Rewriting \bgroup\color{black}${p^2\over 2\mu}$\egroup Using \bgroup\color{black}$L^2$\egroup

We wish to use the \bgroup\color{black}$L^2$\egroup and \bgroup\color{black}$L_z$\egroup operators to help us solve the central potential problem. If we can rewrite \bgroup\color{black}$H$\egroup in terms of these operators, and remove all the angular derivatives, problems will be greatly simplified. We will work in Cartesian coordinates for a while, where we know the commutators.

First, write out \bgroup\color{black}$L^2$\egroup.

\begin{eqnarray*}
L^2&=&(\vec{r}\times\vec{p})^2 \\
&=&-\hbar^2\left[
\left...
...over\partial y}-y{\partial\over\partial x}\right)^2 \right] \\
\end{eqnarray*}


Group the terms.

\begin{eqnarray*}
L^2=&-&\hbar^2\left[
x^2\left({\partial^2\over\partial z^2}...
...er\partial y}
+2z{\partial\over\partial z}\right) \right] \\
\end{eqnarray*}


We expect to need to keep the radial derivatives so lets identify those by dotting \bgroup\color{black}$\vec{r}$\egroup into \bgroup\color{black}$\vec{p}$\egroup. This will also make the units match \bgroup\color{black}$L^2$\egroup.

\begin{eqnarray*}
(\vec{r}\cdot\vec{p})^2&=&-\hbar^2\left(x{\partial\over\parti...
...tial\over\partial y}
+z{\partial\over\partial z}
\right) \\
\end{eqnarray*}


By adding these two expressions, things simplify a lot.

\begin{displaymath}\bgroup\color{black} L^2+(\vec{r}\cdot\vec{p})^2=r^2p^2+i\hbar\vec{r}\cdot\vec{p} \egroup\end{displaymath}

We can now solve for \bgroup\color{black}$p^2$\egroup and we have something we can use in the Schrödinger equation.

\begin{eqnarray*}
p^2&=&{1\over r^2}\left(L^2+(\vec{r}\cdot\vec{p})^2-i\hbar\ve...
...l r}\right)^2
-\hbar^2{1\over r}{\partial\over\partial r} \\
\end{eqnarray*}


The Schrödinger equation now can be written with only radial derivatives and \bgroup\color{black}$L^2$\egroup.

\begin{displaymath}\bgroup\color{black}{-\hbar^2\over 2\mu}\left[{1\over r^2}\le...
... r^2}\right] u_E(\vec{r})+V(r)u_E(\vec{r})=Eu_E(\vec{r})\egroup\end{displaymath}

Jim Branson 2013-04-22