<p>I'm <em>uncertain</em> of the answer; there seems to be some <em>uncertainty</em> involved in the mechanism, as if there were some kind of <em>principle</em> involved ;-)</p>
<p>I'll never get quantum mechanics, but that's the nature of QM; it simply doesn't work the same way we think the &quot;real world&quot; works. I think the challenge question to this question is &quot;Can a photon even have an exact energy to begin with?&quot;</p>
<p>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.</p>
<hr />
<p>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 <span class="math-container">$\Delta \lambda$</span> or frequency <span class="math-container">$\Delta f$</span>.</p>
<p>The energy spread of a group of photons coming out of those devices associated with those spreads is</p>
<p><span class="math-container">$$\Delta E \approx \frac{\Delta \lambda}{\lambda}E \approx \frac{\Delta f}{f}E$$</span></p>
<p>We can <em>calculate</em> a very precise value for the energies of the initial and final states,  but for a single atom do those states have precise energies?</p>
<p>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.</p>
<p>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 <span class="math-container">$\Delta E$</span> is related to the width of  time that that state will exist before decaying <span class="math-container">$\Delta t$</span> by</p>
<p><span class="math-container">$$\Delta E \Delta t = \frac{ħ}{2}$$</span></p>
<p>As @PeterErwin <a href="https://astronomy.stackexchange.com/questions/41858/how-can-a-photon-have-exactly-the-right-energy-to-be-absorbed-by-a-gas-molecule#comment85933_41858">points out</a> this all falls under the <a href="https://en.wikipedia.org/wiki/Uncertainty_principle" rel="nofollow noreferrer">uncertainty principle</a>.</p>
<p>Now as @PeterErwin <a href="https://astronomy.stackexchange.com/questions/41858/how-can-a-photon-have-exactly-the-right-energy-to-be-absorbed-by-a-gas-molecule#comment85934_41858">also points out</a> that atom <em>might not be</em> 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.</p>
<p>So <a href="https://en.wikipedia.org/wiki/Doppler_broadening" rel="nofollow noreferrer">Doppler broadening</a> of the photon due to spread in velocity will contribute to <span class="math-container">$\Delta E$</span> as well.</p>
<p>And <em>what's worse</em> 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 <span class="math-container">$\Delta t$</span> and therefore increases <span class="math-container">$\Delta E$</span>, and is called <a href="https://en.wikipedia.org/wiki/Spectral_line#Pressure_broadening" rel="nofollow noreferrer">collisional broadening</a> or pressure broadening.</p>
<p>I guess that first one I didn't know the name of could be called quasi-static pressure broadening?</p>
<p>Basically, nothing is certain in QM; <strong>there is no such concept as a precise anything when you want to talk about a single transition event by a single atom.</strong> It's all fuzzy, unclear and uncertain. If you want precise, you measure many, many, many times and build up some average.</p>
<p>That's all we've got up here in the macro world, statistical averages of many many many measurements.</p>

