(N+1) enhancement of spontaneous emission, why not (N+1/2)

[Spinnor]

Feynman tells us,

"The probability that an atom will emit a photon into a particular final state is increased by the factor (n+1) if there are already n photons in that state."

4–4 Emission and absorption of photons, http://www.feynmanlectures.caltech.edu/III_04.html

The 1 in (N+1) comes from stimulated emission by the quantum vacuum?

The ground state though has the minimum energy 1/2hν?

Naively I might expect the factor (N+1) to be instead (N+1/2) because the ground state is in some sense only 1/2 the "strength" of a real quantum so its effects might be half as much?

I suspect there is a simple answer to this.

Thanks!

 

[PeterDonis]

The 1 in (N+1) comes from stimulated emission by the quantum vacuum?

No, it comes from Bose statistics. Read through Feynman's derivation of the N+1 factor. Does he talk at all about stimulated emission?

I suspect there is a simple answer to this.

Yes, that your intuitive reasoning is starting from an invalid premise. See above.

 

[Cthugha]

The 1 in (N+1) comes from stimulated emission by the quantum vacuum?

Yes. This point of view has been introduced first by Welton (T.A. Welton, Phys. Rev. 74, 1157 (1948)). Similar ideas can be found in earlier papers by Weisskopf. It is quite interesting that one can equivalently see it as a consequence of radiation reaction. Milonni gave some good insights into this in his somewhat pedagogical paper "Why spontaneous emission?" (Am. J. Phys. 52 (4), 1984). But (as always with a grain of salt) spontaneous emission arises due to something that can be loosely called stimulated emission via the vacuum. The light field is mapped onto harmonic oscillator modes and thus the quadratures of the light field are subject to the same uncertainty as position and momentum in the standard harmonic oscillator.

The ground state though has the minimum energy 1/2hν?

Naively I might expect the factor (N+1) to be instead (N+1/2) because the ground state is in some sense only 1/2 the "strength" of a real quantum so its effects might be half as much?

The ground state energy is calculated in a manner analogous to the classical intensity, which is proportional to the square of the field. The operator for the field is proportional to the difference between photon annihilation and creation operators: $\hat{a}-\hat{a}^\dagger$.
If you square this expression, get the order right and calculate the expectation value for a photon number eigenstate, you will get the typical product of operators that yields the n+1/2-term in energy. That is a simple exercise to do. For transitions, you couple just the field operator (not its square) to the dipole operator. This yields plenty of terms, but after performing the typical approximations, what is left is a operator of the kind $\hat{\sigma}\hat{a}^\dagger$. The first one is the lowering operator for the atom. The second one is the photon creation operator. And this operator has the typical $\sqrt{n+1}$-eigenvalue for the amplitude, which results in the n+1-term, when calculating the expectation value for a transition between different photon number eigenstates. You basically get n+1-like terms when dealing with fields and the n+1/2-like terms when dealing with intensities.

And please do not try to learn anything related to quantum optics from the Feynman lectures. It is a good book for lots of things, but quantum optics did not even really exist when Feynman gave his famous lectures. Time and again it takes me lots of efforts to get my students to unlearn some of the outdated stuff that is presented in the Feynman lectures when it comes to quantum optics.

 

[PeterDonis]

Yes.

I'm not sure that's correct, at least not as a response to the most general version of the OP's question. Stimulated emission is a particular example of Bose statistics, but it's not the only one. So the logical progression would be Bose statistics = N+1 -> stimulated emission, not stimulated emission -> N+1.

 

[vanhees71]

Feynman tells us,

"The probability that an atom will emit a photon into a particular final state is increased by the factor (n+1) if there are already n photons in that state."

4–4 Emission and absorption of photons, http://www.feynmanlectures.caltech.edu/III_04.html

The 1 in (N+1) comes from stimulated emission by the quantum vacuum?

The ground state though has the minimum energy 1/2hν?

Naively I might expect the factor (N+1) to be instead (N+1/2) because the ground state is in some sense only 1/2 the "strength" of a real quantum so its effects might be half as much?

I suspect there is a simple answer to this.

Thanks!

