0

I am aware of two definitions of canonical transformations which I state below.

Definition $1$ We go from old set of $\{q_i,p_i,t\}$ of $2n$ phase space variables to a new set $\{Q(q_i,p_i,t),P(q_i,p_i,t),t\}$ of $2n$ phase space variables such that the determinant of the Jacobian of the transformation is $+1$ and the functional form of the old hamiltonian $H(q,p,t)$ changes to new hamiltonian $K(Q_i,P_i,t)$ but Hamilton's equations in the new variables remain preserved in form $\dot Q_i=\frac{\partial K}{\partial P_i},~ \dot P_i=-\frac{\partial K}{\partial Q_i}$.

Definition $2$ We go from old set $\{q_i,p_i,t\}$ of $2n$ phase space variables to a new set $\{Q(q_i,p_i,t),P(q_i,p_i,t),t\}$ of $2n$ phase space variables such that the the determinant of the Jacobian of the transformation is $=+1$ and the funadamental Poisson bracket relations remain unchanged i.e.$\{Q_i,P_j\}=\delta_{ij};~\{Q_i,Q_j\}=\{P_i,P_j\}=0.$

  • Are these two definitions related? I mean, does 1 imply 2 and 2 imply 1? How to show/see this, if true?
Qmechanic
  • 201,751
Solidification
  • 11,483
  • I reviewed the post with which this has been identified as a duplicate. Honestly, I did not find my answer. That post did not explain if definition 1 implies definition 2 and vice-versa, and if so, how to get one from the other. Could it be reopened? – Solidification Jun 26 '20 at 14:52
  • I am really asking whether $[Q_i,P_j]=\delta_{ij}$ follows as a mathematical consequence, assuming the transformation $Q_i=Q_i(q,p,t), P_i=P_i(q,p,t)$ preserves the Hamilton's equations. – Solidification Jun 26 '20 at 14:57
  • No, that's not true. A counterexample is given in the linked post. – Qmechanic Jun 26 '20 at 19:45

0 Answers0