I Photon absorption for an atomic electron

[Dfault]

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

 

[DrClaude]

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.

 

[Dfault]

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?

 

[kent davidge]

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?

 

[f95toli]

$\Delta t$ 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.

 

[Dfault]

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?

 

[PeterDonis]

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.

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.

 

[Dfault]

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!

 

[PeterDonis]

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.

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.

 

[Dfault]

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?

 

[PeterDonis]

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.

 

[Dfault]

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?

 

[PeroK]

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?

 

[PeterDonis]

Is that right?

Yes.

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.

 

[PeroK]

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?

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[PeroK]

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?

 

[PeterDonis]

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.

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.

 

[PeroK]

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.

 

[PeterDonis]

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.

Unless this is practically or theoretically impossible

It is. See above.

 

[PeroK]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.

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.

 

[PeroK]

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.

 

[PeterDonis]

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.

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.

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.

 

[PeterDonis]

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.

 

[PeroK]

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.

 

[PeterDonis]

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.

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.

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.

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.

I suggest

At this point I think this discussion needs references, not "suggestions", to proceed any further.

 

[PeterDonis]

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.

 

[PeroK]

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.

 

[PeterDonis]

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.

 

[Dfault]

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?

 

[PeterDonis]

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.

 

[Dfault]

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?

 

[Dfault]

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

 

[PeterDonis]

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?

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?

 

[Dfault]

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?

 

[PeterDonis]

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.

 

[Dfault]

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?

 

[PeterDonis]

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.

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.

 

[PeterDonis]

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.

 

[Dfault]

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?

 

[PeterDonis]

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.

 

[Dfault]

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?

 

[PeterDonis]

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.

 

[Dfault]

Okay, that makes sense. Thanks again!

 

[PeterDonis]

Thanks again!

You're welcome!