How in detail the rotation of the electric field component of light during transition through a polarizer takes place? I ask "After a slit, are photons polarized?" and later found two answers for the question "Why does the electric field dominate in light?". From this two answers I have concluded or summarized, that photons which are influenced by sharp edges, are polarised. And the reason for this phenomenon is the interaction between the electric field component of the photon and the surface electrons of the edge.
Now I'm interested in how this happens in detail. May be any quantization during the rotation of the light's electric field takes place? Perhaps the rotation led to a deflection or dissipation of the light?
 A: Polarization rotation due to a polarizer: in the comments I said that a polarizer does not rotate the polarization (a birefringent medium does). It blocks one component, the linear component parallel to its axis. I hold this, but I thought about the following system where a polarization rotation seems to happen due to the action of many polarizers. I never thought about it before, I hope (it's true and) it can help (but sorry, no quantization here, just classical). 
Let's imagine a linear polarization at $0^o$ arriving on a series of $N$ polarizers as in this picture:

each polarizer is rotated by the same angle with respect to the previous one, so their axis spiral from horizontal up to the angle of the last one ($\theta(N)$, $=\pi/2$ in the figure, but you can choose it). Malus law says that the transmission of one polarizer is $I=I_o \cos^2(\alpha)$, where $\alpha$ is the angle between its axis and its input linear polarization. So the total transmission from the series of polarizers will be $T=(\cos^2(\theta(N)/N))^N$, where we can choose the angle of the last one ($\theta(N)$). 
In the next figure I plot $T(N)$ for $\theta(N)=0, \pi/4, \pi/2$. When $\theta(N)=0$, no matter how many polarizers are there, they'll be all parallel and $T=1$ always. For $\theta(N)=45^o$, if there is only one polarizer (i.e. the last one, $N=1$) $T=0.5$. For $\theta(N)=90^o$, if there is only the last polarizer ($N=1$), $T=0$ because it's perpendicular to the polarization. 
 
But in any case, increasing $N$ the transmission goes to 1. So, the polarization is kind of guided by the polarizers, and actually rotates with little losses if $N$ is large. I think this could be  what happens in the liquid crystal displays, where the twisted arrangement of the molecules act as the polarizers here.
This is the closest I could think when you mention rotation by a polarizer.
