[proofplan] We compute the time derivative $\frac{df}{dt}$ of a smooth observable $f(q, p)$ along a solution of Hamilton's equations $\dot{q}_i = \frac{\partial H}{\partial p_i}$, $\dot{p}_i = -\frac{\partial H}{\partial q_i}$ using the chain rule. The resulting expression is exactly the Poisson bracket $\{f, H\}$. [/proofplan] [step:Apply the chain rule to $f(q(t), p(t))$ along a Hamiltonian trajectory] Let $(q(t), p(t))$ be a solution of Hamilton's equations with Hamiltonian $H: \mathbb{R}^{2n} \to \mathbb{R}$: \begin{align*} \dot{q}_i = \frac{\partial H}{\partial p_i}, \qquad \dot{p}_i = -\frac{\partial H}{\partial q_i}, \qquad i = 1, \ldots, n. \end{align*} Let $f: \mathbb{R}^{2n} \to \mathbb{R}$ be a smooth function that does not depend explicitly on time. By the chain rule applied to the composition $t \mapsto f(q(t), p(t))$: \begin{align*} \frac{df}{dt} &= \sum_{i=1}^n \left( \frac{\partial f}{\partial q_i}\, \dot{q}_i + \frac{\partial f}{\partial p_i}\, \dot{p}_i \right). \end{align*} [/step] [step:Substitute Hamilton's equations to obtain the Poisson bracket] Substituting $\dot{q}_i = \frac{\partial H}{\partial p_i}$ and $\dot{p}_i = -\frac{\partial H}{\partial q_i}$ into the chain rule expression: \begin{align*} \frac{df}{dt} &= \sum_{i=1}^n \left( \frac{\partial f}{\partial q_i}\, \frac{\partial H}{\partial p_i} - \frac{\partial f}{\partial p_i}\, \frac{\partial H}{\partial q_i} \right). \end{align*} The right-hand side is precisely the definition of the Poisson bracket of $f$ and $H$: \begin{align*} \{f, H\} = \sum_{i=1}^n \left( \frac{\partial f}{\partial q_i}\, \frac{\partial H}{\partial p_i} - \frac{\partial f}{\partial p_i}\, \frac{\partial H}{\partial q_i} \right). \end{align*} Therefore \begin{align*} \frac{df}{dt} = \{f, H\}. \end{align*} [guided] This result has a clean interpretation: the Poisson bracket $\{f, H\}$ is the directional derivative of $f$ along the Hamiltonian vector field $X_H = \left(\frac{\partial H}{\partial p_1}, \ldots, \frac{\partial H}{\partial p_n}, -\frac{\partial H}{\partial q_1}, \ldots, -\frac{\partial H}{\partial q_n}\right)$. In the compact notation $x = (q, p) \in \mathbb{R}^{2n}$, Hamilton's equations read $\dot{x} = J\, \nabla H(x)$ where $J = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}$, and the chain rule gives \begin{align*} \frac{df}{dt} = \nabla f \cdot \dot{x} = \nabla f \cdot J\, \nabla H = (\nabla f)^\top J\, \nabla H = \{f, H\}. \end{align*} The formula says that the Hamiltonian $H$ generates the time evolution of every observable through the Poisson bracket. This is the classical analogue of the Heisenberg equation $\frac{d\hat{f}}{dt} = \frac{i}{\hbar}[\hat{H}, \hat{f}]$ in quantum mechanics. [/guided] [/step]