# Interference pattern in delayed choice quantum eraser

My question is why in the delayed choice quantum eraser experiment we don't see interference pattern at D0? While it looks like it is similar to the normal double slit experiments. The following image is from Wikipedia: • Are you asking for an explanation of why the delayed choice quantum eraser works? Explaining what you see at D0 required understanding the entire rest of the experiment. – Cort Ammon Aug 8 '17 at 17:07
• No I'm asking why there is no interference pattern at D0 regardless of the rest of experiment. At the end there is a light source, double slits and detector at D0. – user165202 Aug 8 '17 at 17:14
• What do you think would happen if there was an interference pattern and the entangled photon was detected by D3 or D4, either of which would provide path information? – Cort Ammon Aug 8 '17 at 17:18
• Actually I want the answer regardless of the entaglment photon. Why there is no interference? While it looks like the classical double slits experiment. I mean that even as per classical physic there shall be an interference. I know that I'm missing something but what is that ? – user165202 Aug 8 '17 at 17:26
• Well, the purpose of the delayed choice quantum eraser is to demonstrate counterintuitive results which are well explained by quantum mechanics but otherwise very surprising. D0's final pattern is actually easier to arrive at if you look at the joint distributions D01 D02 D03 D04 first, and the sum them to get D0 – Cort Ammon Aug 8 '17 at 17:55

In the usual double-slit experiment the photon is in the superposition of states corresponding to holes A and B. \begin{equation} |DS\rangle=\frac{1}{\sqrt{2}}|A\rangle+\frac{1}{\sqrt{2}}|B\rangle \end{equation} However in this experiment the photon then hits the type II BBO crystal which converts it into two photons with perpendicular polarizations. So the state becomes, \begin{equation} |DQE\rangle=\frac{1}{\sqrt{2}}|A,V\rangle_s\otimes|A,H\rangle_i+\frac{1}{\sqrt{2}}|B,V\rangle_s\otimes|B,H\rangle_i \end{equation} It's actually trickier because the BBO produces Bell state $\frac{1}{\sqrt{2}}|V\rangle\otimes|H\rangle+\frac{1}{\sqrt{2}}|H\rangle\otimes|V\rangle$ but in this experiment we distinguish the photons by polarization so this particular entanglement doesn't play role. Namely the prism sends, say, the one with vertical polarization ("signal") to $D_0$ and the one with horizontal polarization ("idler") to that complex optical system.

The resulting state doesn't factorize into a product of the states of a signal and idler photons $|DQE\rangle\neq|\psi\rangle_s\otimes|\psi\rangle_i$, two photons interfere as a pair but not individually. You may say that the signal photon is marked by a state of the idler photon.

In the formalism that means that we can't assign a pure quantum state (a superposition of some basis quantum states) to a sole photon. Rather if one tries to ignore the idler photon then for the signal photon one gets a so-called mixed state - the $50%$ probability of the state $|A,V\rangle_s$ and the $50%$ probability of the state $|B,V\rangle_s$ without interference of these outcomes.

How then one gets the interference pattern? We use beam splitters to convert idler photons into superpositions. \begin{equation} |A,H\rangle_i\mapsto \frac{1}{\sqrt{2}}|D_4\rangle_i+\frac{1}{\sqrt{2}}\Big(\frac{1}{\sqrt{2}}|D_1\rangle_i+\frac{1}{\sqrt{2}}|D_2\rangle_i\Big), \end{equation} \begin{equation} |B,H\rangle_i\mapsto \frac{1}{\sqrt{2}}|D_3\rangle_i+\frac{1}{\sqrt{2}}\Big(\frac{1}{\sqrt{2}}|D_1\rangle_i-\frac{1}{\sqrt{2}}|D_2\rangle_i\Big) \end{equation} Note that when the photon arrives at $D_1$ or $D_2$ it's impossible to determine the slit it came from, the state is the same. In this case we clear which-path information. And as result when we make this transformation in $|DQE\rangle$ we get, \begin{align} |DQE\rangle\mapsto\frac{1}{2}|A,V\rangle_s\otimes|D_4\rangle_i+\frac{1}{2}|A,V\rangle_s\otimes|D_3\rangle_i \\ +\frac{1}{2}\left(\frac{1}{\sqrt{2}}|A,V\rangle_s+\frac{1}{\sqrt{2}}|B,V\rangle_s\right)\otimes|D_1\rangle_i\\ + \frac{1}{2}\left(\frac{1}{\sqrt{2}}|A,V\rangle_s-\frac{1}{\sqrt{2}}|B,V\rangle_s\right)\otimes|D_2\rangle_i \end{align} you may see that if you measure the idler photon either at $D_1$ or $D_2$ you separate the term with the signal photon now being in superposition (note that the phase shifts are chosen to obtain two different superpositions)

Of course to separate it you need to ignore the events when the idler photon hits $D_3$ or $D_4$. If you sum the statistics at $D_0$ for all four detectors you will recover the initial pattern as if you ignored the idler photon altogether.

• Thank you Can I emphasise your answer and ask again why in the setup thst we have BBO we don't have interference while in the normal double slits setup we have? Is it that in the first one the photon lose its superposition? Can we also explain it classically that the photons from the double slits will be for example phase shifted ? – user165202 Aug 10 '17 at 20:04
• @user165202 No, you can't describe it by phase shifted classical waves. I'll try to explain the main point again. In the usual double slit setup you have one photon which is described by some wavefunction. But when you put BBO the wavefunction describes both photons. You can't assign any wavefunction to the individual photon. – OON Aug 12 '17 at 21:20
• @user165202 I'll probably try to add some analogy to the answer – OON Aug 12 '17 at 21:28