\question
{Shoichi Takahata}
{02/03/12}
{physics2070@yahoo.co.jp}
{5-4}
%question
{
In the privious problem, 
we also have a term which is propotional to ${\mbox{\boldmath $A$}}^{2}$.
In order to neglect this term, this term should be at least 1/10 smaller 
than the linear {\mbox{\boldmath $A$}} term.
What is the corresponding value of the electric field? 
If {\mbox{\boldmath $A$}} gives an electric field above this value,
we should then consider the nonlinear ${\mbox{\boldmath $A$}}^{2}$ effect of light.
}
%answer
{
When monochromatic visible light 
(wavelength ${\lambda}_{\rm light}$, angular frequency $\omega=2\pi c/{\lambda}_{\rm light}$)
enters a graphene layer, 
a $\pi$ electron which exists near the K or K' points 
in the two-dimensional Brillouin zone 
absorbs the light.
The relation between a vector potential {\mbox{\boldmath $A$}} 
and an electric field {\mbox{\boldmath $E$}} of the light 
can be written as:

\begin{equation}
{\mbox{\boldmath $A$}}
=-\frac{i}{\omega} {\mbox{\boldmath $E$}}
=-\frac{{i\lambda}_{\rm light}}{2\pi c} {\mbox{\boldmath $E$}} \label{eq:A}\,,
\end{equation}

by considering the Maxwell equation 
${\mbox{\boldmath $E$}}=-\partial {\mbox{\boldmath $A$}}/\partial t$ 
and ${\mbox{\boldmath $A$}} \propto {\rm exp}(-i\omega t)$. From the Eq.\,(5-4-1), we get

\begin{equation}
{\mbox{\boldmath $A$}}^{2}
=-{\left(\frac{{\lambda}_{\rm light}}{{2\pi c}}\right)}^{2} 
{\mbox{\boldmath $E$}}^{2} \label{eq:A2}\,.
\end{equation}

From the Eq.\,(5.27) in the book, 
the perturbation Hamiltonian $H_{\rm eR}$ for an electron 
in the presence of the vector potential 
is given by 

\begin{eqnarray}
H_{\rm eR}
&=&
\frac{e}{m}\,{\mbox{\boldmath $A$}}\cdot{\mbox{\boldmath $p$}}
+\frac{{e}^2}{2m}\,{\mbox{\boldmath $A$}}^2
\nonumber \\
&=&
-\frac{ie\hbar}{m}\,{\mbox{\boldmath $A$}}\cdot\nabla 
+\frac{{e}^2}{2m}\,{\mbox{\boldmath $A$}}^2
\nonumber \\
&=&
\frac{e\hbar}{m}\,{\mbox{\boldmath $A$}}\cdot{\mbox{\boldmath $k_{e}$}}
+\frac{{e}^2}{2m}\,{\mbox{\boldmath $A$}}^2 \label{eq:HeR}\,,
\end{eqnarray}

where ${\mbox{\boldmath $p$}}
=-i\hbar\nabla=\hbar{\mbox{\boldmath $k_{e}$}}$ 
is a momentum operator of an electorn in vacuum. 
From the Eqs.\,(\ref{eq:A}) and (\ref{eq:A2}), we can obtain the ratio {\it R}
of the absolute value of the term in 
${\mbox{\boldmath $A$}}^{2}$ of the Eq.\,(\ref{eq:HeR})
to the one of {\mbox{\boldmath $A$}} 
as follows:

\begin{equation}
R=\frac{{\displaystyle 
\left| \frac{e^{2}}{2m} {\mbox{\boldmath $A$}}^{2} \right|}}
{{\displaystyle 
\left| \frac{e\hbar}{m}\, {\mbox{\boldmath $A$}}\cdot
{\langle \mbox{\boldmath $k_{e}$} \rangle} \right|}}
=\frac{{\displaystyle 
\frac{e^{2}}{2m}\,{\left(\frac{{\lambda}_{\rm light}}{{2\pi c}}\right)}^{2}}}
{{\displaystyle 
\frac{e\hbar}{m}\,\frac{{\lambda}_{\rm light}}{2\pi c}}} 
\,\frac{|{\mbox{\boldmath $E$}}^{2}|}
{|{\mbox{\boldmath $E$}}\cdot{\langle \mbox{\boldmath $k_{e}$} \rangle}|}
=\frac{e{\lambda}_{\rm light}}{2\pi\hbar c}
\,\frac{|{\mbox{\boldmath $E$}}^{2}|}
{|{\mbox{\boldmath $E$}}\cdot{\langle \mbox{\boldmath $k_{e}$} \rangle}|}
\simeq 
\frac{e{\lambda}_{\rm light}}{2\pi\hbar c}
\,\frac{|{\mbox{\boldmath $E$}}|}{|{\langle \mbox{\boldmath $k_{e}$} \rangle}|} 
\label{eq:R}\,,
\end{equation}\

where $\langle \mbox{\boldmath $k_{e}$} \rangle$ 
is the average of \mbox{\boldmath $k_{e}$}. 
Using Eq.\,(\ref{eq:R}) the corresponding thresfold value of the electric field 
$|{\mbox{\boldmath $E$}}_{\rm th}|$ for $R=1/10$ can be written as:

\begin{equation}
|{\mbox{\boldmath $E$}}_{\rm th}|
=\frac{\pi\hbar c}{5e{\lambda}_{\rm light}}\,
|{\langle \mbox{\boldmath $k_{e}$} \rangle}| \label{eq:Eth}\,.
\end{equation}

The wavenumber of the electoron around the K and K' points
in the graphene layer $|{\langle \mbox{\boldmath $k_{e}$} \rangle}|$ 
can be approximated as follows:

\begin{equation}
|{\langle \mbox{\boldmath $k_{e}$} \rangle}| 
\simeq 
|\mbox{\boldmath $k_{F}$}|
=\frac{m}{\hbar}\,v_{\rm F} \label{eq:ke} \,,
\end{equation} 

where $\mbox{\boldmath $k_{F}$}$ is the Fermi wave vector 
and $v_{\rm F}$ is the Fermi velocity of the $\pi$ electron in the graphene layer. 
From Eq.\,(2.34) in the book, $v_{\rm F}$ is given by

\begin{equation}
v_{\rm F}=\frac{3{\gamma}_0 a_{\rm c-c}}{2\hbar} \label{eq:vF} \,,
\end{equation}

where $a_{\rm c-c}=0.142\,{\rm nm}$ is the nearest neighbor carbon-carbon distance 
and $t=-{\gamma}_0=-3.033 \,{\rm eV}$ is the nearest neighbor transfer integral. 
Using the Eqs.\,(\ref{eq:Eth}), (\ref{eq:ke}) and (\ref{eq:vF}), 
we can obtain the thresfold value of the electric field 
of the monochromatic visible light 
($3.8\le{\lambda}_{\rm light}\le7.5$ in the units of ${10}^{-7}\,{\rm m}$) 
in the units of V$\cdot{\rm m}^{-1}$ as follows:

\begin{equation}
|{\mbox{\boldmath $E$}}_{\rm th}|
=\frac{3{\pi}mc{\gamma}_0 a_{\rm c-c}}{10e\hbar{\lambda}_{\rm light}}\,,
\end{equation}

above which we should consider the nonlinear effect of light. 
In the case of ${\lambda}_{\rm light}=750\,{\rm nm}$, 
$|{\mbox{\boldmath $E$}}_{\rm th}|=2.4\times{10}^{9}\, {\rm V}\cdot{\rm m}^{-1}$.
}