I have been a bit confused recently about the total field operator from a laser source (coherent).
What would be a "good" representation of the quantum state of my laser propagating? (let's say single mode, with some noise, quantum and classical).
I come up with that, that I could find in different thesis or book: $\hat{A}(t) = A_0(t) + \hat{a}$, where $A_0(t) = (1+\chi(t))\alpha e^{i\theta}$, with $\chi(t)$ the stochastic noise of a laser at low frequency, $\alpha$ the amplitude and $\theta$ its phase. $\hat{a}$ is the annihilation operator, to represent the quantum noise.
But originally, I would have used something like that $\hat{A}(t) = A_0(t) + \delta\hat{a}$, with $\delta\hat{a} = \hat{a} - \langle \hat{a}\rangle = \hat{a} - \alpha$, for a coherent state (similar to $\delta x = x - \langle x \rangle$, but I am wondering if it's correct).
Indeed, if you want to compute the current from a photodiode, with $\langle \hat{I} \rangle = q\langle \hat{A}^{\dagger}\hat{A}\hat{A}^{\dagger}\hat{A} + n_e\rangle$, where $n_e$ is the number of electrons you generate independantly from the optical light (so electronic noise) and $q$ is the quantum efficiency, at some point you get weird stuff with the $\delta \alpha$ model where $\langle \delta \hat{a} \delta \hat{a}^\dagger \rangle = -2 |\alpha|^2$ ? so the mean number of these fluctuation would give something which is not equals to 0? EDIT: no it gives 0, i made a stupid mistake.
That's where I am confuse, is the "classical" approximation of fluctuation $\delta x = x - \langle x \rangle$ is wrong, or is it just me who is not intepreting well the physics, or im doing something wrong in the calculation of the expected values for a coherent state?
EDIT: but my question still stands, are both correct interpretation of quantum noise for a single mode laser operator? and how can I check if it's true
