Photon absorption for an atomic electron

In summary: The average of all those measured energies will be close to 10.2 eV, but each individual measurement will be different (and might be higher, or lower)."In summary, an atomic electron can be excited to the next orbital energy state by a photon with more or less energy than the orbital energy gap, as the absorption probability usually follows a Lorentzian centered on the "exact" transition frequency. The excited state has a finite lifetime, resulting in an uncertainty in its energy level, which can't be used to violate conservation of energy laws. The expectation value of the emitted photon's energy when the excited state decays back to the ground state is
  • #1
Dfault
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TL;DR Summary
Can an atomic electron be excited to the next orbital energy state by a photon with more or less energy than the orbital energy gap?
Quick question: let's say we have an atomic electron in the ground state which requires, say, one "unit" of energy* to jump up to the next orbital energy state. If a photon arrives with a bit more or less than this, say 1.00003 or 0.99997 units of energy, is there some finite, non-zero probability that the photon will still be absorbed and the electron will be kicked up into the next orbital energy state? Or does the energy of the incident photon have to be exactly 1.0000000(etc.) units? Thanks!

*So for a Hydrogen atom transitioning from n1=1 to n2=2, for example, this amount of energy would be -13.6*(1 - 1/4) = -10.2 eV
 
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  • #2
Dfault said:
Summary:: Can an atomic electron be excited to the next orbital energy state by a photon with more or less energy than the orbital energy gap?
Yes. The absorption probability usually follows a Lorentzian centered on the "exact" transition frequency (or energy).

This is because the excited state actually doesn't have a precisely defined energy, but has a certain width in energy due to the fact that it has a finite lifetime (it will eventually decay back to the ground state by spontaneous emission). It is related to the time-energy uncertainty relation:
$$
\Delta E \Delta t \sim \hbar/2
$$
unless ##\Delta t \rightarrow \infty## (infinite lifetime), the energy level will have an uncertainty to it.
 
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  • #3
Ah okay, that makes sense. But does this mean that there's a way in which we can use the uncertainty of the excited state's energy level to "cheat" conservation of energy laws? For example, if the expected transition energy for the electron is 10.2 eV but we start firing photons at it with an energy of only 10.199 eV to excite the electron "at a discount" over and over, consistently trying to trigger the transition from the low-side of the Lorentzian distribution, what would happen? When the excited electron falls back down to its ground state by spontaneous emission, will the new photon it emits also (on average, repeating the experiment many times) have an energy of 10.199 eV like the photons that were used to bring about the excited state in the first place? Or will the emitted photons have an average energy of 10.2 eV, consistent with the difference in what we'd expect based on the peak, "most likely" energy values for the ground state and the excited state?
 
  • #4
DrClaude said:
unless ##\Delta t \rightarrow \infty## (infinite lifetime)
at the risk of being warned for invading someone elses thread,

##\Delta t## is the lifetime of exactly what?
 
  • #5
kent davidge said:
##\Delta t## is the lifetime of exactly what?

The state. The easiest example would be an atom in its first excited state, in which case t is the decay time; i.e. roughly the time the electron would typically stay in its excited state before spontaneously decaying back into its ground state.
 
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  • #6
Sorry to necro an old thread, but did anyone ever figure this one out? Does a hydrogenic electron in its first excited state emit a photon with (on average) an expectation value of 10.2 eV when it decays back into its ground state, or is the expectation value of the emitted photon instead equal to the energy that the atom absorbed from the original, incident photon?
 
  • #7
Dfault said:
does this mean that there's a way in which we can use the uncertainty of the excited state's energy level to "cheat" conservation of energy laws?

No. You can't violate conservation of energy.

Dfault said:
if the expected transition energy for the electron is 10.2 eV but we start firing photons at it with an energy of only 10.199 eV to excite the electron "at a discount" over and over, consistently trying to trigger the transition from the low-side of the Lorentzian distribution, what would happen?

