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I have a sketch of the proof of Bell's theorem that goes as follows: Alice measures an observable $A$ along a direction $\vec a$ with possible outcomes $\pm 1$, and Bob does the same for an observable $B$ along $\vec b$ with the same outcomes. Both observables possess the same hidden variable $\lambda$ with normalized probability distribution $\int p(\lambda)d\lambda$.

The expectation value for their measurements once $\vec a$ and $\vec b$ have been fixed is \begin{equation} E(\vec a,\vec b)=\int d\lambda p(\lambda)A(\vec a,\lambda)B(\vec b,\lambda). \end{equation} If we assume that Bob measures in two different directions $\vec b, \vec b'$ without change in the paramter $\lambda$, then \begin{equation}\tag{$*$} E(\vec a,\vec b)-E(\vec a, \vec b')=\int d\lambda(A(\vec a,\lambda)B(\vec b,\lambda)-A(\vec a,\lambda)B(\vec b',\lambda)). \end{equation} My question: how do we get from $(*)$ to the estimate \begin{equation} E(\vec a,\vec b)-E(\vec a, \vec b')\le \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b',\lambda))\right|+ \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b',\lambda))\right| \\ \le 2-| E(\vec a',\vec b')+ E(\vec a',\vec b)|\ ? \end{equation}

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    $\begingroup$ Are you missing a $p(\lambda)$ in (*)? Also, in your final equation, can you confirm that all the primes on $\vec{a}'$ and $\vec{b}'$ are correct? $\endgroup$
    – DaftWullie
    Feb 8 at 7:41
  • $\begingroup$ why do you need integrals, isn't the measure discrete? $\endgroup$
    – Condo
    Feb 8 at 17:59
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    $\begingroup$ Keep in mind that this method of proving Bell's theorem is completely obsolete. The proper way to prove it is to note that the expression is linear on the probabilities, and therefore it is maximized by an extremal point, i.e., a deterministic strategy. You then check all 16 deterministic strategies for the CHSH scenario and then you get the local bound. $\endgroup$ Feb 8 at 19:31
  • $\begingroup$ @DaftWullie Sorry for the late reply. Yes, I was missing $p(\lambda)$ but the primes should be correct. $\endgroup$ Feb 11 at 20:56
  • $\begingroup$ @Condo Good question. I actually don't know... $\endgroup$ Feb 11 at 20:57
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I want to emphasize that you are wasting your time with this historical stuff. But since you insist I'll answer anyway.

First of all, note that the inequality \begin{equation} \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b',\lambda))\right|+ \left|\int d\lambda p(\lambda)(1-A(\vec a', \lambda)B(\vec b,\lambda))\right| \\ \le 2-| E(\vec a',\vec b')+ E(\vec a',\vec b)| \end{equation} doesn't hold (even after correcting the extra prime on the left hand side). The left hand side can reach 4, but the right hand side is at most 2.

Nevertheless, the inequality \begin{equation} E(\vec a,\vec b)-E(\vec a, \vec b')\le 2-| E(\vec a',\vec b')+ E(\vec a',\vec b)| \end{equation} does hold, so we can prove that instead.

First we use the identity \begin{equation} E(\vec a,\vec b)-E(\vec a, \vec b')=\int d\lambda p(\lambda) A(\vec a,\lambda)B(\vec b,\lambda)\big[1-A(\vec a',\lambda)B(\vec b',\lambda)\big]-\int d\lambda p(\lambda) A(\vec a,\lambda)B(\vec b',\lambda)\big[1-A(\vec a',\lambda)B(\vec b,\lambda)\big] \end{equation} Now since $$ \big[1-A(\vec a',\lambda)B(\vec b',\lambda)\big] \ge 0 \quad\text{and}\quad A(\vec a,\lambda)B(\vec b,\lambda) \le 1 $$ we have that $$ \int d\lambda p(\lambda) A(\vec a,\lambda)B(\vec b,\lambda)\big[1-A(\vec a',\lambda)B(\vec b',\lambda)\big] \le \int d\lambda p(\lambda) \big[1-A(\vec a',\lambda)B(\vec b',\lambda)\big].$$ For the second term, note that $$ \big[1-A(\vec a',\lambda)B(\vec b,\lambda)\big] \ge 0 \quad\text{and}\quad -A(\vec a,\lambda)B(\vec b',\lambda) \le 1, $$ so $$-\int d\lambda p(\lambda) A(\vec a,\lambda)B(\vec b',\lambda)\big[1-A(\vec a',\lambda)B(\vec b,\lambda)\big] \le \int d\lambda p(\lambda) \big[1-A(\vec a',\lambda)B(\vec b,\lambda)\big].$$ Putting it all together, we get \begin{align} E(\vec a,\vec b)-E(\vec a, \vec b') &\le \int d\lambda p(\lambda)\big[1-A(\vec a',\lambda)B(\vec b',\lambda)\big] + \int d\lambda p(\lambda) \big[1-A(\vec a',\lambda)B(\vec b,\lambda)\big] \\ &= 2 - E(\vec a',\vec b') - E(\vec a',\vec b) \end{align} Now, repeat the same argument, but starting instead with the identity \begin{equation} E(\vec a,\vec b)-E(\vec a, \vec b')=\int d\lambda p(\lambda) A(\vec a,\lambda)B(\vec b,\lambda)\big[1+A(\vec a',\lambda)B(\vec b',\lambda)\big]-\int d\lambda p(\lambda) A(\vec a,\lambda)B(\vec b',\lambda)\big[1+A(\vec a',\lambda)B(\vec b,\lambda)\big]. \end{equation} You'll see that $$ E(\vec a,\vec b)-E(\vec a, \vec b') \le 2 + E(\vec a',\vec b') + E(\vec a',\vec b). $$ Together with the previous inequality, it implies that \begin{equation} E(\vec a,\vec b)-E(\vec a, \vec b')\le 2-| E(\vec a',\vec b')+ E(\vec a',\vec b)|. \end{equation}

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  • $\begingroup$ Thank you for taking the time. I have a little doubt: why does $p(\lambda)$ 'disappear' in these calculations? $\endgroup$ Feb 24 at 14:59
  • $\begingroup$ You probably forgot to write it down, but really it's irrelevant as due to normalization it just disappears in the final estimate. $\endgroup$ Feb 24 at 15:39
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    $\begingroup$ Sorry, that was just a typo. $\endgroup$ Feb 24 at 15:58

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