Measuring which path in Hardy's paradox, results

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SUMMARY

This discussion centers on the interpretation of Hardy's paradox as revisited by Yakir Aharonov and colleagues, specifically regarding the behavior of particles in a Mach-Zehnder interferometer (MZI). The key finding is that when both detectors in the non-overlapping arm (no+) and the overlapping arm (o-) are present, the detection of both D+ and D- indicates that the joint detector for no+ and o- must have clicked, confirming the collapse of the state to |no+o->. The confusion arises from the expectation of independent probabilities for the other states, which is clarified by the understanding that the joint detector cannot distinguish between certain states, leading to destructive interference at D-.

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  • Understanding of quantum mechanics principles, particularly interference and superposition.
  • Familiarity with Mach-Zehnder interferometer (MZI) configurations and their implications in quantum experiments.
  • Knowledge of particle-antiparticle interactions, specifically positron (e+) and electron (e-) behavior.
  • Basic grasp of quantum measurement theory and state collapse.
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msumm21
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I am reading a paper from Yakir Aharonov and others revisiting Hardy's paradox and I have a question. Let me state the notation first as brief as possible so that we can use it in the discussion:

* a positron e+ goes through either the overlapping arm o+ or the non overlapping arm no+ of a Mach-Zhender interferometer (MZI)
* an electron e- goes through either the overlapping arm o- or the non overlapping arm no- of another MZI
* the two detectors that e+ (resp. e-) may register in after passing the the MZI are C+ and D+ (resp. C- and D-)
* when a single MZI is operated independently without which path measurements the particles always go into the C detector (destructive interference to D) -- C always "clicks"

Now, when the arms of the MZI overlap to give Hardy's experiment, Ahranov et. al claim that, if detectors are placed within no+ and o-, then every situation in which both D+ and D- click, both of the which path detectors also click. I do not understand that. I do understand that, if a detector is in o- ONLY then every time D+ and D- click then o- must have clicked (otherwise e+ would have "interfered with itself" such that it would have to go into C+). However, if both the no+ and o- detectors are there, then I would expect 1/4 of the time there would be annihilation as usual (both e+ and e- enter overlapping arms), but the other 3/4 of the time the particles would collapse to the other 3 combinations no+o-, no+no-, and o+no-with equal probability. And, in each of the latter 3 cases, there would be a 1/4 chance of both particles being registered in their D detectors, right? So why would a measurement of, say no+no-, prevent the D detectors from clicking? Is it somehow impossible to register no+no-? Once the particles "collapse" into the non overlapping arms I'd think their should be no interference and they should independtly have 0.5 probability of going into D (.25 probability of D+D-). I guess I am missing something.
 
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I just reviewed that paper again and I think I caught 2 problems. Maybe someone could confirm if I'm thinking of this correctly now. The two things I missed previously:

1) The author said that the measurements were performed at a spot on the MZIs AFTER annihilation would have occurred -- i.e. after the state collapsed to the superposition |no+o-> + |no+no-> + |o+no->.

2) The detectors in no+ and o- are actually combined into a joint detector which only clicks if the state collapses to |no+o-> and thus can't distinguish between the other two states as I previously thought. So, if this joint detector does not click, then the state only collapses to |no+no-> + |o+no->. In this state, the electron will destructively interfere with itself at the D- detector and hence must enter the C- detector. Therefore, if D+ and D- clicked in a run of this experiment, we immediately know that the joint detector for no+o- must have clicked (otherwise there was no way to get D-).

Correct?
 

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