You are confusing expectation value of energy with actual energy transferred. They're not the same. The fact that you are firing photons with an expectation value for energy of 10.199 eV does not mean that the actual energy transferred when one gets absorbed will be 10.199 eV. In fact you have no way of measuring the actual energy transferred when a photon you fired gets absorbed, except by measuring the energy released when the excited state decays back to the ground state.

So in fact, your implicit assumption that you have two energy measurements that might be different, is wrong. There is only one energy measurement you can have from this whole procedure. That energy measurement might not give a value of exactly 10.2 eV, but if it doesn't, that doesn't mean energy conservation was violated. It just means the energy transferred in that particular run of the experiment was not exactly equal to the expectation value over a large number of runs. Which is exactly what you expect from a quantum process. To verify the expectation value of 10.2 eV, you would make a large number of runs and average the results.
 
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  • #8
Thanks for the reply! Okay, I think that makes sense. So let's say for a second that the following is our procedure:

1.) We've set up a "photon gun" to fire photons at the atom. It's designed to fire photons at 10.199 eV, but on any given run of the experiment, the actual energy of the photon that's fired could be a bit higher or lower than this value. Let's say that we don't measure the energy of this fired photon directly, and instead allow the photon to continue straight to the atom without interacting with anything else along the way;
2.) The atom either absorbs the photon on the first attempt, or else the photon passes through the atom. In the latter case, we use a set of mirrored surfaces to bounce the photon back towards the atom to keep giving the photon another chance to interact with the atom until it is absorbed;
3.) We wait for the atom in its excited state to spontaneously collapse back into the ground state, then measure the energy of the emitted photon. We then repeat the experiment many times and average the results.

If we run that experiment, what value would we get for the average measured energy of the emitted photon? Would it be 10.199 eV, or 10.2 eV? Thanks in advance!
 
  • #9
Dfault said:
It's designed to fire photons at 10.199 eV, but on any given run of the experiment, the actual energy of the photon that's fired could be a bit higher or lower than this value. Let's say that we don't measure the energy of this fired photon directly

If you don't measure it, then it is meaningless to talk about the "actual" energy of the photon.

Dfault said:
If we run that experiment, what value would we get for the average measured energy of the emitted photon? Would it be 10.199 eV, or 10.2 eV?

If the difference in energy levels of the atom is 10.2 eV, then that is the expectation value of the energy of the emitted photon. So that is what you would expect for the average measured energy of the emitted photon over a large number of runs of the experiment.
 
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  • #10
PeterDonis said:
If you don't measure it, then it is meaningless to talk about the "actual" energy of the photon.

Okay, in that case, let's modify step 1 of the procedure so that we do measure the energy of the incident photon each time we run the experiment, then we continue steps 2 and 3 as normal - so each time we run the experiment, we measure both the energy of the photon fired at the atom, and the energy of the photon emitted from the atom. We've established that our energy measurements can vary from one run of the experiment to the next, so let's say the incident photon's energy is measured as 10.20 eV on the first run, 10.21 on the second, 10.18 on the third, etc. In that scenario, would we expect our set of measurements for the emitted photon to also be 10.20 eV on the first run, 10.21 on the second, 10.18 on the third, etc? In other words, are the two energy measurements "coupled" to one another in any given run of the experiment?
 
  • #11
Dfault said:
let's modify step 1 of the procedure so that we do measure the energy of the incident photon each time we run the experiment

Measuring the photon's energy will require having it absorbed by something, which destroys it. So that will stop the experiment.
 
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  • #12
Hmm, okay. So let me see if I understand:

Let's say we start off with a photon gun configured to fire photons with, on average, an energy of 10.199 eV. To confirm that the gun is working as expected, we set up a detector to measure these photons' energy. Running this test many times, we see a Lorentzian distribution of energy values centered about 10.199 eV, as expected.

