Last time, we decomposed $V$ into $Y_{\ell m}$’s, watched the decomposition solve Laplace mode by mode, and got a feel for what $\ell$ and $m$ actually look like. That finishes the toolkit, so from here on it’s applications.

Truncation

Previously, we had

What happens when the sum has to stop? A computer only holds finitely many $c_{\ell m}$. Pick a largest $\ell$, call it $\ell_{\max}$, and keep every shell up to it:

As an operator, truncation is a sum of shell projectors:

Cost: shell $\ell$ holds $2\ell+1$ coefficients, so

What $\ell_{\max}$ Resolves

Previously, we had $\left< Y_{\ell m} \right>_{\varepsilon}-Y_{\ell m}\approx-\frac{\varepsilon^2}{4}\ell(\ell+1)Y_{\ell m}$. How wiggly is that, as a length? Zoomed in far enough, the sphere is flat, so compare a plane wave $\cos kx$ averaged over a ring of radius $s$:

$\cos\psi'$ integrates to $0$ and $\cos^2\psi'$ to $\pi$.

On a sphere of radius $R$, a ring of geodesic angle $\varepsilon$ has $s=R\varepsilon$. Matching the two:

Locally, $Y_{\ell m}$ is a wave that fits $\ell+\frac{1}{2}$ wavelengths around a great circle.

The finest bump a wave can draw is half a wavelength. So the finest bump $f_{\ell_{\max}}$ can draw, as an angle:

So what is a good $\ell_{\max}$? Invert it. To resolve features of angular size $\Delta\theta$:

$\Delta\theta=1^{\circ}=\frac{\pi}{180}$: $\ell_{\max}\approx180$, i.e. $181^2\approx3.3\times10^{4}$ coefficients. Earth’s gravity field (EGM2008) goes to $\ell_{\max}\approx2160$: $\frac{\pi R_{\oplus}}{2160}\approx9\text{ km}$ at the surface. Planck’s maps of the cosmic microwave background go to $\ell\approx2500$: features down to $\frac{\pi}{2500}\text{ rad}\approx4'$. Same rule from the data side: samples spaced by $\Delta\theta$ can’t pin down anything above $\ell_{\max}\approx\frac{\pi}{\Delta\theta}$ (two samples per wavelength).

That’s resolution. How close is $f_{\ell_{\max}}$ to $f$? Previously, $\braket{\ell' m'}{\ell m}=\delta_{\ell\ell'}\delta_{mm'}$, so the error is just the coefficients that got dropped:

$E_{\ell}$ is shell $\ell$’s share of $\|f\|^2$. How fast it dies depends only on how smooth $f$ is.

Finite sum: exact at a finite $\ell_{\max}$. Previously,

so $\ell_{\max}=2$ is exact.

Smooth bump of angular width $\sigma$ (at the pole, so only $m=0$). In flat space, a Gaussian of width $R\sigma$ has Fourier transform $\propto e^{-k^2(R\sigma)^2/2}$, and $k^2R^2=\ell(\ell+1)$ from above:

The $\sqrt{2\ell+1}$ is from $N_{\ell 0}$ and grows too slowly to matter. Demand that the last kept coefficient is down by a factor $\delta$:

$\delta=10^{-3}$ gives $\ell_{\max}\approx\frac{3.7}{\sigma}$, the same size as $\frac{\pi}{\sigma}$. Checked numerically at $\sigma=0.1$: $\ell_{\max}=37$ leaves $\frac{\|f-f_{\ell_{\max}}\|}{\|f\|}\approx7\times10^{-4}$.

Jump: north hemisphere at $+V_0$, south at $-V_0$. Here $E_{\ell}\sim\ell^{-2}$, so the tail goes like $\frac{1}{\ell_{\max}}$ and the relative error like $\ell_{\max}^{-1/2}$:

Halving the error costs $4\times$ the shells. The overshoot at the edge never goes away: it stays at $\approx1.18\,V_0$ for every $\ell_{\max}$ and only gets narrower (Gibbs).

