HW #2
2016145001 김승규
The Lagrangian of a particle in electric and magnetic fields is given as
\[L=\sum_i{\frac12m\dot{x}_i^2}-qV(x_i)+q\sum_i{\dot{x_i}\cdot A_i(x_i)}\\i=1,2,3\]We know that the generalized momentum is calculated by
\[p_k=\frac{\partial{L}}{\partial\dot{q_k}}\]So, Generalized momentum will be
\[p=\begin{bmatrix}\frac{\partial{L}}{\partial{\dot{x_1}}}\\\frac{\partial{L}}{\partial{\dot{x_2}}}\\\frac{\partial{L}}{\partial{\dot{x_3}}}\end{bmatrix}\\=\begin{bmatrix}m\dot{x_1}+qA_1(x_1)\\m\dot{x_2}+qA_2(x_2)\\m\dot{x_3}+qA_3(x_3)\\\end{bmatrix}\]Then, we can set the velocity as
\[\dot{x}=\begin{bmatrix}\frac{p_1-qA_1(x_1)}{m}\\\frac{p_2-qA_2(x_2)}{m}\\\frac{p_3-qA_3(x_3)}{m}\\\end{bmatrix}\]On the other hand, the definition of Hamiltonian is
\[H(p_i,q_i)=\sum_i{p_i\dot{q_i}(p_i,q_i)-L(q_i,\dot{q_i})}\]the left-hand side means that we need to change all the velocity in momentum and position.
Then the Hamiltonian will be
\[H(p_i,q_i)=\sum_i{p_i\cdot(\frac{p_i-qA_i(x_i)}{m})}-\frac{\{p_i-qA_i(x_i)\}^2}{2m}+qV(x_i)-qA_i(x_i)\frac{p_i-qA_i(x_i)}{m}\]Finally, simplify the Hamiltonian,
\[H(p_i,q_i)=\sum_i{\frac{\{p_i-qA_i(x_i)\}^2}{2m}+qV(x_i)}\]