Derivation of the refractive index dependency of the light-matter interaction hamiltonian

Question

I tried to derive a relation of the light-matter interaction Hamiltonian in vacuum and a medium that is characterized by a linear real refractive index but I am not sure if the following is correct or if this approach is valid at all. My result is given in a boxed equation at the bottom but I thought it is sensible to show my approach instead of simply asking whether the result is correct.

The time averaged (over 1 period) energy density $u$ of a electromagnetic wave in a homogenous medium characterized by its permittivity $\varepsilon=\varepsilon_r\varepsilon_0$ is

$$ u = \frac{1}{2}\varepsilon |E_0|^2 $$ where the electric field was assumed to be $\vec E(r,t)=E_0\cos(\vec k\cdot \vec r-\omega t)\hat e$.

Let the electric field amplitude be connected to the vector potential via $$ \vec E(r,t) = -\frac{\partial }{\partial t} \vec A(r,t) $$ Asuming for the vector potential the form $\vec A(r,t) = -A_0\sin(kr-\omega t)\hat e$, we obtain for the amplitudes the relation $ E_0 = \omega A_0. $ Inserting this int the expressoin for the energy density yields $$ u=\frac{1}{2}\varepsilon\omega ^2|A_0|^2 $$ Now we connect this equation to the an expression for the energy density based on $N$ quantized photons in Volume $V$, $$ u = N\frac{\hbar \omega}{V} =\frac{1}{2}\varepsilon\omega ^2|A_0|^2 $$ and rearrange for $A_0$, $$ A_0 = \sqrt \frac{2N\hbar }{V\omega \varepsilon} $$

Now I want to establish a dependency on the refractive index. We have $$n^2 = \varepsilon_r\mu_r$$. We assume that $\mu_r \approx 1$ such that $$ n^2\approx \varepsilon_r $$

Inserting this into the equation via $\varepsilon = \varepsilon_r \varepsilon_0$ for the vector potential amplitude yields, $$ A_0 = \sqrt \frac{2N\hbar }{V\omega \varepsilon_0} \cdot \frac{1}{n} $$

Can I conclude from this the following equations for the light-matter interaction Hamiltonian $$\boxed{\begin{aligned} \hat V_{medium} \propto \frac{e}{m} \hat p\cdot A_{medium} &= \frac{e}{m} \hat p\cdot A_{vac} \frac{1}{n}\\ \approx -\hat \mu \cdot E_{medium} &= - \frac{1}{n} \hat \mu \cdot E_{vac} \end{aligned}}$$

where the second line is the light-matter interaction Hamiltonian within common approximations typically used when describing the interaction of molecules with electromagnetic waves.

This is a follow up question to this.

Answer 1

The derivation is valid but it misses one very important aspect. The electric field at the interacting molecule/system is different from the macroscopic field in the isotropic medium characterized by $n$. A way to model this is the so called cavity factor. This factor establishes a proportionality between the macroscopic field and the local field responsible for the transition. The relation in an isotropic medium is $$ \vec E^{loc} = f \vec E^{macr} $$ The cavity factor is itself a function of the refractive index. There are many models for the exact functional form. In the case of a freely rotating molecule in solution, one can use the so called Empty Spherical Cavity Model, which defines $$ f(n)=\frac{3n^2}{2n^2 +1} $$

Assuming Fermi's Golden rule to calculate a transition rate, one typically has terms of the form $$ \Gamma \propto \langle |\mu\cdot E^{local}|^2\rangle $$ Here we need to replace the local field by the macroscopic field which yields $$ \Gamma \propto f(n)^2\langle |\mu\cdot E^{macr}|^2\rangle $$ in addition to any other factors of $n$ due to the field depenendency as derived in the question.

Things are more complicated if the medium is not isotropic and if the molecule doing the transition is interacting with the solute (for example hydrogen bonds) such that certain orientations are favored. In this case the Empty Spherical Cavity model is no longer the best choice.

An excellent overview of the topic is given here,

Toptygin, D. Effects of the Solvent Refractive Index and Its Dispersion on the Radiative Decay Rate and Extinction Coefficient of a Fluorescent Solute. Journal of Fluorescence 13, 201–219 (2003)