Atom absorbing a photon and emitting a photon afterwards

If an atom absorbs a photon it can only do it if the incoming photon has precisely the energy of the required electron excitation (difference between 2 energy levels of the atom).
The very basic question:
Why can't an electron (bound in an atom) absorb the incoming photon with an energy slightly higher energy than the required excitation and shortly afterwards emit a photon to get rid of the excess energy?
So it would be a bit similar to the Compton effect, just that in the Compton effect the electron is kicked out of the atom.
By this process a continuum of energies could be absorbed (in contradiction to observation)

wacki said:

TL;DR Summary: Why can't an electron absorb the incoming photon with an energy slightly higher energy than the required excitation and shortly afterwards emit a photon to get rid of the excess energy?

If an atom absorbs a photon it can only do it if the incoming photon has precisely the energy of the required electron excitation (difference between 2 energy levels of the atom).
The very basic question:
Why can't an electron (bound in an atom) absorb the incoming photon with an energy slightly higher energy than the required excitation and shortly afterwards emit a photon to get rid of the excess energy?
So it would be a bit similar to the Compton effect, just that in the Compton effect the electron is kicked out of the atom.
By this process a continuum of energies could be absorbed (in contradiction to observation)

The absorption of a photon requires analysis of the quantized EM field. There is actually a range of possible absorption energies centred around each transition energy.

You could search online for more information about the details of the absorption.

Thanks PeroK for your reply.
QED is quite heavy machinery. I was looking for a simpler argument (maybe not as water tight as QED).
There might be a certain breadth of the absorption line due to doppler shift, but let’s keep it simple and let’s assume the real absorption spectrum is discrete (or very close to discrete).
So I still don’t understand why the process I suggested is forbidden, because it would lead to a continuum of absorption lines.

wacki said:

Thanks PeroK for your reply.
QED is quite heavy machinery. I was looking for a simpler argument (maybe not as water tight as QED).
There might be a certain breadth of the absorption line due to doppler shift, but let’s keep it simple and let’s assume the real absorption spectrum is discrete (or very close to discrete).
So I still don’t understand why the process I suggested is forbidden, because it would lead to a continuum of absorption lines.

Perhaps in a way it's simpler in QM. An atom has discrete energy levels that satisfy the Schroedinger equation (SDE). That is what makes the atom quantum mechanical. If an electron could absorb a photon of any wavelength, then you'd need a continuous model, and something other than QM and the SDE.

That wouldn't be quantum mechanics.

When you step up to QED, there is a bit more room for maneouvre - you have a narrow range of wavelengths around each energy level that may be absorbed. The further the energy level is from the mean, the shorter the lifetime of the excited state.

wacki said:

So I still don’t understand why the process I suggested is forbidden, because it would lead to a continuum of absorption lines.

If you look at absorption as an analogy to resonance of a guitar string you can play a loud off-pitch tone and the string will just not respond but just the right pitch will stimulate a resonance.
In the case of atoms, there will be multiple possibles for transitions, some optical and some others.

Raman scattering https://en.wikipedia.org/wiki/Raman_scattering and Two photon excitation https://en.wikipedia.org/wiki/Two-photon_absorption might be topics of your interest.

Guys, thanks for your replies.
The Raman scattering goes a little bit in the direction I suggested, but as far as I understand it, it only works for scattering on the surface of solids and not for gases and it also uses an energy level as intermediate state. But most importantly: The Raman scattering is a special 2 photon scattering that actually exists and I want to understand why the absorption spectrum (of e.g optically thin Hydrogen) involving 2 photons is NOT allowed and is NOT observed in nature (because such an absorption spectrum would be continuous).

Btw, there is no disagreement that the solution of the Schroedinger equation leads to discrete energy levels of the atom (as pointed out by sophiecentaur a classical analogy would be a standing wave of a guitar string). If you assume that the incoming photon is absorbed and after absorption there is no more (outgoing) photon, then the (discrete) excitation energy must equal the energy of the incoming photon (the kinetic energy that the atom receives from the photon is negligible due to momentum conservation). And the result is a discrete absorption spectrum as confirmed by experiments.

At the risk that this might bore people I’ll try one more time:
If there is no outgoing photon after absorption, then the absorption spectrum must be discrete as observed (or close to discrete with very small line breadth).
But if there is an outgoing photon after the absorption, the outgoing photon could take away an arbitrary energy excess of the incoming photon. And the result would be a continuous absorption spectrum in contradiction to observation.
So the question is why is this NOT allowed to have an outgoing photon after the absorption?

wacki said:

So the question is why is this NOT allowed to have an outgoing photon after the absorption?

