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Step #1 Imagine one preforms an electron based Double-slit-experiment and one does so with only one electron being fired at a time.

Step#2 Also included in the experiment is an unobserved alternating opening of the double slits such that there is only one slit open at a time, and that each individual alternating ( the single process of closing of 1 slit and the opening of the other ) occurs during the gap time period that is present between the individual electrons being fired.

Knowledge of which slit each electron goes through is therefore NOT being observed.

Thus does an interference pattern still arise over time here to ?

If so, could you please present any data concerning proof of this in particular experiment, thus in turn eliminating any assumption.

Note:Time is required to accumulate an interference pattern in the double slit experiment, just like it is in the Kim et al Delayed Choice Quantum Eraser experiment of the year 2000.

But in this Delayed Choice Quantum Eraser experiment, what makes the experiment possibly astonishing is that, unlike in the classic double-slit experiment, the choice of whether to preserve or erase the which-path information of the idler was not made until 8 ns after the position of the signal photon had already been measured by detector D0.

Thus this Delayed choice experiment raised questions about time and time sequences.

Two separate events, the position detecting of the signal photon and the determining of the which-path information of the idler photon, occurred at two different times, yet despite being separated by time, both were taken into account to produce a single final outcome.

Thus my above double-slit experiment arises. Does one electron interfere with another electron despite being displaced by time, or does a single electron interfere only with itself.

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    $\begingroup$ Just for clarification, does one slit remain closed for the whole time between emission of the electron and it hitting the detection screen? $\endgroup$ – John Rennie Aug 6 '14 at 5:17
  • $\begingroup$ Yes, that is the case. One slit is closed during an electrons trip from the source to the destination. $\endgroup$ – Sean Aug 6 '14 at 5:18
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    $\begingroup$ If only one slit is open at a time you are just setting up a complicated one slit experiment and the result is "one spot" with maybe some diffraction on sides depending on energy and slit width. so your experiment will get two spots. $\endgroup$ – anna v Aug 6 '14 at 5:22
  • $\begingroup$ The question is to basically prove whether or not this is equivalent to a one slit experiment, or not, thus the intent is to eliminate any immediate assumptions that it is an equivalent via proof of some kind. $\endgroup$ – Sean Aug 6 '14 at 5:26
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/phyopt/mulslid.html . Depending on the distance between your two slits the diffraction tails may interfere, so there will be some difference. It will not show the pattern of the two slit interference though en.wikipedia.org/wiki/… . en.wikipedia.org/wiki/File:Single_slit_and_double_slit2.jpg . It is not worth doing an experiment since there must be thousands around $\endgroup$ – anna v Aug 6 '14 at 5:29
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As you've described it, you would see a simple two slit pattern. The reason being that for any practical implementation of the experiment as you've described it, you would not be able to erase the which path information, as the photons hitting the classical doors would count as a measurement in the practical sense.

But, the greater thought experiment you are proposing is in fact a famous one, due to Wheeler, deemed the delayed choice experiments. And, in the thought experiment sense, if you truly were able to erase the which path information, you would recover the interference pattern. Spooky but true. True not only in theory, but in recent times quantum eraser experiments have actually been carried out and agree with this notion. These experiments rely on beam splitters to handle their erasure, which are workable in the lab.

Moving on to the even spookier, recently (2000), delayed choice quantum eraser experiments have been carried out, wherein you erase the which path information after the electrons have hit the screen. And you still recover the full interference pattern. For full experimental details see: A Delayed Choice Quantum Eraser by Kim, et al. in Phys Rev Letters, [doi][arXiv]

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  • $\begingroup$ I don't follow your reasoning. As I said, "Knowledge of which slit each electron goes through is therefore NOT being observed.", thus the path of each electron is not exposed. $\endgroup$ – Sean Aug 6 '14 at 11:26
  • $\begingroup$ "observation" isn't just whether you record the number in your notebook. Any interaction with a sufficiently large classical device will count. $\endgroup$ – alemi Aug 6 '14 at 16:14
  • $\begingroup$ I don't follow your reasoning. There is no large classical device being used for observation in the experiment which relates to determining which slit each electron goes through. Therefore, there is no which-path information to be erased in the first place. $\endgroup$ – Sean Aug 6 '14 at 20:00
  • $\begingroup$ @Sean: "There is no large classical device being used [...]" -- Well, it's not explicitly mentioned in the OP description, but there's usually a "screen", or "detector screen" involved; i.e. a device which is suitably persistent (itself, as well as in relation to the signal source) throughout a sufficient number of individual trials to allow recording "a pattern" (based on which the most probable number and "geometry of slits in the barrier between source and screen" throughout the set of trials may be determined). [to be contd.] $\endgroup$ – user12262 Aug 6 '14 at 23:57
  • $\begingroup$ @Sean: "[...] which relates to determining which slit each electron goes through." -- If the "pattern" recorded by the screen is described as "(most probably) due to" source and screen having been separated by a (potential) barrier with "two slits available alternatingly/incoherently" (i.e. the described setup, AFAIU) and only if that pattern consists of two well-separated "blobs", then we can unambiguously attribute all individual entries in the pattern to the corresponding signal "having gone through" either one slit, or the other. Otherwise we can't make that distinction as strictly $\endgroup$ – user12262 Aug 6 '14 at 23:59

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