Sanity check: can we define "arbitrary" hamiltonians?

Question

I just want to do a sanity check on my understanding of Hamiltonian mechanics:

My understanding is: For any number $n$, take the phase space $\mathbb R^{2n}$, and take any arbitrary differentiable function $H:\mathbb R^{2n}\to \mathbb R$ to be the Hamiltonian. Then all of the standard results about Hamiltonian mechanics will apply to the system generated by $\dot q=H_p, \dot p=-H_q$ (In particular Liouville's theorem applies, and Noether's theorem applies to the Lagrangian obtained from the Legendre transform). There are no further regularity conditions needed to do Hamiltonian mechanics on this system.

Is this correct?

Answer 1

Yes, any smooth function on the phase space can be Hamiltonian. And to any Hamiltonian corresponds a Hamiltonian vector field $V_H$, such that $$ i_{V_H} \omega = -dH $$ In the simple case of $\mathbb{R}^{2n}$ the symplectic form is $$ \omega = \sum_i d q_i \wedge d p_i $$

Answer 2

1. We only need differentiability of the Hamiltonian $H(q,p,t)$ as long as we stay in the Hamiltonian formulation. In particular, Noether's theorem works with the Hamiltonian action, cf. this Phys.SE post.

2. However to guarantee the existence of a regular Lagrangian formulation in $n$ variables $(q^1,\ldots,q^n)$ (via a Legendre transformation) we need to impose that the Hessian $\frac{\partial^2 H}{\partial p_i\partial p_j}$ has maximal rank, i.e. is invertible.