# The Eigenstates of a Symmetric Operator

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?

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 $$P$$.