# Double slit and detector

• I
Gold Member
Suppose we have a double slit and we fire a photon through it with no photon detector at either slit. We get a interference pattern.

Now we put a photon detector at the left slit. The interference pattern is destroyed, right?

In this last setup, the photon can be located at the left slit. So if the photon went through the left slit, it was detected and the interference is gone. My question is: if the photon went through the right slit, it wasn't detected because we haven't got a detector at the right slit. So how did the photon know it had to go through one of the slits (the right one) if it was never detected, unless it was a wave from the beginning?

Demystifier
Gold Member
So how did the photon know it had to go through one of the slits (the right one) if it was never detected, unless it was a wave from the beginning?
It was a wave from the beginning (or, in Bohmian mechanics, there was both a wave and a point-like particle from the beginning), so there is no problem.

entropy1
If its not detected by the detector at the left slit, then you know it went through the right. In that sense that is the detecting (an inference over the whole experimental set-up). Hence no interference pattern forms. At least that is my understanding.

DrClaude
Mentor
If the detector is perfect, then indeed there is no difference between this set-up and blocking the left slit.

PeroK
Homework Helper
Gold Member
2020 Award
Suppose we have a double slit and we fire a photon through it with no photon detector at either slit. We get a interference pattern.
It's more accurate to say that each photon interacts with the screen according to a probability distribution consistent with a wave function that is itself consistent with the configuration of the double-slit barrier.

Don't forget that the light interacts with first barrier, and that's what determines the wave-function of the light that makes it through to the screen. The intensity of light that gets through is reduced because some photons (perhaps almost all of them) are absorbed by the barrier between and either side of the slits. The photon doesn't "know" to do anything. And, if you don't set up your double-slit carefully, then all your light will simply be absorbed by the first barrier (if the slits are in the wrong place).

Also, if your slits are too far apart, then any beam will either: be totally absorbed by the barrier; all go through one slit; or, all go through the other slit.

Finally, also remember that the incident beam must have some lateral uncertainty: otherwise, no matter how you set up the slits, they would always be too far apart.

So:

1) A beam of photons, described by a wave function with some lateral uncertainty, is directed towards a double slit.

2) Randomly, according to the probabilities associated with this initial wave function, some light is absorbed and some light passes though.

3) The wave function of the light that passes though is effectively a superposition of two wave functions, each associated with one or other slit.

4) The light impacts the screen consistent with this superposition (and exemplifies quantum interference).

If you close one slit (a detector which absorbs a photon is essentially equivalent to closing that slit):

2b) Randomly, more light is absorbed at the double-slit barrier.

2c) The wave function of the light that passes through is effectively a simple wave-function, associated with the only slit that was open.

2d) The light impacts the screen consistent with this wave function, which is different from the first case.

All this about wave and particle behaviour is superfluous and inessential, IMHO.

vanhees71