This is a very good first question. Keep going like this!
- Why there is this confusion and what is the correct answer? I almost sure that Loudon, Zou and Mandel are right.
Let me first clarify the connection between Loudon's classification and the one in the linked answer. For any type of light it is reasonable to assume that the intensity at time $t$ is uncorrelated to the intensity at time $t + \tau$ for very large $\tau$.
$$
\langle I(t) I(t+\tau) \rangle \xrightarrow{\tau \to \infty} \langle I(t) \rangle \langle I (t+\tau) \rangle
$$
Or, phrased with photon detection probabilities, the probability to detect a photon a long time $\tau$ after a photon had been detected at time $t$ is the same as the probability to detect a photon at time $t + \tau$ under any circumstances.
$$
p(t+\tau | t) \xrightarrow{\tau \to \infty} p(t+\tau)
$$
With this one can show that
$$
g^{(2)}(\tau\to\infty) =
\begin{cases}
\lim_{\tau\to\infty} \frac{\langle I(t) I(t+\tau) \rangle}{\langle I(t) \rangle \langle I (t+\tau) \rangle} = \frac{\langle I(t) \rangle \langle I(t+\tau) \rangle}{\langle I(t) \rangle \langle I (t+\tau) \rangle} = 1 \\
\lim_{\tau\to\infty} \frac{p(t) \, p(t+\tau | t)}{p(t) \, p(t+\tau)} = \frac{p(t) \, p(t+\tau)}{p(t) \, p(t+\tau)} = 1
\end{cases}
$$
Assuming that $g^{(2)}(\tau)$ is monotonic (in some cases it isn't), i.e. it goes from its value at $\tau=0$ to its value for $\tau \to \infty$ without wiggles
- $g^{(2)}(0) > g^{(2)}(\tau)$ is equivalent to $g^{(2)}(0) > 1$.
- $g^{(2)}(0) = g^{(2)}(\tau)$ is equivalent to $g^{(2)}(0) = 1$.
- $g^{(2)}(0) < g^{(2)}(\tau)$ is equivalent to $g^{(2)}(0) < 1$.
Therefore both, the definition in Loudon's book and in the linked answer, are correct. I personally prefer the definition which compares the value of
$g^{(2)}(0)$ to
$1$, because there you only need to measure / compute
$g^{(2)}(0)$, not its whole time evolution.
- I understand the math but I don't see completly the physical meaning and how to distinguish the different effects experimentally.
Photon statistics, and whether it's (sub-/super-)Poissonian, talks about the number distribution $P(n)$ of photons detected within a very short time window. Short in comparison to the time at which $g^{(n)}(\tau)$ goes to $1$. Therefore, given the photon statistics $P(n)$, one can calculate $g^{(n)}(0)$. In particular
$$
g^{(2)}(0) = \frac{\langle n \, (n-1) \rangle}{\langle n \rangle^2} \text{.}
$$
Bunched light is typically classical light with intensity fluctuations. For a more detailed description have a look into this question. In short, if you detect a photon at time $t$, chances are high that this is within an intensity maximum (a bunch of photons), therefore the probability to detect another photon shortly afterwards is increased.
Antibunched light is typically the emission from single-photon emitters. An atom / molecule can only emit one photon at a time. So if you just detected one photon from it you know that there can't be another one immediately after it.
Uncorrelated light ($g^{(2)}(\tau) = 1$) is when a detected photon doesn't give you any information about future detection events. See for example this question to see why uncorrelated photons result in a Poissonian number distribution.
Experimentally one measures $g^{(2)}(\tau)$ by time-tagging single photons and then binning the delays between each pair into a histogram. Here is a resource-recommendations question on this.
I understand that bunching means that photons tend to come in bunches on the detector, but super poissonian seems similar because you have $\Delta n^2 = \langle n \rangle + \langle n \rangle^2$. It seems to imply an excess of number of photons.
This is exactly how super-Poissonian light is defined. If the photon statistics have a variance larger than that of the Poissonian distribution (with the same mean value $\bar{n}$) it's called super-Poissonian and for most cases has a $g^{(2)}(0) > 1$. If the variance is below that of a Poissonian distribution it is sub-Poissonian and for most cases has a $g^{(2)}(0) < 1$.
- The Hanbury Brown and Twiss interferometer measure the correlation of intensity
of electric field. This tell us automatically if the source is bunched or anti bunched and super or sub poissonian at the same time? (I guess yes because if we have knowledge of $g^{(2)}(\tau)$, we can know it's value at zero and if it increases or decreases with $\tau$. This is valid only if the definition of Loudon is correct).
As written before, in most cases bunching means super-Poissonian light, antibunching means sub-Poissonian light. You might wonder what these "most cases" don't include. A counterexample is a source which emits perfect 2-photon states $|2\rangle$. There you have
$$
g^{(2)}(0) = \frac{\langle 2 | \hat{n} \, ( \hat{n} - 1 ) | 2 \rangle}{\langle 2 | \hat{n} | 2 \rangle^2} = \frac{2 \cdot 1}{2^2} = \frac{1}{2} < 1 \text{,}
$$
but
$$
\Delta n^2 = 0 < \bar{n} \text{.}
$$
Edit: This example is not specific as it doesn't make any statement about the time-evolution. Still, one finds this in many theory papers. Partially because their calculations don't depend on the dynamics of the system and can therefore be widely applied, partially because it's much easier to calculate $g^{(2)}(0)$ then the full evloution $g^{(2)}(\tau)$. Let me specify a concrete example here:
Imagine a system which is in the steady-state $\psi_{ss} = |2\rangle$, i.e. $2$ photons in a single mode of a cavity, to not deal with flying photon wavepackets. If a photon is detected at time $t$ the system is projected into the new state $\psi(t) = |1\rangle$. From there it will evolve again towards the steady-state. After infinitely long time $\tau \to \infty$ the system is in back in the steady state. This means the probability to detect a second photon right after the first is only half as much as after infinite time, simply because of the expected number of photons in the cavity at these times:
$$
p(t+0|t) = \tfrac{1}{2} p(t+\infty|t)
$$
Together with $g^{(2)}(0) = \frac{1}{2}$ this leads to $g^{(2)}(\infty) = 1$, as it should be for any system.
Now, why is the state $|2\rangle$ not bunched? Because it has no intensity fluctuations which lead to gaps between the bunches. It's like when you compare an infinite forest to a group of trees on a meadow. Seeing the group of trees one would say it's a bunch, but standing in the infinite forest you only know trees everywhere. To make it mathematically precise, take a look at the "group of trees on a meadow" equivalent state: $\sqrt{1-\epsilon} |0\rangle + \sqrt{\epsilon} |2\rangle$. Here the bunches cover a fraction $\epsilon$ of the landscape. For this state
$$
g^{(2)}(0) = \frac{\epsilon \langle 2 | \hat{n} \, (\hat{n}-1) | 2 \rangle}{\left( \epsilon \langle 2 | \hat{n} | 2 \rangle \right)^2} = \frac{2 \epsilon}{\left( 2 \epsilon \right)^2} = \frac{1}{2 \epsilon} \text{.}
$$
For small $\epsilon$ this can lead to insanely high bunching values. Here is a paper in which they measured $g^{(2)}(0) = 21$.
