The Hamiltonian

This has to be worked out in detail, starting from the Lagrangian, which is described in Woolley [8]. The result is

$${\mathcal H}(t) = H(t) + H_{\rm rad}$$

$$H_{\rm rad} = \frac{1}{2}\int d{\bf r} \left[ \varepsilon_0 \mathbf{E}_\perp^2 + \mu_0^{-1} \mathbf{B}^2\right]$$ (3.8)

$$H(t) = \sum_n \frac{1}{2m_n} \left[ {\bf p}_n - q_n \mathbf{A}({\bf r}_n,t)\right] + V_{\rm Coul}+ V_{\bf EP}+V_{\bf gg}$$

$$V_{\rm Coul} \equiv \frac{1}{2}\sum_{n,m}\frac{q_nq_m}{4\pi\varepsilon_0\vert{\bf r}_n-{\bf r}_m\vert}$$

$$V_{\bf EP} \equiv -\sum_n q_n \int d {\bf r} \mathbf{E}_\perp\cdot {\bf g}({\bf r},{\bf r}_n) = - \int d {\bf r} \mathbf{E}_\perp \P_\perp$$

$$V_{\bf gg} \equiv \frac{1}{2\varepsilon_0}\sum_{n,m}q_nq_m\int d {\bf r} {\bf g}_\perp({\bf r},{\bf r}_n) {\bf g}_\perp({\bf r},{\bf r}_m).$$

Up to here everything is completely classical (just classical charges and fields). Quantization of the charge degrees of freedom is done canonically via ${\bf p}_n \to \hat{{\bf p}}_n = -\frac{i}{\hbar}{\bf\nabla}_n$.