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The basis of my confusion is that atomic orbitals, even when described accurately by quantum numbers, have definite energies, meaning they represent energy eigenstates of the wave function. They're stationary states, and thus separable solutions to the Schrödinger equation. The general wave function of an electron, however, is a linear combination of such states, so that general solution does not have a definite energy. Still, we observe discrete spectral lines at specific wavelengths (with some broadening).

Why do we observe spectral lines of specific energy when the wave function of the electrons in an atom are linear combinations of energy eigenstates?

It seems like the spectral lines that are observed are just the energy eigenstates, while, according to quantum mechanics, the wave representation of electrons is a linear combination of those states.

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When an interaction occurs that results in the absorption or emission of a photon, we are not dealing with stationary states. Stationary states are solutions of the time-independent Schrödinger equation, but the absorption and emission of photons is a time-dependent phenomenon.

Spectral lines in fact do not have a definite energy. They have a natural width associated with the time taken for the transition to occur.

The probability that a transition might occur resulting in the absorption or emission of a photon of a given frequency can only be tackled by considering a time-dependent Hamiltonian. What one finds is that transition probabilities become large for frequencies close to the those corresponding to energy differences between stationary states.

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  • $\begingroup$ This is very helpful, where can I find equations showing that the "transition probabilities becomes large for frequencies close to those corresponding to energy differences between stationary states"? $\endgroup$
    – user5341
    Commented Jul 19, 2015 at 18:04
  • $\begingroup$ @user44816 I would just point you to any textbook on quantum mechanics which deals with radiative transitions. $\endgroup$
    – ProfRob
    Commented Jul 19, 2015 at 18:28

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