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When an atom or electron absorbs a photon, before the interaction you have an electron and a photon (possibly virtual). Afterwards, you have an excited electron (or atom). How much time does it take for the change?

According to my simple understanding,the change, when it happens, is instantaneous. It would seem that conservation of energy requires that at an exact time the photon goes away and the electron/atom goes to a higher energy state. This does not seem in the spirit of quantum mechanical variables that are usually associated with an uncertainty.

Is there in fact some delta-t for the event duration?

It seems strange that such a classical mechanics notion of a variable, time, carries over into QM.

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  • $\begingroup$ Good answers to this question could distinguish between absorbtion of a photon coming from an EM field in a Fock state, from the case where the EM field is in a coherent state. The processes aren't entirely different, but distinguishing them could help eliminate a very commonly held belief that the universe is full of single photons flying around and interacting with atoms. Fock states do exist, but usually only in a laboratory. In Nature, we mostly have coherent states. $\endgroup$ – DanielSank Apr 27 '18 at 23:24
  • $\begingroup$ ...continuing... The question of energy conservation makes this question even more tricky because the OP's notion of a photon being there or not there is too coarse for quantum mechanics. In quantum, the photon can be in a superposition of there and not there... $\endgroup$ – DanielSank Apr 27 '18 at 23:32
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It could be argued that the amount of time required to emit a photon is essentially equal to the "coherence length" of the photon. Set up an interferometer, send identical photons through it one at a time, and find out how much path length difference is required to prevent the photons from forming an interference pattern. The photon can be thought of as a wave packet of that length. That path length difference is the coherence length of the photons. The photon can be thought of as a wave packet of that length. It is not unusual these days for a laser to have several meters of coherence length, corresponding to ten or more nanoseconds of travel time.

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  • $\begingroup$ I really like this answer because it gives a nice experimentally measurable way to access the emission time. It might be nice to add something about what affects the coherence length, e.g. the atomic transition's dipole moment, etc. $\endgroup$ – DanielSank Apr 28 '18 at 1:20
  • $\begingroup$ The problem with this interpretation is that in the conventional development of the quantized EM field ("photons") they are monochromatic. A wave packet can't be a photon. $\endgroup$ – garyp Apr 28 '18 at 3:10
  • $\begingroup$ An issue with the idea of a photon being monochromatic is that it would need to be associated with an infinitely long train of waves in order to be monochromatic. There would have to be absolutely no uncertainty in its frequency in order for it to be monochromatic. That just doesn't seem to fit the spirit of quantum mechanics! $\endgroup$ – S. McGrew Apr 28 '18 at 4:31
  • $\begingroup$ How could this be argued? And what do you mean by the "coherence length of the photon"? Defining such a property for a single photon is not obvious, and has many subtleties, i.e. you're just hiding all the difficulties in getting an "emission time" inside the word "coherence length". See also physics.stackexchange.com/q/259116/50583 and its linked questions. $\endgroup$ – ACuriousMind Apr 28 '18 at 9:31
  • $\begingroup$ Of course measurement of a single photon can't reveal much more than its energy, time of arrival, and polarization; and can reveal those values only within the limits set by QM uncertainty. The only way to infer the probability distribution of those values (from which in turn the coherence length can be inferred) is to do measurements on a large number of identically generated photons. The "length of time" associated with emission of a photon would, in the QM context, only be measurable statistically. $\endgroup$ – S. McGrew Apr 28 '18 at 15:25

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