Having confirmed that the photon gun is working as expected, we remove the detector and allow the photons to hit the atom unimpeded, and then measure the energy of the photons emitted from the atom's transition back into the ground state. Repeating this experiment many times, we get a Lorentzian distribution centered about 10.2 eV -- though now we can no longer say that the incident photons had an average energy of 10.199 eV, because for these tests, we had stopped measuring the incident photons' energy directly.

Is that right?
 
  • #13
Dfault said:
Hmm, okay. So let me see if I understand:

Let's say we start off with a photon gun configured to fire photons with, on average, an energy of 10.199 eV. To confirm that the gun is working as expected, we set up a detector to measure these photons' energy. Running this test many times, we see a Lorentzian distribution of energy values centered about 10.199 eV, as expected.

Having confirmed that the photon gun is working as expected, we remove the detector and allow the photons to hit the atom unimpeded, and then measure the energy of the photons emitted from the atom's transition back into the ground state. Repeating this experiment many times, we get a Lorentzian distribution centered about 10.2 eV -- though now we can no longer say that the incident photons had an average energy of 10.199 eV, because for these tests, we had stopped measuring the incident photons' energy directly.

Is that right?

Let me try to rephrase your question. If we ensure the incident photons are on the low side of the expected transition energy, then we will get some absorption and subsequent spontaneous emission. Will these emissions also statistically be on the low side of the expected emission energy?

If yes, then it all seems very logical. If not, then why not?
 
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  • #14
Dfault said:
Is that right?

Yes.

PeroK said:
Will these emissions also statistically be on the low side of the expected emission energy?

No, they won't. The expected emission energy is based on the difference in energy levels in the atom. It has nothing to do with the average energy of photons fired at the atom.
 
  • #15
PeterDonis said:
No, they won't. The expected emission energy is based on the difference in energy levels in the atom. It has nothing to do with the average energy of photons fired at the atom.

Then why can you not input photons with a given average energy into the system and get photons with a higher average energy out?
 
  • #16
PeroK said:
why can you not input photons with a given average energy into the system and get photons with a higher average energy out?

Because, as I said, the average energy of the emitted photons has nothing to do with the average energy of the input photons. It has to do with the difference in energy levels in the atom. If you run the same kind of experiment the OP describes, but with a device that shoots photons whose expectation value of energy is higher than the difference in atom energy levels, then the emitted photons will have lower average energy than the expected energy of the input photons.

In fact, as I pointed out a number of posts ago, the energy of the input photons cannot be measured if they are to be absorbed by the atom. So you cannot say that the average energy of the input photons that get absorbed is the same as the average energy of photons from the source that you measured in order to calibrate the source. So you also cannot say that the average energy of emitted photons is larger (or smaller, if you calibrate the source the way I described in the last paragraph) than the average energy of absorbed photons. All you can say is that the average energy of emitted photons is larger (or smaller) than the expectation value of energy you got from your calibration process. Which just illustrates that the calibration process is a different, separate process from the experiment itself, and you have to be careful not to make statements about the photons in the latter based on the photons in the former that are not supported by QM.
 
  • #17
PeterDonis said:
you cannot say that the average energy of the input photons that get absorbed is the same as the average energy of photons from the source that you measured in order to calibrate the source.

Perhaps it will help to expand on this statement. Suppose that we changed the experiment so that, instead of mirrors reflecting input photons back and forth until they got absorbed, we just had input photons going past the atom one time. Then we would find that not all input photons got absorbed; and if we set things up to measure the energy of the input photons that did not get absorbed, we would find that their average energy was lower than the calibrated expectation value of energy from the source that we got from measurements before the experiment.

This suggests that the mirror reflection process does not actually keep the energy of the input photons exactly the same. The mirrors will jiggle around slightly, so the reflections of the photon off the mirror can change its energy. The absorbing atom itself will jiggle around slightly, so the photon's energy relative to the atom won't be exactly the same as its energy relative to the source (or the mirrors). In other words, even to think of the photon as having a conserved energy from the source to its absorption by the atom, while recognizing that we have not measured this energy so we cannot say what its value is, is not correct.
 
