8.1. Separation of variable in spherical coordinates


# Chapter 8. Separation of variables

## 8.1. Separation of variables in spherical coordinates

### Harmonic polynomials

Consider Laplace equation in spherical coordinates defined by (6.3.7)--(6.3.8) $$\Delta =\partial_\rho^2 + \frac{2}{\rho}\partial_\rho + \frac{1}{\rho^2}\Lambda \label{eq-8.1.1}$$ with $$\Lambda:= \bigl(\partial_{\phi}^2 + \cot(\phi)\partial_\phi \bigr) +\frac{1}{\sin^2(\phi)}\partial_{\theta}^2. \label{eq-8.1.2}$$ 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 Theorem-7.2.2 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.

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

Sketch of the proof.

1. Homogeneous polynomials of degree $l$ form a space of dimension ${l+2 \choose 2}=\frac{(l+2)(l+1)}{2}$.
2. For every homogeneous polynomial $v$ of degree $l-2$ there exists a homogeneous polynomial $u$ of degree $l$ such that $v=\Delta u$. One can prove it easily!
3. Therefore the dimension of the space harmonic homogeneous polynomials of degree $l$ equals ${l+2 \choose 2}-{l \choose 2}=2l+1$.

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 $$\Lambda Y(\phi,\theta)=-l(l+1)Y(\phi,\theta). \label{eq-8.1.5}$$

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 $$\underbracket{\frac{\sin^2(\phi) \bigl(\Phi'' + \cot(\phi)\Phi' \bigr)}{\Phi}+l(l+1)\sin^2(\phi)} + \underbracket{\frac{\Theta''}{\Theta}}=0. \label{eq-8.1.6}$$ 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 $$\sin^2(\phi) \Phi'' +\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi. \label{eq-8.1.9}$$

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

Theorem 2. $\Phi(\phi)=L(\cos(\phi))$ where $L=P^m_l(z)$ is a polynomial for even $m$ and $L=P^m_l(z)$ is a polynomial multiplied by $\sqrt{1-z^2}=: \sin(\phi)$ for odd $m$.

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

Legendre polynomials satisfy ODE $$\bigl((1-z^2)P_l'(z))'+l(l+1)P_l=0 \label{8.1.10}$$ and are defined by $$P_l(z)=\frac{1}{2^ll!}\bigl(\frac{d\ }{dz}\bigr)^l (z^2-1)^l. \label{8.1.11}$$

Associated Legendre polynomials satisfy ODE $$\bigl((1-z^2)P^m_l{}'(z)\bigr)'+\bigl[l(l+1)-\frac{m^2}{1-z^2}\bigr]P^m_l=0 \label{8.1.12}$$ and are defined by $$P_l^m(z)=(-1)^m (1-z^2)^{m/2}\bigl(\frac{d\ }{dz}\bigr)^mP_l(z). \label{8.1.13}$$

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

Remark 1.

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 $\phi,\theta$.
2. Such harmonics for a basis in the linear space of spherical harmonics with fixed $l$;
3. Choice of the polar axis $z$ matters here: selecting other direction bring us a different basis.

### Laplace equation outside the ball

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
$0$ $1/\rho$
$1$ $x/\rho^3$, $y/\rho^3$, $z/\rho^3$
$2$ $xy/\rho^5$, $xz/\rho^5$, $yz/\rho^5$, $(x^2-y^2)/\rho^5$, $(x^2-z^2)/\rho^5$

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

### Applications to the theory of Hydrogen atom

Spherical harmonics play crucial role in the problems with the spherical symmetry, in particular mathematical theory of Hydrogen-like atoms (with $1$-electron): $$-\frac{\hbar^2}{2\mu}\Delta \Psi - \frac{Ze^2}{\rho} \Psi = E\Psi. \label{eq-8.1.14}$$ 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 $$-P'' -\frac{2}{\rho}P' - \frac{\eta}{\rho}P + \frac{l(l+1)}{\rho^2}P = -\alpha^2P \label{eq-8.1.15}$$ 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 $$\rho Q''+2(l+1-\alpha \rho)Q'+\bigl(\eta-2\alpha (l+1)\bigr)Q=0. \label{eq-8.1.16}$$ 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.

### Applications to wave equation in the ball

Consider now 3D-wave equation in the ball $$u_{tt}-c^2 \Delta u=0\qquad \rho \le a \label{eq-8.1.17}$$ with Dirichlet or Neumann boundary conditions. Separating $t$ and the spatial variables $u=T(t) v(x,y,z)$ we get Helmholtz equation $$\Delta v=-\lambda v\qquad \rho \le a \label{eq-8.1.18}$$ with the same boundary condition and $$T''=-c^2 \lambda T. \label{eq-8.1.19}$$ Separating $\rho$ from spherical variables $\phi,\theta$ we get \begin{equation*} \underbracket{\frac{\rho^2 P''+2\rho P'+\lambda \rho^2 P}{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 $$\rho^2 P''+2\rho P'+(\lambda \rho^2 - l(l+1) )P=0. \label{eq-8.1.20}$$ As $\lambda=1$ solutions 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, \label{eq-8.1.21}\\ j'_l (a\sqrt{\lambda}) =0, \label{eq-8.1.22} \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 $jl$ or $jl'$ respectively.