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So far I'm only tasting the quantum mechanics. Haven't gone very deep into the mathematics of it yet.

I read about the double slit experiment, and the weird consequences of it: if you put a detector at the slits you won't have an interference pattern. Since you basically absorbed and recorded the photon then you reemitted another one at that slit: now you have single probability wave emitted at that slit which won't interfere with the another one since it's not there: no interference pattern. At least this is my understanding so far.

Ok, now let's cheat: at the source you detect (absorb) and reemit the photon. You won't measure the reemitted photon again till it hits the detector. So it pass the slits in a fuzzy state and you should have an interference pattern at the detector then. Anyway the photon hits detector at a point somewhere, you record it. If you have atomic clocks you can have interval between the reemission and detection at the screen. Based on the location of the hit and the travel time measured, and knowing the light travels with $c$ the path length travelled can be estimated. From the path length you can deduce which slit it passed through.

Of course nature doesn't like cheaters: you cannot have "which slit" information and "interference pattern" (went through both slits) at the same time. So my thoughts:

  • I think nature cannot prohibit to have a low rate photon source: you can put some lead screen in front of your radioactive gamma source, to reduce the rate. So you should be able to measure travel times occasionally when the next photon comes late enough.

  • Nature cannot mess with travel times too since from that would mean light would't go woth $c$ but slower or faster - the latter would have nasty consequences.

  • Another thing I can think of is that there will be enough uncertainty in detector response times or the reemitting apparatus so I cannot reliably measure the travel times. But this would mean that this uncertainty depends on the distance between the slits: the bigger the distance the bigger possible difference in travel times to a point on the detector through the slits. It's very hard to believe the the sole fact of changing the slit spacing can affect uncertainty of the detector response times...

So what do you think will happen? Do you know of experiment that tested this?

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  • $\begingroup$ If you know when the photon left accurately enough then its frequency isn't well enough defined anymore and the interference pattern will be washed out. If the photon is emmited within a time interval Delta t, then the frequency spectrum is given by the Fourier transofrm of the function f(t) which is 1 for t between 0 and Delta t and zero otherwise. $\endgroup$
    – Count Iblis
    Commented Jun 30, 2014 at 17:29

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In addition to the normal uncertainty relation between position and momentum, there is also a relation between the uncertainties of when an event took place and the energy involved. \begin{equation} \Delta E \Delta t \le \frac{\hbar}{2}\end{equation} The experiment you have proposed involves measuring the time of flight of the photon very accurately, in particular making a very accurate measurement of when it was emitted. This means that the energy of the photon must be very uncertain.

Now Planks formula allows us to relate the energy of a photon with its frequency \begin{equation} E = hf\end{equation} So an uncertain energy will mean an uncertain frequency. Photons going through the apparatus without a well defined frequency will wash out the interference pattern on the screen, so you still cannot cheat nature.

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  • $\begingroup$ Not it's all clear. Nature wins again as always. $\endgroup$
    – Calmarius
    Commented Jun 30, 2014 at 19:21
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    $\begingroup$ I think this answer is missing some quantitative analysis. Clearly there will be an uncertainty in frequency, but will it be enough to completely wash off the interference pattern? Suppose that the maximal tolerable frequency uncertainty is $\Delta f$ (to preserve interference patterns). Then using Heisenberg's uncertainty relations we have the minimal time uncertainty $\Delta t = \frac{1}{4\pi\Delta f}$. So if we use time measurements with uncertainty $\Delta t$, the interference pattern is still observed. We may then consider slits where the distance $d$ is larger than $c\Delta t$. $\endgroup$
    – Tob Ernack
    Commented Jul 31, 2019 at 20:52
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    $\begingroup$ In that case there will be again a measurable time delay between slit 1 and slit 2, so we obtain which-way information again. So I think we need a more careful analysis of the amount of uncertainty present in the system, and how much uncertainty is required for interference to remain present. $\endgroup$
    – Tob Ernack
    Commented Jul 31, 2019 at 20:54

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