  • #18
PeterDonis said:
Perhaps it will help to expand on this statement. Suppose that we changed the experiment so that, instead of mirrors reflecting input photons back and forth until they got absorbed, we just had input photons going past the atom one time. Then we would find that not all input photons got absorbed; and if we set things up to measure the energy of the input photons that did not get absorbed, we would find that their average energy was lower than the calibrated expectation value of energy from the source that we got from measurements before the experiment.

This suggests that the mirror reflection process does not actually keep the energy of the input photons exactly the same. The mirrors will jiggle around slightly, so the reflections of the photon off the mirror can change its energy. The absorbing atom itself will jiggle around slightly, so the photon's energy relative to the atom won't be exactly the same as its energy relative to the source (or the mirrors). In other words, even to think of the photon as having a conserved energy from the source to its absorption by the atom, while recognizing that we have not measured this energy so we cannot say what its value is, is not correct.
Let's leave the mirrors out of it! Let's assume:

1) We can produce a beam of almost monochromatic photons that have a negligible proportion above the midpoint emission energy in question.

2) We fire the beam, one by one, at a hydrogen atom. Occasionally one is absorbed. We know that the average energy of the absorbed photons must be below the emission midpoint.

3) We get the occasional emitted photon with an average of the emission midpoint.

If this is the case, there must be another factor. I can see the practical difficulties of conducting this experiment and perhaps one of the stages 1-3 is invalid in some way. Perhaps it's simply not possible to keep track of photons in this way?
 
  • #19
PeroK said:
We know that the average energy of the absorbed photons must be below the emission midpoint.

No, we don't. All we know is that the average energy of all photons emitted by the source must be below the expected energy of emission. But that does not mean that the average energy of the subset of photons emitted by the source that are absorbed must be below the expected energy of emission.

In fact, as I said before, if we measure the average energy of all photons not absorbed (by, for example, placing a detector well past the absorbing atom), we will find that it is lower than the average energy of all photons emitted by the source. And this would suggest that, if we were able to measure the average energy of the photons that were observed (which we can't), we would find that it was higher than the average energy of all photons emitted by the source--which in turn makes it quite possible that the average energy of photons absorbed is the same as the average energy of photons emitted by the atom after it absorbs a photon.

PeroK said:
perhaps one of the stages 1-3 is invalid in some way

Yes, stage 2 is, or at least the claim you make at the end of it. See above.
 
  • #20
PeterDonis said:
No, we don't. All we know is that the average energy of all photons emitted by the source must be below the expected energy of emission. But that does not mean that the average energy of the subset of photons emitted by the source that are absorbed must be below the expected energy of emission.
What if none of the source photons has more than the average emission energy? To be precise: We have a large number of source photons in a very narrow energy spectrum, which overlaps the absorption spectrum (hence entails significant absorption and subsequent emission), but has a statistically insignificant number of source photons above the average emission energy.

Unless this is practically or theoretically impossible, then we have significant absorption of photons almost all below the average emission energy. At face value, this is a possible statistical scenario. One could even postulate an almost perfectly monochromatic source entirely within the lower half of the spectrum for likely absorption.

If a photon is absorbed, and it has more or less than the precise energy at the centre of the absorption spectrum, then the resulting atom must be in a superposition of the ground and excited states. This would be an alternative way to describe why the excited state does not have a perfectly defined energy: it's usually/always actually a superposition of ground and excited states. Then, when it decays back to the ground state it must emit a photon of the energy equal to the difference between the superposed excited state and the ground state.

That suggests that the energy of the emitted photon is determined by the energy of the excited superposition, which in turn is determined by the absorption process.
 
