Question 7.2: Construct the Hamiltonian and Hamilton’s equations for a cha......

In Cartesian coordinates we have

L =\frac{m}{2}\left(\dot{x}^{2} +\dot{y}^{2}+\dot{z}^{2}\right) -e\phi \left(r,t\right) +\frac{e}{c}v\cdot A\left(r,t\right),   (7.18)

whence

p_{x}=\frac{\partial L}{\partial \dot{x} }=m \dot{x}+\frac{e}{c} A_{x}, p_{y}=\frac{\partial L}{\partial \dot{y} }=m \dot{y}+\frac{e}{c} A_{y}, p_{z}=\frac{\partial L}{\partial \dot{z} }=m \dot{z}+\frac{e}{c} A_{z}.  (7.19)

Thus,

v =\frac{1}{m}\left(p −\frac{e}{c} A\right)  (7.20)

and, consequently,

H = v · p − L =\frac{1}{2m}\left(p −\frac{e}{c} A\right)^{2}+e\phi.  (7.21)

This Hamiltonian is the total energy if \phi and A do not explicitly depend on time (E and B static fields). Hamilton’s equations take the form

\dot{r}= \frac{\partial H}{\partial p }= \frac{1}{m }\left(p −\frac{e}{c} A\right),  (7.22a)

\dot{p}=-\frac{\partial H}{\partial r }\equiv  – \nabla H = \frac{e}{mc}\left[\left(p −\frac{e}{c} A\right) \cdot \nabla A+ \left(p −\frac{e}{c} A\right)\times \left(\nabla \times A\right)\right] – e \nabla \phi,  (7.22b)

where we have used ∇(G · G) = 2(G · ∇)G + 2G × (∇ ×G) with G = p − (e/c)A.