No, the "+1" comes from spontaneous emission. It's one of the most simple proofs for the necessity of quantization of the radiation field in addition to the quantization of particles. The importance of spontaneous emission has been discovered by Einstein, when he rederived Planck's Law for the black-body radiation by a kinetic approach of statistical physics. He realized that in addition to absorption and induced emission (both proportional to the intensity of the radiation) he had to introduce spontaneous emission.

In modern QT it's described most easily with the method of field quantization, which most naturally described processes, where the number of particles (or rather quanta!) is not conserved. To understand, why field quantization is necessary to describe spontaneous emission, simply think about the hydrogen atom. When treating it in the usual QM1 way you describe the electron around the proton quantum mechanically and the electromagnetic field classically as the Coulomb field of the proton, in which the quantum electron is moving. Solving for the energy eigenstates, which are the stationary states of the atom, you find the well-known ground and excited states. Within this model, by construction not only the ground state but also all excited states are stable, i.e., have infinite lifetime, i.e., a lonely hydrogen atom once excited will stay excited forever.

Now you use the complete QFT to describe the hydrogen atom. In Coulomb gauge (which is most appropriate for this bound-state problem) you can start by solving for the electron moving in the Coulomb field, giving you the description as if the em. field is not quantized, but then you can solve for the rest of the em. interaction with perturbation theory, and then you'll see that there are socalled "radiative corrections", which among other things give rise to spontaneous emission, i.e., if you prepare a hydrogen atom in an excited energy eigenstate (of the unperturbed problem) the radiation corrections inevitably lead to a certain non-zero transition probability to a hydrogen atom at a lower energy eigenstate and a photon, i.e., the spontaneous creation of a photon. It's one of the most stringent manifestations of the fact that there are quantum fluctuations of the electromagnetic field. Only the groundstate of the system is really stable, i.e., if the system is prepared in the energy eigenstate of the lowest possible energy there cannot be spontaneously emitted any photons since energy conservation forbids it. So in the full description of the hydrogen atom only the ground state is stable, having infinite lifetime.

Then there are of course also other radiative corrections, giving rise to slight shifts in the energy eigenvalues. The calculation of these socalled Lamb shifts of the hydrogen levels leads to one of the most accurate agreements between theory and experiment ever, but that's another story.

Although it is not as easy to argue for field quantization as unfortunately many textbooks suggest: The introductory chapter on QM 1 textbooks usually contains the false historical claims that the photoelectric or Compton effects necessitate the introduction of photons, but that's not true; on the level of accuracy treated in this introductory chapter you don't need to quantize the em. field, but the semiclassical method where only the particles are quantized by the em. field treated still classically, is sufficient. The observed spontaneous emission, however, is the most simple proof for the necessity of treating also the em. field as quantum system.

 

[Cthugha]

So the logical progression would be Bose statistics = N+1 -> stimulated emission, not stimulated emission -> N+1.

I am not sure I get your point. The question was whether the "+1" spontaneous emission term can be interpreted as stimulated emission via the vacuum. Yes, this is one way to look at it (IF one knows what he is doing and has an idea of how field quantization works - sticking to a somewhat mystical vision of "quantum fluctuations do it" without having a clear vision of what that means is of course not helpful).

Feynman's derivation is a nice trick that gives the reader the right impression without actually demonstrating it. If you really wanted to calculate spontaneous emission this way, you end up having to evaluate equation 4.17 for n=0, which is not defined in a sensible manner. At this point one really needs to include spontaneous emission in a rigid way. But without any doubt, it is great pedagogical skill by Feynman to hide this point so well that students will get the basic idea without worrying about the non-trivial physics of spontaneous emission, which will typically be over their head at this point of the lecture.

 

[vanhees71]

No! Although zillions of popular-science writers claim that the vacuum fluctuates, it doesn't. The field fluctuates, but not the vacuum!

 

[Cthugha]

No! Although zillions of popular-science writers claim that the vacuum fluctuates, it doesn't. The field fluctuates, but not the vacuum!

Sure, but I never claimed that. This is the reason why I explicitly mentioned in my first post that it is the field operator that will give rise to the terms that go like n+1, while the terms proportional to energy would not.

 

[PeterDonis]

I am not sure I get your point.

Another way of putting it would be that the argument you give in the last part of post #3 does not just apply to the particular case under discussion; it applies to any process involving bosons. But stimulated emission is only one of many possible such processes; so saying the +1 is "due to stimulated emission" doesn't seem like a good general way of putting it, since it only applies to that particular process; whereas saying the +1 is due to Bose statistics makes it clear that it applies to all processes involving bosons.

