Refractive index and light-matter interaction Hamiltonian

Question

I am wondering if the light-matter Hamiltonian obtains a dependency on the refractive index if we insert our system into a homogeneous medium that can be characterized by a scalar refractive index $n$.

Lets assume we work in the dipole approximation and treat the field classically. In this case the coupling term to a electromagnetic wave reduces to $$ H_{vac}(t) = -e \mathbf{\hat r} \cdot \mathbf E_{vac}(t)\\ \mathbf E_{vac}(t) =\boldsymbol{\epsilon} E_0\cos(\omega t) $$ where $E_0$ is the scalar amplitude of the electric field, $\boldsymbol \epsilon$ is the normalized polarization vector and $\omega$ is the angular frequency of the incident wave.

What happens in a medium with $n$ ? Are the Hamiltonians connected by a simple proportionality factor of $n$, like this $$ H_{medium}(t) \propto n^\alpha H_{vac}(t). $$ If this is appropriate, which values takes $\alpha$ and how is it derived ? How does $n$ enter the equation ?

Answer 1

I think the only way in which the refractive index enters the picture is through the amplitude of the field at a point. This makes intuitive sense, as the light-matter interaction depends on the amplitude and the frequency of the light, and $\omega$ does not depend on the refractive index.

The oscillating electric field will give rise to a polarization field - the strength of this field can be computed by treating the atom as a driven harmonic oscillator (https://www.feynmanlectures.caltech.edu/II_32.html). Note that the $\epsilon_0 \epsilon_r$ is dependent on the frequency of the incident radiation.

In the dipole model, the Hamiltonian assumes this form because the potential energy depends on the separation distance and the total field between the electron and the nucleus ($V = qE$). The polarization field, $\vec{P}$ is produced as a result of the separation between the electron and the nucleus, which is brought about by the effect of the incident light. However, when we talk about the polarization field within a material at a point, it is not the field produced by the atom at that point, but by all the other surrounding particles (see the part in the reference about the Clausius-Mossotti equation). So the total field is the sum of $\vec{P}$ and $\vec{E_{rad}}$.

Since the Schroedinger equation is linear, assuming $\epsilon_r$, and therefore $n$, is uniform within the material, $\hat H$ is proportional to the electric displacement, $\vec{D} = \epsilon_r \epsilon_0 \vec{E}$, and since $n = \sqrt{\epsilon_0 \epsilon_r}$, this would make $\alpha = 2$.