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Vector Spherical Harmonics and Multipoles

Recall that

\begin{displaymath}
{\bf L} = -i \mbox{\boldmath$r$}\times \mbox{\boldmath$\nabla$}.
\end{displaymath} (12.38)

This is an ``orbital'' rotation operator. In systems with spin it is more convenient in many cases to define a ``total'' rotation operator that adds the orbital rotation operator to a ``spin'' rotation operator (defined below). Since total angular momentum (as opposed to orbital angular momentum) is a relativistically invariant quantity that appears ``naturally'' in covariant kinematics, we are inspired to find a representation that is
  1. A vector function of its coordinates.
  2. Simultaneous eigenfunctions of $J^2$, $L^2$, and $J_z$.
  3. Possessed of certain desirable properties we will derive.

Actually, figuring out something like this the first time is not quite so easy; it is full of false starts and exploring alternatives. After the fact, however, it is clear that this is the correct choice. It is also extremely useful in quantum theory.

The total rotation operator is

\begin{displaymath}
{\bf J} = {\bf L} + {\bf S}
\end{displaymath} (12.39)

where
\begin{displaymath}
{\bf S} = i {\bf I} \times
\end{displaymath} (12.40)

is the ``spin'' operator.

Aside: The Spin Operator

S in this expression is a tensor operator. It (like all operators) has no meaning by itself. It is, however, quite different from the scalar operators you are used to. Among other things, when S operates on a vector A, it generates a new vector that points in a different direction. Let us see this.

In the definition of S, I is the identity tensor (unit diagonal matrix) and it is crossed into whatever sits on its right. To understand its action, let us evaluate its cartesian components acting on some vector A:

$\displaystyle S_x {\bf A}$ $\textstyle =$ $\displaystyle i {\bf I}_x \times {\bf A} = i \hat{x} \times {\bf A}$ (12.41)
$\displaystyle S_y {\bf A}$ $\textstyle =$ $\displaystyle i \hat{y} \times {\bf A}$ (12.42)
$\displaystyle S_x {\bf A}$ $\textstyle =$ $\displaystyle i \hat{z} \times {\bf A}$ (12.43)

or (e. g.)
\begin{displaymath}
S_z {\bf A} = i (A_x \hat{y} - A_y \hat{x}).
\end{displaymath} (12.44)

Note that the action of a component of S on a vector A shifts the direction of A to a direction perpendicular to both S and the component. Only by considering the action of all the components can the total vector action of S on A in a given direction be evaluated.

There are several important properties of S. The first is to note that it has the form of an angular momentum operator with a special action on vectors. If we form $S^2$ and evaluate its action on A:

$\displaystyle S^2 {\bf A}$ $\textstyle =$ $\displaystyle - \Big\{ \hat{x} \times (\hat{x} \times {\bf A}) +
\hat{y} \times (\hat{y} \times {\bf A}) + \hat{z} \times (\hat{z} \times {\bf
A}) \Big\}$  
  $\textstyle =$ $\displaystyle - \{ {\bf A} - 3{\bf A} \}$  
  $\textstyle =$ $\displaystyle 2 {\bf A} = s(s + 1){\bf A}$ (12.45)

for $s = 1$. $S^2$ acting on any vector produces 2 times the same vector, telling us that a vector has ``spin angular momentum'' of 1. Note that this connection is universal. In field theory a ``vector boson'' has spin 1. In electrodynamics (quantum or classical) the ``vector radiation field'' has spin one.

The spin operator thus formed is more general, because its action can be extended to higher rank tensors. (2nd rank tensor) gravitational fields have spin 2. Scalar (0th rank tensor) fields have spin 0. To treat more general cases, however, we have to work with tensor indices explicitly and you'll see enough of that in the section on relativity. Feel free to study this matter further. Louck and Biedenharn's book (Encycl. of Math Phys., see me for ref.) contains a much deeper discussion of this entire subject.

