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
  • #36
And to clarify, I mean "get shifted down" from the energy distribution we'd expect, not "get shifted down" from the energy they had previous to passing through the atom, since we've already established that we can't talk about the energy of anyone photon before it's measured
 
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  • #37
Dfault said:
can you explain how the photons at detector #1 wind up with a lower energy distribution than the initial calibration?

Um, because they're the ones that aren't absorbed by the atom?

Dfault said:
the energy distribution we'd expect

Why would you expect the subset of photons that aren't absorbed by the atom to have the same energy distribution as the set of all photons from the photon gun?
 
  • #38
PeterDonis said:
Why would you expect the subset of photons that aren't absorbed by the atom to have the same energy distribution as the set of all photons from the photon gun?

Imagine for a second that you do the calibration test with no atom present; you observe the calibration energy distribution as a result. Now imagine you add the atom and measure the energy distribution of the photons which pass through it and don't get absorbed; this time, you get a lower energy distribution. I was always under the impression that if a photon doesn't get absorbed by the atom, then from the photon's perspective, it was as if the atom had never existed at all - that the photon "ghosts" through the atom as if it weren't even there, with no possibility for interaction between the two. But evidently, that's not the case: if the presence of the atom can downshift the energy distribution of a set of photons even if they're not getting absorbed by the atom, then there must be some kind of interaction between the atom and these photons. Is that right?
 
  • #39
Dfault said:
if the presence of the atom can downshift the energy distribution of a set of photons even if they're not getting absorbed by the atom, then there must be some kind of interaction between the atom and these photons. Is that right?

No. Again, you are misdescribing what is happening. All that is happening is that the set of photons that do not get absorbed by the atom is different from the set of all photons emitted by the photon gun. There is no reason to expect the energy distribution of the former to be the same as the energy distribution of the latter. There is no need for the photons that do not get absorbed by the atom to be influenced by the atom at all. All that is needed is for the set of photons that do not get absorbed by the atom to be different from the set of all photons emitted by the photon gun.

This aspect of it actually doesn't require any quantum mystery at all. It can be understood purely in classical terms. For example, consider this analogy: you roll a 6-sided die 1000 times. You pick out all the die rolls that are 6 and call them the "absorbed" rolls. Obviously the average of all the die rolls that are not "absorbed" will be lower than the average of all 1000 die rolls. That's not because of any mysterious "interaction" going on; it's just a simple consequence of taking a subset of the die rolls instead of all of them.
 
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  • #40
Okay, let me see if I can articulate a little better where my confusion is coming from:

  1. We assume that photons, as they’re fired from the photon gun on their way towards the atom, have no predetermined energy value: there’s no “nametag” or “stamp” on the photons to indicate what energy value each one has, no information anywhere on the photons themselves that could indicate what anyone photon’s energy is.
  2. The photons arrive at the atom. Of the photons that get absorbed, the [absorption-> emission -> detection at detector #2] process (which I’ve condensed into a single step for the sake of brevity) defines the energies of these photons: it stamps a particular energy value on the photons that were absorbed, emitted, and detected at detector #2. We’re free to do this, because the incident photons had no particular energy when they arrived at the atom: their energy values were a “blank slate,” and so the atom could impose whatever energy value on each of the photons that it wanted to.
  3. What about the photons which pass through the atom? Does the atom “stamp” a definite energy value on them, too?
    1. If so, then there is a non-absorptive atom-photon interaction; we said earlier that this doesn’t happen
    2. If not, then the photons arrive at detector #1 in exactly the same state as they were when they were first fired from the photon gun: with their energy values in a “blank slate” form, with the exact values of each photon’s energy determined only by the interaction between the photon and detector #1. But if that were the case, then we’d see the same energy distribution that we had in the initial calibration curve; we said that this doesn’t happen, and that the energy distribution at detector #1 would actually be lower than the calibration curve, because most of the “higher-energy” photons got absorbed by the atom, leaving behind most of the “lower-energy” photons to pass through the atom and reach detector #1. But then that assertion contradicts our assumption in step number 1, namely, that the photons do not a priori have a definite energy value on their way from the photon gun to the detector: no single one of the non-absorbed photons can, on its way from the atom to detector #1, know that it’s one of the “lower-energy” photons, because nothing happened in its journey from the photon gun to detector #1 which would have defined it to be a “lower-energy photon.” Unless the atom is communicating to the non-absorbed photons: “I’m not absorbing you, therefore I’m defining you to have a lower energy,” but again, that would imply a non-absorptive atom-photon interaction, which we said doesn’t happen. Alternatively, we could try to argue that there were “high-energy” and “low-energy” photons all along from the moment each photon was fired from the photon gun, but then that would contradict our assumption in step number 2, in which we said that the [absorption-emission-detection at detector #2] process was free to impose a particular energy value upon the “absorbed/emitted/detected at detector #2” photons since they didn’t have a definite energy value prior to that point.

