8.1. Separation of variable in spherical coordinates

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Chapter 8. Separation of variables

8.1. Separation of variable in spherical coordinates

  1. Laplace equation in the ball
  2. Laplace equation outside of the ball
  3. Applications to the theory of Hydrogen atom
  4. Applications to wave equation in the ball

Laplace equation in the ball

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.

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

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): \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.

Applications to wave equation in the ball

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.

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