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?

  • $\begingroup$ You referred to a measurement but didn't specify what observable is being measured. What observable $\hat O$ do you have in mind? Does it satisfy the condition $\langle\psi|\hat O|\psi\rangle=\langle \hat P\psi|\hat O|\hat P\psi\rangle$ for all $\hat P$ and all $\psi$? $\endgroup$ Oct 10, 2020 at 3:12
  • $\begingroup$ I had in mind an observable for which $|a\rangle$ and $|b\rangle$ are both eigenstates. $\endgroup$
    – Lory
    Oct 10, 2020 at 6:52
  • $\begingroup$ In that case, have you tried to find an explicit example of an observable $\hat O$ that has $|a\rangle$ and $|b\rangle$ as eigenstates and that also satisfies the condition $\langle\psi|\hat O|\psi\rangle=\langle \hat P\psi|\hat O|\hat P\psi\rangle$ for all $\hat P$ and all two-particle states $\psi$? $\endgroup$ Oct 10, 2020 at 13:36

1 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.

  • $\begingroup$ Thanks, but can we say that after the measurement, what was previously taken as an entangled state is now taken as a mixed state? For the indistinguishability after the measurement seems to be a classical one. $\endgroup$
    – Lory
    Oct 10, 2020 at 6:54

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