CC BY-SA 4.0
2021-03-08T21:45:38.667Z
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.
Peter Erwin
CC BY-SA 4.0
2021-03-08T21:55:12.927Z
@PeterErwin you would calculate the average effect over a billion incident photons using a thermal distribution yes, but for a single transition where we don't know which atom will be involved nor which way it was going, I'm not yet convinced that no quantum mechanical treatment is called for. Question is about a single event.
uhoh
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2021-03-09T00:40:27.973Z
@uhoh ‘s answer to the question about the photon energy is completely wrong.
MeL
CC BY-SA 4.0
2021-03-09T00:51:04.340Z
@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*.
uhoh
CC BY-SA 4.0
2021-03-09T01:09:40.267Z
@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."?
Connor Garcia
CC BY-SA 4.0
2021-03-09T13:58:17.320Z
@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.
Peter Erwin
CC BY-SA 4.0
2021-03-09T15:33:43.803Z
@PeterErwin While you’re describing how people most often *encounter* these concepts analyzing data of a large number of photons from a dense gas, let’s keep focus on the OP's question about a single excitation of a single atom. That atom will certainly have an uncertainty in its velocity, and that distribution might be characterizable by a temperature. Yes most people usually use these concepts in the macroscopic world and ensemble averages, but that doesn’t mean you can’t still talk about it in the context of this question. The question isn't "How do astronomers...?"
uhoh
CC BY-SA 4.0
2021-03-09T19:36:35.723Z
@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.)
Peter Erwin
CC BY-SA 4.0
2021-03-09T23:29:03.553Z
@PeterErwin you really can. While we more often talk about temperature when characterizing an ensemble average of kinetic energies (and therefore velocities), we can also use it to characterize a time average of kinetic energies and probability for ONE atom to have a given energy and therefore velocity. All that's necessary is that the atom is known to be in contact with other atoms and that that population is in thermal equilibrium. Thermal/Doppler CAN be treated microscopically a single-event phenomenon in answers about quantum mechanics, even though astronomers need ensembles for day-to-day
uhoh
CC BY-SA 4.0
2021-03-09T23:42:43.017Z
@PeterErwin As an aside, I'm going to go ahead and "put my head in the lion's mouth" and write a question in Physics SE that's related to all of this. I'll ping you when I do. And though this is about the internal excitation degrees of freedom and not related here, you might find it interesting: [Has a concept of temperature ever been defined in the context of a single atom?](https://chemistry.stackexchange.com/q/125177/16035)
uhoh
CC BY-SA 4.0
2021-03-08T21:33:45.793Z
2021-03-08T21:39:10.410Z
2021-03-08T21:39:10.410Z
uhoh
<p>The Physics SE answer (or the part quoted) was incorrect. The photon does not have to have &quot;precisely&quot; the right energy to cause a transition. The reality is that there is a non-zero probability of causing a transition at <em>all</em> photon energies, but the probability distribution is <em>sharply peaked</em> at the energy we calculate to be the energy difference between the two energy eigenstates of the absorbing atom/molecule.</p>
<p>There are a number of effects that cause this broadening of the frequency response of a bunch of atoms/molecules to photon energy.</p>
<p>The transitions have a &quot;natural width&quot; 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 <em>continuous</em> frequency spectrum.</p>
<p>Related to this is collisional broadening. Transitions can be interrupted/truncated by collisions, again resulting in a broadened frequency response.</p>
<p>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.</p>
<p>The application of electromagnetic fields can also broaden and shift the frequency response.</p>

