# Hamiltonian of a charged particle in a magnetic field coincides with that of a free particle?

The Hamiltonian of a charged particle of charge $q$ in an arbitrary magnetic field $\textbf{B}$ is given by $$H=\frac{(\textbf{p}-q\textbf{A})^2}{2m}\tag{1}$$ where $\textbf{p}$ is the canonical momentum given by $\textbf{p}=m\textbf{v}+q\textbf{A}$. One can also write (1) as $$H=\frac{\boldsymbol{\pi}^2}{2m}\tag{2}$$ where $\boldsymbol{\pi}=m\textbf{v}$ is the kinetic momentum.

I find it strange that the Hamiltonian of a charged particle in a magnetic field is identical to that of a free particle (because the free particle Hamiltonian is also given by $H_{free}=\frac{\textbf{p}^2}{2m}$ with $\textbf{p}=m\textbf{v}$). What am I missing?

• As functions on phase space, the two Hamiltonians are different. I don't see why it is relevant that they coincide on-shell ($p = m\dot{x}$ only holds on-shell!) since the dynamics lie in the Poisson brackets with it (or Hamilton's equations, if you wish), which are off-shell quantities. – ACuriousMind Apr 6 '17 at 19:18

The value of the Hamiltonian on-shell is the same; this is simply the fact that magnetic fields do no work and hence don't affect the energy. Of course, the equations of motion aren't the same because $\pi$ is not the canonical momentum, $p$ is, and it's the canonical momenta that go into Hamilton's equations.
I guess you have ignored a constant, which does not affect dynamic equations but it reflects the Physical meaning. The Exact Hamiltonian for a charged particle travels in the electromagnetic field is $$H=c[m^2c^2+(\vec{P}-e\vec{A})^2]^{1/2}+e\Phi$$ If only magnetic is, $$H=c[m^2c^2+(\vec{P}-e\vec{A})^2]^{1/2}=c[m^2c^2+\vec{\pi}^2]^{1/2}=E$$ Clearly, Hamiltonian represents the Total energy of the system, which reflects energy consevation, or say magnetic doesn't contribute Work to the system.