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.

../../_images/fixed-points-eigenvalues.png

Fig. 1.35 Eigenvalues of Jacobian and fixed points. Source: Stability Analysis for ODEs by Marc R. Roussel


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