author hackbard Thu, 14 Feb 2013 15:49:44 +0000 (16:49 +0100) committer hackbard Thu, 14 Feb 2013 15:49:44 +0000 (16:49 +0100)

index 5b79972..306d235 100644 (file)
@@ -208,20 +208,40 @@ This can be shown by rewriting the $LS$ operator

J=L+S \Leftrightarrow J^2=L^2+S^2+2LS \Leftrightarrow
LS=\frac{1}{2}\left(J^2-L^2-S^2\right)
-\text{ ,}

+and corresponding eigenvalue
+\begin{eqnarray}
+j(j+1)-l(l+1)-s(s+1)&=&
+(l\pm\frac{1}{2})(l\pm\frac{1}{2}+1)-l^2-l-\frac{3}{4} \nonumber\\
+&=&
+l^2\pm\frac{l}{2}+l\pm\frac{l}{2}+\frac{1}{4}\pm\frac{1}{2}-l^2-l-\frac{3}{4}
+\nonumber\\
+&=&\pm(l+\frac{1}{2})-\frac{1}{2}=\left\{\begin{array}{rl}
+l & \text{for } j=l+\frac{1}{2}\\
+-(l+1) & \text{for } j=l-\frac{1}{2}
+\end{array}\right.
+\text{ ,}
+\end{eqnarray}
which, if used in equation~\eqref{eq:solid:so_bs2}, gives the same (diagonal) matrix elements
-\begin{align}
+\begin{eqnarray}
\bra{l\pm\frac{1}{2},m\pm\frac{1}{2}}V(\vec{r})
-\ket{l\pm\frac{1}{2},m\pm\frac{1}{2}}&=
+\ket{l\pm\frac{1}{2},m\pm\frac{1}{2}}&=&
\bar{V}_l(\vec{r})+V^{\text{SO}}_l(\vec{r})
-\frac{1}{2}\left(l(l+1)-j(j+1)-\frac{3}{4}\right) \\
-&= \bar{V}_l(\vec{r})+\frac{1}{2}V^{\text{SO}}_l(\vec{r})
+\frac{1}{2}\left(l(l+1)-j(j+1)-\frac{3}{4}\right) \nonumber\\
+&=&\bar{V}_l(\vec{r})+\frac{1}{2}V^{\text{SO}}_l(\vec{r})
\left\{\begin{array}{rl}
--\left(l+\frac{3}{2}\right) & \text{ for } j=l+\frac{1}{2}\\
-\left(l-\frac{1}{2}\right) & \text{ for } j=l-\frac{1}{2}
+l & \text{for } j=l+\frac{1}{2}\\
+-(l+1) & \text{for } j=l-\frac{1}{2}
+\end{array}\right. \nonumber\\
+&=&\frac{1}{2l+1}\left(lV_{l,l-\frac{1}{2}}(\vec{r})+
+                       (l+1)V_{l,l+\frac{1}{2}}(\vec{r})\right)+\nonumber\\
+&&+\frac{1}{2l+1}\left\{\begin{array}{rl}
+l\left(V_{l,l+\frac{1}{2}}(\vec{r})-V_{l,l-\frac{1}{2}}(\vec{r})\right) &
+ \text{for } j=l+\frac{1}{2}\\
+-(l+1)\left(V_{l,l+\frac{1}{2}}(\vec{r})-V_{l,l-\frac{1}{2}}(\vec{r})\right) &
+ \text{for } j=l-\frac{1}{2}
\end{array}\right.
-\end{align}
+\end{eqnarray}
as equation~\eqref{eq:solid:so_bs1}

\text{ .}