The Lorentz Force from the Classical Hamiltonian

In this section, we wish to verify that the Hamiltonian

\begin{displaymath}\bgroup\color{black} H={1\over 2m}\left(\vec{p}+{e\over c}\vec{A}\right)^2-e\phi \egroup\end{displaymath}

gives the correct Lorentz Force law in classical physics. We will then proceed to use this Hamiltonian in Quantum Mechanics.

Hamilton's equations are

\dot{q}&=&{\partial H \over \partial p} \\
\dot{p}&=&-{\partial H \over \partial q} \\

where \bgroup\color{black}$\vec{q}\equiv\vec{r}$\egroup and the conjugate momentum is already identified correctly \bgroup\color{black}$\vec{p}\equiv\vec{p}$\egroup. Remember that these are applied assuming q and p are independent variables.

Beginning with \bgroup\color{black}$\dot{q}={\partial H/\partial p}$\egroup, we have

{d\vec{r}\over{dt}}&=&{1\over m}\left( \vec{p} + {e\over c}\ve...
...{p} + {e\over c}\vec{A} \\
\vec{p}&=&m\vec{v}-{e\over c}\vec{A}

Note that \bgroup\color{black}$\vec{p}\neq m\vec{v}$\egroup. The momentum conjugate to \bgroup\color{black}$\vec{r}$\egroup includes momentum in the field. We now time differentiate this equation and write it in terms of the components of a vector.

\begin{displaymath}\bgroup\color{black}{dp_i\over dt} = m {dv_i\over{dt}}-{e\over c}{dA_i\over{dt}.}\egroup\end{displaymath}

Similarly for the other Hamilton equation (in each vector component) \bgroup\color{black}$\dot{p_i}=-{\partial H\over\partial x_i}$\egroup, we have

\begin{displaymath}\bgroup\color{black}{dp_i\over dt}=\dot{p_i}=-{e\over {mc}}\l...
...over\partial x_i} +e\vec{\partial\phi\over\partial x_i}.\egroup\end{displaymath}

We now have two equations for \bgroup\color{black}${dp_i\over dt}$\egroup derived from the two Hamilton equations. We equate the two right hand sides yielding

\begin{displaymath}\bgroup\color{black}ma_i = m{dv_i\over dt} = -{e\over mc}\lef...
...ial\phi\over\partial x_i}
+ {e\over c}{d A_i\over dt}. \egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black}ma_i = -{e\over mc}\left(m\vec{v}\right)
...ial\phi\over\partial x_i}
+ {e\over c}{d A_i\over dt}. \egroup\end{displaymath}

The total time derivative of \bgroup\color{black}$A$\egroup has one part from \bgroup\color{black}$A$\egroup changing with time and another from the particle moving and \bgroup\color{black}$A$\egroup changing in space.

\begin{displaymath}\bgroup\color{black}{d\vec{A}\over{d t}}={\partial\vec{A}\ove...
...tial t}} +

so that

\begin{displaymath}\bgroup\color{black}F_i=ma_i = -{e\over c}\vec{v}\cdot{\parti...
+ {e\over c}\left(\vec{v}\cdot\vec{\nabla}\right)A_i. \egroup\end{displaymath}

We notice the electric field term in this equation.

\begin{displaymath}\bgroup\color{black} e {\partial\phi\over{\partial x_i}} + {e\over c}{\partial A_i\over{\partial t}}
= -eE_i\egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black}F_i = ma_i = -{eE_i}+{e\over c}\left[-\ve...
+ \left(\vec{v}\cdot\vec{\nabla}\right)A_i\right]. \egroup\end{displaymath}

Let's work with the other two terms to see if they give us the rest of the Lorentz Force.

\begin{displaymath}\bgroup\color{black}{e\over c}\left[\left(\vec{v}\cdot\vec{\n...
...rtial x_j}}
- {\partial A_j\over{\partial x_i}}\right] \egroup\end{displaymath}

We need only prove that

\begin{displaymath}\bgroup\color{black}\left(\vec{v}\times \vec{B}\right)_i=v_j ... x_i}} -
{\partial A_i\over{\partial x_j}} \right). \egroup\end{displaymath}

To prove this, we will expand the expression using the totally antisymmetric tensor.

\begin{displaymath}\bgroup\color{black}\left(\vec{v}\times \vec{B}\right)_i
...l x_m}} \left(\varepsilon_{mnk}\varepsilon_{jki}\right) \egroup\end{displaymath}

\begin{displaymath}\bgroup\color{black} = -v_j {\partial A_n\over{\partial x_m}}...
...tial x_i}} - {\partial A_i\over{\partial x_j}} \right). \egroup\end{displaymath}


So we have

\begin{displaymath}\bgroup\color{black}F_i=-eE_i-{e\over c}\left(\vec{v}\times\vec{B}\right)_i\egroup\end{displaymath}

which is the Lorentz force law. So this is the right Hamiltonian for an electron in a electromagnetic field. We now need to quantize it.

Jim Branson 2013-04-22