# Parity Operator

by tshafer
Tags: operator, parity
 P: 42 We're working on the parity operator in my second semester quantum mechanics class and there is one point I am confused about, either in the definition of degeneracy or in the parity operator itself. We talked about a theorem whereby the parity operator and the Hamiltonian cannot share simultaneous eigenkets (or, alternatively, wave functions) if there is a degeneracy in the Hamiltonian, regardless of whether or not parity and the Hamiltonian commute. However, I thought that the Hydrogen wave functions have a definite parity (going as whether $$\ell$$ is even or odd), even though the Hydrogen spectrum is highly degenerate ignoring corrections. What am I missing? Thanks! Tom
 Sci Advisor P: 779 Parity Operator If $|\phi\rangle$ is a normalized eigenket of H, then so is $P|\phi\rangle$ if H and P commute. Furthermore, these states are degenerate. Therefore, any linear combination of these states are also eigenstates of H. In particular: $$|\pm\rangle=\frac{1}{\sqrt{2}}\left(|\phi\rangle\pm P|\phi\rangle\right)$$ are degenerate eigenstates with definite parity. So you should always be able to chose eigenstates that have definite parity. There must be more to this theorem...