# Photon absorption for an atomic electron

• I

## 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?

## Main Question or Discussion Point

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

Last edited:

Related Quantum Physics News on Phys.org
DrClaude
Mentor
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.

Spinnor, PeroK, Dfault and 1 other person
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?

unless ##\Delta t \rightarrow \infty## (infinite lifetime)

##\Delta t## is the lifetime of exactly what?

f95toli
Gold Member
##\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.

kent davidge
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
Mentor
2019 Award
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
Mentor
2019 Award
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
Mentor
2019 Award
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
Homework Helper
Gold Member
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?

Dfault
PeterDonis
Mentor
2019 Award
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
Homework Helper
Gold Member
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
Mentor
2019 Award
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
Mentor
2019 Award
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
Homework Helper
Gold Member
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
Mentor
2019 Award
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
Homework Helper
Gold Member
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
Mentor
2019 Award
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
Homework Helper
Gold Member
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
Mentor
2019 Award
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
Mentor
2019 Award
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
Homework Helper
Gold Member
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.