# Justifying assumptions for deriving Poissonian photon statistics

Suppose you have a perfect monochromatic (angular frequency $\omega$) laser beam in vacuum, such that it is well described classically by $$E(t,x)=E_0\sin(\omega t-kx+\phi)$$ with the $x$-coordinate along the beam, where $E_0, \phi$ do not depend on time and $k=\omega/c$.

I am reading the book "Quantum Optics" by Mark Fox. Fox derrives that such a beam must have Possonian photon statistics (i.e. the probability that a fixed segment of the beam contains exactly $N$ photons follows a poisson distribution). To do so, he makes a few assumptions that I have trouble justifying. I was hoping you could help me. The derrivation goes as follows:

Take a segment of the beam of length $L$ that has a well-defined time independent average number of photons in it, $\overline{N}$. Fox wants to take the length large enaugh, so that the number $\overline{N}$ can be taken to be an integer, which I do agree should be possible to reasonable precision, however I see no need for it (That is, I do see why $L$ should be assumed large, but no reason why one would want $\overline{N}$ to be an integer).

The next step is to subdivide the beam segment into $n$ sub-segments of equal length, which are so small, that the probability of finding a photon in an indivicual sub-segment is negligible. Fox now claims that the probability $P(N)$ to find exactly $N$ photons in the segment of length $L$ is given by the binomial distribution $$P(N)=\binom{n}{N}\left( \frac{\overline{N}}{n}\right )^N \left(1- \frac{\overline{N}}{n}\right )^{n-N}.$$ He then simply takes the limit $n\to\infty$ to arrive at the Poisson distribution.

This derivation seems a bit like a trick to me intuitively. If anyone knows a better one, I'd love to hear it. As far as I can see, the assumptions necessary to arrive at the binomial distribution above are as follows:

• The probability of finding two photons in a sub-segment is zero.
• The photons are distributed along the beam randomly.

I can see how the first assumption is justified by "the probability of finding a photon in an indivicual sub-segment being negligible". However, it would be interesting to examine how much of an effect it has on the end result, if one allows a very small such probability. I'm unsure how to approach that question, since the expression $\overline{N}/n$ for the probability can no longer be used.

My doubts mainly concern the second assumption: The photons are distributed completely randomly. That includes (very unlikely) scenarios where, if you have an extremely long beam segment, the detection rate could vary noticably with time, enaugh to be registered by a photon counter. That would be inconsistend with the initial assumption of a perfectly coherent, homogenous beam. How does one arrive at this conclusion (photons distributed perfectly randomly) just from knowing what the best classical description (i.e. my $E(t)$ formula) of the laser beam is?

Thank you very much in advance for your answers and comments. (linked to unanswered question How do we show that photons generated by a constant electric current are distributed according to a Poisson distribution?)

The problem that you could have 2 photons in one sub-segment is negligible, because the Poisson distribution arises as $n \to \infty$ limit. In this limit $\frac{P(2)}{P(1)} \to 0$.
In a coherent monochromatic wave the quantum mechanical state of the field $\left| \alpha \right\rangle$ must be an eigenfunction of the electric field operator $\hat{E}$ in order to have a "classical" field amplitude. $\hat{E}$ is (up to some constants) equal to the photon annihilation operator $\hat{a}$, which acts on an $n$-photon state $\left| n \right\rangle$ in the following way: $$\hat{a} \left| n \right\rangle = \sqrt{n} \left| n-1 \right\rangle \\ \hat{a} \left| 0 \right\rangle = 0$$ Eigenstates of the photon annihilation operator expressed in the photon number basis are $$\left| \alpha \right\rangle = \exp \left( - \frac{ \left| \alpha \right|^2 }{2} \right) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \left| n \right\rangle .$$ The probability distribution of this state to find $n$ photons is then a Poisson distribution: $$P(N) = \left| \left\langle n | \alpha \right\rangle \right|^2 = \exp \left( - \left| \alpha \right|^2 \right) \frac{\left| \alpha \right|^{2n}}{n!} = \exp \left( - \left\langle n \right\rangle \right) \frac{\left\langle n \right\rangle^n}{n!}$$
Indeed, the detection rate varies with time, which can be seen by calculating the variance of the photon number: $$\left\langle n^2 \right\rangle - \left\langle n \right\rangle^2 = \left\langle n \right\rangle$$ This is the so-called shot noise. For very intense coherent states its contribution relative to the overall intensity $\left\langle n \right\rangle$ vanishes: $$\frac{\sqrt{\left\langle n^2 \right\rangle - \left\langle n \right\rangle^2}}{\left\langle n \right\rangle} = \frac{1}{\sqrt{\left\langle n \right\rangle}}$$