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Continuing from my previous (1, 2) questions on Brunner et al.'s paper on Bell nonlocality.

Again, we have the following standard Bell experiment setup:

Bell experiment setup

where independent inputs $x,y \in \{0, 1\}$ decide the measurement performed by Alice & Bob on quantum state $S$ with outcomes $a,b \in \{-1, 1\}$. We say $a$ and $b$ are correlated (not independent) if:

$P(ab|xy) \ne P(a|x)P(b|y)$

which is a lazy physicist's way of writing:

$P[A = a \cap B = b | X = x \cap Y = y] \ne P[A = a | X = x] \cdot P[B = b | Y = y]$

Where $A, B, X, Y$ are discrete random variables and $a,b,x,y$ some specific elements from the sets defined above.

I wanted to check this basic (in)equality with some simple example values, so I considered the following:

  • $S = |++\rangle$, a non-entangled quantum state
  • If $X = 0$, Alice measures with $\sigma_z$; if $X = 1$, she measures with $\sigma_x$
  • If $Y = 0$, Bob measures with $\sigma_x$; if $Y = 1$, he measures with $\sigma_z$

Since $S$ is not an entangled state, we can write out the following probability tables:

$\begin{array}{|c|c|c|} \hline x & a & P(a|x) \\ \hline 0 & 1 & 0.5 \\ \hline 0 & -1 & 0.5 \\ \hline 1 & 1 & 1 \\ \hline 1 & -1 & 0 \\ \hline \end{array}$ $\begin{array}{|c|c|c|} \hline y & b & P(b|y) \\ \hline 0 & 1 & 1 \\ \hline 0 & -1 & 0 \\ \hline 1 & 1 & 0.5 \\ \hline 1 & -1 & 0.5 \\ \hline \end{array}$

We then expect $P(ab|xy) = P(a|x)P(b|y)$ for all the values of $a,b,x,y$. The problem is I don't know how to calculate the LHS of that equation! I can make the following table:

$\begin{array}{|c|c|c|c|c|c|} \hline x & y & a & b & P(a|x)P(b|y) & P(ab|xy) \\ \hline 0 & 0 & 1 & 1 & 0.5 \cdot 1 = 0.5 & ? \\ \hline 0 & 0 & 1 & -1 & 0.5 \cdot 0 = 0 & ? \\ \hline 0 & 0 & -1 & 1 & 0.5 \cdot 1 = 0.5 & ? \\ \hline 0 & 0 & -1 & -1 & 0.5 \cdot 0 = 0 & ? \\ \hline 0 & 1 & 1 & 1 & 0.5 \cdot 0.5 = 0.25 & ? \\ \hline 0 & 1 & 1 & -1 & 0.5 \cdot 0.5 = 0.25 & ? \\ \hline 0 & 1 & -1 & 1 & 0.5 \cdot 0.5 = 0.25 & ? \\ \hline 0 & 1 & -1 & -1 & 0.5 \cdot 0.5 = 0.25 & ? \\ \hline 1 & 0 & 1 & 1 & 1 \cdot 1 = 1 & ? \\ \hline 1 & 0 & 1 & -1 & 1 \cdot 0 = 0 & ? \\ \hline 1 & 0 & -1 & 1 & 0 \cdot 1 = 0 & ? \\ \hline 1 & 0 & -1 & -1 & 0 \cdot 0 = 0 & ? \\ \hline 1 & 1 & 1 & 1 & 1 \cdot 0.5 = 0.5 & ? \\ \hline 1 & 1 & 1 & -1 & 1 \cdot 0.5 = 0.5 & ? \\ \hline 1 & 1 & -1 & 1 & 0 \cdot 0.5 = 0 & ? \\ \hline 1 & 1 & -1 & -1 & 0 \cdot 0.5 = 0 & ? \\ \hline \end{array}$

But cannot figure out how to fill in the value of $P(ab|xy)$. How do I do that (without using the values of $P(a|b)P(b|y)$)?

I would then like to perform the same exercise with the CHSH setup:

  • $S = |\Psi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$
  • If $X = 0$, Alice measures with $\sigma_z$; if $X = 1$, she measures with $\sigma_x$
  • If $Y = 0$, Bob measures with $\sigma_z$ rotated $\frac{\pi}{8}$ radians counter-clockwise around the y-axis; if $Y = 1$, he measures with $\sigma_z$ rotated $\frac{\pi}{8}$ radians clockwise around the y-axis

How would we then write out the above three probability tables? I guess we probably wouldn't be able to easily write out the first two, but we can with the third?

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I think you're doing things a little bit backwards. You probably shouldn't be calculating $P(a|x)$ or $P(b|y)$ in advance, because you're simply trying to ask:

Given a set of $\{P(ab|xy)\}$, do there exist assignments to $P(a|x)$ and $P(b|y)$ that satisfy $P(ab|xy)=P(a|x)P(b|y)$ for all $a,b,x,y$?

So, how do you evaluate the probability of getting answers $a$ and $b$ when you make measurements $x$ and $y$? Let's assume your two observables for measurement choices $x$ and $y$ are $\vec{n}\cdot\vec{\sigma}$ and $\vec{m}\cdot{\sigma}$ respectively. Then you have projectors for each of the 4 possible outcomes described by $ab$ with $$ P_{ab}=\frac{1}{4}(\mathbb{I}+(-1)^a\vec{n}\cdot\vec{\sigma})\otimes(\mathbb{I}+(-1)^b\vec{m}\cdot\vec{\sigma}). $$

So, what's $P(ab|xy)$? $$ P(ab|xy)=\langle\psi|P_{ab}|\psi\rangle $$

For example, with $|\psi\rangle=|++\rangle$, and $x=y=0$ (meaning $X$ measurements, as specified in the question), then $$ P(00|00)=\langle++|\frac{1}{4}(\mathbb{I}+X)\otimes(\mathbb{I}+X)|++\rangle=1. $$

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