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I have a problem I cannot solve on my own. I have given two states $\psi_1$ and $\psi_2$ and an Operator $O$ such that

$P \psi_1 = \epsilon_1 \psi_2$, $P \psi_2 = \epsilon_2 \psi_2$ and $POP^{-1} = \epsilon_3 P$ where $P$ is the parity operator and $\epsilon$ consists of the numbers -1 and +1. For which cases is $< \psi_1 | O | \psi_2 > = 0$?

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closed as off-topic by John Rennie, Sebastian Riese, Gert, user81619, Bill N Nov 11 '15 at 22:56

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The situation is impossible and therefore doesn't happen.

When you claim to have two states $\psi_1$ and $\psi_2$ then I'll assume they are linearly independent, otherwise they aren't really two different states. Then you claim you have an Operator $O$ such that $POP^{-1} = \epsilon_3 P$ where $P$ is the parity operator and further that $P \psi_1 = \epsilon_1 \psi_2$ and $P \psi_2 = \epsilon_2 \psi_2.$ and that $\epsilon$ consists of the numbers -1 and +1.

But if $P \psi_1 = \epsilon_1 \psi_2$ then $P (\epsilon_1\psi_1)= \epsilon_1^2 \psi_2=\psi_2$ and similarly if $P \psi_2 = \epsilon_2 \psi_2$ then $P (\epsilon_2\psi_2)= \epsilon_2^2 \psi_2=\psi_2.$ So we have two distinct vectors mapping to the same vector.

Which means $P (\epsilon_1\psi_1-\epsilon_2\psi_2)= 0.$ Which means there can't be a $P^{-1}$ since $$\begin{align}0&=P^{-1}(0)\\&=P^{-1}(P (\epsilon_1\psi_1-\epsilon_2\psi_2))\\&=\epsilon_1\psi_1-\epsilon_2\psi_2,\end{align}$$ but $0\neq\epsilon_1\psi_1-\epsilon_2\psi_2,$ since the states are linearly independent and the $\epsilon$ consists of the numbers -1 and +1.

So there are no cases where you have two states, some $\epsilon$ and an $O$ and a $P$ that satisfy your assumptions.

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