# Textbook tries to argue stimulated emission from boson behaviour?

1. Jan 3, 2012

### nonequilibrium

Hello,

"Quantum Mechanics" by Basdevant and Dalibard tries to qualitatively deduce stimulated emission of atoms shined upon with some light by using Bose Einstein statistics.

Imagine a certain photon in eigenstate n and if we turn on a potential v temporarily, the chance of it ending up in eigenstate m (after turning off the potential v), is alpha.

The "theory of N identical bosons" (I have to give it a name) tells us that if we do the same but now with that one photon in a batch of identical photons already in eigenstate m (before turning on the potential v), then the probability of them all eventually being in eigenstate m, is much larger than alpha (after turning off the potential).

So in a certain sense, BE statistics indeed gives a sort of stimulated transition in some cases.

The book, however, immediately goes on to state
I find this explanation rather vague, more specifically I don't understand how the bold follows from the previous: the case of stimulated emission seems to talk about the creation of a photon, whereas the previous was talking about the transition of a photon... Also, I don't know what the temporary potential is in this case.

I realize there are other ways to explain stimulated emission. But what I'm interested in is understanding the above explanation, or hearing that this explanation is rubbish.

Thank you!

2. Jan 3, 2012

### Cthugha

The textbook is rather vague, but correct. However,it should be clarified that Bose-Einstein statistics does not mean a Bose-Einstein photon number distribution in this case, but just the property of bosons that the wavefunction (or rather probability amplitudes in the case of photons) does not change sign in permutations.

That means that for every situation where you have several indistinguishable ways of getting from the initial to the final situation, you need to sum up the probability amplitudes for all of these processes and square afterwards, while you just sum the square of the individual probability amplitudes for distinguishable ways. This means that for bosons any process where photons end up being in a state with many indistinguishable photons is statistically enhanced. This is true for stimulated emission as well as the more general bosonic final state stimulation.

This is also at the heart of the Hanbury Brown-Twiss effect and the Hong-Ou-Mandel effect. The quantum explanation in terms of HBT (along with some easy math) has been given by Ugo Fano in "Quantum Theory of Interference Effects in the Mixing of Light from Phase-Independent Sources", American Journal of Physics -- August 1961 -- Volume 29, Issue 8, pp. 539.

By the way this also gives an intuitive approach to the Pauli exclusion principle. For fermions the wavefunction changes sign for permutations, so that the probability amplitudes for two fermionic particles to end up in the same state interfere destructively instead of constructively like it is for bosons.

3. Jan 3, 2012

### nonequilibrium

It seems like your answer to my question is in your 2nd paragraph, but I don't really understand your wording, e.g. "indistinguishable ways".

So the bold text (in the OP) follows from what precedes it? Good. And is it an application of what precedes it, or merely an analogy? If it's an application: what is the potential v in the case of the stimulated emission of light? And what are the bosons? ("the photons" seems like an obvious answer, but photons are being created, whereas the bosons are not)

4. Jan 4, 2012

### Cthugha

The bosons considered here are indeed the photons. There is no potential as you do not need one for emission processes. You just care about the initial state, the final state and the probability amplitudes for the way to get from one to the other. If the initial state is an eigenstate you need some potential to go from the initial to the final state. Excited states of e.g. an atom in contact with the vacuum field are not eigenstates of the system and therefore emission processes will happen without applying an external potential. Alternatively you can consider the intrinsic coupling between the atom and the electromagnetic field as the potential needed here. Check the Jaynes-Cummings model for details.

Indistinguishable ways have just the same meaning like in the double slit experiment. You have some initial state (light source emitting photons towards the double slit), a final state (photon detection events at the screen) and several possible ways to get from the initial to the final state (slit 1 or slit 2). If the ways to get from the initial to the final state are indistinguishable, the probability amplitudes will interfere. Otherwise they will not. The same is true for processes involving many photons. Imagine you have an excited state and many photons around that could cause a stimulated emission process. If you cannot distinguish in the end which photon caused the process, all of the probability amplitudes will interfere constructively and you will have a huge probability for stimulated emission. If you can distinguish which photon caused the process, e.g. if all the photons arrive from different directions, the probability amplitudes will not interfere.