# Parity transformation in quantum mechanics

It was written in a book that if parity commutes with Hamiltonian and for some operator $$\hat O$$ if $$P\hat O P^{-1} = -\hat O$$ then $$\langle\hat O\rangle = 0$$.

I know how to show $$\langle\hat O\rangle = 0$$ using the condition $$P\hat O P^{-1} = -\hat O$$. But I do not understand from where this condition ($$P\hat O P^{-1} = -\hat O$$) comes and how it proves that this process is parity conserving?

Apply parity operator from the right side ($$P^{-1}P=I$$). Then $$PO=-OP$$. This means $$PO+OP=0$$ and the Parity operator is anti-commuting with operator $$O$$. This operator can be for instance momentum operator which anti-commutes with parity operator. When an operator anti-commutes with parity then the operator has odd-parity (if commutes it is called even parity). In my opinion, from the given information we cannot understand whether parity is conserved or not. For instance, you need something like this: parity of plus charged pion is odd. Then after the decay of plus charged pion, the products should satisfy this odd parity. I hope this helps.