How does a coherent state excite charge carriers across a semiconductor band gap?

Question

In standard semiconductor theory, we have a valence and conduction band energetically separated by band gap energy $E_g$.

A photon with energy $E>E_g$ can excite an electron-hole pair from the valence to the conduction band where the energy difference $E-E_g$ is transferred to the kinetic energy of the electron-hole pair.

In contrast to the single-photon absorption, the two-photon absorption is (usually) strongly suppressed, suggesting that the single-photon absorption is the main process for transferring energy from the electromagnetic field to a semiconductor.

So what happens when we illuminate a semiconductor not with a single photon but with a coherent state?

We know that a coherent state can be written as a coherent superposition of $n$-photon states $$ \vert\alpha\rangle = e^{-\vert\alpha\vert^2/2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} \vert n\rangle $$ which has mean photon number $\overline{n}=\vert\alpha\vert^2$.

Following our initial reasoning, the absorption of two-and-more-photon states should be highly suppressed, and we would expect the photon

$$ \vert1\rangle \langle1\vert\alpha\rangle = \alpha e^{-\vert\alpha\vert^2/2} \vert1\rangle $$

to "solely" excite the electron-hole pair accross the band-gap.

Is this reasoning sensible?

On the other hand, from a semi-classical perspective, the photocurrent (or number of electron-pairs) should be proportional to the intensity which in turn is proportional to the mean photon number $\overline{n}$.

What is the mechanism to "pick-out" the mean-photon number rather than the photon state?

Have there been investigations on this subjects?

Update 1:

Discussing the issue with colleges, I gained the following new insight: Absorption of the light field inside the photodiode is a function of the detector depth, meaning, the coherent state loses "intensity" or "photons" while propagating through the detector atoms.

If we assume a linear coupling between the light and photoelectron fields, then there is a finite probability (per unit time, per unit depth) that a photon is destroyed and an electron-hole pair is excited (in that sense we would have a beam-splitter-like interaction between the annihilation operator of the light field and the creation operator of the photoelectrons).

In that sense, it might be helpful to think about the coherent state "loosing" one photon of its photon mean while passing through the detector. For a sufficiently deep detector, the probability to find the light field vacuum at the "bottom" of the detector should asymptotically approach 1.

Answer 1

Given a photon state $|n\rangle$, the absorption of one photon means that the state is changed to $|n-1\rangle$. This is to point out that $|1\rangle$ is not the only state in the superposition, from which a photon can be taken.

Precise form of coupling beween light and a semiconductor can be found in textbooks: Kittel and Ashkroft&Mermim would be the first place to look; a good source specifically devoted to this subject is Electrodynamics of the semiconductor band edge by Balslev and Stahl; another good source is the reviews by Axt&Mukamel and Rossi&Kuhn. It's crude form is something like: $$ H_{int} = \sum_{k, q}M_{k,q}c_{k+q}^\dagger v_k b_q + h.c., $$ where $c_k^\dagger$ is the operator creating electron of momentum $k$ in the conduction band, $v_k$ is the operator destroying and electron in the valence band, and $b_q$ is the photon destruction operator. Quite often one would introduce hole operators as $$ h_{-k}^\dagger = v_k. $$ (For I have omitted here spin indices and polarization, although these are important for real calculations, to account correctly for the selection rules.)

The action of the photon annihilation operator on a coherent state gives us again the coherent state: $$ b_q|\alpha_q\rangle = \alpha_q|\alpha_q\rangle. $$