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Upon proper excitation, single molecules (e.g. fluorescent dyes, quantum dots, etc.) spontaneously emit single photons, namely Fock states $|1\rangle$. These states have no statistical uncertainity over the photon number. However, experimental measure of fluorescence (say, an image of fluorescence microscopy) is well-known to be affected by shot-noise, namely the photons collected by the detector follow a Poisson distribution. This distribution is the same of that of a coherent state $|\alpha\rangle$. Why is this happening? What is causing the transition from number-state to an (almost) coherent-state in an ensemble of single-photon emitters?

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  • $\begingroup$ Counterexample: dye laser. You can get as much coherence as you wish, you just have to set up your physical system inside a cavity with enough optical feedback. $\endgroup$ Oct 7, 2022 at 9:38
  • $\begingroup$ @FlatterMann do you mind elaborating? I don't get what you are trying to say. $\endgroup$ Oct 7, 2022 at 9:42
  • $\begingroup$ The point is that the states of the electromagnetic field depend not just on the emitter but also on the boundary conditions (laser resonator). Indeed, in an active medium of sufficiently large dimensions and sufficient optical transparency we wouldn't need any boundary conditions like mirrors. Stimulated emission will always lead to a coherent state far enough away from a fluctuation and that means that there is always some level of correlation in a system with more than one molecule. That is, of course, independent of the absorber statistics. $\endgroup$ Oct 7, 2022 at 9:51

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Upon proper excitation, single molecules (e.g. fluorescent dyes, quantum dots, etc.) spontaneously emit single photons, namely Fock states |1⟩.

Most substances have a quantum yield that is much smaller than unity, so it would probably be more accurate to describe the light emitted as a density matrix in which, say, 90% of the probability is zero photons, and 10% is the one-photon state. This can be described by a binomial distribution $B(1,p)$, where $p=0.1$. Because $p\ll 1$, this distribution is actually well approximated by a Poisson distribution with mean $p$, although the approximation is not exact. (It can't be exact, because the Poisson distribution will have some probability for $n\ge 2$, in fact for arbitrarily large $n$, which is impossible based on conservation of energy.)

These states have no statistical uncertainity over the photon number.

So this is not quite true. The mean is $p$, and the standard deviation is $\sqrt{p(1-p)}$.

However, experimental measure of fluorescence (say, an image of fluorescence microscopy) is well-known to be affected by shot-noise, namely the photons collected by the detector follow a Poisson distribution.

Experimentally, your detector has some efficiency, which is also probably much less than one. This reduces $p$ by an additional factor. But at the same time, the detector counts photons for a certain time window, and you can see photons from lots of different excitation events. The sum of a large number of low-$p$ binomials is well approximated by a Poisson distribution.

This distribution is the same of that of a coherent state |α⟩. Why is this happening? What is causing the transition from number-state to an (almost) coherent-state in an ensemble of single-photon emitters?

But there is no close logical connection here. The different molecules are all emitting photons independently, so their radiation is not coherent. Poisson distributions arise in lots of real-world situations (insurance claims, etc.), most of which have nothing to do with coherent photon states.

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  • $\begingroup$ Thank you very much! I like and agree with most of your answer, except for the last part. With a "coherent state" I don't mean spatially or temporally coherent light, but I am referring to the coherent quantum state (see physics.stackexchange.com/questions/300034/…). Those states can be described as a superposition of number states with Poisson weight. Maybe they have nothing to do with my question (this is why I wrote "almost coherent"), but maybe there is a relationship. $\endgroup$ Oct 6, 2022 at 17:01

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