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I'm studying the linear electric susceptibility, using Schroedinger equation and perturbation theory of the interaction potential

$$V=-\mu \cdot E$$

and the book arrive to an expression where

$$\mu_{mn}=\langle \psi_m\mid\hat\mu\mid\psi_n\rangle$$ is a "parameter" of the problem.

My question is: which theory describes and calculates $\hat\mu$?

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What do you mean "which theory describes" ? This is the standard electric dipole moment in Electrodynamics. – Chris Gerig Dec 6 '11 at 11:39
Ok, so you are saying that I can use the multipolar expansion to explicitly obtain $\mu$? – J. C. Leitão Dec 6 '11 at 11:47
up vote 1 down vote accepted

The idea is this. You will start from an action in the form

$$S_I=\int d^4x j_\mu A^\mu$$

and for a system of charge you will write down

$$j_\mu=(\rho(x),\rho(x)\dot{\bf x}).$$

An integration by part in time will do the job producing the contribution

$$S'_I=-\int d^4x \rho(x){\bf x}\cdot {\bf E}$$

using the standard relation ${\bf E}=-\partial{\bf A}/\partial t-\nabla\phi$ and using is made of the continuity equation. You can see classically that the contribution is the standard dipolar interaction. With a set of pointlike charges, $\rho(x)=\sum_aq_a\delta^3(x-x_a)$, you can integrate in the volume and you will be left with the interaction term

$$V=-\sum_aq_a{\bf x}_a\cdot{\bf E(x_a,t)}.$$

This translate as is into quantum mechanics and finally

$${\hat\mu}_{mn}=\sum_aq_a\langle\psi_m|{\bf x}_a|\psi_n\rangle.$$

In this case we are assuming that the e.m. field is an external field.

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