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From an answer to this question, https://physics.stackexchange.com/questions/281660/how-does-an-electron-absorb-or-emit-light,

Absorption of a photon will occur only when the quantum energy of the photon precisely matches the energy gap between the initial and final states of the system. (the atom or a molecule as a whole) i.e., by the absorption of a photon, the system could access to some higher permissible quantum mechanical energy state. If there is no pair of energy states such that the photon energy can elevate the system from the lower to the upper energy state, then the matter will be transparent to that radiation.

Given that the photon's energy is proportional to the photon's electromagnetic frequency, and that the frequency is subject to tiny Doppler shifts due to differences in velocity between the emitting and absorbing molecules, how can a photon have precisely the same energy as the exact quantum state change for a gas molecule? Does the photon energy only have to be very close to the quantum state change for a molecule to absorb it? If so, what happens to the extra(or lesser) energy absorbed by the molecule? Does this margin contribute to the width of observed absorption spectra in radio astronomy?

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  • $\begingroup$ In practice, the energy required is variable due to the Uncertainty Principle; see e.g. physics.stackexchange.com/questions/239250/… $\endgroup$
    – Peter Erwin
    Mar 8, 2021 at 19:42
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    $\begingroup$ The observed width of absorption (or emission) lines is mostly due to the combination of pressure broadening and Doppler broadening (from thermal motions of individual atoms/ions, from turbulent motions in gas, etc.). $\endgroup$
    – Peter Erwin
    Mar 8, 2021 at 19:46
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    $\begingroup$ Although the focus is on hydrogen only, I think you could perhaps use (the first part of) chapter 3 in this thesis (I apologize for the blatant self-promotion, and I'll remove this comment if it's deemed out of place). $\endgroup$
    – pela
    Mar 9, 2021 at 8:20
  • $\begingroup$ Isn't semantic at play here? Let aside indetermination, nothing involves the atom from the inner. It just "reads" the photons having different frequency because of its motion. But still there is the same match as in the boxed bold sentence (amended as per the incipit of Prof Rob answer here as well his comment in the thread we are redirect to. I feel that who wrote that is aware, and the sentence should be taken with intrinsic broadening in mind, btw). $\endgroup$
    – Alchimista
    Mar 9, 2021 at 10:24
  • $\begingroup$ @pela I found chapter 3 extremely helpful, though I got a "404" error when trying to access the paper for from link, so I used arxiv.org/abs/1012.3175 instead. $\endgroup$
    – Connor Garcia ♦
    Mar 9, 2021 at 17:01

2 Answers 2

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The Physics SE answer (or the part quoted) was incorrect. The photon does not have to have "precisely" the right energy to cause a transition. The reality is that there is a non-zero probability of causing a transition at all photon energies, but the probability distribution is sharply peaked at the energy we calculate to be the energy difference between the two energy eigenstates of the absorbing atom/molecule.

There are a number of effects that cause this broadening of the frequency response of a bunch of atoms/molecules to photon energy.

The transitions have a "natural width" because the transitions take a finite amount of time. This is encapsulated in a form of the famous uncertainty principle, which means that when viewed over a short time period, there is a fuzziness to energy levels in the atom/molecule. A classical analogy is to try and define the frequency of a damped harmonic oscillator. The larger the damping, the quicker the oscillation dies away and the broader is the continuous frequency spectrum.

Related to this is collisional broadening. Transitions can be interrupted/truncated by collisions, again resulting in a broadened frequency response.

A photon of fixed frequency (note that having a group of photons at fixed frequency is equally impossible, for similar reasons at the emitting end) will encounter atoms/molecules travelling with different velocities and as a result their frequency responses will be Doppler shifted by the appropriate amount.

The application of electromagnetic fields can also broaden and shift the frequency response.

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  • $\begingroup$ I've been told my answer is completely wrong. If you have a chance to skim it could you let me know if you see any glaring errors? Thanks! $\endgroup$
    – uhoh
    Mar 9, 2021 at 13:12
  • $\begingroup$ @uhoh it is wrong as you are mixing quite different thing. But you are right that if you are after the position and momentum of a single atom, sure they still satisfy the indeterminacy principle. Double slit exps are done with much bigger molecules, at the end. But on top of it, there will be a thermal broadening (in a sample). I modestly point to the fact that most of the question is semantic. As also ProfRob says although in a different way. The photon matches with the actual atom. I wouldn't say that your ans is totally wrong. It just merged two unrelated things. $\endgroup$
    – Alchimista
    Mar 10, 2021 at 11:51
  • $\begingroup$ @Alchimista please feel free, if you like, to explain under my question not why you would not include them together in an answer, nor how you might have answered differently, but specifically what precisely is wrong. I think I have expressed things differently than is most commonly done, but I don't see anything that is actually wrong in my answer post. Thanks! $\endgroup$
    – uhoh
    Mar 10, 2021 at 13:07
  • $\begingroup$ @uhoh I do not have to add anything to the comment above. As I said, there isn't really a question to be answered in this thread, at least as formulated. The photon that matches, it matches. Doppler broadening of the absorption spectrum has nothing to do with some uncertainty within the atom. Intrinsic broadening does. In other words, the atom absorbs what is needed, that appears under different frequencies just to our detector. $\endgroup$
    – Alchimista
    Mar 10, 2021 at 14:31
  • $\begingroup$ Basically the very first two comments by Peter Erwin under the question said all. $\endgroup$
    – Alchimista
    Mar 10, 2021 at 14:40
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I'm uncertain of the answer; there seems to be some uncertainty involved in the mechanism, as if there were some kind of principle involved ;-)

