Why are the wavefunctions of the excited states so symmetric?

Question

The excitation from a lower level to an excited state happens through absorption of a photon. But the photon comes from a certain direction. I would expect that the wavefunction of the excited state has to mirror that circumstance and has some non-symmetry in direction of the incoming photon. Even if the photon is 'bigger' than the atom it has at least an impulse in some direction and one could think that that impulse should be visible in the pattern of the wave function, which would made it nonsymmetric. Can there be a transient wavefunction which reflects the impulse of the photon and rapidly migrates to the known excited state wavefunction?

Answer 1

When a particle (for example an atom) absorbs a photon, it gets a very small momentum kick in the direction in which the photon was traveling. Usually this isn't large enough to make an appreciable difference, but if the particle is cold enough then this can be observed experimentally. See How does one account for the momentum of an absorbed photon? for a more detailed description of this.

In terms of the wavefunction, it's important to specify what part of the wavefunction you are looking at, whether that is the position wave function or the atomic orbital. The position wavefunction will be affected by this momentum kick. The orbital transition will only have to do with the frequency of the photon (and in a more complicated picture, the polarization of the light), and therefore should not depend on the direction of the photon.

Answer 2

Upon absorption of a photon of a particular set of energies, an electron in a particular atomic orbital transitions to an orbital with energy higher than the original orbital. The difference between the energies of the two orbitals is equal to the energy of the absorbed photon. The photon does not affect the ("shape of the") wave function corresponding to any particular atomic orbital. Rather, it merely provides the energy necessary for the transition.