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If I have some photon detector, say a CCD, how do I estimate the error introduced by shot noise correctly? Typically, sources found on the internet say that the shot noise is the square root of the number of electrons counted (multiple sources possible, but here's one example: http://hamamatsu.magnet.fsu.edu/articles/ccdsnr.html).

What is then the origin of the shot noise? Is it in the process when photons generate electrons within the semiconductor? Say 10 photons may cause 1 electron, or sometimes 0 or 2 - and hence the Poisson distribution of electrons?

Other sources describe shot noise resulting from varying numbers of photons arriving at the detector. So sometimes 10 photons arrive, and sometimes just 9 or even 11 - and hence the Poisson distribution of electrons, that results from this variation in number of photons?

Is it both? The Wikipedia Article for Shot Noise describes both, shot noise due to quantization of light (photons) as well as of electric current (electrons).

I wonder because if I want to estimate the error introduced by shot noise in terms of standard deviation, then the unit of measure is important.

If the origin is in photons, where 10 photons generate 1 electron on average (i.e. a quantum efficiency of 10 %), but the number of photons varies over time, the shot noise is given as:

$$ \begin{align} \sigma_\mathrm{shot, photons} &= \sqrt{N_\mathrm{photons}}\\ \sigma_\mathrm{shot, electrons} &= \sqrt{N_\mathrm{photons}} / 10 \end{align} $$

Example: A signal of 100 photons has a shot noise standard deviation of 10. This converts to an avergae signal of 10 electrons with a shot noise standard deviation of 1. If I would estimate now the shot noise based on the number of electrons, I would estimate a standard deviation of \$\sqrt{10}\$ - which would be wrong.

If the origin is in conversion to electrons, where 10 photons generate approximately 1 electron, but rather a Poisson distribution around 1, then the shot noise is given as:

$$ \sigma_\mathrm{shot, electrons} = \sqrt{N_\mathrm{electrons}} = \sqrt{N_\mathrm{photons}/10} $$

If I now estimate the shot noise for a signal averaging to 10 electrons, I'd get again \$\sqrt{10}\$ - which would be correct if photons to electron counts conversion was the origin.

An example with Python:

>>> import numpy as np
>>> photon_count = 100 + 10*np.random.randn(10000)
>>> photon_count.std()
10.022125370282811
>>> electron_count = photon_count/10
>>> electron_count.std()
1.002212537028281
>>> np.sqrt(electron_count.mean())  # estimating shot noise as sqrt of number of electrons yields wrong result
3.1664560020385544

Note: I've asked this question over at Signal Processing Stackexchange, and have been hinted to ask here. Please let me know whether the cross-post is fine or not. Thanks!

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  • \$\begingroup\$ Are you talking about high levels of photons or just a few. Because at very low levels the dominant effect changes, dramatically. \$\endgroup\$
    – jonk
    Sep 22, 2021 at 4:15
  • \$\begingroup\$ Good question @jonk, thanks! As I'm interested in a theoretical understanding rather than practical relevance, let's say I'm talking about sufficient levels of photons that shot noise is dominating the noise of the photon flux, even if its a small effect then. hope that's an ok answer ;) \$\endgroup\$
    – hintze
    Sep 22, 2021 at 4:18
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    \$\begingroup\$ The reason I asked is because I've observed, personally, the effect of "flocking" (gathering into the same quantum state) of photons at low levels of current generation in diode detectors. This became observable at around \$600\:\text{aA}\$ or thereabouts and soon overtook shot noise computed by the basic \$\sqrt{2 q B I}\$ formula. \$\endgroup\$
    – jonk
    Sep 22, 2021 at 4:22
  • \$\begingroup\$ When thinking about shot noise right now, my head is taken to more recent work I've enjoyed, reading about electronic shot noise (electrons, not photons) that occurs and can be detected and analyzed in modern nanoscale conductors where classical mechanics breaks down. It's not just about photons. I don't have the time to cover what I'd want. So I'd prefer to refer you this page at MIT, under II.3. Theory of Shot Noise where the subject is covered quite well. \$\endgroup\$
    – jonk
    Sep 22, 2021 at 4:34

1 Answer 1

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Is it both? The Wikipedia Article for Shot Noise describes both, shot noise due to quantization of light (photons) as well as of electric current (electrons).