Recipe: pick $\ell_{\max}\approx\frac{\pi}{\Delta\theta}$ from the smallest feature, then check the dropped tail $\sum_{\ell>\ell_{\max}}E_{\ell}$ against the tolerance. Smooth data converges fast; jumps converge slowly and never pointwise.

Why Cut at a Full Shell

“Full shell” means the cut is in $\ell$ only: every kept $\ell$ keeps all $2\ell+1$ values of $m$. The alternative is an $m$-cut: also drop every $|m|>M$.

Why not cut in $m$? Previously, rotations about $x$ and $y$ shuffle the $m$’s within a shell without changing $\hat{\Lambda}$. In symbols, a rotation $\hat{R}$ commutes with $\hat{\Lambda}$, so it can’t move a ket out of its shell:

So $\hat{R}\hat{P}_{\ell}=\hat{P}_{\ell}\hat{R}$ for every shell, and summing over $\ell\leq\ell_{\max}$ gives $\hat{R}\hat{T}_{\ell_{\max}}=\hat{T}_{\ell_{\max}}\hat{R}$. Truncate-then-rotate is rotate-then-truncate: the blur doesn’t care where $z$ points.

An $m$-cut keeps the single projectors $\ket{\ell m}\bra{\ell m}$ only for $|m|\leq M$. Those commute with $\hat{L}_z$ alone, i.e. with rotations about $z$. Previously, picking a $z$-axis was a free choice; an $m$-cut makes that choice show up in the answer. Previously, $m$ counts wavelengths around a circle of latitude. That circle has circumference $2\pi\sin\theta$, so the finest detail along $\psi$ (as an arc on the unit sphere) is

Near the poles $\sin\theta\to0$, so a small $M$ resolves everything. At the equator it takes $M\approx\ell_{\max}$. Same spot, different blur: with $\ell_{\max}=8$, full shells blur a spot at the pole and a spot on the equator identically; adding $|m|\leq2$ leaves the pole spot alone (only $m=0$ is nonzero at the pole, since $P_{\ell}^{|m|}(\cos\theta)\propto\sin^{|m|}\theta$) and smears the equator spot along the equator.

Other Geometries

Previously, the sphere’s modes came from its two boundary conditions: periodicity in $\psi$ gave $m$, finiteness at the poles gave $\ell$. What changes for a different boundary? The coordinates that fit it, the operator, and so the family of modes. The moves stay the same.

Bessel Functions

Circular drumhead of radius $a$, clamped at the rim. A time-harmonic $u=U e^{-i\omega t}$ turns the wave equation into Helmholtz:

Polar coordinates $(\rho,\psi)$:

Previously, $\hat{\Lambda}=\frac{1}{\sin\theta}\partial_{\theta}(\sin\theta\,\partial_{\theta})+\frac{1}{\sin^2\theta}\partial_{\psi}^2$. Same shape with $\sin\theta\to\rho$, so the same tensor-product split, with $\hat{A}\to\frac{1}{\rho}\partial_{\rho}(\rho\,\partial_{\rho})$ and $\hat{W}\to\frac{1}{\rho^2}$.

$\psi$ slot: unchanged. $\braket{\psi}{m}=e^{im\psi}$, and periodicity gives $m\in\mathbb{Z}$.

$\rho$ slot, at fixed $m$:

The same centrifugal barrier as before, $\frac{m^2}{\rho^2}$ instead of $\frac{m^2}{\sin^2\theta}$. Eigenvalue problem $\hat{D}_m^{(\rho)}R=-k^2R$. Multiply by $\rho^2$ and set $s=k\rho$:

This is Bessel’s equation. Solutions: $J_m(k\rho)$, and a second one that goes like $\rho^{-|m|}$ ($\ln\rho$ for $m=0$) at the center. Infinite at the center, so chuck it, same as $Q_{\ell}^{m}(\cos\theta)$.

