Description of quantum state of entangled photons after polarizer I'm wondering if anyone can help me understand how a polarizer changes the quantum state of two polarization-entangled photons. I haven't found a clear description in the literature.
Suppose you have two polarization-entangled photons A and B in the following state:
\begin{equation}
\Phi=\frac{1}{\sqrt{2}}\bigl(\left|H_{A},H_{B}\right\rangle + \left| V_{A},V_{B}\right\rangle\bigr)
\end{equation}
Suppose the photon A passes through a polarizer oriented at the +45 angle. 
Does that convert $ H_{A} $ in the equation above to $ \frac{1}{\sqrt{2}}\bigl(\left|+45\right\rangle\bigr) $, which is $ \frac{1}{2}\bigl(\left|H_{A} + V_{A}\right\rangle\bigr) $, and $ V_{A} $ also to $ \frac{1}{2}\bigl(\left|H_{A} + V_{A}\right\rangle\bigr) $? I used $ \frac{1}{\sqrt{2}} $ because only a half of the inputlight passes through the polarizer.
Is, therefore, the resulting state the following, or did I misunderstand it completely?
\begin{equation}
\Phi=\frac{1}{\sqrt{2}}\bigl(
\frac{1}{2}\bigl(\left|H_{A} + V_{A}\right\rangle\bigr) \bigotimes \left|H_{B}\right\rangle 
+ \frac{1}{2}\bigl(\left|H_{A} + V_{A}\right\rangle\bigr) \bigotimes \left| V_{B}\right\rangle\bigr)
\end{equation}
\begin{equation}
\Phi=\frac{1}{2\sqrt{2}}\bigl(
\left|H_{A},H_{B}\right\rangle
+ \left|V_{A},H_{B}\right\rangle
+ \left|H_{A},V_{B}\right\rangle
+ \left|V_{A},V_{B}\right\rangle
\bigr)
\end{equation}
 A: Your final result is correct up to a normalization. Another way to see this is that a polarizer at  a $45^\circ$ angle would simply be a projector onto the $|+_A\rangle \equiv \frac1{\sqrt2}(|H_A\rangle+|V_A\rangle)$ state for the $A$ photon. i.e. the state of the photon after going through it is simply $|+_A \rangle \langle+_A| \times |\text{initial state} \rangle$ in the Hilbert space of photon $A$. So the action of putting only photon $A$ through the $45^\circ$ polarizer on the joint Hilbert space of both photons is simply described by the transformation:
$$|+_A\rangle\langle+_A|\otimes \mathbb 1_B$$ 
Now simply act this transformation on your initial state to get the final state:
$$|\psi\rangle = \Big(|+_A\rangle\langle+_A|\otimes \mathbb 1_B \Big)\frac1{\sqrt 2} \Big(|H_A H_B \rangle + |V_A V_B \rangle \Big)$$
$$=\frac1{\sqrt 2}|+_A \rangle \langle +_A|H_A \rangle \otimes 1|H_B \rangle+\frac1{\sqrt 2}|+_A \rangle \langle +_A|V_A \rangle \otimes 1|V_B \rangle$$
Using our definition of $|+_A \rangle$ gives:
$$|\psi \rangle=\frac12|+_A \rangle \otimes |H_B \rangle + \frac12|+_A \rangle \otimes |V_B \rangle$$
$$= \frac1{2\sqrt 2}|H_A H_B \rangle + \frac1{2\sqrt 2}|V_A H_B \rangle + \frac1{2\sqrt 2}|H_A V_B \rangle+\frac1{2\sqrt 2}|V_A V_B \rangle$$
Which is identical to your result. However, note that the projector $|+_A \rangle \langle +_A|$ doesn't keep the norm of the initial state vector, it in fact multiplies its norm by $\langle +_A | \psi_0 \rangle$. So you do need to normalize the resulting state at the end:
$$|\psi \rangle = \frac1{2}\Big[ |H_A H_B \rangle + |V_A H_B \rangle + |H_A V_B \rangle+|V_A V_B \rangle \Big]$$
EDIT:
As flippiefanus points out in his comment, the above state is nothing but:
$$|\psi \rangle = |+_A \rangle \otimes |+_B \rangle$$
So the initial entanglement of the two photons puts the second photon in the $|+ \rangle = \frac1{\sqrt 2}(|H \rangle + |V \rangle)$ state as well, even though the polarizer only acted on the first photon.
