Neutrino Interaction with Matter
Flavor Basis
In terms of formalism, vacuum oscillations is already a Rabi oscillation at resonance with oscillation width $\omega_\vv \sin 2\theta_\vv$. As derived, neutrino oscillations in matter are determined by Hamiltonian in flavor basis $$ \begin{equation} H^{(\ff)} = \left(- \frac{1}{2} \omega_\vv \cos 2\theta_\vv +\frac{1}{2}\lambda(x) \right)\sigma_3 + \frac{1}{2} \omega_\vv \sin 2\theta_\vv \sigma_1, \end{equation} $$ with the Schröding equation $$ \begin{equation} i \partial_x \Psi^{(\ff)} = H^{(\ff)} \Psi^{(\ff)}. \end{equation} $$ To make connections to Rabi oscillations, we would like to remove the changing $\sigma_3$ terms, using a transformation $$ \begin{equation} U = \begin{pmatrix} e^{-i \eta (x)} & 0 \\ 0 & e^{i \eta (x)} \end{pmatrix}, \end{equation} $$ which transform the flavor basis to another basis $$ \begin{equation} \begin{pmatrix} \psi_e \\ \psi_x \end{pmatrix} = \begin{pmatrix} e^{-i \eta (x)} & 0 \\ 0 & e^{i \eta (x)} \end{pmatrix} \begin{pmatrix} \psi_{a} \\ \psi_{b} \end{pmatrix}. \end{equation} $$ The Schrodinger equation can be written into this new basis $$ \begin{equation} i \partial_x (T \Psi^{(r)}) = H^{(\ff)} T\Psi^{(r)}, \end{equation} $$ which is simplified to $$ \begin{equation} i \partial_x \Psi^{(r)} = H^{(r)} \Psi^{(r)}, \end{equation} $$ where $$ \begin{equation} H^{(r)} = - \frac{1}{2}\omega_\vv \cos 2\theta_\vv \sigma_3 + \frac{1}{2} \omega_\vv \sin 2\theta_\vv \begin{pmatrix} 0 & e^{2i\eta(x)} \\ e^{-2i\eta(x)} & 0 \\ \end{pmatrix}, \end{equation} $$ in which we remove the varying component of $\sigma_3$ elements using $$ \begin{equation} \frac{d}{dx}\eta(x) = \frac{\lambda(x)}{2}. \end{equation} $$ The final Hamiltonian would have some form $$ \begin{equation} H^{(r)} = - \frac{1}{2}\omega_\vv \cos 2\theta_\vv \sigma_3 + \frac{1}{2} \omega_\vv \sin 2\theta_\vv \begin{pmatrix} 0 & e^{i\int_0^x \lambda(\tau)d\tau + 2i\eta(0)} \\ e^{-i\int_0^x \lambda(\tau)d\tau - 2i\eta(0)} & 0 \\ \end{pmatrix}, \end{equation} $$ where $\eta(0)$ is chosen to counter the constant terms from the integral.
For arbitrary matter profile, we could first apply Fourier expand the profile into trig function then use Jacobi-Anger expansion so that the system becomes a lot of Rabi oscillations. Any transformations or expansions that decompose $\exp{\left(i\int_0^x \lambda(\tau)d\tau\right)}$ into many summations of $\exp{\left( i a x + b \right)}$ would be enough for an Rabi oscillation interpretation. As for constant matter profile, $\lambda(x) = \lambda_0$, we have $$ \begin{equation} \eta(x) = \frac{1}{2} \lambda_0 x. \end{equation} $$ The Hamiltonian becomes $$ \begin{equation} H^{(r)} = - \frac{1}{2}\omega_\vv \cos 2\theta_\vv \sigma_3 + \frac{1}{2} \omega_\vv \sin 2\theta_\vv \begin{pmatrix} 0 & e^{i\lambda_0 x} \\ e^{-i\lambda_0 x} & 0 \\ \end{pmatrix}, \end{equation} $$ which is exactly a Rabi oscillation. The resonance condition is $$ \begin{equation} \lambda_0 = \omega_\vv \cos 2\theta_v. \end{equation} $$
Instantaneous Matter Basis
Neutrino oscillations can be calculated in instantaneous matter basis, where the Schrödinger equation is transformed to instantaneous matter basis by applying a rotation $U$, $$ \begin{equation} i \partial_x \left( U\Psi^{(\mm)} \right)= H^{(\ff)} U\Psi^{(\mm)}, \end{equation} $$ where $$ \begin{equation} U = \begin{pmatrix} \cos \theta_m & \sin \theta_m \\ -\sin\theta_m & \cos \theta_m \end{pmatrix}. \end{equation} $$ With some simple algebra, we can write the system into $$ \begin{equation} i \partial _x \Psi^{(\mm)} = H^{(\mm)}\Psi^{(\mm)} , \end{equation} $$ where $$ \begin{equation} H^{(\mm)} = U^\dagger H^{(\ff)} U - i U^\dagger \partial_x U. \end{equation} $$ By setting the off-diagonal elements of the first term $U^\dagger H^{(\ff)} U$ to zero, we can derive the relation $$ \begin{equation} \tan 2\theta_m = \frac{\sin 2\theta_\vv}{\cos 2\theta_\vv - \lambda/\omega_\vv}. \end{equation} $$ Furthermore, we derive the term $$ \begin{equation} i U^\dagger \partial_x U = - \dot\theta_m \sigma_2. \end{equation} $$ We can calculate $\dot\theta_m$ by taking the derivative of $\tan 2\theta_m$, $$ \begin{equation} \frac{d}{dx} \tan 2\theta_m = \frac{2}{\cos^2 2\theta_m} \dot\theta_m, \end{equation} $$ so that $$ \begin{align} \dot\theta_m &= \frac{1}{2} \cos^2 (2\theta_m) \frac{d}{dx} \tan 2\theta_m \\ & = \frac{1}{2} \frac{(\cos 2\theta_\vv - \lambda/\omega_v)^2}{ (\lambda/\omega_v)^2 + 1 - 2\lambda \cos 2\theta_\vv /\omega_\vv } \frac{d}{dx} \frac{\sin 2\theta_\vv}{\cos 2\theta_\vv - \lambda/\omega_\vv} \\ & = \frac{1}{2} \frac{(\cos 2\theta_\vv - \lambda/\omega_v)^2}{ (\lambda/\omega_v)^2 + 1 - 2\lambda \cos 2\theta_\vv /\omega_\vv } \frac{\sin 2\theta_\vv}{(\cos 2\theta_\vv - \lambda/\omega_v)^2} \frac{1}{\omega)v} \frac{d}{dx} \lambda(x) \\ & = \frac{1}{2} \sin 2\theta_m \frac{1}{\omega_m} \frac{d}{dx} \lambda(x). \end{align} $$