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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 ?

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  • $\begingroup$ Why is there a subscript 'vac' on the Hamiltonian? When considering the light-matter Hamiltonian, how do you suppose it is in vacuum? $\endgroup$
    – Kksen
    Jul 21, 2021 at 17:19
  • $\begingroup$ @KamKahSen The electromagnetic field is typically treated as freely propagating, which is the same as if it was propagating in vacuum. The $vac$ subscript refers to this freely propagating field. Perhaps this wasn't the best choice for a subscript but I think it's meaning is clear from context. $\endgroup$
    – Hans Wurst
    Jul 21, 2021 at 18:54
  • $\begingroup$ I see. But a medium will only affects the phase velocity of the electromagnetic wave,isn't it? $\endgroup$
    – Kksen
    Jul 22, 2021 at 0:22

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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 $\alpha = 2$.

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  • $\begingroup$ I would accept your answer if you can give a reference or produce the equations that show that the amplitude gains a dependency on $n$. $\endgroup$
    – Hans Wurst
    Jul 17, 2021 at 7:16
  • $\begingroup$ @HansWurst I have updated my answer but I will try to find more references. My issue is that the refractive index is usually derived by putting the quantum stuff aside, however, that quantum stuff is exactly what we want to find out about when trying to write down the Hamiltonian here. $\endgroup$
    – Lili FN
    Jul 17, 2021 at 9:09
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    $\begingroup$ I have looked for a clear reference myself in the past and my lack of finding one was the very reason why I asked here. I still think there should be a concise way to derive it from the vector potential and by showing how the vector potential attains its dependency on $n$, then everything follows since the approximations don't change anything regarding the constants. $\endgroup$
    – Hans Wurst
    Jul 17, 2021 at 9:18

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