It may seem that with such a peculiar structure, $S_z$ can have no eigenvectors. This is not the case. You should verify that

$\displaystyle \chi_1^1$ $\textstyle =$ $\displaystyle -\frac{1}{\sqrt{2}} \big(\hat{x} + i \hat{y} \big)$ (12.46)
$\displaystyle \chi_1^0$ $\textstyle =$ $\displaystyle \hat{z}$ (12.47)
$\displaystyle \chi_1^{-1}$ $\textstyle =$ $\displaystyle \frac{1}{\sqrt{2}} \big(\hat{x} - i \hat{y} \big)$ (12.48)

are eigenvectors such that
\begin{displaymath}
S_z \chi_1^{m_s} = m_s \chi_1^{m_s}
\end{displaymath} (12.49)

for $m_s = -1,0,1$ and
\begin{displaymath}
S^2 \chi_1^{m_s} = s(s+1) \chi_1^{m_s}
\end{displaymath} (12.50)

for $s = 1$. You should also verify the commutation relations for the components of S, that is, show that
\begin{displaymath}
{\bf S} \times {\bf S} = i {\bf S}
\end{displaymath} (12.51)

making it a ``true'' rotation/angular momentum operator.

In addition, we will need to use the operators

\begin{displaymath}
J^2 = J_x J_x + J_y J_y + J_z J_z,
\end{displaymath} (12.52)


\begin{displaymath}
J_z = L_z + S_z
\end{displaymath} (12.53)

(etc.) and
\begin{displaymath}
L^2 = L_x L_x + L_y L_y + L_z L_z
\end{displaymath} (12.54)

so that
\begin{displaymath}
J^2 = L^2 + 2 + 2i {\bf L} \times
\end{displaymath} (12.55)

which can be proven as follows.

Consider its action on A (as usual):

$\displaystyle J^2 {\bf A}$ $\textstyle =$ $\displaystyle \big\{ L^2 + S^2 + 2 {\bf L} \cdot {\bf S} \big\} {\bf
A}$  
  $\textstyle =$ $\displaystyle \big\{ L^2 + 2 + 2i \big[ L_x (\hat{x} \times ~~~) + L_y (\hat{y}
\times ~~~) + L_z (\hat{z} \times ~~~) \big] \big\} {\bf A}$  
  $\textstyle =$ $\displaystyle \big\{ L^2 + S^2 + 2 i ({\bf L} \times ~~~) \big\} {\bf A}$ (12.56)

where the meaning of the latter expression is hopefully now clear.

Then we define the vector spherical harmonics ${\bf Y}_{j,\ell}^m$ by:

$\displaystyle J^2 \mbox{\boldmath$Y$}_{j,\ell}^{m}$ $\textstyle =$ $\displaystyle j(j+1) \mbox{\boldmath$Y$}_{j,\ell}^{m}$ (12.57)
$\displaystyle L^2 \mbox{\boldmath$Y$}_{j,\ell}^{m}$ $\textstyle =$ $\displaystyle \ell(\ell+1) \mbox{\boldmath$Y$}_{j,\ell}^{m}$ (12.58)
$\displaystyle J_z \mbox{\boldmath$Y$}_{j,\ell}^{m}$ $\textstyle =$ $\displaystyle m \mbox{\boldmath$Y$}_{j,\ell}^{m} .$ (12.59)

Note that in order for the latter expression to be true, we might reasonably expect the vector spherical harmonics to be constructed out of sums of products of spherical harmonics and the eigenvectors of the operator $S_z$ defined above. This is the vector analogue of constructing a spinor wavefunction in quantum theory.

