Wave function collapse of photon state transformed by beamsplitter $\newcommand{\ket}[1]{\left|#1\right>}$I am trying to make sense of the math behind transforming the state of a photon with a beam splitter.
Suppose I have a beamsplitter crystal which split by photon into two paths, $\ket{U}$ and $\ket{D}$, say up and down. By example horizontal polarization would be sent to path $U$ and horizontal polarization to $D$.
In term of transformation on a single photon this gives the transformation $T$:
$$\ket{0} \rightarrow \ket{U}$$
$$\ket{1} \rightarrow \ket{D}$$
Where $\ket{0}$ is my horizontal polarization component and $\ket{1}$, my vertical polarization component.
Suppose I sent the following diagonally polarized photon through my crystal.
$$\ket\psi = \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1}$$
Applying the transformation I get
$$T\ket\psi = \frac{1}{\sqrt{2}} \ket{U} + \frac{1}{\sqrt{2}} \ket{D}$$
Which basically means my photon is half up, half down.
Now, on the down path, I will put a detector and on the up path I will apply the inverse transformation $T^{-1}$, say by using another beamsplitter crystal.
If I put a detector on the down path, this should make my photon collapse to either $\ket{U}$ or $\ket{D}$. So if the photon collapses to state $\ket{D}$ and I apply my inverse transformation, I get
$$ T^{-1}\ket{D} = \ket{1}$$
Is this correct? Do I actually get a vertically polarized photon one time out of two because I detected the photon on the $D$ path? Or is my intepretation wrong and a photon passing through $T$ and $T^{-1}$ will comeback out as $\ket\psi$, but only one time out of two?
Which is the correct interpretation?
 A: 
Which basically means my photon is half up, half down.

Ah, let's be careful with our language here. It's an equal amplitude superposition of up and down. Better to be precise.
Now, we also want to be precise about what $T$ and $T^{-1}$ do. $T$ takes a photon's polarization information and turns it into path information (upper path and lower path, I guess?). So a full-on $T^{-1}$ would be something that turns the path information into polarization information. I guess you can rejoin the paths and have a particular beamsplitter and if you set it up correctly it would then indeed be a proper inverse of $T$, turning path information into polarization information.
So, what will you get? Since you have a detector, you either measure the photon down there, in which case it's detected and gone and all you get is a beep in the detector. Or you don't measure the photon down there. In this case, you know that it's in $|U\rangle$. That means the wavefunction collapses and we're in state 
$|U\rangle$ and not in a superposition anymore. Putting that state through the inverse transform will then give you $|0\rangle$. 
In summary, half of the time you find a photon down at the detector, and half of the time a photon in state $|0\rangle$ will show up at the final beam splitter.
