Cause of spontaneous emission?

[sol47739]

TL;DR Summary

I have some questions about spontaneous emission and if it is purely due to quantum fluctuations or if something called radiation reaction is involved as I encountered in some literature.

I am reading this chapter 3 from the book called The Quantum Vacuum by P.Milonni.(Attached in the pdf, look at chapter 3.2 Spontaneous emission)There they say that spontaneous emission is due to both quantum fluctuations and radiation reaction. They say the transitions induced by the quantum fluctuations gives only the half of the Einstein A coefficient for spontaneous emission and in order to account for the full we need to consider radiation reaction.

„Consider now the effect of radiation reaction. As shown in Appendix A, the radiation reaction field is responsible for the rate 2 e 2 a 2 / 3 c 3 at which an oscillating charge loses energy to the electromagnetic field. For oscillation at frequency ωσ with amplitude x0 this rate is…“What exactly do they mean with radiation reaction here? I mean the electron is bound to the nucleus and the position probability is calculated by the Schrödinger wavefunction, so I have difficulty understanding what is exactly oscillating. Of course the Schrödinger orbital resembles the wavefunticons for three dimensional oscillations but in the quantum case they represent a probability function where we can encounter the electron instead.So what are they talking about for oscillation? From my understanding it seems as if they mean that the electron moving around in the orbital will have a oscillating motion and that when the charge accelerates in this oscillating motion it will according to classical electromagnetism emit electromagnetic radiation, but then my understanding from what they write, the half A coefficient due to radiation reaction would be emitting continuously all the time, which we know doesn’t happen, otherwise we are back at the level of why an electron orbiting the nucleus doesn’t emit radiation while spiraling down to the nucleus, just that it somehow just emitted half of the energy and the rest was due to the familiar quantum jumps.So since we know the emission of photons from atoms are an all or none process, I personally can’t put the cause of spontaneous emission on anything else than the quantum fluctuations, and that a quantum fluctuation of the right frequency would induce a spontaneous emission to occur. Why does it in their calculating seem that the quantum fluctuations just is responsible for half of the emission rate?They mention that their derivation is semiclassical can this be the cause for the inconsistencies in the reasoning?In another paper by B. Fain Spontaneous Emission vs. Vacuum Fluctuations.(I also attached this paper) In this paper they conclude the following:„Thus we come to the quite obvious conclusion that, in order to achieve noncontroversial results for the spontaneous emission, we must treat both the matter and the field according to the quantum theory. Another conclusion which stems from the example of the harmonic oscillator interacting with the radiation is that nonspontuneous emission is purely a quantum effect owing to the vacuum fluctuations, but the absence o/ this emission in the grou~,d state is purely a quantum el]cot which is due to the vacuum fluctuations.“

„From the above several conclusions follow.

1) Spontaneous emission exists both in classical and in quantum theories.

2) The semi-classical approach, when matter is treated quantum-me- chanically and radiation in the classical way, does not give correct results at lowlying levels of matter.

3) The crucial role of the vacuum fluctuations emerges in the ground state of matter. The stability o/ the ground state (i.e. the fact that it does not radiate) is purely a quantum eJ]evtwhivh is due to the vacuum ]luvtuations. „
They don't mention anything about radiation reaction.

So to summarize what I wonder: Is the correct answer that the only cause of spontaneous emission are quantum fluctuations?
In the chapter by Milonni why do they talk about radiation reaction and what do they actually mean with it, what does it refer to? If the radiation reaction idea is false why do they only come to half the A coefficient for quantum fluctuations? Is it because they treat it somewhat semiclassically? As they themselves say „On the basis of this simplistic semiclassical analysis, therefore, we have arrived at the conclusion that..“. And does this have to do with something they say in the other paper by B.Fain that one need to treat both the radiation and the matter quantum mechanically otherwise it will result in wrong results especially for the lower energy levels? Another thing that makes me think Milonni reasoning is wrong comes from what I read about Cavity QED, which literally modifies the emission properties of atoms by the modification of the quantum vacuum. And in most literature I have read about Cavity QED I haven’t seen radiation reaction being mentioned.

