# Determining whether $P(ab|xy)$ factorizes in Bell experiments

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:

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?

## 1 Answer

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 $$P(00|00)=\langle++|\frac{1}{4}(\mathbb{I}+X)\otimes(\mathbb{I}+X)|++\rangle=1.$$