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First off - I am not an expert. I am intrigued by quantum physics and have a layman's understanding at best. I've looked at other questions, but either the answers are not answering my question, or I've failed to understand the answer (entirely possible). If the latter, then if I could trouble you for a simple explanation, I would appreciate it.

I was reading this article. In the "Quantum Eraser" section at the bottom, there is the following text:

Figure 6.5 shows a Bell-state quantum eraser, named after John Bell. It illustrates the application of the following steps:

  1. a laser fires photons into a Beta Barium Borate (BBO) crystal;
  2. the crystal entangles some of the photons; and then
  3. entangled photons travel to two different detectors: A and B.

Placed between the crystal and detector B is a double-slit, like in the previous experiments. Immediately in front of detector A is a polarizing filter that can be rotated. Figure 4.5 showed an experiment using sunglasses to see the effects of rotating a polarizer. Those same effects apply here.

Bell-state Quantum Eraser

The Bell-state quantum eraser has one more feature: each slit is covered by a substance that filters the polarization of a photon. Consequently, the left-hand slit will receive photons with a counter-clockwise polarization, and the right-hand slit will pass photons with a clockwise polarization.

Note: Polarization does not affect interference patterns.

Initially, neither detector shows an interference pattern. Since we control the polarization of photons passing through the slits and we know the polarization accepted by each slit, we can deduce which way the photons travelled (counter-clockwise through the left; clockwise through the right). Thus no interference patterns are detected.

However, if we rotate the polarizing filter in front of detector A so that the polarizations of the photons that hit the detector are the same (that is, we can no longer distinguish between clockwise and counter-clockwise polarizations), then the interference pattern appears at both detectors!

How do the photons arriving at detector B know that the polarizations have been "erased" at detector A?

I find this result fascinating, and disturbing. By rotating the filter at A, I can change the interference pattern at B. In fact, if I was in charge of the filter at A and a friend was watching detector B, they could tell when I rotated the filter right? If so, then isn't this communication?

If A and B are spatially separated by many light years, wouldn't this even be FTL communication?

Is there something wrong with the article I linked? Perhaps I am misunderstanding it, but it seems pretty clear to me.

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  • $\begingroup$ Related post by OP: physics.stackexchange.com/questions/223392/… $\endgroup$ – Norbert Schuch Dec 11 '15 at 13:49
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    $\begingroup$ Did you read the wikipedia article, especially the section "Does delayed choice violate causality", en.wikipedia.org/wiki/… $\endgroup$ – Norbert Schuch Dec 11 '15 at 13:50
  • $\begingroup$ @NorbertSchuch Yes. However, I don't understand how that is applicable in this test. By looking at detector B, you can either see an interference pattern, or not. There are no "idler photons" or "subsets of signal photons" in this experiment. $\endgroup$ – Trenin Dec 11 '15 at 14:47
  • $\begingroup$ @NorbertSchuch Yes - I posted these both. Answers to my first one seemed to be concerned with the details of the experiment. I thought it easier to start with a new post which clarified an experiment rather than try to put all this in the old post. $\endgroup$ – Trenin Dec 11 '15 at 14:49
  • $\begingroup$ Nothing which is done in front of detector B changes the interference pattern at Detector A. And yes, you are right: if this werent the case you would easily have FTL communication. $\endgroup$ – lalala Feb 25 '19 at 10:52
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Let's denote the photon going through the left slit by $\vert L\rangle_B$, and the one going through the right slit by $\vert R\rangle_B$. The photon pair is prepared in the entangled state $$ \vert L\rangle_A\vert L \rangle_B + \vert R \rangle_A\vert R\rangle_B\ . $$

If we now measure the $A$ photon in the $\vert L\rangle$, $\vert R\rangle$ basis, we obtain the which-way information and no interference pattern is observed.

Let us now consider the case where we erase the which-way information, i.e., measure in the basis $\vert L\rangle \pm \vert R\rangle$. Let us first consider the $+$ outcome. Then, the $B$ photon is in the state $\vert L\rangle + \vert R\rangle$ and the two paths will interfere. This is, the photon will be observed at a position $x$ with a probability $p_+(x)$ which describes an interference pattern.

Let us now consider the $-$ outcome. Then, the photons are in the state $\vert L\rangle-\vert R\rangle$, i.e., as compared to the $+$ outcome, the photon going through the right slit will experience a phase shift of $\pi$. This phase shift will lead to a distribution $p_-(x)$ of photons on the screen which is shifted by half a spacing with respect to $p_+(x)$ (i.e., dark and light are interchanged).

Now in order to observe the interference pattern, $B$ will have to detect many (or at least more than one) photon. However, for each photon, the outcomes $+$ and $-$ will be distributed randomly: the overall pattern observed by $B$ will thus be distributed according to $q(x)=(p_+(x)+p_-(x))/2$. Since the two distributions are shifted by half a spacing, the distribution $q(x)$ observed by $B$ will show no interference pattern, just as if $A$ had measured in the which-way basis.

(Note that if $B$ records the position for each photon and $A$ and $B$ later compare their data, they will indeed find that the photons with a $+$ outcome and the photons with a $-$ outcome individually did form an interference pattern; however, since the outcomes were random $B$ had no way of identifying these patterns prior to learning the outcomes.)

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  • $\begingroup$ So are you saying that the experiment in the linked article is wrong? i.e. the detector at B will never show an interference pattern? $\endgroup$ – Trenin Dec 11 '15 at 18:56
  • $\begingroup$ What if you let 1000 photons through, rotate the polarizer then let another 1000 through. Can you detect the interference pattern turning on and off as an aggregate of many results? $\endgroup$ – Quantumplate Dec 11 '15 at 19:07
  • $\begingroup$ Found another article about bell-state quantum erasure. Erasing the which-way information at one detector causes the interference pattern to reappear at the other detector. Even if the erasure happens after the photons are detected (i.e. delayed quantum erasure). This means that if I were responsible for the polarizer at one detector, you could see if I applied it at the other detector by the presence or absence of the interference pattern. To me, this is communication. $\endgroup$ – Trenin Dec 11 '15 at 19:14
  • $\begingroup$ If I understand you correctly, you are saying that yes the interference pattern is there, but only for half the photons. When all the photons are present, the interference pattern is muddled because the interference pattern of the other half of the photons fills in all the dark spots. However, neither of these experiments say that is the case. Both of them say that the interference pattern is readily seen depending on the action taken at the other detector. $\endgroup$ – Trenin Dec 11 '15 at 19:16
  • $\begingroup$ @Trenin The beginning of your last comment summarizes it accurately. -- Note that none of the sources you cite are scientific articles reporting original research, just some summary which might have left out important points. Have you tried looking for the original papers? $\endgroup$ – Norbert Schuch Dec 11 '15 at 19:20
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I find this result fascinating, and disturbing. By rotating the filter at A, I can change the interference pattern at B.

It's not like that. Interference pattern at B is observed only by ignoring all of the photons whose entangled twins never reached A (only coincident photons are counted).

It's easy to see that by rotating polariser in front of A you change which photons never reach A (and which thus you choose to ignore at B).

By rotating polariser at A you just stop considering the photons that are giving interference pattern at B and start considering those that don't.

So not FTL possible because guy at A must still tell B by normal channels which photos to ignore to either see or not see interference pattern.

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