Do you know a formula for the integral of a product of 4 spherical harmonics?

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SUMMARY

The discussion focuses on calculating the integral of a product of four spherical harmonics, specifically using Cartesian coordinates and symmetric traceless tensors of SO(3). The integral formula for products of spherical harmonics is established, with specific coefficients derived for cases involving one, two, and three harmonics. The conversation emphasizes the utility of computational tools like Mathematica, Maple, and Octave for performing these calculations, especially when dealing with complex angular momentum scenarios.

PREREQUISITES
  • Understanding of spherical harmonics and their properties
  • Familiarity with symmetric traceless tensors of SO(3)
  • Knowledge of integral calculus on spherical surfaces
  • Experience with computational tools such as Mathematica or Maple
NEXT STEPS
  • Learn how to derive integrals of products of spherical harmonics using Cartesian coordinates
  • Explore the Clebsch-Gordan coefficients and their application in angular momentum coupling
  • Investigate the use of Mathematica for automating spherical harmonic integrals
  • Study the molecular electronic structure theory as outlined by Helgaker, Joergensen, and Olsen
USEFUL FOR

Physicists, mathematicians, and computational scientists involved in atomic physics and quantum mechanics, particularly those working with spherical harmonics and angular momentum calculations.

andresordonez
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Hi, this may seem like something I should ask in the math forums but, as I came into this problem in atomic physics I'm confident that this is a question more appropriate here than in the math forums.

So far I've been only able to find the common integral of a product of three spherical harmonics.

Any kind of help (formulas, bibliography, etc. ) is welcome!
 
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I don't know of a good reference, but I can outline a method that will work. First of all, you want to start with the spherical harmonics written in terms of Cartesian coordinates

Y^I = C^I_{i_1\cdots i_k} x^{i_1}\cdots x^{i_k}.

The x^i are one of the coordinates x,y,z, while the C^I_{i_1\cdots i_k} are symmetric traceless tensors of SO(3). k counts the degree, so it's the \ell quantum number of the harmonic. The index I ranges over the 2k+1 different symmetric traceless tensors, or alternatively over the same number of harmonics of degree k. So I is analogous to the m quantum number of the harmonic.

To compute integrals of harmonics, we need the formula

\int_{S^2} x^{i_k} \cdots x^{i_{2m}} = 4\pi \frac{2^m}{(2m+1)!} \left( \delta^{i_1 i_2} \cdots \delta^{i_{2m-1} i_{2m}} + \text{perms} \right). ~~~(*)

I didn't compute this very carefully, but considered a few cases and guessed the coefficient. It's been years since I've had to use this for anything and can't remember if there's a trick to do it cleanly.

Now to compute integrals of products of harmonics, we just multiply (*) by factors of C^I_{i_1\cdots i_k} and compute the number of ways that the C\text{s} can be contracted. For 2 of them,

\int_{S^2} Y^{I_1} Y^{I_2} = 4\pi \frac{2^k}{(2k+1)!} k! \delta^{I_1I_2},

while for 3 factors

\int_{S^2} Y^{I_1} Y^{I_2} Y^{I_3}= 4\pi \frac{2^{\Sigma/2}}{(\Sigma+1)!} \frac{k_1!}{\alpha_1!}\frac{k_2!}{\alpha_2!}\frac{k_3!}{\alpha_3!} \langle C^{I_1}C^{I_2}C^{I_3}\rangle,

where

\Sigma = k_1 + k_2 + k_3,

\alpha_i = \frac{\Sigma}{2} - k_i,

and

\langle C^{I_1}C^{I_2}C^{I_3}\rangle = C^{I_1}_{i_1\cdots i_{\alpha_2+\alpha_3}} {C^{I_2i_1\cdots i_{\alpha_3}}}_{j_1\cdots j_{\alpha_1}} C^{I_3i_{\alpha_3+1}\cdots i_{\alpha_2+\alpha_3}j_1\cdots j_{\alpha_1}}.

I've never tried to work out the product of 4, but it seems straightforward if you can reproduce these formulas. I've also never tried to convert any of these formulas to ones with Y_{\ell m}s, but I explained part of the translation, so that's certainly possible to do.
 
The only contribution to the integral comes from that combinations of the Y's with L=0.
So you could alternatively use the Clebsch Gordan formula repeatedly to find all the ways to couple the l's to L=0.
 
I recommend the Cartesian approach fzero described mainly because you can actually check the intermediate values and the Cartesian formulas. I think it's the only way which is straight-forward. If you need formulas for expanding solid harmonics in terms of Cartesians, these are given in Molecular Electronic Structure theory of Helgaker, Joergensen and Olsen.

Clebsch-Gordan coefficients become incredibly messy if many angular momenta are involved, and additionally you have the problems of phase conventions, order of components, real vs complex spherical harmonics, etc. Unless you are willing to re-calculate the formulas yourself for your concrete Ylm definition, your chances of obtaining correct results by simply taking any textbook or published formulas are rather slim (even if the published formulas do not contain errors... unfortunatelly not something you can count on). When I need CB coefficients, I actually calculate them from the Cartesian expansions as described by fzero, because this appears to be the least painful way.
 
Maybe Mathematica, Maple, Octave or even the internet can do the calculation for you.
I just tried:
Integrate[SphericalHarmonicY[1,1,x,y]*SphericalHarmonicY[2,1,x,y]*SphericalHarmonicY[2,-2,x,y]*SphericalHarmonicY[1,0,x,y],{ x,-Pi/2,Pi/2},{y,-Pi,Pi}]
in
http://www.wolframalpha.com.
Here is the result:
 

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DrDu said:
Maybe Mathematica, Maple, Octave or even the internet can do the calculation for you.
I just tried:
Integrate[SphericalHarmonicY[1,1,x,y]*SphericalHarmonicY[2,1,x,y]*SphericalHarmonicY[2,-2,x,y]*SphericalHarmonicY[1,0,x,y],{ x,-Pi/2,Pi/2},{y,-Pi,Pi}]
in
http://www.wolframalpha.com.
Here is the result:

Oh, right... this is of course the best approach if the maximum angular momentum of the spherical harmonics is known beforehand and not too large: run it through mathematica and tabulate everything.
 

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