The question was whether the "+1" spontaneous emission term can be interpreted as stimulated emission via the vacuum. Yes, this is one way to look at it

One way to look at it for that particular process, yes. But what about other processes involving bosons, to which the same argument applies?

 

[Cthugha]

Another way of putting it would be that the argument you give in the last part of post #3 does not just apply to the particular case under discussion; it applies to any process involving bosons. But stimulated emission is only one of many possible such processes; so saying the +1 is "due to stimulated emission" doesn't seem like a good general way of putting it, since it only applies to that particular process; whereas saying the +1 is due to Bose statistics makes it clear that it applies to all processes involving bosons.

Yes, there are plenty of effects, where this feature arises. Stimulated emission, photon bunching, bosonic final state stimulation. You name it. The problem with attributing this property directly to Bose statistics is that this connection has been done wrong or inadequately in many old (and sometimes also new) textbooks. If you think of "Bose statistics" in terms of the spin statistics theorem and what happens with respect to exchange symmetry and commutation relations, this works fine. And then plenty of texts go on to say that Bose-Einstein statistics apply to bosons, investigate the equilibrium photon gas and use "Bose statistics" as a synonym for particles following Bose-Einstein statistics. This is wrong. For example lasers are highly non-equilibrium systems, which create light fields that do not follow Bose-Einstein statistics. The photon number distribution is Poissonian. Accordingly, there is no photon bunching. However, of course laser light will still induce stimulated emission. Still, plenty of texts introduce photon bunching and stimulated emission as "basically the same" and a consequence of Bose-Einstein statistics. That is incorrect.
And also I would like to note that the emphasis on the quote should be on the other part. The "due to stimulated emission" is way less important than the "by the quantum vacuum" part. This is a general effect of fields with non-zero field variances in their ground states.

One way to look at it for that particular process, yes. But what about other processes involving bosons, to which the same argument applies?

Milonni's 1994 book on the quantum vacuum gives a good explanation of how general this is.
Well, of course any bosonic system can be mapped onto the harmonic oscillator and the vacuum state of any bosonic field will be similar. Their Wigner functions will of course look pretty similar and the non-zero field variances will of course also always have a similar effect. If one emphasizes Bose statistics, one emphasizes discrete variables. If one emphasizes the fields, one emphasizes continuous variables. As it is always the fields that fluctuate and not the occupation numbers, I always try to emphasize the continuous variable point of view, when it comes to questions like this.
Yes, it is unfortunate that some particle physics guys go ahead and tell obscure stories about virtual particles entering and leaving reality, but that does not change the fact that the properties of vacuum states of fields are pretty general.

 

[Spinnor]

No! Although zillions of popular-science writers claim that the vacuum fluctuates, it doesn't. The field fluctuates, but not the vacuum!

When I think of a vacuum I think of some region in empty space with zero field excitations, no neutrinos, electrons, protons ect and no excitations of the electromagnetic field. What is left is the ground state of all the fields of the standard model, that is the vacuum to me. Why can't we equate the vacuum with the ground state? I am sure you have written many posts about this, can you point me there.

Thanks

 

[Cthugha]

I realized that my last post was a bit too brief to understand, so I think I should comment on this question a bit more in detail to make it a bit easier to understand what I mean. My problem with the following statement:

But stimulated emission is only one of many possible such processes; so saying the +1 is "due to stimulated emission" doesn't seem like a good general way of putting it, since it only applies to that particular process; whereas saying the +1 is due to Bose statistics makes it clear that it applies to all processes involving bosons.

is that I do not think that it is as general as people sometimes consider it to be. Let me explain that in some detail. What is happening microscopically in an emission process? Assuming emission from some typical dipole transition of the hydrogen atom, we already know that both the excited state and the ground state are stable, when just considering the Hamiltonian of bare hydrogen. These states do not have any time-dependent charge distributions that could radiate. A state that will have some time-dependent spatial probability distribution for the electron is a superposition between the ground state and an excited state of different parity, e.g. a superposition of 2p and 1s. However, in order to put the excited state into a superposition state, one needs a perturbation. This is easily done by an external field. This coupling is the quantized version of the classical dipole coupling $-q\vec{r}\vec{E}$.

