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Consider the following question and its solution:

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My question is concerning the solution of $a_{nm}$. Surely if the energy eigenstates are orthogonal then $a_{nm}$ must be equal to zero. WHy is this not the case?

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closed as off-topic by ZeroTheHero, Jon Custer, AccidentalFourierTransform, Cosmas Zachos, Chris May 21 '18 at 2:58

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Orthogonality of the energy eigenstates $|u_n \rangle$ means that for $m \neq n$

$$\langle u_m | u_n \rangle = \int u^*_m u_n dx = 0 $$

It doesn't require that

$$\langle u_m | F | u_n \rangle = \int u^*_m F(x) u_n dx = 0 $$ for any operator F.

So, in general, the quantity $$a_{nm} = \frac{\langle u_m | \delta V(x) | u_n \rangle}{E_n - E_m} $$

is not necessarily zero.

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