# Integral of two Wigner d-functions

I want to integrate two general Wigner d-functions $d_{k q}^{\ell}(\theta)$. There is a simple integral in case magnetic numbers $(q,k)$ are the same:

$$\int_0^\pi d\theta~\sin \theta ~d_{kq}^{\ell}(\theta) d_{kq}^{\ell'}(\theta) = \frac{2}{2\ell + 1} \delta_{\ell \ell'}.$$

I'm trying to find a closed expressions for the case where each Wigner d-functions has arbitrary momenta:

$$\int_0^\pi d\theta~\sin \theta ~d_{kq}^{\ell}(\theta) d_{k'q'}^{\ell'}(\theta) = ~?~.$$

Also does there exist a generating function for $d_{k q}^{\ell}(\theta)$? I was hoping to use it to compute this type of integrals.

P.S. I'm aware it is basically a math question, but given that it is tightly related to the quantum mechanics and angular momentum theory I thought it would be more appropriate to ask it here.

• Have you considered simple cases of WP? Dec 6 '17 at 15:17
• Not quite sure what you mean by WP? I tried few simpler cases of integrals and tried to use the explicit representation (that you have linked) but was not very successful in solving the resulting integrals. With the expansion that JamalS suggested it might now be easier.
– z.v.
Dec 6 '17 at 15:35
• WP means WikiPedia. The classic text of L C Biedenharn and J D Louck on Angular momentum has a gallimaufry of such relations. Dec 6 '17 at 16:35

A possible avenue that may be fruitful is to expand the product of Wigner $d$-matrices in terms of Clebsch-Gordan coefficients or Wigner 3-$j$ symbols, so that your integral becomes,

$$\int_0^\pi \sin\theta \, d^\ell_{kq}(\theta)d^{\ell'}_{k'q'} \mathrm d\theta = \sum_{jps}(-1)^{-2j+2\ell'-k'-q'}(2\ell'+1) \times \\ \times \begin{pmatrix} j &\ell' & \ell' \\ p& k' & -k' \end{pmatrix}\begin{pmatrix} j &\ell' & \ell' \\ p& q' & -q' \end{pmatrix}\int_0^\pi \sin\theta\, d^j_{ps}(\theta)\mathrm \, d \theta$$

reducing the problem to a general integral of one $d$-matrix and then finding the summation. You should be able to find the required identities to complete everything in, Quantum Theory of Angular Momentum by Varshalovich, Moskalev and Khersonskii.

To give a slightly different twist and expand a bit on the answer of @JamalS:

Use $$d^\ell_{kq}(\theta)d^{\ell'}_{k'q'}(\theta) =\sum_{L} C^{LK}_{\ell k;\ell' k'}C^{LQ}_{\ell q;\ell' q'} d^L_{KQ}(\theta)$$ with $K=k+k'$, $Q=q+q'$, and $C^{LK}_{\ell k;\ell' k'}$ a Clebsch-Gordan coefficient. The sum over $L$ runs from $\vert \ell-\ell'$ to $\ell+\ell'$ in steps of 1.

There remains to use the explicit expression \begin{align} d^J_{MM'}(\theta)&=(-1)^{J-M'} \sqrt{(J+M)!(J-M)!(J+M')!(J-M')!}\\ &\times \sum_\nu (-1)^\nu \frac{\left(\cos\frac{\theta}{2}\right)^{M+M'+2\nu} \left(\sin\frac{\theta}{2}\right)^{2J-M-M'-2\nu}}{\nu!(J-M-\nu)!(J-M'-\nu)!(M+M'+\nu)!} \tag{1} \end{align} and integrate. In (1) the sum extends to all those $\nu$ for which each the arguments in the denominator of the sum is positive.

Finally note that \begin{align} \int_0^\pi\,d\theta \sin\theta \left(\textstyle\cos\frac{\theta}{2}\right)^a \left(\sin\textstyle\frac{\theta}{2}\right)^b &=2\int_0^\pi \,d\theta \cos\left(\textstyle\frac{\theta}{2}\right) \sin\left(\textstyle\frac{\theta}{2}\right) \left(\textstyle\cos\frac{\theta}{2}\right)^a \left(\sin\textstyle\frac{\theta}{2}\right)^b\, ,\\ &=4\int_0^1 dx\, (1-x^2)^{a/2}x^{b+1}\, . \end{align}

• Tnx! So $\nu$ here are such that $a$ and $b$ factors will always be positive I guess? The last integral can then be expressed in simple gamma functions.
– z.v.
Dec 6 '17 at 16:49