Talk:Solid harmonics

Article Checklist for "Solid harmonics"

Workgroup category or categories	Physics Workgroup, Mathematics Workgroup [Categories OK]
Article status	Developing article: beyond a stub, but incomplete
Underlinked article?	Yes
Basic cleanup done?	No
Checklist last edited by	--Paul Wormer 04:18, 22 August 2007 (CDT)

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This article (my own article from Wikipedia) gave LaTeX errors which I worked around. Added to WP version section on relation to regular and irregular solid harmonics and references. --Paul Wormer 04:18, 22 August 2007 (CDT)

Hi Paul. Is the definition of the orbital angular momentum $\mathbf {L}$ (second displayed equation) correct? The factor $\hbar$ seems out of place. -- Jitse Niesen 08:01, 24 August 2007 (CDT)

Hallo Jitse, I don't see what is wrong. Classically :

$\mathbf {L} =\mathbf {r} \times \mathbf {p}$

QM:

$\mathbf {p} \rightarrow -i\hbar \mathbf {\nabla } ,\qquad \mathbf {r} \rightarrow \mathbf {r}$

What do you think it should be? --Paul Wormer 09:21, 24 August 2007 (CDT)

I may well be misunderstanding what the article is saying. My issue is that you say first that, in spherical coordinates,

$\nabla ^{2}={\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r-{\frac {L^{2}}{r^{2}}}$.

If I try to substitute

$\mathbf {L} =-i\hbar \,(\mathbf {r} \times \mathbf {\nabla } ),$

then I get

$\nabla ^{2}={\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r+{\frac {\hbar ^{2}}{r^{2}}}(\mathbf {r} \times \mathbf {\nabla } )\cdot (\mathbf {r} \times \mathbf {\nabla } ).$

But the Laplacian should be independent of hbar. Perhaps the problem is in the first formula, and it should be

$\nabla ^{2}\Phi (\mathbf {r} )=\left({\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r-{\frac {L^{2}}{\hbar ^{2}r^{2}}}\right)\Phi (\mathbf {r} )=0,\qquad \mathbf {r} \neq \mathbf {0} .$

I hope you can make sense of it. It's ten years ago that I did this stuff, and I never did it properly. -- Jitse Niesen 09:53, 24 August 2007 (CDT)

Looking at the dimensions (sorry, I should have started with that; but I'm only a mathematician), it's indeed the formula for the Laplacian that's weird (or, more likely, me!). -- Jitse Niesen 10:09, 24 August 2007 (CDT)

You are right wrt to the Laplacian. My problem is that I 've worked my whole life with atomic units ($\hbar =1$). I have to be more careful with hbar, I'm a bit sloppy with it. I'm working on an article on spherical harmonics right now and there I specifically stated that I take hbar = 1. Maybe you can have a look at this article and also at Legendre polynomials, which is more up your street? --Paul Wormer 10:35, 24 August 2007 (CDT)