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Background: I'm currently doing self-study of quantum physics by following university level quantum physics lectures on YouTube (e.g. the YouTube MIT 8.04 Quantum Physics I, Spring 2013 or Stanford Quantum Lectures by Leonhard Susskind, plus supplemental YouTube courses like DrPhysicsA, with the occasional popular level YouTube on specific subjects). My classic math and physics is somewhat robust (high-school) but rusty (about 30 years ago). However, I'm trying to understand quantum physics not merely on popular YouTube level, so feel free to throw in some math.

I recently came across the Quantum Eraser experiments and various popular explanations and in one article by Sean Carroll, he says that "Entanglement of any sort kills interference."

What do we see at the screen when we do this with many electrons? A smoothed-out distribution with no interference pattern, of course. Interference can only happen when two things contribute to exactly the same wave function, and since the two paths for the traveling electrons are now entangled with the recording electrons, the left and right paths are distinguishable, so we don’t see any interference pattern. In this case it doesn’t matter that we didn’t have honest decoherence; it just matters that the traveling electrons were entangled with the recording electrons. Entanglement of any sort kills interference.

So, without going down the full rabbit hole of the quantum eraser experiment (and the associated wild claims), is it true that an experimental setting that involves entanglement and (potential) interference, there will be no interference, even if the actual path is not measured/observed?

Or, in other words, is it true that the experimental setup below (or similar) will never show interference, even if A/B are not measuring devices or actual observers of any kind?

enter image description here

Or am I subject to a basic misunderstanding here?

Norbert Schuch
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Terminality
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2 Answers2

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The short answer is: entangled particles do not exhibit traditional interference effects unless they are coherent. But making one coherent negates the entanglement.

Were that not true, FTL signaling would be possible. Below a reference that supports this from recent Nobel laureate Zeilinger, see figure 2:

https://courses.washington.edu/ega/more_papers/zeilinger.pdf

DrChinese
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Apparently the answer is: It depends. I was looking at the Zeilinger paper from the other answer with it's mention of the "recent experiment (Dopfer, 1998)":

In this experiment, photon 2 passes a double slit while the other, photon 1, can be observed by a detector placed at various distances behind the Heisenberg lens which plays exactly the same role as the lens in the gamma-ray microscope discussed by Heisen- berg (1928) and extended by Weizsacher (1931). If the detector is placed at the focal plane of the lens, then registration of a photon there provides information about its direction, i.e., momentum, before entering the lens. Thus, because of the strict momentum correlation, the momentum of the other photon incident on the double slit and registered in coincidence is also well de- fined. A momentum eigenstate cannot carry any posi- tion information, i.e., no information about which slit the particle passes through. Therefore, a double-slit in- terference pattern for photon 2 is registered conditioned on registration of photon 1 in the focal plane of the lens.

I then went on to look at the Dopfer PhD Thesis and apparently there are experiment setups where you can have entangled particles and still register an interference pattern behind the double slit, depending on what measurement that is done in the other arm (as outlined in the quote above).

They even did a version with delayed measurement with the same results.

This means that the fact that entangled particles exist, does not automatically in all cases cancel the interference.

Terminality
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