How Does Adding a Second Pair of Slits Affect Interference Patterns?

[Paul Howard A]

Standard slit lamp experiment with detectors at the slits turned on. This creates two beams without an interference pattern at the screen. Now, impose a second pair of slits (without detectors) in the path of one of the beams. What will be the pattern on the screen?

I'm confident this has been done. Could someone give me a reference please?

 

[DaveC426913]

There will be an interference pattern. What were you expecting? I don't see how the second light is relevant.

 

[BvU]

Interesting casus. I've heard about detecting which slit an electron goes through (it indeed destroys the interference pattern) but I wonder how you would detect a photon without hindering it ?

My bet is on the "re"-appearance of an interference pattern: this one beam doesn't know its history in the preceding path.

But I'm curious to hear better arguments !

 

[Paul Howard A]

I, too, would speculate that a diffraction pattern would reemerge. However, If the detector measures a particle, then has not the wave function of the photon collapsed? Does the wave function then reappear after a photon passes through the first slit? I assume it does but wondered if it had been tested.

 

[Nugatory]

I've heard about detecting which slit an electron goes through (it indeed destroys the interference pattern) but I wonder how you would detect a photon without hindering it ?

A clever and complicated technique is to entangle the photon (the "signal") with another one (the "idler") before it reaches the slit, and then find a measurement on the idler photon that will produce a different result according to which slit the signal photon went through. However, it is sufficient to put a polarizer behind each slit - if the two polarizers are at right angles there will be no interference.

 

[Nugatory]

I, too, would speculate that a diffraction pattern would reemerge. However, If the detector measures a particle, then has not the wave function of the photon collapsed? Does the wave function then reappear after a photon passes through the first slit?

In the basic version of the double slit experiment, you send a particle towards a barrier with two slits in it, the particle interacts with the barrier and ends up in a superposition of "went left" and "went right". These two interfere at the screen and we get an interference pattern. On the other hand, if we put detectors in the slits the particle interacts with the detectors as well as the barrier, and the particle ends up in either the state "went left" (and the left-hand detector triggered) or the state "went right" (and the right-hand detector triggered). Either way, there's no superposition to produce an interface pattern at the screen.

The situation you're asking about is no different. All we have to do is think of the slit in the first barrier as the source of the particle that we're sending towards the double slit in the second barrier, and we're back to the standard version of the thought experiment.

 

[Paul Howard A]

Thank you all.