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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?
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