In textbooks, the violation of Bell's inequality is usually demonstrated for spin-1/2 systems in the state $$| \psi \rangle = \frac{1}{\sqrt{2}} \left( | +- \rangle - | -+ \rangle \right).$$

Most experiments are done with photons and there, usually the state $$| \psi \rangle = \frac{1}{\sqrt{2}} \left( | ++ \rangle + | -- \rangle \right)$$ is used.

Is this just a random convention or do these state have special properties compared to the other Bell states? Do the the states correspond to each other somehow if we take the differences between the spin-1/2 case and the photon polarization case into account?

| cite | improve this question | | | | |
  • $\begingroup$ What do you mean by "spin-1/2 photon states?" A photon has integer spin. $\endgroup$ – user4552 Nov 8 '19 at 22:13
  • $\begingroup$ @Ben Crowell: I didn't write "spin-1/2 photon states". Maybe my phrasing was a bit misleading and I should have written "Most experiments aren't done with spin-1/2 systems but with photons". $\endgroup$ – Marc Nov 12 '19 at 11:55

Any maximally entangled state will give the same amount of violation (modulo choosing suitable measurement directions).

You can see that this must be the case because any maximally entangled state can be made into another via local operations. This means that given any pair of maximally entangled states $|\Psi\rangle,|\Phi\rangle$, there is always a local unitary $U$ such that $$|\Psi\rangle=(I\otimes U)|\Phi\rangle.$$ You can check this is the case for your examples with $U=X\equiv\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Local operations do not affect the amount of entanglement shared by two parties. If you have a set of measurements corresponding to an inequality for $|\Phi\rangle$, the second party can just apply $U$ on their share of the state before doing the same set of measurement, and the same results and correlations will be observed.

| cite | improve this answer | | | | |
  • $\begingroup$ Thanks for this interesting point! I'm still not completely satisfied because I don't see the reason why different states are used. If I were to perform a Bell test experiment why wouldn't I use the first state of my question above? Why does one state seem more obvious for theoretical analyses and the other for experiments? $\endgroup$ – Marc Nov 12 '19 at 11:58
  • $\begingroup$ it doesn't, they are fully equivalent for both theoretical and experimental purposes. It's totally up to convention what physical state you associate to "$|0\rangle$" and "$|1\rangle$". From a theoretical point of view, these states have the exact same kind of structure, up to a redefinition of what the kets represent. It makes sense to consider them differently only if more than one is used at the same time $\endgroup$ – glS Nov 12 '19 at 12:08

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.