# Monopole Spherical Harmonics

I was following Yakov Shnir his book about magnetic monopoles, there they derive the monopole spherical harmonics. I will sketch the derivation briefly,

The starting equation is the eigenvalue problem of the total angular momentum,

\begin{align} \hat{\textbf{J}}^2 Y_{\mu lm}(\theta, \phi) &= \hbar^2 \Big\{\frac{-1}{\sin^2 \theta} \Big[ \sin \theta \frac{\partial}{\partial \theta} \big( \sin \theta \frac{\partial}{\partial \theta} \big) + \big( \frac{\partial}{\partial \varphi} - i\mu (1-\cos \theta )\big)^2 \Big]+ \mu^2\Big\} Y_{\mu lm}(\theta, \phi)\\&= \hbar^2\lambda Y_{\mu lm}(\theta, \phi). \end{align}

The $\theta$-parameter can be replaced by the relation $x=\cos\theta$ and the $\phi$ dependence can be eliminated by $Y_{\mu lm}(\theta, \phi)= \exp(i(\mu+m)\phi) P(\theta)$,

$$\label{eq: diff eq angular part1} \Big\{ -(1-x^2)\frac{d}{dx^2} + 2x \frac{d}{dx} + \frac{(m+\mu x)^2}{1-x^2} +\mu^2 \Big\} P(x) = \lambda P(x).$$

To solve the problem we want to rewrite it as the Euler's hypergeometric function. To obtain that form $P(x)$ has to be rewritten as,

$$P(x) = (1-x)^{-\frac{\mu +m}{2}} (1+x)^{-\frac{\mu -m}{2}} F(x),$$

equation (\ref{eq: diff eq angular part1}) will then become,

$$(1-x^2) \frac{d^2F}{dx^2} + 2 ( m+(\mu-1)x) \frac{dF}{dx} + (\mu - \mu^2 + \lambda)F = 0.$$ If now the transformation $z= (1+x)/2$ is applied, then we have rewritten the differential equation as the Euler's hypergeometric differential equation,

\begin{align} z(1-z) F'' &+ \big\{ m-\mu + 1 + 2z(\mu-1) \big\} F' + (\mu -\mu^2 + \lambda) F \\ & = z(1-z) F'' + \big\{ c-(a+b+1)z \big\}F' -abF = 0, \end{align}

where $c=m-\mu+1$,$~ab = \mu^2 -\mu -\lambda$ and $a+b+1 = 2(1- \mu)$. The solution of this differential equations is the hypergeometric function $_2F_1(a;b;c;z)$ which is defined as, $$_2F_1(a;b;c;z) = \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!} \equiv \sum_{n=0}^{\infty} \alpha_n z^n.$$

Now my question is the following, they consider that $a$ is negative integer such that the sum terminates, my question now is why? Because if the sum not terminates then will the convergence radius be 1, but since $\theta$ is between $0$ and $\pi$ and by definiton of $z$ you still have a bijection from $z$ to the range of $\theta$.

This indeed doesn't say anything about being square integrable but how could one see this from the hypergeometric function/series? In fact I am searching for an analytical argument why the quantum numbers only differ in integers?

• First of all you have $z=\cos(\theta)$, so the radius of convergence being one is ok. Second, there exists indeed simple argument why the spherical harmonic is a fixed order polynomial in $\cos(\theta)$, in fact the order is equal to $l$ if i'm not mistaken. You need to work out a couple more formulae, fistly that $\lambda = l(l+1)$, and then it seems to me things will only work out if $\mu$ is an integer which can also be justified by single-valuedness under $\phi -> \phi+2\pi$. – Kostas Feb 17 '19 at 1:55
This is not a direct answer to the question asked, but rather a different approach to the problem. It's much easier to derive the monopole harmonics from a geometric consideration of the Hopf map ${\rm Hopf}: S^3\to S^2$, which is the same thing as the Hopf bundle $\pi: {\rm SU}(2) \to {\rm SU}(2)/{\rm U}(1)$. Then one immediately sees monopole spherical harmonics coincide with the ${\rm SU}(2)$ represenation matrices. Recall that the ordinary spherical harmonics are given in terms of these matrices as $$Y^L_m(\theta, \phi)= \sqrt{\frac{2L+1}{4\pi}}[D^L_{m,0}(\theta, \phi,\psi)]^*$$ where, on the LHS, $\theta$ $\phi$ are the spherical polar angles on $S^2$. On the RHS $\theta$, $\phi$, $\psi$ are the Euler-angle co-ordinates on $S^3$. The Hopf map simply takes $(\theta, \phi, \psi)$ to $(\theta, \phi)$. Because $$D^L_{m,n}(\theta, \phi,\psi)= <L,m|\exp(-i\phi J_z)\exp(-i\theta J)\exp(-i \psi J_z)|L,n>$$ the RHS does not depend on $\psi$ when $n=0$.
For a monople of strength $\int B d(Area)= 4\pi \Lambda$, where $4\pi \Lambda$ must be an integer multiple of $2\pi$, the monopole harmonics are
$${\mathcal Y}^J_{m, \Lambda} (\theta, \phi,\psi)= \sqrt{\frac{2J+1}{4\pi}}[D^J_{m\Lambda }(\theta,\phi,\psi)]^*$$ Now both sides depend on $\psi$, so one must choose a $\psi$ for each pair $(\theta, \phi)$. This is just a choice of gauge. For a derivation and explanantion of all this see page 278 in my online lecture notes https://courses.physics.illinois.edu/phys509/sp2017/bmaster.pdf.