Where does $k$ get quantized? The sphere had no edge, so $\ell$ came from finiteness. The drum has one:

$j_{mn}$ is the $n$-th zero of $J_m$, with no closed form: $j_{01}\approx2.405$, $j_{11}\approx3.832$, $j_{21}\approx5.136$, $j_{02}\approx5.520$. On the sphere the eigenvalues were integers, $-\ell(\ell+1)$. Here $\frac{\omega_{mn}}{\omega_{01}}=1.59,\ 2.14,\ 2.30,\dots$ are not: the overtones aren’t harmonics, which is why a drum has no clear pitch.

Orthogonality, by the same argument as before, with weight $\rho$ instead of $\sin\theta$:

$\rho=0$ kills the lower end and the clamp $J_m(k_{mn}a)=0$ kills the upper. On the sphere, $\sin\theta$ killed both.

Nodal lines: $|m|$ diameters from $\cos(m\psi)$ and $n-1$ circles from the zeros of $J_m$ inside the rim. Previously: $\ell-|m|$ latitude circles and $|m|$ meridians.

Laplace inside a cylinder: the same $\rho$ and $\psi$ modes, and the leftover coordinate $z$ gets one ODE per mode, $\partial_{z}^2Z=k_{mn}^2Z$, so $\sinh(k_{mn}z)$ and $\cosh(k_{mn}z)$. Previously, $r$ got $r^{\ell}$ and $r^{-(\ell+1)}$.

Mathieu Functions

Squash the drum into an ellipse with foci at $(\pm d,0)$. Elliptic coordinates $(\mu,\nu)$:

$\mu=\text{const}$: confocal ellipses ($\mu=0$ is the segment between the foci). $\nu=\text{const}$: confocal hyperbolas. $\nu$ is periodic, like $\psi$. Far away, $\cosh\mu\approx\sinh\mu\approx\frac{e^{\mu}}{2}$, so $d\cosh\mu\approx d\sinh\mu\to\rho$ and $\nu\to\psi$: polar coordinates come back.

Laplacian:

Helmholtz, multiplied through by $d^2\left(\sinh^2\mu+\sin^2\nu\right)$, with $\sinh^2\mu=\frac{\cosh2\mu-1}{2}$ and $\sin^2\nu=\frac{1-\cos2\nu}{2}$:

Separate, $U=R(\mu)N(\nu)$, with separation constant $\alpha$:

These are Mathieu’s equation and the modified Mathieu equation.

What’s new? $q$ contains $k$. On the sphere and on the drum, the $\psi$ eigenvalue $m^2$ never knew the frequency. Here the angular eigenvalue $\alpha$ depends on $k$ through $q$, so the angular modes change shape with frequency: the two slots no longer come apart completely.

$\nu$ slot: $N(\nu+2\pi)=N(\nu)$ only for special $\alpha$. $\alpha=\alpha_n^{c}(q)$: even solutions $\mathrm{ce}_n(\nu,q)$. $\alpha=\alpha_n^{s}(q)$: odd solutions $\mathrm{se}_n(\nu,q)$. At $q=0$: $\partial_{\nu}^2N+\alpha N=0$, so $\mathrm{ce}_n=\cos n\nu$, $\mathrm{se}_n=\sin n\nu$ and $\alpha=n^2$. Previously, the $\psi$ slot: $e^{im\psi}$ with eigenvalue $m^2$.

$\mu$ slot: the clamp at the rim $\mu=\mu_0$ picks $q$, and with it $k=\frac{2\sqrt{q}}{d}$:

Both conditions have to hold at the same $q$, so the $\nu$ and $\mu$ problems get solved together, not one after the other.

Degeneracy splits. On the circle, $J_m(ka)=0$ doesn’t care whether the $\psi$ part is $\cos m\psi$ or $\sin m\psi$: same $k$, a degenerate pair (a rotation turns one into the other). The ellipse has no rotation symmetry, $\alpha_n^{c}(q)\neq\alpha_n^{s}(q)$ for $q>0$, so $\mathrm{ce}_n$ and $\mathrm{se}_n$ ring at different $k$. Computed numerically, for an ellipse with the unit circle’s area (semi-axes $1.3$ and $\frac{1}{1.3}$): the pair at $k=3.83$ splits into $3.54$ (nodal line cutting the long axis in half) and $4.53$ (nodal line along the long axis). The fundamental moves from $2.40$ to $2.56$.

