A canonical transformation is defined as a transformation such that afterwards Hamilton's equations still hold.

It can then be shown that this requirement implies that canonical transformations are necessarily generated by some function $Q$ which acts on the coordinates via \begin{align} q \to q' &= q + \epsilon \frac{\partial Q}{\partial p} \\ p \to p' &= p - \epsilon \frac{\partial Q}{\partial q} \tag{1} \end{align} (See for example, page 102 in Tong's notes.)

A symmetry in the Hamiltonian formalism is defined as a transformation which is generated by a function $Q$ for which $$\{Q,H\}=0 \tag{2} $$ holds, where $H$ denotes the Hamiltonian.

Are symmetries solely defined by Eq. (2) or do they need to fulfill additionally Eq. (1)? In other words, are symmetries necessarily canonical transformations?

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    $\begingroup$ Briefly, eq. (2) has nothing to do with the definition of canonical transformations. $\endgroup$ – Qmechanic Feb 20 at 12:43
  • $\begingroup$ @Qmechanic cool thanks. But is there any error in my reasoning above? Because clearly I have a problem finding a conceptual difference between the two. Both concepts seem to map solutions of the EOM to solutions of the EOM and are therefore describe what I would call symmetries. $\endgroup$ – jak Feb 20 at 12:55
  • $\begingroup$ @Qmechanic Moreover, isn't the statement "eq. (2) has nothing to do with canonical transformations" a bit too strong? At least I would think that symmetries are canonical transformations which additionally fulfil the condition $\{Q,H\}=0$ $\endgroup$ – jak Feb 20 at 13:13
  • $\begingroup$ I modified the comment a bit. $\endgroup$ – Qmechanic Feb 20 at 15:18
  • $\begingroup$ @Qmechanic I modified the question a bit. $\endgroup$ – jak Feb 20 at 15:37

Any reversible variable change maps solutions of the EOM to solutions of the EOM.

EOM themselves have N integrals of motion: N different combinations of the variables that equal to constants, for example, to the N initial data, whatever whether you manage to integrate them explicitly or not.


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