In quantum optics, coherent light with constant frequency, phase, and amplitude shows poissonian photon number statistics:

$$P(n) = \frac{\bar{n}^{n}}{n!}e^{-\bar{n}}.$$

A well-known result for Poisson distributions is that their variance equals their average and therefore their standard deviation is equal to the square root of the average:

$(\Delta n)^2 = \bar{n}$ and $\Delta n = \sqrt{\bar{n}}$.

Now, if you were to attenuate a coherent light beam, your average n would go down, let's say by a factor x. For as far as I understand, when scaling a probability distribution, the average and the standard deviation get scaled by that factor. So a 'new' light beam would have the following properties in terms of an 'old' light beam:

$\bar{n}_2 = x \bar{n}_1$ and $\Delta n_2 = x \Delta n_1 = x \sqrt{\bar{n}_1}$.

The variance of the attenuated beam is then the following:

$(\Delta n_2)^2 = x^2 n_1$.

This means that now $\bar{n}_2 \ne (\Delta n_2)^2$, indicating that this light is not Poissonian anymore. However, this would make it impossible to create coherent light, since there is always some sort of attenuation in any set-up or light source. Furthermore, books like "Quantum optics, an introduction" by Mark Fox, still speak of Poissonian statistics, even after attenuation or inefficient detection. This indicates that the scaling of the probability distribution explained above is incorrect.

So my question is, how can these two concepts be combined? Why do the properties of light not get scaled like other probability distributions?

  • $\begingroup$ "For as far as I understand, when scaling a probability distribution, the average and the standard deviation get scaled by that factor." --> Why do you say this? This kind of scaling might make sense for a Gaussian distribution (depending on the context), but for a Poisson distribution it doesn't work, as you indeed demonstrate in your question. My immediate reaction is that the contradiction is resolved by just removing this assumption, scaling the mean by a factor, and keeping the distribution as Poisson. But, I'm not sure why you assumed this so maybe I'm missing something. $\endgroup$
    – Andrew
    Commented Feb 18, 2022 at 18:01
  • $\begingroup$ Hi Andrew, the assumption is indeed an assumption that could be wrong. It is based on me running some scripts in python experimenting with datasets. Simply creating a poison distribution and then scaling the whole array and looking at the variance, std and mean of the result. $\endgroup$
    – julian
    Commented Feb 18, 2022 at 18:11
  • $\begingroup$ I made an example: onecompiler.com/python/3xts8bwxt $\endgroup$
    – julian
    Commented Feb 18, 2022 at 18:25

1 Answer 1


The key is how to physically describe attenuation. For a coherent state of light $|\alpha\rangle$ with $\bar{n}=|\alpha|^2$, the amplitude $\alpha$ simply gets multiplied by a scale factor when undergoing attenuation: $$|\alpha\rangle\to|\alpha\cos\theta\rangle.$$ The scale factor always has magnitude less than one, which is why I wrote it as $\cos\theta$. Then, all of the other properties of coherent states are maintained: they are new coherent states with average energy $\bar{n}=|\alpha|^2\cos^2\theta$, and they still have Poissonian statistics.

These formulas may remind you of something like the Malus law for polarizers: polarized light going through a polarizer has a probability $\cos^2\theta$ of getting transmitted, where $\theta$ is the relative angle between the incident light's polarization and the polarizer's optimal transmission axis. This is exactly an attenuation process for a coherent state of light with a particular polarization, so it tells you that a coherent state with a given polarization can be attenuated by passing through a polarizer optimized for some other polarization, and the resulting state will still be coherent and polarized but with a smaller intensity.

The mathematics for this process follow from attenuation being described by beam splitters. In quantum optics, these move around the energies and phases between two coherent states. So in the case of a single input coherent state, there will be some probability $\cos^2\theta$ of transmission by the beam splitter and the remaining probability $\sin^2\theta$ describes the reflected portion of the beam. If you ignore the reflected part, this describes attenuation that reduces the intensity by $\cos^2\theta$.

To learn the details of how this works, one must investigate the mathematics of beam splitters, which are surely covered in the books mentioned. With some phase convention, coherent states will transform as $$|\alpha\rangle\otimes|0\rangle\to|\alpha\cos\theta\rangle\otimes|\alpha\sin\theta\rangle.$$

  • 1
    $\begingroup$ Thank you for your answer. I think now that you say it, it also intuitively makes sense. The most important is to think at the beamsplitter again in terms of probability it seems. When I make a script and pass all the photons through the beamsplitter with a certain probability, I indeed get again poisson statistics! Very interesting, thank you! $\endgroup$
    – julian
    Commented Feb 18, 2022 at 20:46
  • $\begingroup$ Here I did that, in case people have the same confusion: onecompiler.com/python/3xtsh8d63 $\endgroup$
    – julian
    Commented Feb 18, 2022 at 20:47
  • $\begingroup$ @julian that's exactly the way to go. But with quantum, we really have to worry about probability amplitudes, because sometimes you'll be able to have interference terms (eg if you have something other than the vacuum coming in to one port of a beam splitter). $\endgroup$ Commented Feb 18, 2022 at 21:47
  • 1
    $\begingroup$ With quantum, the intuitive approach always hides some dangers it seems indeed. I will review the chapters on beamsplitters, thank you for your help! $\endgroup$
    – julian
    Commented Feb 19, 2022 at 9:20

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