 

[Demystifier]

https://opg.optica.org/josab/abstract.cfm?uri=josab-2-3-447

 

[Demystifier]

I think the Milonni's writings are not wrong, but are unnecessarily confusing. In particular, in the comment [1] on the Puri's paper I linked above, Milonni writes:
"Thus I disagree with Puri's "assertion that the physical interpretation of the natural processes should not depend on the choice of our mathematical procedures."
... The more physical interpretations we can adduce for one and the same
phenomenon, the richer is our understanding of that phenomenon!"

More importantly, Milonni agrees with Puri that relative contributions of radiation reaction and vacuum fluctuations depend on operator ordering. Explicitly, Milonni writes [1]: "In particular, one can emphasize the role of either radiation reaction or vacuum-field fluctuations, depending on how the operators are ordered."

[1] P. W. Milonni, Comment on "Spontaneous emission: vacuum fluctuations
or radiation reaction" by R. R. Puri, J. Opt. Soc. Am. B/Vol. 2, No. 12/December 1985

 

[sol47739]

I think the Milonni's writings are not wrong, but are unnecessarily confusing. In particular, in the comment [1] on the Puri's paper I linked above, Milonni writes:
"Thus I disagree with Puri's "assertion that the physical interpretation of the natural processes should not depend on the choice of our mathematical procedures."
... The more physical interpretations we can adduce for one and the same
phenomenon, the richer is our understanding of that phenomenon!"

More importantly, Milonni agrees with Puri that relative contributions of radiation reaction and spontaneous emission depend on operator ordering. Explicitly, Milonni writes [1]: "In particular, one can emphasize the role of either radiation reaction or vacuum-field fluctuations, depending on how the operators are ordered."

[1] P. W. Milonni, Comment on "Spontaneous emission: vacuum fluctuations
or radiation reaction" by R. R. Puri, J. Opt. Soc. Am. B/Vol. 2, No. 12/December 1985

Thanks for your answer! I see that it is different depending on how you order the operators and that Milonni agrees with Puri about this. The remaining question I have is what is the radiation reaction physically referring to in this case? Since radiation reaction in the normal sense is thought of as being the influence on the charge itself when electromagnetic waves are emitted from an accelerating charge. Therefore I wonder where in the atom do they consider some radiation reactions to occur? And how is this thought compatible with the picture that emission of photons from an atom is an all or none process and not continuous as in the classical case of an accelerating charge? And how would the radiation reaction be causing emission?

 

[Demystifier]

The remaining question I have is what is the radiation reaction physically referring to in this case?

Radiation reaction refers to terms coming from interaction between charges and electromagnetic potential. Note that the electromagnetic potential is an operator, not a classical field, so it is non-zero even when the classical radiation is absent. More precisely, if we denote by $A$ the electromagnetic potential operator and by $j$ the charged current operator, then we may have $\langle A\rangle=0$ (the absence of classical radiation), but $\langle Aj\rangle \neq 0$ (the presence of radiation reaction).

So the claim that spontaneous emission is caused by "radiation reaction" really means that it is caused by quantum interaction between charges and electromagnetic potential, which is a rather intuitive picture.

 

[sol47739]

Radiation reaction refers to terms coming from interaction between charges and electromagnetic potential. Note that the electromagnetic potential is an operator, not a classical field, so it is non-zero even when the classical radiation is absent. More precisely, if we denote by $A$ the electromagnetic potential operator and by $j$ the charged current operator, then we may have $\langle A\rangle=0$ (the absence of classical radiation), but $\langle Aj\rangle \neq 0$ (the presence of radiation reaction).

So if I understand correctly that is indeed also a form of quantum fluctuation? Since the commutator is non-zero? To physically interpret it, am I correct in the following reasoning? The electromagnetic potential varies in space limited by the uncertainty principle as other types of quantum fluctuations. And the radiation reaction in this case would refer to the electromagnetic potential fluctuations "accelerating" the charge temporarily and this temporarily accelerating charge will be influenced by that and emit a virtual photon which then imparts on it's momentum, but all of this happenings within the virtual regime so one could draw a Feynman diagram of it.

As you pointed out in your previous answer what Milonni wrote " "In particular, one can emphasize the role of either radiation reaction or vacuum-field fluctuations, depending on how the operators are ordered." If I understand this correctly now both radiation reaction and "normal" vacuum fluctuations are both due to the quantum vacuum, they would just represent different paths in a Feynman diagram, and you can get the same result regardless of if you emphasize radiation reaction or normal quantum fluctuations.

Did I understand it correctly?

