[NOTES/CM-04006] Hamiltonian for a Charged Particle

The Lagrangian for a charged particle in electromagnetic field is (using 1,2,3 notation for vectors, \(\vec x=(x_1,x_2,x_3)\)) \begin{eqnarray}\Label{EQ01} L &=&\frac{1}{2}m (\vec x\cdot \vec x) -(e/c) (\vec A\cdot \vec x) + e\phi(\vec x)\\ &=&\frac{1}{2}m \sum_k \dot x_k^2 - (e/c)\sum_k A_k \dot x_k - e\phi(\vec x)\Label{EQ02} \end{eqnarray}
where \(\vec A\) and \(\phi\) are vector and scalar potentials. The momenta conjugate to the position \(\vec r\) are given by \begin{equation} p_k = \pp[L]{\dot x_k} = m \dot x_k +(e/c) A_k,\quad k=1,2,3 \Label{EQ03} \end{equation} The Hamiltonian is given by \begin{equation}\Label{EQ04} H=\sum p_k\dot x_k -L = \frac{1}{2}m \dot{\vec x}^{\,2} + e\phi \end{equation} In the right hand side the velocities must be eliminated and the Hamiltonian is to be expressed in terms of canonical variable \((x_k, p_k)\). Solving for velocities we have \[\dot x_k =\frac{p_k -(e/c) A_k}{m}.\] The Hamiltonian \(H\) in \EqRef{EQ04} can now be expressed in terms of \(\vec x\) and \(\vec p\). The final answer is given by \begin{equation} H(\vec x, \vec p) = \frac{1}{2m}(\vec p- (e/c) \vec A)^2 + e\phi(\vec x) \end{equation}