How Does a Photon Know Which Slit to Pass Through in a Double Slit Experiment?

[entropy1]

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]

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.

 

[StevieTNZ]

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]

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

 

[PeroK]

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]

I think to understand this example (which I find quite difficult to begin with, but Feynman famously starts his lectures on QT this way, so many textbook others do so too), it's a good idea to think about a concrete setup to determine which-way information or not, which can be described in a simple way. It's very complicated to discuss in detail the scenario, where you put a detector close to the double slit and discuss how the interactions of the photons with this detector destroy the interference pattern (which forms of course only when many equally prepared photons are used, because each photon simply leaves one spot at the registration screen).

A very elegant way out is to encode the which-way information in each of the photons themselves. One way is to start with linearly polarized photons. Let's say the photon momentum is in $z$ direction and the polarization in $x$-direction to be definite. Now just bombarding the double slit with many such photons you get an interference pattern, and indeed if the setup is chosen such that it is possible to get an interference pattern at all (i.e., the photon distribution must be wide enough to cover both slits) it is indeed impossible to say, even in principle, through which slit the photon went.

Now you can gain which-way information by putting quarter-wave plates in each slits, the one directed in $+45^{\circ}$, and the other in $-45^{\circ}$ direction. Without going into the physical details you only need to know that this is a device which makes out of an $x$-polarized photon a left- (right-) circularly (helicity $\pm 1$) polarized photon, respectively. In other words, a photon going to one slit gets with certainty left-circurlarly polarized and going through the other slit right-circularly polarized. That marks each photon with a distinctive property (i.e., the corresponding polarization states are perpendicular to each other since $\langle h=+1|h=-1 \rangle=0$) depending through which slit it came, i.e., the which-way information is 100% entangled with the helicity of each photon behind the double-slit. Now since the polarization states are prependicular there's no interference term and thus no interference pattern occurs on the screen when repeated with many photons. Of course, that indeed is due to the local interaction of each photon with the quarter-wave plates positioned in the slits.

Quantum theory now claims that this is a generic thing, i.e., no matter how you gain (or only the putative possibility to gain!) the which-way information destroys the interference pattern. So far there's no example, where this prediction is invalidated, i.e., nobody has ever found any clever way to gain complete which-way information without destroying completely the interference pattern.