I'll never get quantum mechanics, but that's the nature of QM; it simply doesn't work the same way we think the "real world" works. I think the challenge question to this question is "Can a photon even have an exact energy to begin with?"

We have to let go of our own feel for the physical world when we want to learn about how to explain quantum mechanical phenomena.

The only thing you know about the photon is the general conditions under which it arose. You can use a light bulb and a grating, or a narrow spectrum laser, but those only tell you some window in wavelength $\Delta \lambda$ or frequency $\Delta f$.

The energy spread of a group of photons coming out of those devices associated with those spreads is

$$\Delta E \approx \frac{\Delta \lambda}{\lambda}E \approx \frac{\Delta f}{f}E$$

We can calculate a very precise value for the energies of the initial and final states, but for a single atom do those states have precise energies?

Well the ground state of an atom at rest, or at least at very very cold temperature (e.g. laser cooled and trapped), the state where no other transitions are possible or very likely can be pretty well defined, but that upper state is not.

The upper state will have a lifetime, it will last a while and then decay again. We don't know when. Quantum mechanics tells us that if it has a finite lifetime, then it will also have a finite spread in frequencies. The given excited state in a single given atom will have a spread in energies where it can be excited, and the width of that energy $\Delta E$ is related to the width of time that that state will exist before decaying $\Delta t$ by

$$\Delta E \Delta t = \frac{ħ}{2}$$

As @PeterErwin points out this all falls under the uncertainty principle.

Now as @PeterErwin also points out that atom might not be in a near zero velocity, super cool laser cooled trap, but bouncing around in a gas somewhere. That means it will have some uncertainty in speed and direction. Until the transition happens, we don't know which atom it will be, or how fast or which way it will be moving.

So Doppler broadening of the photon due to spread in velocity will contribute to $\Delta E$ as well.

And what's worse is that when atoms collide they screw up each other's quantum states, both by affecting each other's wave functions temporarily (I don't know the name of this effect) and more importantly by shortening the lifetime of the excited state; if you kick it, it just might decay sooner by emitting a photon or by transferring energy to the other atom. That decreases $\Delta t$ and therefore increases $\Delta E$, and is called collisional broadening or pressure broadening.

I guess that first one I didn't know the name of could be called quasi-static pressure broadening?

Basically, nothing is certain in QM; there is no such concept as a precise anything when you want to talk about a single transition event by a single atom. It's all fuzzy, unclear and uncertain. If you want precise, you measure many, many, many times and build up some average.

That's all we've got up here in the macro world, statistical averages of many many many measurements.

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    $\begingroup$ Doppler broadening isn't really a QM effect due to "uncertainty"; it's a population effect due to the fact that different atoms will be moving with different velocities, and so incoming photons will be redshifted or blueshifted different amounts with respect to the atoms. $\endgroup$
    – Peter Erwin
    Mar 8, 2021 at 21:45
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    $\begingroup$ @MeL I'd ask "In what way" but "completely wrong" makes that impossible. And yet there's nothing useful, insightful, helpful or actionable in your comment. Therefore by itself it lacks any ascertainable credibility. $\endgroup$
    – uhoh
    Mar 9, 2021 at 0:51
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    $\begingroup$ @MeL It might be helpful if you were more specific or, even better, posted an alternative answer. Do you dispute Uhoh's claim, that for high energy state atoms: "The upper state will have a lifetime, it will last a while and then decay again."? $\endgroup$
    – Connor Garcia ♦
    Mar 9, 2021 at 1:09
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    $\begingroup$ @uhoh Thermal/Doppler broadening is not a single-event phenomenon. It is the result of the superposition of many Doppler-shifted absorption or emission events (each of which will be intrinsically "QM-broadened" by "natural-width" and "collisional" effects), due to an ensemble of atoms/ions/molecules moving with different (e.g., thermally distributed) velocities. $\endgroup$
    – Peter Erwin
    Mar 9, 2021 at 13:58
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    $\begingroup$ @uhoh If you want to talk about single excitations of single atoms, then just don't mention "thermal broadening"; it has nothing to do with that. (You can't really talk about the "temperature" of a single atom; it's a thermodynamic property of an ensemble.) $\endgroup$
    – Peter Erwin
    Mar 9, 2021 at 19:36

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