# Physically unacceptable solutions for the QM angular equation

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. $$\Box$$