# How does optical phase modulation produce photons with different frequencies?

The classical description of electro-optic modulators is an index of refraction that depends on the applied voltage. For example, for a sine modulation $\sin(\Omega t)$, a monochromatic laser of frequency $\omega$ would get an additionnal phase $\varphi\propto\sin(\Omega t)$. This results in sidebands in the spectrum at $\omega-\Omega$ and $\omega+\Omega$.

Now, what is the interpretation of this phenomenon in terms of photons? A photon with initial frequency $\omega$ will end up at $\omega-\Omega$ or $\omega+\Omega$. How can the time-variation of the refractive index create new photon frequencies? Is it a non-linear effect similar to second-harmonic generation? If yes, it could be explained by an interaction such as $\hbar\omega + \hbar\omega \rightarrow \hbar (\omega-\Omega ) + \hbar(\omega+\Omega)$?

EDIT: A corollary to the original question. I shake my hand very fast in front of a laser beam, what happens to the photons? Do they get chopped in shorter photons? Instead of my hands, I could use a super-fast chopper. I would see photons with new frequencies (the sidebands) because of this modulation. How come the incident photons get a different energy?

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One mathematical approach (which I personally don't like because it doesn't say whats physically going on) involves looking at the electric field of the beam. This is the Fourier transform of the spectrum of the beam, so for a monochromatic beam with angular frequency = $\omega$

$E(x,t) = A\cos\left(kx - \omega t\right)$

After applying a sine-wave chopper at angular frequency $\gamma$:

$E(x,t) = A\cos(kx - \omega t)cos(\gamma t)$

Use the convolution theorem to transform this to a spectrum without picking up your pencil:

Two frequencies active now: $\omega \pm \gamma$

For a square-wave* chopper:

$E(x,t) = A\cos(kx - \omega t)\,\textrm{square}(\gamma t)$

Spectrum contains many components now (centered on a maximum at $\omega$), since the spectrum of a perfect* square wave has infinite terms.

* I put an asterisk on square-wave, since it must be slightly round at the corners rather than perfectly square. The reason for this is that if the chopper could switch from transparent to opaque instantly across the whole beam, then special relativity would scream at you since information (chopper on/off) would be moving faster than light.

I had quite a discussion with a professor at Oxford about a problem similar to this, and I'm currently working with a friend on a more "real-world" explanation and understanding of fast choppers. I'll type up some more detail another day when I'm more awake.

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The classical, Fourier interpretation, of the sidebands are fine, but that still does not give an explanation on how it makes individual photons to change energy. –  fffred Jul 31 '13 at 21:35
How individual photons change energy (or even if that phrase actually makes physical sense) is still quite a grey area as far as I know. The mathematical explanation is the best that we have, unfortunately... although I would actually love it if someone else now proved me wrong about that! –  Mark K Cowan Aug 1 '13 at 1:19
My feeling is that the photon, in presence of a chopper, is in a quantum state like $\left|\omega-\Omega\right\rangle+\left|\omega+\Omega\right\rangle$. So the average energy is still $\hbar \omega$. However I don't know how to get to this kind of state. –  fffred Aug 1 '13 at 1:47

A photon is an elementary particle , a building block of the Standard Model. Elementary particles follow quantum mechanics and not classical physics trajectories, once one has been able to isolate one of them and follow it course.

A light beam which optics works with, microscopically emerges from the congruence of zillions of photons, each moving at the velocity of light and with point dimensions within the Heisenberg uncertainty principle, HUP. An individual photon cannot be chopped, it is either there or it is not. Lubos Motl, who contributes here, has an article in his blog on how classical electromagnetic fields emerge out of an ensemble of photons. One could in principle use similar mathematics for any kind of light beam but it would be as stupid as digging a well with a surgical knife. Classical EM works beautifully and QM is necessary only when paradoxes and anomalies appear, to explain them.

So the photon manifestation is not useful for your observation except to explain changes in frequency on the elementary particle level. These can happen:

1) within the HUP, but due to the smallness of h would not be observable macroscopically

2) to interactions at the quantum level that again build up from the ensemble a coherent light beam.

If one is ambitious enough one should examine the collective atomic/molecular field that induces the change in the index of refraction, the generic higher order Van der Waals fields, and consider the interactions: compton scattering with the field; or excitation from a low to a higher energy level in the induced spectrum of the WdW field, and subsequent decay to a lower level than initial, etc .

That is how interactions work at the micro elementary particle level. That a change in frequency has been seen means an interaction, that a coherent beam emerges means that there is a coherent mechanism in the medium that allows for the rebuild/emergence of a different frequency beam.

Do they get chopped in shorter photons?

Absolutely not. The photon is there and interacts in the detector, or not. It is an elementary particle. You can chop beams because no matter how fast you try to chop them they are composed of zillions of photons. Can you chop water down the stream and consider you are chopping individual molecules?

For edification of readers of this, there exist experiments where photons appear individually one by one building up the two slit experiments showing the interference slowly emerging. The smallest chop of a beam is a photon at a time.

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I am not convinced that the observation of sidebands emanates solely from a large ensemble of photons. You can do the experiment with one photon. Sometimes it will be measured at $\omega+\Omega$ and sometimes at $\omega-\Omega$. The same thing works for one photon in a cubic "box", where we know only discrete values of the photon energy are permitted. The interaction of the photon with the box is not discussed in detail: it is only a boundary condition (the photon is reflected). –  fffred Jul 31 '13 at 22:33
Did you read the blog entry I linked in my answer? All classical fields emerge out of ensembles of individual photons. An individual photon is an entire elementary particle and it can only interact with the rules of quantum mechanics. In a way it is similar to how thermodynamic quantities emerge from quantum statistical mechanics. Did you watch the two slit experiment I linked to? where each individual photon follows the qm probability? –  anna v Aug 1 '13 at 3:30
I read the blog entry and I know already about the two slit experiment, but I still don't see how that answers the question. What happens if you send photons one by one through the chopper? You should see sidebands, with some probability gaining or loosing energy (average energy is the same). Sorry in case I missed your point. –  fffred Aug 1 '13 at 4:36
You will see sidebands but the value will not be ω+Ω or -. There will be an indeterminacy, similar to the one seen in the accumulation of the interference pattern when the photons go one by one. There will be a delta(ω+Ω). It is the coherent "summation" of the ensemble of photons that eliminates the delta indeterminacy and adds up to the classical accurate formulae. –  anna v Aug 1 '13 at 4:56
If I understand your comment correctly, you are explaining why a $\omega+\Omega$ photon is actually not exactly $\omega+\Omega$, but only when there are many of them you have $\omega+\Omega$ in average. Fine. Now, how come they get to $\omega+\Omega$ whereas they started at $\omega$? –  fffred Aug 1 '13 at 5:16