Left over energy when an electron moves to a higher energy level?

[ncstebb]

Hi All,

I'm pondering some questions around photon absorption causing bound electrons to move to higher energy levels.

When a photon strikes an atom, what will happen if the photon has more than enough energy to make an electron move to the first excited state, but not enough for it to move to the second. Surely it will still be absorbed? But what happens to the extra energy? Would it be re-emitted? Wouldn't this result in a broader range of frequencies being emitted than we actually see?​

Many thanks for your help.

 

[diazona]

No, I don't think the photon would be absorbed at all in that situation. It would just go on through without interacting with the atom. This is why you can see discrete spectral lines in the absorption spectrum of a gas, rather than every frequency above a certain one being absorbed.

 

[ncstebb]

Thanks diazona, in hindsight that was a dumb question. It's a difficult thing to explain though. It's as if the atom can check how much energy the photon has before it decides to absorb it??

I expect a complete explanation of this is pretty complex.

 

[diazona]

Not a dumb question at all

I would say this, though: you shouldn't think of it in terms of the atom checking the photon's energy and then making a decision to absorb it or not. For one thing, atoms don't make decisions; but also the "check" and the "decision" are not separate - there's just "absorb if possible."

In a slightly more technical sense, in quantum mechanics, all these events are governed by probabilities. So there's some probability that the photon will be absorbed by the atom, and some probability that it won't. If you do the math, I believe you can calculate that the probability of the photon being absorbed is zero unless the final energy of the bound electron would match the energy of one of the atom's excited states.

A complete explanation, of course, would require quite a bit of math (and it could be fairly complex too - I'm not sure offhand exactly what would be involved).

 

[Petr Mugver]

In general, starting from the two free-particle states af the photon and the atom, different processes are possible (different kinds of absorbtion, different kinds of scattering), and they all involve conservation of momentum and energy. This means that in your case no absorbtion can occur, so there will be scattering (Compton).

 

[phyzguy]

In general, starting from the two free-particle states af the photon and the atom, different processes are possible (different kinds of absorbtion, different kinds of scattering), and they all involve conservation of momentum and energy. This means that in your case no absorbtion can occur, so there will be scattering (Compton).

Compton scattering only occurs with free electrons, not bound electrons. In the case described by the OP, scattering cannot occur in a way that conserves energy and momentum, so the incoming photon is unaffected. As diazona said, this is why there are spectral lines.

 

[Petr Mugver]

So you say that a photon cannot be scattered from an atom?

 

[granpa]

Thanks diazona, in hindsight that was a dumb question. It's a difficult thing to explain though. It's as if the atom can check how much energy the photon has before it decides to absorb it??

I expect a complete explanation of this is pretty complex.

the energy of the photon depends only on its frequency so yes the atom 'knows' how much energy it will have when it is absorbed

due to superposition the electron is, during the brief period while the atom is absorbing the photon, partially in the ground state and partially in the higher state at the same time.

 

[diazona]

the energy of the photon depends only on its frequency so yes the atom 'knows' how much energy it will have when it is absorbed

due to superposition the electron is, during the brief period while the atom is absorbing the photon, partially in the ground state and partially in the higher state at the same time.

Actually as far as I know, there is no evidence to indicate that absorption takes a finite amount of time. At any instant of time, you either have a photon and an atom in the ground state, or no photon and an atom in the excited state. So the transition between the two is, as far as we can tell, instantaneous.

Of course, if you know of an experiment that shows otherwise, please do post a reference.

 

[Dr Lots-o'watts]

Actually as far as I know, there is no evidence to indicate that absorption takes a finite amount of time. At any instant of time, you either have a photon and an atom in the ground state, or no photon and an atom in the excited state. So the transition between the two is, as far as we can tell, instantaneous.

Of course, if you know of an experiment that shows otherwise, please do post a reference.