Hermite-Gaussian Functions

2D quantum harmonic oscillator, natural units:

Tensor-product split, as for $\hat{\Lambda}$:

Previously, $\hat{\Lambda}=(\hat{A}\otimes\hat{1})-(\hat{W}\otimes\hat{L}_z^2)$: the $\theta$ slot carried $\hat{W}=\frac{1}{\sin^2\theta}$, which is why $\hat{D}_m$ depended on $m$. Here there is no cross term. Each slot is diagonalized alone, the labels never talk, and $E=E_{n_x}+E_{n_y}$.

1D slot, $\hat{h}\phi=E\phi$. Peel off the Gaussian, $\phi=h(x)\,e^{-x^2/2}$:

The $\frac{1}{2}x^2h$ from the potential cancels the $-\frac{1}{2}x^2h$ from the second derivative.

Power series, $h=\sum_k a_kx^k$:

If it never stops, $\frac{a_{k+2}}{a_k}\to\frac{2}{k}$, the same ratio as $e^{x^2}=\sum_j\frac{x^{2j}}{j!}$. Then $\phi\sim e^{x^2}e^{-x^2/2}=e^{x^2/2}$: not normalizable. Previously, the Legendre series diverged at the poles until $\lambda=\ell(\ell+1)$ cut it off. Same fix: stop the series at $k=n$ (and kill the other parity, as with $a_1=0$ before):

The polynomial that’s left is the Hermite polynomial $H_n(x)$.

The 2D modes and energies:

Nodal lines: $n_x$ vertical and $n_y$ horizontal, a checkerboard of lobes. Shell $N=n_x+n_y$ holds $N+1$ degenerate modes. The potential is round, so rotations about the origin shuffle modes within a shell, as rotations shuffled $m$ within an $\ell$-shell.

Laser beams: the paraxial wave equation for a beam along $z$, $\nabla_{\perp}^2u+2ik\,\partial_{z}u=0$, has the form of the 2D Schrödinger equation with $z$ playing time. The Hermite-Gaussian patterns are the $\mathrm{TEM}_{n_xn_y}$ transverse modes of a laser cavity: they keep their shape as they propagate, widening with the beam and picking up a phase proportional to $n_x+n_y+1$ (the Gouy phase). The shell index again.

Spheroidal Wave Functions

Stretch the sphere into a cigar. Prolate spheroidal coordinates $(\xi,\eta,\psi)$, foci at $z=\pm d$, with $\rho$ the distance from the $z$-axis:

$\xi\geq1$ and $-1\leq\eta\leq1$. $\xi=\text{const}$: prolate spheroids. $\eta=\text{const}$: hyperboloids. $\psi$ is untouched. Far away, $\xi\gg1$: $d\xi\to r$ and $\eta\to\cos\theta$. Spherical coordinates come back.

Laplacian:

Helmholtz, multiplied through by $d^2(\xi^2-\eta^2)$, with $\braket{\psi}{m}=e^{im\psi}$ in the $\psi$ slot and

Separate, $U=R(\xi)\,S(\eta)\,e^{im\psi}$, with separation constant $\lambda$:

One equation on two ranges: $-1\leq\eta\leq1$ for the angular functions $S_{mn}(c,\eta)$, $\xi\geq1$ for the radial ones. $c=kd$ compares the focal distance with the wavelength.

Recognize it? At $c=0$, put $\eta=\cos\theta$. Previously, the $\theta$ equation:

So $S_{mn}(0,\cos\theta)=P_{n}^{|m|}(\cos\theta)$ and $\lambda=n(n+1)$. At $c=0$ the wave can’t tell the spheroid from a sphere.