In addition, we normalize these orthogonal functions so that they are orthonormal as a dot product. This will allow us to use them to construct projections.

\begin{displaymath}
\int \mbox{\boldmath$Y$}_{j,\ell}^{m \ast} (\theta,\phi) \c...
...\phi) d\Omega = \delta_{jj'} \delta_{\ell \ell'} \delta{m m'}
\end{displaymath} (12.60)

We now need to derive the properties of these functions. We begin by applying $J^2$ to $\mbox{\boldmath$Y$}_{j,\ell}^{m}$

\begin{displaymath}
J^2 \mbox{\boldmath$Y$}_{j, \ell}^{m} = \left \{ L^2 + 2 + 2i {\bf L} \times \right \}
\mbox{\boldmath$Y$}_{j,\ell}^{m}
\end{displaymath} (12.61)

so that we get
\begin{displaymath}
2 i {\bf L} \times \mbox{\boldmath$Y$}_{j,\ell}^{m} = \left...
...ll(\ell+1) - 2
\right \} \mbox{\boldmath$Y$}_{j,\ell}^{m} .
\end{displaymath} (12.62)

Most of the later results will be based on this one, so understand it completely.

If we take ${\bf L} \cdot$ of both sides of (12.62), use a vector identity and recall that ${\bf L} \times {\bf L} = i {\bf L}$ we get:

\begin{displaymath}[j(j+1) - \ell(\ell+1)]{\bf L} \cdot \mbox{\boldmath$Y$}_{j,\ell}^{m} = 0.
\end{displaymath} (12.63)

Similarly, we form the vector product of L with both sides of (12.62):

\begin{displaymath}
\left \{ j(j+1) - \ell(\ell+1) - 2 \right \} {\bf L} \times...
...L} \times ({\bf L} \times \mbox{\boldmath$Y$}_{j,\ell}^{m}) .
\end{displaymath} (12.64)

To reduce this further, we must use the operator vector identity (which you should prove)
\begin{displaymath}
{\bf L} \times ({\bf L} \times {\bf V}) = {\bf L}({\bf L} \cdot {\bf V}) +
i {\bf L} \times {\bf V} - L^2 {\bf V}
\end{displaymath} (12.65)

and eliminate the ${\bf L} \times {\bf Y}$ using (12.62). One gets:
$\displaystyle \left [ j(j+1) - \ell(\ell+1) \right ] \left [j(j+1) - \ell(\ell+1) - 2
\right ] \mbox{\boldmath$Y$}_{j,\ell}^{m}$ $\textstyle =$ $\displaystyle \hfill$  
    $\displaystyle 4\ell(\ell+1) \mbox{\boldmath$Y$}_{j,\ell}^{m} - 4 {\bf L} ({\bf L} \cdot
\mbox{\boldmath$Y$}_{j,\ell}^{m} .$ (12.66)

If we eliminate the ${\bf L} \cdot {\bf Y}$ (using the result above) we get the characteristic equation that is a constraint on the possible values of $j$ and $\ell$:

\begin{displaymath}
x^3 - 2x^2 - 4\ell(\ell+1) x = 0
\end{displaymath} (12.67)

where
\begin{displaymath}
x = j(j+1) - \ell(\ell+1)
\end{displaymath} (12.68)

by definition. The solutions to this factorizable cubic are:

\begin{displaymath}j = \ell, \ell+1, \ell-1, -\ell-1, -\ell-2, -\ell . \end{displaymath}

We only need to consider the solutions with positive $j$ in this problem as the others are not independent in this case. Since $\ell \ge 0$ we only need consider the first three possibilities.

Solutions with $j = \ell$

Then $x = 0$ and

\begin{displaymath}
j(j+1) \mbox{\boldmath$Y$}_{jj}^{m} = {\bf L}({\bf L} \cdot \mbox{\boldmath$Y$}_{jj}^{m})
\end{displaymath} (12.69)

from the third equation above. If we take the dot product of L with this relation, we get
\begin{displaymath}
L^2 ({\bf L} \cdot \mbox{\boldmath$Y$}_{jj}^{m}) = j(j+1)({\bf L} \cdot \mbox{\boldmath$Y$}_{jj}^{m})
\end{displaymath} (12.70)

and we thus see that ${\bf L} \cdot \mbox{\boldmath$Y$}_{jj}^{m} \propto Y_{j,m}$ and so:
\begin{displaymath}
\mbox{\boldmath$Y$}_{jj}^{m} = \frac{1}{\sqrt{j(j+1)}} {\bf L} Y_{j,m}
\end{displaymath} (12.71)

(!) where we have normalized the result.

