In short: A laser at frequency $\omega$ gets phase-modulated at $\Omega$ (e.g. with an electro-optical modulator), such that the lowest order sidebands appear at $\omega\pm\Omega$. If this laser beam is detected with a photodiode (with sufficient bandwidth), will there be any measurable beat notes, e.g. at $\Omega$ or $2\Omega$?

My considerations: (based on the description of the Pound-Drever-Hall technique in E.D. Black's laser frequency stabilization article, p.82)

The electric field of a phase-modulated laser beam can be written as \begin{equation} E=E_0 e^{i(\omega t+\beta \sin(\Omega t))} \end{equation} with some modulation depth $\beta$. This can be expanded with Bessel $J_m(x)$ functions: \begin{equation} E=E_0 e^{i\omega t}\sum_{m =-\infty}^{\infty}J_m(\beta)e^{im\Omega t}\approx E_0 \left(J_0(\beta)e^{i\omega t}+J_1(\beta)e^{i(\omega+\Omega) t}-J_1(\beta)e^{i(\omega-\Omega) t} \right) \end{equation} where higher sidebands have been neglected and $J_{-m}(x)=(-1)^mJ_{m}(x)$ is used.

The optical power is given by: \begin{align} P\propto \left|\mathrm{Re}(E) \right|^2 & =E_0^2 J_0^2 \cos(\omega t)^2-4E_0^2 J_0 J_1 \cos(\omega t)\sin(\omega t)\sin(\Omega t)+4 E_0^2 J_1^2 \sin(\omega t)^2\sin(\Omega t)^2\\ & = P_c \cos(\omega t)^2-4 \sqrt{P_c P_s} \cos(\omega t)\sin(\omega t)\sin(\Omega t)+4 P_s\sin(\omega t)^2\sin(\Omega t)^2 \end{align} where $P_c$ and $P_s$ are the carrier power and the (single)sideband power, such that $P_c+2P_s \approx |E_0|^2$.

Since the fast oscillations at $\omega$ won't be resolved by the photodiode, the power measured at the photodiode should be the time average of $P$ (i.e. $\lim_{T\to\infty}\frac{1}{T}\int_0^T P\mathrm{d}t$), where $\Omega t$ is treated as a constant: \begin{equation} P_\mathrm{PD}\propto \frac{1}{2} P_c+2 P_s \sin(\Omega t)^2 = P_c + P_s - P_s \cos{2\Omega t}. \end{equation} According to this calculation a beat note at $2\Omega$ would be expected. However...

...a measurement I made with a diode laser modulated via current modulation shows a frequency component at $\Omega$ (only).

  • $\begingroup$ Modulating the diode laser current doesn't produce pure (or ideal) phase modulation does it? $\endgroup$ Jul 16 '20 at 11:00
  • $\begingroup$ It's said to be prone to residual amplitude modulation (which would appear at $\Omega$ I guess), but in principle a modulation of the diode current results in a modulation of the emitted frequency and therefore creates sidebands just like an EOM. I have also successfully used this method to generate a decent PDH error signal. $\endgroup$
    – faber
    Jul 16 '20 at 14:05
  • $\begingroup$ You’ve missed all of the DC cross terms in your expression for $P \propto \text{Re}\left(|E|^2\right)$. $\endgroup$
    – Jagerber48
    Jul 16 '20 at 15:37
  • $\begingroup$ $\mathrm{Re}(E)=E_0 J_0 \cos(\omega t)-2E_0 J_1 \sin(\omega t)\sin(\Omega t)$, no? Sorry, I don't see it. $\endgroup$
    – faber
    Jul 16 '20 at 16:03

As you realize, you’re actually doing amplitude modulation, but it doesn’t matter; there will be side bands either way.

So why aren’t you observing the side bands? Likely they’re there, but you have too much noise in your measurement. Or perhaps you’re laser isn’t all that narrow-band, and its own spectrum is swallowing up the sidebands. Or maybe your modulation depth is actually really weak. Or there could be some silly mistake where, e.g., your optical spectrum analyzer is set with insufficient resolution.

Can you verify that these types of issues aren’t getting in your way? You could start by checking the power modulation with a fast photodiode and oscilloscope. Then switch your oscilloscope with an (electrical) spectrum analyzer (or have your scope perform an FFT). Finally, look at the optical spectrum. Confirm that the baseline around the laser line is sufficiently low to see the expected side bands. Confirm the resolution of the spectrometer. Then start averaging spectra.

Let us know how it goes!


I didn't check the math, but just took an FFT of a simulated signal ($\omega/(2\pi)=1~\mathrm{MHz},\Omega=\omega/10, \beta=1$), i.e. $|\mathcal F\{|E(t)|^2\}(f)|$:

Frequency spectrum of power

There is a comb of side bands at frequency intervals $2\Omega$ around the double carrier frequency, but assuming that your photodiode is not fast enough to see the carrier frequency (1 MHz in the calculation, but it should really be in the PHz range for visible light), there is nothing to see aroud the modulation frequency.

If you observe a sideband at $2\pi f=\Omega$, it's likely that what your EOM does is something more complicated than $E=E_0 e^{i(\omega t + \beta \sin(\Omega t))}$. For example, the photodiode may be picking up the periodic blueshift/redshift of the light or the EOM causes some parasitic attenuation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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