At $c>0$, the extra $c^2\eta^2$ acts as a potential: the eigenvalue problem is

The Legendre modes don’t diagonalize it: $\eta P_{n}^{|m|}(\eta)$ only reaches $P_{n\pm1}^{|m|}(\eta)$, so $\eta^2$ couples $n$ to $n\pm2$. Rediagonalize:

Only $\ell-n$ even. The matrix is diagonal, $\ell(\ell+1)$, plus $c^2$ times a band ($\ell\leftrightarrow\ell\pm2$). Previously, we diagonalized $\hat{\Lambda}$ in $\ket{\ell m}$; now the operator has changed, so the basis gets rotated within each $m$-block. $m$ stays a good label because $\psi$ is untouched.

Large $c$: the potential $c^2\eta^2$ squeezes $S$ toward $\eta=0$. There $1-\eta^2\approx1$, and for $m=0$:

Rescale, $t=\sqrt{c}\,\eta$:

That’s the oscillator from before, so $\frac{\lambda}{2c}=n+\frac{1}{2}$:

Numerically at $c=14$: $\lambda=13.2,\ 40.2,\ 66.0,\ 90.6$ against $c(2n+1)=14,\ 42,\ 70,\ 98$. Low $n$ fits best, since the approximation needs the mode to stay near $\eta=0$. Spheroidal functions interpolate: Legendre at $c=0$, Hermite-Gaussian as $c\to\infty$.

Oblate spheroids (a squashed sphere): $c\to ic$ flips the potential to $-c^2\eta^2$, which pushes $S$ toward the poles $\eta=\pm1$ instead.

Where they show up: waves scattering off or radiating from elongated bodies (antennas, needles); the $\mathrm{H}_2^{+}$ ion, whose two nuclei sit at the foci; and Slepian’s prolate spheroidal functions in signal processing, where the $S_{0n}$ are the band-limited signals most concentrated in a time window.

Laguerre Polynomials

Hydrogen, atomic units:

Previously, $\hat{\nabla}^2=\frac{1}{r^2}\partial_{r}(r^2\partial_{r})+\frac{1}{r^2}\hat{\Lambda}$ and $\mel{\ell m}{\hat{\Lambda}}{\Psi}=-\ell(\ell+1)\braket{\ell m}{\Psi}$. Sandwich $\hat{H}\ket{\Psi}=E\ket{\Psi}$ with $\bra{\ell m}$:

The same radial operator as Laplace, plus a potential and an energy. $\frac{\ell(\ell+1)}{2r^2}$ is the centrifugal barrier again.

Peel off both ends. Near $r=0$ the barrier wins, and previously that gave $r^{\ell}$. Far away, $-\frac{1}{2}\partial_{r}^2\approx E$ with $E<0$ gives $e^{-\kappa r}$, $\kappa=\sqrt{-2E}$:

What’s left for $u$:

Power series, $u=\sum_kb_ks^k$:

If it never stops, $\frac{b_{k+1}}{b_k}\to\frac{1}{k}$, like $e^{s}$. Then the mode goes like $e^{-\kappa r}e^{2\kappa r}=e^{\kappa r}$: not normalizable. Same fix, third time. Stop the series at $k=n_r$:

The polynomial that’s left is the associated Laguerre polynomial $L_{n-\ell-1}^{2\ell+1}(s)$, with $s=\frac{2r}{n}$:

Nodes: $n-\ell-1$ spheres from the zeros of $L_{n-\ell-1}^{2\ell+1}$, plus the $\ell$ angular nodal surfaces of $Y_{\ell m}$ from before: $n-1$ in total.

Degeneracy: $E$ depends on $n$ alone, and shell $n$ holds

The same count as $(\ell_{\max}+1)^2$ in Truncation, with $\ell_{\max}=n-1$. Rotations explain the $2\ell+1$ within each $\ell$. The degeneracy across $\ell$ is special to $\frac{1}{r}$ (an extra conserved quantity, the Runge-Lenz vector).

