Physically unacceptable solutions for the QM angular equation

Question

I'm reading Griffiths's Introduction to Quantum Mechanics 3rd ed textbook [1]. On p.136, the author explains:

But wait! Equation 4.25 (angular equation for the $\theta$-part) is a second-order differential equation: It should have two linearly independent solutions, for any old values of $\ell$ and $m$. Where are all the other solutions? (One is related to the associated Legendre function.) Answer: They exist of course, as mathematical solutions to the equation, but they are physically unacceptable because they blow up at $\theta=0$ and/or $\theta=\pi$ (see Problem 4.5).

In problem 4.5, I can find that the function $A\ln[\tan (\theta/2)]$ satisfies the $\theta$ equation for $\ell=m=0$. And such function blows up at $\theta=0$ and $\theta=\pi$.

But why such function is physically unacceptable? For the wave function to be physically acceptable, it fundamentally needs to be square-integrable. And $\ln[\tan (\theta/2)]$ does actually!

$$\int_{0}^\pi [\ln[\tan (\theta/2)]]^2\sin\theta \text d\theta= \frac{\pi^2}6$$

For the well-behaved function case, it makes sense to set the function condition 'finite' and 'square-integrable' equivalently. In this case, although $\ln[\tan (\theta/2)]$ blows up at $\theta=0$ and $\theta=\pi$, it is still square-integrable tamed by $\sin \theta$ term. So it can be normalized to satisfy the Born's statistical interpretation. But the author says such function is physically unacceptable so I wonder why.

Reference

Griffiths, D. J.; Schroeter, D. F. Introduction to Quantum Mechanics 3rd ed; Cambridge University Press, 2018. ISBN 978-1107189638.

Answer 1

We are in principle trying to solve the angular TISE problem$^1$ $$ \vec{\bf L}^2Y~=~\hbar^2\ell(\ell+1)Y, \qquad {\bf L}_zY~=~\hbar m Y, $$ on the unit 2-sphere $\mathbb{S}^2$. However, we are using a "tropical" coordinate system $(\theta,\phi)$ that is singular at the north & south poles $\theta=0,\pi$. Hence, we should strictly speaking also solve the TISE in mathematically well-defined "arctic/antarctic" coordinate neighborhoods of the north & south poles, respectively, and see if we can glue the local solutions together into a global solution on $\mathbb{S}^2$. Not surprisingly$^2$, the "arctic/antarctic" coordinate solutions have no singularities at the poles. So the gluing is not possible if the tropical $(\theta,\phi)$ coordinate solution displays singularities at $\theta=0,\pi$, i.e. such singularities are physically unacceptable.

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$^1$ Here we stick to the differential-geometric formulation using wavefunctions. There is of course also a well-known algebraic formulation using ladder operators, which we will not address here.

We can assume wlog that $\ell\geq 0$. The single-valuedness of the wavefunction $Y$ implies that the constant $m\in\mathbb{Z}$ is an integer. Its range $|m|$ is bounded by $\ell$ for physical reasons. In particular it follows that for fixed $\ell$, the number of independent tropical solutions are finite.

$^2$ After all the $Y$ solutions should maintain $SO(3)$ covariance. Recall that the tropical solutions $Y$ have no singularities or discontinuities at internal points. In fact they are smooth maps in the interior. This can e.g. be derived by a bootstrap argument a la what is done in my Phys.SE answer here. A formulation using weak solutions doesn't change the main conclusion.

An arctic/antarctic solution should then be a linear combination of the finitely many $90^{\circ}$-rotated tropical solutions for the corresponding problem with ${\bf L}_z$ replaced by, say, ${\bf L}_x$. A finite sum cannot develop internal singularities. $\Box$

Answer 2

As mentioned in the answers to How to know if a wave function is physically acceptable solution of a Schrödinger equation? one should also require square-integrability of higher-order derivatives. In your case this already fails for the first derivative.