# Stimulated emission

1. Oct 23, 2009

### shakespeare86

Hi.
I can't see why the photon created by stimulated emission in a common laser should be in phase and with the same polarization of the original photon.
Thank you.

2. Oct 23, 2009

### Bob_for_short

Because:

1) The energy difference is the same (the same photon energy is emitted),

2) The driving force makes electron oscillate and thus radiate in phase,

3) The resonance conditions are the most favorable for such photon polarizations.

Last edited: Oct 23, 2009
3. Oct 23, 2009

### shakespeare86

It's surely OK that the new photon has got the same energy and momentum as the original one.
I just can't see why they have the same phase and momentum.

I could make my question clearer in this way: why a spontaneous emitted photon has phase and polarization uncorrelated with the original photon while a stimulated emitted photon is identical to the original photon?

I've serious problems thinking at an electron vibrating, anyway, even if I did, didn't get how it could solve the problem.

4. Oct 23, 2009

### shakespeare86

sorry, i meant I can't see why they have the same phase and polarization.

5. Oct 23, 2009

### Bob_for_short

A real photon is an excited state of the corresponding photon oscillator. Photons oscillators come together with charges. Photon oscillators differ with energies, momenta and polarizations. When there is no external driving force for electrons, the photon population numbers grow with time due to spontaneous transitions. In presence of an external driving force the resonance conditions occur for the same photons as the external filed so their populations grow better (an additional source term) and the other populations get depleted due to the conservation law for the total number of emitted photons.

Last edited: Oct 24, 2009
6. Oct 23, 2009

### shakespeare86

Now I see better, thanks.
This is a qualitative but very strong argument.

Anyway I was trying to perform a calculation with the quantized electromagnetic field A, using the dipole approximation as the interaction hamiltonian, H=-DE, E=-dA/dt, and trying to calculate the probability transition to the first order between a state with just one photon with momentum p and polarization r, and a state with two photons, where the second has got different polarization s and momentum k.
Actually I see that the momentum should be the same, thanks to the Fermi golden rule, but don't see where I should get the information about the polarization and the phase.

Anyway, thanks for the answers, now I've at least a qualitative picture of the physics.

7. Oct 24, 2009

### sokrates

This is one of the best qualitative descriptions of SE, I have seen. Could you elaborate more on that?
Do you mean that the stimulated emission is a collective, statistical effect?

8. Oct 24, 2009

### Bob_for_short

Some indications on how charges are "coupled" to the electromagnetic filed are given in my publications ("electronium" description).

9. Oct 24, 2009

### sokrates

Your publications are too technical for me to dig out the piece of insight I am looking for, specifically OP's question (why does SE have to produce equal phase photons?)...

What you mean is not very clear in your second response to the OP, perhaps it'd be better if you wanted to briefly clarify that.

10. Oct 27, 2009

### Bob_for_short

This is another way to say that the population evolution equation dN(k,e)/dt for a given photon direction k and polarization e has a "pumping term" due to an external field.

11. Oct 27, 2009

### Manchot

This is something that used to bother me, until I realized that it is energy quantization that requires that the emitted photon be either completely in-phase (emission) or completely out-of-phase (absorption). In other words, if you could add a photon of arbitrary phase to a mode, you could change its amplitude continuously, and the mode's energy could be changed arbitrarily. Oscillator quantization tells us that this cannot be the case.

12. Oct 27, 2009

### Bob_for_short

It is not so. The energy can take any "intermediary" value between N and N+1 photons just because the number of photons is uncertain in case of certain phase. There is an uncertainty relationship between the pnoton number N in a wave and the sine of the wave phase φ: ∆N∆sinφ >= ћ. When the number of photons is uncertain (a coherent state) the field state is not an eigenfunction of the Halimtonian - it has an average energy but not a certain energy (not an eigenvalue).

13. Oct 27, 2009

### Manchot

^ I see what you mean, but if you're talking about stimulated emission, doesn't it make more sense to talk about a field prepared in a number state rather than a coherent state? That is, the statement that "the energy increased by $\hbar \omega$" doesn't make a whole lot of sense unless you're in an energy eigenstate to start with.

14. Oct 27, 2009

### Bob_for_short

It depends on regime, I think. If we speak of a permanent laser emission, for example, then a certain frequency (phase) implies uncertain number of photons. There is an average number but it has a dispersion.

If we speak of a transient state, avalanche of photons, the beginning of emission, then the number N(t) is a varying (increasing) function of time and the phase is not yet well established.

15. Oct 28, 2009

### conway

Can we not simply say that the emitted photon has a random phase with respect to the incident photon? Then we can always resolve the phase difference into an in-phase and and out-of-phase component. Then the in-phase component represents stimulated emission and the out-of-phase component is spontaneous emission.

16. Nov 3, 2009

### conway

In my last post, I proposed that we more or less define the in-phase component as stimulated emission and the out-of-phase component as spontaneous emission. I've been watching the "views" ticker and the thread has drawn over a hundred views since I posted but no one else has responded. So I'm going to elaborate on my previous thoughts by pointing out that in terms of purely classical electrodynamics, an oscillator emits more strongly when it is in phase with an existing ambient field, by precisely the factor given by Einstein's A and B coefficients.

17. Nov 23, 2009

### conway

The views ticker was at two or three hundred when I entered this thread, and since I posted it's gone up to over 750. But no one else has commented. So I'm going to try to stir the pot again by pointing out that you can't have much of a meaningful discussion of stimulated emission as it occurs in lasers without getting into the question of population inversion. Which hasn't been discussed so far.

18. Nov 24, 2009

### shakespeare86

Yes, but the main problem, at least mine, was to understand why the emitted radiation is in phase with the stimulating radiation.
Actually don't think we could speak about the phase of a photon. Anyway the question is perfectly well defined from a classical point of view.

19. Nov 24, 2009

### sokrates

We cannot think of a phase of a -single- photon, but we usually think in terms of a 1-particle equation anyway...

The phase of the wavefunction represents a collection of photons but we model them as having one, average wavefunction.

So still, your initial question is a valid one, also quantum mechanically.

The correct insight emerges, I think, if we think of the present photon, and the excited electron as a collection, rather than single entities.

Then the present photon is not just some random guy scattering off of a -single- excited electron, but is actually a SOURCE.