Answer for HW_11

1 正则变换

题为相空间的旋转变换

[\begin{matrix} Q \\ P \end{matrix}] = [\begin{matrix} \cos t & - \sin t \\ \sin t & \cos t \end{matrix}] [\begin{matrix} q \\ p \end{matrix}]

易证

[Q, P] = [q \cos t, p \cos t] + [- p \sin t, q \sin t] = [q, p] = 1

所以为正则变换。

根据

p = - Q \sin t + P \cos t = - (q \cos t - p \sin t) \sin t + P \cos t

得到

\frac{\partial F_{2} (q, P)}{\partial q} = p (q, P) = - q \tan t + \frac{P}{\cos t}

同理

\frac{\partial F_{2}}{\partial P} = Q = \frac{q}{\cos t} - P \tan t

采用折线积分，

F_{2} (q, P; t) = \int (- q \tan t + \frac{P}{\cos t}) d q + f (P, t) = - \frac{q^{2}}{2} \tan t + \frac{P q}{\cos t} + f (P, t)

\frac{\partial F_{2}}{\partial P} = \frac{q}{\cos t} + f^{'} (P, t)_{P} = \frac{q}{\cos t} - P \tan t

得到 $f (P, t) = - \frac{P^{2}}{2} \tan t + φ (t)$ ，即 $F_{2} (q, P; t) = - \frac{q^{2} + P^{2}}{2} \tan t + \frac{P q}{\cos t} + φ (t)$ 。则哈密顿函数变为

\tilde{H} (Q, P; t) = H + \frac{\partial F_{2}}{\partial t} = φ^{'} (t)

正则方程

\dot{P} = - \frac{\partial \tilde{H}}{\partial Q} = 0, \dot{Q} = \frac{\partial \tilde{H}}{\partial P} = 0

记 $Q (t) \equiv Q_{0}, P (t) \equiv P_{0}$ ，根据变换有：

q (t) = Q_{0} \cos t + P_{0} \sin t, p (t) = - Q_{0} \sin t + P_{0} \cos t

于是 $q (0) = Q_{0}, p (0) = P_{0}$ ，得到

q (t) = q (0) \cos t + p (0) \sin t, p (t) = - q (0) \sin t + p (0) \cos t

采用第一类生成函数 $F (q, Q, t)$ ，采用爱因斯坦求和约定，对于 $n$ 维情形：

[Q_{α}, Q_{β}] = [P_{α}, P_{β}] = 0

是易证的，这里略去，下面证明 $[Q_{α}, P_{β}] = δ_{α β}$ ，已知

P_{β} = - \frac{\partial F}{\partial Q_{β}}

两边对 $p_{γ}$ 求导

\frac{\partial P_{β}}{\partial p_{γ}} = - \frac{\partial^{2} F}{\partial Q_{β} \partial Q_{σ}} \frac{\partial Q_{σ}}{\partial q_{γ}}

两边对 $q_{γ}$ 求导

\frac{\partial P_{β}}{\partial q_{γ}} = - \frac{\partial^{2} F}{\partial Q_{β} \partial q_{γ}} - \frac{\partial^{2} F}{\partial Q_{β} \partial Q_{σ}} \frac{\partial Q_{σ}}{\partial q_{γ}}

于是

\begin{aligned} [Q_{α}, P_{β}] & = \frac{\partial Q_{α}}{\partial q_{γ}} \frac{\partial P_{β}}{\partial p_{γ}} - \frac{\partial Q_{α}}{\partial p_{γ}} \frac{\partial P_{β}}{\partial q_{γ}} \\ = - \frac{\partial Q_{α}}{\partial q_{γ}} \frac{\partial^{2} F}{\partial Q_{β} \partial Q_{σ}} \frac{\partial Q_{σ}}{\partial q_{γ}} + \frac{\partial Q_{α}}{\partial p_{γ}} (\frac{\partial^{2} F}{\partial Q_{β} \partial q_{γ}} + \frac{\partial^{2} F}{\partial Q_{β} \partial Q_{σ}} \frac{\partial Q_{σ}}{\partial q_{γ}}) \\ = - (\frac{\partial Q_{α}}{\partial q_{γ}} \frac{\partial Q_{σ}}{\partial q_{γ}} - \frac{\partial Q_{α}}{\partial p_{γ}} \frac{\partial Q_{σ}}{\partial q_{γ}}) \frac{\partial^{2} F}{\partial Q_{β} \partial Q_{σ}} + \frac{\partial Q_{α}}{\partial p_{γ}} \frac{\partial^{2} F}{\partial Q_{β} \partial q_{γ}} \\ = - [Q_{α}, Q_{σ}] \frac{\partial^{2} F}{\partial Q_{β} \partial Q_{σ}} + \frac{\partial Q_{α}}{\partial p_{γ}} \frac{\partial^{2} F}{\partial Q_{β} \partial q_{γ}} \\ = 0 + \frac{\partial Q_{α}}{\partial p_{γ}} \frac{\partial p_{γ}}{\partial Q_{β}} \\ = δ_{α β} \end{aligned}

成立。

题目给出单位矩阵的辛形式保持不变：

Ω = d p_{1} \land d q_{1} + d p_{2} \land d q_{2} = d P_{1} \land d Q_{1} + d P_{2} \land d Q_{2} = Ω^{'}

那么

\begin{aligned} Ω \land Ω & = (d p_{1} \land d q_{1} + d p_{2} \land d q_{2}) \land (d p_{1} \land d q_{1} + d p_{2} \land d q_{2}) \\ = (d p_{1} \land d q_{1}) \land (d p_{1} \land d q_{1}) + (d p_{1} \land d q_{1}) \land (d p_{2} \land d q_{2}) + (d p_{2} \land d q_{2}) \land (d p_{1} \land d q_{1}) + (d p_{2} \land d q_{2}) \land (d p_{2} \land d q_{2}) \\ = (d p_{1} \land d q_{1}) \land (d p_{2} \land d q_{2}) + (d p_{2} \land d q_{2}) \land (d p_{1} \land d q_{1}) \\ = 2 \cdot (d p_{1} \land d q_{1}) \land (d p_{2} \land d q_{2}) \\ = 2 d τ \end{aligned}

即相空间体积元也保持不变，注意到

d Q_{1} \land d Q_{2} \land d P_{1} \land d P_{2} = | \frac{\partial (Q_{1}, Q_{2}, P_{1}, P_{2})}{\partial (q_{1}, q_{2}, p_{1}, p_{2})} | \cdot d q_{1} \land d q_{2} \land d p_{1} \land d p_{2}

所以雅可比行列式为 1：

| \frac{\partial (Q_{1}, Q_{2}, P_{1}, P_{2})}{\partial (q_{1}, q_{2}, p_{1}, p_{2})} | = 1

大二时写过这题。用直角坐标算会比我的计算更简单，