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Mar10-12, 09:58 AM
Sci Advisor
P: 3,571
Quote Quote by mrandersdk View Post
Is it because the ground state is no vibration at all, which is not threated in this problem? If this is the case then it is clear that it should have the full symmetry of the problem e.g. A1g. But do the state "no vibration" exist in quantum mechanics?
Yes, exactly. that's what I wrote in post #7.
After introduction of the normal coordinates, the qm problem reduces to that of a product of harmonic oscillator eigenstates. The ground state is a product of gaussian functions and can easily be seen to be totally symmetric. The states with one quantum correspond (up to a normalization constant) to one where the ground state function is multiplied by q_i, the coordinate of the ith normal coordinate which transforms according to one of the irreps you wrote down.