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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, 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
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