# Delayed Choice Quantum Eraser communication

1. May 27, 2013

### gespex

Hi all,

Okay, I'm writing this post keeping in mind that FTL communication (or even backward in time communication) are impossible. However, I fail to understand why this would be impossible using the delayed choice quantum eraser, so my question is: where is the error in my logic here?

Imagine a copy of the delayed choice quantum eraser, except that the mirrors that decide whether the which-path information of the photon will be erased or not is not a beamsplitter but an actual mirror that can be added or removed.
In the traditional experiment, one would add or remove this mirror and inspect the joint detection rate to find interference or the lack thereof, depending on the presence of the mirror.

For photons that don't get entangled we'll see an interference pattern, and at some point there will be no (or very little) photons due to destructive interference. If the mirrors are removed, the which-path information is removed and the entangled photons will, too, form an interference pattern. Again, at the same point there will be no (or very little) photons.
If we add the mirrors the which path information is obtained and no interference pattern will emerge. So again measuring at the same point, there will be a few photons.

Conclusion (according to my logic): If measured at the point where destructive interference happens, the presence or absence of the mirrors can be detected. According to the logic of the delayed choice quantum eraser, the decision of whether the mirrors should be added or not can be made after the measurement is made, allowing backward in time communication.

Where's the error in my logic here?

2. May 27, 2013

### Cthugha

No, this is not what you will see. All the interference patterns in this experiment are only visible in coincidence counts. Basically what you measure without coincidence counting is not one interference pattern, but the superposition of many of them, each single one corresponding to a different phase difference at the slits. The superposition of these will correspond to no pattern at all. You now need the coincidence counting to throw away all the detections which give which-way information and to single out only one of these phase differences (this depends on the position of the second detector).
If you move the second detector around, you will also notice that you will get a different interference pattern in coincidence for each detector position.

3. Sep 4, 2013

### Zafa Pi

Phase differences

In order not to see interference at D0 the phase difference at D1 and D2 must be 180°. Why is it impossible to introduce devices to make the phase difference 0? Of course a paradox would arise if it were possible.

4. Sep 5, 2013

### Cthugha

This is a bit difficult to answer without knowing what you are exactly referring to. Are you talking about the Kim et al. version of the DCQE experiment?

If so, 180° phase shift comes from the beam splitter you need to overlap the two beams (Bsc in the wikipedia image). Either beam 1 gets reflected and beam 2 transmitted (at D1) or beam 2 gets reflected and beam 1 transmitted (at D2). The phase difference between transmission and reflection now fixes the phase difference of 180°. Simple phase retarders or similar methods do not work because there is no position where you can change the phase of one beam alone. Before the beam splitter you can change the phase of beam 1 or beam 2 only (which changes the phase of both the reflected and transmitted parts of these beams). After the beam splitter you can only change the phase of reflected beam 1 and transmitted beam 2 or reflected beam 2 and transmitted beam 1 together.

5. Sep 5, 2013

### Zafa Pi

Elegant reply, thanks. You interpreted my question perfectly.
Certainly the key phase shifting must happen at BSc (where else?).
The only part where I remain confused is your sentence: "The phase difference between transmission and reflection now fixes the phase difference of 180°."
There are both reflections and transmissions coming off either side of BSc yet the spot at D0 is shifted by 180° depending on either the D1 side or the D2 side. Yet the spot happens before the idlers get to BSc. The entangling of the idler and signal must be crucial.
Maybe I need to see the math, unless you have a clear way to remove the (my) fog.

6. Sep 5, 2013

### Cthugha

I posted a somewhat simplified version of the math quite a while ago on these forums:Ancient post from 2009.

The style is somewhat broken now, but I hope it is still readable. Maybe this short sketch helps. Otherwise I am afraid you need to work out the math.