In cavity quantum electrodynamics why is $langle hat{a} rangle$ called the cavity amplitude?

Question

What is the physical meaning behind $langle hat{a} rangle$ (where $hat{a}$ is the annhilation operator)?
I was thinking that it represents $sqrt{n}$ (where $n$ is the number of photons in the system) since $hat{n} = hat{a}^{dagger} hat{a}$ (where $hat{n}$ is the number operator).
Why is $langle hat{a} rangle$ referred to as the cavity amplitude in cavity quantum electrodynamics?

Gomboc · Answer

The electric and the magnetic field operators can be constructed from the ladder-operators. The real part of the expectation value is proportional to the classical field amplitude, so I guess that is the reason.

Answered by Gomboc on February 3, 2021

Wolpertinger · Answer

Much like the electromagnetic field in classical electrodynamics $mathbf{E}$, the expectation value $langlehat{a}rangle$ is a complex number. It thus corresponds to something like an amplitude, whereas the number operator expactation value $langlehat{a}^daggerhat{a}rangle$ corresponds to something like the intensity.
The correspondence can also be formalized since in quantum electrodynamics, we have
$$mathbf{E} sim hat{a} + hat{a}^dagger ,.$$
For details see any textbook on quantum optics or this wikipedia article as an easily accessible source.

In cavity quantum electrodynamics why is $langle hat{a} rangle$ called the cavity amplitude?

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