# Which spin-1/2 photon states can be used in the context of Bell's inequality violations?

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?

• What do you mean by "spin-1/2 photon states?" A photon has integer spin. – user4552 Nov 8 '19 at 22:13
• @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". – Marc Nov 12 '19 at 11:55

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.
• 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 – glS Nov 12 '19 at 12:08