Neutron interferometry

[stax]

Hi,

 

I have a question on neutron interferometry. If I have a setup like in this figure , I would expect to measure the same signal at both detectors. However, this actually does not seem to be the case, as claimed for example here.

 

My analysis of the situation would be:

 

for the upper detector:

interference of a two beams, one is reflected only at the mirror, which adds an addition pi to the phase accumulated during propagation. The other beam is reflected three times and picks up the phase change phi induced by the sample, so the phase difference between the two beams should be 2*pi + phi = phi

 

for the lower detector:

upper beam is reflected twice and picks up phase difference from sample. Lower beam is also reflected twice. Phase difference between the two beams is again phi.

 

Where did I go wrong? They claim on the page I linked and in every other source, that the signal change in the two detectors has opposite sign. How does this work?

 

[edit]: btw. same problem when I use x-rays in this setup

 

[edit2]: an idea to resolve this problem: the Mach-Zehnder-interferometer is basically the same setup. There's just one significant difference. On reflection at the inside of the glass, there is no phase change, and so the phase differences measured by the two detectors always differ by pi. This is exactly what I would need to explain the result in the neutron interferometer. I just can't see, where I can introduce a reflection without phase change...

[Klaynos]

Are your ray's drawn correctly? Because I'm sure that momentum can't be conserved like that?

[stax]

Momentum doesn't need to be conserved, at least not for the rays alone since the whole setup is fixed and will take up everything that's left over. It's simple Bragg-reflection. The only thing is that I would expect two additional beams going straight through the "mirror". Or do I misunderstand your statement?

[Klaynos]

I'm talking about the two beams after the splitter. But it has been a long time since I looked at this.

 

I can't recall ever seeing and incident beam of anything do have a -ve trasmission angle, even with bragg reflections

[stax]

I'm pretty sure that it is correct like this. The only thing neglected in this drawing is the fact that all the beams have a lateral extension and that the Splitter, Mirror and Analyzer have a certain thickness. Maybe a better drawing would show one beam going straight through the BS and the other one branching off of the former right in the middle of the BS. But then you'd have to take into account the refraction index of the material for neutrons and so on. I'm pretty sure there's no problem here, all the sources I could find draw basically the same picture.

[swansont]

It's a Bonse-Hart interferometer (though topologically it is the same as a Mach-Zehnder), and these are transmission materials, so there is no reflection — it's diffraction. The interferometer is made of a crystal, probably Si, and what you are seeing are (some of the) diffraction orders off of the crystal planes. The analyzer plane can translate, so it would block or not block the resultant beam, depending on whether the interference was constructive or destructive. Then you do things to one arm of the device to introduce a phase change, and measure it.

 

Diffraction means no phase change. It's all based on path length.

 

A similar apparatus is used for atom interferometry, with the gratings etched so they have gaps in them (I used silicon nitride for the ones I made)

[stax]

Hm, I just returned from the library, read an article bei H. Rauch, where he shows that there is a phase shift. But even if he's wrong, without a phase shift I don't understand the result as well. Because then again, both detectors should detect the same signal, and not opposite ones.

 

Is it possible that there is a phase change here and it plays a role that we're dealing with fermions, so that the 2\pi-phase shift shifts the neutron only in its "-"-Position and there's destructive interference? I'm pretty sure it shouldn't, but I just can't think of another solution.

 

If you look at this setup without the sample in it, would you expect the detectors to measure the same signal?