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The Hamiltonian

This has to be worked out in detail, starting from the Lagrangian, which is described in Woolley [8]. The result is
$\displaystyle {\mathcal H}(t)$ $\displaystyle =$ $\displaystyle H(t) + H_{\rm rad}$  
$\displaystyle H_{\rm rad}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int d{\bf r} \left[ \varepsilon_0 \mathbf{E}_\perp^2 + \mu_0^{-1} \mathbf{B}^2\right]$ (3.8)
$\displaystyle H(t)$ $\displaystyle =$ $\displaystyle \sum_n \frac{1}{2m_n} \left[ {\bf p}_n - q_n \mathbf{A}({\bf r}_n,t)\right] + V_{\rm Coul}+
V_{\bf EP}+V_{\bf gg}$  
$\displaystyle V_{\rm Coul}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{2}\sum_{n,m}\frac{q_nq_m}{4\pi\varepsilon_0\vert{\bf r}_n-{\bf r}_m\vert}$  
$\displaystyle V_{\bf EP}$ $\displaystyle \equiv$ $\displaystyle -\sum_n q_n \int d {\bf r} \mathbf{E}_\perp\cdot {\bf g}({\bf r},{\bf r}_n)
= - \int d {\bf r} \mathbf{E}_\perp \P_\perp$  
$\displaystyle V_{\bf gg}$ $\displaystyle \equiv$ $\displaystyle \frac{1}{2\varepsilon_0}\sum_{n,m}q_nq_m\int d {\bf r} {\bf g}_\perp({\bf r},{\bf r}_n)
{\bf g}_\perp({\bf r},{\bf r}_m).$  

Up to here everything is completely classical (just classical charges and fields). Quantization of the charge degrees of freedom is done canonically via \bgroup\color{col1}$ {\bf p}_n \to \hat{{\bf p}}_n = -\frac{i}{\hbar}{\bf\nabla}_n$\egroup.



Subsections
next up previous contents index
Next: Coulomb Gauge Up: Gauge invariance for many Previous: Charge and current densities,   Contents   Index
Tobias Brandes 2005-04-26