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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).

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    $\begingroup$ Modulating the diode laser current doesn't produce pure (or ideal) phase modulation does it? $\endgroup$ Commented Jul 16, 2020 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
    Commented Jul 16, 2020 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
    Commented Jul 16, 2020 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
    Commented Jul 16, 2020 at 16:03

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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.

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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!

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Your analysis looks correct. Like Gilbert said, noise is likely producing a signal at $\Omega$. Depending on the shielding and components in the EOM, stray field from the device could produce an emf that introduces noise at that driving frequency. You can try blocking the light going into the photodiode to see of the noisy signal at $\Omega$ remains. That said, if you connect the photodiode to a spectrum analyzer, focus the SA's span around 2$\Omega$, and set the resolution bandwidth very narrow you should be able to see a spike at the expected frequency.

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In the OP $P = |Re(E)|^2$ should be $P = |E|^2 = EE^*.$ Applying the proper conjugation results in the DC terms dropping out (without the DC terms, you get negative powers, which is unphysical), and also the measured $\Omega$ beat note, which emerges from the cross-terms between the $J_0$ and $J_1$ components.

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  • $\begingroup$ The complex representation $E$ corresponds to a physical electrical field $\mathrm{Re}(E)$ (or $\mathrm{Im}(E)$ up to a constant phase shift), so the intensity is something like $I=cn\varepsilon_0 1/2|\mathrm{Re}(E)|^2$. When using $EE^*$, fast oscillating terms are neglected, which is equivalent to $|\mathrm{Re}(E)|^2$ only when averaging over fast-oscillating terms (up to a factor $1/2$). In most applications, this averaging is irrelevant or even desired, because you often anyways only (can) measure the time average, but the full picture is only given by the actual physical electrical field. $\endgroup$
    – faber
    Commented Oct 17, 2022 at 8:22
  • $\begingroup$ If you include complex conjugates in your expression for $E$, then it will be real-valued and thus there is no need to disambiguate the "physical" part. $EE^*$ then gives you both DC terms and high-frequency components as expected since $EE^* = |E_0|^2(e^{i\omega t} + e^{-i \omega t})(e^{-i\omega t} + e^{i \omega t}) = |E_0|^2(2 + e^{2i\omega t} + e^{-2i\omega t})$. Going back to your original post though, $|Re(E)|^2$ and $|E|^2$ are very clearly algebraically different, and it's not correct to equate the two. $\endgroup$
    – Shovelhead
    Commented Nov 2, 2022 at 3:15
  • $\begingroup$ If you include the complex conjugate already in the field, you have the same thing as me (up to a constant phase and factor). You would do something of pattern $E_A=E_0/2\left(e^{i\omega t}+e^{-i\omega t}\right)=E_0\cos{\omega t}$ and calculate $P\propto E_A (E_A)^*=|E_A|^2$, while I would write $E_B=E_0e^{i\omega t}$ and say that the $\mathrm{Re}(E_B)=E_0\cos{\omega t}$ is my 'physical field' and $P\propto|\mathrm{Re}(E_B)|^2$. $\endgroup$
    – faber
    Commented Nov 3, 2022 at 6:55
  • $\begingroup$ Yes, you are correct, however taking $Re(E_B)$ doesn't take advantage of the complex algebra as the $E_A$ expression does. P.S. I should apologise, I just now realised I misread your original post - I assumed from the citation of the PDH paper you were probing a cavity, however I now realise that you were measuring the beam directly, in which case the sideband interference terms should cancel and no $\Omega$ component should appear as you correctly supposed. Incidentally, the complex representation makes this sideband cancellation explicit, whereas the trig algebra disguises it somewhat. $\endgroup$
    – Shovelhead
    Commented Jan 10, 2023 at 6:39
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Phase modulation of a laser beam expressed by a formula: $E=E_0(e^{(i\omega t + i \sin (\Omega T)})$ (for example a modulation introduced with an electro-optic modulator) cannot produce any beat note, as $I \propto |E|^2$ and in this case clearly: $|E|^2 = E_0(e^{(i\omega t + i \sin (\Omega T))}) \cdot E_0(e^{-(i\omega t + i \sin (\Omega T))}) = E_0^2 $, so no temporal modulation should be visible.

However, when one is modulating a laser diode with a seeding current two effects come into play:

  1. Phase modulation due to change in the refraction index of the medium - described by the aforementioned formula;
  2. The fluctuations of the light amplitude with a frequency of applied modulation, as the laser beam intensity is directly dependent on a driving current in the case of laser diodes.

OP likely observed the latter effect, however, it would not be surprising to pick up with an amplifier chain a stray RF frequency present in the lab.

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