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Suppose I have a three-level system with $E_0$ the ground level, $E_1$ the intermediate and $E_2$ the upper level. In thermal equilibrium they will have a certain probability distribution according to the Boltzmann Statistic, in a laser one needs a population inversion, but that doesn't matter for my question.

My question is this: 3-level laser uses the $E_1 \rightarrow E_0$ transition because the $E_2$ level decays quickly (by design), i.e., emission of an $E_1$ photon is stimulated with an $E_1$ photon. But is there also a stimulated emission of energy $E_1$ for an incident photon of energy $E_2-E_1$? My thinking is that a photon energy of $E_2-E_1$ kicks the electron from $E_1$ in the upper state $E_2$, which will then decay, and every once in a while it should decay into the ground level, not back into the $E_1$ state. Is that correct? And if that is so, is the momentum of the emitted photon with energy $E_2$ aligned with the momentum of the incident photon of energy $E_2-E_1$?

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A photon of energy $E_2-E_1$ can cause stimulated emission from the state at $E_2$ to the state at $E_1$. It cannot cause stimulated emission from the state at $E_2$ to the ground state. The photon produced in stimulated emission has the same energy and momentum as the stimulating photon.

Photons of energy $E_2-E_1$ can stimulate various nonlinear effects, such as two-photon absorption from the ground state to the level at $E_1$. But it can't induce the direct transition that you imagine.

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  • $\begingroup$ But why doesn't the E_2 state decay with some probability into the ground state after the E_1 level absorbed an E_2-E_1 photon? I mean, what's the difference between any electron in E_2 (which would decay with some probability into E_1 or E_0) and an electron that got there by absorbing an E_2-E_1 photon, which you say will only decay into E_1? $\endgroup$ – WIMP Jun 17 '14 at 13:44
  • $\begingroup$ @WIMP "Decaying w/ probability" is normal behavior, not stimulated emission. The whole point (so to speak) of stimulated emission is the phase, polarization, time, and wavelength matching that occurs. $\endgroup$ – Carl Witthoft Jun 17 '14 at 14:04
  • $\begingroup$ Can you point to what I wrote that you take to mean "an electron that got there by absorbing an E_2-E_1 photon, which you say will only decay into E_1"? If I said something like that, then I need to edit my answer. It doesn't matter how the electron got into the state at $E_2$. $\endgroup$ – garyp Jun 17 '14 at 14:42
  • $\begingroup$ I think I know what you mean, it seems to be a matter of terminology. Let us forget 'stimulated emission' for a moment. A particle in E_1 can absorb a photon of energy E_2-E_1, right? That photon can subsequently decay into E_0, not only into E_2, right? Depending on how much E_0 is populated, this will result in an energy gain, right? Alas, I suspect the so emitted photons are not correlated. I am asking for something that shows this suspicion is either wrong or right. $\endgroup$ – WIMP Jun 18 '14 at 11:30
  • $\begingroup$ Regarding your question "Can you point to what I wrote that you take to mean "an electron that got there by absorbing an E_2-E_1 photon, which you say will only decay into E_1"?" You didn't write that - I did. That's my question. I assumed you were trying to answer my question. $\endgroup$ – WIMP Jun 18 '14 at 11:31
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WIMP you wrote "A particle in $E_1$ can absorb a photon of energy $E_2-E_1$, right?" - Yes

"That photon can subsequently decay into $E_0$, not only into $E_2$, right?" - No, it will most likely decay into $E_1$, then decay to $E_0$.

"Depending on how much $E_0$ is populated, this will result in an energy gain, right?" - The decay to $E_0$ is an energy loss equal the the energy of the photon emitted. If all the particles are in $E_0$ then they have zero (point) energy.

"Alas, I suspect the so emitted photons are not correlated. I am asking for something that shows this suspicion is either wrong or right." - If the decay is caused by stimulated emission then the emitted photon will be coherent with the stimulated photon. If the emitted photon is produced by spontaneous emission then it will be emitted in any direction.

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  • $\begingroup$ I've thought about this for a while and think I've answered my own question. It is basically right what I wrote, just that a) while the emission is stimulated in some sense, it's not what is technically referred to as stimulated emission and b) one can't make net gain with this because the probability is tiny since the ground level is already occupied. I'm willing to mark your answer as right if you correct your second answer, it doesn't make sense. Yes, the photon can decay to E_0, even though it is (in the configuration I laid out) unlikely to do so. $\endgroup$ – WIMP Aug 1 '14 at 9:42

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