# How does Raman scattering not break energy quantization?

I understand how Raman scattering works and I understand that it can only be clearly explained if you think about light as a wave instead of a particle (otherwise you have to start talking about "virtual states" which I think are not the most clear and accurate way of describing it).

But, for some reason, I never thought that it implies that the energy of the photons change when the (anti)Stokes scattering happens, which would be against having a defined photon energy quantization.

I could understand this if the photons are actually being absorbed while new ones are created, but that's not how I think it works.

So, what am I missing here?

• What you are missing is that Raman scattering is a nonlinear optical phenomenon – Lewis Miller Oct 3 '17 at 15:28
• @LewisMiller That actually doesn't help me to understand. Could you be more specific on what you mean? – cinico Oct 3 '17 at 15:31
• The simplest interactions between radiation and matter are linear in the E&M field strength. That is where your ideas of photon energy quantization originate. Raman scattering can only occur if there are higher powers of the E&M field in the interaction (called nonlinear optical phenomenon). Your statement "if the photons are actually being absorbed while new ones are created" is a valid interpretation in that case. The answer by @EmilioPisanty is correct but I thought my comment might be easier for you to understand. Emilio is saying the same thing in a more rigorous way. – Lewis Miller Oct 4 '17 at 2:55
• @LewisMiller Thanks. Together with the answer of EmilioPisanty, your comment was quite helpful. The wrong concept I had, which led to this question, was that photons could not change their energy (a new photon would have to be created). I now understand that this interpretation was incorrect. – cinico Oct 4 '17 at 8:11

Like any good quantum mechanical phenomenon, Raman scattering admits both a wave picture and a particle picture. You might not like either of them, though.

• As far as the photon picture goes, it's simple: you have a scattering process that implements a unitary evolution $\hat U$ that takes a single photon $|1;\nu_0\rangle$ of energy $h\nu_0$ and the medium in its ground state $|g\rangle$ and produces a final state with a nonzero amplitude to have a lower-frequency photon $|1;\nu_0-\nu_v\rangle$ and the medium in its excited state $|e\rangle$, or in other words an evolution with a nonzero amplitude $$\langle e|\langle 1;\nu_0-\nu_v| \ \hat U \ |g\rangle |1;\nu_0\rangle\neq 0,$$ or more pictorially

Now, if that big monolithic unitary $\hat U$ looks too mysterious, then you can apply the tools of leading-order perturbation theory to the problem (which, in the usual regime, can essentially be taken to be exact) and it will give you a nice expression in terms of virtual transitions. You don't like those? Too bad, because they're essential features of the description.

• If you want to see this in terms of wave mechanics, and you're OK with confining yourself exclusively to the classical aspects of the EM emission, then the thing to look like is the medium itself: it is interacting with the field, and the field is inducing a coherent oscillation between the $|g\rangle$ and $|e\rangle$ states, which then amplitude-modulates the medium's oscillations, so that they're no longer monochromatic, and when you look at the oscillations of the medium's polarization, it develops AM sidebands which are precisely the Raman lines.

Now, the wave picture has a couple of problems. For one, you're pushing under the rug the description of the reaction of the medium to the light, even without the Raman-induced vibrational coherence, and if you want to correctly describe that polarization... you need virtual transitions, same as always. More importantly, though, you become blind to quantum properties of the emission, such as the fact that the Raman photons will normally be entangled with the pump photons, in experimentally-relevant ways that can only be explained in the photon picture. You can try to run away from quantum mechanics, if you want, but the only thing that you'll achieve is shut off doors that would otherwise stay open.

• I was trying to avoid the virtual state description because I thought it was a simplistic overview of the problem and not a consequence of a quantum mechanics description (I don't know why I thought that). My question was exactly related to the fact that I could not confine this problem simply to a classical wave mechanics description, and I didn't know what I was missing. I understood from your answer that I would have to dive into perturbation theory to understand the virtual states. Does this imply that the scattered photon is a new particle, thus not breaking the energy quantization? – cinico Oct 3 '17 at 15:29
• @cinico The "new" in "new particle" is neither here nor there in quantum mechanics - it's not in the math and it's entirely an interpretational issue. There was a photon before, and there's a photon afterwards, with a different energy. Is the photon "new"? Contrast this to a situation where, say, the photon energy stays constant but the momentum changes (points in a different direction) after the interaction (a.k.a. reflecting off of a mirror); is that a "new" photon? – Emilio Pisanty Oct 3 '17 at 15:32
• While that way of looking into things makes sense, wouldn't that make the concept of "a photon has a indivisible energy" useless? – cinico Oct 3 '17 at 15:44
• I'm not sure where you're getting that concept; photons are perfectly divisible. What does happen is that each photon's energy is proportional to its frequency, which is all you need for e.g. Planck's blackbody-spectrum manipulations. Most normal electromagnetic interactions are linear, and linear processes are unable to couple modes of different frequencies, so the photon energy stays constant for that class of interactions. However, there's plenty of nonlinear processes (SPDC, 2nd & 3rd harmonic generation, SFG, DFG, HHG, Raman scattering) that don't have that restriction. – Emilio Pisanty Oct 3 '17 at 15:51
• I don't know why I had such wrong idea in my head, but I think it's clear now. Thank you Emilio. – cinico Oct 4 '17 at 7:48

I think the issue here is that you believe there is a "quantified and indivisible photon energy quantification" which is not true. The equation for photon energy is: $$E = h f$$ Which, since any frequency $f$ is possible (as far as we know), means that $E$ can be of any value. What you might be thinking of is the fact that if I am counting energy in a particular mode -- at a single frequency -- then the energy must come in "lumps" of $hf$. But this doesn't mean we can't absorb a photon at one energy and emit it at any other.

• That's not what I meant. I don't have any doubts about what you said. I was referring to the fact that you could not divide the energy of a single photon, e.g. reducing the energy of the same photon from 2 eV to 1 eV. – cinico Oct 3 '17 at 15:13