User:Nik

Foreword

I use \(Q\:/\:P\) for the transformed system instead of \(\widetilde{q}\:/\:\widetilde{p}\) because it's easier to write in Latex.

Problem

Let \( \Phi \in C^\infty(\mathbb{R}^n) \) have the property that the system \( p_i = \frac{\partial}{\partial q_i} \Phi (q, Q) \) has a unique smooth solution \( Q = Q(q,p) \).

Define \( P_i(q,p) = - \frac{\partial}{\partial Q_i} \Phi (q, Q) | _{Q= Q(q,p)} \)

Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that \( \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} \)

Show that:

I) \( \{Q_i(q,p), Q_j(q,p)\} = 0 \)

II) \( \{P_i (q,p), P_j(q,p)\} = 0 \)

III) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)

Solution

Important equations

$$ \underbrace{\frac{\partial p_i}{\partial q_k}}_{0} = \frac{\partial}{\partial q_k}(\frac{\partial \Phi}{\partial q_i}(q, Q)) $$ $$ = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) $$ $$ \Rightarrow 0 = \frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) + \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) $$ $$ \tag{1} \Rightarrow \boxed{ -\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) } $$

$$ \underbrace{\frac{\partial p_i}{\partial p_k}}_{\delta_{ik}} = \frac{\partial}{\partial p_k}(\frac{\partial \Phi}{\partial q_i}(q, Q)) $$ $$ = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) $$ $$ \tag{2} \boxed{ \Rightarrow \delta_{ik} = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) } $$

$$ \frac{\partial P_i}{\partial q_k} = \frac{\partial}{\partial q_k}(-\frac{\partial \Phi}{\partial Q_i}(q, Q)) $$ $$ = \sum_{s=1}^{n}(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) $$ $$ \tag{3} \boxed{ \Rightarrow \frac{\partial P_i}{\partial q_k} = -\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q) - \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) } $$

$$ \frac{\partial P_i}{\partial p_k} = \frac{\partial}{\partial p_k}(-\frac{\partial \Phi}{\partial Q_i}(q, Q)) $$ $$ = \sum_{s=1}^{n}(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) $$ $$ \tag{4} \Rightarrow \boxed{ \frac{\partial P_i}{\partial p_k} = -\sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) } $$

Solution I)

$$ \{Q_i, Q_j\} = \sum_{k=1}^{n}(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}) $$

Solution II)

$$ \{P_i, P_j\} = \sum_{k=1}^{n}(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}) $$

Solution III)

$$ \{Q_i, P_j\} = \sum_{k=1}^{n}\left( \frac{\partial Q_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial P_j}{\partial q_k} \right) $$ Now we put in the results from equation (3) and (4) $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)\right) - \frac{\partial Q_i}{\partial p_k}\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)\right)\right) $$ Now we multiply out and separate all the sums. We can rearrange it like this because all the parts we put inside are independent on the running variable. $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) - \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$ Now we rearrange again such that we can build a new Poisson bracket $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q) \underbrace{\sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial Q_s}{\partial q_k} - \frac{\partial Q_i}{\partial q_k}\frac{\partial Q_s}{\partial p_k}\right)}_{=\{Q_i, Q_s\}=0}\right) $$ $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial p_k}{\partial Q_j} \right) = \frac{\partial Q_i}{\partial Q_j} = \delta_{ij} $$