  • #21
PeroK said:
We have a large number of source photons in a very narrow energy spectrum, which overlaps the absorption spectrum (hence entails significant absorption and subsequent emission), but has a statistically insignificant number of source photons above the average emission energy.

This is not possible; it's self-contradictory. You can't have significant overlap between the source energy spectrum and the absorption energy spectrum, while at the same time having a statistically insignificant number of source photons above the average emission energy. The emission energy spectrum is the same as the absorption energy spectrum, because they both are based on the same energy level transition in the same atom.

PeroK said:
Unless this is practically or theoretically impossible

It is. See above.
 
  • #22
PeterDonis said:
This is not possible; it's self-contradictory. You can't have significant overlap between the source energy spectrum and the absorption energy spectrum, while at the same time having a statistically insignificant number of source photons above the average emission energy. The emission energy spectrum is the same as the absorption energy spectrum, because they both are based on the same energy level transition in the same atom.
It's not self-contradictory. To take an idealisation:

The absorption and emission spectrums go from ##1-2## units, say. The source photons all have energy ##1.25##. We have significant absorption at ##1.25## energy units, whereby we cannot have an emission average of ##1.5## energy units.
 
  • #23
PeroK said:
To take an idealisation:

Can you show me actual math that says your idealization is theoretically possible? You can't just wave your hands and say spectra are whatever you want. You have to show that a source that can produce your claimed source spectrum, and an atom that has your claimed absorption and emission spectrum, are theoretically possible.

Your claimed source spectrum is a point value, which AFAIK is theoretically impossible; there has to be some spread. Your claimed absorption/emission spectrum is a uniform distribution over a finite interval, which AFAIK is also theoretically impossible; it has to be peaked about the energy difference between the two energy levels.

With distributions that are theoretically possible, each one will be peaked about some nominal value and will have some spread. As the peak values (source and absorption/emissions) get further apart, the rate of absorption (in the absence of mirrors) goes down; and as the spreads in the distributions get narrower, for a fixed difference in peak values, the rate of absorption also goes down.
 
  • #24
PeroK said:
the energy of the emitted photon is determined by the energy of the excited superposition

More precisely, it's determined by the energy distribution of the excited state.

PeroK said:
which in turn is determined by the absorption process

No. The wave function of the excited state has nothing whatever to do with the absorption process. It has to do with the structure of the atom itself, which determines the energy levels, and the lifetime of the state. Neither of those depend on the absorption process. The only thing the absorption process affects is the probability of absorption.
 
  • #25
PeterDonis said:
More precisely, it's determined by the energy distribution of the excited state.
No. The wave function of the excited state has nothing whatever to do with the absorption process. It has to do with the structure of the atom itself, which determines the energy levels, and the lifetime of the state. Neither of those depend on the absorption process. The only thing the absorption process affects is the probability of absorption.

The energy of the excited state must be a) determined by the energy of the absorbed photon and b) reflected in a superposition of energy eigenstates. In that sense, the energy distribution of the excited state is determined by the energy distribution of the absorbed photons.

The emission energy must equate to the absorption energy in each specific case.

And, I suggest, if you could control the source photos sufficiently, you must be able to generate an atypical emission spectrum, mirroring the distribution of the source photons.
 
  • #26
PeroK said:
The energy of the excited state must be a) determined by the energy of the absorbed photon

No. First of all, you can't talk about the energy of the absorbed photon if you don't measure it. Second, you have things backwards: what you should say is that the photon won't get absorbed unless its energy matches the energy of the excited state. Absorbing a photon can't change the energy levels of the atom.

PeroK said:
b) reflected in a superposition of energy eigenstates

Yes, but that superposition is determined, as I said before (and as @DrClaude said way earlier in this thread), by the structure of the atom and the lifetime of the state, neither of which are affected by the absorption of a photon.

PeroK said:
The emission energy must equate to the absorption energy in each specific case.

