# 正則變換

## 定義

$q_i=q_i(Q_1,\ Q_2,\ \dots,\ Q_N,\ t)\ ,\qquad\qquad\qquad\qquad i=1,\ 2,\ 3,\ \dots ,\ N$

$q_i=q_i(Q_1,\ Q_2,\ \dots,\ Q_N,\ P_1,\ P_2,\ \dots,\ P_N,\ t)\ ,\qquad\qquad\qquad\qquad i=1,\ 2,\ 3,\ \dots ,\ N$
$p_i=p_i(Q_1,\ Q_2,\ \dots,\ Q_N,\ P_1,\ P_2,\ \dots,\ P_N,\ t)\ ,\qquad\qquad\qquad\qquad i=1,\ 2,\ 3,\ \dots ,\ N$

$\dot{\mathbf{q}} =~~\frac{\partial \mathcal{H}}{\partial \mathbf{p}}$
$\dot{\mathbf{p}} = - \frac{\partial \mathcal{H}}{\partial \mathbf{q}}$

$\dot{\mathbf{Q}} =~~\frac{\partial \mathcal{K}}{\partial \mathbf{P}}$
$\dot{\mathbf{P}} = - \frac{\partial \mathcal{K}}{\partial \mathbf{Q}}$

## 實際用處

$\mathcal{H}=\mathcal{H}(\mathbf{q},\ \mathbf{p},\ t)$

$\frac{\partial \mathcal{H}}{\partial q_i}=0$

$\dot{p}_i= - \frac{\partial \mathcal{H}}{\partial q_i}=0$

## 生成函數方法

$\delta\int_{t_{1}}^{t_{2}}\left[\mathbf{p}\cdot\dot{\mathbf{q}} - \mathcal{H}(\mathbf{q},\ \mathbf{p},\ t)\right]dt=0$
$\delta\int_{t_{1}}^{t_{2}}\left[ \mathbf{P}\cdot\dot{\mathbf{Q}} - \mathcal{K}(\mathbf{Q},\ \mathbf{P},\ t) \right]dt=0$

$\sigma \left[ \mathbf{p} \cdot \dot{\mathbf{q}} - \mathcal{H}(\mathbf{q},\ \mathbf{p},\ t) \right] = \mathbf{P} \cdot \dot{\mathbf{Q}} - \mathcal{K}(\mathbf{Q},\ \mathbf{P},\ t) + \frac{dG}{dt}$

$\sigma\left[\mathbf{p}\cdot\dot{\mathbf{q}} - \mathcal{H}\right]=\mathbf{P}'\cdot\dot{\mathbf{Q}}' - \mathcal{K}\,'+\frac{dG\,'}{dt}$

$\frac{\partial \mathcal{K}}{\partial \mathbf{P}}=\alpha\frac{\partial \mathcal{K}\,'}{\partial \mathbf{P}'}=\alpha\dot{\mathbf{Q}}'=\dot{\mathbf{Q}}$
$\frac{\partial \mathcal{K}}{\partial \mathbf{Q}}=\beta\frac{\partial \mathcal{K}\,'}{\partial \mathbf{Q}'}= - \beta\dot{\mathbf{P}}'= - \dot{\mathbf{P}}$
$\mathbf{p}\cdot\dot{\mathbf{q}} - \mathcal{H}=\alpha\beta(\mathbf{P}'\cdot\dot{\mathbf{Q}}' - \mathcal{K}\,'+\frac{dG\,'}{dt})=\mathbf{P}\cdot\dot{\mathbf{Q}} - \mathcal{K}+\frac{dG}{dt}$(1)

### 第一型生成函數

$G=G_{1}(\mathbf{q},\ \mathbf{Q},\ t)$

$\mathbf{p} \cdot \dot{\mathbf{q}} - \mathcal{H}(\mathbf{q},\ \mathbf{p},\ t) = \mathbf{P} \cdot \dot{\mathbf{Q}} - \mathcal{K}(\mathbf{Q},\ \mathbf{P}, t) + \frac{\partial G_{1}}{\partial t} + \frac{\partial G_{1}}{\partial \mathbf{q}} \cdot \dot{\mathbf{q}} + \frac{\partial G_{1}}{\partial \mathbf{Q}} \cdot \dot{\mathbf{Q}}$

$\mathbf{p} = ~~\frac{\partial G_{1}}{\partial \mathbf{q}}$(2)
$\mathbf{P} = -\frac{\partial G_{1}}{\partial \mathbf{Q}}$(3)
$\mathcal{K} = \mathcal{H} + \frac{\partial G_{1}}{\partial t}$(4)

