From all the demonstrations Iv been able to find of Heisenberg's double slit experiment, whenever an observer tries to "see" which slit an electron passes through it collapses the wave function. My question is what would happen when you have two observers or detectors trying to detect the same electron, will they see the same thing? does the wave function only collapse to one possibility no matter who or what the observer?

  • 2
    $\begingroup$ Yes, it only collapses to one possibility. If this weren't true, there would be a probability $P> 1$ to find the electron! $\endgroup$ – Danu Sep 2 '14 at 9:17
  • $\begingroup$ Can the observers observe each other? $\endgroup$ – PyRulez Jan 5 '16 at 1:21

If you are saying that one electron (at a time) is used in the experiment, than one electron (at a time) must be observed at all circumstances, basically by definition.

When no detectors are used to determine which slit the electron passed through, you still detect one electron on the screen, the probabilistic spatial distribution of which is such that when a lot of electrons are used you see the interference pattern.

Placing a detector on one of the slits basically forces the electron to "answer" the question: "which slit was traversed by the electron?". Placing two detectors, one per slit, doesn't give much more information on this regard: you are still "asking the same question" to the electron.

The correlation of the detection events of the two detector is an interesting point, though. They will never see the same thing, because when one of them detects the electron, surely the other does not. This is one example of intensity correlation (or anticorrelation if you want), and is a crucial point for the determination of the quantum nature of the particle under consideration. An experimental evidence of the quantum nature of photons, for example, is based on this intensity correlation effect, see for example Hanbury Brown and Twiss effect and photon antibunching for more on the subject.

  • $\begingroup$ Nice answer, (+1) (Should be Hanbury Brown and Twiss, Hanbury Brown was one guy.) $\endgroup$ – George Herold Sep 2 '14 at 16:18

The observation either results in the electron being there, or not being there. But once the electron has been observed going through one slit, it cannot also show up at the other slit. That would require there to be more than one electron.

On the other hand, if the observer had an apparatus that resulted in 50-50 probability of observing an electron if it was there , then the situation is different - because "not observing an electron" would not be the same thing as saying "the electron was not here".

That might make for an interesting thought experiment...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.