I have a question regarding the spherical Bessel functions. Two standard relations in the literature are
$$\sum_{\ell=0}^{\infty} (2\ell + 1) j_{\ell}^{2}(x) = 1$$
$$\sum_{\ell=0}^{\infty} (2\ell + 1) [j'_{\ell}(x)]^{2} = {1 \over 3}$$
However, I cannot find their derivation. I need to calculate higher order analogs of these relations such as
$$\sum_{\ell=0}^{\infty} (2\ell + 1) [j''_{\ell}(x)]^{2} = f(x)$$
etc, where $f(x)$ is possibly some function of $x$. I am not sure where to start. I would be much obliged if somebody could suggest a way forward, maybe by explaining how the first two relations are derived. I have tried using the recursion relations, which swap derivatives for products of $\ell$, $\ell + 1$ and $j_{\ell}$, $j'_{\ell}$. This does not seem to help though, it recasts the problem into the calculation of terms such as
$$\sum_{\ell=0}^{\infty} (2\ell + 1) \ell^{2}(\ell + 1)^{2} [j_{\ell}(x)]^{2}$$
$$\sum_{\ell=0}^{\infty} (2\ell + 1) \ell(\ell + 1) [j'_{\ell}(x)]^{2}$$
Anyway, any suggestions would be great, thank you in advance!
Best Answer
General expressions can be obtained by using the addition theorem \begin{equation} \frac{\sin w}{w}=\sum_{n=0}^{\infty}(2n+1)j_{n}\left(v\right)j_{n}\left(u\right)P_{n}\left(\cos\alpha\right) \end{equation} where $w=\sqrt{u^2+v^2-2uv\cos\alpha}$ and $P_n$ denotes the Legendre polynomial of degree $n$. Choosing $\alpha=0$ we have $P_n(\cos\alpha)=1$ and thus \begin{equation} \frac{\sin\sqrt{(u-v)^2}}{\sqrt{(u-v)^2}}=\sum_{n=0}^{\infty}(2n+1)j_{n}\left(v\right)j_{n}\left(u\right) \end{equation} which can be written simply \begin{equation} \frac{\sin(u-v)}{u-v}=\sum_{n=0}^{\infty}(2n+1)j_{n}\left(v\right)j_{n}\left(u\right) \end{equation} Now, by taking the limit of the above identity for $v\to u$, one obtains \begin{equation} \sum_{n=0}^{\infty}(2n+1)\left[j_{n}(u)\right]^2=1 \end{equation} If we differentiate successively the identity with respect to $u$ and to $v$, \begin{equation} \frac{\sin \left(u -v \right)}{u -v}+\frac{2 \cos \left(u -v \right)}{\left(u -v \right)^{2}}-\frac{2 \sin \left(u -v \right)}{\left(u -v \right)^{3}}=\sum_{n=0}^{\infty}(2n+1)j'_{n}\left(v\right)j'_{n}\left(u\right) \end{equation} Taking the limit for $v\to u$ we have \begin{equation} \sum_{n=0}^{\infty}(2n+1)\left[j'_{n}(u)\right]^2=\frac13 \end{equation} This operation can be generalized by remarking that, to obtain the series of the $p$th derivative \begin{equation} S_p=\sum_{n=0}^{\infty}(2n+1)\left[j^{(p)}_{n}\left(u\right)\right]^2 \end{equation} one can differentiate successively $p$ times $\sin(u-v)/(u-v)$ wrt to $u$ and to $v$ and take the limit of the result for $v\to u$ \begin{equation} S_p=\lim_{v\to u}\frac{\partial^{2p}}{\partial u^p\partial v^p}\frac{\sin(u-v)}{(u-v)} \end{equation} But, for a function $f$ \begin{align} \frac{\partial^{2p}}{\partial u^p\partial v^p}f(u-v)&=\left.(-1)^p\frac{\partial^{2p}}{\partial x^{2p}}f(x)\right|_{x=u-v}\\ \lim_{v\to u}\frac{\partial^{2p}}{\partial u^p\partial v^p}f(u-v)&=\left.(-1)^p\frac{\partial^{2p}}{\partial x^{2p}}f(x)\right|_{x=0} \end{align} As \begin{equation} \frac{\sin x}{x}=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k+1)!} \end{equation} it comes directly \begin{equation} \sum_{n=0}^{\infty}(2n+1)\left[j^{(p)}_{n}\left(u\right)\right]^2=\frac{1}{2p+1} \end{equation}