# Mechanism for a photon to impart momentum to an atom [duplicate]

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?

• 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. – flaudemus Feb 22 '19 at 7:39
• If the linked duplicate does not answer your question, you should edit this post to explain what you're missing from it. – Emilio Pisanty Feb 22 '19 at 8:26
• 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. – Emilio Pisanty Feb 22 '19 at 8:31
• @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. – Mathews24 Feb 22 '19 at 15:09
• 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). – Emilio Pisanty Feb 22 '19 at 15:23

• 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. – Mathews24 Feb 22 '19 at 15:16