In the classic Young double slit experiment, with slits labeled as "A" and "B" and the detector screen "C", we put a detector with 100% accuracy (no particle can pass through the slit without the detector noticing) on slit B, leaving slit A unchecked. What kind of pattern should we expect on the detector C? Probably the right question is: knowing that a particle hasn't been through one of the slits makes the wavefunction collapse, precipitating in a state in which the particle passed through the other slit?
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$\begingroup$ Google "Quantum eraser", you may find something interesting... $\endgroup$– ManishearthCommented Mar 22, 2012 at 8:00
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4 Answers
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Dreelich, you might want to get hold of a copy of the Feynman Lectures on Physics and take a look at Vol. III, Chapter 37, Section 1-6 "Watching the electrons". The sections leading up to that one are also relevant.
In addition to being a great read (well, if you like that sort of thing, but that's likely a safe assumption in this forum!), Section 1-6 confirms what you said: Watching the electron go through the slit makes it behave classically in terms of how it hits the screen. Interestingly, Feynman detested the phrase "wave function collapse." His approach was always to look at the start and end of a process and calculate the probabilities for each end point.
answered Mar 22, 2012 at 22:58
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It obviously depends on what the detector does with the particles: Heisenberg uncertainty only mandates that it does something, but not what.
In reality, pretty much the only detectors with 100% accuracy are those that completely "swallow" the particle, i.e. solid-state detectors in which the particles get stuck. So the result would be the same as with slit A alone, as particles passing through B never actually hit the screen.
answered Mar 21, 2012 at 20:23
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$\begingroup$ What about if we put, instead of the controlled slit, we put a detector linked to another particle gun, shooting a particle every time the detector "clicks", in the same pattern as the "one slit" experimental setting? Is that a reasonable simulation of a 100% efficient detector? $\endgroup$– DreelichCommented Mar 22, 2012 at 1:17
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$\begingroup$ Yes. But those "simulated" particles know nothing about the phase of the particles going through A, so there's no interference. $\endgroup$ Commented Mar 22, 2012 at 1:47
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Well, I am sorry to say that there exists a recent experiment that records which slit the electrons went through in a manner so as not to disturb the wavefunction too much. It is recorded in the two slit article of the wikipedia. This is the third time I am copying the pictures from there, to physics.se
But an experiment performed in 1987 produced results that demonstrated that information could be obtained regarding which path a particle had taken, without destroying the interference altogether. This showed the effect of measurements that disturbed the particles in transit to a lesser degree and thereby influenced the interference pattern only to a comparable extent. And in 2012, researchers finally succeeded in correctly identifying the path each particle had taken without any adverse effects at all on the interference pattern generated by the particles.[24] In order to do this, they used a setup such that particles coming to the screen were not from a point-like source, but from a source with two intensity maximas.
So this is a cumulative experiment that shows an interference pattern from a beam of electrons even though the path of the electrons is known and demonstrates the entirely probabilistic nature of the underlying quantum mechanical solutions.
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If the way the particle goes leaves a record by any means whatsoever, (whether that record is examined by you or not, so long as you could examine it) and as long as the record still exists at the moment the particle is detected at the screen then the interference effect is destroyed for that particle. (see video of leonard suskind's lectures: quantum entanglement 1 lecture 6 and 7 on Stanford continuing education website).
Weak measurements seem to contradict this as noted below which reports that
" an experiment performed in 1987 produced results that demonstrated that information could be obtained regarding which path a particle had taken, without destroying the interference altogether"
This is wrong! The measurements made in that paper do not show which path a particle had taken. They deduce from their measurements "average trajectories". They also state that no single particle can travel on one of these trajectories! And they reaffirm that the Heisenberg uncertainty principle is not violated in their experiment (as it would have to be if the which way information is obtained without destroying the interference).
What is an average trajectory then? It is like the way that an average value of an observable can be defined even for a state which is not an eigenfunction of that observable so that a measurement produces different eigenvalues of the observable each time and we can only give a probability for each one. The average value is then different from each of the actually observed measurements. (Only when the state is an eigenfunction of the observable will the average be the same as the measured value and then there is no collapse of the wavefunction associated with measurement).
