# Orthonormal Basis integration and Kronecker delta

Given that this integral I'm trying to solve is $$\frac{2}{\pi}\sum^{\infty}_{l=0}\sum^{l}_{m=-l}\int_{r=0}^{\infty}\int_{k=0}^{\infty} R_{nl}(r)b_{lm}(k)j_{l}(kr)k^2 r^2 \int_{\theta = 0}^{\pi}\int_{\phi=0}^{2\pi} Y^*_{lm}(\theta,\phi)Y_{l'm'}(\theta, \phi)\sin\theta \,\mathrm d\phi\,\mathrm d\theta\,\mathrm dk\,\mathrm dr$$ where $R_{nl}(r)$ is the Radial Wave Solution of the Hamiltonian of the Hydrogen atom, $b_{lm}(k)$ is a spectral function that depends on $k$, $j_l(kr)$ is the Spherical Bessel Function and $Y_{lm}(\theta, \phi)$ is the Spherical Harmonics. Given the orthonormal relations of the Spherical Harmonics, the integral should technically then becomes $$\frac{2}{\pi}\sum^{\infty}_{l=0}\sum^{l}_{m=-l}\int_{r=0}^{\infty}\int_{k=0}^{\infty} R_{nl}(r)b_{lm}(k)j_{l}(kr)k^2 r^2 dkdr \delta_{ll'}\delta_{mm'}$$ I have a very simple question which is how does this integral usually proceed from here with the sums and Kronecker delta given that they are summing over two different indices, do the sums not just collapse to one term where $l=l'$ and $m=m'$?

• in a word: yes. Basically do the $m$ sum first to pick up the $m=m’$ term and then the sum over $\ell$ to pick up the $\ell=\ell’$ term. – ZeroTheHero Aug 21 '18 at 1:16
• But how does that affect the l's and m's of the $R_{nl}(r)$, $b_{lm}(k)$ and $j_l (kr)$? – user3613025 Aug 21 '18 at 1:37
• @ZeroTheHero That should have been posted as an answer – David Z Aug 21 '18 at 3:02

The Kronecker Delta in a sum acts like the Dirac Delta function in an integral. So basically you can drop the sums and remove the delta terms. Your integral will be the integrand without the Kroneker Deltas. You technically can make the $l$s and $m$s primed if you wanted to, but that's unneeded primes (unless the context requires the primes).