A Triple Product Integral Identity for Vector Spherical Harmonics

Last updated 2021-05-21 01:08:20 SGT

As everyone knows, the product of three spherical harmonics can be related to the Wigner [3j] symbols (and therefore Clebsch-Gordan coefficients) in the form

[\int \mathrm d \Omega\ Y_{l_1}^{m_1} Y_{l_2}^{m_2} Y_{l_3}^{m_3} = \sqrt{(2 l_1 + 1)(2 l_2 + 1)(2 l_3 + 1) \over 4\pi}\left(\begin{array}{ccc} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3\end{array}\right) \left(\begin{array}{ccc} l_1 & l_2 & l_3 \\ 0 & 0 & 0\end{array}\right).\tag{1}]

Lately, however, I have been extensively using vector spherical harmonics, for which I have not been able to find equivalent identities to this scalar version. In particular I have been confronted with situations where I have needed to perform integrals of the form

[\int \mathrm d \Omega\ Y_1 (\nabla Y_2) \cdot (\nabla Y_3), \tag{2}]

where here I am schematically using 1, 2, 3 to stand in for the relevant multi-indices.

To proceed, let us first consider how one recovers the orthogonality relation of the poloidal vector spherical harmonics [\boldsymbol{\Psi}_l^m = r \nabla Y_l^m]. Limiting our attention to the unit sphere, we have [\nabla_3 = \partial_r + {1 \over r} \nabla_2] (although [Y_l^m] has no [r] dependence in any case). Therefore

[\int \mathrm d \Omega\ \boldsymbol{\Psi}_l^m \cdot \boldsymbol{\Psi}_{l'}^{m'} = \int \mathrm d \Omega \ \nabla Y_l^m \cdot \nabla Y_{l'}^{m'},]

where the derivative here is understood to be purely horizontal (acting only on angles). We now integrate by parts to obtain

[\int \mathrm d \Omega\ \nabla \cdot \left(Y_{l'}^{m'} \nabla Y_l^m\right) - \int \mathrm d \Omega\ Y_{l'}^{m'} \nabla^2 Y_l^m.]

The first term vanishes by Stokes' Theorem as it is the integral of a 2-divergence over the sphere (which has no boundary). For the second term, we recall the definition of the spherical harmonics is that they are the eigenfunctions of the Laplacian: [\left(\nabla^2 + l(l+1)\right)Y_{l}^{m} = 0]. Thus we obtain

[\int \mathrm d \Omega \ \nabla Y_l^m \cdot \nabla Y_{l'}^{m'} = l(l+1) \int \mathrm d \Omega \ Y_l^m Y_{l'}^{m'} = l(l+1) \delta_{l,l'}\delta_{m,m'} \equiv L^2 \delta_{l,l'}\delta_{m,m'},]

as expected.

Let us now turn our attention to the triple product integral, eq. 2. Again integrating by parts, we obtain

[\begin{aligned}I_{123} \equiv \int \mathrm d \Omega\ &Y_1 (\nabla Y_2) \cdot (\nabla Y_3) \\&= \int \mathrm d \Omega\ \nabla \cdot \left(Y_1 Y_2 (\nabla Y_3)\right) - \int \mathrm d \Omega\ Y_2 (\nabla Y_1) \cdot (\nabla Y_3) - \int \mathrm d \Omega\ Y_2 Y_1 \nabla^2 Y_3\\&= - \int \mathrm d \Omega\ \left[Y_2 (\nabla Y_3) \cdot (\nabla Y_1)\right] + l_3(l_3 + 1) \int \mathrm d \Omega\ Y_2 Y_1 Y_3 \\ &= -I_{231} + L_3^2 F_{123}\end{aligned}]

Again, the integral of the 2-divergence vanishes, and we are left with 2 terms. The first is an integral of similar form to eq. 2, but with the multi-indices having undergone a cyclic permutation (and with a sign change). The second is the scalar triple integral in eq. 1, which does not depend on the ordering of the multi-indices. Effecting this cyclic permutation twice more, we arrive at

[I_{123} = (L_3^2 - L_1^2 + L_2^2) F_{123} - I_{123} \implies I_{123} = \left(L_3^2 - L_1^2 + L_2^2 \over 2\right) F_{123}.]

In full:

[\int \mathrm d \Omega\ Y_{l_1}^{m_1} (\nabla Y_{l_2}^{m_2}) \cdot (\nabla Y_{l_3}^{m_3}) = \left(l_3(l_3 + 1) - l_1(l_1 + 1) + l_2(l_2 + 1) \over 2\right) \int \mathrm d \Omega\ Y_{l_1}^{m_1} Y_{l_2}^{m_2} Y_{l_3}^{m_3}.]

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