How is intensity defined for quantized EM fields? Classically intensity is defined as 
$$ I \equiv  \frac{1}{2} c \epsilon_0 E^2, $$
but when you perform a second quantization this definition becomes a bit ambiguous since the $E^2$ could be interpreted as $\vert E^2 \vert $ or as $ \vert E \vert^2$.
So my question is simply: Which is these definitions is correct?
The relevance becomes clear when you consider that $ \vert E \vert^2=0$ for Fock states (including the vacuum state) while $\vert E^2 \vert $ is not.
See e.g. wikipedia for an example of this definition:
https://en.wikipedia.org/wiki/Intensity_(physics)
 A: Classical you work the intensity of a wave out by integrating the energy arriving over a known area in the course of one cycle, and then divide by the period and the area. The form you exhibit is correct for harmonic waves in SI units if the $E$ is the maximum amplitude of the electric field. But for a general normally-incident plane-wave  you are looking for 
$$\begin{align*}
I &\equiv \frac{1}{T} \int_{t=0}^T u(\mathbf{x},t) \,\mathrm{d}t \\
&= \frac{1}{T} \int_{t=0}^T \epsilon_0 \left[ E(\mathbf{x},t) \right]^2 \,\mathrm{d}t \,,
\end{align*}$$ 
where I've written $u$ for the energy density of the electromagnetic fields and $T$ for the period of the wave. I've also taken $\mathbf{x}$ as constant and left out a uninteresting factor of $A/A$. (Of course you can do the integral in space if you prefer.) 
To recover the expression you exhibited, just write
$$ E(\mathbf{x},t) = E \sin \left(\mathbf{k}\cdot\mathbf{x} - \omega t \right) \,,$$
and evaluate the integral.
Anyway, the kernel of the integral is your guide for how to interpret the $E^2$ in performing the second quantization. You want the average of the square (which is not zero) rather than the square of the average (which is zero).
