Can this classical double-slit experiment duplicate loss of interference?

[JesseM]

Not sure whether to put this in the QM forum or the classical forum, you'll see why--

I was thinking about the http://grad.physics.sunysb.edu/~amarch/ version of the double slit experiment, in which the information about which slit the particle went through is obtained in a somewhat unusual way; instead of actually measuring the particle as it goes through the slits, two polarizing filters at different angles are placed in front of the slits, so that only photons of one polarization can get through the left slit, and only photons of a different polarization can get through the right. As described in the link:

[PLAIN]http://grad.physics.sunysb.edu/~amarch/PHY5656.gif

To make the "which-way" detector, a quarter wave plate (QWP) is put in front of each slit. This device is a special crystal that can change linearly polarized light into circularly polarized light. The two wave plates are set so that given a photon with a particular linear polarization, one wave plate would change it to right circular polarization while the other would change it to left circular polarization.

With this configuration, it is possible to figure out which slit the s photon went through, without disturbing the s photon in any way. Because the s and p photons are an entangled pair, if we measure the polarization of p to be x we can be sure that the polarization of s before the quarter wave plates was y. QWP 1, which precedes slit 1, will change a y polarized photon to a right circularly polarized photon while QWP 2 will change it to a left circularly polarized photon. Therefore, by measuring the polarization of the s photon at the detector, we could determine which slit it went through. The same reasoning holds for the case where the p photon is measured to be y. The following table provides a summary.

This experiment involves measuring both members of an entangled pair, which is important for the later "erasure" phase of the experiment. But for the sake of this post, suppose we just sent non-entangled photons through the double slit. Without the polarizing filters (the "quarter wave plates") you should see interference on the screen, while with the filters in place you won't see an interference pattern, correct?

If so, I was wondering if the correspondence principle means that even if you analyze this experiment in the context of classical electromagnetism, you would still have the result that with no filters you'd see an interference pattern (in the intensity of the waves hitting the screen, since obviously there are no individual photons in classical EM), while with the filters you would see the interference destroyed. My thinking was that the classical limit should be approached when you send large numbers of photons through the slits, and in QM the intensity at a given point on the screen for a large number of photons is proportional to the probability that any individual photon lands at that point. I asked my professor, who said that probably this wouldn't work because the size of the slits still needs to be very small in order to get an interference pattern, and the correspondence principle may require that you're dealing with large distances along with large numbers of particles. But my professor hadn't done any detailed calculations here, or seen this specific problem analyzed before...does anyone here have an opinion?

 

[Greg Bernhardt]

@JesseM did you find any more insight on this topic?