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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 :

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

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  • $\begingroup$ Have you checked this integral in Gradshteyn&Ryzhik? $\endgroup$ Jun 21 at 7:25
  • $\begingroup$ 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. $\endgroup$ Jun 21 at 13:07

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