$2N+1$ 個方程式設定了變換 $(\mathbf{q},\ \mathbf{p}) \rightarrow (\mathbf{Q},\ \mathbf{P})$ ，步驟如下：

$\mathbf{p}=\mathbf{p}(\mathbf{q},\ \mathbf{Q},\ t)$

$\mathbf{Q}=\mathbf{Q}(\mathbf{q},\ \mathbf{p},\ t)$(5)

$\mathbf{P}=\mathbf{P}(\mathbf{q},\ \mathbf{Q},\ t)$

$\mathbf{P}=\mathbf{P}(\mathbf{q},\ \mathbf{p},\ t)$(6)

$2N$ 個函數方程式 (5) 、(6) ，可以逆算出 $2N$ 個函數方程式

$\mathbf{q}=\mathbf{q}(\mathbf{Q},\ \mathbf{P},\ t)$
$\mathbf{p}=\mathbf{p}(\mathbf{Q},\ \mathbf{P},\ t)$

$\mathcal{K} =\mathcal{K}(\mathbf{Q},\ \mathbf{P},\ t)$

### 第二型生成函數

$G = - \mathbf{Q}\cdot\mathbf{P}+G_{2}(\mathbf{q},\ \mathbf{P},\ t)$

$\mathbf{p} = \frac{\partial G_{2}}{\partial \mathbf{q}}$ ，
$\mathbf{Q} = \frac{\partial G_{2}}{\partial \mathbf{P}}$
$\mathcal{K} = \mathcal{H} + \frac{\partial G_{2}}{\partial t}$

### 第三型生成函數

$G = \mathbf{q} \cdot \mathbf{p} + G_{3}(\mathbf{p}, \mathbf{Q}, t)$

$\mathbf{q} = -\frac{\partial G_{3}}{\partial \mathbf{p}}$
$\mathbf{P} = -\frac{\partial G_{3}}{\partial \mathbf{Q}}$
$\mathcal{K} = \mathcal{H} + \frac{\partial G_{3}}{\partial t}$

### 第四型生成函數

$G = \mathbf{q} \cdot \mathbf{p} - \mathbf{Q} \cdot \mathbf{P} + G_{4}(\mathbf{p}, \mathbf{P}, t)$

$\mathbf{q} = -\frac{\partial G_{4}}{\partial \mathbf{p}}$
$\mathbf{Q} = ~~\frac{\partial G_{4}}{\partial \mathbf{P}}$
$\mathcal{K}= \mathcal{H} + \frac{\partial G_{4}}{\partial t}$

### 實例 1

$G_{1} = \mathbf{q} \cdot \mathbf{Q}$

$\mathbf{p} = ~~\frac{\partial G_{1}}{\partial \mathbf{q}} = \mathbf{Q}$
$\mathbf{P} = -\frac{\partial G_{1}}{\partial \mathbf{Q}} = -\mathbf{q}$

$\mathcal{K}(\mathbf{Q},\ \mathbf{P},\ t)=\mathcal{H}(\mathbf{q},\ \mathbf{p},\ t)$

### 實例 2

$G_{2} \equiv \mathbf{g}(\mathbf{q};\ t) \cdot \mathbf{P}$

$\mathbf{Q}=\frac{\partial G_{2}}{\partial \mathbf{P}} =\mathbf{g}(\mathbf{q};\ t)$

## 不變量

### 辛條件

$\boldsymbol{\xi}^T=[q_1,\ q_2,\ q_3,\ \dots,\ q_N,\ p_1,\ p_2,\ p_3,\ \dots,\ p_N]$

$\dot{\boldsymbol{\xi}}=\boldsymbol{\Omega}\frac{\partial \mathcal{H}}{\partial \boldsymbol{\xi}}$

$\dot{\boldsymbol{\Xi}}=\boldsymbol{\Omega}\frac{\partial \mathcal{K}}{\partial \boldsymbol{\Xi}}$

$\boldsymbol{\Xi}=\boldsymbol{\Xi}(\boldsymbol{\xi},\ t)$ 關於時間 $t$ 的導數，

$\dot{\boldsymbol{\Xi}}=\mathbf{M}\dot{\boldsymbol{\xi}}+\frac{\partial \boldsymbol{\Xi}}{\partial t}$

$\mathbf{M}\dot{\boldsymbol{\xi}}+\frac{\partial \boldsymbol{\Xi}}{\partial t}=\boldsymbol{\Omega}\frac{\partial \mathcal{K}}{\partial \boldsymbol{\Xi}}=\boldsymbol{\Omega}\frac{\partial \mathcal{H}}{\partial \boldsymbol{\Xi}}+\boldsymbol{\Omega}\frac{\partial^2 G_1}{\partial \boldsymbol{\Xi}\ \partial t}$ ;

