# Multipole expansion and spectral decomposition

We can always write a Hermitian operator in the form of its spectral decomposition:

$$\hat{A}=\sum_i \lambda_i | \chi_i(\boldsymbol{r})\rangle\langle\chi_i(\boldsymbol{r}')|$$

where $$\lambda_i$$ are the eigenvalues and $$\chi_i$$ the eigenvectors of the operator. The corresponding integral kernel can be expanded as well:

$$\hat{A}f(\boldsymbol{r})=\int A(\boldsymbol{r},\boldsymbol{r}') f(\boldsymbol{r}') d\boldsymbol{r}'$$

with

$$A(\boldsymbol{r},\boldsymbol{r}')=\sum_i \lambda_i \chi_i(\boldsymbol{r})\chi_i(\boldsymbol{r}')$$

So far so good. Now let's consider the Poisson equation:

$$\Delta g = u$$

which can be solved by inverting the Laplace operator:

$$g = \Delta^{-1}u$$

The corresponding kernel to $$\Delta^{-1}$$ is just the Coulomb kernel or Green's function for a spherical problem with Dirichlet BC at infinity $$\frac{1}{|\boldsymbol{r}-\boldsymbol{r}'|}$$:

$$\Delta^{-1}f = \int \frac{1}{|\boldsymbol{r}-\boldsymbol{r}'|} f(\boldsymbol{r}')d\boldsymbol{r}'$$

Now we should be able to express the Coulomb kernel using the eigenfunctions of $$\Delta^{-1}$$ with the spectral decomposition introduced above. Is this what leads us somehow to the multipole/Laplace expansion of the Coulomb kernel given as:

$$\frac{1}{|\boldsymbol{r}-\boldsymbol{r}'|}=\sum_{lm} (-1)^m I_l^{-m}(\boldsymbol{r})R_l^m(\boldsymbol{r}')$$

, where $$I$$ and $$R$$ are the solid harmonics?

In comparing this to the expression above

$$A(\boldsymbol{r},\boldsymbol{r}')=\sum_i \lambda_i \chi_i(\boldsymbol{r})\chi_i(\boldsymbol{r}')$$

I do not get why suddenly in this expansion we have a product of two different functions. And also are $$I$$ and $$R$$ really eigenfunctions of the inverse Laplacian? Hope someone can help my problems in connecting these two concepts.

This is covered, for example, in J.D. Jackson, Classical Electrodynamics, section 3.12 Eigenfunction expansions for Green functions.

Essentially, a separated solution for the eigenfunctions will have your angular $$\ell$$, $$m$$ indices and a radial index $$k$$. That is for your case, you could have spherical harmonics and spherical Bessel functions $$\chi \sim j_\ell(k r) Y_{\ell m}(\theta,\phi)$$ with a sum over $$\ell$$ and $$m$$ and an integral over $$k$$ to produce the Fourier transform of $$k^{-2}$$ that gives Coulomb's law.

The integral over $$k$$ can be done analytically and depends on whether $$r or $$r>r'$$. This gives the multipole expansion. The more usual way, as Jackson explains, is to separate variables and then enforce Laplace's equation. In that way, the radial Green function equation is solved directly, and the sum is over the indices of the first two separated variables only.

Jackson's section shows how this triple sum is equal to the double sum in particular cases. For your case, in spherical coordinates the usual separation of variables in $$\theta$$ and $$\phi$$ give spherical harmonics, and then the $$r$$ equation is usually solved to get the Green function as $$$$\frac{1}{|\vec r- \vec r'|} = 4\pi \sum_{\ell m} \frac{1}{2\ell + 1} \frac{r^\ell_<}{r^{\ell+1}_>} Y^*_{\ell m}(\theta' ,\phi') Y_{\ell m}(\theta,\phi)$$$$ which is Jackson's Eq. 3.70.
As noted above in spherical coordinates eigenfunctions of the Laplacian are proportional to $$j_\ell(kr)Y_{\ell m}(\theta,\phi)$$ with eigenvalue $$-k^2$$. Normalizing to get the correct delta function (see Jackson Eq. 3.112 for the relevant integral), your spectral representation for this is $$$$\frac{1}{|\vec r- \vec r'|} = 4\pi \left [ \frac{2}{\pi} \int_0^\infty dk k^2 \sum_{\ell m} \frac{j_\ell(kr') Y^*_{\ell m}(\theta' ,\phi') j_\ell(kr) Y_{\ell m}(\theta,\phi)}{k^2} \right ]$$$$ where the $$k^2$$ from the completeness and the $$k^2$$ from the eigenvalue cancel to give $$$$\frac{1}{|\vec r- \vec r'|} = 8 \sum_{\ell m} Y^*_{\ell m}(\theta' ,\phi') Y_{\ell m}(\theta,\phi) \int_0^\infty dk j_\ell(kr') j_\ell(kr)$$$$ One way to do the integral is to use the usual contour integration for these problems. First note the integral is even in $$k$$, so integrate from $$-\infty$$ to $$\infty$$ and divide by 2. The integrand is analytic on the real axis, so we can change the contour to integrate around the origin with a small semicircle in the lower half plane. Next split the $$2j_\ell(x) = h_\ell^{(1)}(x)+h_\ell^{(2)}(x)$$ and note that $$h_{\ell}^{(1)} \sim e^{ix}/x$$ and $$h_\ell^{(2)}\sim e^{-ix}/x$$ for large $$x$$. We close the contour such that a large semicircle at infinity is zero. Since the integrand for the four terms has poles only at the origin, only those terms that close in the upper half plane contribute. These terms are proportional to $$h_\ell^{(1)}(k r_>)j_\ell(k r_<)$$ where we can combine the $$h_\ell$$ functions back to a $$j_\ell$$ for the smaller of $$r$$ and $$r'$$. Expanding around the $$k$$ origin, using Jackson 9.88, $$$$h_\ell^{(1)}(kr_>)j_\ell(kr_<) = -\frac{i}{k(2\ell+1)} \frac{r_<^\ell}{r_>^{\ell+1}} +...$$$$ where the $$...$$ part is not singular. $$2\pi i$$ times the residue gives $$$$\frac{1}{|\vec r- \vec r'|} = 4\pi \sum_{\ell m} \frac{1}{2\ell + 1} \frac{r^\ell_<}{r^{\ell+1}_>} Y^*_{\ell m}(\theta' ,\phi') Y_{\ell m}(\theta,\phi)$$$$ as before. I didn't check my algebra carefully, so there may be a few algebra mistakes, but this should show you how it works.