Why is the quantum operator for a photodetector $\hat{E}^{(+)} \hat{E}^{(-)}$ and not just $|\hat{E}|^2$? According to this resource, the quantum operator for a photodetector is $\hat{E}^{(+)} \hat{E}^{(-)}$.
In classical physics a photodetector measures the square of the electric field: $|\hat{E}|^2$. So why is it different here? When we quantize the electric field, the E-field operator becomes $\hat{E} = \hat{E}^{(+)} + \hat{E}^{(-)}$. So I would expect we just have to plug it into our classical operator to obtain:
$|\hat{E}^{(+)} + \hat{E}^{(-)}|^2 = |\hat{E}^{(+)}|^2 + |\hat{E}^{(-)}|^2+\hat{E}^{(+)}\hat{E}^{(-)}+\hat{E}^{(-)}\hat{E}^{(+)}$
Somehow the quantum version of a photodetector is just one of these four terms. I haven't found any justification for this.
 A: Light-matter coupling is determined by the general physical principles, see minimal coupling. The expression that you are referring to is just one of many forms of this coupling between the light and the charged particles (electrons in this case) that we used in calculations. What we calculate in the end is the absorption rate, which, as we know already from the Fermi Golden rule, is proportional to the square of the matrix element, which means that the photodetector with coupling proportional to $E$ will measure $|E|^2$. $|E^{(\pm)}|$ can be ignored, since they describe the processes in which two photons are emitted or absorbed.
Update
In view of the discussion that followed in the comments, let me expand my answer.
Classical EM detection
In classical macroscopic electrodynamics detection is performed by a device which is very slow, as compared to the period of the oscillations in the EM wave. A device responding linearly to the electric field would, after averaging, produce zero result. One therefore needs to use a non-linear element - such as a (photo)diode. The simples non-linearity is quadratic: talking square of a field and averaging over a long time produces finite result:
$$
E(t)=E_0\cos(\omega t)\Rightarrow \begin{cases}
\lim_{T\rightarrow+\infty}\frac{1}{T}\int_0^Tdt E(t)\rightarrow 0,\\
\lim_{T\rightarrow+\infty}\frac{1}{T}\int_0^Tdt|E(t)|^2\rightarrow \frac{E_0^2}{2}
\end{cases}
$$
Quantum EM detection
Quantum mechanics explains what happens at microscopic level. The coupling between the electromagnetic field and the detector can be written, e.g., in the dipole approximation, as
$$
H_{int}=-\mathbf{d}\cdot\mathbf{E}
$$
In a photodetector this coupling results in electrons being excited from the valence to the conduction band and thus creating an elecytron current, which is what we (amplify and) measure. The measured current is thus proportional to the rate of electron transitions, which can be estimated using the Fermi Golden rule,a nd is proportional to the square matrix element of the interaction, and hence to the square of the electric field:
$$
\Gamma\propto |\langle f|H_{int}|i\rangle|^2
$$
Matrix element
The initial and final states in the matrix element are those with an electron in the valence band and $n$ photons, and the electron in the conduction band and $n-1$ photons:
$$
|i\rangle=|v;n\rangle,|f\rangle=|c;n-1\rangle
$$
(The momentum of photon is much smaller than that of electrons and is usually neglected, so that the momentum of electron remains unchanged - it merely moves from one band to another).
In decomposition of the electric field, $E=E^{(+)}+E^{(-)}$, $E^{(+)}$ is the operator increasing photon number, while $E^{(-)}$ is decreasing the photon number. This means that
$$
\langle n-1|E^{(+)}+E^{(-)}|n\rangle = \langle n-1|E^{(-)}|n\rangle = \langle n|E^{(+)}|n-1\rangle^*.
$$
moreover, since $E^{(\pm)}$ destroy or create only one photon, i.e.,
$$\langle m|E^{(-)}|n\rangle =0\text{ if } m\neq n-1,$$
we can write
$$
\Gamma\propto |\langle n-1|E^{(-)}|n\rangle|^2=
\langle n|E^{(+)}|n-1\rangle\langle n-1|E^{(-)}|n\rangle=\\
\sum_m \langle n|E^{(+)}|m\rangle\langle m|E^{(-)}|n\rangle =
\langle n|E^{(+)}E^{(-)}|n\rangle.
$$
That is, saying that the absorption is proportional to $E^{(+)}E^{(-)}$ is in agreement with saying that it is proportional to $|E|^2$.
A: A reason one cannot use $\hat E^2$ for the photo detection, is because $\left<\hat E^2\right>\neq 0$ for the ground state. Depending on how you look at it, it is due to either vacuum fluctuations, or the Heisenberg uncertainty principle applied to the quadratures of the field. Since “no light”  corresponds to this ground state (the vacuum), the photodetection cannot be given by  $\hat E^2$.
To go more into details, the electric field is a complex quantity, corresponding to two real quantities, the quadratures $Q$ and $P$ defined by
$$
\begin{align}
  Q&=\frac{E^+ + E^-}{2} & P & =\frac{E^+ - E^-}{2 i}.
\end{align}
$$
These are the quantities which correspond to quantum observables. Since they
do not commute, they are linked by an uncertainty relation. The energy of the field, both quantumly and classically is given by:
$$
\begin{align}
  Q^2 + P^2 &=\frac{(E^+ + E^-)^2 - (E^+ - E^-)^2}{4}\\ 
     &=\frac{E^+ E^- + E^- E^+}{2}.
\end{align}
$$
Classically, this indeed corresponds to $|E|^2=E^+E^-=E^-E^+$, but quantumly, we have $\hat E^- E^+ = E^+ E^- + c$, where $c$ is a constant given by the commmutataor of $E^+$ and $E^-$. This leads to an energy operator $E^+ E^- + \frac c2$. The photodetector operator just removes the constant $\frac c2$ — sometimes informally called the vacuum’s half photon — since a photodectero only sees the energy difference with the ground state.
