Expression for Hamiltonian vector field!

Question

How does one prove that the Hamiltonian vector field has the following expression, what is the reasoning? \begin{equation} X_H=\sum ^n_{i=1}\frac{\partial H}{\partial q_i}\frac{\partial }{\partial p_i}+\sum ^n_{i=1}\frac{\partial H}{\partial p_i}\frac{\partial }{\partial q_i} \end{equation} In pretty much every text I read it is "obvious" or "implied" it takes that form if we use Cartan's formula and the Lie derivative. I am familiar with both symplectic geometry and the basics of differential geometry. I am sorry if it is either obvious or basic.

Answer 1

Let $H$ be the Hamiltonian. By definition, $X_H$ is the unique vector field which satisfies $$\omega(X_H, Y) = dH(Y)$$ for all vector fields $Y$. Equivalently, $\iota_{X_H}\omega = dH$.

By the uniqueness of $X_H$, it suffices to check that the formula you quoted does indeed satisfy $\omega(X_H, Y) = dH(Y)$ for all vector fields $Y$. I leave this for you to check; once you check this, you're done, that's it.

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However, if you want to see how to actually derive the formula, here's how to do that. In Darboux coordinates, we can write $$X_H = \sum a^i \frac{\partial}{\partial p_i} + \sum b_i \frac{\partial}{\partial q_i}$$ for some functions $a_i$ and $b_i$.

On the one hand, $$dH = \sum \left( \frac{\partial H}{\partial p_i}\,dp_i + \frac{\partial H}{\partial q_i}\,dq_i\right).$$ On the other hand, $$\iota_{X_H}\omega = \left( \sum a^i \frac{\partial}{\partial p_i} + \sum b_i \frac{\partial}{\partial q_i} \right) \,\lrcorner\,\left( \sum dp_i \wedge dq_i \right) = \sum (a^i\,dq_i - b^i\,dp_i).$$ Equating $dH = \iota_{X_H}\omega$ gives that \begin{align} a^i & = \frac{\partial H}{\partial q_i} \\ b^i & = -\frac{\partial H}{\partial p_i}, \end{align} as claimed.