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In the quantum eraser experiment, why is it necessary to have both D1 and D2? Why not do away with D2 altogether and, as with the standard Mach–Zehnder interferometer, make the photons from BSA and BSB then interfere with each other or otherwise intersect so that they both end up at D1? As I understand it, this should still erase the which-way information.

I already know that sending information back in time with the quantum eraser is considered impossible because the interference patterns from D1 and D2 complement each other. But if it is indeed possible to eliminate D2, how does the resulting pattern look on a single which-path-erasing detector, and if a single interference pattern is formed, how is causality preserved?

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So long as your experiment doesn't 'leak' photons, then removing only one of the detectors will not cause any effect, as its information can be reconstructed from the rest of the counts.

In other words, if you assume that the detecting efficiencies of $D_0$, $D_1$, $D_2$, $D_3$ and $D_4$ are perfect, and that photons have no way of getting absorbed by the optics or otherwise lost, if you remove (say) $D_1$ and you observe a count at $D_0$ but you don't see a count at $D_2$, $D_3$ or $D_4$, then you can conclude that the other member of the pair would have been detected by $D_1$ had the detector been present. And, in that case, you will see an interference pattern appear in the $x$ dependence of $D_0$ counts for the counts where $D_2$ fired and for the counts where none of the left-hand side detectors fired.

That assumption is, of course, imperfect, and real experiments will have some degree of loss, which will act to the detriment of the contrast in the interference observed in the "missing-count" pattern.

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