Hong Ou Mandel interference and bell basis measurment

Question

It is well known that using Hong Ou Mandel interference in polarization one can only detect 2 out of the 4 bell states($|\psi^+\rangle$ and $|\psi^-\rangle$ can be detected but $|\phi^+\rangle$ and $|\phi^-\rangle$ can not). A general setup for this would look something like

where one utilizes polarization qubits. Both channels A and B possess one photon each and can be entangled in order to create the bell states, such as $|\phi^+\rangle = \frac{1}{2}(|HH\rangle + |VV\rangle)$. Such a state can be created/analysed using the interference created due to beam splitter. For more information you can take a look here and here.

Now let's say one has the state(assume that it is normalized) $$|\text{state}\rangle = a|\psi^+\rangle + b|\psi^-\rangle + c|\phi^+\rangle + d|\phi^-\rangle$$ What is the probability that upon measurement of the state we get $|\psi^+\rangle$ or $|\psi^-\rangle$ as the output? My confusion stems from the fact that we can not measure $|\phi^+\rangle$ and $|\phi^-\rangle$ states with the given setup and hence, the probability of measuring those is 0, how does this impact the probability of the measurable bell states.

Just to let you know, this is not a part of some assignment but relates to my older question where they perform conditional measurement on bell states. The conditional measurement part in the hyperlinked question is achieved through hong ou mandel interference in a different paper, so I am primarily trying to link the conditional measurement to the experimental setup.

Answer 1

The beam splitter acts as $a_H\to (a_H+b_H)/\sqrt{2}$, $a_V\to (a_V+b_V)/\sqrt{2}$, $b_H\to (a_H-b_H)/\sqrt{2}$, $b_V\to (a_V-b_V)/\sqrt{2}$ if we used the balanced beam splitter convention. Afterward, the polarizing beam splitters split each polarization spatially, so that a state like $(a_H b_V)^\dagger |\mathrm{vac}\rangle$ will register a click at D1 and D4, $(a_H b_H)^\dagger |\mathrm{vac}\rangle$ will register a click at D1 and D3, and so on. Using this on various initial states allows us to find some differences.

Namely, take each of the Bell states, propagate it through the first beam splitter, and see which detectors will click. The $\phi^\pm$ states transform as $$|\phi^\pm\rangle=\frac{a_H^\dagger b_H^\dagger\pm a_V^\dagger b_V^\dagger}{\sqrt{2}}|\mathrm{vac}\rangle\to \frac{a_H^{\dagger 2}-b_H^{\dagger 2}\pm a_V^{\dagger 2}\mp b_V^{\dagger 2}}{2\sqrt{2}}|\mathrm{vac}\rangle.$$ Just from looking at the click patterns, it is impossible to distinguish between these two states! They each will register two photons at any of the four detectors with equal probability (the probability amplitudes differ, but the probabilities themselves do not); for example, the probability of detecting two photons at D2 is $||\pm\frac{a_V^{\dagger 2}}{2\sqrt{2}}|\mathrm{vac}\rangle||^2=1/4$.

This is in contrast to the $|\psi^\pm\rangle=\frac{a_H^\dagger b_V^\dagger\pm a_V^\dagger b_H^\dagger}{\sqrt{2}}|\mathrm{vac}\rangle$ states, which transform as $$|\psi^+\rangle\to \frac{a_H^{\dagger}a_V^\dagger-b_H^{\dagger}b_V^\dagger}{\sqrt{2}}|\mathrm{vac}\rangle$$ and $$|\psi^-\rangle\to \frac{-a_H^{\dagger}b_V^\dagger+a_V^{\dagger}b_H^\dagger}{\sqrt{2}}|\mathrm{vac}\rangle.$$ Thus, for $\psi^+$, either detectors D1 and D2 will both register a photon or detectors D3 and D4 will both register a photon, while for $\psi^-$ either detectors D1 and D4 will both register a photon or detectors D2 and D3 will both register a photon.

If you take all of these results together, it means that any probability that your state starts in a particular Bell state translates into probabilities that a certain pattern of detectors will register photons. The two Bell states that "cannot be measured" just lead to a detection pattern that cannot be distinguished from each other, but these states can indeed be distinguished from the other two that can be measured. In other words, the measurement gives three outcomes: state is $\psi^+$, state is $\psi^-$, or state is either of $\phi^\pm$.

An extra layer on top of these results is that some detectors can't distinguish between different numbers of photons arriving, and photons may be lost along the way, so sometimes people only consider events in which two different detectors register the incidence of a photon. In that case, only results $\psi^+$ and $\psi^-$ are possible outcomes. The other input states all do not lead to such successful events and are discarded - that's where the remaining probability goes.