# How should I get the expectation value of $r^2$ in the hydrogen atom?

I'm having trouble finding this expectation value: $$\langle r^2\rangle=\frac{C_{n,l}}{\alpha^5}\int_0^{\infty}e^{-x}x^{2l+4}[L^{2l+1}_{n-l-1}(x)]^2dx$$

Where $$x=\alpha r$$ and $$\alpha=\frac{2}{na_o}$$.

I'm using recurrence relations of the associated Laguerre Polynomials... So, far I've used the following RRs: $$L^m_n(x)=L^{m+1}_n(x)-L_{n-1}^{m+1}(x)$$ $$xL^m_n(x)=(2n+m+1)L^{m}_n(x)-(n+m)L_{n-1}^{m}(x)-(n+1)L^m_{n+1}(x)$$

And the orthogonality conditions :

$$\int_0^{\infty}e^{-x}x^mL^m_p(x)L^m_n(x)dx=\frac{(n+m)!}{n!}\delta_{m,n}$$ $$\int_0^{\infty}e^{-x}x^{m+1}[L^m_n(x)]^2dx=\frac{(n+m)!}{n!}(2n+m+1)$$

• Have you checked this integral in Gradshteyn&Ryzhik? Jun 21 at 7:25
• There is a recursion known as the Pasternach relation to solve for this. See also Balasubramanian, S. "A simple derivation of the recursion relation for $\langle r^n\rangle$ in energy eigenstates." American Journal of Physics 68.10 (2000): 959-960. Jun 21 at 13:07