I'm aware of the solid spherical harmonics functions, which are basically the surface spherical harmonics $Y^m_{\ell}(\theta,\varphi)$ with an additional monomial term along the radial direction:

$R^m_{\ell}(\mathbf{r}) \equiv \sqrt{\frac{4\pi}{2\ell+1}}\; r^\ell Y^m_{\ell}(\theta,\varphi)$

$I^m_{\ell}(\mathbf{r}) \equiv \sqrt{\frac{4\pi}{2\ell+1}} \; \frac{ Y^m_{\ell}(\theta,\varphi)}{r^{\ell+1}}$

These solid harmonics obey known translation formulae for a shift of the reference frame along a vector $\mathbf{a}$:

$R^m_\ell(\mathbf{r}+\mathbf{a}) = \sum_{\lambda=0}^\ell\binom{2\ell}{2\lambda}^{1/2} \sum_{\mu=-\lambda}^\lambda R^\mu_{\lambda}(\mathbf{r}) R^{m-\mu}_{\ell-\lambda}(\mathbf{a})\; \langle \lambda, \mu; \ell-\lambda, m-\mu| \ell m \rangle$, $I^m_\ell(\mathbf{r}+\mathbf{a}) = \sum_{\lambda=0}^\infty\binom{2\ell+2\lambda+1}{2\lambda}^{1/2} \sum_{\mu=-\lambda}^\lambda R^\mu_{\lambda}(\mathbf{r}) I^{m-\mu}_{\ell+\lambda}(\mathbf{a})\; \langle \lambda, \mu; \ell+\lambda, m-\mu| \ell m \rangle$,

where the terms in angled brackets are Clebsch-Gordan coefficients.

Solid spherical harmonics have been studied extensively because they arise as natural solutions of the Laplace equation in spherical coordinates. I am interested in going beyond the Laplace equation, however, and consider harmonics where the monomial term is no longer tied to the degree $\ell$ of $Y^m_{\ell}(\theta,\varphi)$:

$R^m_{\ell,k}(\mathbf{r}) \propto r^k Y^m_{\ell}(\theta,\varphi)$

$I^m_{\ell,k}(\mathbf{r}) \propto \frac{ Y^m_{\ell}(\theta,\varphi)}{r^k}$

Apart from issues of orthogonality, would it be possible to derive translation formulae for these generalised solid harmonics? It would be tempting to think that the formulae above would still apply to the new harmonics but that seems unlikely to be true. Much more likely, a summation over $k$ would also be needed.

Thank you.

  • 1
    $\begingroup$ I am a newbie so I honestly don't know if this should be a MathSE question, just so I know in future. $\endgroup$
    – user139561
    Dec 19, 2016 at 19:00
  • $\begingroup$ @ user139561 Of course it is a MathSE question, and the OP should move it there himself. $\endgroup$ Oct 31, 2017 at 22:13

1 Answer 1


You can use the (generalized) binomial theorem and the original result to compute something similar. First, define $$R^m_{l,k}(\mathbf{r})\equiv\sqrt{\frac{4\pi}{2l+1}} r^k\, Y_{l,m}(\theta,\phi),$$ which we can rewrite as $$R^m_{l,k}(\mathbf{r})=r^{k-l}\, R^m_l(\mathbf{r}),\qquad \text{with} \qquad R^m_l(\mathbf{r})\equiv \sqrt{\frac{4\pi}{2l+1}} r^l\, Y_{l,m}(\theta,\phi).$$ Using Newton's generalized binomial theorem, one has that $$ \rvert \mathbf{r}+\mathbf{a} \rvert^{n} = ((r+a)^2+2ra(\cos\theta-1))^{n/2}=\sum\limits_{p=0}^{\infty}\left(\begin{matrix} n/2 \\ p \end{matrix}\right)T_{p,n}^\pm(\mathbf{r},\mathbf{a}) P_{p,n}^{\pm}(r,a),$$ where $$ T_{p,n}^{\pm}(\mathbf{r},\mathbf{a}) = \begin{cases} (2ra)^{n/2-p}(\cos\theta-1)^{n/2-p}, & \quad\text{for}\quad +\\ (2ra)^{p}(\cos\theta-1)^{p}, & \quad\text{for}\quad - \end{cases},\qquad P_{p,n}^{\pm}(r,a) = \begin{cases} (r+a)^{2p} & \quad\text{for}\quad +\\ (r+a)^{n-2p} & \quad\text{for}\quad - \end{cases}.$$ where the $\pm$ denotes the sign of $2ra(\cos\theta-1) - (r+a)^2$ and $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{a}$. Moreover, $$ (r+a)^{2p}=\sum\limits_{j=0}^{2p} \left(\begin{matrix} 2p \\ j \end{matrix}\right)r^{2p-j}a^{j}, \qquad (r+a)^{n-2p}=\sum\limits_{\substack{0\leq j \leq n\\ 0\leq i \leq \infty}} \left(\begin{matrix} n \\ j \end{matrix}\right)\left(\begin{matrix} -2p \\ i \end{matrix}\right)t_{i,p}^{\pm}(r,a) r^{n-j}a^{j}, $$ where $$ t_{i,p}^\pm(r,a) = \begin{cases} r^{-2p-i}a^i, &\quad\text{for}\quad +\\ a^{-2p-i}r^i, &\quad\text{for}\quad - \end{cases},$$ with $\pm$ denoting the sign of $r-a$. This shows that $P_{p,n}^\pm(r,a)$ can be expressed as a finite $(+)$ or infinite $(-)$ power series in $r$ and $a$. Then, one has $$\begin{aligned} R^m_{l,k}(\mathbf{r}+\mathbf{a})&=\rvert \mathbf{r}+\mathbf{a} \rvert^{k-l} R^m_l(\mathbf{r}+\mathbf{a})\\ &= R^m_l(\mathbf{r}+\mathbf{a}) \sum\limits_{p=0}^{\infty}\left(\begin{matrix} (k-l)/2 \\ p \end{matrix}\right)T_{p,k-l}^\pm(\mathbf{r},\mathbf{a}) P_{p,k-l}^{\pm}(r,a). \end{aligned}$$ Collecting the powers of $r$ and $a$ one gets $$ (+):\, r^{n/2+p-j}a^{n/2-p+j},\qquad (-):\, r^{n-j\pm(i+p)}a^{j\pm(i+p)},\quad\text{(the }\pm\text{ in the exponents denotes the one of } t_{i,p}^\pm(r,a)),$$ which can be paired with the appropriate $R^\mu_\lambda(\mathbf{r})$ or $R^{m-\mu}_{l-\lambda}(\mathbf{a})$ in the decomposition of $R^m_l(\mathbf{r}+\mathbf{a})$ to write an analog decomposition in terms of sums of products of $R_{\lambda,\kappa}^{\mu}(\mathbf{r})$ and $R_{l-\lambda,k-\kappa}^{m-\mu}(\mathbf{a})$ (or the $I_{l,m}^k(\mathbf{r})$ which are defined similarly) with $T_{p,n}^{\pm}(\mathbf{r},\mathbf{a})$, together with the binomial coefficients throughout, playing a similar role as the Clebsch-Gordan coefficients.

P.S.:"Apart from issues of orthogonality [...]" This is what you should focus on. Find an algebraic structure that admits representations on your set of functions. Then, all those kinds of question are drastically more approachable.


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