- #36
Marcin
- 41
- 2
@PeterDonisPeterDonis said:The "travel time" itself is not the critical parameter; it's the uncertainty in the travel time.
Consider: the two different classical "travel times" are ##(a + l_1) / c## and ##(a + l_2) /c##. The difference between them is ##(l_2 - l_1) / c##. So if we measure the times of emission at the source and arrival at the detector, we have two possible cases:
Case 1: The time resolution of our measurements is smaller than ##(l_2 - l_1)/c##. In that case, as @DrChinese has pointed out, measuring the travel time is equivalent to measuring which slit the photon went through, so there is no interference. But you've said you're not interested in this case; you're interested in the case where there is interference.
Case 2: The time resolution of our measurements is larger than ##(l_2 - l_1)/c##. In that case, the time measurements do not distinguish which slit the photon went through, and there is interference. But that has nothing to do with the travel time itself; it's only the large enough uncertainty in the travel time that allows the interference to occur.
Or, to put it in a more provocative way: for interference to occur at the detector, the travel time of the photon must be uncertain enough that it is in a superposition of having gone through both slits; but that in turn means that it is in a superposition of having left the source at two different times, which differ by ##(l_2 - l_1)/c##. The time of leaving the source has to be uncertain to that extent for interference to occur.
Let's assume, that we can get 3 possible, statistical results with ##Δt=(l_2 - l_1)/c##:
##t=(a+l_1)/c±Δt##
##t=(a+(l_1 + l_2)/2)/c±Δt##
##t=(a+l_2)/c±Δt##
Would they be noticeably different? For the sake of argument, let's assume that ##Δt## is slightly greater.
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