Pauli Matrices and Rotations
Given a rotation
\[\begin{split}U = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin\theta & \cos \theta \end{pmatrix},\end{split}\]
its effect on Pauli matrices are
\[\begin{split}U^\dagger \sigma_3 U &=\cos 2\theta \sigma_3 + \sin 2\theta \sigma_1 \\
U^\dagger \sigma_1 U & = -\sin 2\theta \sigma_3 + \cos 2\theta \sigma_1.\end{split}\]
1.1.2.1. Flavor Basis
Vacuum Oscillations
Vacuum oscillations is already a Rabi oscillation at resonance with oscillation width \(\omega_v \sin 2\theta_v\).
Neutrino oscillation in matter has a Hamiltonian in flavor basis
\[H^{(f)} = \left(- \frac{1}{2} \omega_v \cos 2\theta_v +\frac{1}{2}\lambda(x) \right)\sigma_3 + \frac{1}{2} \omega_v \sin 2\theta_v \sigma_1.\]
The Schroding equation is
\[i \partial_x \Psi^{(f)} = H^{(f)} \Psi^{(f)}.\]
To make connections to Rabi oscillations, we would like to remove the changing \(\sigma_3\) terms, using a transformation
\[\begin{split}T = \begin{pmatrix} e^{-i \eta (x)} & 0 \\ 0 & e^{i \eta (x)} \end{pmatrix},\end{split}\]
which transform the flavor basis to another basis
\[\begin{split}\begin{pmatrix} \psi_e \\ \psi_x \end{pmatrix} = \begin{pmatrix} e^{-i \eta (x)} & 0 \\ 0 & e^{i \eta (x)} \end{pmatrix} \begin{pmatrix} \psi_{a} \\ \psi_{b} \end{pmatrix}.\end{split}\]
The Schrodinger equation can be written into this new basis
\[i \partial_x (T \Psi^{(r)}) = H^{(f)} T\Psi^{(r)},\]
which is simplified to
\[i \partial_x \Psi^{(r)} = H^{(r)} \Psi^{(r)},\]
where
\[\begin{split}H^{(r)} = - \frac{1}{2}\omega_v \cos 2\theta_v \sigma_3 + \frac{1}{2} \omega_v \sin 2\theta_v \begin{pmatrix}
0 & e^{2i\eta(x)} \\
e^{-2i\eta(x)} & 0 \\
\end{pmatrix},\end{split}\]
in which we remove the varying component of \(\sigma_3\) elements using
\[\frac{d}{dx}\eta(x) = \frac{\lambda(x)}{2}.\]
The final Hamiltonian would have some form
\[\begin{split}H^{(r)} = - \frac{1}{2}\omega_v \cos 2\theta_v \sigma_3 + \frac{1}{2} \omega_v \sin 2\theta_v \begin{pmatrix}
0 & e^{i\int_0^x \lambda(\tau)d\tau + 2i\eta(0)} \\
e^{-i\int_0^x \lambda(\tau)d\tau - 2i\eta(0)} & 0 \\
\end{pmatrix},\end{split}\]
where \(\eta(0)\) is chosen to conter the constant terms from the integral.
For arbitary 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.
Let’s discuss the constant matter profile, \(\lambda(x) = \lambda_0\). Thus we have
\[\eta(x) = \frac{1}{2} \lambda_0 x.\]
The Hamiltonian becomes
\[\begin{split}H^{(r)} = - \frac{1}{2}\omega_v \cos 2\theta_v \sigma_3 + \frac{1}{2} \omega_v \sin 2\theta_v \begin{pmatrix}
0 & e^{i\lambda_0 x} \\
e^{-i\lambda_0 x} & 0 \\
\end{pmatrix},\end{split}\]
which is exactly a Rabi oscillation. The resonance condition is
\[\lambda_0 = \omega_v \cos 2\theta_v.\]
1.1.2.2. Instanteneous Matter Basis
Neutrino oscillation in matter has a Hamiltonian in flavor basis
\[H^{(f)} = \left(- \frac{1}{2} \omega_v \cos 2\theta_v +\frac{1}{2}\lambda(x) \right)\sigma_3 + \frac{1}{2} \omega_v \sin 2\theta_v \sigma_1.\]
The Schroding equation is
\[i \partial_x \Psi^{(f)} = H^{(f)} \Psi^{(f)},\]
which can be transformed to instantaneous matter basis by applying a rotation \(U\),
\[i \partial_x \left( U\Psi^{(m)} \right)= H^{(f)} U\Psi^{(m)},\]
where
\[\begin{split}U = \begin{pmatrix} \cos \theta_m & \sin \theta_m \\ -\sin\theta_m & \cos \theta_m \end{pmatrix}.\end{split}\]
With a little algebra, we can write the system into
\[i \partial _x \Psi^{(m)} = H^{(m)}\Psi^{(m)}\]
\[H^{(m)} = U^\dagger H^{(f)} U - i U^\dagger \partial_x U.\]
By setting the off-diagonal elements of the first term \(U^\dagger H^{(f)} U\) to zero, we can derive the relation
\[\tan 2\theta_m = \frac{\sin 2\theta_v}{\cos 2\theta_v - \lambda/\omega_v}.\]
Furthermore, we derive the term
\[i U^\dagger \partial_x U = - \dot\theta_m \sigma_2.\]
We can calculate \(\dot\theta_m\) by taking the derivative of \(\tan 2\theta_m\),
\[\frac{d}{dx} \tan 2\theta_m = \frac{2}{\cos^2 2\theta_m} \dot\theta_m,\]
so that
\[\begin{split}\dot\theta_m &= \frac{1}{2} \cos^2 (2\theta_m) \frac{d}{dx} \tan 2\theta_m \\
& = \frac{1}{2} \frac{(\cos 2\theta_v - \lambda/\omega_v)^2}{ (\lambda/\omega_v)^2 + 1 - 2\lambda \cos 2\theta_v /\omega_v } \frac{d}{dx} \frac{\sin 2\theta_v}{\cos 2\theta_v - \lambda/\omega_v} \\
& = \frac{1}{2} \frac{(\cos 2\theta_v - \lambda/\omega_v)^2}{ (\lambda/\omega_v)^2 + 1 - 2\lambda \cos 2\theta_v /\omega_v } \frac{\sin 2\theta_v}{(\cos 2\theta_v - \lambda/\omega_v)^2} \frac{1}{\omega)v} \frac{d}{dx} \lambda(x) \\
& = \frac{1}{2} \sin 2\theta_m \frac{1}{\omega_m} \frac{d}{dx} \lambda(x).\end{split}\]