View Single Post
Mar8-12, 10:12 AM
P: 230

My question is about the ground state of vibrations for a solid. I'm working with graphite and have found out that for k=0 (The Gamma symmetry point), the vibrational modes can be decomposed into irreducible represenations in the following way

Vibration = 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g

To calculate which modes are raman active I need to know the ground state symmetry. Is this always the totally symmetric represenation (A1g) even though it does not belong to the vibrational modes? I think i heard someone mention that the ground state needs to be treated seperately for some reason, so that it is not a problem that A1g not is in the vibrational modes.

I've sone the calculation with A1g as the ground state in this case i get that only E2g is raman active (which is the right result)

Doing the calculation with E2g (which has an energy going towards 0 for k=0) I get the wrong result, so I guess A1g must be the groudn state symmetry.

So my questions are:
- Why do we need to treat the ground state seperately?
- If not which is the ground state of graphite?

Additional question:
- I've also read that the ground state usually have the full symmetry of the solid. When is this not the case?
Phys.Org News Partner Physics news on
Researchers study gallium to design adjustable electronic components
Gadolinium-based material that can be cooled by varying magnetic field
Experiments explain why some liquids are 'fragile' and others are 'strong'