So it seems that every path leads us to a contradiction; where did we make a mistake?
 
  • #41
Dfault said:
We assume that photons, as they’re fired from the photon gun on their way towards the atom, have no predetermined energy value

Yes, but they do have an energy distribution. That is, they have a wave function, and that wave function can be expressed as a function of energy, i.e., for every value of energy the function gives a probability amplitude--a complex number whose squared modulus is the probability that a measurement of the photon's energy will give that value. This distribution is what we measure in the initial calibration of the photon gun.

Dfault said:
The photons arrive at the atom.

No. The photons do not have a well-defined position or trajectory in space, and do not "arrive at the atom". They are either absorbed by the atom, or they aren't. You cannot say that the ones that are not absorbed by the atom still "passed through it" or "were influenced by it" or anything else. All you can say is that they were not absorbed.
 
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  • #42
Dfault said:
If not, then the photons arrive at detector #1 in exactly the same state as they were when they were first fired from the photon gun

No, they don't. The experiment with the atom present is a different physical configuration than the experiment with the atom not present, and that makes a difference in the wave function of the photons at detector #1.

With no atom present, we can take the wave function of the photons at detector #1 (which is the only detector in this version of the experiment) to be identical with the wave function of the photons at the photon gun. Strictly speaking, this is not correct for any real experiment; in any real experiment there will be some nonzero probability that a photon hits the wall instead, or otherwise fails to reach the detector. But we are idealizing away all those complications, and with that idealization, every photon emitted by the photon gun reaches the detector in this version, and the wave function of the photon is unchanged.

With the atom present, however, the wave function of the photons at detector #1 is no longer the same as the wave function of the photons at the photon gun, because there is now a nonzero probability amplitude for each photon to be absorbed by the atom. From the standpoint of detector #1, this is no different from a nonzero probability amplitude for each photon to hit the wall instead of detector #1, but unlike that case, we can't idealize away the nonzero probability amplitude for each photon to be absorbed by the atom, because that's precisely what we want to experiment with--what happens when the atom is present. So we have to take that nonzero probability amplitude into account when analyzing the experiment.

However, it's important to understand what this does not mean. It does not mean that photons that hit detector #1, and are not absorbed by the atom, still "interacted with the atom" in some way and had their energy changed (any more than a photon in the no-atom version that doesn't hit the wall on its way to detector #1 still had to "interact with the wall" in some way). All it means is that, if we want to analyze what happens when each photon is emitted, we have to take the photon wave function emitted by the photon gun and split it into two pieces: one piece to represent the possibility that the photon will be absorbed by the atom, and the other piece to represent the possibility that the photon will not be absorbed by the atom and will hit detector #1. And both of these pieces of the wave function have different energy distributions than the wave function as a whole. The first piece, roughly speaking, has the same energy distribution as the energy level transition in the atom; and the second piece, roughly speaking, has whatever is "left over" in the photon gun energy distribution once the first piece is subtracted. So the second piece will have a lower expected energy than the photon gun distribution as a whole, and that is what we will observe at detector #1.