We have at last found something recognizable. This is precisely the combination of spherical harmonics and L we found in our brief excursion into multipoles! We see that we could have written the (e. g.) magnetic solution as

$\displaystyle \mbox{\boldmath$E$}_L^{(M)}$ $\textstyle =$ $\displaystyle g_\ell(kr) \sqrt{\ell(\ell+1)} \mbox{\boldmath$Y$}_{\ell \ell}^{m}$ (12.72)
$\displaystyle \mbox{\boldmath$B$}_L^{(M)}$ $\textstyle =$ $\displaystyle - \frac{i}{\omega} \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$}_L^{(M)} .$ (12.73)

With just a little more work (later) we will be able to obtain the curl part as a general result, which will really simplify life for us. It is a trivial exercise (left for the reader) to verify that

\begin{displaymath}
J_z \mbox{\boldmath$Y$}_{jj}^{m} = m \mbox{\boldmath$Y$}_{jj}^{m}.
\end{displaymath} (12.74)

One simply plugs in the explicit form of $J_z$ and commutes the resultant $L_z$ with L to cancel the ``spin'' part.

Solutions with $j \ne \ell$

If $j \ne \ell$, we see from the equation after (12.62) that ${\bf L}
\cdot {\bf Y} = 0$. To go further we have to go back to (12.62) and follow a different line. If we multiply both sides by $\hat{r} \cdot$ and $\hat{r} \times$,

\begin{displaymath}
\left [j(j+1) - \ell(\ell+1) -2 \right ] \hat{r} \cdot \mbo...
...hat{r} \cdot {\bf L} \times \mbox{\boldmath$Y$}_{j \ell}^{m}
\end{displaymath} (12.75)

and
\begin{displaymath}
\left [j(j+1) - \ell(\ell+1) -2 \right ] \hat{r} \times \mb...
...{r} \times ({\bf L} \times \mbox{\boldmath$Y$}_{j \ell}^{m})
\end{displaymath} (12.76)

We can reduce these with the vector identities

\begin{displaymath}
\hat{r} \cdot ({\bf L} \times {\bf A}) = 2i\hat{r} \cdot {\bf A} - {\bf L}
\cdot (\hat{r} \times {\bf A})
\end{displaymath} (12.77)

and
\begin{displaymath}
\hat{r} \times ({\bf L} \times {\bf A}) = {\bf L}(\hat{r} \cdot {\bf A}) +
i \hat{r} \times {\bf A} .
\end{displaymath} (12.78)

You should get
\begin{displaymath}
\left [ j(j+1) - \ell(\ell+1) + 2 \right ] \hat{r} \cdot \m...
...f L} \cdot (\hat{r} \times \mbox{\boldmath$Y$}_{j \ell}^{m})
\end{displaymath} (12.79)

and
\begin{displaymath}
\left [ j(j+1) - \ell(\ell+1) \right ] \hat{r} \times \mbox...
...
2i {\bf L} (\hat{r} \cdot \mbox{\boldmath$Y$}_{j \ell}^{m}).
\end{displaymath} (12.80)

Finally, if we plug the second of these into the first and eliminate the cross product, we get the scalar equation:
\begin{displaymath}
\frac{1}{4} \left [ j(j+1) - \ell(\ell+1) \right ] \left [ ...
...m}) = L^2 (\hat{r} \cdot
\mbox{\boldmath$Y$}_{j \ell}^{m}) .
\end{displaymath} (12.81)

This implies that $(\hat{r} \cdot \mbox{\boldmath$Y$}_{j \ell}^{m})$ is a spherical harmonic: that is a constant $\times Y_{k,m}$. What? This is not obvious to you? Well, just this once:

\begin{displaymath}
\left [ \frac{j(j+1) - \ell(\ell+1)}{2} \right] \left [ \frac{j(j+1) -
\ell(\ell + 1)}{2} + 1 \right ] = k(k+1)
\end{displaymath} (12.82)

This has the solutions
  1. $k = \left [ \frac{j(j+1) - \ell(\ell+1)}{2} \right ]$

  2. $k = \left [ \frac{j(j+1) - \ell(\ell+1)}{2} \right ] - 1$.

