sin\varphi|=\frac{1}{2}|\sin\phi\sin\theta|$ and the maximum value is $\frac{1}{2}|\sin\phi\sin\theta|+1.$ This latter bound is then maximum for $\phi=\theta=\pi/2$ and the maximum value is $3/2$. It is therefore reached for $\varphi=\pi/4$.

 John S. Bell. On the Einstein Podolsky Rosen paradox. Physics, 1(3):195–200, 1964. (pdf)

• Thank you, very interesting. But the proof is only for the specific case of spin 1/2 It is still interesting to know what is the Tsirelson's bound for quantum correlations in general. Will it always be 1.5 ??(The known Tsirelson's bound for CHSH inequality was originally shown for any quantum correlations) – Jacob Oct 3 '17 at 10:04
• I mean, is there any general proof that does not use spin properties, but is generally correct for any three quantum variables (each of which can have the value 1 or minus 1) ?? – Jacob Oct 3 '17 at 10:18
• The CHSH inequality only works for observables with $\pm 1$ outcomes as far as I know, and since the upper bound is $2\sqrt{2}$, as opposed to $3/2$ for Bell's original one, there is clearly no universal value valid for any inequality. I really don't see how such an universal value could hold without some constraints on how to build such inequalities. In any case, your question only asked about the original Bell's inequality, not about a general case. – user154997 Oct 3 '17 at 10:54
• To clarify what I mean, I would replace $P(\vec{a},\vec{b})$ by $\langle \psi | A(\vec{a})B(\vec{b}) | \psi \rangle$ where $\psi$ can be any quantum state. – user154997 Oct 3 '17 at 12:42
• You have nothing to be sorry for, it was interesting to see your calculation anyway. If you have proof of what I'm looking for of course I'd love to see. (By the way, I asked Tsirelson himself this question, and he answered that CHSH inequality is more fundamental, and Bell's original inequality is interesting only for historical reasons, so he is not so interested in finding the limit to Bell's inequality .Still, I'm interested in it ...) – Jacob Oct 3 '17 at 14:02

Tsirelson's bound for CHSH works only with matrix sum. If one uses Kronecker sum instead then the eigenvalues are -4,-2,0,2,4. So the average of CHSH can reach 4 in tensor quantum mechanics.