Laguerre-Gaussian: the 2D oscillator from before, in polar coordinates:

$\psi$ slot: $e^{im\psi}$. $\rho$ slot: $-\frac{1}{2}\hat{D}_m^{(\rho)}+\frac{1}{2}\rho^2$, with $\hat{D}_m^{(\rho)}$ from the drum. Peel off $\rho^{|m|}$ at the center and $e^{-\rho^2/2}$ far away, and the series stops the same way:

Same energies as the Hermite-Gaussians with $N=2p+|m|=n_x+n_y$. They are two bases of the same shells, the way $Y_{\ell m}$ about different $z$-axes are two bases of one $\ell$-shell. Shell $N=2$, with real $\psi$ parts:

In optics, Laguerre-Gaussian beams are the “donut” modes. The $e^{im\psi}$ makes them eigenkets of $\hat{L}_z$, so each photon carries orbital angular momentum $m\hbar$.

Poisson’s Equation

With a charge density $\rho(\vec{r})$:

Sandwich with $\bra{\ell m}$, as for Laplace:

The same radial operator, now driven: each mode is pushed only by its own piece of the charge. The modes still don’t talk.

Point charge $q$ at $\vec{r}\,'=r'\hat{n}'$. Previously, $\braket{\hat{n}'}{\hat{n}}=\frac{\delta(\theta-\theta')}{\sin\theta}\,\delta(\psi-\psi')$. Insert the completeness of $\ket{\ell m}$:

So $\rho=q\,\frac{\delta(r-r')}{r^2}\braket{\hat{n}}{\hat{n}'}$ has one coefficient per mode:

Off the shell $r=r'$, the right side vanishes and it’s Laplace, with $r^{\ell}$ and $r^{-(\ell+1)}$ from before. Finite at $0$ inside, at $\infty$ outside, continuous at $r'$:

The $\delta$ fixes $C_{\ell}$. Multiply by $r^2$ and integrate across $r=r'$; only the derivative jumps:

Sum the modes and compare with Coulomb, $\varphi=\frac{q}{4\pi\varepsilon_0|\vec{r}-\vec{r}\,'|}$:

Charge on the $z$-axis: previously, pointing $z$ along the symmetry axis kills every $m\neq0$. With $Y_{\ell 0}^*(\hat{z})=\sqrt{\frac{2\ell+1}{4\pi}}$ and $Y_{\ell 0}(\hat{n})=\sqrt{\frac{2\ell+1}{4\pi}}P_{\ell}(\cos\theta)$:

Interpretation: the charge acts like boundary data on the sphere $r=r'$. Each mode is Laplace’s $r^{\ell}$ inside and $r^{-(\ell+1)}$ outside, and the charge is the kink that joins them.

Multipole expansion: for $r$ outside all the charge, $r_<=r'$ and $r_>=r$. Superposing point charges:

$\ell=0$ is the total charge, $\ell=1$ the dipole, $\ell=2$ the quadrupole. The $\ell$-th term dies as $r^{-(\ell+1)}$, so far away the lowest nonzero $\ell$ wins. Previously, from far away the sphere at $V_0\cos^2\theta$ looked like a point charge. Check at $\ell=0$, with $q_{00}=\frac{Q}{\sqrt{4\pi}}$ and $Y_{00}=\frac{1}{\sqrt{4\pi}}$:

Other PDEs

Same sandwich, different equation. In each one, $\bra{\ell m}$ turns the angular part into $-\ell(\ell+1)$, and the leftover coordinate gets one ODE per mode.

Heat on a spherical shell of radius $R$, diffusion constant $D$:

Sandwich:

The leftover coordinate is $t$. High $\ell$ die first, so diffusion is a soft low-pass filter where truncation was a hard one. $\ell=0$ never decays (eigenvalue $0$), so the sphere settles to the average of the starting temperature. A hot spot at the north pole, $f(0)=\braket{\hat{n}}{\hat{z}}$, with $Y_{\ell m}^*(\hat{z})=\delta_{m0}\sqrt{\frac{2\ell+1}{4\pi}}$:

At short times this is a Gaussian of angular width $\sigma=\frac{\sqrt{2Dt}}{R}$: the same $e^{-\ell(\ell+1)\sigma^2/2}$ as the smooth bump in Truncation.