Quantizing this is pretty simple. Using minimal substitution one can either express it as a coupling of the vector potential to the electron momentum or of the electric field to the dipole operator. The latter case is easier because it is the literal "translation" of the classical case to the quantum formalism. One finds that the dipole part transforms to: $q\vec{r}\rightarrow\vec{d}\sigma^\dagger+\vec{d}^\ast \sigma$, where $\vec{d}$ corresponds to the dipole moment and $\sigma$ and $\sigma^\dagger$ are the lowering and raising operators for the two-level system. More important is the operator for the electric field, which is given by $\hat{E}=c(\hat{a}-\hat{a}^\dagger)$, where c is a factor summing up constants and geometric parameters and $\hat{a}$ and $\hat{a}^\dagger$ are the photon annihilation and creation operators. Putting all constants into the light-matter coupling constant g, the total expression yields $\hat{H}_{int}=\hbar g (\hat{\sigma}^\dagger \hat{a} -\hat{\sigma}^\dagger \hat{a}^\dagger -\hat{\sigma}\hat{a}+\hat{\sigma}\hat{a}^\dagger)$.
These four terms are easy to interpret: photon absorption while the atom goes to the excited state, photon emission while the atom goes to the excited state, photon absorption while the atom goes to the ground state and photon emission while the atom goes to the ground state. Two of these processes do not conserve energy and are only relevant for huge coupling strengths. The other two describe typical absorption and emission. In the absence of any losses or other perturbations, the superposition state will oscillate back and forth emitting a photon and absorbing it again until all eternity. In the case of spontaneous emission, this disturbance comes pretty regularly or the perturbing field is not constant in time, so just the one term describing the atom going to the ground state and emitting a photon becomes relevant. This leaves us with the following terms for the rate of an optical transition:
$\langle g,n+1|g \hat{\sigma}\hat{a}^\dagger |e,n\rangle=\langle g,n+1|g \sqrt{n+1} |g,n+1\rangle$.
The absolute square of this yields the transition rate, so this is $g^2 (n+1)$, which is the typical result. The $\sqrt{n+1}$-factor comes from the definition of the photon creation operator and without any doubt this is a consequence of Bose statistics. But what is often lost in explaining spontaneous emission is that the photon number does not enter because there is a photon number operator involved somewhere, but because of the electric field operator. The physics behind this is easy to grasp. The field perturbs the atom and turns the excited state into the superposition state, which will end up in the ground state emitting a photon eventually. Larger perturbations via larger fields result in larger transition rates. There is nothing in this equation that really distinguishes between stimulated and spontaneous emission. Any field will do, whether it is a field arising due to the vacuum state or a field arising due to a Fock state. The physics is the same. The difference between stimulated emission and spontaneous emission is more or less a historical one because depending on the level of approximation used one of these terms survives, while the other does not, but there is no intrinsic physical difference between spontaneous and stimulated emission. And in this sense it is perfectly adequate to describe spontaneous emission as stimulated emission via the vacuum.
Is a description in terms of Bose statistics more general? Well, any transition due to stimulation/perturbation via a bosonic field will show this dependence because the field operator includes the creation operator. However, other processes, such as non-linear processes or quadrupole transitions and similar processes do not necessarily show this scaling, unless the transition depends linearly on the perturbation by the bosonic field.
Now one might also argue that there is a difference between spontaneous and stimulated emission because for the vacuum field the perturbation is obviously caused by fluctuations. However, this is true as well for any Fock state as all Fock states have a vanishing expectation value of the electric field. So with respect to this, there is also no essential difference between spontaneous and stimulated emission.
However, the main point why I dislike the simple (n+1) description of stimulated emission is that it only really works well for photon number states as their phase is completely undefined. Here you have this simple scaling between field and photon number that allows you to express the emission rate as a function of photon number. For light fields that do not have a well defined photon number, one cannot simply use the mean photon number to determine the stimulated emission rate. This is something that is not directly clear from the (n+1) description. Consider for example squeezed vacuum. Here the orthogonal field quadratures have drastically different variances. Accordingly, the stimulated emission rate will depend strongly on the relative phase orientation of the light field with respect to the atom (see e.g. https://journals.aps.org/prx/abstract/10.1103/PhysRevX.6.031004, the paper is open access). This shows that it is indeed the field that matters and not the photon number. This is often overlooked and something I like to emphasize.