CC BY-SA 4.0
2021-03-09T13:12:50.213Z
I've been told my answer is [completely wrong](https://astronomy.stackexchange.com/questions/41858/how-can-a-photon-have-exactly-the-right-energy-to-be-absorbed-by-a-gas-molecule/41865#comment85954_41860). If you have a chance to skim it could you let me know if you see any glaring errors? *Thanks!*
uhoh
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2021-03-10T11:51:07.230Z
@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.
Alchimista
CC BY-SA 4.0
2021-03-10T13:07:00.497Z
@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!*
uhoh
CC BY-SA 4.0
2021-03-10T14:31:38.157Z
@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.
Alchimista
CC BY-SA 4.0
2021-03-10T14:40:32.247Z
Basically the very first two comments by Peter Erwin under the question said all.
Alchimista
CC BY-SA 4.0
2021-03-11T03:17:18.130Z
@Alchimista and those have been addressed. The treatment may be unfamiliar because in astronomy these things are usually taught in an *applied way* "this is what you do to analyze astronomical data". When people see it in a different context the first reaction is "No no no! It cant' be right because it' *looks different* than how I learned it!"
uhoh
CC BY-SA 4.0
2021-03-11T09:55:21.543Z
@uhoh you can insist as you like. In all cases the photons have the right (within the indetermination limits) energy to excite the atom. All the rest is what appears to us in an absorption plot. A transition at x nm is a transition at x nm. If we see the atom absorb y it has nothing to do with a confused atom. That is the spirit of the question. The atom reads the photons and takes whatever looks x for it. It is not, as the OP intended, that it absorb different energies depending on its motion. That is apparent to us.
Alchimista
CC BY-SA 4.0
2021-03-11T10:00:03.833Z
@Alchimista I can't wrap my head around anthropomorphized QM. I do understand that this is a useful way to discuss it as we folks in the macro world want to project things on to the behavior of things we know about, but I can't do it. It's likely some others will find you description much more accessible than I. But if we do the experiment that the OP suggests, photon-by-photon, tuning the wavelength as we go, we'll see a broader distribution in the transition probabilities if the atom is bouncing around in a hot vacuum chamber than a cold chamber and that's Doppler broadening *sans ensemble*
uhoh
CC BY-SA 4.0
2021-03-11T10:05:20.853Z
@Alchimista and if the atom is colliding with the chamber walls with a characteristic time shorter than the half-life of the excited state, then *I believe* that we'll see collisional broadening effects as well. It's just not the way astronomers hear about these things first, as they are working on giant clouds of gas or even plasma very far away, *passively* rather than doing a controlled experiment, and looking at static spectra recording zillions of events.
uhoh
CC BY-SA 4.0
2021-03-11T10:07:14.677Z
Well *There is nothing to antropomorohisises! *. If I say the atom read 1) I treat it like a detector 2) it means that the photon has an actual energy for that atom. The absorbed photon have the energy that corresponds to a transition. Dot. Nothing change between an atom at rest and one moving. Dot. The Doppler broadening is apparent to us. Is that so difficult?
Alchimista
CC BY-SA 4.0
2021-03-11T10:11:31.993Z
@Alchimista atoms "reading" photons is too much for me and nobody mentioned atoms being "confused" until your comment. I won't walk down that road and there's no more to be said about it.
uhoh
CC BY-SA 4.0
2021-03-11T12:08:31.473Z
I really have no words. At least you finally come to a comment in which uncertainty and Doppler effect are kept separate. I wasn't overly interested to this question, which is answered by the accepted ans as well by the 2 comments at the very top of the thread. I've answered to your comment pointing to a clear fact. If you have resentment, ask who said your answer is totally wrong. I have no idea as for I can't really understand it. But I see that your are mixing uncertainty and an effect caused by different systems of references.
Alchimista
CC BY-SA 4.0
2021-03-09T08:51:14.090Z
2021-03-09T08:51:14.090Z
ProfRob
<p>From an answer to this question, <a href="https://physics.stackexchange.com/questions/281660/how-does-an-electron-absorb-or-emit-light">https://physics.stackexchange.com/questions/281660/how-does-an-electron-absorb-or-emit-light</a>,</p>
<blockquote>
<p><strong>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)</strong>
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.</p>
</blockquote>
<p>Given that the <a href="https://en.wikipedia.org/wiki/Photon_energy" rel="nofollow noreferrer">photon's energy is proportional to the photon's electromagnetic frequency</a>, and that the frequency is subject to tiny Doppler shifts due to differences in velocity between the emitting and absorbing molecules, <strong>how can a photon have precisely the same energy as the exact quantum state change for a gas molecule?</strong>  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?</p>