 

[Demystifier]

Did I understand it correctly?

You are right that it is also a kind of quantum fluctuation. However, it is conceptually wrong to think of quantum fluctuations as random changes in time. Also, a quantum fluctuation is not necessarily related to non-commutativity. For example, for a harmonic oscillator in ground state we have $\langle x^2\rangle \neq 0$, but $x$ commutes with itself and nothing changes with time.

 

[Demystifier]

In fact, the true fundamental reason for spontaneous emission is this. When the atom is initially in an excited state without radiation, this state is the Hamiltonian-eigenstate of the atom Hamiltonian alone, but not a Hamiltonian-eigenstate of the full Hamiltonian involving also the interaction of atom with the EM field. Since this is not a Hamiltonian eigenstate of the full Hamiltonian, the true state in the Schrodinger picture has an oscillatory dependence on time, so at a later time it has a finite probability to be in a different state of atom+radiation. This fundamental picture, however, is obscured by the fact that we compute perturbatively, i.e. don't understand the system well from the fundamental non-perturbative point of view.

 

[Demystifier]

If I understand this correctly now both radiation reaction and "normal" vacuum fluctuations are both due to the quantum vacuum,

Yes, but here "vacuum" means initial state without photons.

they would just represent different paths in a Feynman diagram,

There are no "paths" in Feynman diagram*, perhaps you meant different Feynman diagrams?

and you can get the same result regardless of if you emphasize radiation reaction or normal quantum fluctuations.

That part is true.

*Feynman diagram, Feynman path integral, and Feynman method of computing integrals appearing in Feynman diagrams are 3 very different concepts.

 

[sol47739]

Yes, but here "vacuum" means initial state without photons.

There are no "paths" in Feynman diagram*, perhaps you meant different Feynman diagrams?

That part is true.

*Feynman diagram, Feynman path integral, and Feynman method of computing integrals appearing in Feynman diagrams are 3 very different concepts.

I meant Feynman diagram yes, here I tried to draw what I understand from this in form of some sort of Feynman diagram. Is this correct or am I really far away from understanding the radiation reaction in this case?

If this is somewhat correct I wonder what the the "electromagnetic potential operator photon"(the brown in the picture) would be addressed with in a Feynman diagram? Or is it that one could think of it as some positrons and electron pairs producing an electromagnetic field nearby which then accelerates the electron?

 

[sol47739]

You are right that it is also a kind of quantum fluctuation. However, it is conceptually wrong to think of quantum fluctuations as random changes in time. Also, a quantum fluctuation is not necessarily related to non-commutativity. For example, for a harmonic oscillator in ground state we have $\langle x^2\rangle \neq 0$, but $x$ commutes with itself and nothing changes with time.

But am I correct that one could say that the radiation reactions origin lies in the quantum vacuum too? If so through which crucial process in the radiation reaction contributes to spontaneous emission? Is it somewhat close to what I talk about in the following message?

I meant Feynman diagram yes, here I tried to draw what I understand from this in form of some sort of Feynman diagram. Is this correct or am I really far away from understanding the radiation reaction in this case? View attachment 314090
If this is somewhat correct I wonder what the the "electromagnetic potential operator photon" would be addressed with in a Feynman diagram? Or is it that one could think of it as some positrons and electron pairs producing an electromagnetic field nearby which then accelerates the electron?

Another question that arose is in the case we are treating the phenomena of spontaneous emission in terms of radiation reaction + vacuum fluctuations = A Einstein coefficient for spontaneous emission. I know the A Einstein coefficient refers to the probability for a transition to occur. But I wonder how should I interpret it? For one single emission do we need both a vacuum fluctuation and a radiation reaction? Or is it rather that 50 % of the emissions happens due to radiation reactions and 50 % due to vacuum fluctuations, or other percentages depending on how you order the operators as we talked about earlier. The question is in other do we need either a radiation reaction OR a vacuum fluctuation for on particular emission to occur or do we need both, and both contribute equally to one particular emission?

 

[sol47739]

In fact, the true fundamental reason for spontaneous emission is this. When the atom is initially in an excited state without radiation, this state is the Hamiltonian-eigenstate of the atom Hamiltonian alone, but not a Hamiltonian-eigenstate of the full Hamiltonian involving also the interaction of atom with the EM field. Since this is not a Hamiltonian eigenstate of the full Hamiltonian, the true state in the Schrodinger picture has an oscillatory dependence on time, so at a later time it has a finite probability to be in a different state of atom+radiation. This fundamental picture, however, is obscured by the fact that we compute perturbatively, i.e. don't understand the system well from the fundamental non-perturbative point of view.

