$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

Consider Laplace equation in spherical coordinates defined by
(6.3.7)--(6.3.8)
\begin{equation}
\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho +
\frac{1}{\rho^2}\Lambda
\label{eq-8.1.1}
\end{equation}
with
\begin{equation}
\Lambda:=
\bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2.
\label{eq-8.1.2}
\end{equation}
Let us plug $u=P(\rho)Y(\phi,\theta)$ into $\Delta u=0$: \begin{equation*}
P''(\rho)Y(\phi,\theta) + \frac{2}{\rho}P' (\rho)Y(\phi,\theta) + \frac{1}{\rho^2} P(\rho)\Lambda Y(\phi,\theta)=0
\end{equation*}
which could be rewritten as
\begin{equation*}
\frac{\rho^2 P''(\rho) + \rho P' (\rho)}{P(\rho)}+
\frac{\Lambda Y(\phi,\theta)}{Y(\phi,\theta)}=0
\end{equation*}
and since the first term depends only on $\rho$ and the second only on
$\phi, \theta$ we conclude that both are constant:
\begin{align}
&\rho^2 P'' +2\rho P' = \lambda P,\label{eq-8.1.3}\\[3pt]
&\Lambda Y(\phi,\theta)=-\lambda Y(\phi,\theta).
\label{eq-8.1.4}
\end{align}
The first equation is of Euler type and it has solutions $P:=\rho^l$ iff
$\lambda= l(l+1)$. However if we are considering ball, solution *must* be infinitely smooth in its center due to some general properties of Laplace equation and this is possible iff $l=0,1,2,\ldots$ and in this case $u$ must be a polynomial of $(x,y,z)$.

**Definition 1.**
Such polynomials are called *harmonic polynomials*.

One can prove

**Theorem 1.**
Harmonic polynomials of degree $l$ form
$(2l+1)$-dimensional space.

Table 1

$l$ | Basis in the space of harmonic polynomials |
---|---|

$0$ | $1$ |

$1$ | $x$, $y$, $z$ |

$2$ | $xy$, $xz$, $yz$, $x^2-y^2$, $x^2-z^2$ |

$3$ | $x^3-3xz^2$, $y^3-3yz^2$, $xz^2-xy^2$, $yz^2-yx^2$, |

$xyz$, $x^2z-y^2z$, $2z^3-3x^2z-3y^2z$ |

Then \begin{equation} \Lambda Y(\phi,\theta)=-l(l+1)Y(\phi,\theta). \label{eq-8.1.5} \end{equation}

**Definition 2.**
Solutions of $\Lambda v=0$ are called *spherical harmonics*.

To find spherical harmonics we apply method of separation of variables again: $Y(\phi,\theta)=\Phi(\phi)\Theta(\theta)$. Recalling (\ref{eq-8.1.2}) we see that \begin{equation} \underbracket{\frac{\sin^2(\phi) \bigl(\Phi'' + \cot(\phi)\Phi' \bigr)}{\Phi}+l(l+1)\sin^2(\phi)} + \underbracket{\frac{\Theta''}{\Theta}}=0. \label{eq-8.1.6} \end{equation} Therefore again both terms in the left-hand expression must be constant: \begin{align} &\sin^2(\phi) \bigl(\Phi'' +\cot(\phi)\Phi' \bigr) = -\bigl(l(l+1)\sin^2(\phi) -\mu \bigr)\Phi, \label{eq-8.1.7}\\[3pt] &\Theta''=-\mu\Theta. \label{eq-8.1.8} \end{align} The second equation is easy, and keeping in mind $2\pi$-periodicity of $\Theta$ we get $\mu=m^2$ and $\Theta = e^{-im\phi}$ with $m=-l,1-l,\ldots,l-1,l$ (for $|m|>l$ we would not get a polynomial).

Therefore (\ref{eq-8.1.7}) becomes \begin{equation} \sin^2(\phi) \Phi'' +2\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi, \label{eq-8.1.9} \end{equation}

One can prove that $\Phi$ is a polynomial of $\cos(\phi)$:

**Theorem 2.**
$\Phi(\phi)=L(\cos(\phi))$.

Such polynomials are called Legendre polynomials as $m=0$ and Associated Legendre polynomials as $m\ne 0$.

Therefore we number spherical harmonics by $l,m$: we have $Y_{lm}$ with $l=0,1,\ldots$ and $m=-l,1-l,\ldots,l-1,m$.

**Remark 1.**

- We are talking now about spherical harmonics with separated $\phi,\theta$; linear combination of spherical harmonics with the same $l$ but different $m$ is again a spherical harmonic albeit without separated $\phi,\theta$.
- Such harmonics for a basis in the linear space of spherical harmonics with fixed $l$;
- Choice of the polar axis $z$ matters here: selecting other direction bring us a different basis.

