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Molecules with dipole moments

  1. Oct 10, 2006 #1


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    It can be shown that if a hamiltonian is invariant under space inversion and if an eigenstate is non-degenerate, then the state is either even or odd in each position coordinate. So if there is a perturbation given by a uniform electric field, which has:

    [tex]H_1 = - \vec E \cdot \left( \sum_i q_i \vec r_i \right) [/tex]

    then there will be no energy shift to first order, since the square of the wavefunction is even in each coordinate, so the integral over each position coordinate will vanish, and so the expectation value of H_1 will be zero. But there are molecules with permanent dipole moments, like water, and these will have energy shifts linear in the electric field, so there must be some degeneracy.

    I'm also told that if the electric field is below a certain magnitude, the shift will no longer be linear. I'm guessing this is because the degeneracy is not perfect, and once <H_1> is of the order of the splitting the degeneracy is no longer important. I'm having a hard time thinking about eigenstates of an entire molecule, and how there could be slight degeneracies. Can anyone help me understand this?
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  3. Oct 11, 2006 #2


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    I think I've got it. The electrons in the H2O molecule can tunnel from one H atom to another, and this introduces a small tunelling splitting (ie, between a symmetric and an antisymmetric combination of s-orbitals around each atom). Moving the electron from one H atom to another is equivalent to rotating the atom, and so the magnitude of the splitting is on the order of the rotational energy levels of the molecule (this part sounds sketchy). When the perturbation is much larger than this splitting, the states look degenerate, so there can be a dipole moment, while when the perturbation is less than the splitting, the states look non-degenerate and so the energy shift is no longer linear in the electric field. Can someone let me know if this sounds right?
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