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Angular momentum and spherical harmonics

The angular part of the Laplace operator $\nabla^2$ can be written:

\begin{displaymath}
\frac{1}{r^2} \left\{ \frac{1}{\sin
\theta}\frac{\partial }...
...rac{\partial^2 }{\partial \phi^2} \right\} = -\frac{L^2}{r^2}
\end{displaymath} (12.1)

Eliminating $-r^2$ (to solve for the $L^2$ differential equation) one needs to solve an eigenvalue problem:

\begin{displaymath}
L^2 \psi = e \psi
\end{displaymath} (12.2)

where $e$ are the eigenvalues, subject to the condition that the solution be single valued on $\phi \in [0,2\pi)$ and $\theta \in
[0,\pi]$.

This equation easily separates in $\theta,\phi$. The $\phi$ equation is trivial - solutions periodic in $\phi$ are indexed with integer $m$. The $\theta$ equation one has to work at a bit - there are constraints on the solutions that can be obtained for any given $m$ - but there are many ways to solve it and at this point you should know that its solutions are associated Legendre polynomials $P_{\ell,m}(x)$ where $x = \cos\theta$. Thus the eigensolution becomes:

\begin{displaymath}
L^2 Y_{\ell m}= \ell(\ell+1)Y_{\ell m}
\end{displaymath} (12.3)

where $\ell = 0,1,2...$ and $m = -\ell, -\ell+1,...,\ell-1,\ell$ and is typically orthonormal(ized) on the solid angle $4\pi$.

The angular part of the Laplacian is related to the angular momentum of a wave in quantum theory. In units where $\hbar = 1$, the angular momentum operator is:

\begin{displaymath}
\mbox{\boldmath$L$}= \frac{1}{i} (\mbox{\boldmath$x$}\times \mbox{\boldmath$\nabla$})
\end{displaymath} (12.4)

and
\begin{displaymath}
L^2 = L_x^2 + L_y^2 + L_z^2
\end{displaymath} (12.5)

Note that in all of these expressions ${\bf L}, L^2, L_z$, etc. are all operators. This means that they are applied to the functions on their right (by convention). When you see them appearing by themselves, remember that they only mean something when they are applied, so $\mbox{\boldmath$\nabla$}$'s out by themselves on the right are ok.

The $z$ component of $L$ is:

\begin{displaymath}
L_z = -i\frac{\partial}{\partial \phi}
\end{displaymath} (12.6)

and we see that in fact $Ylm$ satisfies the two eigenvalue equations:
\begin{displaymath}
L^2 Y_{\ell m}= \ell(\ell+1)Y_{\ell m}
\end{displaymath} (12.7)

and
\begin{displaymath}
L_z Y_{\ell m} = m Y_{\ell m}
\end{displaymath} (12.8)

The $Ylm$'s cannot be eigensolutions of more than one of the components of $\mbox{\boldmath$L$}$ at once. However, we can write the cartesian components of L so that they form an first rank tensor algebra of operators that transform the $Y_{\ell m}$, for a given $\ell$, among themselves (they cannot change $\ell$, only mix $m$). This is the hopefully familiar set of equations:

$\displaystyle L_+$ $\textstyle =$ $\displaystyle L_x + i L_y$ (12.9)
$\displaystyle L_-$ $\textstyle =$ $\displaystyle L_x - i L_y$ (12.10)
$\displaystyle L_0$ $\textstyle =$ $\displaystyle L_z$ (12.11)

The Cartesian components of $\mbox{\boldmath$L$}$ do not commute. In fact, they form a nice antisymmetric set:
\begin{displaymath}[L_i,L_j]= i \epsilon_{ijk} L_k
\end{displaymath} (12.12)

which can be written in the shorthand notation
\begin{displaymath}
{\bf L} \times {\bf L} = i {\bf L}.
\end{displaymath} (12.13)

Consequently, the components expressed as a first rank tensor also do not commute among themselves:

\begin{displaymath}[ L_+ , L_- ]= 2 L_z
\end{displaymath} (12.14)

and
\begin{displaymath}[ L_\pm , L_z ]= \mp L_\pm
\end{displaymath} (12.15)

but all these ways of arranging the components of $\mbox{\boldmath$L$}$ commute with $L^2$:
\begin{displaymath}[L_i,L^2]= 0
\end{displaymath} (12.16)

and therefore with the Laplacian itself:
\begin{displaymath}[\nabla^2,L_i]= 0
\end{displaymath} (12.17)

which can be written in terms of $L^2$ as:
\begin{displaymath}
\nabla^2 = \frac{1}{r} \frac{\partial^2}{\partial r^2}(r \:) -
\frac{L^2}{r^2}
\end{displaymath} (12.18)

As one can easily show either by considering the explict action of the actual differential forms on the actual eigensolutions $Y_{\ell m}$ or more subtly by considering the action of $L_z$ on $L_\pm Y_{\ell \ell}$ (and showing that they behave like raising and lower operators for $m$ and preserving normalization) one obtains:

$\displaystyle L_+ Y_{\ell m}$ $\textstyle =$ $\displaystyle \sqrt{ (\ell - m) (\ell + m + 1)} \: Y_{\ell, m+1}$ (12.19)
$\displaystyle L_- Y_{\ell m}$ $\textstyle =$ $\displaystyle \sqrt{ (\ell + m) (\ell - m + 1)} \: Y_{\ell, m-1}$ (12.20)
$\displaystyle L_z Y_{\ell m}$ $\textstyle =$ $\displaystyle m Y_{\ell m}$ (12.21)

Finally, note that L is always orthogonal to r where both are considered as operators and r acts from the left:

\begin{displaymath}
{\bf r} \cdot {\bf L} = 0.
\end{displaymath} (12.22)

You will see many cases where identities such as this have to be written down in a particular order.

Before we go on to do a more leisurely tour of vector spherical harmonics, we pause to motivate the construction.


next up previous contents
Next: Magnetic and Electric Multipoles Up: Vector Multipoles Previous: Vector Multipoles   Contents
Robert G. Brown 2013-01-04