# Question about the shift of patterns in the Delayed Choice Quantum Eraser

In the double-slit Delayed Choice Quantum Eraser experiment by Scully with the installation in the figure below, as in Wikipedia, when a photon hits D4 or D3 the intricated photon hits D0 with a particle pattern as the path way was detected.

When a photon hits D1 or D2 the intricated photon hits D0 with a interferencial pattern as the path information was erased.

However You need the Coincidental counter to compare the datas to obtain those informations as depending if the Photon hits D1 or D2 The interferencial pattern is shifted by half a wave length. Consequences are that on D0 without coincidental counter it looks like a particle pattern because of the superposition of those 2 shifted patterns.

My question is: Why does this phase shift occur? What causes this shift?

To restore the signal photon (the one that hits D0) to a single-particle non-entangled state that can show useful interference, you need to project the idler photon onto a superposition state, and this is what the D1/D2 detector pair does: it measures on the superposition-state basis $$\bigg\{ |\text{D1}\rangle=\frac{|\mathrm{left}\rangle+|\mathrm{right}\rangle}{\sqrt{2}}, |\text{D2}\rangle = \frac{|\mathrm{left}\rangle-|\mathrm{right}\rangle}{\sqrt{2}} \bigg\},$$ where the difference between a D1 and a D2 detection is a change in the relative phase of the superposition. This, in turn, is caused by the change in whether the beam that's reflected at the $$\rm BS_C$$ beam splitter is reflected from the side of the air (which incurs a $$\pi$$ phase) versus from the side of the glass (which doesn't incur a reflection phase).