View Full Version : stimulated emission
shakespeare86
Oct23-09, 05:57 PM
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.
Bob_for_short
Oct23-09, 06:23 PM
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.
shakespeare86
Oct23-09, 06:34 PM
Thank you for answering.
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.
shakespeare86
Oct23-09, 06:39 PM
sorry, i meant I can't see why they have the same phase and polarization.
Bob_for_short
Oct23-09, 06:48 PM
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.
shakespeare86
Oct23-09, 07:12 PM
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.
sokrates
Oct24-09, 01:44 AM
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 they 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.
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?
Bob_for_short
Oct24-09, 03:36 PM
This is ... qualitative descriptions of SE... Could you elaborate more on that?
Some indications on how charges are "coupled" to the electromagnetic filed are given in my publications ("electronium" description).
sokrates
Oct24-09, 10:03 PM
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.
Bob_for_short
Oct27-09, 05:57 AM
...Stimulated emission stems from the fact that the photon is a boson, and there is an enhanced probability for a photon to be created in a state where there are already some photons. In the case of a laser (almost a classical field, with definite phase), there are lots of photons in a given mode, thus the stimulated emission is the dominant process.
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.
Manchot
Oct27-09, 08:33 AM
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.
Bob_for_short
Oct27-09, 09:02 AM
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).
Manchot
Oct27-09, 06:35 PM
^ 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.
Bob_for_short
Oct27-09, 07:15 PM
^ 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.
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.
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.
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.
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.
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