I wonder if it's as follows. You will only get an absorption line if a particular wavelength, alone, is absorbed. all other wavelengths will go steaming past. If the incident beam is very high energy, many different wavelengths will cause the emission + left - over energy( i.e. many emitted wavelengths). So there wouldn't be a specific wavelength absorbed, measurably and no particular absorption line.
Does that logic make sense?

wacki said:

At the risk that this might bore people I’ll try one more time:
If there is no outgoing photon after absorption, then the absorption spectrum must be discrete as observed (or close to discrete with very small line breadth).
But if there is an outgoing photon after the absorption, the outgoing photon could take away an arbitrary energy excess of the incoming photon. And the result would be a continuous absorption spectrum in contradiction to observation.
So the question is why is this NOT allowed to have an outgoing photon after the absorption?

You would agree that observed discrete light energy series, e.g. the Balmer, the Paschen and the Lyman, decline continuous light spectrum which classical physics suggests.
So light of smaller than gap energy go through and has nothing to do with the system at 100 percent ? No, QM predicts such a case with quntitative evaluation of probability, which is much lower than the normal case as we expect. As an example two photon micsoscopy which uses photons of half energy of the gap excitation
[tex]2\hbar\omega=E_g[/tex]
is now used in brain science and other field of biology to see deep inside. The life forms get less damage from double ##\hbar\omega## photon than from a single ##\hbar 2 \omega## photon. Laser technology can increase the probability to take place by non linear effect.

QM belongs to study of probability. You may calculate your case of not-justfit-absorption-emission process but would get result of very small probability, which is negligible if we are not prepared with advanced technology and special environment to enhance it.

wacki said:

But if there is an outgoing photon after the absorption, the outgoing photon could take away an arbitrary energy excess of the incoming photon.

If you look at the operation of a 'simple' optically pumped laser laser, you flood the laser material with light. Some is absorbed and a large number of atoms are 'pumped' into this state. There then follows a number of energy states, the penultimate of which is a metastable state (it sits there, waiting to be stimulated by a passing photon. So there you have multiple states as a result of one incident photon. BUT the secret is in the probability of each transition, which causes a lot of atoms to be 'trigger able' to release their emitted photons in step.

The "outgoing photon" has a very well defined energy.

Look up fluorescence.
The electron still needs a place to go. I am not understanding your quandry here.

hutchphd said:

Look up fluorescence.
The electron still needs a place to go. I am not understanding your quandry here.

If the incident photon is high enough energy it will ionise the atom - no emissions until an electron hits another atom.
If less than that, it may just go past and not interact. Or it will be absorbed and kick an electron into a high energy state, if one exists. Then the decay can be in one or more stages. This implies an absorption line and one or more emission lines.

I don't see a quandary here either.

I think I see what the OP is asking.

We have a description of virtual absorption. A photon has slightly higher energy than an allowed absorption. It is close enough to the allowed energy that it can be briefly absorbed into a virtual state. However, in a very short time the energy uncertainty resolves and it becomes clear that the energy was wrong. The photon is re-emitted with the original photon energy leaving the electron back in its initial state.

I believe what the OP is asking is why does the virtual state always fall back down to the initial state? Why can’t the virtual state decay to the upper state giving off a very low energy photon?

Clearly not allowed. Carrying away the little bit of excess energy works for energy conservation, but energy is not the only thing that must be conserved. For photon emission, the upper and lower states must differ in a particular way. You can state it as selection rules for angular momentum or you can say that the net difference in the two orbitals must be an electric dipole. Either way, giving off a photon when transitioning from a virtual state to the almost identical allowed state is forbidden.

Yes, this is indeed my problem. You’ve described it through an intermediate virtual state, which makes sense to me. I also like your argument for ruling out the process: Energy conservation is not enough and of course you need to respect conservation of angular momentum as well. I have to admit that the violation of angular momentum in the suggested process is not totally obvious to me.
I have to refresh my memory about selection rules and J conservation, but I think I have something to go on now. So many thanks!

Btw, thanks for all your replies! And sorry that I phrased my question not more precisely, such that most of you didn’t see where my problem was.

Why do we hear so much about Raman scattering for molecular vibrational levels and so little about Raman scattering from atoms? Raman scattering has different selection rules than dipole radiation, but what are allowed Raman scattering lines for a hydrogen atom?
Also, most talk of Raman scattering is for the case the shift is small (the vibrational levels of a few thousand/cm) and the remaining/"scattered" frequency large (tens of thousands/cm). Why don´t we hear much about Raman scattering where the remaining frequency is small compared to the shift? What does the cross-section of Raman scattering do on the large shift, small remaining frequency limit?

