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I'm trying to understand modulation transfer spectroscopy in simple terms.

For those unfamiliar with it, this article gives a very good summary. To sum it up, two counterpropagating beams, a pump and a probe beam, are sent on an atomic vapor. In the lab frame, both beams have the same frequency (let's call it $\omega$), except that the pump beam is modulated (typically via an EOM), resulting in two sidebands with frequencies $\omega_\pm = \omega \pm \omega_m$.

When the sub-Doppler resonance condition between the pump sidebands and the probe is satisfied, sidebands appear on the probe beam, through a four-wave mixing process. The beating signal between the probe and its sidebands is measured on a photodiode, and, to put it simply, has sub-Doppler features (the exact expression of the signal is written in the article above, but is of little interest here).

My goal is to explain very simply why the four-wave mixing process that "gives" sidebands to the probe beam can only happen when the pump sideband and the probe beam are interacting with an atom at rest. However, as a graduate student whose background is mostly in atomic physics, I have only faint memories about non-linear optics, and I'm not sure how to think properly about the problem.

I'll try to explain below how I tried to reason so far. First, a very simple representation of the four waves I'm trying to mix to obtain sidebands on the probe :

The wavevectors $\vec{k}_+$ and $\vec{k}_-$ are associated to the sidebands of the pump, with $\vec{\delta k} = \frac{\omega_m}{c}$. I'm thinking about the problem with an atom at rest, as it's supposed to be the only velocity class for which the four-wave mixing happens. In that case, $\vec{k}_\text{probe} = -\vec{k}_\text{pump}$ in the frame of reference of the atom.

I'm trying to find which four-wave mixing process will create a sideband on the probe, which should then have the wavevector $-\vec{k}_+$ or $-\vec{k}_-$. Let's choose $-\vec{k}_-$ for instance. I'm thus looking for a way to combine three wavevectors $\vec{k}_1$, $\vec{k}_2$ and $\vec{k}_3$, choosen among the four on my drawing above so that $\vec{k}_1 + \vec{k}_2 +\vec{k}_3 = -\vec{k}_-$.

One solution would be $\vec{k}_1 = \vec{k}_+$ and $\vec{k}_2 = \vec{k}_3 = \vec{k}_\text{probe}$. Indeed:

\begin{align} \vec{k}_+ + \vec{k}_\text{probe} + \vec{k}_\text{probe} &= \vec{k}_\text{pump} + \vec{\delta_k} - 2 \vec{k}_\text{pump}\\ &= - (\vec{k}_\text{pump} - \vec{\delta_k}) \\ &= -\vec{k}_- \end{align}

However, let's now write the conservation of energy :

\begin{equation} \omega + \omega = \omega_+ + \omega_- \end{equation}

The way I interpret it is that, in the process to generate a $-\vec{k}_-$ wave described above, two photons were absorbed from the probe beam, and one was emitted through stimulated emission by a $-\vec{k}_+$ wave, which doesn't exist (yet).

Is my reasoning correct so far, i.e. can I indeed say that the process above won't create a $-\vec{k}_-$ wave from my four initials beams?

How can I find the processes that will create a $-\vec{k}_-$ wave among the $\sim \frac{8^3}{3!}$ (this is not an integer, sorry for the bad combinatorics!) possible combinations of 3 of my 4 wavevectors?

