# Shot noise in optics

There are many explanations to be found about shot noise in optics, but the answers I find are incompatible. There are three ways shot noise in optics is explained.

(Note that according to Wikipedia, in general, shot noise is a type of noise which can be modeled by a Poisson process.)

1. It is the noise purely arising form (vacuum) fluctuations of the EM-field. For example, the book of Gerry and Knight states that "In an actual experiment, the signal beam is first blocked in order to obtain the shot-noise level." I guess the number of photons you would detect in this way follows a Poissonian distribution, hence the name shot noise'. (For context, see screenshot of relevant section below - courtesy of Google Books)

2. It is due to 'the particle nature of light'. Semi-classically, a low intensity laser beam will emit photons following a Poisson distribution. If the beam is incident on a photon detector, this detector will receive a fluctuating number of photons per time bin (according to the Poissonian). Thus the intensity (~number of photons per time bin) will fluctuate. These fluctuations are the shot noise'.

3. A laser beam emits a coherent state $$|\alpha \rangle$$. The probability to find $$n$$ photons upon measurement follows the poisson distribution, $$P(n)=|\langle n | \alpha \rangle|^2= \frac{\bar n}{n!}e^{-\bar n}$$ with $$\bar n = |\alpha|^2= \langle \alpha | a^\dagger a | \alpha \rangle$$ the average number of photons. Thus there is shot noise in the number of photons. (Here $$| n \rangle$$ is in the Fock basis but $$|\alpha \rangle$$ is in the coherent state basis.)

So what is shot noise? Can you have multiple sources of shot noise, throw them all on one heap and call the combination 'the shot noise'. Then how can you 'measure the shot noise level' as in 1, or 'measure at the shot noise level'?

Explanation 1 is incompatible with 2 and 3, for both 2 and 3 will cause no photons at all to be counted in the vacuum state. (The vacuum state is the coherent state with $$\alpha=0$$.)

• I looked up the quote from Gerry and Knight and, as I suspected, when the signal beam is blocked, there is a strong local oscillator (LO) beam present on the photodetector (PD). I assume the power of the LO beam incident on the PD is high enough that the beam's inherent shot-noise dominates the PD noise. But (if I read your post correctly) you're assuming that the quote implies there is no beam on the photodetector when the signal beam is blocked? Commented Dec 21, 2018 at 11:59
• I think it means that, when the input beam is blocked, there are still coming photons from that port due to vacuum fluctuations. True, there are also photons coming from the LO beam, but the LO is only used as a means to measure the photons coming from the input beam. Commented Dec 21, 2018 at 12:45
• The LO beam has inherent shot-noise and since it is stipulated that the LO beam is strong, the LO shot-noise dominates. See, for example, this: "If the signal port is blocked, the difference of photocurrents exhibits the shot noise level of the local oscillator beam, even if the actual noise level of that beam is different." Commented Dec 21, 2018 at 12:58
• If you block all light on a detector, we normally call that dark noise rather than shot noise. However one way to explain dark noise is as shot noise due to black body radiation from objects around the detector. Commented Dec 21, 2018 at 15:44
• Georg, as I review some of the squeezed light literature, it appears that the shot-noise measured with just the LO on the PDs (signal port blocked) corresponds to the vacuum noise as 'revealed' by the shot-noise in the LO beam, e.g., "The measured vacuum noise level corresponds to the LO’s (electronically amplified) shot noise level." Commented Dec 21, 2018 at 16:32

The explanations that are provided in points 2 and 3 give a good description of the shot noise in light. They are consistent with each other (right?).

The situation in point 1 is probably related to shot noise in detectors. This is electronic shot noise and not related to light. All detectors have a dark current, which exists even without the presence of light. Dark currents produce shot noise in the detector.

Then how can you 'measure the shot noise level' as in 1

For the balanced homodyne detector (BHD) described in the relevant section of Gerry and Knight, and with the signal port blocked, the sum of the two photodetector (PD) outputs will have noise contributions from the local oscillator (LO) field shot-noise as well as other laser related noise contributions, e.g., relative intensity noise (RIN).

However, the non-shot-noise contributions are common-mode and so the difference of the two PD outputs will (ideally) have only the LO shot-noise.

or 'measure at the shot noise level'?

At the PD difference output, amplitude variations of the signal as well as the shot-noise scale with the LO power. Increasing the LO power generally improves the signal-to-noise ratio (SNR) when other sources of noise, e.g., electronic noise in the PDs, dominate.

However, if the shot-noise contribution of the LO dominates the other noise sources, increasing the LO power does not improve the SNR; the sensitivity of the apparatus is then shot-noise limited. This is why, in figure 7.12 of Gerry and Knight, a strong coherent field is specified as being injected into the LO port.

While I am no expert in squeezed light and do not understand what the "local oscillator" is, I can tell you about shot noise in more general contexts:

In general, any Poissonian process will show the "shot noise" behaviour. Consider this: if you roll steel balls through a pipe at some rate, say, 9 balls/s on average, you will not count exactly 9 balls/s in any given interval. This can fluctuate a bit. If the balls are coordinated to be equally spaced, you will see 9 ± 1 balls/s.

But if the arrival of one has no correlation to the arrival of the next (uncoordinated balls), this will be a Poisson process, with the standard deviation given by the square root of the number of balls you counted, so you will see 9 ± 3 balls/s. If you count for 4 seconds, you will have seen 36 balls, so the noise will be $$\sqrt{36} = 6$$. Essentially, if you count the number of balls in many random intervals such that the average number of events in that interval is $$N$$, the standard deviation of the distribution you obtain will be given by $$\sqrt{N}$$.

The same logic applies to photons. Thus:

1. If, when the signal beam is blocked, no light falls on the detector, this may be to estimate the detector dark shot noise. Photodiodes always have dark current, which can be due to thermal processes mimicking the same pathway by which photons cause current in the diode. If there are, say, 1000 such events per second, leading to a corresponding dark current, then the associated dark noise will be $$\sqrt{1000} \approx 30$$ events/s. You would then also be able to account for the electronic read noise, which may follow the same process.

I think the other answers have already done a good job of covering 2 and 3.

In general there is noise in light itself, called photon shot noise (different from electronic shot noise), i.e. if a bulb or laser is lit and has a flux of 1M photons per second it really is releasing 1M +- square root of 1M or 1000 photons per second, one std deviation. So that's typically 999,000 or 1001,000 photons actually measured. Light itself is statistical in its emission. You can't take one atom, stimulate it and not know precisely when the photon will emit. Yes 2 sources can mix, i.e. 2 bulbs gives 2,000,000 photons with a std error of 2000 but a single bulb emitting 2,000,000 gives a shot noise of 1414.