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Consider an orthonormal set $|\psi_n\rangle$ in the energy basis. Clearly $\langle\psi_n|\psi_m\rangle=\delta_{m,n}$. Does this tell me anything about $\langle\psi_n|\hat{X}|\psi_m\rangle$ or for that matter $|\langle\psi_n|\psi_m\rangle|^2$?

The logical part of my brain is saying no, but the handwavy wavy intuition part of my brain that occasionally is necessary for solving some homework problems says yes. I should be able to say something about $|\psi_n\rangle$'s in the x basis.

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For $\langle \psi_n | X |\psi_m\rangle$ you can't assume anything. These are totally free.

But $|\langle \psi_n |\psi_m\rangle|^2=\langle \psi_n |\psi_m\rangle \overline{\langle \psi_n |\psi_m\rangle} = \delta_{mn}^2=\delta_{mn}$

Note for clarity that there is no sum in the second expression despite repeated indices. Also, the first equality comes from $|z|^2 = z \bar{z}$.

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  • $\begingroup$ Thanks! I didn't mean $|\langle\psi_n|\psi_m\rangle|^2$. Oops! However, I realized that what I was thinking of didn't make any sense when I go to write it out correctly! Anyway, Thanks again! $\endgroup$
    – EEH
    Commented Apr 25, 2023 at 23:16

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