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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. $\endgroup$ – Carl Witthoft Jan 30 '14 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 '14 at 15:18
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    $\begingroup$ electron, photon, ocean liner, neutrino: same rules apply regardless of hte particle $\endgroup$ – Carl Witthoft Jan 30 '14 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 '14 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? $\endgroup$ – NeutronStar May 28 '14 at 20:10
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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 '14 at 16:46
  • $\begingroup$ We could, and that would constitute observing the second slit $\endgroup$ – Jim Jan 30 '14 at 18:48
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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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