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In the double slit experiment, when a detector is added to tell which hole the photon went through the interference pattern disappears. Would there be an interference pattern if the data was sent to a computer, but never recorded? What if it was recorded but destroyed before anyone looked at it?

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  • $\begingroup$ There's no way to convert something into data without measuring it. You'll always destroy the interference pattern. $\endgroup$ – QuantumBrick Nov 13 '14 at 23:35
  • $\begingroup$ Maybe I'm wrong, but I like to think of it as interaction rather than measurement. So if anything immediately interacts with the particle after it exits one of the slits, there can be determination. It's not human observation that collapses. $\endgroup$ – Lee Louviere Nov 19 '14 at 16:10
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Yes and no. "Sending the data to a computer, then destroying it" is probably too complex an operation to let the state of a photon produce the same interference pattern again.

Yet, experiments in the spirit of your idea have indeed been performed, by playing around with entangled photons, sending one through the slit and using the other to obtain information about the path taken. They are called quantum eraser experiments: (quoting from the description of such an experiment from Wikipedia)

First, a photon is shot through a specialized nonlinear optical device: a beta barium borate (BBO) crystal. This crystal converts the single photon into two entangled photons of lower frequency, a process known as spontaneous parametric down-conversion (SPDC). These entangled photons follow separate paths. One photon goes directly to a detector, while the second photon passes through the double-slit mask to a second detector. Both detectors are connected to a coincidence circuit, ensuring that only entangled photon pairs are counted. A stepper motor moves the second detector to scan across the target area, producing an intensity map. This configuration yields the familiar interference pattern.

Next, a circular polarizer is placed in front of each slit in the double-slit mask, producing clockwise circular polarization in light passing through one slit, and counter-clockwise circular polarization in the other slit. This polarization is measured at the detector, thus "marking" the photons and destroying the interference pattern.

Finally, a linear polarizer is introduced in the path of the first photon of the entangled pair, giving this photon a diagonal polarization. Entanglement ensures a complementary diagonal polarization in its partner, which passes through the double-slit mask. This alters the effect of the circular polarizers: each will produce a mix of clockwise and counter-clockwise polarized light. Thus the second detector can no longer determine which path was taken, and the interference fringes are restored.

What happens is that though the paths through the slits are in principle distinguishable, no interaction (in particular, no macroscopic interaction that could cause decoherence or whatever you believe happens during a measurement) is actually taking place that would depend on the path the photon takes after the second polarizer is introduced. "Recording data", as you propose, would change that, and so that cannot work.

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  • $\begingroup$ @julianfernandez: Well, I'm not 100% certain that some mad scientist may one day come up with something like that since I'm not an expert in quantum information, but in my naive view, to "record data" you'd have to measure the polarization of the photon (or anything else of its properties). That's ordinarily rather impossible without destroying the photon (and thus, entanglement). It could be that "non-destructive weak measurement" of photon polarization I've lately heard of could achieve that, but I'm very reluctant to pronounce on that until I've seen the experiment. $\endgroup$ – ACuriousMind Nov 13 '14 at 23:56
  • $\begingroup$ well, at least you are correct on a standard measurement (recording), I just found that article that put in a complementary answer $\endgroup$ – Wolphram jonny Nov 14 '14 at 0:06
  • $\begingroup$ Note, however, that in quantum eraser experiments the interference pattern is never restored in the total pattern of particles on the screen, only in the coincidence count between particles on the screen and entangled twins that went to a particular detector, as I discussed in this answer. It seems that if the particles going through the slits interact with other particles in a way that would in principle allow for a determination of which slit they went through, that's sufficient to destroy interference in the total pattern on the screen. $\endgroup$ – Hypnosifl Nov 14 '14 at 1:59
  • $\begingroup$ So, what you're saying, is that if you introduce a method of measurement, the interference is removed, got that. However, if you introduce a method that disguises the measurement so that it is indistinguishable, the interference comes back. So, Since the measurement was subsequently hidden, its not measurable at the end, and thus the measurement was erased? $\endgroup$ – Lee Louviere Nov 19 '14 at 16:03
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To add a concrete example to ACuriousMind answer's, consider the Schrodinger's cat experiment. An analysis of an actual experiment found that measurement alone (for example by a Geiger counter) is sufficient to collapse a quantum wave function before there is any conscious observation of the measurement. See this paper for more details http://web.archive.org/web/20061130173850/http://www.ensmp.fr/aflb/AFLB-311/aflb311m387.pdf

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Sending data to a computer is not possible without detection, and a detection for sending to some computer is a macroscopic intervention, that introduces decoherence between the two components of the wave-function, that which passed through one hole, and that with passed through the other. But the issue is simpler. For detecting which hole the photon passed through, a detector has to be placed quite close to the holes, i.e. where the two branches of the wave-function that which passed through one hole and that which passed through the other, are well-separated. The interference pattern appears there where the branches superpose, and in this region there is no more memory through which hole the photon passed. In terms of operators, the operator "which hole" does not commute with the operator "minimum or maximum" s.t. s.t. no memory of the "which hole" is retained in the interference pattern.

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protected by Qmechanic May 25 '17 at 15:43

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