# 四元数和转动

$$$q = \begin{pmatrix}\;z & w \\ -w^* & \;z^*\end{pmatrix} .$$$

$$$C = \begin{pmatrix} \;\; a & b \\ - b & a \end{pmatrix}.$$$

$$$z=a \mathbf I + b \mathbf J$$$

$$\mathbf J ^ 2 = - \mathbf I.$$

$$$q = \begin{pmatrix} z & w \\ -w^* & z^* \end{pmatrix} = z_ R \mathbf I + z_ I \mathbf J_ 3 + w_ R \mathbf J_ 2 + w_ I \mathbf J_ 1,$$$

$\begin{pmatrix} \cos \theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} = \cos \theta \mathbf I - \sin \theta \mathbf J.$

## 四元数的几个好玩的性质

1. 在加法，乘法和实数乘法下面是 closed.
2. 模是两个复数的模之和。

$$$q^\dagger q= q q^\dagger = \| q \|^2 I.$$$

\begin{align} q(x) & = x^0 - i \vec{\boldsymbol\sigma} \cdot \vec x \\ & = \begin{pmatrix} x^0 - i x^3 & -x^2 -ix^1 \\ x^2 -i x^1 & x^0 + i x^3 \end{pmatrix}. \end{align}

## 转动

$$$R \circeq \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix},$$$

$$$R \circeq \cos\theta - i \sin\theta,$$$

$$$R\circeq e^{-i \theta}.$$$

$$$z_1 z_2=(\rho_1 e^{i\phi_1}) (\rho_2 e^{i\phi_2})= \rho_1\rho_2 e^{i(\phi_1+\phi_2)}$$$

1. 长度增加了 $\rho_1$ 倍
2. 角度转动了 $\phi_1$

$$$e^{\frac{\theta}{2} \hat u} = \cos \frac{\theta}{2} + \hat u \sin \frac{\theta}{2}.$$$

$\mathbf q = \lvert q \rvert \left( \cos \theta \mathbf I + \sin\theta \frac{ u_1 \mathbf J_1 + u_2 \mathbf J_2 + u_3 \mathbf J_3 }{ \lvert u_1 \mathbf J_1 + u_2 \mathbf J_2 + u_3 \mathbf J_3 \rvert } \right).$

$\mathbf q = \lvert q \rvert \left( \cos \theta \mathbf I + \sin\theta \hat{\mathbf u} \right).$

$\mathbf q = \lvert q \rvert e^{ \theta \hat {\mathbf u} }.$

## Hamiltonian

$$$\mathbf H = \vec h\cdot \vec{\boldsymbol \sigma} = -i \vec h \cdot \vec{ \mathbf J }.$$$

$$$\mathbf U_0 \mathbf H\mathbf U_0^\dagger =(\cos \theta \mathbf I + \hat{\mathbf{H}}\sin\theta) \mathbf{H}(\cos \theta \mathbf I + \sin \theta \hat{\mathbf{H}}) ^\dagger = \mathbf{H}.$$$

## 中微子物理

$H \to \mathbf H_f = \frac{\omega}{2}\begin{pmatrix} -\cos 2\theta & \sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{pmatrix} = -\frac{\omega}{2}( \cos 2\theta \boldsymbol\sigma_3 + \sin 2\theta \boldsymbol\sigma_1 )$

$\mathbf H_p = - \frac{\omega}{2} \boldsymbol \sigma_3$

By OctoMiao

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