When I think of a vacuum I think of some region in empty space with zero field excitations, no neutrinos, electrons, protons ect and no excitations of the electromagnetic field. What is left is the ground state of all the fields of the standard model, that is the vacuum to me. Why can't we equate the vacuum with the ground state? I am sure you have written many posts about this, can you point me there.

What is your background in physics? Do you know why the ground state of the harmonic oscillator has a finite energy because position and momentum cannot both be exactly zero due to uncertainty? When quantizing the light field, it is mapped to the harmonic oscillator with the two quadratures of the light field playing the role of position and momentum. So in other words, the vacuum is the ground state, but just as position and momentum cannot both be exactly zero simultaneously, not all components of the electromagnetic field cannot be zero simultaneously. And accordingly you can never switch off emission completely.

 

[Spinnor]

What is your background in physics?

As far as a masters in physics, but that was a while ago.

I was confused by Dr. Vanhees comment below,

No! Although zillions of popular-science writers claim that the vacuum fluctuates, it doesn't. The field fluctuates, but not the vacuum!

To my understanding the an empty vacuum is the ground state, so I was hoping to get my thinking corrected with regards to Dr. Vanhees comment.

Normally I reread my posts several times before posting. What I meant to write was, "the 1 in (N+1) comes from spontaneous emission by the quantum vacuum" but I think you figured out what I meant.

The 1 in (N+1) comes from stimulated emission by the quantum vacuum?

I understand the basics of the harmonic oscillator and how it is similar with the electromagnetic field. So the ground state of any mode of the electromagnetic field has an energy of 1/2 hν. So it is this ground state that is responsible for spontaneous emission, yes? It seemed to me that because the ground state had half as much energy as a single photon of energy hν it would only be half as effective at causing spontaneous emission,simple thinking I know. There may be no simple way for you to explain this, I will keep studying.

I guess you can only say so much about the ground state of the electromagnetic field?

Thanks!

 

[Spinnor]

There may be no simple way for you to explain this, I will keep studying.

This paper may be of some help? http://www.famaf.unc.edu.ar/~vmarconi/moderna1/emision_estimulada_AJP.pdf

 

[vanhees71]

As far as a masters in physics, but that was a while ago.

I was confused by Dr. Vanhees comment below,
To my understanding the an empty vacuum is the ground state, so I was hoping to get my thinking corrected with regards to Dr. Vanhees comment.

Normally I reread my posts several times before posting. What I meant to write was, "the 1 in (N+1) comes from spontaneous emission by the quantum vacuum" but I think you figured out what I meant.

Yes, and that's wrong. By construction in relativistic QFT the vacuum is the ground state, and there cannot be anything emitted since there's no energy there for it. So there must be something present to emit something. An example is an excited atom (i.e., an atom prepared in the energy eigenstate which is not its ground state). Then according to the fluctuations of the quantized radiation field, there is spontaneous emission, and this process gives rise to the 1 in (N+1), where N is the number of photons present in the initial state. The N in this formula is the process of induced emission, which survives the semiclassical limit, i.e., it's also present when the quantum field is not quantized. The effect of spontaneous emission, postulated by Einstein as early as 1917 in his kinetic derivation of Planck's black-body-radiation law, is thus from the point of view of modern QT one of the arguments for the necessity of field quantization, i.e., going beyond the semiclassical approximation (matter quantized, radiation field treated classically).

I understand the basics of the harmonic oscillator and how it is similar with the electromagnetic field. So the ground state of any mode of the electromagnetic field has an energy of 1/2 hν. So it is this ground state that is responsible for spontaneous emission, yes? It seemed to me that because the ground state had half as much energy as a single photon of energy hν it would only be half as effective at causing spontaneous emission,simple thinking I know. There may be no simple way for you to explain this, I will keep studying.

I guess you can only say so much about the ground state of the electromagnetic field?

Thanks!

