\question
{Shoichi Takahata}
{02/03/12}
{physics2070@yahoo.co.jp}
{5-3}
%question
{
When we consider the Hamiltonian in the presence of a vector potential, expand the Hamiltonian 
and retain the linear term in $\mbox{\boldmath $A$}$. 
This corresponds to a perturbation Hamiltonian for the electron-photon coupling constant.
Use the Coulomb gauge ${\rm div}\mbox{\boldmath $A$}=0$ when you obtain this result.
}
%answer
{
We consider an electron in the presence of a vector potential \mbox{\boldmath $A$}. 
Since the charge of an electron is $-e$, the form of the Hamiltonian for the electron is given by

\begin{eqnarray}
H
&=&
\frac{1}{2m}(\mbox{\boldmath $p$}+e\mbox{\boldmath $A$})^2+V(\mbox{\boldmath $r$})
\nonumber \\
&=&
\frac{1}{2m}\,(-i\hbar\nabla+e\mbox{\boldmath $A$})^2+V(\mbox{\boldmath $r$})
\nonumber \\
&=&
\frac{1}{2m}\,
(-{\hbar}^{2}\triangle
-ie\hbar{\bf A}\cdot\nabla
-ie\hbar\nabla\cdot\mbox{\boldmath $A$}
+{e}^2\mbox{\boldmath $A$}^2)
+V(\mbox{\boldmath $r$}) \label{eq:H(1)} \,.
\end{eqnarray}

Since $\nabla$ in the third term in the third row of Eq.\,(\ref{eq:H(1)}) operates 
not only \mbox{\boldmath $A$} but wavefunctions, 
we can obtain the following equation:

\begin{equation}
\nabla\cdot\mbox{\boldmath $A$}
={\rm div}\mbox{\boldmath $A$}
+\mbox{\boldmath $A$}\cdot\nabla
=\mbox{\boldmath $A$}\cdot\nabla \label{eq:operation} \,,
\end{equation}

if we use the fact that 
\mbox{\boldmath $A$} satisfies the Coulomb gauge ${\rm div}\mbox{\boldmath $A$}=0$. 
Substituting Eq.\,(\ref{eq:operation}) into Eq.\,(\ref{eq:H(1)}), 
the form of the Hamiltonian can be written as:

\begin{equation}
H
=H_0
-\frac{ie\hbar}{m}\,\mbox{\boldmath $A$}\cdot\nabla
+\frac{{e}^2\mbox{\boldmath $A$}^2}{2m} \label{eq:H(2)} \,,
\end{equation}

where $H_0$ is defined by

\begin{equation}
H_0=-\frac{{\hbar}^2}{2m} \triangle+V(\mbox{\boldmath $r$}) \,.
\end{equation}

From Eq.\,(\ref{eq:H(2)}), 
we can obtain the linear term in \mbox{\boldmath $A$} of the Hamiltonian $H_{\rm eR}$ 
as follows:

\begin{equation}
H_{\rm eR}=-\frac{ie\hbar}{m}\,\mbox{\boldmath $A$}\cdot\nabla \label{eq:H(eR)}\,.
\end{equation}

}