TL;DR: How can radiation pressure conserve energy, if we consider the case where the atom absorbs all the Energy of the incident photon via its newly excited electron, and stills gains additional kinetic energy from recoil?

So I have a naive question again. Let's imagine a H atom at rest in its fundamental state, and let's say that it has a Bohr frequency $\omega_0 = \frac{E_1-E_0}{h}$ where $E_1-E_0$ is the difference between the fundamental and the first excited state of the electron.

I would like to pose 2 different but similar problems to illustrate the same thing

Problem 1:

Now, let's say that we throw a photon to the H atom. It has a frequency $\omega_0$ to tickle that sweet spot of resonance for absorption, and let's say our photon does get absorbed by the H atom.

If I look at a naive conservation of energy, I'd say; we lose a photon whose Energy was $h\omega_0$, but the electron gained the same Energy: seems fair to me, I won't complain.

But... Thing is; we have to conserve Momentum as well. Luckily, the dispersion relation for light is not a model of complexity, so we can easily say that our photon must transfer a momentum $\textbf{p}=h\textbf{k} \implies |\textbf{p}|=h\omega/c$

OK, but then, doesn't the hydrogen system acquires a kinetic Energy $Ec=\frac{p^2}{2m}=\frac{h^2\omega^2}{2mc} $?

But then...where does this energy come from? I mean, the electron already gave all of its energy in order to promote the electron to a higher orbital...

Problem 2:

Actually, I realize that it gets worse than that; if I have a photon of frequency $\omega < \omega_0$ (but "not too far"), it still has a certain probability to promote the electron to a higher orbital (it can be estimated through time-dependent perturbation theory). In this case, the conservation of Energy breaks down even without considering the recoil of the atom: a photon with $E<h\omega_0$ is enough to give the electron the Energy $h\omega_0$ it needs to sit in the first excited state.

What am I missing here? My best guess would have something to do with the time-energy uncertainty relation $\Delta E \Delta t > \hbar/2$. Indeed, the electron is not going to stay forever in the excited state. I don't know enough about spontaneous emission theory (I didn't study the quantization of the electric field yet, and what I know of stimulated absorption/emission comes from a perturbative treatment of the interaction between the atom and the EM wave). But I think there must be some kind of time constant $\tau$ associated with the spontaneous deexcitation.

Maybe this characteristic time can reconcile with the "deviation from energy conservation" $\Delta E$ (which was $\frac{h^2\omega^2}{2mc}$ in our first case) in a way that the product remains "of the order of $\hbar/2$"? So basically, the energy conservation would be violated but on a short enough time, so who cares? I'm being deliberately very vague here because I don't really have all the elements to be more precise and it might be flat out wrong

  • $\begingroup$ Thank you, this is an interesting consideration. All the more interesting because.... I think I'm not familiar with the idea of a width for the Energy levels of an atom. In my mind they were strictly quntified to a discrete set of integer values (eigenvalues of the hamiltonian), but I did not know that there is an acceptable width. Could you please point me out to a reference showing how to get/estimate that width? A quick google search with probably inappropriate keywords got me nowhere $\endgroup$ Feb 23, 2019 at 8:47
  • $\begingroup$ @BarbaudJulien width of energy level is a figurative term. Of course, negative eigenvalues of the hydrogen atom Hamiltonian are discrete single real numbers. The width appears already in the sense that interaction of the atom manifests more observable details than just energy eigenvalues, such as line width and line shape. One unremovable source of (emission) line width is the phenomenon of spontaneous emission; the atom in excited state won't stay there indefinitely, but gradually changes into state of lower <E> and finally to the ground state. $\endgroup$ Feb 23, 2019 at 18:27
  • $\begingroup$ @anna v Thanks for the links ! I think this indeed answers my question pretty well. And it seems the most fundamental reason behind the broadening of the levels is Indeed a form of the time-Energy uncertainty, so I Don't feel too off-tracks here :-P. If you want to upgrade your comment to an answer, I am ok to accept it $\endgroup$ Feb 25, 2019 at 10:12
  • $\begingroup$ Thanks for the question! I came here with the same one (and also asking me why I didn't learn about it at the university). $\endgroup$ Feb 7, 2021 at 20:51

3 Answers 3


I explained this in depth in How does one account for the momentum of an absorbed photon?, but here's the short of it.

  1. The kinetic energy comes from the photon itself, i.e. the photon energy $\hbar\omega$ needs to account for all of the energy changes in the absorbing system, and that includes the change in the electron's state as well as the kinetic energy gained by the center-of-mass motion.

    Generally speaking, this is typically a negligible amount: the center-of-mass kinetic energy, while nonzero, will often be smaller than the electronic transition energy by a factor of the order of $10^{-9}$. Unless you've got all of those nine significant figures in stability in your laser frequency, thermal Doppler shifts, the intrinsic atomic linewidth, and everything else in your experiment, it's pointless to worry about that minuscule rounding error.

    (On the other hand, if you do have that much precision, and you often have much more, then you absolutely need to make sure you're tuning your laser to account for that recoil shift.)

  2. if I have a photon of frequency $\omega < \omega_0$ (but "not too far"), it still has a certain probability to promote the electron to a higher orbital (it can be estimated through time-dependent perturbation theory).

    That's incorrect. By saying "I have a photon of frequency $\omega$", you're (implicitly or not) committing to a monochromatic perturbation, and if you do that time-dependent perturbation theory with $\omega$ below the electronic transition energy $\omega_0$ you will get a transition probability of exactly zero.