Only in the sense that if an absorption and emission takes place, the only energy we can measure is the emission energy, so that's the only information we have about the absorption energy. Since we can't measure the absorption energy separately from the emission energy, we can't talk about the absorption energy having a specific value independently of the emission energy.

In short, this scenario is simply not a good scenario for checking conservation of energy.
 
  • #27
PeroK said:
I suggest, if you could control the source photos sufficiently, you must be able to generate an atypical emission spectrum, mirroring the distribution of the source photons.

Can you find any reference that supports this suggestion? Either a textbook or paper showing a theoretical derivation, or an experiment showing the phenomenon itself?

I strongly suspect that you can't, because I don't think it's possible. I think all that will happen, as I said before, is that the rate of absorption will drop off as the overlap between the expected source distribution and the absorption/emission spectrum decreases. But the emission spectrum itself won't change.
 
  • #28
PeterDonis said:
Can you find any reference that supports this suggestion? Either a textbook or paper showing a theoretical derivation, or an experiment showing the phenomenon itself?

I strongly suspect that you can't, because I don't think it's possible.
That's a good question. It may be practically very difficult and I don't see that the original question is important enough that it's an experiment that demands to be done.

Nevertheless, the key (to the question as posed) is that the excited states - although theoretically precise energy states - in practice must always be a superposition of states. That's where the distribution of the absorption and emission spectra originates. What I suggest we don't have is a well-defined (unique) first excited state that emits photons according to an emission spectrum when it decays. A unique excited state would always emit a photon with a precise energy when decaying to the (unique) ground state. Instead the first excited state is a collection of superpositions of energy eigenstates, determined by the energy of the absorbed photon and centred on the precise first excited eigentstate.

Just to emphasise this: I suggest that an atom is never practically in that precise first excited state.

That explains things as far as I can see. Although, yes it would be good to see some official corroboration of that hypothesis.
 
  • #29
PeroK said:
the excited states - although theoretically precise energy states - in practice must always be a superposition of states.

A superposition of energy eigenstates, yes.

However, the same is true of the photon states: the photons themselves are not in energy eigenstates unless they are measured. The measurement of an emitted photon gives you a single energy value (more precisely, a single frequency or wavelength, but that corresponds to a single energy value). But other than that, you cannot think of the photon as having a single definite energy, because it doesn't.

PeroK said:
That's where the distribution of the absorption and emission spectra originates.

Saying "there is a distribution of absorption and emission spectra" is just restating "the atom, when excited, is in a superposition of energy eigenstates" in different words. It's not explaining why there is a distribution of absorption and emission spectra.

The reason why there is a distribution of absorption and emission spectra is, as @DrClaude pointed out many posts ago, that excited states can spontaneously emit photons and decay back to the ground state (or to a lower energy excited states). That is why they cannot be stationary states: because stationary states can't change.

PeroK said:
What I suggest we don't have is a well-defined (unique) first excited state that emits photons according to an emission spectrum when it decays.

Wrong. We do have a well-defined excited state; it just isn't an energy eigenstate. Not all well-defined states are energy eigenstates.

PeroK said:
A unique excited state would always emit a photon with a precise energy when decaying to the (unique) ground state.

Nope. It is impossible to have a state that will always emit a photon with a single precise energy, because such a state would have to have a definite energy itself, and would therefore be a stationary state, which cannot emit a photon at all since it cannot change.

You are simply failing to apply basic QM to the fact that the excited state is not a stationary state.

PeroK said:
I suggest

At this point I think this discussion needs references, not "suggestions", to proceed any further.
 
  • #30
PeroK said:
A unique excited state would always emit a photon with a precise energy when decaying to the (unique) ground state.

Perhaps it will help to expand on my "no" answer to this a bit. When textbooks talk about "energy levels" of atoms, they are actually talking about eigenstates of the Hamiltonian of an isolated atom, considering no other degrees of freedom at all. But such a treatment cannot even model absorption or emission of photons, since the photons are other degrees of freedom not included in the isolated atom Hamiltonian.

