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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: enter image description here

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  • $\begingroup$ 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. $\endgroup$
    – Cort Ammon
    Aug 8, 2017 at 17:07
  • $\begingroup$ 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. $\endgroup$
    – MSH
    Aug 8, 2017 at 17:14
  • $\begingroup$ 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? $\endgroup$
    – Cort Ammon
    Aug 8, 2017 at 17:18
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    $\begingroup$ 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 ? $\endgroup$
    – MSH
    Aug 8, 2017 at 17:26
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    $\begingroup$ It is worth noting that the photons going to D0 are not normal photons. They are entangled photons. We give them the title "entangled" rather than just calling them plain ol' photons because they behave differently enough for physicists to give them a title. $\endgroup$
    – Cort Ammon
    Aug 8, 2017 at 18:00

2 Answers 2

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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 the 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.

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  • $\begingroup$ 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 ? $\endgroup$
    – MSH
    Aug 10, 2017 at 20:04
  • $\begingroup$ @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. $\endgroup$
    – OON
    Aug 12, 2017 at 21:20
  • $\begingroup$ @user165202 I'll probably try to add some analogy to the answer $\endgroup$
    – OON
    Aug 12, 2017 at 21:28
  • $\begingroup$ Actually, the reason for no directly visible interference patterns at any detector is the destruction of photon at BBO. Their wave function collapses completely. The superposition of two slits path becomes one slit path. From there afterwards, only one slit diffraction pattern is possible unless there is a new double slit or beam splitter-like device introduced later down the path. $\endgroup$ Jun 21, 2023 at 15:05
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@MSH it's similar to the normal double slit experiment when you watch which slit it goes through. The fact that you later erase the which-way information doesn't affect what's detected at $D_0$ by the no-signaling theorem, and a related theorem about unitary operators on just the idler photon.

The only way to get interference back is to use the coincidence detector to look at a subset of detections at $D_0$. The name "quantum eraser", and the emphasis on erasing which-way information, is unfortunate as it obscures the role of filtering, which is essential.

Sabine Hossenfelder has a good video on it, and Sean Caroll a blog post.

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  • $\begingroup$ But why do interference patterns with the idler photon at D1 and D2 cancel each other? They have to cancel each other, otherwise there would be an interference pattern visible without filtering. $\endgroup$
    – cuckoo
    Jan 8 at 2:05

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