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Let $q,p$ denote the position and momentum. Consider a transform generated by $g$:

$q' = q + \epsilon \{q,g\}---(1a)$

$p' = p + \epsilon \{p,g\}---(1b)$

Then:

$\{q',p'\} = \{q,p\}+o(\epsilon^2)+\epsilon \{\{q,g\},p\}+\epsilon \{q,\{p,g\}\}$

The last two terms vanish since:

$\{p,\{g,p\}\}+\{q,\{p,g\}\} = -\{g,\{q,p\}\}$

And $\{g,\{q,p\}\} = \{g,1\} = 0$

Therefore this transform generated by $g$ is indeed a canonical transform.

In the equation above we use the fact that $\{q,p\}=1$. However, this does not hold in general if $p,q$ denotes some other variable with non-constant commutator. In those cases, the commutator is not preserved!:

$\{q',p'\} \ne \{q,p\}---(2)$

But we already know that (1a) and (1,b) are canonical transform, which should mean that every commutator should be preserved, even if $q,p$ are not position and momentum, which contradict (2)!

What is going on here?? Is it a problem caused by the higher-order term $o(\epsilon^2)$?

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More generally, a finite symplectomorphism is formally of the form

$$ f~~\mapsto ~~f^{\prime} ~=~e^{-\varepsilon \{g,\cdot\}}f ~=~ f- \varepsilon \{g,f\} + \frac{\varepsilon^2}{2} \{g,\{g,f\}\} +o(\varepsilon^3),\tag{A}$$ where $f=f(q,p,t)$ and $g=g(q,p,t)$ are functions. One may use the Jacobi identity to prove that the finite transformation (A) respects the Poisson bracket: $$ e^{-\varepsilon \{g,\cdot\}}\{f_1,f_2\}~=~\{e^{-\varepsilon \{g,\cdot\}}f_1,e^{-\varepsilon \{g,\cdot\}}f_2\} .\tag{B}$$ Returning to OP's question: If a Poisson bracket $\{f_1,f_2\}$ is not a constant, it might transform.

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