Rewriting ${p^2\over 2\mu}$ Using $L^2$

We wish to use the $L^2$ and $L_z$ operators to help us solve the central potential problem. If we can rewrite $H$ 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 $L^2$.

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

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 $\vec{r}$ into $\vec{p}$. This will also make the units match $L^2$.

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

By adding these two expressions, things simplify a lot.

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

We can now solve for $p^2$ 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 $L^2$.

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