# Quantum mechanics perturbation and the orthogonality of energy states [closed]

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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?

## 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 $m \neq n$
$$\langle u_m | u_n \rangle = \int u^*_m u_n dx = 0$$
$$\langle u_m | F | u_n \rangle = \int u^*_m F(x) u_n dx = 0$$ for any operator F.
So, in general, the quantity $$a_{nm} = \frac{\langle u_m | \delta V(x) | u_n \rangle}{E_n - E_m}$$