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Picture a simple hydrogen atom with one electron which is bound to a proton (nucleus). When trying to impart momentum to the atom, we may specify the photon wavelength to be $\lambda = 121.57$ nm to access the 1s-2s electronic transition. But in this picture, where exactly does the photon's momentum go and how does it get there?

We know that the electron can only occupy quantized states and the transition is instantaneous. The photon energy has decreased the electric potential energy between the proton-electron pair by $hc/\lambda$. But do not the quantized orbits occupied by electrons have well-defined momentum, too? Since the electron's momentum now that it has gone from 1s to 2s may not necessarily be the photon's total original momentum, $h/\lambda$, where does the rest of this momentum go? And is this momentum transfer an instantaneous process as well?

I believe a partial answer is that the entire atom has momentum imparted to it, but what is the mechanism to make the nucleus start moving if the photon was wholly absorbed by just the electron and the electron cannot simply drift off its quantized orbit (which would then pull on the nucleus via Coulomb interaction)? Perhaps answering this question may help elucidate the above: is imparting momentum by photons with wavelengths corresponding to particular transitions more efficient than photons with arbitrary wavelength?

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marked as duplicate by Emilio Pisanty electromagnetism Feb 22 at 8:25

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    $\begingroup$ Perhaps it helps to emphasize here, that we need to distinguish linear momentum and angular momentum here. The electrons in the atom have an angular momentum, but no linear momentum in the rest frame of the nucleous. The momentum transferred from the photon to the electron can be two-fold: there is linear momentum transfer, and there can be angular momentum transfer. $\endgroup$ – flaudemus Feb 22 at 7:39
  • $\begingroup$ If the linked duplicate does not answer your question, you should edit this post to explain what you're missing from it. $\endgroup$ – Emilio Pisanty Feb 22 at 8:26
  • $\begingroup$ As for the process being instantaneous - no, it's not. That's a handy mental picture for some situations, but it has no real basis in reality, starting with the fact that it's incompatible with the time-energy uncertainty principle. $\endgroup$ – Emilio Pisanty Feb 22 at 8:31
  • $\begingroup$ @EmilioPisanty To clarify, is there a relevant time-scale to the momentum transfer process besides that given by the uncertainty principle? And yes, it does mostly answer my question coupled with the below response. $\endgroup$ – Mathews24 Feb 22 at 15:09
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    $\begingroup$ The momentum-transfer process is intrinsically tied to the electronic transition and it doesn't have any timescale of its own. However, quantifying "when" an electronic transition happens, or how long it takes, is a tricky and subtle question to even formulate correctly - and it is also one of the places where cutting-edge work is happening (for recent examples, see e.g. this and this papers, though note that those are for bound-to-continuum transitions, and bound-bound transitions are different in several key ways). $\endgroup$ – Emilio Pisanty Feb 22 at 15:23
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First of all let us make it clear that the atom is a quantum mechanical entity, and there are no orbits around the center of mass of electron-nucleus, which look like planetary orbits. There are orbitals, quantum mechanical probability locations , i.e. after many measurements for the hydrogen atom, this is where the electron (dot) will be found:

orbital

As far as momentum goes, conservation of momentum implies that the whole atom acquires a momentum when a photon transition happens with the correct ( within a width) frequency.

It is the atom that is interacting with the photon, not the individual electron.

If the photon does not have an energy level frequency, it may interact with the neutral atom through scattering off spill over fields, transferring momentum and energy to the whole atom, with very low probability. In various energy regions there are different names for this. Special calculations are needed , for example here, according to the energy of the photons.

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  • $\begingroup$ @Farcher I found the arxiv $\endgroup$ – anna v Feb 22 at 8:42
  • $\begingroup$ That was very helpful. Could you possibly expand upon this statement: "It is the atom that is interacting with the photon, not the individual electron." For example, if I have a Rydberg atom where the n = 137 state of hydrogen has an atomic radius of approximately $d = 1 \mu$m, then by causality it will take a finite amount of time, $\tau \approx d/c$, for the photon to interact with the nucleus if it encounters the electron directly first, correct? Will not momentum be imparted to the electron first in this picture? I'm just trying to better understand the sequence of events here. $\endgroup$ – Mathews24 Feb 22 at 15:16
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    $\begingroup$ The atom is one quantum mechanical entity.We usually talk of the electron changing levels but it is the whole atom that changes energy level. In the orbitals we see the electron spot because the nucleus is so small that its orbitals will map to the center anyway. The electron is not separable from the nucleus except probabilistically. It encounters the atom, and there is a probability for transition or scatter (if not with a frequency of an energy level).. the correct theory can be seen here arxiv.org/abs/1606.00939 . also in the link given as duplicate $\endgroup$ – anna v Feb 22 at 16:00

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