Once you include the interaction Hamiltonian between the atom and the EM field, the states that textbooks call "energy levels" of atoms are no longer eigenstates of the full Hamiltonian. But they are still eigenstates of the isolated atom Hamiltonian. In other words, there is still a perfectly good way of identifying unique states corresponding to each "energy level" of the atom. The states so identified simply aren't eigenstates of the full Hamiltonian once the degrees of freedom corresponding to the EM field and the atom-field interaction terms in the Hamiltonian are included.

So there are indeed "unique excited states", but those states do not always emit a photon with a single precise energy. When textbooks talk about the "energy difference" between two energy levels as a single number, what they mean is that the difference between the two eigenvalues of the isolated atom Hamiltonian is a single number. But that does not mean that a "unique excited state" would always emit photons whose energy was that single number. That single number is not a number that comes from the full Hamiltonian with the atom-field interaction included, and only that full Hamiltonian can describe absorption and emission processes.
 
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  • #31
PeterDonis said:
Wrong. We do have a well-defined excited state; it just isn't an energy eigenstate. Not all well-defined states are energy eigenstates.

I think that summarises the two options:

a) The first excited state is a single well-defined state.

b) The first excited state is an infinite collection of different states, all with approximately the theoretical first excited energy.
 
  • #32
PeroK said:
I think that summarises the two options

They aren't two options. Your option 2) is simply wrong. See my post #30 just now (looks like we cross posted).

I strongly suggest that you take a step back and think things through before posting again.
 
  • #33
Hi all, thanks for the insights. This has helped to clarify things, though now it's introduced a new question to me that I can't make heads or tails of: it would seem that the photons know, as they leave the photon gun, whether or not they'll encounter an atom at some point along their way towards the detector: when you fire the photons with the atom in front of the detector, the photons are retroactively discovered to have an energy distribution consistent with that of the atomic transition; when you fire the photons without the atom in front of the detector, the photons are retroactively discovered to have an energy distribution consistent with that of the initial photon-gun calibration. Is this true? If so, how can the photons possibly "know" what obstacles they may-or-may-not encounter along their path at the moment they're created, regardless of how distant those obstacles may be to their point of creation at the photon gun?
 
  • #34
Dfault said:
when you fire the photons with the atom in front of the detector, the photons are retroactively discovered to have an energy distribution consistent with that of the atomic transition; when you fire the photons without the atom in front of the detector, the photons are retroactively discovered to have an energy distribution consistent with that of the initial photon-gun calibration. Is this true?

No. You are misdescribing what would actually be observed. Here's what would actually be observed:

If you fire the photons with no atom in front of the detector, all of the photons from the photon gun reach the detector, and the energy distribution seen at the detector is consistent with that of the initial calibration of the photon gun.

If you fire the photons with the atom present, you need two detectors: one to detect the photons from the photon gun that are not absorbed by the atom, and the other to detect photons emitted by the atom after the atom absorbs a photon from the photon gun. Call these detector #1 (photons not absorbed by atom) and detector #2 (photons emitted by atom after an absorption).

The energy distribution seen by detector #2 will be consistent with that of the atomic transition. The energy distribution seen by detector #1 will have a lower average energy (assuming that the average energy of the atomic transition is higher than the average energy of the photon gun calibration) than the energy distribution expected from the initial calibration of the photon gun. If you combine the data from both detectors, the energy distribution in the combined data will be consistent with the initial calibration of the photon gun.

All of the above is ignoring real-world complications that will introduce noise into the data.
 
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  • #35
Hmm, okay. It makes sense that the combined data looks the same as the calibration data, and it makes sense that the data from detector #2 is consistent with the energy distribution for the atomic transition. But can you explain how the photons at detector #1 wind up with a lower energy distribution than the initial calibration? What is it that happens to the photons, as they pass through the atom without getting absorbed, that causes their energy distribution to get shifted down?
 

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