# 1.1.8. Flavor Isospin Method¶

## 1.1.8.1. Flavor Isospin Applied¶

Using flavor isospin, the equation of motion is written as

$\frac{d\vec s}{dx} = \vec{s} \times \vec H,$

where

$\begin{split}\vec s = \begin{pmatrix} \mathrm{Re}(\psi_1^*\psi_2) \\ \mathrm{Im}(\psi_1^*\psi_2) \\ (\lvert \psi_1 \rvert^2 - \lvert \psi_2 \rvert^2)/2. \end{pmatrix}\end{split}$

Background Matter Basis

In background matter basis the Hamiltonian vector is

$\begin{split}\vec H = \begin{pmatrix} \delta \lambda(x) \sin 2\theta_m \\ 0 \\ \omega_m - \delta \lambda(x) \cos 2\theta_m \end{pmatrix}.\end{split}$

For two perturbations, we write it as

$\begin{split}\vec H = \begin{pmatrix} 0 \\ 0 \\ \omega_m \end{pmatrix} + \begin{pmatrix} \delta \lambda_1(x) \sin 2\theta_m \\ 0 \\ - \delta \lambda_1(x) \cos 2\theta_m \end{pmatrix} + \begin{pmatrix} \delta \lambda_2(x) \sin 2\theta_m \\ 0 \\ - \delta \lambda_2(x) \cos 2\theta_m \end{pmatrix}.\end{split}$

The initial condition is

$\begin{split}\Psi(0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix},\end{split}$

which corresponds to a flavor isospin vector

$\begin{split}\vec s(0) = \frac{1}{2} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.\end{split}$

T-basis

In this basis, the Hamiltonian is

$\begin{split}H_1 &= -\frac{\omega_m}{2} \sigma_3 - \frac{\delta \lambda}{2} \sin 2\theta_m \begin{pmatrix} 0 & e^{2i\eta_1(x)} \\ e^{-2i\eta_1(x)} & 0 \end{pmatrix} \\ & = -\frac{\omega_m}{2} \sigma_3 +\frac{\delta \lambda}{2} \sin 2\theta_m \sin 2\eta_1(x) \sigma_2 - \frac{\delta \lambda}{2} \sin 2\theta_m \cos 2\eta_1(x) \sigma_1,\end{split}$

or

$\begin{split}H_2 &= - \frac{\delta \lambda}{2} \sin 2\theta_m \begin{pmatrix} 0 & e^{2i\eta_2(x)} \\ e^{-2i\eta_2(x)} & 0 \end{pmatrix} \\ &= \frac{\delta \lambda}{2} \sin 2\theta_m \sin 2\eta_2(x) \sigma_2 - \frac{\delta \lambda}{2} \sin 2\theta_m \cos 2\eta_2(x) \sigma_1,\end{split}$

where the background is removed from diagonal elements in $$H_1$$ but not in $$H_2$$.

The corresponding vectors are

$\begin{split}\vec H_1 = \begin{pmatrix} \delta\lambda \sin 2\theta_m \cos 2\eta_1(x) \\ -\delta\lambda \sin 2\theta_m \sin 2\eta_1(x)\\ \omega_m \end{pmatrix},\end{split}$

and

$\begin{split}\vec H_2 = \begin{pmatrix} \delta\lambda \sin 2\theta_m \cos 2\eta_2(x) \\ -\delta\lambda \sin 2\theta_m \sin 2\eta_2(x)\\ 0 \end{pmatrix}.\end{split}$

Given the initial condition in background matter basis

$\begin{split}\Psi(0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix},\end{split}$

we have to apply the T transformation to get the initial condition in the T-basis

$\begin{split}\Psi_1(0) &= \begin{pmatrix} e^{i \eta_1 (x)} & 0 \\ 0 & e^{-i \eta_1 (x)} \end{pmatrix}\Psi(0) = \begin{pmatrix} e^{i \eta_1 (x)} \\ 0 \end{pmatrix} \\ \Psi_2(0) &= \begin{pmatrix} e^{i \eta_2 (x)} & 0 \\ 0 & e^{-i \eta_2 (x)} \end{pmatrix}\Psi(0) = \begin{pmatrix} e^{i \eta_2 (x)} \\ 0 \end{pmatrix},\end{split}$

which correspond to flavor isospin vectors

$\begin{split}\vec s_1(0) = \vec s_2(0) = \vec s(0) = \frac{1}{2} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix},\end{split}$

since the T transformation is unitary.

