The Hamiltoninian function always verifies $H(-p,q)=H(p,q)$?

Question

Is it true that the Hamiltoninian function of a system always verifies $H(-p,q)=H(p,q)$? If $H$ is the sum of the potential energy plus the kinetic energy then it seems true. But is there a possible case where this does not hold?

Answer 1

It is not true in the case of the electromagnetic Hamiltonian for a charged particle in an electromagnetic field: $$H = \frac{(\mathbf{p}-q\, \mathbf{A})^2}{2m} + q\phi, $$ where $\mathbf{A}$ is the vector (Magnetic) potential, $\phi$ is the scalar (Electric) potential, and $q$ and $m$ are the particle's charge and its mass, respectively.

Similarly, you could construct Hamiltonians in other scenarios possibly involving other forces that do not have ''even parity'' with respect to momentum.

Answer 2

We can take a free particle, with $T = \frac{1}{2} m \dot{x}^2$ & $U = 0$, and redefine our coordinates so that $x \equiv q + u t$ for some constant $u$. In terms of these new coordinates, it works out that $\mathcal{L} = \frac{1}{2} m (\dot{q} - u)^2$, and that $$ \mathcal{H} = \frac{p^2}{2m} + p u. $$ This Hamiltonian does not have the desired symmetry under $p \to - p$.