Thanks for shining light on another way of thinking of it. So if I understand this it is that from the Schrödinger picture there is a non-zero probability that the system will at some time be in a state of atom+emitted photon. Is the problem we not knowing the fundamental non-perturbative point of view related to the "physical process of how the emission happens"? Radiation reaction or vacuum fluctuation and their combinations? Or am I wrong about this?

 

[Demystifier]

So if I understand this it is that from the Schrödinger picture there is a non-zero probability that the system will at some time be in a state of atom+emitted photon.

Yes.

Is the problem we not knowing the fundamental non-perturbative point of view related to the "physical process of how the emission happens"?

The problem is that we don't know how to solve the interacting system exactly.

 

[sol47739]

Yes.

The problem is that we don't know how to solve the interacting system exactly.

Okey, I see, do you have any comments on this:

I meant Feynman diagram yes, here I tried to draw what I understand from this in form of some sort of Feynman diagram. Is this correct or am I really far away from understanding the radiation reaction in this case? View attachment 314090
If this is somewhat correct I wonder what the the "electromagnetic potential operator photon"(the brown in the picture) would be addressed with in a Feynman diagram? Or is it that one could think of it as some positrons and electron pairs producing an electromagnetic field nearby which then accelerates the electron?

But am I correct that one could say that the radiation reactions origin lies in the quantum vacuum too? If so through which crucial process in the radiation reaction contributes to spontaneous emission? Is it somewhat close to what I talk about in the following message?Another question that arose is in the case we are treating the phenomena of spontaneous emission in terms of radiation reaction + vacuum fluctuations = A Einstein coefficient for spontaneous emission. I know the A Einstein coefficient refers to the probability for a transition to occur. But I wonder how should I interpret it? For one single emission do we need both a vacuum fluctuation and a radiation reaction? Or is it rather that 50 % of the emissions happens due to radiation reactions and 50 % due to vacuum fluctuations, or other percentages depending on how you order the operators as we talked about earlier. The question is in other do we need either a radiation reaction OR a vacuum fluctuation for on particular emission to occur or do we need both, and both contribute equally to one particular emission?

Or is it that we just know that spontaneous emission happens because of quantum physics and it's probabilistic nature, which can be seen from the Schrödinger picture? But although we can formulate it in terms of some perturbation, radiation reaction and vacuum fluctuations, we can't know for sure the true origin of it just that it is a non zero probability in terms of quantum physics?

 

[Demystifier]

Or is it that we just know that spontaneous emission happens because of quantum physics and it's probabilistic nature, which can be seen from the Schrödinger picture?

Yes.

 

[vanhees71]

Quantum theoretically you have a Hamiltonian (in Coulomb gauge)
$$\hat{H}=\hat{H}_{\text{el}} + \hat{H}_{\text{em}} + \hat{H}_{\text{int}}.$$
Here $\hat{H}_{\text{el}}$ is the Hamiltonian describing the electrons in the Coulomb potential of the nucleus and the Coulomb potentials of their mutual interaction. The eigenvalue problem of this Hamiltonian give (besides the free motion of its center of mass) and the bound states of the atom (as well as the scattering states, but these we are not interested in). $\hat{H}_{\text{em}}$ is the free Hamiltonian of the radiation field, and $\hat{H}_{\text{int}}=-\hat{J}_{\mu} \hat{A}^{\mu}$ the interaction between the electrons with the em. radiation field, which you treat with perturbation theory, and this gives you transition matrix elements for the transition of a state $|E_n,\{N_{\gamma}(\omega)=0\}_{\omega}=0 \rangle$ to $|E_{n'},N_{\gamma}(\omega_{nn'})=1,\{N_{\gamma}(\omega)=0 \}_{\omega \neq \omega_{nn'}}$, i.e., the transition from an excited energy level $E_n$ to a lower state $E_{n'}$ under emission of a photon with frequency $\omega_nn'=(E_n-E_{n'})/\hbar$. This is spontaneous emission, which you get (only!) when treating the em. field as a quantum field. It's the most simple argument for the necessity of field quantization!