# 1.1.2. MSW Effect Revisted¶

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

$\begin{split}\dot\theta_m &= \frac{1}{2} \cos^2 (2\theta_m) \frac{d}{dx} \tan 2\theta_m \\ & = \frac{1}{2} \frac{(\cos 2\theta_v - \lambda/\omega_v)^2}{ (\lambda/\omega_v)^2 + 1 - 2\lambda \cos 2\theta_v /\omega_v } \frac{d}{dx} \frac{\sin 2\theta_v}{\cos 2\theta_v - \lambda/\omega_v} \\ & = \frac{1}{2} \frac{(\cos 2\theta_v - \lambda/\omega_v)^2}{ (\lambda/\omega_v)^2 + 1 - 2\lambda \cos 2\theta_v /\omega_v } \frac{\sin 2\theta_v}{(\cos 2\theta_v - \lambda/\omega_v)^2} \frac{1}{\omega)v} \frac{d}{dx} \lambda(x) \\ & = \frac{1}{2} \sin 2\theta_m \frac{1}{\omega_m} \frac{d}{dx} \lambda(x).\end{split}$

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