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)

Hbar

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)

Okay, at least I now understand the first bit. However, I still feel uncomfortable with physical constants floating around in this article. I know where you're coming from and I've only seen these functions in QM, so I'm happy to concede if you think it's better to leave them in. But this is how I'd write the first couple of sentences (caution: probably contains some sign errors):

Introducing r, θ, and φ for the spherical polar coordinates of the 3-vector r, we can write the Laplace equation in the following form $\nabla ^{2}\Phi (\mathbf {r} )=\left({\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r+{\frac {1}{r^{2}}}(\mathbf {r} \times \nabla )^{2}\right)\Phi (\mathbf {r} )=0,\qquad \mathbf {r} \neq \mathbf {0} .$ The differential operator $\mathbf {r} \times \nabla$ equals the orbital angular momentum $\mathbf {L} =-i\hbar \,(\mathbf {r} \times \mathbf {\nabla } )$ up to a constant. It is known that spherical harmonics Yml are eigenfunctions $(\mathbf {r} \times \nabla )^{2}$: $(\mathbf {r} \times \nabla )^{2}Y_{\ell }^{m}=-\ell (\ell +1)Y_{\ell }^{m}.$

I'll definitely have a look at Legendre polynomials. I regret that I've learnt so little about special functions. Oh, I suppose you read on my user talk page that \begin{align} now works (thanks to Greg). -- Jitse Niesen 23:27, 24 August 2007 (CDT)

Laplace operator

Hi Jitze, I start a new section, that indentation is driving me crazy. I believe I see your point: a mathematical article should stick to math and not be partly physics. Only in passing one can refer to the physics meaning of some math result. I agree with that. I have only one problem with your reasoning (a friend of mine calls this a matter of Kindergarten): your way of writing the Laplace operator is a bit foreign to me. The way I got to this result is: from Laplace-Beltrami in spherical polars and orbital angular momentum in spherical polars, comparison gives "your" Laplacian. I have seen your way before in Messiah p. 347 (that book is now in Dover, if you don't own it already it is worth buying). So, I would prefix your text with Messiah's outline of the way of writing the Laplacian. What do you think? PS. I would write the square of a vector(operator) always as a dot product. --Paul Wormer 03:03, 25 August 2007 (CDT)