Quantum Mechanics - Orthonormal Basis integration and Kronecker delta

Question

Given that this integral I'm trying to solve is $$\frac{2}{\pi}\sum^{\infty}_{l=0}\sum^{l}_{m=-l}\int_{r=0}^{\infty}\int_{k=0}^{\infty} R_{nl}(r)b_{lm}(k)j_{l}(kr)k^2 r^2 \int_{\theta = 0}^{\pi}\int_{\phi=0}^{2\pi} Y^*_{lm}(\theta,\phi)Y_{l'm'}(\theta, \phi)\sin\theta d\phi d\theta dkdr$$ 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 $m=m'$?

Answer 1

They do indeed. Put another way: If you integrate a linear superposition of orthonormal basis functions with one of the basis functions, you get the coefficient of that basis function.