Since we already know that $j = \ell \pm 1$, we can investigate these two cases explicitly. The positive solutions (in both cases) are easily seen to be $k = j$. We can then construct the complete solutions, since

\begin{displaymath}
\mbox{\boldmath$Y$}_{j,\ell}^{m} = \hat{r} (\hat{r} \cdot \...
...} \times
(\hat{r} \times \mbox{\boldmath$Y$}_{j, \ell}^{m} )
\end{displaymath} (12.83)

is an identity (related to the symmetric/antisymmetric decomposition and hence worth proving) and since we have already shown that
\begin{displaymath}
\hat{r} \times \mbox{\boldmath$Y$}_{j,\ell}^{m} = 2 i \left...
...{-1} {\bf L} (\hat{r} \cdot \mbox{\boldmath$Y$}_{j,\ell}^{m})
\end{displaymath} (12.84)

with $(\hat{r} \cdot \mbox{\boldmath$Y$}_{j,\ell}^{m})$ a constant times $Y_{\ell,m}$. We get:
\begin{displaymath}
\mbox{\boldmath$Y$}_{j,\ell}^{m} = (constant) \left \{ \hat...
...\ell+1)]^{-1} (\hat{r} \times {\bf L}) \right \} Y_{\ell,m} .
\end{displaymath} (12.85)

An exercise will be to verify the normalization of the final solutions:

$\displaystyle \mbox{\boldmath$Y$}_{j,j-1}^{m}$ $\textstyle =$ $\displaystyle - \frac{1}{\sqrt{j(2j+1)}} [-j \hat{r} + i \hat{r}
\times {\bf L}] Y_{\ell,m}$ (12.86)
$\displaystyle \mbox{\boldmath$Y$}_{j,j+1}^{m}$ $\textstyle =$ $\displaystyle - \frac{1}{\sqrt{(j+1)(2j+1)}} [(j+1)\hat{r} + i
\hat{r} \times {\bf L}] Y_{\ell,m}.$ (12.87)

You must also verify that they satisfy the equation for $J_z$.

Finally, you are probably wondering why we have bothered to learn all of this about the $j \ne \ell$ cases in the first place. It is because

$\displaystyle i \mbox{\boldmath$\nabla$}\times (\mbox{\boldmath$Y$}_{jj}^{m} f(r))$ $\textstyle =$ $\displaystyle \sqrt{\frac{j+1}{2j+1}}
\Big[(j+1) \frac{f}{r} + \frac{df}{dr}\Big] \mbox{\boldmath$Y$}_{j,j-1}^{m} \hfill$  
    $\displaystyle \quad \quad + \quad \sqrt{\frac{j}{2j+1}} \Big[-j \frac{f}{r} +
\frac{df}{dr} \Big] \mbox{\boldmath$Y$}_{j,j+1}^{m} .$ (12.88)

The action of the curl mixes the vector spherical harmonics. In fact, it acts to shift $j$ by one in any permitted direction (see handout sheet). Therefore, in order to evaluate the entire EM field and express it compactly, one must use the notation of the vector spherical harmonics. You should prove this, and at leat one of the divergence equations for homework. You will need to get the components of the v.s.h. along and transverse to $\hat{r}$ in order to do the vector algebra.

This is not too bad, but (as we shall see) it is not the best we can do. By carefully defining a particular set of multipolar solutions, we can make our notation itself do almost all the work of doing the curls, etc. so that all we have to do at either end is translate a paticular problem into and out of the notation with the formal solution in hand. Next time we will do just that as we develop the Hansen Multipolar Solutions.


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