CC BY-SA 4.0
2021-03-08T19:42:49.660Z
In practice, the energy required is variable due to the Uncertainty Principle; see e.g. https://physics.stackexchange.com/questions/239250/how-precise-must-the-energies-match-for-absorption-of-photons/239302
Peter Erwin
CC BY-SA 4.0
2021-03-08T19:46:00.080Z
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.).
Peter Erwin
CC BY-SA 4.0
2021-03-09T08:20:46.117Z
Although the focus is on hydrogen only, I think you could perhaps use (the first part of) chapter 3 in [this thesis](https://ui.adsabs.harvard.edu/abs/2010PhDT.......254L) (I apologize for the blatant self-promotion, and I'll remove this comment if it's deemed out of place).
pela
CC BY-SA 4.0
2021-03-09T10:24:52.863Z
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).
Alchimista
CC BY-SA 4.0
2021-03-09T17:01:30.073Z
@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 https://arxiv.org/abs/1012.3175 instead.
Connor Garcia
CC BY-SA 4.0
2021-03-09T23:15:58.757Z
Yeah, the original doc was migrated to [here](https://www.nbi.ku.dk/english/theses/phd-theses/peter-laursen/peter_laursen.pdf) by the publisher (university), but the arXiv version is the same. I should contact ADS and have them fix the link and merge the two versions. Anyway, glad you liked it :)
pela
CC BY-SA 4.0
2021-03-11T10:48:48.467Z
Your two comments combined are actually the answer.
Alchimista
<light><radio-astronomy><photons><electromagnetic-spectrum>
Does a photon need to have EXACTLY the right energy to be absorbed by a gas molecule?
Connor Garcia
# Does a photon need to have EXACTLY the right energy to be absorbed by a gas molecule?

(Asked by: username_0 on 2021-03-08)

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](https://en.wikipedia.org/wiki/Photon_energy), 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?

**Tags:** light, radio-astronomy, photons, electromagnetic-spectrum
> **Comments:**
> **username_1**: In practice, the energy required is variable due to the Uncertainty Principle; see e.g. https://physics.stackexchange.com/questions/239250/how-precise-must-the-energies-match-for-absorption-of-photons/239302  
> **username_1**: 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.).  
> **username_2**: Although the focus is on hydrogen only, I think you could perhaps use (the first part of) chapter 3 in [this thesis](https://ui.adsabs.harvard.edu/abs/2010PhDT.......254L) (I apologize for the blatant self-promotion, and I'll remove this comment if it's deemed out of place).  
> **username_3**: 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).  
> **username_0**: @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 https://arxiv.org/abs/1012.3175 instead.  
> **username_2**: Yeah, the original doc was migrated to [here](https://www.nbi.ku.dk/english/theses/phd-theses/peter-laursen/peter_laursen.pdf) by the publisher (university), but the arXiv version is the same. I should contact ADS and have them fix the link and merge the two versions. Anyway, glad you liked it :)  
> **username_3**: Your two comments combined are actually the answer.  

---

## 2 answers

### Answer by username_6 on 2021-03-09. Score: 6 | Accepted answer

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.

> **Comments:**
> **username_4**: I've been told my answer is [completely wrong](https://astronomy.stackexchange.com/questions/41858/how-can-a-photon-have-exactly-the-right-energy-to-be-absorbed-by-a-gas-molecule/41865#comment85954_41860). If you have a chance to skim it could you let me know if you see any glaring errors? *Thanks!*  
> **username_3**: @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.  
> **username_4**: @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!*  
> **username_3**: @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.  
> **username_3**: Basically the very first two comments by Peter Erwin under the question said all.  
> **username_4**: @Alchimista and those have been addressed. The treatment may be unfamiliar because in astronomy these things are usually taught in an *applied way* "this is what you do to analyze astronomical data". When people see it in a different context the first reaction is "No no no! It cant' be right because it' *looks different* than how I learned it!"  
> **username_3**: @uhoh you can insist as you like. In all cases the photons have the right (within the indetermination limits) energy to excite the atom. All the rest is what appears to us in an absorption plot. A transition at x nm is a transition at x nm. If we see the atom absorb y it has nothing to do with a confused atom. That is the spirit of the question. The atom reads the photons and takes whatever looks x for it. It is not, as the OP intended, that it absorb different energies depending on its motion. That is apparent to us.  
> **username_4**: @Alchimista I can't wrap my head around anthropomorphized QM. I do understand that this is a useful way to discuss it as we folks in the macro world want to project things on to the behavior of things we know about, but I can't do it. It's likely some others will find you description much more accessible than I. But if we do the experiment that the OP suggests, photon-by-photon, tuning the wavelength as we go, we'll see a broader distribution in the transition probabilities if the atom is bouncing around in a hot vacuum chamber than a cold chamber and that's Doppler broadening *sans ensemble*  
> **username_4**: @Alchimista and if the atom is colliding with the chamber walls with a characteristic time shorter than the half-life of the excited state, then *I believe* that we'll see collisional broadening effects as well. It's just not the way astronomers hear about these things first, as they are working on giant clouds of gas or even plasma very far away, *passively* rather than doing a controlled experiment, and looking at static spectra recording zillions of events.  
> **username_3**: Well *There is nothing to antropomorohisises! *. If I say the atom read 1) I treat it like a detector 2) it means that the photon has an actual energy for that atom. The absorbed photon have the energy that corresponds to a transition. Dot. Nothing change between an atom at rest and one moving. Dot. The Doppler broadening is apparent to us. Is that so difficult?  
> **username_4**: @Alchimista atoms "reading" photons is too much for me and nobody mentioned atoms being "confused" until your comment. I won't walk down that road and there's no more to be said about it.  
> **username_3**: I really have no words. At least you finally come to a comment in which uncertainty and Doppler effect are kept separate. I wasn't overly interested to this question, which is answered by the accepted ans as well by the 2 comments at the very top of the thread. I've answered to your comment pointing to a clear fact. If you have resentment, ask who said your answer is totally wrong. I have no idea as for I can't really understand it. But I see that your are mixing uncertainty and an effect caused by different systems of references.  

