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Good Afternoon,

By definition, an observable $O$ for a system of N identical particles is symmetric just in case $\langle\psi|O|\psi\rangle = \langle\psi|P^{\dagger}OP|\psi\rangle$ for any permutation operator $P$ in the permutation group $S_{N}$. (Or, equivalently, $[P, O] = 0$ for any $P$ in $S_{N}$).

Now, it seems that if $O$ is $not$ symmetric in the above sense, then the eigenstates of $O$ cannot be symmetric as well. (To say that a state vector is symmetric is to say that $\langle\psi|O|\psi\rangle = \langle P\psi|O|P\psi\rangle$ for any relevant $P$.)

This looks intuitively true to me, but is there a way to prove this?

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Assume that $O$ is bounded with point spectrum so that it admits a Hilbert basis of eigenvectors $\psi_n$ and, in the strong operator topology, $$O = \sum_n a_n |\psi_n\rangle\langle \psi_n|.$$ If every eigenvector is also eigenvector of each said $P$, then it is easy to see that (notice that the eigenvalues are $\pm 1$) $P$ commute with $O$. That is equivalent to saying that if $O$ does not commute with some $P$, then $O$ must admit at least an eigenvector that is not eigenvector of that $P$.

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