Dipole - Atomic Dipoles

Atomic Dipoles

A non-degenerate (S-state) atom can have only a zero permanent dipole. This fact follows quantum mechanically from the inversion symmetry of atoms. All 3 components of the dipole operator are antisymmetric under inversion with respect to the nucleus,

where is the dipole operator and is the inversion operator. The permanent dipole moment of an atom in a non-degenerate state (see degenerate energy level) is given as the expectation (average) value of the dipole operator,

\langle \mathfrak{p} \rangle = \langle\, S\, | \mathfrak{p} |\, S \,\rangle,

where is an S-state, non-degenerate, wavefunction, which is symmetric or antisymmetric under inversion: . Since the product of the wavefunction (in the ket) and its complex conjugate (in the bra) is always symmetric under inversion and its inverse,

\langle \mathfrak{p} \rangle = \langle\, \mathfrak{I}^{-1}\, S\, | \mathfrak{p} |\, \mathfrak{I}^{-1}\, S \,\rangle = \langle\, S\, | \mathfrak{I}\, \mathfrak{p} \, \mathfrak{I}^{-1}| \, S \,\rangle = -\langle \mathfrak{p} \rangle

it follows that the expectation value changes sign under inversion. We used here the fact that, being a symmetry operator, is unitary: and by definition the Hermitian adjoint may be moved from bra to ket and then becomes . Since the only quantity that is equal to minus itself is the zero, the expectation value vanishes,

\langle \mathfrak{p}\rangle = 0.

In the case of open-shell atoms with degenerate energy levels, one could define a dipole moment by the aid of the first-order Stark effect. This gives a non-vanishing dipole (by definition proportional to a non-vanishing first-order Stark shift) only if some of the wavefunctions belonging to the degenerate energies have opposite parity; i.e., have different behavior under inversion. This is a rare occurrence, but happens for the excited H-atom, where 2s and 2p states are "accidentally" degenerate (see this article for the origin of this degeneracy) and have opposite parity (2s is even and 2p is odd).

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