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I am studying eigenfunction methods to solve Fokker-Planck equations and I got stuck with a calculation that is related to some typical problems in QM. In particular, the radial part of an hydrogen atom can be characterized through a Laguerre equation of the type , $$x y''+(1-x)y+\lambda y =0.$$ In undergraduate notes, I've always seen statements like "the solution to Laguerre equations are Laguerre polynomials". However, this is a second order ODE which has two linearly independent solutions, as stated in mathworld.wolfram, $$ y(x) = c_1 L_\lambda (x) + c_2 U(-\lambda,1,x).$$ Where $L$ stands for the Laguerre polynomial and $U$ is the second standard solution to Kummer's equation . What is the reasoning in QM to discard the confluent hypergeometric function $U$? I.e. Why should $c_2=0$?

Qmechanic
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Javi
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1 Answers1

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This is related to the fact that the solutions in the textbook are written only for the discrete spectrum of energies. We are interested in the normalized solutions, i.e. such that \begin{equation*} \int d^3x |\Psi|^2 \end{equation*} were finite. However, if you take arbitrary energies one of the solutions will have singularity at $r=0$ and another one will grow with $r\rightarrow+\infty$. This will break down the normalizability.

For the discrete spectrum of energies you will find the solutions that have good behavior both at $r=0$ and $r\rightarrow+\infty$. This solutions will be normalizable. But the second solution will in contrast have bad behavior both at $r=0$ and for $r\rightarrow+\infty$.

OON
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