# 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?

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.