$\mathbf{M}\dot{\boldsymbol{\xi}}=\boldsymbol{\Omega}\frac{\partial \mathcal{H}}{\partial \boldsymbol{\Xi}}$

$\mathcal{H}=\mathcal{H}(\boldsymbol{\xi})$ ，所以，

$\frac{\partial \mathcal{H}}{\partial \boldsymbol{\Xi}}=\frac{\partial \boldsymbol{\xi}}{\partial \boldsymbol{\Xi}}\frac{\partial \mathcal{H}}{\partial \boldsymbol{\xi}}=(\mathbf{M}^{-1})^T\frac{\partial \mathcal{H}}{\partial \boldsymbol{\xi}}= - (\mathbf{M}^{-1})^T\boldsymbol{\Omega}\dot{\boldsymbol{\xi}}$

$\mathbf{M}= - \boldsymbol{\Omega}(\mathbf{M}^{-1})^T\boldsymbol{\Omega}$

$\mathbf{M}^T= - \boldsymbol{\Omega}\mathbf{M}^{-1}\boldsymbol{\Omega}$
$\mathbf{M}^T\boldsymbol{\Omega}=\boldsymbol{\Omega}\mathbf{M}^{-1}$

$\mathbf{M}^T\boldsymbol{\Omega}\mathbf{M}=\boldsymbol{\Omega}$

### 基本帕松括號不變量

$\big[f,\ g\big]_{(\mathbf{q},\ \mathbf{p})} = \sum_{i=1}^{N} \left( \frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} - \frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}} \right)$

$\big[f,\ g\big]_{\boldsymbol{\xi}}= \left(\frac{\partial f}{\partial \boldsymbol{\xi}}\right)^T \boldsymbol{\Omega}\ \frac{\partial g}{\partial \boldsymbol{\xi}}$

$\big[q_i,\ q_j\big]_{\boldsymbol{\xi}}=\big[p_i,\ p_j\big]_{\boldsymbol{\xi}}=0$
$\big[q_i,\ p_j\big]_{\boldsymbol{\xi}}= - \big[p_i,\ q_j\big]_{\boldsymbol{\xi}}=\delta_{ij}$

$\big[\boldsymbol{\xi},\ \boldsymbol{\xi}\big]_{\boldsymbol{\xi}}=\boldsymbol{\Omega}$

$\big[\boldsymbol{\Xi},\ \boldsymbol{\Xi}\big]_{\boldsymbol{\xi}}=\left(\frac{\partial \boldsymbol{\Xi}}{\partial \boldsymbol{\xi}}\right)^T \boldsymbol{\Omega}\ \frac{\partial \boldsymbol{\Xi}}{\partial \boldsymbol{\xi}}$

$M=\frac{\partial \boldsymbol{\Xi}}{\partial \boldsymbol{\xi}}$

$\big[\boldsymbol{\Xi},\ \boldsymbol{\Xi}\big]_{\boldsymbol{\xi}}=\mathbf{M}^T\boldsymbol{\Omega}\mathbf{M}$

$\big[\boldsymbol{\Xi},\ \boldsymbol{\Xi}\big]_{\boldsymbol{\xi}}=\boldsymbol{\Omega}$

### 帕松括號不變量

\begin{align}\big[f,\ g\big]_{\boldsymbol{\xi}} & =\left(\frac{\partial f}{\partial \boldsymbol{\xi}}\right)^T \boldsymbol{\Omega}\ \frac{\partial g}{\partial \boldsymbol{\xi}} \\ & = \left(\frac{\partial \boldsymbol{\Xi}}{\partial \boldsymbol{\xi}}\frac{\partial f}{\partial \boldsymbol{\Xi}}\right)^T \boldsymbol{\Omega}\ \frac{\partial \boldsymbol{\Xi}}{\partial \boldsymbol{\xi}}\frac{\partial g}{\partial \boldsymbol{\Xi}} \\ & = \left(\frac{\partial f}{\partial \boldsymbol{\Xi}}\right)^T M^T \boldsymbol{\Omega}M\frac{\partial g}{\partial \boldsymbol{\Xi}}\ \ _\circ \\ \end{align}

$\big[f,\ g\big]_{\boldsymbol{\xi}}=\left(\frac{\partial f}{\partial \boldsymbol{\Xi}}\right)^T\boldsymbol{\Omega}\ \frac{\partial g}{\partial \boldsymbol{\Xi}}=\big[f,\ g\big]_{\boldsymbol{\Xi}}$

## 參考文獻

1. ^ Goldstein, Herbert. Classical Mechanics 3rd. United States of America: Addison Wesley. 1980: pp. 384. ISBN 0201657023 （English）.