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I'm trying to understand how polarizers affect bunched photons. Or, more generally, how a projection operator affects a two-photon state corresponding to photon bunching.

Toy example: Imagine you have two photons distinguishable by their orthogonal polarizations H and V. They enter a balanced non-polarizing beamsplitter each through a different input port. As they are distinguishable, I believe there is some probability for them to exit via separate output ports vs. bunched together via the same output port (subindices A and B correspond to two output ports):

$ |\Psi\rangle = \frac{1}{2}(|H_{A}\rangle|V_{B}\rangle + i|H_{A}\rangle|V_{A}\rangle + i|H_{B}\rangle|V_{B}\rangle - |V_{A}\rangle|H_{B}\rangle) $

What happens to this state if you put a horizontally oriented polarizer into the output port A?

Especially, how does it affect the component $ i|H_{A}\rangle|V_{A}\rangle $ (two photons exiting from the same output port, but having orthogonal polarizations)? I can see three alternatives, but I don't know which one is correct:

  • $ (|H_{A}\rangle\langle H_{A}|)(i|H_{A}\rangle|V_{A}\rangle) = i|H_{A}\rangle|V_{A}\rangle $
  • $ (|H_{A}\rangle\langle H_{A}|)(i|H_{A}\rangle|V_{A}\rangle) = i|H_{A}\rangle $
  • $ (|H_{A}\rangle\langle H_{A}|)(i|H_{A}\rangle|V_{A}\rangle) = 0 $

Logically, I would expect the #2 to be true, but it't be great to see someone's step-by-step calculations leading to this (or different) result.

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  • $\begingroup$ Look at the 2nd answer in the post below for some great intuition. I think photon antibunching is more of a statistical phenomenon, i.e. many photons are required and that your question with single photos may not work with the anti bunching theory. physics.stackexchange.com/questions/257581/… $\endgroup$ Jun 1, 2019 at 2:09

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First let me clear up some possible confusion. The situation that you are describing is the Hong-Ou-Mandel effect with distinguishable photons. Although the term photon-bunching is often associated with it, one should perhaps add that here the bunching is associated with output ports and not bunching in time, for instance.

Now to address your question. The polarizer causes a loss of photons, so one needs to implement a loss mechanism, which means that we need to trace out the lost degree of freedom. You seemed to have tried it, but one needs to do it with the entire state and not just one factor of it.

Focussing of the V polarization at port A, one can write the state before the polarizer in terms of two parts

$$ |\Psi\rangle = \tfrac{1}{2}(|H_{A}\rangle|V_{B}\rangle + i|H_{B}\rangle|V_{B}\rangle) + \tfrac{1}{2}(i|H_{A}\rangle - |H_{B}\rangle)|V_{A}\rangle $$

To trace out the lost photon we need to do

$$\langle {\rm vac}|\Psi\rangle \langle\Psi|{\rm vac}\rangle + \langle V_{A} |\Psi\rangle \langle\Psi|V_{A}\rangle $$

$$= \tfrac{1}{4}(|H_{A}\rangle|V_{B}\rangle + i|H_{B}\rangle|V_{B}\rangle) (\langle H_{A}| \langle V_{B}| - i\langle H_{B}| \langle V_{B}|) + \tfrac{1}{4} (i|H_{A}\rangle - |H_{B}\rangle)(-i\langle H_{A}| - \langle H_{B}|)$$

Hope it helps.

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  • $\begingroup$ Practically speaking, how do you calculate $ i|H_{B}\rangle|V_{B}\rangle\langle H_{A}|\langle V_{B}| $? Is it $ i\langle V_{B}|H_{B}\rangle\langle V_{B}|V_{B}\rangle\langle H_{A}| $? Sorry, I'm only used to more "symmetric" terms (ABAB and not BBAB, if it makes sense). $\endgroup$
    – triclope
    Jun 3, 2019 at 14:50
  • $\begingroup$ The order of the bras and kets are important. One is not supposed to flip them around, unless one performs a trace. The way I expressed the last expression implies a density operator that is mixed. One can multiply out all the terms, but none should form an inner product with another. $\endgroup$ Jun 4, 2019 at 4:21

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