Coriolisが無い場合 - keisuke araki's project

Coriolisが無い場合

動機：Coriolisが無いときにわざと回転と外部磁場がフルの固有ベクトルで行列を作るとどうなるか？

内積（同じ$ s ）

$ \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize \AR{r}{ \sigma s \lambda^s \bm\phi_\sigma(-\vec{k}) \\ \DS \sigma k \bm\phi_\sigma(-\vec{k}) }^T \AR{cc}{ 0 & - \bm B_0 \times \phantom{\big|}\!\! \\ \phantom{\big|}\!\! - \bm B_0 \times & \alpha \bm B_0 \times } \AR{r}{ \sigma s \lambda^s \bm\phi_\sigma(\vec{k}) \\ \DS \sigma k \bm\phi_\sigma(\vec{k}) }

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize \AR{r}{ \sigma s \lambda^s \bm\phi_\sigma(-\vec{k}) \\ \DS \sigma k \bm\phi_\sigma(-\vec{k}) }^T \AR{r}{ - \sigma k (\bm B_0 \times \bm\phi_\sigma(\vec{k}) ) \\ \DS - \sigma s \lambda^{s} (\bm B_0 \times \bm\phi_\sigma(\vec{k}) ) + \alpha \sigma k (\bm B_0 \times \bm\phi_\sigma(\vec{k}) ) }

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize (\sigma s \lambda^{s})(\sigma k) \frac{\bm B_0 \cdot \vec k}{\sigma k} + (\sigma s \lambda^{s})(\sigma k) \frac{\bm B_0 \cdot \vec k}{\sigma k} - \alpha (\sigma k)^2 \frac{\bm B_0 \cdot \vec k}{\sigma k}

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize (\sigma s \lambda^{s})\bm B_0 \cdot \vec k + (\sigma s \lambda^{s}) \bm B_0 \cdot \vec k - \alpha (\sigma k) \bm B_0 \cdot \vec k

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize \sigma \bm B_0 \cdot \vec k \big[ s \lambda^{s} + s \lambda^{s} - \alpha k \big]

内積（異なる$ s ）

$ \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize \AR{r}{ - \sigma s \lambda^{-s} \bm\phi_\sigma(-\vec{k}) \\ \DS \sigma k \bm\phi_\sigma(-\vec{k}) }^T \AR{cc}{ 0 & - \bm B_0 \times \phantom{\big|}\!\! \\ \phantom{\big|}\!\! - \bm B_0 \times & \alpha \bm B_0 \times } \AR{r}{ \sigma s \lambda^s \bm\phi_\sigma(\vec{k}) \\ \DS \sigma k \bm\phi_\sigma(\vec{k}) }

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize \AR{r}{ - \sigma s \lambda^{-s} \bm\phi_\sigma(-\vec{k}) \\ \DS \sigma k \bm\phi_\sigma(-\vec{k}) }^T \AR{r}{ - \sigma k (\bm B_0 \times \bm\phi_\sigma(\vec{k}) ) \\ \DS - \sigma s \lambda^{s} (\bm B_0 \times \bm\phi_\sigma(\vec{k}) ) + \alpha \sigma k (\bm B_0 \times \bm\phi_\sigma(\vec{k}) ) }

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize (- \sigma s \lambda^{-s})(\sigma k) \frac{\bm B_0 \cdot \vec k}{\sigma k} + (\sigma s \lambda^{s})(\sigma k) \frac{\bm B_0 \cdot \vec k}{\sigma k} - \alpha (\sigma k)^2 \frac{\bm B_0 \cdot \vec k}{\sigma k}

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize ( - \sigma s \lambda^{-s})\bm B_0 \cdot \vec k + (\sigma s \lambda^{s}) \bm B_0 \cdot \vec k - \alpha (\sigma k) \bm B_0 \cdot \vec k

$ = \newcommand{\AR}[2]{\left[\begin{array}{#1}#2\end{array}\right]}\newcommand{\DS}{\displaystyle} \footnotesize \sigma \bm B_0 \cdot \vec k \big( - s \lambda^{-s} + s \lambda^{s} - \alpha k \big)

行列の成分

$ \big<+\big|\hat\Omega\big|+\big> $ = \footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{\lambda^2+1} \big[ \lambda + \lambda - \alpha k \big]

$ \big<-\big|\hat\Omega\big|-\big> $ = \footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{\lambda^{-2}+1} \big[ -\frac1\lambda -\frac1\lambda - \alpha k \big] $ =\footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{1 + \lambda^{2}} \big[ - \lambda - \lambda - \lambda^2\alpha k \big]

$ \big<+\big|\hat\Omega\big|-\big> $ = \footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{\sqrt{\lambda^2+1}\sqrt{\lambda^{-2}+1}} \big( \lambda - \frac1\lambda - \alpha k \big) $ = \footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{\lambda^2+1} \big( \lambda^2 - 1 - \lambda\alpha k \big)

$ \big<-\big|\hat\Omega\big|+\big> $ = \footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{\sqrt{\lambda^2+1}\sqrt{\lambda^{-2}+1}} \big( - \frac1\lambda + \lambda - \alpha k \big) $ = \footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{\lambda^2+1} \big( - 1 + \lambda^2 - \lambda \alpha k \big)

traceは$ =\footnotesize \frac{\sigma \bm B_0 \cdot \vec k}{1 + \lambda^{2}} \big[ \lambda + \lambda - \alpha k - \lambda - \lambda - \lambda^2\alpha k \big] $ = \footnotesize \sigma \bm B_0 \cdot \vec k \big( - \alpha k \big)

determinantのコア部分は$ \footnotesize \big[ \lambda + \lambda - \alpha k \big] \big[ - \lambda - \lambda - \lambda^2\alpha k \big] - \big( \lambda^2 - 1 - \lambda\alpha k \big) \big( - 1 + \lambda^2 - \lambda \alpha k \big) $ = \footnotesize \big[ \lambda + \lambda - \alpha k \big] \big[ - \lambda - \lambda - \lambda^2\alpha k \big] - \big[ ( \lambda^2 - 1 )^2 - 2 \lambda\alpha k ( \lambda^2 - 1 ) + (\lambda \alpha k)^2 \big] $ = \footnotesize - 4 \lambda^2 - 2 \lambda^3 \alpha k + 2 \lambda \alpha k + (\lambda\alpha k)^2 - ( \lambda^2 - 1 )^2 + 2 \lambda\alpha k ( \lambda^2 - 1 ) - (\lambda \alpha k)^2 $ = \footnotesize - 4\lambda^2 - ( \lambda^2 - 1 )^2 $ = \footnotesize - ( \lambda^2 + 1 )^2

determinantは$ \footnotesize (\frac{\sigma \bm B_0 \cdot \vec k}{\lambda^2+1})^2( -(\lambda^2+1)^2 ) $ = \footnotesize -(\sigma \bm B_0 \cdot \vec k)^2

よって固有方程式は$ C = \sigma \bm B_0 \cdot \vec k とおくと$ X^2 + C \alpha k X - C^2 = 0 $ \Longrightarrow $ (\frac{X}{C})^2 + \alpha k \frac{X}{C} - 1 = 0 なので、固有値は$ \frac{X}{C} = -\frac{\alpha k}{2} \pm \sqrt{(\frac{\alpha k}{2})^2+1}

だから$ \big<s'\big|\hat\Omega\big|s\big> を対角化してしまうと結局$ \tilde\Omega = 0 の固有モードを導いてしまう。

$ \bm B_0 = \bm0 ，$ \bm\Omega_0 \ne \bm0 の場合は