In the quantum eraser double slit experiment, does the photon (or wavefunction) pass through one slit or both slits when different polarizers are placed over the slits?
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$\begingroup$ See the chapter "Which-Way Marker" in the "technical paper", cited by Wikipedia. $\endgroup$– TrimokCommented Dec 26, 2013 at 12:16
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$\begingroup$ Nobody has any idea? $\endgroup$– Spiros252Commented Jul 6, 2014 at 18:09
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2 Answers
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According to quantum physics, when certain different polarizers are placed over the slits in the double-slit experiment (for instance, one vertical and one horizontal polarizer, or one circular clockwise and one circular counter-clockwise), thus "marking" each photon with which-way information, the photon indeed passes through only one slit, resulting in no interference. If you cover up one of the slits, you'll observe the very same absence of interference.
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$\begingroup$ That's right, a single photon can only go through one of the slits. There needs to be thousands of individual photons and both slits need to remain open in order for there to be a double slit fringe pattern. $\endgroup$ Commented May 9, 2016 at 5:15
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$\begingroup$ Don't forget, Bill, that in the quantum eraser experiment many photons are fired off, even if one at-a-time, and both slits remain open. Yet, when you add to the photons the "which-way" information of which slit they travel through (by polarizing all the light traveling through the two slits in two certain different ways), you lose the double-slit interference pattern. $\endgroup$ Commented May 9, 2016 at 5:45
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$\begingroup$ That's because the polarizers interfere with the overall Fringe pattern. There's more to it than just photons making it through a slit. $\endgroup$ Commented May 9, 2016 at 6:28
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$\begingroup$ Of course the polarizers interfere with the overall fringe pattern, Bill. They make the pattern disappear. Your statement has to be more specific to be meaningful. $\endgroup$ Commented May 9, 2016 at 6:46
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$\begingroup$ The photons need to interact with the slit and the edges completely. With the polarizers there this cannot be done. It has nothing to do with knowing which slit a photon went through. Also photons reflecting back from the detection screen will interact with those slits and contribute to the fringe pattern eventually and polarizers would interfere with this too. $\endgroup$ Commented May 9, 2016 at 7:03
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Quantum mechanics is a theory that can only predict probability distributions. It cannot predict trajectories. It is ruled by differential equations which have as solutions the wavefunctions, and the complex conjugate square of the wavefunction gives the probability of a specific, photon, electron, to be at (x,y,z,t) given the boundary conditions of the problem.
The importance of the boundary conditions is demonstrated clearly in this experiment.
In the experiment you refer to, both the boundary conditions and the initial states are changed during the experiment and it is not surprising that the interference disappears when the boundary condition constrains the slit through which a photon passed. The fact that by further manipulation an interference pattern appears, is again due to the wavefunction's probabilistic waving.
Change of boundary conditions changes the wavefunctions, the boundary conditions are not only "photon scatter from two slits" , but "photon scatter from two slits and all paraphernalia like polarizers upstream and downstream". If one really wanted to solve for the specific quantum mechanical problem all these should be taken into account when picking the wavefunction that describes the probability of the photon scatter.
Single photon double slit experiments for classrooms can be seen here.
It is the photon wavefunction that has a probability for the photon to pass through a slit, and has a sinusoidal dependence that gives interference patterns in the accumulation. The photon, when detected is detected as a point particle interaction with a screen or a ccd or a photomultiplier. One has never seen experimentally the signal of an elementary particle spread in space. Thus it is not the photon that is waving, but its probability of manifesting at (x,y,z,t).
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$\begingroup$ If I'm not mistaken, Anna, you just disposed of the quantum eraser experiment. $\endgroup$ Commented May 9, 2016 at 6:01
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$\begingroup$ @DavidReishi I do not know what I am disposing; I am just stating very basic postulates that have to do with quantum mechanical solutions. To look at probability distributions and make up narratives for what is mathematically calculable in principle is misleading. $\endgroup$– anna vCommented May 9, 2016 at 7:19