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Does anyone know how to evaluate integrals of them following form: $$\int_{0}^{\pi} \int_{0}^{2 \pi} (\cos{\phi})^{n_1} (\sin{\phi})^{n_2} (\cos{\theta})^{n_3} (\sin{\theta})^{n_4}Y^m_l Y^p_q Y^i_j \sin{\theta}\; d\phi\; d\theta.$$

Where $Y_l^m$ is the spherical harmonic of degree $l$ and order $m$. The symbols $n_1$, $n_2$, $n_3$, $n_4$ $l$, $m$, $q$, $p$, $i$, and $j$ are integers. For any given combination of these integers it is not hard to evaluate the integral, but I need to find all results for a large number of different values of these integers. So many in fact that even tools like Mathematica become to slow.

My current idea is to make some kind of selection criteria for when the integral should be zero, based on symmetry considerations, and then evaluate the non-zero integrals by help of Mathematica. However, if anyone knows any result that would simplify my efforts that would be great.

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  • $\begingroup$ Does my answer helped you out in any way or another, it would be nice to accept it ;-) $\endgroup$ – kvantour Dec 23 '19 at 15:25
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We are solving the integral:

$$I=\int_{0}^{\pi} \int_{0}^{2 \pi} (\cos{\phi})^{n_1}(\sin{\phi})^{n_2}(\cos{\theta})^{n_3}(\sin{\theta})^{n_4}Y^u_p Y^v_q Y^w_r \sin{\theta}\, d\phi\, d\theta.$$

Due to the nature of the spherical harmonics, which can be written as a product of associated Legendre polynomials and an exponential :

$$Y_l^m(\theta,\phi)=\sqrt{\frac{2l+1}{4\pi}\cdot\frac{(l-m)!}{(l+m)!}}P_l^m(\cos\theta)e^{im\phi},$$

you can split the integral in two parts:

$$\Phi(n_1,n_2,u+v+w)=\int_0^{2\pi}(\cos\phi)^{n_1} (\sin\phi)^{n_2} e^{i(u+v+w)\phi}d\phi,$$

$$\Theta(n_3,n_4,p,q,r,u,v,w)=\int^1_{-1} x^{n_3}(1-x^2)^\frac{n_4}{2} P_p^u(x)P_q^v(x)P_r^w(x)dx.$$

The first integral $\Phi$, can be solved by converting the cosine and sine into their exponential form and using the Binomial theorem and taking into account that

$$\int_0^{2\pi} e^{im\phi}d\phi=2\pi\delta_{m,0}.$$

Thus,

$$\Phi(n_1,n_2,m)=\frac{2\pi}{2^{n_1+n_2}i^{n_2}}\sum_{k_1=0}^{n_1}\sum_{k_2=0}^{n_2}(-1)^{n_2-k_2}\binom{n_1}{k_1}\binom{n_2}{k_2}\delta_{n_1+n_2-m,2(k_1+k_2)}.$$

This gives you your first selection rules:

  • if $n_1+n_2-m$ is odd, then $\Phi=0$,
  • if $n_1+n_2-m<0$, then $\Phi=0$,
  • if $-m > n_1+n_2$, then $\Phi=0$.

To solve the integral of $\Theta$, you should try to reduce it to the form :

$$\int^1_{-1}P_p^u(x)P_q^v(x)P_r^w(x)dx,$$

which has a closed-form expression. It is related to 3j-symbols/Clebsch-Gordan coefficients and is known as Gaunt's Formula (See Dong S.H., Lemus R., (2002), "The overlap integral of three associated Legendre polynomials", Appl. Math. Lett. 15, 541-546.).

The integral of $\Theta$ can be reduced to this form using the recurrence relations of the associated Legendre polynomials:

$$ P_{l-1}^m(x) - P_{l+1}^m(x) = (2l+1)\sqrt{1-x^2}P_l^{m-1}(x)$$

$$ (2l+1)xP_l^m(x) = (l-m+1)P^m_{l+1}(x)+(l+m)P^m_{l-1}(x)$$

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  • $\begingroup$ Excellent answer. $\endgroup$ – Giuseppe Negro Mar 12 '18 at 15:49
  • $\begingroup$ This explains it well, I found the paper you cite and came up with a similar solution. I will post the last parts and we can merge them into one later. I have worked out a few expressions that could be useful for others. I will post soon. $\endgroup$ – HaakonA Apr 26 '18 at 17:49
  • $\begingroup$ @Kvantour thanks for your answer, but do you have a sign error in your kronecker delta? If n1+n2-m is strictly positive then +2(k1+k2) would have to be negative, which is a contradiction. should it be -2(k1+k2) ? $\endgroup$ – walczyk Oct 24 '19 at 1:06
  • $\begingroup$ @walczyk you are correct. I've updated the text and the selection rules. $\endgroup$ – kvantour Oct 24 '19 at 7:44

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