# Quantum Eraser Interference Patterns

1. Apr 17, 2013

### msumm21

I'm not clear as to how the overall interference pattern (or lack thereof) observed in the quantum eraser experiments (http://grad.physics.sunysb.edu/~amarch/ [Broken]) ends up being the same if the eraser is present or not present. My thinking follows.

Say N entangled photons are sent through the apparatus without the eraser, and then N with the eraser.

Configuration 1: Eraser is NOT present
In this configuration some M (<N) photons will end up being detected with appropriate coincidence conditions satisfied. Roughly M/2 of these cases will correspond to the photon knowingly passing through one slit and resulting in a pattern without interference, call it pattern A, on the screen. Similarly, about M/2 cases correspond to the photon knowingly passing through the other slit and resulting in pattern B. In the remaining N-M cases we can't determine which-path info and we get the usual interference pattern, call it pattern C.

Configuration 2: Eraser is present
In this configuration which-path info is never available and hence pattern C is the result after all N photons pass through the apparatus.

So, the OVERALL pattern in configuration 1 should be (M/(2N))A + (M/(2N))B + ((N-M)/N)C, but the overall pattern in configuration 2 should be C. If the overall patterns are the same then:

(M/(2N))A + (M/(2N))B + ((N-M)/N)C = C
now moving the RHS to the LHS:
(M/(2N))A + (M/(2N))B - (2M/(2N))C = 0
now multiplying by 2N/M
A + B = 2C

This seems impossible to me. Suppose the screen is setup at a distance behind the slits where the maxima in pattern C occurs on the line normal to the screen and midway between the two slits (as appears to be the case in publication by Walborn, Cunha, Padua, ...). I don't think the pattern A+B can have a maxima in that location.

I must be missing something here.

Last edited by a moderator: May 6, 2017
2. Apr 18, 2013

### Cthugha

First, as a disclaimer: The patterns are visible only in coincidence counts, not by just looking at the screen or analyzing just the detections in one arm. I am not sure, whether you already know that, but this is a common source of confusion in this experiment.

Following up on that, there is no single pattern C, but a slightly different coincident count pattern for every possible position of the detector in the p-arm (where the double slit is NOT located). The position of the detector acts roughly as a filter for the relative phase difference of the light field at the two slits of the double slit.

You can imagine that as a situation similar to having a light source which is not centered between the two slits illuminate a double slit. Depending on the position of your light source, you will see a different interference pattern. If you shine light from many directions at once, they will superpose to no pattern at all. The entangled light source basically does the same and emits photons under very different angles and from a rather large surface (well, large for optical emitters). So you will not be able to see any pattern directly. However, the coincidence detection counting circuit now acts similar to a filter. All photons arriving at the p-side under some well defined angle will also arrive under another well defined angle at the double slit. This will give some pattern in the coincidence counts as you do not have all the possible angles present anymore. However, as you move the detector at the p-side around, you will pick a different angle and will get a different interference pattern. The sum of all these is the overall or total pattern (and in summary no interference pattern at all).

3. Apr 18, 2013

### msumm21

To simplify this discussion, can we say the position of the detector in the p-arm is fixed?

In configuration 2 I meant for the pattern C to be the OVERALL pattern detected at the screen, after all N photons strike the screen (counting all photons regardless of coincidences, ...). Presumable this must converge to some fixed pattern as N approaches infinity.

Are you saying the pattern I called "C" in config 1 is not the same as the pattern I called "C" in config 2 (as N->infinity)?

Thanks

4. Apr 18, 2013

### Cthugha

Sure, we can do that. It does not make things much easier though.

The overall pattern that you will see in both configurations is just a broad lump. It looks like the image you can find in the link directly below the sentence "The coincidence counts were tallied at each detector location, as before, and it was found that indeed the interference pattern was gone". The two patterns you call "C" should be the same, if I get your description right.

Also:

Why not? The sum of two Gaussians has a maximum in the middle between them as long as their width is broader than the distance between the maxima of the single Gaussians. You can easily draw this yourself to verify it. In the experiment, the distance between the slits is 300 micrometers, so the distributions are easily broader than that.

5. Apr 18, 2013

### msumm21

OK, at least one problem I had was that I thought C was an interference pattern, not a single modal "lump" (is it a Gaussian?). I thought that because pattern C is associated with experiments where no which-path information is available. Among these experiments with no which-path information, what is it specifically that leads to the lack of an interference pattern (or how are they post-selecting cases to get an interference pattern)? Maybe they are post selecting cases in which the p-arm photon hit it's detector in a specific location, related to the point you made in your first post? Something else?

Thanks again!

6. Apr 18, 2013

### Cthugha

It is simply the lack of coherence which leads to the lack of an interference pattern. If you just shine any light on a double slit, you will not see an interference pattern, if the length scale over which the light is coherent is smaller than the distance of the slits. Try for example placing a light bulb in front of a double slit and you will see what I mean. You can easily increase the spatial coherence of your light field by filtering it. This is why some of the first double slit experiments using light from the sun used a pinhole before shining the light on the slit. A pinhole gives you an artificial point source which has high spatial coherence.

Entangled light is typically also pretty incoherent in terms of simple spatial coherence. However, entanglement allows you to create a nonlocal version of the pinhole. Putting a small detector in the second path will only detect the photons a pinhole of the same size would let through. This detected photon fraction has higher spatial coherence than the total light field. As you have entangled light, the corresponding set of photons on the other side will also have higher spatial coherence. When you do coincidence counting, you can therefore see a double slit pattern.

If you move the detector on the other side, you will get a different pattern in coincidence counting. This is equivalent to moving the pinhole around in a "standard" double slit experiment using light from the sun.

7. Apr 18, 2013

### msumm21

Could you get around this issue by placing something like the pinhole well in front of the quarter wave plates on the s-arm. I.e. reapply my whole argument above but with this new pinhole added to both configurations so that we're only seeing photons that made it through the pinhole? Or maybe that destroys the polarization-entanglement of the photons?