# Is my understanding of canonical transformations flawed?

Consider a system described by Hamilton's equations $$\dot{q}_i=\frac{\partial H}{\partial p_i}=\{q_i,H\}, \quad \dot{p}_i=-\frac{\partial H}{\partial q_i}=\{p_i,H\}.\tag{1}$$ I want to prove that a time-independent transformation of the form $$q_i\to Q_i(q,p), \quad p_i\to P_i(q,p)\tag{2}$$ preserves Hamilton's equations provided $$\{Q_i,Q_j\}=0,\qquad \{Q_i,P_j\}=\delta_{ij},\qquad\{P_i,P_j\}=0, \tag{3}$$ i.e. if the transformation is canonical.

However, I find that $$\dot{Q}_i=\sum_k\left[\frac{\partial Q_i}{\partial q_k}\dot{q}_k+\frac{\partial Q_i}{\partial p_k}\dot{p}_k\right]=\{Q_i,H\},\tag{4}$$ and similarly, $$\dot{P}_i=\sum_k\left[\frac{\partial P_i}{\partial q_k}\dot{q}_k+\frac{\partial P_i}{\partial p_k}\dot{p}_k\right]=\{P_i,H\}\tag{5}$$ without using (3)!

Does it mean that any transformation of the form (2) is canonical?

But as far as I knew, only a subset of transformations of type (1) which satisfies (3) preserve the form of Hamilton's equations i.e. are canonical. But it turns out that I am able to reproduce Hamilton's equations in the transformed variables without using the restriction that (3), rather trivially. Did I make any mistakes?

• It is really a problem of bad notation: The Poisson bracket you are using are with respect to $q,p$ but not $Q,P$. So even though you get the time evolution as a Poisson bracket it does not imply the Hamiltonian equation of motion. Hint: Try to find the condition such that $\{\cdot,\cdot\}_{q,p} = \{\cdot,\cdot\}_{Q,P}$ May 29 at 6:10
• One way to see why what you've done isn't so surprising: You've just computed two particular cases of the general fact that $\dot{G} = \{G, H\}$ for any $G(p,q)$ with $\frac{\partial G}{\partial t} = 0$. Arguably the whole point of the Hamiltonian formulation is finding a function $H$ and brackets $\{ \cdot , \cdot \}$ such that $\frac{d}{dt} = \{\cdot, H\}$ as operators. We turn the analysis problem of taking a derivative into the algebra problem of calculating a bracket. May 29 at 17:44

Hint: In symplectic notation $$z^I~=~(q^i,p_i)\qquad\text{and}\qquad Z^I~=~(Q^i,P_i),$$ and assuming no explicit time dependence in the transformation $$Z^I=f^I(z)$$, OP has shown that \begin{align}\dot{Z}^I~\stackrel{\text{chain rule}}{=}~&~ \frac{\partial Z^I}{\partial z^J}\dot{z}^J\cr ~\stackrel{\text{old Ham. eqs.}}{=}&~ \frac{\partial Z^I}{\partial z^J}J^{JK}\frac{\partial H}{\partial z^K}\cr ~=~~~~~~&\{Z^I,H\}_z,\end{align}\tag{4/5} where $$J^{JK}$$ is the symplectic unit. However the task is instead to show the new Hamilton's equations $$\dot{Z}^I~=~ J^{IJ}\frac{\partial H}{\partial Z^J}~=~\{Z^I,H\}_{\color{red}{Z}}\tag{new Ham. eqs}$$ with the help of the symplectic condition $$J^{IL}~=~\{Z^I,Z^L\}_z~=~\frac{\partial Z^I}{\partial z^J}J^{JK}\frac{\partial Z^L}{\partial z^K}.\tag{3}$$