This is also an old-fashioned view. If you were right, there'd by an infinite amount of energy present due to this socalled zero-point energy. The zero-point energy is an artifact of a naive treatment of the operator-ordering problem when quantizing the electromagnetic field and constructing the energy-momentum four-vector operator via the field operators (a la Noether's theorem). If something diverges, it's wrong, and thus must be corrected. It's the most simple example for the necessity of renormalization. In this case it's very simple: The diverging sum of zero-point energies, wherever truncated with an energy-momentum cut-off to make the sum finite, just adds a constant to the total energy of any system and thus is (within special relativity) unobservable. That's why you can subtract it without changing the physics, and usually you do just that, i.e., you define the vacuum state to have energy eigenvalue 0. Concerning the operator level, i.e., for the proper definition of the Hamiltonian of the radiation field, this subtraction is justified since it's proportional to the unit operator and thus commuting trivially with all other operators, i.e., the subtracted and finite Hamiltonian generates the same time evolution of the field operators in the Heisenberg picture and thus defines the same dynamics of the quantum system as the ill-defined naive Hamiltonian containing the diverging contributions from the zero-point energies. For the case of the free radiation field, that's all you have to do to define the theory properly. You get a complete system of relativistic QFT, i.e., one possible realization of the proper orthochronous Lorentz group in terms of a massless spin-1 field, which can be used to describe the quantized electromagnetic field and thus free photons.

 

[Cthugha]

It seemed to me that because the ground state had half as much energy as a single photon of energy hν it would only be half as effective at causing spontaneous emission,simple thinking I know. There may be no simple way for you to explain this, I will keep studying.

This puzzles me a bit as there is little I can add to what I said before. The condensed version: I emphasized that spontaneous emission happens VIA the vacuum and not BY the vacuum. The role of external fields in the emission process is to provide the perturbation that renders the excited state unstable. They do not deposit any energy in the process. Really, all you need to know about the vacuum is that you get non-zero field variances and then everything else falls into place.

 

[Spinnor]

This is also an old-fashioned view. If you were right, there'd by an infinite amount of energy present due to this socalled zero-point energy.

Isn't one possible solution to this problem supersymmetry? Isn't that what supersymmetry is supposed to do, cancel one infinity with another?

So you don't believe that zero point energy of the electromagnetic field is really there in a vacuum? My opinion doesn't carry much weight here (and it should not ) but I don't see how the zero point energy of the electromagnetic field does not exist.

Thank you for your time!

 

[vanhees71]

No, I don't believe that zero-point energy has any physical significance within special-relativistic QFT. You cannot measure absolute energies.

There's also another intuitive argument why the idea of a physical significance of vacuum fluctuations in a naive sense promoted by some (unfortunately not only popular-science books): To test some claim about the physical world you have to observe it, i.e., to make an experiment and measure effects of an assumed "mechanism", and you cannot do this without introducing an experimental setup and measurement devices, and that implies that there is something (matter) present and thus no vacuum.

An example is the Casimir effect, which very often is derived simply by counting modes of the em. fields between two metallic (uncharged) plates and the vacuum without any plates. As it turns out, a careful analysis shows that this calculation, based on using simply boundary conditions according to the presence of ideally conducting plates (as in classical electromagnetics), is the infinite-coupling-constant limit of a proper treatment of QED. This proper treatment shows that indeed the Casimir effect is nothing else than the result of the quantum fluctuations of the electrons within the plates, i.e., it's a kind of van der Waals force, and that's also how the Casimir effect has been derived in the first place by Casmir himself.

Of course, the term "vacuum fluctuations" or "vacuum polarization" also has a strict formal meaning in QFT. It refers to radiative corrections of the photon's dispersion relation incoded in the photon self-energy. It is symbolized by a Feynman diagram with two (amputated) photon legs. The phenomenological consequences of such radiative corrections are indeed profound and lead to the astonishing success of relativistic QFT (particularly QED). It's, e.g., an important part of the Lamb shift of the hydrogen spectral lines, which was the main motivation for the 2nd attempt to make sense of relativistic field quantization in 1948. The 1st attempt was by the early pioneers, most importantly by Jordan and Dirac.

For a very detailed account of how to rectify the misleading concepts about vacuum fluctuations and, related to it, the issue of "virtual particles" see Arnold Neumaier's corresponding Insight articles:

https://www.physicsforums.com/insights/misconceptions-virtual-particles/
https://www.physicsforums.com/insights/physics-virtual-particles/