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  • $\begingroup$ I don't feel qualified enough to post a full answer, so I give you 2 comments. 1- FWM is $k_1+k_2=k_3+k_4\Rightarrow k_1=k_3+k_4-k_2$. For a $-\vec{k}_-$ (i'd prefer to call it $k_{probe-}$) a possible path is $k_{probe-}=k_{pump}+k_{probe}-k_{pump-}\Rightarrow k_{probe-}=0-k_{pump-}$. I am not sure if this helps, but wanted to clarify. Your equation is for either up- or down-conversion ($k_{out}=k_{in1}+k_{in2}+k_{in3}$). 2- The fourth photon does not need to exist for its creation (it is not stimulated emission). As long as your FWM phase-matching and energy conservation is there (1/2) $\endgroup$ Commented Jan 21, 2022 at 19:53
  • $\begingroup$ (2/2) There will be the creation of a new photon to satisfy the conservation of energy and momentum. Actually, you can even have non-phase matched signals if your interaction length is smaller than the coherence length of the process. But that is not important here. $\endgroup$ Commented Jan 21, 2022 at 19:56
  • $\begingroup$ Thanks for your comments, and sorry for replying so late. I will try to address them both : 1°) Not being familiar with non-linear optics, I'm curious about the difference you make between $k_1 + k_2 = k_3 + k_4$ and anything else, for instance $k_1 + k_2 + k_3 + k_4 = 0$. What is the convention for linear optics? For me, as long as they're vectors, it's all the same. I wrote $k + k_{probe}+ k_{probe} = k_{probe-}$ in my own question simply to highlight that the leftside terms were supposed to be already existing beams, and the rightside one was the one which was created. (1/?) $\endgroup$
    – Banjo
    Commented Jan 27, 2022 at 11:59
  • $\begingroup$ I agree with the possible path you describe, which I arrived to myself a bit after posting my initial question. The problem I have with it is that it doesn't perfectly satisfy the conservation of energy. I'll adress that at the end of my replies. 2°) Yes, I agree, I had indeed understood that the point of FWM is to create that fourth photon from 3 existing beams. You made me realize that my question is poorly worded, and reading it 2 weeks later I'm having trouble understanding exactly what I was asking myself. I think I was mostly wondering whether my way of thinking about the problem (2/?) $\endgroup$
    – Banjo
    Commented Jan 27, 2022 at 12:16
  • $\begingroup$ was correct. As in : in order to find which are the correct paths to create that $k_{probe-}$ beam, should I simply try to combine three of my four beams (pump with its two sidebands + probe) and see which of these combinations = $k_{probe-}$ while also satisfying the conservation of energy. Reading your replies, I assume this is indeed the correct way? (3/?) $\endgroup$
    – Banjo
    Commented Jan 27, 2022 at 12:26

2 Answers 2

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enter image description here

So, the picture I drew shows the energy conservation cases. Each arrow correspond to a certain $\hbar\omega$ and if it points up its being absorbed, if down is being emitted. I did not bother making the arrows of different sizes for the up- and down- conversions. Strictly speaking, yes, all of these can be called four-wave mixing, however is more typical to call the 3rd case four-wave mixing. The $k$ vector treatment for each case I already told you in the comments to your question. So, for up- or down- conversion you have that the case where the $k$-vector of the single transition has to be equal to the sum of the other three, while in four wave mixing you have that the sum of the two in one direction equal the sum of the other two.

I now realize that I rushed my comment the other time and I have been thinking hard about this and there might be something that I have been missing and I would need to dig into the papers to understand how exactly this modulation transfer works. I couldn't yet figure out how it really works in simple words. It might be that the way this works is that you first generate all fields also in the probe direction, as the phase mismatch is small, and after all fields are present you can seed the process effectively with all 6 different combination of photons. What this entails is that in the end, after the full cell propagation, everything is phase-matched, but not because it was initially, but cascaded into existence. Because But of course this is just loud talk, like my first comment, this is a technique which I do not have experience with and to which I do not feel very comfortable answering.

Lastly, just a general remark, lets assume that in your example case of the 3 incident waves you had figured out how everything works. Out of the 3 photons used to generate the fourth, one of them will be amplified. So you have the case where you have 2 photons absorbed, one emitted due to the present of the 3rd (which amplifies that photon count) and the generation of a new photon.

EDIT: So, I skimmed through the papers the other day and it seems that the MHz modulation on top of such high frequencies means that the difference in $k$ vector is negligible. What this means is that although there is a slight phase-mismatch, the coherence length is many times the size of the cell, so its practically the same as the pump being either way just pump with the sidebands glued very close together. Now, this is of course easy to measure as a beat signal, but in terms of energy, its a very small addition.

So in the end you consider the four-wave mixing diagram that I shown here, for the energy where there you can mix all the photons with the added $\pm\Delta E$ to see what are the end results while disregarding the phase matching. From what I understood the momenta from molecular motion is many times greater that the one applied by the modulation which then in the end means that (slightly phase-mismatched) phase-matching only occurs for the sub-doppler case.

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Intuitive explanation leaving out explicit reference to nonlinear optics:

The pump modulates the population In the excited/ground states. The absorption of the probe is related to the ground state population, so the modulation of the pump is mapped, through modulation of ground state population onto the probe.

Regarding sub Doppler feature: it’s the same as for normal saturated absorption spectroscopy (even without fm modulation, remember, the modulation is only needed to convert the absorptive sub Doppler feature into a dispersive signal that you can lock to). If the pump/probe are on resonance then the effect is clearly strongest for atoms at rest. If the pump/probe are detuned then the pump will be on resonance for some velocity class of atoms with velocity $v$. But the probe will be Doppler detuned by $2kv$ from this class of atoms, so the effect is suppressed. The strongest effect occurs when both beams are resonant with the same atoms, and this only happens when both beams are on resonant and atoms are not moving.

Edit: one more things helps non-moving atoms. There are more of them than any other velocity class because of the Boltzmann distribution.

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