To understand how spherical harmonics apply to integer spin but not half integer spin


by Cruikshank
Tags: angular momentum, spherical harmonics, spin 1/2
Cruikshank
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#1
Jul4-11, 07:10 PM
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1. The problem statement, all variables and given/known data
I want to understand spherical harmonics. I want to really grok them deeply. I want to be able to visualize them and understand them.

I'm the sort who can't take anything on faith, especially where quantum mechanics is concerned. So I want to understand angular momentum so I can really see what spin 1/2 means in contrast. I need to see the craziness proven.

What should I know, to learn this? What should I read? I consider myself fairly capable with calculus and differential equations. I've taken several more math classes. I've worked nearly all the problems in the first five chapters of Bransden and Joachain's Introduction to Quantum Mechanics, and more than half in the angular momentum chapter, but I am greatly dissatisfied.

I have tried Arfken and Weber, and got a bit overloaded, but I will try that book again. I am also looking at Fourier Series and Orthogonal Functions by Harry F. Davis. Do people have other sources that they prefer?
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bcrowell
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#2
Jul4-11, 11:50 PM
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This was originally posted in the hw forum, but really isn't a hw question, so I moved it here.
alxm
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#3
Jul5-11, 03:30 AM
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I don't see what spherical harmonics have to do with spin or quantum mechanics specifically.

dextercioby
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#4
Jul5-11, 01:14 PM
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To understand how spherical harmonics apply to integer spin but not half integer spin


Spherical harmonics are related to ORBITAL angular momentum, and not to the SPIN angular momentum. And knowing the general mathematical theory of them is useful for quantum mechanics...

The problem whether orbital angular momentum operators have only positive integer eigenvalues is a totally different issue and I remember seeing a quite recent post in this (sub)forum by prof. Hendrik van Hees (user <vanhees71>) containing a nice algebraic proof why 1/2, 3/2, ... are not valid numbers for orbital angular momentum.
Cruikshank
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#5
Jul5-11, 03:08 PM
P: 64
Thank you. I went to the other thread and read it through, but most of the commentary was missing the point of my question and the other poster's question, so I summarized as clearly as I could in a comment in that thread. I'm still new at this so I'm not sure whether copying that post here would be okay, but here's the...line at the top of the browser (is "URL" the name of that thing?):
http://www.physicsforums.com/showthr...nhees71&page=3
dextercioby
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#6
Jul5-11, 03:59 PM
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I mean this thread below, not the one you pointed to.

http://www.physicsforums.com/showthr...gular+momentum
Bill_K
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#7
Jul5-11, 06:10 PM
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Spherical harmonics are related to ORBITAL angular momentum, and not to the SPIN angular momentum.
Spin spherical harmonics are related to TOTAL angular momentum. They are eigenfunctions of J, Jz and L, defined as sums of products of orbital and spin wavefunctions:

Ψjℓm(θ,φ) = ∑ (ℓ m μ | ℓ j m) Yℓm(θ,φ) χμ

where ( ... | ) are Clebsch-Gordan coefficients and χμ, μ = are spin states.
Cruikshank
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#8
Jul5-11, 09:28 PM
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Thank you dextercioby. I could not understand that thread at all. It appears that I have to study a whole lot of group theory (covering group??) before I can try again to understand spherical harmonics. (There's always another prerequisite, isn't there? I don't know how any of you actually *get* to the topics that interest you. At this rate I will die of old age before I get through undergraduate quantum mechanics.)
alxm
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#9
Jul6-11, 10:17 AM
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You don't need group theory to understand spherical harmonics. I think you're mixing up the concepts of spin and angular momentum with spherical harmonics.

If you want to understand spherical harmonics, you need a book on partial differential equations. Learn about various PDEs that have solutions, and the derivations of the various special functions such as Legendre Polynomials, Bessel functions, Hankel functions, and so also Spherical Harmonics. They're just the solution to the angular part of Laplace's equation in spherical coordinates. As such, they form a set of angular-momentum eigenfunctions in quantum mechanics the same way plane waves form the eigenfunctions for linear momentum.

That's really all there is to it.
samalkhaiat
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#10
Jul6-11, 02:24 PM
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Actually it is good question! It seems that the algebra of angular-momentum does not reveal why the orbital part cannot have half-integer values. However, one can provide many arguments against it:
1) on completely mathematical ground (Sturm-Liouville) we know that {[itex]Y_{\ell m}(\theta , \phi )[/itex]} forms a complete set if and only if [itex]\ell[/itex] and m are integers.
2) for half-integer [itex]\ell[/itex], and therefore for half-integer m, the wave function would not be single-valued: under [itex] 2\pi [/itex] rotation, we would have
[tex]\Psi (r, \theta , \phi ) \sim e^{im(2\pi)} = -1[/tex]
3) we know that for [itex]\ell = 0, 1, 2, …[/itex],
[tex]Y_{\ell , \ell }(\theta , \phi ) \sim e^{i\ell \phi } \sin^{\ell} \theta[/tex]
Suppose (pushing our luck) it is possible to put
[tex]\ell = m = \frac{1}{2}[/tex]
that is
[tex]Y_{1/2,1/2} \sim e^{i \phi /2} \sqrt{\sin \theta }[/tex]
Since
[tex]Y_{\ell , m-1}(\theta , \phi ) \sim L^{-}Y_{\ell , m}(\theta , \phi),[/tex]
where
[tex] L^{-}= e^{-i \phi} (- \partial_{\theta} + i \cot \theta \partial_{\phi}),[/tex]
we would then get
[tex]Y_{1/2,-1/2}(\theta , \phi ) \sim e^{-i \phi /2} \cot \theta \sqrt{\sin \theta }[/tex]
Clearly, this expression is singular at [itex]\theta = 0, \pi [/itex].
4) we would also need to impose
[tex]L^{-}Y_{1/2, -1/2}(\theta , \phi ) = 0[/tex]
but this leads to
[tex]Y_{1/2 , -1/2} \sim e^{-i\phi /2} \sqrt{\sin \theta },[/tex]
but this contradicts the expression we obtained in (3).
So, one concludes that orbital angular-momentum can only have integer [itex]\ell[/itex]-values. The real challenge is to prove that within the so(3) algebra!


sam
dextercioby
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#11
Jul6-11, 02:47 PM
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Sam, point 1) is enough. It says it all. Angular momentum operators need to be (essentially) self-adjoint.


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