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?
Best Answer
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.