Probably this has an obvious answer but I did not manage to find it...

When a photon of the appropriate frequency excites an atom, causing an electron transition, it acts as a “driving force”. In doing so, does it behave as a “sinusoidal” driving force or rather as a “sharp impulse”? If the answer is the latter, does it act as a (non-repetitive) delta function or a (periodic) Dirac comb?

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    $\begingroup$ neither? absorption and emission is quantum phenomena, not classical. The "driving force" analogy (that I haven't heard of BTW) is, at most, just an heuristic picture. $\endgroup$ Dec 1, 2016 at 23:37
  • $\begingroup$ @ AccidentalFourierTransform: Hm… The thing is that in a classical system, if you ask “will the system be excited at this frequency”, people will reply with another question: “is your stimulus sinusoidal or sharp?” That is basic, because the way to mathematically model the answer depends on that. Ok, quantum world is different and the question is not asked here. So that must be because its mathematical formulation does not assume any of the two models, can we be positive about that? $\endgroup$
    – Sierra
    Dec 2, 2016 at 1:20

1 Answer 1


I'm guessing the "sharpness" you're alluding to - rather like the persistent, but in many ways inaccurate, notion of a quantum "leap" or "jump" is a manifestation of nonunitary quantum measurement; see also the companion discussion of the quantum measurement problem.

When the quantum system you refer to is undisturbed ("unmeasured" or "unobserved"), it comprises a one-photon state of the electromagetic field coupled with the electronic states in question of the atom in question. The one photon electromagnetic field state can be any one photon state - totally delocalized (a plane, sinusoidal wave), somewhat localized, "sharp" whatever. But the coupled system's quantum state evolves smoothly with time, changing significantly, in this case, over time scales that are typically of the order of nanoseconds (for the case of allowed transitions). I say more about this smooth evolution in this answer here.

But when you observe the atom - or, more precisely, detect its emitted light - you're making measurement (usually in this case with a photodetector) with a "sharp" yes / no answer: the atom has / hasn't already absorbed / emitted the light in question. You can see that the "sharpness" comes wholly from quantum measurement.

The electromagnetic field unitary state can be as "sharp" or as "smooth" as you like: its simply that its initial state governs the probability distribution of when the photodetector is going to detect the transition. If the EM state is "smooth", spread out, sinusoidal and delocalized, then you will observe a wide variety of detection times as you repeat your measurement. More "bunched up" EM field states will lead to a narrower probability distribution of detection times.

  • $\begingroup$ This is all interesting... But I am not sure if it attempts at answering the question and, if so, how it answers it. By "not attempting to answer" I mean what AccidentalFourierTransform did: he said that a quantum system does not follow classical concepts and that is it. Do you adhere to that or by contrast are you attempting to provide an answer? If the latter, is the answer by chance that it depends...? $\endgroup$
    – Sierra
    Dec 4, 2016 at 11:54
  • $\begingroup$ @Sierra I don't think that "driving force" or other classical analogies are accurate, as with AccidentalFourierTransform's comment. I am trying to describe what the quantum situation is: the photon coupling can be "sharp" or "sinusoidal" or almost anything else, depending on the exact state of the EM field. What state it is in exactly a given situation will affect the probabilities and statistics of what you measure. The transition happens in quantum superposition, as I describe further in the answer I reference. You only measure the emission with a quantum measurement, which is what gives ... $\endgroup$ Dec 4, 2016 at 12:20
  • $\begingroup$ ... the appearance of a "jump" $\endgroup$ Dec 4, 2016 at 12:21
  • $\begingroup$ Ok, I understand your answer better now. You're referring to how the transition happens and how it is afterwards measured. But maybe I should have specified that my question concerns not so much how a system absorbs the energy but when it does. As I understand it, a classical one always absorbs the energy sinusoidally; the difference is that it doesn't absorb a sinusoidal influence at 50% of its natural frequency, because half of the way it'd do negative work, but it does absorb sharp impulses repeated at that frequency if made in phase with its oscillation. $\endgroup$
    – Sierra
    Dec 4, 2016 at 12:46

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