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Suppose you have a non-polarizing beamsplitter (NPBS) with 2 outputs: A and B. You place a polarizing beamsplitter (PBS) at the output A, and want to follow the evolution of the quantum state after this step. (Let's say it's just the first of many transformations before a detector.)

Let's say the state after the NPBS is: $$ \frac{1}{2}(|H_{A}\rangle + i|H_{B}\rangle + |V_{A}\rangle + i|V_{B}\rangle) $$

The state after the PBS is: $$ (|H_{A}\rangle\langle H_{A}| - |V_{A}\rangle\langle V_{A}|)\frac{1}{2}(|H_{A}\rangle + i|H_{B}\rangle + |V_{A}\rangle + i|V_{B}\rangle) $$

My question is how to treat the terms $ i|H_{B}\rangle $ and $ i|V_{B}\rangle $ during the calculation.

Can the state be rewritten as the following? $$ (|H_{A}\rangle\langle H_{A}|\otimes 1_{B} - |V_{A}\rangle\langle V_{A}|\otimes 1_{B})\frac{1}{2}(|H_{A}\rangle + i|H_{B}\rangle + |V_{A}\rangle + i|V_{B}\rangle) $$

Unfortunately, this would in practice remove all the "B" terms from the description of the state. For example, if we look at what would happen to the $ i|H_{B}\rangle $ part: $$ (|H_{A}\rangle\langle H_{A}|\otimes 1_{B} - |V_{A}\rangle\langle V_{A}|\otimes 1_{B})i|H_{B}\rangle = i|H_{B}\rangle - i|H_{B}\rangle = 0 $$

This doesn't seem right. So, how do you rewrite the state after the PBS in terms of both A and B?

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  • $\begingroup$ a better way to do this is to apply the unitary operation corresponding to the BS on the $A$ mode, and then if you want to see what happens when you don't look at the $B$ mode you do the partial trace $\endgroup$ – glS Jul 3 at 7:57
  • $\begingroup$ @glS Yes, but what if I'm interested in the full description of the state: both A and B modes? $\endgroup$ – triclope Jul 3 at 14:32
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    $\begingroup$ ah, I see what you mean. Where did you take that expression for the PBS though? That certainly doesn't look correct. If $A$ and $B$ denote the spatial modes, the PBS should change the spatial mode conditionally on the polarisation, which is not what the operator you wrote (I mean $|H_A\rangle\langle H_A|-\lvert V_A\rangle\langle V_A|$) does. $\endgroup$ – glS Jul 3 at 17:08
  • $\begingroup$ @glS What is the correct operator for a PBS? $\endgroup$ – triclope Jul 3 at 19:54
  • $\begingroup$ If there are two spatial modes $A$ and $B$, a PBS among these two modes would be something like $\newcommand{\ketbra}[2]{\lvert #1\rangle\!\langle #2\rvert}\ketbra{H}{H}\otimes I+\ketbra{V}{V}\otimes(\ketbra{A}{B}+\ketbra{B}{A})$, which changes the spatial mode conditionally to the polarisation mode. However, I am a bit confused by your description. You say that the PBS is applied to the first output $A$. Doesn't this mean that there are then three output modes? Two output of the PBS and the third the other output of the BS? $\endgroup$ – glS Jul 3 at 21:28
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Because the PBS operates only on the first output, it implements the transformation $$ \begin{cases} |H,A\rangle&\to&|H,A_1\rangle,\\ |V,A\rangle&\to&i|V,A_2\rangle,\\ |H,B\rangle&\to&|H,B\rangle, \\ |V,B\rangle&\to&|V,B\rangle, \end{cases}$$ that is, it leaves unchanged photons on the spatial mode $B$, and operates as a CNOT between spatial mode and polarisation on the spatial mode $A$.

The output state is therefore $$\frac{1}{2}\left( |H,A_1\rangle + i|V,A_2\rangle + i|+,B\rangle \right),$$ where $|+\rangle\equiv\frac{1}{\sqrt2}(|0\rangle+|1\rangle)$.

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