If a photon has a frequency in the order of a radio wave, and it was absorbed by an antenna, wouldn't the absorption take a finite time?

Scaling down to atomic scale, I would expect the time of absorption to be on the order of the inverse of the photon frequency (corresponding period T = 1/f).

 

[diazona]

If a photon has a frequency in the order of a radio wave, and it was absorbed by an antenna, wouldn't the absorption take a finite time?

With just one photon, I wouldn't think so. But like I said (or, well, implied), I'm not completely sure, so it'd be nice to get a reference on this.

 

[granpa]

emission certainly take a finite amount of time. Why wouldn't absorption?

 

[diazona]

emission certainly take a finite amount of time. Why wouldn't absorption?

Does it? I certainly agree that emission is basically a time-reversed version of absorption, however I don't know that there is any evidence that emission takes a finite time either. As far as I know, both processes are considered to be instantaneous.

 

[granpa]

People that make very accurate atomic clocks specifically look for transitions that take longer.

 

[diazona]

I think that may be referring to something like the half-life of the excited state - that would be the ~average amount of time until the transition happens, not a duration for the transition itself. Can you provide a reference with more specific information?

 

[granpa]

No they want transitions that take longer because the wave train has more cycles and therefore its closer to being a single freq
the spectral line is thinner and that makes for more accurate clocks.

 

[Dr Lots-o'watts]

In non-linear optics, there is a phenomena called two-photon absorption. When the density of photons is sufficient, it is possible for two of them to be absorbed by a state, such that the electron gains a total energy equal to the sum of each photon's. There is a time window in which the second photon has to arrive for this process to happen. I can't explicitly express how long is this delay at the moment. My literature speaks in terms of cross-sections and rates, but I suspect one could work this out to define a time-delay. I'll try to look into it more.

 

[diazona]

No they want transitions that take longer because the wave train has more cycles and therefore its closer to being a single freq
the spectral line is thinner and that makes for more accurate clocks.

Sounds like the energy-time uncertainty principle at work. It makes sense that you want the wavefunction to have a wide extent in time to fix the frequency to high precision, but that still doesn't necessarily mean the transition between quantum states is itself an extended process. I would really like to see something from a paper or textbook about this.

@Dr Lots-o'watts: I'll be interested to see what you find.

 

[granpa]

I've already answered your question.
Its clear to me that you have your own ideas and you will never ever admit that I am right no matter what I show you.

good day to you sir

 

[diazona]

I've already answered your question.
Its clear to me that you have your own ideas and you will never ever admit that I am right no matter what I show you.

How can this be clear to you, when you haven't shown me anything at all? If you insist on believing that I have my mind made up, I can't stop you, but I assure you that you would be wrong about that. If you had actually addressed my question in which I asked you to provide a reference, this conversation could be going in a whole different direction.

 

[K^2]

It's as if the atom can check how much energy the photon has before it decides to absorb it??

No. It's like coupled oscillators. They still interact even if frequencies are wrong, but there is no net energy transfer.

Imagine you have a weight suspended by a string. You are holding the free end. You can make the weight oscillate by moving your hand side to side. Try moving it at too high a frequency, you'll make the weight bounce around a bit, but you wouldn't be able to build up a descent swing. Same thing.

As far as time required for absorption/emission, it is finite. It is, however, described in field theory as instantaneous, but having undetermined time. It's the same thing, physically.

 

[amaresh92]

No, I don't think the photon would be absorbed at all in that situation. It would just go on through without interacting with the atom. This is why you can see discrete spectral lines in the absorption spectrum of a gas, rather than every frequency above a certain one being absorbed.

so according to you there is some energy limit for the each photon to be absorbed?

 

[K^2]

There isn't an upper limit. A really high energy photon can knock an electron out completely, which is almost always an allowed state. So there is no problem.

The problem is with absorbing a photon with energy slightly higher than what's required to transfer an electron to a higher energy state. It has to be just the right amount.

