 Solution

For heralded Bell-pair projection with two polarization parity filters, a Hadamard gate followed by computational-basis detection is a measurement in the diagonal polarization basis $|\pm ⟩=(|H⟩\pm |V⟩)/\sqrt{2}$. It can therefore be realized directly by a polarizing beam splitter at $45^{\circ }$ with a photodetector in each output: its transmitted and reflected clicks label $+$ and $-$. No separate Hadamard gate plate is required for this measurement.

Conditioning on one photon in each output of the first polarizing beam splitter implements the even $Z$-parity filterThe $45^{\circ }$ polarizing beam splitter similarly implements the even $X$-parity filter ${\Pi }_X^{46}=(I+X_4{X}_6)/2$. These are ideal coherent filters, with the optical phases chosen consistently. The all-zero photodetector record projects the upper outputs onto $⟨++{|}_{25}$ and the lower outputs onto $⟨00{|}_{46}$. The two effective bras areTherefore the unnormalized retained state isFor example, the simultaneous Bell contractions require the three pair bits to coincide, leaving only the $000$ and $111$ terms of the original pair expansion. Each surviving $|00{⟩}_{13}$ or $|11{⟩}_{13}$ amplitude is $1/(8\sqrt{2})$. ConsequentlyThis probability is conditioned on production of all three input pairs, but not yet on successful polarizing beam splitter routing. The routing probability itself is $⟨{\Pi }_Z^{25}{\Pi }_X^{46}⟩=1/4$: each nonidentity expectation vanishes on the relevant initial reduced state. Thus, if the question instead regards the routing postselection as already performed,These two denominators describe different conditioning. Neither includes the probability that the three crystals actually emit the required pairs, photodetector inefficiency or undesired multipair events.