Modes

For each mode of the multi-frequency case, the Hamiltonian is

$\begin{split}H = \frac{1}{2}\begin{pmatrix} 0 & B_N e^{i(n_i k_i -\omega_m)x} \\ B_N^* e^{-i(n_i k_i -\omega_m)x} & 0 \end{pmatrix},\end{split}$

where $$B_N$$ is either real or pure imaginary,

$\begin{split}B_N &= -(-i)^{\sum_a n_a} \tan 2\theta_m \left( \sum_a n_a k_a \right) \left( \prod_a J_{n_a}\left( \frac{A_a}{k_a}\cos 2\theta_m \right) \right)\\ & = - \tan 2\theta_m \left( \sum_a n_a k_a \right) \left( \prod_a J_{n_a}\left( \frac{A_a}{k_a}\cos 2\theta_m \right) \right) e^{-i \sum_a n_a \pi/2}\\ & = \rho_{N} e^{-i \sum_a n_a \pi/2}.\end{split}$

The Hamiltonian vector is

$\begin{split}\vec H = \begin{pmatrix} \rho_N \cos\left( (n_i k_i -\omega_m)x - \sum_a n_a \pi/2 \right) \\ -\rho_N \sin\left( (n_i k_i -\omega_m)x - \sum_a n_a \pi/2 \right) \\ 0 \end{pmatrix}.\end{split}$

## 1.1.8.2. Equilibrium Points, Linear Stability Analysis, and Limit Cycles¶

In background matter basis, the equation of motion is

$\begin{split}\frac{d}{dx}\begin{pmatrix} s_1 \\ s_2 \\ s_3 \end{pmatrix} = \begin{pmatrix} s_1 \\ s_2 \\ s_3 \end{pmatrix} \times \begin{pmatrix} \delta \lambda(x) \sin 2\theta_m \\ 0 \\ \omega_m - \delta \lambda(x) \cos 2\theta_m \end{pmatrix}.\end{split}$

Such a system is still not easy to solve. However, we can use phase portrait to get some information.

The fixed points are obtained by setting $$\vec s\times \vec H = 0 = \frac{d}{dx}\vec s$$. Even though in general we need to obtain the fixed points first before infering the linear stability, this is not needed since this equation is linear to $$\vec s$$.

The Jacobian is obtained

$\begin{split}J_{mn} & = \frac{d (\vec s\times \vec H)_m}{ds_n} \\ & = \begin{pmatrix} 0 & H_3 & -H_2\\ -H_3 & 0 & H_1 \\ H_2 & -H_1 & 0 \end{pmatrix},\end{split}$

which comes from the result

$\begin{split}\vec s\times \vec H = \begin{pmatrix} s_2 H_3 - s_3 H_2 \\ s_3 H_1 - s_1 H_3 \\ s_1 H_2 - s_2 H_1 \end{pmatrix}.\end{split}$

Plugin in the Hamiltonian in background matter basis, the eigenvalues of this Jacobian are

$\begin{split}& 0 \\ & -\frac{1}{\sqrt{2}} \sqrt{ - ( A_1 \sin (k_1 x) -\omega_m )^2 + 2 A_1 \omega_m \sin (k_1 x) (1 - \cos 2\theta_m) }\\ & \frac{1}{\sqrt{2}} \sqrt{ - ( A_1 \sin (k_1 x) -\omega_m )^2 + 2 A_1 \omega_m \sin (k_1 x) (1 - \cos 2\theta_m) }.\end{split}$

For $$- ( A_1 \sin (k_1 x) -\omega_m )^2 + 2 A_1 \omega_m \sin (k_1 x) (1 - \cos 2\theta_m) > 0$$, the eigenvalues have real parts, which means the system is a saddle point arround such equilibrium points.

| Created with Sphinx and . | | Index |