How does stimulated emission work in the process of light amplification?

[Albarok]

I understand the concept of stimulated emission and how it works as light amplification, but a certain technicality in its process eludes me. How is the inciting photon actually interacting with the electron that falls to a lower energy level?

In every physical interaction that I know of there is some sort of an energy exchange. When two particles collide they change direction/momentum/velocity. When two waves collide you get interference. When an EM wave interacts with a certain material, part of the wave is reflected and its direction and polarization may change.

During absorption the incoming photon gives all of its energy to the electron, and that's why the electron increases in energy level. What happens during stimulated emission? The inciting photon didn't lose any energy and it didn't experience any sort of change to its properties, so where is the "cost" that it has to pay for stimulating the electron? Where is the indication that that photon had just interacted with something?

 

[DrClaude]

The interaction is between the emitter and the electromagnetic field. The probability of emission depends on the excitation of the field, so the higher number of photons there are, the more probable it is that an additional photon will be emitted. There doesn't need to be a "cost" for the field.

Another way to see it is as a scattering event. You have an incoming field interacting with an emitter, say an excited atom. The state of the field and of the atom after the scattering event depends (of course) on the state of both systems before the interaction, and the final state of each can be affected. Nothing mysterious there.

 

[Albarok]

If you're saying that it's like scattering, that means that the photon does give some energy to the excited atom in order to vibrate it and cause its fall. I guess the energy exchange can be negligible, but still something happens there doesn't it?

 

[DrClaude]

If you're saying that it's like scattering, that means that the photon does give some energy to the excited atom in order to vibrate it and cause its fall. I guess the energy exchange can be negligible, but still something happens there doesn't it?

Your thinking is much too classical. This is a quantum mechanical process, so you can't talk about making the atom "vibrate." You can only look at it in terms of of initial and final states, there is no "in-between" things happening.

 

[bhobba]

The following may help:
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

The phenomena of spontaneous emission is totally inexplicable in standard QM where the electron should be in a stationary state. The answer is you must go to quantum field theory to explain it. Doing this the electron is coupled to the universal quantum EM field that pervades all space and is not really in a stationary state. You can treat this like a perturbation and apply Fermi's Golden Rule to give the probabilities of things like emission, absorption etc:
https://en.wikipedia.org/wiki/Fermi's_golden_rule

Thanks
Bill

 

[Vanadium 50]

The phenomena of spontaneous emission is totally inexplicable in standard QM where the electron should be in a stationary state.

I'm going to disagree here, since it was first explained by Einstein in 1916, working in standard QM.

 

[bhobba]

I'm going to disagree here, since it was first explained by Einstein in 1916, working in standard QM.

Of course true. I suppose the view you take depends what yuo mean by explain. Einstein gave some rules, not the full quantum explanation that had to await Dirac quantizing the the EM field. Einstein's rules follow from that. However the why of those rules is an interesting story in itself and hopefully the OP will gain some insight from the link I posted on What is a Photon. It also important to understand even 'experts' here have slightly different takes on things.

Thanks
Bill