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I have done an elementary calculation that appears to be giving me a simple (intuitive) formula for an arbitrary Wigner D-matrix. I can't seem to find this formula anywhere.

In the following, $\mathscr{R}$ represents an active rotation, $\Omega$ is solid angle, and $|\Omega\rangle$ are an orthonormal basis.

$$\begin{align} D^{l}_{m,m'}(\mathscr{R})&=\langle l,m'|\mathscr{R}|l,m\rangle\\ &=\int d\Omega\,\langle l,m'|\mathscr{R}|\Omega\rangle\langle\Omega|l,m\rangle\\ &=\int d\Omega\,\langle l,m'|\mathscr{R}\Omega\rangle\langle\Omega|l,m\rangle\\ &=\int d\Omega\, \,Y_l^{m'*}(\mathscr{R}\Omega)\,Y_l^m(\Omega) \end{align}$$

So it's a convolution of spherical harmonics. Is this right? I know my calculation is a bit cavalier but they seem more-or-less sound.

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I don't see anything cavalier about this calculation.(*) You can verify it from another direction using the rotation law for the (conjugate of the) spherical harmonic: \begin{equation} \bar{Y}^{m'}_\ell(\mathscr{R}\Omega) = \sum_{m''} D^{(\ell)}_{m'',m'}(\mathscr{R})\, \bar{Y}^{m''}_\ell(\Omega). \end{equation} Insert this into your expression and use orthogonality of the spherical harmonics to find your result the other way around: \begin{align} \int d\Omega\, \,Y_l^{m'*}(\mathscr{R}\Omega)\,Y_l^m(\Omega) &= \sum_{m''} \left[ \int d\Omega\, D^{(\ell)}_{m'',m'}(\mathscr{R})\, \bar{Y}^{m''}_\ell(\Omega)\, Y_l^m(\Omega) \right] \\ &= \sum_{m''} \left[ D^{(\ell)}_{m'',m'}(\mathscr{R})\, \int d\Omega\, \bar{Y}^{m''}_\ell(\Omega)\, Y_l^m(\Omega) \right] \\ &= \sum_{m''} \left[ D^{(\ell)}_{m'',m'}(\mathscr{R})\, \delta_{m, m''} \right] \\ &= D^{(\ell)}_{m,m'}(\mathscr{R}). \end{align} Of course, the logic works exactly the same if you go backwards in this sequence, so now that you see how it's done, you can also use this as a derivation of your result.

The bigger question is why this is useful. Usually, an expression like your last integral would arise when you're trying to evaluate the spherical-harmonic modes of a function in a rotated coordinate system. But we already know how to rotate modes. If you're suggesting that this might be an easier way to actually calculate the $D$ matrices, or just a simpler way to define them, I'm skeptical. But it's certainly another valid way to think of them, if you like.


(*) My one quibble is that I'm not sure that the order of your $m, m'$ indices is standard. Then again, I've never seen much consistency at all about any conventions involved in rotations. I've tried to stick with your ordering as best I can. But if you see swapper indices, that's probably why.

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Your expression relies on the relation $$ D^\ell_{m0}(\alpha,\beta,0)= \sqrt{\frac{4\pi}{2\ell+1}} Y_\ell^{m*}(\beta,\alpha) $$ so that you get \begin{align} &\int \sin\beta d\beta\,d\alpha \frac{2\ell+1}{4\pi} D^\ell_{m'0}(\omega\cdot(\alpha,\beta,0))D^{\ell*}_{m0}(\alpha,\beta,0)\, ,\\ &=\sum_{k} \int \sin\beta d\beta\,d\alpha \frac{2\ell+1}{4\pi} D^\ell_{m'k}(\omega)D^\ell_{k0}(\alpha,\beta,0))D^{\ell*}_{m0}(\alpha,\beta,0)\, ,\\ &=\sum_{k}D^\ell_{m'k}(\omega)\frac{2\ell+1}{4\pi} \int \sin\beta d\beta\,d\alpha D^\ell_{k0}(\alpha,\beta,0))D^{\ell*}_{m0}(\alpha,\beta,0)\, \end{align} The integral is proportional to $\delta_{km}$ so the result is indeed $D^{\ell}_{m'k}(\omega)$.

Given there are closed form expressions for $D^{\ell}_{m'k}(\omega)$, and given it is defined quite intuitively by $$ D^{\ell}_{m'k}(\omega) = \langle \ell m'\vert R(\omega)\vert \ell k\rangle $$ I'm not sure what is gained by expressing this as a convolution.

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