2
$\begingroup$

I have been trying to understand Gisin's paper, Bell's inequality holds for all non-product states. As I've worked through the proof he provides for his theorem, I've run into two things that simply seem wrong. Can someone please clarify either by showing me where I'm making a mistake or confirming that his paper is wrong? I list my concerns below.

First, Gisin states the theorem assuming $$P(a, b) = \langle (2a-1) \otimes (2b-1) \rangle_\psi, $$ but never specifies what $a$ and $b$ are. He calls them projectors, which seemingly implies $a = | a \rangle \langle a | = \mathbf{a} (\mathbf{a}^\dagger)$, $b = | b \rangle \langle b | = \mathbf{b} (\mathbf{b}^\dagger)$. But then $(2a-1) \otimes (2b-1)$ is a 9x9 matrix, while $\psi$ is (presumably) 4D, making the calculation of $\langle (2a-1) \otimes (2b-1) \rangle_\psi = \mathbf{\psi}^\dagger [(2a-1) \otimes (2b-1)] \mathbf{\psi}$ impossible.

Replacing $\rho$ with $\psi^*\psi$ in Bell's definition of $P$ (equation 2 in his paper On the Einstein Podolsky Rosen Paradox), we get $$P(\mathbf{a}, \mathbf{b}) = \int \psi^*(\lambda) A(\mathbf{a}, \lambda) B(\mathbf{b}, \lambda) \psi(\lambda) \,d\lambda\,, $$ similar to the analytic expression for the expectation value of the spin correlation operator (sorry if that's not what it's actually called—I don't know it's official name). In bra-ket notation it would be $$P(\mathbf{a}, \mathbf{b}) = \langle (\mathbf{a} \cdot \mathbf{\sigma}) \otimes (\mathbf{b} \cdot \mathbf{\sigma}) \rangle_\psi. $$ Using this expression for $P$ instead of the one Gisin provides I can follow most of the proof, and the fact that he himself mentions this value on page 202 (though in a context that seems unrelated to $P$) is a hint that maybe this is what he intended. But then his proof doesn't actually prove his theorem (involving $P = \langle (2a-1) \otimes (2b-1) \rangle$)—it proves a similar theorem involving $P = \langle (\mathbf{a} \cdot \mathbf{\sigma}) \otimes (\mathbf{b} \cdot \mathbf{\sigma}) \rangle$. So what are $a$, $a'$, $b$, and $b'$? Are they supposed to relate to $\mathbf{a} \cdot \mathbf{\sigma}$ and $\mathbf{b} \cdot \mathbf{\sigma}$? Why does he use $2a-1$ and $2b-1$?

Second, Gisin states, as the final step of his proof, that $$|P(a, b) - P(a, b')| + P(a', b) + P(a', b') = 2(1+4|c_1 c_2|)^{-\frac{1}{2}} > 2.$$ However, for all values of $c_1$ and $c_2$, I get $$2(1+4|c_1 c_2|)^{-\frac{1}{2}} \leq 2.$$ What is going on here?

All help is appreciated.

$\endgroup$
  • $\begingroup$ Isn't this not what is normally referred to as Gisin's theorem? I thought when people referred to Gisin's theorem, they meant a no-go theorem for nonlinear quantum mechanics, Gisin, "Weinberg's non-linear quantum mechanics and supraluminal communications," unige.ch/gap/quantum/publications:bib:gisin1990 $\endgroup$ – Ben Crowell May 10 at 3:25
  • $\begingroup$ @BenCrowell You could be right. When I have talked about this with my research advisor, he has offhandedly called it Gisin's theorem, so I assumed that was what it was known as. If you have a suggestion for a better title, I'd appreciate it. $\endgroup$ – The Ledge May 10 at 15:52
2
$\begingroup$

After some more research I was able to find satisfactory answers to my questions.

First answer: I was able to find that the projection operators $a$ and $b$ can be written as $$ a = \frac{1}{2}(1 + \mathbf{a}\cdot\mathbf{\sigma} ),\quad b = \frac{1}{2}(1 + \mathbf{b}\cdot\mathbf{\sigma} ), $$ from which easily follows $$ 2a - 1 = \mathbf{a} \cdot \mathbf{\sigma},\quad 2b - 1 = \mathbf{b} \cdot \mathbf{\sigma}. $$ Then, $\langle (2a - 1) \otimes (2b - 1) \rangle $ is equivalent to $ \langle \mathbf{a} \cdot \mathbf{\sigma} \otimes \mathbf{b} \cdot \mathbf{\sigma} \rangle, $ so Gisin's proof is consistent with the statement of his theorem.

Second answer: Applying the same assumptions as Gisin to relate the vectors $\mathbf{a},$ $\mathbf{a}',$ $\mathbf{b},$ and $\mathbf{b}'$ to the angles $ \alpha, $ $ \alpha', $ $ \beta, $ and $ \beta', $ we have $$ S = |P(a, b) - P(a, b')| + P(a', b) + P(a', b') = |\cos\beta - \cos\beta'| + 2|c_1 c_2|(\sin\beta + \sin\beta'). $$ Using $ \sin\beta = (1 - \cos^2\beta)^{-\frac{1}{2}} $ to solve $ \frac{\partial}{\partial \beta} S = 0 $ for $ \cos\beta, $ we see that $S$ is maximized when $$ \cos\beta = -\cos\beta' = (1 + 4|c_1 c_2|^2)^{-\frac{1}{2}}. $$ Using $ \cos\beta = (1 - \sin^2\beta)^{-\frac{1}{2}} $ to solve this equation for $ \sin\beta $ and $ \sin\beta', $ we obtain $$ \sin\beta = \sin\beta' = \Bigg(1 + \frac{1}{1 + 4|c_1 c_2|^2}\Bigg)^{-\frac{1}{2}} $$ $$ \implies S = \frac{2}{(1 + 4|c_1 c_2|^2)^{-\frac{1}{2}}} + 4|c_1 c_2|\Bigg(1 - \frac{1}{1 + 4|c_1 c_2|^2}\Bigg)^{-\frac{1}{2}} = 2(1 + 4|c_1 c_2|^2)^{-\frac{1}{2}} > 2. $$ This final expression differs from Gisin's result by a power of $-1$ outside the parentheses and a power of $2$ inside. This looks like it's just a typo on Gisin's part.

$\endgroup$

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.