Cutter Ketch said:

We have a description of virtual absorption.

I read this in the preliminary brief question at the top of what Google offered in response to my search:

"A more natural distinction between real and virtual photons is that a real photon is one that transfers empirically detectable energy, while virtual photons do not.

Would that be enough to allow what the OP was unhappy about? The slight difference in energy is small and not measurable as the radiated photons would be in all directions from the experimental sample.

wacki said:

Why can't an electron (bound in an atom) absorb the incoming photon with an energy slightly higher energy than the required excitation and shortly afterwards emit a photon to get rid of the excess energy?

It can, but the probability for that happening is smaller when the energy difference (between photon energy and required energy) is larger.

https://physics.stackexchange.com/q...-an-atom-interact-with-a-photon-off-resonance

Demystifier said:

It can, but the probability for that happening is smaller when the energy difference (between photon energy and required energy) is larger.

https://physics.stackexchange.com/q...-an-atom-interact-with-a-photon-off-resonance

I really don’t think it can. The critical point is “excess energy”. He’s not asking about virtual absorption and then re-emitting the the same photon energy as described in your link. He is suggesting the same process, but where only the tiny bit of excess energy is re-emitted and the electron is left in the upper allowed state. That can’t happen.

Cutter Ketch said:

I really don’t think it can. The critical point is “excess energy”. He’s not asking about virtual absorption and then re-emitting the the same photon energy as described in your link. He is suggesting the same process, but where only the tiny bit of excess energy is re-emitted and the electron is left in the upper allowed state. That can’t happen.

Can you give a theoretical explanation why this can't happen?

Demystifier said:

Can you give a theoretical explanation why this can't happen?

Indeed. There are 3 processes under consideration for A+hν:

1. A+hν→A+hν - Rayleigh scattering
2. A+hν→A*+hν* - Raman scattering
3. A+hν→A* absorption

As I mentioned, most discussion of Raman scattering is for the case where ν* >>Δν. But what does Raman scattering cross-section do near the limit ν*→0?

Demystifier said:

Can you give a theoretical explanation why this can't happen?

It seems to me that the underlying question is whether an atom can transition between two energies that are technically part of the same energy level? Let's assume that we are talking about the ground state and the first excited state and the energy difference is ##E##. The basic model is that a photon of energy ##E_{\gamma}## is absorbed (where ##E_{\gamma} \approx E##) and a photon of energy ##E_{\gamma}## is then emitted (with the mean time between absorption and emission being a function of the difference between ##E_{\gamma}## and ##E##). This represents the basic transition between ground state and first excited state and back again.

The question, as I understand it, is whether we can have a three-stage process:

A photon of energy ##E_{\gamma}## is absorbed and (almost immediately) a photon of energy ##E_{\epsilon} \approx E_{\gamma} - E## is emitted. Then, later, a second photon of energy ##E_{\gamma} - E_{\epsilon}## is emitted. The first emission, therefore, would effectively represent a transition between energies within the narrow energy spectrum of the first excited state!

Does the theory of QED actually forbid this?

PeroK said:

A photon of energy ##E_{\gamma}## is absorbed and (almost immediately) a photon of energy ##E_{\epsilon} \approx E_{\gamma} - E## is emitted. Then, later, a second photon of energy ##E_{\gamma} - E_{\epsilon}## is emitted. The first emission, therefore, would effectively represent a transition between energies within the narrow energy spectrum of the first excited state!

Does the theory of QED actually forbid this?

If the atom doesn't change state, definitely! For one, angular momentum wouldn't be conserved.

I believe the process is not strictly forbidden for a complicateed system but will have a vanishingly small cross-section in practice,for a host of reasons that flow from the detailed dynamics. The computations would be, er, interesting.
Perhaps the probabilities of this event are easier to contemplate (as small) if one considers the time-reversed sequence??

wacki said:

TL;DR Summary: Why can't an electron absorb the incoming photon with an energy slightly higher energy than the required excitation and shortly afterwards emit a photon to get rid of the excess energy?
...
By this process a continuum of energies could be absorbed (in contradiction to observation)

The only difficulty is the fact that the electron has no available state that
keeps it bound to its atom or in its material. In X-ray absorption, it is common
for the photoelectric absorbed energy to be too great for binding the electron, so it is emitted
from the material (Auger electrons) or goes on a trajectory in the locale
of its origin (X-ray absorption fine structure can be studied by generating
such outgoing wave electrons).

A ballistic outgoing electron will typically ionize several atoms as it
travels, at circa 30 eV of energy per ion. This is not a simple process, and the
resulting radiations (fluorescence) are not subjects for easy analysis or modeling.