3
$\begingroup$

Say I have a carrier laser (optical) frequency $\omega_c$: $E=E_0 e^{i\omega_c t}$.

I propagate it through an electro-optical modulator that modulates the phase by $\beta \sin\Omega t$: $E = E_0 e^{i\omega_c t + i\beta \sin(\Omega t)}$.

If $\beta \ll 1$, the field can be expanded into:

$$ E \propto e^{i\omega_c t} + e^{i(\omega_c+\Omega) t} + e^{i(\omega_c-\Omega) t} , $$

where the new $\omega_c\pm\Omega$ are the sidebands.

Question:

The laser emits photons at energy $\hbar\omega_c$. After the modulation, are there actually photons at energies $\hbar(\omega_c\pm\Omega)$?

$\endgroup$
3
  • $\begingroup$ If you put the beam into a spectrometer what would you see? Are those not real photons? $\endgroup$
    – Jon Custer
    Oct 18, 2019 at 4:05
  • $\begingroup$ What else could they possibly be? $\endgroup$
    – knzhou
    Dec 28, 2019 at 0:36
  • $\begingroup$ I am puzzled by second-harmonic generation devices needing a non-linear effect to generate photons of a different frequency from the incident one. An EOM since to be able to do this too easily. $\endgroup$
    – SuperCiocia
    Dec 28, 2019 at 0:42

1 Answer 1

1
$\begingroup$

First of all, I believe the answer to the question in the title is yes, they are real photons. As Jon Custer mentioned, in a spectrometer there would be photons detected at the frequencies $\omega_c \pm \Omega$. Moreover, if you make the EOM modulation depth very strong, e.g. $\beta = 2.405 =$ a root of first Bessel function, you can completely extinguish the carrier: there are no photons of $\omega_c$ left.

The distinction between creating sidebands and so-called 'non-linear effects' (eg. second-harmonic generation in optics) I would explain as follows.

High-harmonic generation (HHG)

The products of HHG are photons which have a large energy mismatch compared to the initial photons. Therefore, you need a large power (strong non-linearity) to create them. Similarly, in an electrical circuit, you can create harmonics using a mixer, a highly non-linear element.

Sidebands

The products of creating sidebands usually have very similar energies as the initial photons, essentially no energy has to be added or taken away from the initial photons. Therefore a 'small' non-linearity is enough to prepare them. The EOM does act as a non-linear element, the non-linearity given by the response of a polarising beam splitter on a turning polarisation, which is sinusoidal, that is, non-linear. (You can create sidebands with an EOM operating on the linear slope the sine and then you only get sidebands at $\pm\Omega$ but no higher harmonics than that.) Similarly, in an electrical circuit you can simply have the input voltage of a voltage-controlled oscillator (VCO) oscillate and you will get some sidebands at $n\Omega$. But the bandwidth of this input voltage is not usually enough to create multiples of the carrier frequency $\omega_c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.