Xyius said:
So my first thought is, "I know coherent state photons exhibit a Poissonian distribution, thus a stimulated emission from a laser emits quantum coherent states because it also exhibits this distribution."
Ok, the thorough definition of coherence is given by a correlation function that factorizes in all orders. However, this is not really helpful for getting a basic understanding. One can loosely interpret the normalized nth-order correlation function as the conditional probability of detecting the nth photon after you have already detected n-1 photons. It is normalized with respect to the case of statistically independent photons.
Now one can show that the Poissonian distribution has several features which match what you expect for the coherent state. The Poissonian distribution is the distribution for statistically independent events. Therefore it is not surprising that it gives you a normalized correlation function that factorizes. Also, if you take the coherent state and apply the photon annihilation operator to it, you will find that the average photon number does not change. This is also matching the statistical independence requirement.
So basically, yes, it is somewhat like the duck test: You expect to have some properties and stimulated emission shows these properties. There is some more to it, but that is somewhat deep. The lecture by Glauber which was linked earlier really is a very good start to introduce yourself to the topic.
The one point you need to keep in mind is the timescale. The correlation functions needs to factorize at timescales shorter than the coherence time of the light in order for the state to be a coherent one. On long timescales any light field will look like a Poissonian one. This is because photons from a single mode are only indistinguishable within their coherence time.
naima said:
before detecting one photon the Poissonian probability for k photons is \lambda^k e^{-\lambda}/k!
we have p_0 + p_1 + p_2 + ... = 1
once we have found one photon we get other probabilities p'_0 p'_1 p'_2 .with
p'_k.= p_{k+1}/(1 - p_0)
I do not see that conditional probabilities remain constant
That is not really a conditional probability. You would have to calculate the photon number distribution in the case that one photon was already detected (so p_0 is out, for example). Let me roughly sketch the math for the equal-time second-order correlation function. It is defined as:
g^{(2)}(0)=\frac{\langle n (n-1) \rangle}{\langle n \rangle^2}.
Here, the n is the instantaneous photon number. This quantity gives you the normalized probability to detect a second photon after you already detected one (the n-1-term comes from normal ordering and represents the destruction of a photon by the first detection process). It is the relative probability compared to the case of a light field with the same mean photon number, but photon detection events which are statistically completely independent. In that case the expected number of detections would be \langle n \rangle^2 as shown in the denominator.
You can express the instantaneous photon number n as the combination of the mean value and some fluctuation around it: n=\langle n \rangle +\delta n.
That gives you:
g^{(2)}(0)=\frac{\langle \langle n \rangle^2 + 2 \langle n \rangle \delta n - \langle n \rangle + \delta n^2 -\delta n \rangle}{\langle \langle n \rangle +(\delta n) \rangle^2}
The expectation value of the fluctuation is of course 0, so all terms linear in the fluctuation will vanish. That leaves you with:
g^{(2)}(0)=\frac{\langle n \rangle^2 - \langle n \rangle + \langle (\delta n)^2 \rangle}{\langle n \rangle^2}.
Or in short:
g^{(2)}(0)= 1 -\frac{1}{\langle n \rangle}+\frac{\langle (\delta )n^2 \rangle}{\langle n \rangle^2}.
A value of 1 gives you an unchanged conditional probability. Values above 1 describe a tendency for photons to bunch (like for sunlight). Values below 1 represent a tendency for photons to avoid each other (like for single photon sources).
Now the first term is always 1. The second term is always negative and represents the destruction of a photon in the first detection process. The third term is always positive and describes a tendency to bunch in terms of the fluctuations present in the light field. The central quantity \langle (\delta n)^2 \rangle is the variance of the photon number distribution.
So in order to get a value of 1 and an unchanged conditional probability, you need the second and the third term to cancel exactly. As one can easily see, this will happen when the variance of the distribution is exactly equal to the mean value of the photon number. This is exactly the case for the Poissonian distribution. It is also the smallest relative variance you can get from a classical light field. All light fields showing a smaller relative variance are inherently non-classical.