Angular Momentum

For the common problem of central potentials \bgroup\color{black}$V(r)$\egroup, we use the obvious rotational symmetry to find that the angular momentum, \bgroup\color{black}$\vec{L}=\vec{x}\times \vec{p}$\egroup, operators commute with \bgroup\color{black}$H$\egroup,

\begin{displaymath}\bgroup\color{black}[H,L_z]=[H,L_x]=[H,L_y]=0\egroup\end{displaymath}

but they do not commute with each other.

\begin{displaymath}\bgroup\color{black}[L_x,L_y]\neq 0\egroup\end{displaymath}

We want to find two mutually commuting operators which commute with \bgroup\color{black}$H$\egroup, so we turn to \bgroup\color{black}$L^2=L_x^2+L_y^2+L_z^2$\egroup which does commute with each component of \bgroup\color{black}$L$\egroup.

\begin{displaymath}\bgroup\color{black}[L^2,L_z]=0\egroup\end{displaymath}

We chose our two operators to be \bgroup\color{black}$L^2$\egroup and \bgroup\color{black}$L_z$\egroup.

Some computation reveals that we can write

\begin{displaymath}\bgroup\color{black}p^2={1\over r^2}\left(L^2+(\vec{r}\cdot\vec{p})^2-i\hbar\vec{r}\cdot\vec{p}\right).\egroup\end{displaymath}

With this the kinetic energy part of our equation will only have derivatives in \bgroup\color{black}$r$\egroup assuming that we have eigenstates of \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}

The Schrödinger equation thus separates into an angular part (the \bgroup\color{black}$L^2$\egroup term) and a radial part (the rest). With this separation we expect (anticipating the angular solution a bit)

\begin{displaymath}\bgroup\color{black}u_E(\vec{r})=R_{E\ell}(r) Y_{\ell m}(\theta,\phi)\egroup\end{displaymath}

will be a solution. The \bgroup\color{black}$Y_{\ell m}(\theta,\phi)$\egroup will be eigenfunctions of \bgroup\color{black}$L^2$\egroup

\begin{displaymath}\bgroup\color{black}L^2 Y_{\ell m}(\theta,\phi)=\ell(\ell+1)\hbar^2 Y_{\ell m}(\theta,\phi)\egroup\end{displaymath}

so the radial equation becomes

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

We must come back to this equation for each \bgroup\color{black}$V(r)$\egroup which we want to solve.

We solve the angular part of the problem in general using angular momentum operators. We find that angular momentum is quantized.

\begin{displaymath}\bgroup\color{black}L_zY_{\ell m}(\theta,\phi)=m\hbar Y_{\ell m}(\theta,\phi) \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}L^2 Y_{\ell m}(\theta,\phi)=\ell(\ell+1)\hbar^2 Y_{\ell m}(\theta,\phi)\egroup\end{displaymath}

with \bgroup\color{black}$\ell$\egroup and \bgroup\color{black}$m$\egroup integers satisfying the condition \bgroup\color{black}$-\ell\leq m\leq\ell$\egroup. The operators that raise and lower the $z$ component of angular momentum are

\begin{displaymath}\bgroup\color{black}L_\pm=L_x\pm iL_y\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} L_\pm Y_{\ell m}=\hbar\sqrt{\ell(\ell+1)-m(m\pm 1)}Y_{\ell(m\pm 1)}\egroup\end{displaymath}

We derive the functional form of the Spherical Harmonics \bgroup\color{black}$Y_{\ell m}(\theta,\phi)$\egroup using the differential form of the angular momentum operators.

Jim Branson 2013-04-22