4
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.