How are the field operator and quantum state after a beam splitter and a polarizing beam splitter individually? How are the field operator $\hat{a}$, $\hat{a}^\dagger$ and the quantum state (like coherent state $|\alpha>$, Fock state $|n>$) changed after a beam splitter and a polarizing beam splitter individually?
for example, two coherent states $|\alpha>$ and $|\beta>$ come into a beam splitter. $\hat{a}_{out1}=r\hat{a}_{\alpha}+it\hat{a}_{\beta}$, and $\hat{a}_{out2}=it\hat{a}_{\alpha}+r\hat{a}_{\beta}$, where $r^2+t^2=1$, and the output states are two coherent states with $\hat{a}_{out1}$ and $\hat{a}_{out2}$ operators separately. Is this correct?
But for a polarizing BS, a $\hat{a}=\cos{\frac{\theta}{2}}a_H+e^{i\phi}\sin{\frac{\theta}{2}}a_V$ comes into a HV PBS, what's the output? $\hat{a}_{out1}=\cos{\frac{\theta}{2}}a_H$ and $\hat{a}_{out2}=e^{i\phi}\sin{\frac{\theta}{2}}a_V$? If this is correct, $\hat{a}_{out1}$ and $\hat{a}_{out2}$ don't satisfy the commutation because of the sin and cos, which is $[\hat{a}_{out1},\hat{a}_{out1}^\dagger]=1$ and $[\hat{a}_{out2},\hat{a}_{out2}^\dagger]=1$. So is this wrong?
 A: I think one reason why confusion arises in trying to model a polarizing beam splitter (PBS) as quantum-optical device is because of the term beam splitter. However, functionally speaking, a PBS is a polarizer and is therefore not a reversible device. A single photon in the state $\alpha |H\rangle + \beta |V\rangle$ impinging upon a polarizer (say aligned along the horizontal polarization) will get transmitted with an $|\alpha|^2$ probability or absorbed with a $|\beta|^2$ probabilty. This clearly means the photon number is not conserved, or that the transformation by a polarizer cannot be unitary.
Now of course, an ideal (and lossless) PBS does not absorb the vertically-polarized
photon. In the above example, it would be available in another mode; say, at the
reflection port (if the horizontally polarized photons were obtained at the transmission
port).
One has then two choices: either neglect this other mode (which would then be exactly
like a polarizer) if it is not of interest, or consider it independently. I
emphasize on the independence because this could not be done for a normal beam splitter.
(This is also the reason why you cannot obtain interference in a PBS).
The mode relations in the Heisenberg picture (obtained by substituting $\eta = 1$ in equation 5.13 of this article on quantum physics of simple optical instruments) for an ideal PBS could be written as:
$$
\hat{c}_H = \hat{b}_H
$$
$$
\hat{c}_V = \hat{a}_V
$$
$$
\hat{d}_H = \hat{a}_H
$$
$$
\hat{d}_V = \hat{b}_V
$$
with $a$ and $b$ denoting the input modes, and $c$ and $d$ denoting the output modes, as also illustrated in this figure.

Note that $\eta = 1$ essentially stood for $\,\eta_H = \eta_V = 1$ so one could model a realistic PBS using $\eta_H < 1$ and $\eta_V < 1$ in equation 5.13 and derive the mode relations accordingly.
Given an arbitrarily polarized single photon prepared in mode $a$ (state $\alpha |H_a\rangle + \beta |V_a\rangle$), the evolution of this state through the ideal PBS would be described by:
$\alpha |H_a\rangle + \beta |V_a\rangle =  (\alpha a_H^{\dagger} + \beta a_V^{\dagger}) |0\rangle \rightarrow (\alpha d_H^{\dagger} + \beta c_V^{\dagger})|0\rangle = \alpha |H_d\rangle + \beta |V_c\rangle$
(I haven't used the hats on the operators to prevent cluttering).
You can see that the photon is a superposition of two different spatial modes $c$ and $d$ now.
A: Of course $\hat{a}_{out1}$ and $\hat{a}_{out2}$ are O.K. We have the commutation rules 
(I) $ \ [a_H, a^{\dagger}_H] = 1, \ \ \ [a_V, a^{\dagger}_V] = 1$, and $[a_H, a^{\dagger}_V] = 0. $ 
These are the commutation rules that the outputs have to satisfy, and they do satisfy.
What I think that confuses you is that you treated wrongly the beam-splitter (BS). You'd better consider a single beam landing on it, and see what happens. Otherwise you cannot compare with what happened at the polarization beam-splitter (PBS), because there you also considered a single input beam.
The output of the BS for a single beam is
(II) $ \ \hat a _{\alpha} \to t\hat a _{\alpha ,1} + ir \hat a _{\alpha ,2}$,
with $t$ is the transmission amplitude, real, and $ir$ the reflection amplitude, imaginary, and $r^2 + t^2 = 1$. I shortened "out1" and "out2" into 1 and 2.
The two outputs of the beam, although spacely separated, remain a coherent quantum superposition; if you bring them on a new BS, let's call it BS', and see that the path-lengths to BS' be equal except for a phase-shift of $\pi$ on the path of the reflected wave, they will emerge through one single output. 
(III) $ \ t\hat a_{\alpha ,1} - ir \hat a_{\alpha ,2} \to t(t \hat a_{\alpha ,1 \to 2'} + ir \hat a_{\alpha ,1 \to 1'}) - ir(t\hat a_{\alpha ,2 \to 1'} + ir\hat a_{\alpha , 2 \to 2'})$
$ = (t^2 + r^2) \hat a_{\alpha ,2'} = \hat a_{\alpha ,2'} . $
The same with the PBS. The two outputs satisfy the commutation rules (1). As to the transformation at the PBS, it preserves the coherent superposition, as the BS,
(IV) $ \ \hat{a}=\cos{\frac{\theta}{2}}a_H + e^{i\phi}\sin{\frac{\theta}{2}}a_V .$
If you will input the two branches, now spacely separated, into the two input faces of a PBS', and take care of the path-lengths to PBS' and (if necessary) phase-shifts, the two components will exit through one single output face.
(Maybe my statement with exiting through a single output face of PBS', puzzles you. But, the transformation at the PBS is reversible as is the transformation at BS. The PBS' reverses the action of the PBS, i.e. reunites the two beams into one, as did BS'.) 
