Consider the single-electron double-slit setup, where an electron gun fires a single electron, and a dot is observed on the detector screen behind the slits. We do not place any detector in the slits.

Suppose that we record the time at which the electron was fired, and the time at which the dot appeared on the screen. We repeat the same experiment for a large number of electrons, so that we hopefully obtain the typical interference pattern from the statistical distribution predicted by Schrödinger's equation.

My question is, what is the observed relationship between the time delay (detection time $-$ firing time) and the position of the dot on the screen?

Is the time delay always the same for all particles that reach the same point on the screen, or is there some variance, perhaps a discrete/continuous set of time delay values? Are the time delays consistent with the speed of the electron and the path length going through either one of the slits?

I think that in the wave analogy (solving Schrödinger's equation), the single electron's wave function will behave like two interfering spherical ripples after passing the two slits. The two ripples should propagate with the same velocity (in the sense of radius increase over time). However, for points on the screen outside the axis of symmetry, the distance between one of the slits is smaller, so you'd expect the ripple from one of the slits to reach the point before the other ripple. Hence there are two distinct time values $t_1$ and $t_2$ for which $|\psi(\mathbf{x}, t)|^2$ is nonzero at that point. The detection event at point $\mathbf{x}$ on the screen will therefore either occur at time $t_1$ or $t_2$, because any other time the probability density is $0$ assuming the Born rule.

Knowing that the time delay is $t_1$ or $t_2$ should be enough information to provide which-way information without having detectors in the slits. But then this seems to contradict the presence of the interference pattern, which is supposed to vanish when which-way information is available. So I am not sure if this is the correct prediction for the observed delay time behavior.

I found another question which is similar but it asks about photons instead of electrons. I also did not find the accepted answer convincing enough. I explained my concerns in a comment under that answer. Perhaps a more quantitative reason why the energy uncertainty is so large that the interference pattern has to vanish if time measurements are performed would be good.

  • $\begingroup$ Nice try. No, you are not thinking in the quantum realm, you seem to think that the electron has possible definitive paths. Yet it is a wave interfering with itself... $\endgroup$
    – Jon Custer
    Jul 31 '19 at 21:12
  • $\begingroup$ But its wavefunction still has a time dependence, and for any given point on the screen, there can be a few time values where it is nonzero. This can be discussed without treating the electron as a classical particle. As an analogy, imagine that you look at the "ripple in a tank" setup. The initial ripple hits the two slits, and two new spherical ripples are produced. They reach the same point on the edge at different times. $\endgroup$
    – Tob Ernack
    Jul 31 '19 at 21:15
  • $\begingroup$ And then you need to add in Heisenberg as well... $\endgroup$
    – Jon Custer
    Jul 31 '19 at 21:17
  • $\begingroup$ I can see how Heisenberg can be important, but the question is to show it explicitly. As I mentioned in the other question, a maximally tolerable uncertainty $\Delta f$ can be achieved with time measurements of uncertainty $\Delta t = \frac{1}{4\pi\Delta f}$. We can then choose the slit distance to be large enough for the time difference between two paths to be larger than $\Delta t$, hence experimentally observable. For electrons, the speed would be smaller than $c$ so the distance can be more reasonable. $\endgroup$
    – Tob Ernack
    Jul 31 '19 at 21:20
  • $\begingroup$ Perhaps there is an issue in that the maximal $\Delta f$ actually depends on $d$ somehow, then I can't just set $d$ independently from $\Delta f$. But that would be something that should be explained in the answer. $\endgroup$
    – Tob Ernack
    Jul 31 '19 at 21:24

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