Question

Explain poission bracket condition for canonical transformation.

Answer 1

Poission bracket is invariant under the canonical transformation is

[f,g]p,q = [f,g]_P,Q

Poission bracket

$[Q,P] = -(\frac{ \partial Q}{ \partial p})_q (\frac{ \partial P}{ \partial q})_Q =(\frac{ \partial Q}{ \partial p})_q \frac{\partial^2F} {\partial q \partial Q} =(\frac{\partial Q} {\partial p})_q (\frac{\partial p}{\partial Q})_q =1$

This is used to show general poission bracket is independent of the parametrization of phase space