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Mar8-12, 12:45 PM
P: 230
Ok thanks.

I think I have understood how to find the selections rules.

Could you explain why the vibrational modes only describes the exited modes and not the ground state. I guess it is like that because A1g is not contained in the vibrational modes of graphite (2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g).

I thinks it's is logical as you say that the ground state is fully symmetrical (guess you could think of it as all the atoms being at their equilibrium sites, maybe with some caution), but as i said I'm not quit sure why the ground state is not in the problem, let me explain:

If I count the dimensions of the representations 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g i get 12, which I expected because there are 4 atoms in the unit cell each with three degrees of freedom. The representations tells me that I can devide these eigenfunction into sets that transform among each other, but the fully symmetric representation is not present, that is no of the eigenfunctions of the problem has the full symmetry of the problem, and hence is not the ground state. This is what i mean by it seems like the ground state has to be treated seperately.

I guess somehow that the representation theory approach only describes exited modes, but i fail to see why that is?

Hope that explains my problem a bit clearer.