    On the other hand, if you have a realistic pulse, with a finite time duration $\Delta t$, then you no longer have a monochromatic perturbation, and your guess

    My best guess would have something to do with the time-Energy uncertainty relation $\Delta E \Delta t > \hbar/2$

    is exactly the solution. Having a finite time duration gives your pulse a finite frequency bandwidth, and this allows you to have a central frequency below the transition while still having some of your spectrum overlap with the transition frequency, causing a nonzero effect on the atom.

  • $\begingroup$ Dear Emilio, why the photon "pushes" the atom? Is the electromagnetic field of the exited electron interacting with the photon one? $\endgroup$ Feb 7, 2021 at 21:04
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    $\begingroup$ @LeopoldoSanczyk Ask separately - old comment threads are not the place to start new questions. But as feedback, the question doesn't make that much sense. In a basic sense, it is the electromagnetic field of the photon exerting forces on the charged particles in the atom -- what else could it be? In a more detailed sense, the in-depth explanation is already linked inside this answer. $\endgroup$ Feb 8, 2021 at 11:32
  • $\begingroup$ Of course, however I already read the detailed explanation, and that was not clear to me until now. No one here seems talking about the 'why' of radiation pressure, just taking it for granted and thinking in scenarios, And as you said, very few textbooks include any of this (In mine, I found 1 page in 2 thousands). Such as the author, I learned in the quantum mechanics course that the energies of the absorbed and emitted photon were "the same". Now thanks to your answer I know is "almost the same" and from your feedback, I'm sure is because electromagnetic interaction and not anything else. $\endgroup$ Feb 9, 2021 at 6:02

There is no need to relax conservation of energy (or momentum). If you evaluate the kinetic energy of the atom implied, it is much lower than the excitation energy of the transition (by a factor $\hbar\omega_0/(mc^2)$ which is at most $13.6/2=7$ eV compared to $932$ MeV). So, major part of $\hbar \omega$ goes to excite the atom, and much smaller energy goes to net kinetic energy of the excited atom. There is no requirement that all energy $\hbar \omega$ has to be spent on the excitation; from measurements we know that hydrogen gas absorbs mostly from a thin range of frequencies, some smaller, some higher than $\omega_0$. The effect of kinetic energy increase/decrease depends on the velocity of the atom and manifests as Dopplerian line width broadening.

Regarding the problem 2, energy conservation does not break even here. It is true that radiation of lower frequency than the transition frequency $\omega_0$ can excite the atom/molecules. But there is no requirement that this transition has to be modeled as absorption of a single quantum of energy $\hbar \omega_0$. First, there are always some components of EM field at higher frequencies than $\omega_0$: in the quantum model, this corresponds to a small probability that a quantum $\hbar\omega\gt\hbar\omega_0$ will be absorbed. Second, even if the field was perfectly monochromatic (which in practice it never is), the atom can simply absorb more than one quantum.

  • $\begingroup$ Thank you for your answer ! Regarding the first part; I understand that not all the Energy has to go to the excitation (if $h\omega>h\omega_0$ for example). But if we consider as a thought experiment a photon of Energy exactly $h\omega_0$, then all this Energy is necessary to Promote the electron to the next level, and nothing remains for kinetic Energy? unless we consider a width for the Energy level as Anna v suggests. As to the second part of your answer, considering a multi-photon absorption is an interesting answer that can probably explain a part of my doubts. $\endgroup$ Feb 23, 2019 at 8:38
  • $\begingroup$ But what if we consider a monochromatic EM of frequency $\lambda\omega_0$ with \lambda taken ~1 but irrational so that no integer number of photon can add-up to a multiple of $h\omega_0$ ? I think considering the width of the level will be necessary again here? $\endgroup$ Feb 23, 2019 at 8:39
  • $\begingroup$ In the framework of photons, I guess the interaction may or may not happen, there is no guarantee for either. If it does not, the photon is unchanged. If it does, then there have to be more photons whose total energy is greater than the minimum required. The excess energy may be radiated away by the atom or absorbed in other degrees of freedom (nuclei oscillations). $\endgroup$ Feb 23, 2019 at 18:24

You are missing that the energy levels of an atom have a width. For the photon to be absorbed by the atom, both energy and momentum have to be conserved, energy can be accommodated by the width. The center of the line widths gives the energy that the photon has to have in the center of mass photon-atom. There are special relativity transformations that can give the equivalent energy of the photon for atom at rest .

In the study of transitions in atomic spectra, and indeed in any type of spectroscopy, one must be aware that those transitions are not precisely "sharp". There is always a finite width to the observed spectral lines.

One source of broadening is the "natural line width" which arises from the uncertainty in energy of the states involved in the transition. This source of broadening is important in nuclear spectra,


A typical lifetime for an atomic energy state is about $10^{-8}$ seconds, corresponding to a natural linewidth of about $6.6 \times 10^{-8}$ eV.

Heisenberg's uncertainty principle will give a corresponding narrow width for the line.

Then there is the broadening due to atomic motions:

For atomic spectra in the visible and uv, the limit on resolution is often set by Doppler broadening. With the thermal motion of the atoms, those atoms traveling toward the detector with a velocity $v$ will have transition frequencies which differ from those of atoms at rest by the Doppler shift.

The same will be true for photons approaching to interact with the given atoms. The photon, interacting with the atom in a transition between energy states will fit within the width of the energy level, solving simultaneously for momentum conservation and energy conservation. If the solution is not within the width, there will be no raising or lowering of the energy level. It will be a simple scattering problem.


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