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Near the end of chapter 2 of R. Shankar's book Principles of Quantum Mechanics, he talks about two consequences of invariance of the Hamiltonian under a regular canonical transformation. My problem is with the second consequence and it's proof (pg 99 and 103 respectively) :

If $H$ is invariant under the regular canonical, but not necessarily infinitesimal, transformation $(q,p) \to (\bar{q}, \bar{p})$ and if $(q(t),p(t))$is a solution to the equations of motion, so is the transformed (translated, rotated, etc) trajectory $(\bar{q}(t), \bar{p}(t))$.

In his proof of this consequence he tries to show that $\bar{q}$ and $\bar{p}$ satisfy $$\dot{\bar{q}} = \frac{\partial H}{\partial \bar{p}}$$ $$\dot{\bar{p}} = - \frac{\partial H}{\partial \bar{q}}$$ using the invariance of the Hamiltonian $$H(q,p) = H(\bar{q}, \bar{p}).$$ But why should the invariance of the Hamiltonian matter at all? If the transformation $(q,p) \to (\bar{q}, \bar{p})$ is canonical then by definition of canonical transformations, $\bar{q}$ and $\bar{p}$ should already satisfy Hamilton's equations. Shouldn't this be a consequence of just canonical transformations in general rather than being a consequence of the invariance of the Hamiltonian?

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You need to verify that because you don't necessarily know if $ \dot {\bar q } = \frac{dH}{d\bar p } $ or the other equation. The underlying reason is you don't know if the Poisson brackets will end up being the same. He states in page 98 that canonical only refers to equations 2.7.18, so you "only" know that $ \{ \bar q_{j}, \bar q_{k} \} = \{ \bar p_{j}, \bar p_{k} \} = 0 $ as well as $ \{ \bar q_{j}, \bar p_{k} \} = \delta_{jk}$. So really you need to do some back-tracking just in case.

Shankar notes that if it were passive, you could assume what you wrote. This is at the bottom of page 103.

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  • $\begingroup$ But that's not how he initially defined canonical transformations. In page 94 he defines canonical transformations as those transformations for which $(\bar{q}, \bar{p})$ satisfies the Hamilton's equation 2.7.10 . Then using this definition he derived 2.7.18 . So only knowing 2.7.18 already implies that they satisfy Hamilton's equation. $\endgroup$ – Brain Stroke Patient May 16 at 18:02
  • $\begingroup$ I'm unsure why he changes the definition, need to think more about that. However, if you accept this change, and you try to go backwards from 2.7.18 to 2.7.17, you would realize that if you want $ \dot{\bar q} = dH/d\bar p $ or something, you would like $ H $ to be in terms of $ \bar q$ and $ \bar p$. In 2.7.17, the Hamiltonian is still in terms of the old variables. Then you need to assume that $ H(\bar p , \bar q ) = H (p, q ) $ (which you can) and then proceed to go back up through the proof (or back down, if you want Shankar's way). $\endgroup$ – anon.jpg May 16 at 18:19

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