I have to solve the radial Schrodinger equation for a particle subjected to the potential:

$$V(r)= \dfrac{A}{r^2}-\dfrac{B}{r}$$

Where $r$ is the radial component (spherical coordinates) and $A,B>0$ While I was uncoupling the equations it was useful to define the following constants/function: $$k^2\dot{=}\dfrac{2mE}{\hslash^2} \ \ a\dot{=}\dfrac{2Am}{\hslash^2} \ \ b\dot{=}\dfrac{2Bm}{\hslash^2} \ \ \sigma\dot{=}[l(l+1)-a] \text{ and } u(r)\dot{=}R(r)$$ where $R(r)$ is the radial part of the variable separation and $l$ is the $L^2$ correspondent quantum number. So, I`ve found the following diffential equation (I'll denote u' as differentiation with respect to r):


As I want my result to be square integrable function I`d impose that $$\lim _{r\rightarrow\infty}r^2R^2(r)=0 \Rightarrow u(\infty)=0$$ But I'm not really sure if, for the $r=0$ condition, I should take $$|u(0)|<\infty \quad\text{ or }\quad u(0)=0~?$$

Which choice of those last are more suitable for physical means? Is there a single choice? Would 'em be the same of hydrogen atom?

  • 3
    $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/183045/2451 & physics.stackexchange.com/q/134719/2451 & physics.stackexchange.com/q/291222/2451 The reason is the same: The boundary condition $u(r\!=\!0)~=~0$ follows because the wave function should be normalizable, and the kinetic energy finite. $\endgroup$
    – Qmechanic
    Jun 24, 2017 at 13:46
  • 2
    $\begingroup$ Maybe I didn't made myself clear: why doesn't both of the last boundary conditions I suggested satisfy square integrability and finite energy? That's what I can`t see mostly and couldn't find an answer Is there a theorem of something which states that the physically acceptable boundary conditions are unique? $\endgroup$
    – Janov
    Jun 24, 2017 at 14:37

1 Answer 1


The question of what boundary conditions are admissible for a Shroedinger equation with singular endpoints was studied by Weyl. For each possible boundary condition you need to find out whether one or or both of the linearly independent solutions are square integrable near the endpoint. To do this you only need consider one possible eigenvalue --- as the answer is the same for all. If only one is normalizable you are the limit point case. If both are you are in the limit circle case. In the latter you have a one parameter family of possible boundary conditions. An example of the latter occurs for the $l=0$ radial Schrodinger equation in 3d with no potential. Possible solutions with $E=0$ are $\psi_1(r)=1$ and $\psi_2(r)=1/r$. Despite the fact that the second solution diverges near the origin, both are normalizable. A suitable boundary condition would be to set $$ \psi(r)\sim 1-\frac{a_s}{r} $$ as $r\to 0$ where $a_s$ is a quantity called the scattering length. Physically $a_s$ arises as the effect of a scatterer at the origin whose size is much smaller that the wavelength of the incoming wave. It therefore appears as a boundary condition rather than an explicit potential. For example this is how we parametrize scattering of atoms in Bose condensed atomic gasses. A more complete analysis was made by Von Neumann who introduced the terms deficiency indices and self-adjoint extensions. This is too complicated to describe properly here, so I suggest that you Google these terms and chase up some references.

To start with , I suggest that you look at chapter 4 (for deficiency indices) and chapter 8 (for Weyl's theorem) of Goldbart and Stone's "Mathematics for Physics." (I like this book because I am a co-author). For Weyl's theorem, see Ivar Stackgold's "Boundary Values of Mathematica Physics" Vol 1. For both Weyl and Von Neumann, see Richtmeyer's "Principles of Advanced Mathematical Physics" Vol 1. The relevant chapters of my book can be found as part of the lecture on the University of Illinois course page for Physics 508:


  • $\begingroup$ Do you know the name of this work? Seems pretty interesting. I couldn`t find . $\endgroup$
    – Janov
    Jun 24, 2017 at 21:17
  • $\begingroup$ I'll edit my answer to add some references. $\endgroup$
    – mike stone
    Jun 25, 2017 at 5:47

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