14
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ The condition for a quantum mechanical wave function is square integrability. That's not the condition in E&M. $\endgroup$
    – Buzz
    Aug 27 '20 at 8:18
  • $\begingroup$ Shouldn't the wave function be continuous as well for it to be physically acceptable? $\endgroup$ Aug 27 '20 at 8:22
5
$\begingroup$

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.

--

$^1$ 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. $\Box$

$\endgroup$
8
  • 1
    $\begingroup$ 1. Is the term "arctic/antarctic coordinates" a well-defined term of art referring to a unique coordinate system? Or does it just refer to any coordinate system whose domain includes a neighbourhood of the North/South pole? $\endgroup$ Aug 28 '20 at 12:48
  • 1
    $\begingroup$ In any case, I find this answer less clear than I would like this argument to be. 2. By 'singularity', does this mean a discontinuity, or does it include any place where the solution goes to infinity? If the latter, why exactly are square-integrable singularities (even on a smooth coordinate system) ruled out? $\endgroup$ Aug 28 '20 at 12:50
  • 3
    $\begingroup$ Let me summarize. We are actually soving $L^2 \psi =0$. Where $L^2$is selfadjoint so that $\psi$ must belong to its selfadjointness domain. $L^2$ is the Laplace operator on $S^2$ and the domain of selfadjointness is the second Sobolev space. Due to elliptic regularity, every solution must be classic and smooth. Smoothness is referred to the smooth atlas on $S^2$. If you pass form the spherical coordinates to local smooth coordnates around each pole you see that our candidate solution is not smooth. Therefore it cannot belong to the selfadjointness domain of $L^2$. $\endgroup$ Aug 28 '20 at 16:28
  • 2
    $\begingroup$ I think it is correct, but all that cannot be understand simply declaring that the candidate solution is "physically unacceptable". If $L^2$ were replaced by a non-elliptic operator all the reasoning would be false. I think the comment of the book is very misleading... $\endgroup$ Aug 28 '20 at 16:31
  • 2
    $\begingroup$ I find these sort of statement (I am referring to the book) quite dangerous and OP's question very healthy: he/she is completely right to raise this issue. I believe these statements produce a drift towards completely herroneous feelings. When I was student I spent hours to anayze and prove wrong similar ideas. Some authors do not know why some statement is true and it would be much safer for all the community to avoid to suggest misconceptions. Here the point is "elliptic regularity" (thanks Qmechanic). Is this so directly related to physics? $\endgroup$ Aug 28 '20 at 16:42
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.