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:
PeterDonis said:
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.
Spinnor said:
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.
) but I don't see how the zero point energy of the electromagnetic field does not exist.
