\question
{Shotaro Maruyama}       
{2/3/12}          
{shotaro.maruyama@cmpt.phys.tohoku.ac.jp}  
{2-16}
{For a given $(n,m)$ SWNT, show the expression of $\vec{T} = (t_1 , t_2)$ as a function of $n $ and $m$. What is the length of $\vec{T}$ ?}
{
If we apply the condition that 
\begin{eqnarray}
\vec{T} \cdot \vec{C_h} = 0,
\end{eqnarray}
we get
\begin{eqnarray}
(t_1\vec{a_1}+t_2\vec{a_2}) \cdot (n\vec{a_1}+m\vec{a_2}) = 0,
\end{eqnarray}
where {$\vec{a_1}=(\frac{\sqrt{3}}{2},\frac{1}{2})a$}  and  {$\vec{a_2}=(\frac{\sqrt{3}}{2},\frac{-1}{2})a$}. Eq.\,(2-16-2) can be expressed by
\begin{eqnarray}
(m+2n)t_1+(n+2m)t_2=0.
\end{eqnarray}
Thus we get
\begin{eqnarray}
t_1:t_2=(n+2m):-(m+2n).
\end{eqnarray}
The condition that {\rm gcm}$(t_1,t_2)=1$ gives
\begin{eqnarray}
t_1&=&\frac{n+2m}{d_R},\nonumber \\
t_2&=&\frac{m+2n)}{d_R},
\end{eqnarray}
where\ $d_R={\rm gcm}$$(n+2m,m+2n)$. The length of $\vec{T}$ is given by using the formula  $|\vec{C_h}|=a\sqrt{n^2+m^2+nm}$:
\begin{eqnarray}
T
& = &
|\vec{T}|
\nonumber \\
&= &
\sqrt{t_1^2a^2+t_2^2a^2+t_1t_2a^2}
\nonumber \\
&= &
\frac{a}{d_R}\sqrt{(n+2m)^2+(m+2n)^2-(n+2m)(m+2n)}
\nonumber \\
& = &
\frac{a}{d_R}\sqrt{3(n^2+m^2+nm)}
\nonumber \\
& = &
\frac{\sqrt{3}C_h}{d_R}.
\end{eqnarray}
}