Now, as I said before, this particular experiment doesn't actually have any real quantum weirdness about it. What I have said above can be understood reasonably well as just a simple classical process where the atom preferentially absorbs photons of higher energy, leaving those of lower energy to be detected at detector #1. There are no quantum interference effects involved. But it is still true that you cannot attribute a definite trajectory to each photon, you cannot say that photons not absorbed by the atom still interacted with it, and you cannot say that each photon has a definite energy when it is emitted by the photon gun. So it is still true that classical reasoning, if applied too literally, can lead you astray.
 
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  • #43
Ah okay, that makes a lot more sense. I think my main problem was in assuming that just because the photons from the photon gun didn’t have a well-defined energy (i.e. energy eigenvalue), that meant that the photons didn’t carry any information at all about their energy. But if, like you said, the photons fired from the photon gun do carry information on what distribution of energy levels they might have, then that resolves a lot of the issues I was having. I shouldn’t be calling the photons from the photon gun “blank slate” photons, because they do carry information about their energy distribution, just not a mandate on which particular energy they’ll end up having when detected.

That did lead me to a new complication for a few minutes, but I think I’ve resolved it: in imagining the photons fired from the photon gun as carrying a distribution of energy levels, at first I imagined that, for the detection stage, the universe has to go through a two-step process where first it “rolls a die” to determine whether a given photon is absorbed by the atom, and then, if the photon is absorbed/re-emitted/detected at detector #2, then the universe has to roll a second “die” to determine what energy value the detected photon has. It seemed strange to me that, of the photons that the universe rolled “yes” on for the first roll (i.e. the photons that got absorbed by the atom), it would just so happen that these photons would also roll high for the second dice toss (i.e. determining what energy value the absorbed photons had). But in retrospect, I suppose this isn’t really a coincidence at all, since there aren’t really two separate dice-tosses going on here: rather, there’s only one dice-toss, since we determine both the energy of the photon and whether it was absorbed by the atom at the same time, namely, when the photon hits one of the detectors. If the photon hits detector #2, we learn at that moment both that the photon had been absorbed by the atom and that it has (averaged across many experiments) a higher energy than the calibration mean; likewise, when a photon hits detector #1, we learn at that moment both that the photon had not been absorbed by the atom and that it has (averaged across many experiments) a lower energy than the calibration mean. So there’s no “conspiracy” going on here at all, since the information about absorption probability and the information about energy distribution at the detectors are joined at the hip. Is that right?
 
  • #44
Dfault said:
Is that right?

This seems like a reasonable description in words, yes. Bear in mind that any description in words is only going to be an approximation. The true description is in the math. But it is true that in the math, there is only one "die roll", as you describe it: each run of the experiment has only one result (either the atom absorbs the photon, re-emits a photon, and detector #2 detects it, or the atom does not absorb the photon and detector #1 detects it), so nature only has to make one "decision" for each run, and that "decision" includes both which of the two possible results happens--which detector detects a photon--and what actual number that result yields for the energy that is detected.
 
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  • #45
Great! Thanks for all your help. One last question: if we built a sort of "black box" around the atom and detector #2, so that we have the photon gun firing photons into the box and detector #1 detecting photons that exit the box on the opposite side, can we use the energy distribution at detector #1 to infer whether there's an atom inside of the box? For example, if there's no atom inside, I'd expect to see the calibration curve at detector #1, but if there is an atom, I'd expect to see a lower-than-calibration-mean energy distribution. Is that right?
 
  • #46
Dfault said:
f we built a sort of "black box" around the atom and detector #2, so that we have the photon gun firing photons into the box and detector #1 detecting photons that exit the box on the opposite side, can we use the energy distribution at detector #1 to infer whether there's an atom inside of the box?

You can compare the distribution measured at detector #1 with the calibrated distribution of the photon gun to infer whether there is something inside the box that can absorb photons, yes. You won't necessarily be able to tell exactly what the something is.
 
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  • #47
Okay, that makes sense. Thanks again!
 
  • #48
Dfault said:
Thanks again!

You're welcome!
 

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