Waves on a sphere (a vibrating shell, the surface of a star), wave speed $v$:

Each shell rings at one frequency, all $2\ell+1$ values of $m$ together. Helioseismology labels the Sun’s oscillation modes by this $\ell$.

Helmholtz in 3D (sound in a spherical room, a particle in a ball):

Solutions: the spherical Bessel function $j_{\ell}(kr)$, and a second one that blows up at $r=0$. Near the center the barrier beats $k^2$, and $j_{\ell}(kr)\propto(kr)^{\ell}$: Laplace’s $r^{\ell}$ from before. With $U=0$ on $r=R$, $k_{\ell n}=\frac{(n\text{-th zero of }j_{\ell})}{R}$. For $\ell=0$, $j_0(kr)=\frac{\sin kr}{kr}$, so $k_{0n}=\frac{n\pi}{R}$ exactly.

Any central potential $V(r)$: the same $Y_{\ell m}$, and the radial equation gets $\frac{\ell(\ell+1)}{2r^2}+V(r)$. Hydrogen was $V=-\frac{1}{r}$.

Conclusion: Diagonalize and Decouple

Everything in this series is one trick: find the operator the problem is secretly diagonal in, and the PDE falls apart into independent pieces.

The recipe, as it played out on the sphere: (1) Pick the operator acting on the coordinates the boundary data lives on: $\hat{\Lambda}$ on $(\theta,\psi)$. (2) Diagonalize it. The boundary conditions decide which modes are allowed: periodicity in $\psi$ gave $m$, finiteness at the poles gave $\ell$. (3) The leftover coordinate gets one ODE per mode, with the eigenvalue as a parameter: $r$, with $r^{\ell}$ and $r^{-(\ell+1)}$. (4) Fix the coefficients by projection: Fourier’s trick on $V$ gave $\braket{\ell m}{V}$.

References

1. R. L. Herman, An Introduction to Mathematical Physics via Oscillations, Chapter 9, “Oscillations in Higher Dimensions,” University of North Carolina Wilmington (2012). PDF, licensed CC BY-NC-SA 3.0.
2. F. W. J. Olver et al., eds., NIST Digital Library of Mathematical Functions, Chapter 10: Bessel Functions, Chapter 18: Orthogonal Polynomials, Chapter 28: Mathieu Functions and Hill’s Equation, and Chapter 30: Spheroidal Wave Functions.
3. J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley (1999), Chapters 3 and 4.
4. A. E. Siegman, Lasers, University Science Books (1986).
5. N. K. Pavlis, S. A. Holmes, S. C. Kenyon, and J. K. Factor, “The development and evaluation of the Earth Gravitational Model 2008 (EGM2008),” Journal of Geophysical Research 117, B04406 (2012). doi:10.1029/2011JB008916
6. Planck Collaboration, “Planck 2018 results. I. Overview and the cosmological legacy of Planck,” Astronomy & Astrophysics 641, A1 (2020). doi:10.1051/0004-6361/201833880, arXiv:1807.06205
7. D. Slepian and H. O. Pollak, “Prolate spheroidal wave functions, Fourier analysis and uncertainty – I,” Bell System Technical Journal 40, 43–63 (1961). doi:10.1002/j.1538-7305.1961.tb03976.x
8. L. Allen, M. W. Beijersbergen, R. J. C. Spreeuw, and J. P. Woerdman, “Orbital angular momentum of light and the transformation of Laguerre-Gaussian laser modes,” Physical Review A 45, 8185 (1992). doi:10.1103/PhysRevA.45.8185

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We started by treating functions on a sphere as vectors and ended up with drums, lasers, hydrogen, and heat, all from diagonalizing one operator and letting the PDE fall apart into independent pieces. Hopefully, the next time a $Y_{\ell m}$ shows up, we both know what the hell we’re looking at. This concludes my blog post series on spherical harmonics.