The Indistinguishability Postulate and the Copenhagen Interpretation

Question

The indistinguishability postulate states that any (normalized) state vector $|\psi\rangle$ for a system of $N$ identical particles should satisfy $\langle \psi | \hat{O} | \psi \rangle = \langle \hat{P} \psi | \hat{O} |\hat{P} \psi \rangle$ for any observable for the system $\hat{O}$ and any permutation $\hat{P}$ in the permutation group $S_{N}$.

But suppose that there are two photons of which composite state is given by, say, $\frac{1}{\sqrt{2}}(|a\rangle |b\rangle + |b\rangle |a\rangle)$. According to the Copenhagen interpretation, this state will collapse into $|a\rangle|b\rangle$ or $|b\rangle|a\rangle$ if measured. Since both $|a\rangle|b\rangle$ and $|b\rangle|a\rangle$ do not satisfy the above-mentioned postulate, I have difficulty understanding how the Copenhagen interpretation can be consistent with the indistinguishability postulate.

Should one understand the indistinguishability postulate as applying only to ‘pre-measured’ states?

Answer 1

This is similar to a question and answer from the other day.

If $a$ and $b$ represent locations, and the left and right factors in the tensor product represent arbitrary labels for the two photons, then if the system is in the state $\frac{1}{\sqrt{2}}(|a\rangle |b\rangle + |b\rangle |a\rangle)$ and you look for a photon at $a$, you are guaranteed to find one. Since the measurement has only one possible outcome, there is no collapse. If you could tell that the photon you found at $a$ was the "left" one, then the state would collapse to $|a\rangle |b\rangle$, but you can't tell since they're indistinguishable.