 

[diazona]

Yeah, that's what I meant. Unless the photon has enough energy to knock the electron out completely, it will only be absorbed at specific frequencies.

I'm still really curious as to why everyone's saying that the absorption/emission takes a finite amount of time. As far as I can remember, in what I've done it's always been considered instantaneous (at least in the sense that there isn't any specific time at which the atom would be in a superposition of the ground and excited states, assuming it started out in the ground state).

 

[Dr Lots-o'watts]

... what you find.

"According to the uncertainty principle, population can reside in a virtual level for a time interval of the order of hbar/deltaE, where deltaE is is the energy difference between the virtual level and the nearest real level." - R. Boyd, NLO 3rd ed.

It is during this short time that a second photon can join the first in making the electron hop to the next level. (teamwork!)

It seems to me that the transition from a level to a random (virtual or not) level may be instantaneous, but it may be a short delay before the final, stable (vs virtual) transition takes place. The idea of an electron "trying out" virtual levels before settling on a real transition may not be so wrong. Hopefully, I can get this sorted out more rigorously.

 

[Cthugha]

I'm still really curious as to why everyone's saying that the absorption/emission takes a finite amount of time. As far as I can remember, in what I've done it's always been considered instantaneous (at least in the sense that there isn't any specific time at which the atom would be in a superposition of the ground and excited states, assuming it started out in the ground state).

That is one of the basic result of quantum optics. There are two good arguments showing that the emission process takes a finite amount of time:

1) Accessing the problem experimentally, you can take single photon sources and take autocorrelation traces using a Michelson interferometer. If you find some finite coherence time, this means that the photon itself has some finite "duration" and therefore also the emission process should have some finite duration. This has been done for example for single quantum dots in C. Santori et al., "Indistinguishable photons from a single-photon device", Nature 419, 594-597 (10 October 2002).

2) If you assume the case of strong light-matter interaction (for example an atom inside a cavity), you will find that when using a quantum mechanical treatment for the atom and for the light field, the eigenstates of the coupled system are superposition states of (atom excited, no photon present) and (atom in ground state, 1 photon present) and you get vacuum Rabi oscillations. Now you can increase the decoherence rate and will find that the Rabi oscillations will vanish as the coupling becomes weaker than the decoherence because the superposition just decoheres. The limit of a single atom emitting in free space is the reached by choosing even weaker coupling (or larger decoherence) resulting in a small, but finite time for the emission. Roughly speaking, this timescale reflects in the coherence time of the emitted light. See anything the web has to offer on the Jaynes-Cummings model for more details.

 

[f95toli]

No they want transitions that take longer because the wave train has more cycles and therefore its closer to being a single freq
the spectral line is thinner and that makes for more accurate clocks.

No, they are looking for levels that have longer half-lives. It is just the usual (classical) time-frequency "uncertainty principle" at work: A random process with a long half-life will have a narrow spectral with and vice versa (this is just a mathematical result from Fourier analysis, it has nothing as such to do with QM, hence "classical").

(I am not an expert in this field, but I work with people who are, so I have a fairly good idea why they are working to replace Cs clocks etc)
.

 

[JDługosz]

Yeah, that's what I meant. Unless the photon has enough energy to knock the electron out completely, it will only be absorbed at specific frequencies.

I'm still really curious as to why everyone's saying that the absorption/emission takes a finite amount of time. As far as I can remember, in what I've done it's always been considered instantaneous (at least in the sense that there isn't any specific time at which the atom would be in a superposition of the ground and excited states, assuming it started out in the ground state).

As I recall, there is a period of time where the old state and new state are in superposition. The probability shifts from 100-0 to 0-100 over some time. So, "finite time". But, at any instant within that time, you will only detect it to be in one state or the other, so you will see the absorption as having occurred yet or not; no funny intermediate state. So, there is an instantaneous cut-off where your observations change from seeing one state to seeing the other. So, "instantaneous".