Derivation of Hamiltonian of charged particle in EM field

Question

I attempt to derive the Hamiltonian of a charged particle in an electromagnetic field but I come up with an additional term. Could you help me check my process?

Using SI units, the Lorentz force on a particle with charge $e$ is $$ \mathbf{F} = e\mathbf{E} + e\mathbf{v\times B}. $$ We may write $\mathbf{E}$ and $\mathbf{B}$ as $$ \mathbf{E} = -\nabla\psi - \frac{\partial\mathbf{A}}{\partial t} $$ and $$ \mathbf{B} = \nabla\times\mathbf{A}. $$ In Einstein's tensor notation, these equations may be written as $$ F_i = eE_i + e\varepsilon_{ijk}v_jB_k, $$ $$ E_i = -\frac{\partial \psi}{\partial x_i} - \frac{\partial A_i}{\partial t} $$ and $$ B_i = \varepsilon_{ijk}\frac{\partial}{\partial x_j}A_k. $$ Thus, substituting these last equations on Lorentz's force law, $$ F_i = -e\frac{\partial \psi}{\partial x_i} - e\frac{\partial A_i}{\partial t} + e\varepsilon_{ijk}\varepsilon_{klm}v_j\frac{\partial}{\partial x_l}A_m. $$ I perform a permutation $\varepsilon_{ijk} = \varepsilon_{kij}$ and use the well-known relation between the Levi-Civita symbol and the Kronecker delta $$ \varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} $$ to obtain $$ F_i = e\left(-\frac{\partial\psi}{\partial x_i} - \frac{\partial A_i}{\partial t} + v_j\frac{\partial A_j}{\partial x_i} - v_j\frac{\partial A_i}{\partial x_j}\right). $$ I identify $$ \frac{dA_i}{dt} = \frac{\partial A_i}{\partial t} + v_j\frac{\partial A_i}{\partial x_j}, $$ and hence obtain $$ \frac{d}{dt}\left(p_i + eA_i\right) = e\left(-\frac{\partial\psi}{\partial x_i} + v_j\frac{\partial A_j}{\partial x_i}\right), $$ where $p_i$ is the $i$th component of momentum and I applied Newton's second law.

The additional term $v_j\partial A_j/\partial x_i$ is the one that seems out of place and stops me from writing this as an equation derivable from Hamilton's equations of motion. Should this term vanish for some reason?

Answer 1

I actually came upon my answer as I finished posting my question, so I might as well share it. The problem was that I was confusing canonical momentum with kinetic momentum. The momentum derivative in Newton's law refers to kinetic momentum $md\mathbf{q}/dt$, and the one in the charged particle Hamiltonian is the conjugate momentum, obtained by Legendre transform of the Lagrangian.

To show this, we consider that it is well-known that the Hamiltonian of the charged particle in an EM field is $$ H = \frac{1}{2m}\left(p_j - eA_j\right)^2 + e\psi. $$ Hamilton's equations of motion are $$ \frac{dq_i}{dt} = \frac{\partial H}{\partial p_i} \qquad \frac{dp_i}{dt} = - \frac{\partial H}{\partial q_i}. $$ From the first one, we obtain $$ \frac{dq_i}{dt} = \frac{p_i}{m} - e\frac{A_i}{m}. $$ From the second one $$ \frac{dp_i}{dt} = \frac{e}{m}\left(p_j - eA_j\right)\frac{\partial A_j}{\partial q_i} - e\frac{\partial\psi}{\partial q_i} $$. Combining these to eliminate the canonical momentum $p_i$, we obtain $$ \frac{d}{dt}\left(m\frac{dq_i}{dt} + eA_i\right) = e\left(\frac{dq_j}{dt}\frac{\partial A_j}{\partial q_i} - \frac{\partial\psi}{\partial q_i}\right) $$, which is the expression I obtained from writing Lorentz's force law in terms of the vector and scalar potentials in my original post, and thus proves that the well-known Hamiltonian is indeed correct, although this procedure would not have been possible had we not known the form of the Hamiltonian a priori.

I hope this helps other people who were stuck with the same problem.