# Band structure and diffraction intensity

1. Apr 17, 2013

### Talker1500

Hello,

Is there an established relationship between the diffraction intensity that a sample produces with its band structure?

Thanks

2. Apr 17, 2013

### ZapperZ

Staff Emeritus
I can't see why there should be, since diffraction probes the crystal structure, while the band structure is a property of the electronic energy-momentum states/bands. Off the top of my head, the diffraction intensity is more tied to the form factor.

Zz.

3. Apr 17, 2013

### DrDu

The diffraction intensity is determined by the Patterson function which is the autocorrelation function of charge density. Obviously charge density can be expressed in terms of the band occupation.

4. Apr 18, 2013

### Talker1500

Thanks for the answer, could you explain that a little? I'm not referring to diffraction patterns, but diffraction intensity, i.e., if I use a microscope to observe a sample, the intensity of the image I see due to diffracted light will depend on the band structure?

Last edited: Apr 18, 2013
5. Apr 22, 2013

### M Quack

Looks like ZapperZ was talking about neutron diffraction, and DrDu about x-ray diffraction.

What kind of diffraction are you talking about? Visible light? That would depend on the index of refraction and the geometrical shape of your diffracting object. The band structure would affect the real and imaginary (absorption) parts of the index of refraction.

6. Apr 23, 2013

### Talker1500

I'm referring to visible light diffraction. The sample I'm using is 2D, it's in a slide, and the sample itself is 2D, it's a layer of silicon atoms.

7. Apr 23, 2013

### ZapperZ

Staff Emeritus
OK, now I'm confused.

What experimental technique are you using here? Visible light has wavelengths of the order of 100's of nm. The lattice constant of a typical material is of the order of 10's of Angstrom. Not only are you not able to get any kind of structural info with that kind of a light source, I'm not sure you can get at the band structure either. Phonon modes via optical diagnostics, maybe, but that is an optical transmission/reflection experiment, not "diffraction".

Zz.

8. Apr 24, 2013

### Talker1500

I'm just using standard optical microscopy in dark field mode. I know I cannot get structure info with this kind of light, but I can't explain the intensity I'm seeing, so I'm looking into more exotic explanations.

9. Apr 28, 2013

### M Quack

I think that in this case "diffraction" is probably the wrong term to use. Diffraction implies coherent scattering from a regular, periodic structure (or at least a structure with a prominent Fourier component, Quasicrystals diffract very nicely).

In your case, you are looking at scattered light. That can be reflected off a surface, refracted by volumes with oblique surfaces, or diffracted by periodic structures like gratings.

Reflection and refraction depend on the index of refraction of the material. In general, this is wavelength dependent and complex, i.e. direction and amplitude of the refracted light depend on the wavelength. This wavelength dependence is determined by the electronic structure of the material, e.g. the band gap. AFAIK, however, calculating the index of refraction from the band structure and optical phonon energies is not straight forward.

I am not sure this is of any help...

10. Apr 28, 2013

### Talker1500

You're right, I should have said scattering instead of diffraction. The thing is, in dark field, I can only see the light scattered by the edges of the sample, the light diffracted and reflected by the volume is blocked before arriving to the condenser. So basically, I'm wondering why I'm able to see at all my sample, when it's only a 2D array of atoms and the light only scatters with the edges, with such a thin sample I should not be able to see it. I thought that maybe some resonance was coming into play.

11. Apr 29, 2013

### M Quack

In dark field imaging you block out all rays that are in line with the source.
You only see light that is scattered by a certain minimum angle. Beyond that,
the (secondary) source of the observed light is not restricted.

This source could be scattering off edges, but also (isotropic) fluorescence
or whatever other scattering.

It may well be that usually refracted light is not observed because the angle wrt the
incident beam is too small and the refracted light gets blocked.

12. Apr 29, 2013

### Talker1500

At the beginning I thought that the edges of the sample interacted with the incoming photon waves and affected the interference patterns, creating some kind of boost effect, but I wasn't sure this was the correct approach, that's why I started to think that maybe the electronic bands played an important role in this and then I opened this thread asking for guidance. Maybe the first approach is more accurate.