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I am having a difficult time solving this. Say that electrons are emitted from a source S at a very slow rate. If both slits S1 and S2 are observed, we would have roughly 50% probability of detecting an electron at one of the two slits. The interference pattern is lost and the intensity distribution will appear as the sum of two individual sources: I = I1 + I2.

But what if only one slit (S1) is observed? The observed slit (S1) will appear to produce a normal distribution, but what about the unobserved slit? This experiment has been performed with individual electrons, so we know that if both S1 and S2 are unobserved the intensity distribution contains an oscillating term for each electron. Does concluding that an electron must have passed through the unobserved slit count as an observation, and therefore destroy the interference pattern?

Edit: changed the source to electrons

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  • $\begingroup$ How would you propose to detect a photon at S1 and yet allow it to propagate to the detection plane? Tho' I recall recent experiments showing indirect observation of a particle in PathOne but its quantum states are in Path2. $\endgroup$ Jan 30, 2014 at 15:05
  • $\begingroup$ @CarlWitthoft I changed the source particles, but the question remains. Don't quantum states exist in both S1 and S2 for an interfering electron before observation? $\endgroup$
    – user38594
    Jan 30, 2014 at 15:18
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    $\begingroup$ electron, photon, ocean liner, neutrino: same rules apply regardless of hte particle $\endgroup$ Jan 30, 2014 at 15:59
  • $\begingroup$ The example my textbook gives (Zettili) is placing a light source by the slits and scattering light over the electrons. A Geiger counter is used at the detection wall. I assumed that the same electrons that scattered the light are propagating to the wall--is this assumption wrong? $\endgroup$
    – user38594
    Jan 30, 2014 at 16:48
  • $\begingroup$ What do you mean by observing and not observing specific slits? Are you not looking at one of the slits? Are you covering them? $\endgroup$ May 28, 2014 at 20:10

3 Answers 3

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An interesting question, but by its own nature, the true answer is unknowable. If we managed to make some predictions about what happens to the unobserved slit, we would need to still experimentally verify them. But how could we? Any experiment that involves observing one slit and having zero knowledge about what happens at the other would necessarily require us to never observe the slit of interest. So we could never gain the data to verify our predictions. Unfortunately, when quantum mechanics says something is unknown, it usually means we can never know it or even guess at it.

To answer your last question, if you track when an electron is detected and the slit you are observing had none pass by, then deducing that it must have come through the other slit does count as observing it.

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  • $\begingroup$ But could we not subtract the intensity of the observed slit from the total, and then determine exactly what was produced by the remaining half of the electrons? Timing the detection at the observed slit and at the detection plane would allow this, as seen in delayed choice experiments. $\endgroup$
    – user38594
    Jan 30, 2014 at 16:46
  • $\begingroup$ We could, and that would constitute observing the second slit $\endgroup$
    – Jim
    Jan 30, 2014 at 18:48
  • $\begingroup$ If we put a photomultiplier in one slit and other slit is open what pattern we get? $\endgroup$
    – Sancol.
    Feb 22 at 16:53
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    $\begingroup$ @Sancol. a regular single-slit pattern. Observing which slit the photon goes through breaks the magic, unfortunately $\endgroup$
    – Jim
    8 hours ago
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The electrons of the unobserved slit produce fringes on the screen. Once more, electrons near an edge produce fringes too. That happens because there is an interaction between the edges of the slit plate (correct the material of the edges). Young saw fringes, thought about waves and later we got geometrical equations, satisfying what we see on the screen.

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As far as I know, if there is only one which way detector, which can only detect electrons/photons passing through one of the slits, the interference on the slit will disappear. The same particle distribution is obtained as if two detectors were operating and both slits were observed. The slits do not cause the distribution, but the electrons. And if the device is such that it can be determined which slit it passes through, the interference disappears. This arrangement is such that if an electron named Alice arrives at the screen, there can be two cases:

A) either it has been previously detected by the one which way detector, then it is determined to have passed through that slit

B) it was not detected, in which case it is determined that it passed through the other slit.

Assuming, of course, that the detector is operating at 100% efficiency, so there are no eletrons that have passed through it and yet it has not measured them. The matter of not fully efficient which way measurements is another matter. The answer I am suggesting is even clearer in the case of an experiment with a Mach-Zehnder device using a similar principle. There, also, the quality change in the signal measured by the output detectors is caused by whether there is a which way measurement en route or not. But it is clear that only one detector in one branch is enough to do this, and that is how the corresponding experiments are usually performed.

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