User:Nik

Foreword

I use \(Q\:/\:P\) 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) \( \{Q_i(q,p), Q_j(q,p)\} = \{P_i (q,p), P_j(q,p)\} = 0 \)

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

Attempts

- Use of hamilton's equations from script page 59. I asked Raisa if we are allowed to use things from the script without proof, no answer yet (31.12.14 - 21:00) The problem there was, that you have to use a hamiltonian from the new coordinate system. May work, haven't checked that exactly.

- Use of simply the chain rule and insert all the things. Some solutions i saw with that said that \( \frac{\partial \Phi}{\partial p} = 0 \) but since \(\Phi \) depends on \(Q \) whicht depends on \(p \) I am not convinced with that.

- Use of simply the chain rule as before. At some point of this solution you should show that \( \frac{\partial Q_a}{\partial q_j}\frac{\partial}{\partial p_j}\frac{\partial}{\partial Q_j}\Phi (q, Q) = 0 \) which nobody has until now.