In most array sensors, shot noise comes from photon arrivals, so technically it is photons. However, it does not matter, you'll get the same answer either way since there is a 1:1 relationship between photons and photoelectrons, at least in a conventional CCD detector without photoelectron gain.

Note that if you have a detector with photoelectron gain such as a photomultiplier tube then you actually get significant shot noise from both electrons and photons.

If the origin is in photons, where 10 photons generate 1 electron, but the number of photons varies over time, the shot noise is given as:

Your premise does not make sense. While each photon may or may not be detected (due to quantum efficiency being less than 100%), you cannot have 10 photons generate 1 electron. Each photon either generates an electron or is not detected. Photon arrivals are independent events. Are perhaps thinking of the sensor's A/D gain, which is the relationship between the number of photoelectrons and the A/D counts?

The crux is, in order to detect 1 photon, on average, with a QE of 10 %, I'll need 10 photons to arrive.

QE encompasses losses due to reflection, finite fill factor, and photons passing through the pixel without being absorbed. If you want to count photons that miss the detector elements or bounce back at the source without being absorbed as having "arrived", you also need to include the photons that bounce off the lens, the photons that miss the lens, the photons from the sun that miss the Earth, the photons from other stars that miss the solar system, extragalactic photons that miss the galaxy, ...

Clearly that is a very, very large number of photons. But since these are photon that were never detected, they also don't matter. If something goes through the detector without being absorbed or misses the camera entirely, you can ignore it for purposes of calculating shot noise.

And then you say, "a measurement can only encompass information from the thing you measured". As I measure 10 electrons, I estimate the shot noise uncertainty in terms of standard deviation correctly by taking the square root of my electron count. Or not?

Outside of very specific quantum optical experiments, one cannot measure a photon without absorbing it and collecting its photoelectron. So all of these photons missing your pixel are not measured.

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    \$\begingroup\$ So what does this mean, then? You say: "shot noise comes from photon arrivals" With a QE of 10 %, 100 Photons arrive, with an uncertainty of 10 Photons. On average, I'll get 10 electrons out of those 100 Photons, with standard deviation of 1 electron. Thus, I estimate the shot noise uncertainty in terms of standard deviation correctly by dividing my electron count by the QE, taking the square root, and multiplying again with QE. Or not? \$\endgroup\$
    – hintze
    Sep 23, 2021 at 8:21
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    \$\begingroup\$ And then you say, "a measurement can only encompass information from the thing you measured". As I measure 10 electrons, I estimate the shot noise uncertainty in terms of standard deviation correctly by taking the square root of my electron count. Or not? \$\endgroup\$
    – hintze
    Sep 23, 2021 at 8:22
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    \$\begingroup\$ The crux is, in order to detect 1 photon, on average, with a QE of 10 %, I'll need 10 photons to arrive. Thus, my measurement actually does encompass more than the 1 photoelectron that I measured. It tells me that approximately 10 photons arrived at my detector. \$\endgroup\$
    – hintze
    Sep 23, 2021 at 8:24
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    \$\begingroup\$ I think the point of the question is that if photons arrive according to a Poisson distribution, and then a fraction are converted to electrons you get a different amount of noise than if the Poisson distribution arises from or after that conversion process. The difference is a factor of sqrt(QE). Multiplying a Poisson distribution by a constant makes it no longer a Poisson distribution. \$\endgroup\$
    – Matt
    Sep 24, 2021 at 2:23
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    \$\begingroup\$ @Matt I think the answer here is that "Arrive" and "are converted to electrons" are not distinct processes. The amount of shot noise you get is the square root of the number of detected photons. From the perspective of the detector, if you have 100 photoelectrons, your shot noise is 10. This is true if you have a QE of 100% and had 100 total photons, or a QE of 1% and 10,000 photons. Thus the QE has no effect on your shot noise for a given number of detected photons. \$\endgroup\$ Sep 24, 2021 at 2:31

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