# Quantum shot-noise and the fluctuation dissipation theorem

Classically, shot noise observed in the signal generated by a laser incident on a photodiode is explained as being due to the quantization of light into photons, giving rise to a Poisson process. In quantum optics, on the other hand, the shot noise is said to arise from interference with the vacuum field, which leaks in at points of optical loss (absorption).

Meanwhile, mechanical oscillators are subject to the fluctuation-dissipation theorem, which says, roughly, that the thermal excitement of the various mechanical modes of the oscillator are proportional to the dissipation in that mode. In the context of electronics, Johnson noise is an example of this effect.

I recently read a thought-provoking statement (in this paper) that these effects are really "the same"; just as thermal motion arises at points of mechanical dissipation, shot noise arises at points of optical dissipation.

Is this analogy physically/theoretically significant?

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The principle is the same in both pictures. I'm not sure how to answer "is this analogy significant?", since I don't think it's an analogy at all. It's the same phenomena, as explained below.

Any time you have a mechanism for dissipation (a coupling between the "system" and a "bath") that coupling mechanism will give rise to back-action of the bath the system (i.e. fluctuations/noise). This is true whether the bath is in a thermal state or a vacuum state. Quantum optics can treat both cases; the usual fluctuation-dissipation formula only works in the limit that the bath is well approximated by a classical thermal state (i.e. kT >> the energy of 1 quanta of excitation).

And just to clarify one point: when you say "the shot noise is said to arise from interference with the vacuum field", that's not always the whole story. Noise can come from lots of places. It can be in the original state of your optical field, in which case it will be observed even if you don't have lossy mirrors/detectors. But the idea is that if you DO have loss, then even if you start with a sub-shot-noise beam, you'll pretty quickly approach the shot noise limit.

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Thanks for the response! Can you show how to treat a lossy mechanical oscillator using the quantum optics formalism? –  nibot Feb 22 '11 at 16:13
For that specific case (damped oscillator vs. quantum electromagnetic field), the QM treatment of the simple harmonic oscillator is mathematically equivalent to the QM treatment of a single mode of the EM field. Can I show how? Maybe, but that's more work and typing than I care do do. Look at whatever section of your quantum optics book quantizes the field. I'd bet money they're using photon creation and annihilation operators with the same commutation relations as the raising and lowering operators of the harmonic oscillator. –  Anonymous Coward Feb 22 '11 at 17:08

Nibot,

I strongly suggest you read Noise and Fluctuations by D.K.C. MacDonald. It has lots of great discussions related to thermal noise. That's where most of this answer comes from.

You are probably used to the Fluctuation Dissipation Theorem (FDT) written in a form similar to the way Nyquist derived Johnson noise on a resistor: $$<\delta V_f^2> = 4RkT df$$ Which is the variance in the Voltage squared, in terms of $R$ the resistence, $kT$ the temperature, and $df$ the measurement bandwidth. (Alternatively divide both sides by $df$ and the quantity is the power spectral density).

But there is another form of Nyquist's theorem for when $hf \approx kT$, i.e. valid in the quantum regime. $$<\delta V_f^2> = 4R\left( \frac{hf}{2}+\frac{hf}{e^{hf/kT}-1}\right) df$$ You should be able to convince yourself that this reduces to the standard Nyquist formula in the appropriate limit.

Using this form of the theorem, and considering a charged particle which oscillates in vacuum, there is a damping back-reaction of the electromagnetic field given by the Larmour formula: $$\vec{E} = -\frac{(2\pi f)^2}{6\pi \epsilon_0 c^3} \dot{ \vec{p}},$$ for the electric field $\vec{E}$ and dipole $\vec{p}$. So with analogy to the Nyquist formula, $<\delta V^2_f>$ describes the electric field fluctuations, and $R=\frac{(2\pi f)^2}{6 \pi \epsilon_0 c^3}$. Surprisingly, plugging this into the quantum Nyquist theorem reproduces the blackbody radiation spectrum! The FDT never ceases to amaze!

Note that my quantum FDT includes a zero point energy term, which is a bit controversial, because it also predicts a blackbody spectrum which has a zero point energy term, which can't be observed directly.

Now, I must admit I tried and failed to derive the shot noise formula from the blackbody spectrum with the zero point energy term added, but because shot noise is often attributed to zero point energy fluctuations of the EM field, it feels like this represents the same thing physically. I think my math skillz just weren't cutting it.

I guess one thing to realize is that these optical measurments are working in the $hf\gg kT$ limit while usually thermal noise is concerned with the opposite limit. But imagine an interferometer working with 10 $\mu$m light, where the room temperature thermal spectrum is large. This interferometer would be primarily concerned with thermal fluctuations entering ports, rather than quantum ones!

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