Are electron energy transitions instantaneous

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Naty1
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I've been wondering if it is thought energy transitions in an atom are instantaneous or not...as when a photon is absorbed or emitted.

Whatever the answer, is the same case with transitions in molecules?

Seems like creation and annihilation operators might describe what happens, but a quick scan of Wiki did not provide an obvious answer.

It seems the center of a wave packet DOES suffer a delay when reflected from a potential discontinuity while there is no apparent delay with a classical particle. Is this a clue?
 
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I think there is an important difference between treating the entire process as instantaneous (which would not work), versus treating the various contributing amplitudes as instantaneous (which I believe is how the theory does it). Just as in the two-slit problem, we must add the amplitudes over all the possible paths (as in the Feynman path integral approach), and each one has a definite propagation time associated with it, but this does not mean that the whole process of a particle going through two slits has a definite time associated with it. Indeed, this would seem to connect with the uncertainty in the arrival time of the particle, which also relates to the "wave packet" issue you raise. A wave packet already reflects a sum over possible histories, in a sense. If we cannot tell how long something takes to happen without precisely measuring the time, and possibly changing the happening, then how can we say it is instantaneous?
 
Yes, an atomic transition in which an electron emits a photon is instantaneous. The electron is pointlike, and the electron-photon interaction takes place at a single point. A photon cannot be emitted gradually! However as Ken_G says, the interaction can occur at different places and different times, and the amplitudes for each possibility must be summed over.

The situation is not the same for a molecule, or even am atomic nucleus. Since these are extended objects, an electromagnetic transition will not be instantaneous. It will involve a readjustment, which takes a finite amount of time.