Consider solutions of the Laplace equation for $\rho>0$ decaying as $\rho\to \infty$. Since spherical harmonics are already defined we have $\lambda=-l(l+1)$ and then $P=\rho^{k}$ with $k<0$ satisfying $k(k+1)=l(l+1)$ which implies that $k=-1-l$. In particular we get from Table 1

Table 2.

$l$ | Basis in the space of homogeneous harmonic functions |
---|---|

$0$ | $1$ |

$1$ | $x/\rho^3$, $y/\rho^3$, $z/\rho^3$ |

$2$ | $xy/\rho^5$, $xz/\rho^5$, $yz/\rho^5$, $(x^2-y^2)/\rho^5$, $(x^2-z^2)/\rho^5$ |

with $\rho=(x^2+y^2+z^2)^{1/2}$.

Spherical harmonics play crucial role in the problems with the spherical symmetry, in particular mathematical theory of Hydrogen-like atoms (with $1$-electron): \begin{equation} -\frac{\hbar^2}{2\mu}\Delta \Psi - \frac{Ze^2}{\rho} \Psi = E\Psi. \label{eq-8.1.10} \end{equation} Here $\hbar$ is a Planck constant, $-Ze$ is the charge of the nucleus, $e$ is the charge of electron, $\mu$ is its mass, $E<0$ is an energy level.

After separation of variables we get $\Psi = P(\rho)Y_{lm}(\phi,\theta)$ with $P$ satisfying \begin{equation} -P'' -\frac{2}{\rho}P' - \frac{\eta}{\rho}P + \frac{l(l+1)}{\rho^2}P = -\alpha^2P \label{eq-8.1.11} \end{equation} with $\eta= 2\mu Ze^2 \hbar^{-2}$, $\alpha= (-2E\mu )^{\frac{1}{2}}\hbar^{-1}$.

Solutions are found in the form of $e^{-\alpha\rho}\rho^l Q(\rho)$ where $Q(\rho)$ is a polynomial satisfying \begin{equation} \rho Q''+(2l+2-2\alpha \rho)+(\eta-2\alpha)\rho -2\alpha l)Q=0 \label{eq-8.1.12} \end{equation} It is known that such solution (polynomial of degree exactly $n-l-1$, $n=l+1,l+2,\ldots$) exists and is unique (up to a multiplication by a constant) iff $2\alpha (n-1)+ 2\alpha -\eta=0$ i.e. $\alpha= \frac{\eta}{2n}$ and also $l\le n-1$. Such polynomials are called Laguerre polynomials.

Therefore $E_n =- \frac{\kappa}{n^2}$ (one can calculate $\kappa$) and has multiplicity $\sum_{l=0}^{n-1} \sum_{m=-l}^l 1= \sum_{l=0}^{n-1} (2l+1)=\frac{1}{2}n(n+1)$.

**Remark 2.**
We see that $E_n$ are very degenerate. Different perturbations decrease or remove degenerations splitting these eigenvalues into clusters of less degenerate or non-degenerate eigenvalues.

Consider now 3D-wave equation in the ball
\begin{equation}
u_{tt}-c^2 \Delta u=0\qquad \rho \le a
\label{eq-8.1.13}
\end{equation}
with Dirichlet or Neumann boundary conditions. Separating $t$ and the spatial variables $u=T(t) v(x,y,z)$ we get *Helmholtz equation*
\begin{equation}
\Delta v=-\lambda v\qquad \rho \le a \label{eq-8.1.14}
\end{equation}
with the same boundary condition and
\begin{equation}
T''=-c^2 \lambda T
\label{eq-8.1.15}
\end{equation}
Separating $\rho$ from spherical variables $\phi,\theta$ we get
\begin{equation*}
\underbracket{\frac{\rho^2 P''+2\rho P'+\lambda P}{\rho^2 P}}+
\underbracket{\frac{\Lambda Y }{Y}}=0
\end{equation*}
and therefore both selected expressions must be $\mu$ and $-\mu$ respectively. So $Y(\phi,\theta)$ is a spherical harmonic and $\mu=l(l+1)$. Then
\begin{equation}
\rho^2 P''+2\rho P'+(\lambda \rho^2 - l(l+1) )P=0.
\label{eq-8.1.16}
\end{equation}
As $\lambda=1$ Ssolutions are *spherical Bessel functions* $j_l$ and $y_l$ which are called spherical Bessel functions of the 1st kind and of the 2nd kind, respectively, and the former are regular at $0$.

So $P=j_l(\rho\sqrt{\lambda})$ and for $u$ to satisfy Dirichlet or Neumann boundary conditions we need to impose the same conditions to $P$ resulting in \begin{gather} j_l (a\sqrt{\lambda}) =0,\\ j'_l (a\sqrt{\lambda}) =0,\end{gather} and then $\lambda = z_{l,n}^2a^{-2}$ and $\lambda = w_{l,n}^2a^{-2}$ respectively where $z_{l,n}$ and $w_{l,n}$ are $n$-th zero of $j_l$ or $j_l'$ respectively.