1
$\begingroup$

In electrostatics, the polarization vector $\vec{P}$ is introduced as the sum of dipoles in materials. But from the viewpoint of microscopic, the dipole is ill-defined due to all materials are discretized. So how can we understand the origin of the polarization vector $\vec{P}$ with consideration of the nature of the discretization of the real materials?

$\endgroup$

1 Answer 1

2
$\begingroup$

You are right, that the medium is made of quantum systems which have to be treated accordingly. When you consider every individual molecule/atom in your material the induced polarization at that individual molecule/atom is given by the induced dipole moment $d(t)$, which is well defined via

$$ d(t) = \langle \psi(t) | \hat d | \psi(t) \rangle, $$

where $\hat d$ is the dipole operator. Thus the polarization of your medium is

$$ P (t) = \sum_i N_i \langle \psi_i(t) | \hat d | \psi_i(t) \rangle, $$

where we have introduced the number density of the molecules/atoms $N_i$ in the state $|\psi_i(t)\rangle$ ($N_i$ is typically Boltzmann-distributed). And the sum $\sum_i$ accounts for the initial incoherent distribution of different states to describe the overall polarization averaged over this distribution.

At a side note, I also want to mention that the material may be discrete, but, in classical electrostatics your polarization, as the sum of dipoles, is already discrete. A dipole essentially consist of two opposite charges at some distance, which in this sense is already a discrete distribution of charges. And microscopically the induced dipole is well defined through the expectation value of your dipole operator. I hope that this addresses your concern.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.