# The Indistinguishability Postulate and the Copenhagen Interpretation

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?

• 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 $\psi$? Oct 10 '20 at 3:12
• I had in mind an observable for which $|a\rangle$ and $|b\rangle$ are both eigenstates. Oct 10 '20 at 6:52
• 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 $\psi$? Oct 10 '20 at 13:36

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.