---

### Answer by username_4 on 2021-03-08. Score: 1

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](https://astronomy.stackexchange.com/questions/41858/how-can-a-photon-have-exactly-the-right-energy-to-be-absorbed-by-a-gas-molecule#comment85933_41858) this all falls under the [uncertainty principle](https://en.wikipedia.org/wiki/Uncertainty_principle).

Now as @PeterErwin [also points out](https://astronomy.stackexchange.com/questions/41858/how-can-a-photon-have-exactly-the-right-energy-to-be-absorbed-by-a-gas-molecule#comment85934_41858) 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Spectral_line#Pressure_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.

> **Comments:**
> **username_1**: 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.  
> **username_4**: @PeterErwin you would calculate the average effect over a billion incident photons using a thermal distribution yes, but for a single transition where we don't know which atom will be involved nor which way it was going, I'm not yet convinced that no quantum mechanical treatment is called for. Question is about a single event.  
> **username_5**: @uhoh ‘s answer to the question about the photon energy is completely wrong.  
> **username_4**: @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*.  
> **username_0**: @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."?  
> **username_1**: @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.  
> **username_4**: @PeterErwin While you’re describing how people most often *encounter* these concepts analyzing data of a large number of photons from a dense gas, let’s keep focus on the OP's question about a single excitation of a single atom. That atom will certainly have an uncertainty in its velocity, and that distribution might be characterizable by a temperature. Yes most people usually use these concepts in the macroscopic world and ensemble averages, but that doesn’t mean you can’t still talk about it in the context of this question. The question isn't "How do astronomers...?"  
> **username_1**: @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.)  
> **username_4**: @PeterErwin you really can. While we more often talk about temperature when characterizing an ensemble average of kinetic energies (and therefore velocities), we can also use it to characterize a time average of kinetic energies and probability for ONE atom to have a given energy and therefore velocity. All that's necessary is that the atom is known to be in contact with other atoms and that that population is in thermal equilibrium. Thermal/Doppler CAN be treated microscopically a single-event phenomenon in answers about quantum mechanics, even though astronomers need ensembles for day-to-day  
> **username_4**: @PeterErwin As an aside, I'm going to go ahead and "put my head in the lion's mouth" and write a question in Physics SE that's related to all of this. I'll ping you when I do. And though this is about the internal excitation degrees of freedom and not related here, you might find it interesting: [Has a concept of temperature ever been defined in the context of a single atom?](https://chemistry.stackexchange.com/q/125177/16035)  

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