Consider the radial differential equation

$$\bigg( - \frac{d^2}{dr^2} + \frac{(\ell+\frac{d-3}{2})(\ell+\frac{d-1}{2})}{r^2} + V(r) + m^2 \bigg) \phi_\ell (r) = \lambda\ \phi_\ell (r),$$

which I've obtained by solving the Schrodinger equation in $d$ dimensions using the method of separation of variables.

Now, the boundary condition that I have been given is $$\phi_\ell (r) \sim r^{\ell+\frac{d-1}{2}},\ r \rightarrow 0.$$

However, I was expecting the boundary conditions $$\phi_\ell (0) = 0,\ \phi^{\prime}_{\ell} (0) = 1.$$

Does anybody have an idea if there is a relation of some sort between the two sets of boundary conditions in the context of the given differential operator?


1 Answer 1


I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE.

The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See also this Phys.SE post. (Of course we might have to discard solutions for other reasons, see e.g. Section III below.)

II) The regular power behavior $$\phi_{\ell}~\sim~ r^{\ell+\frac{d-1}{2}}, \qquad r~\to~ 0,\tag{1} $$ seems to be deduced under the assumption that the kinetic term dominates over the potential term $V(r)$ for $r\to 0$, i.e. by considering the free radial TISE.

III) More generally, since the radial TISE is a 2nd order ODE, there are actually two independent branches, where (1) is one of them. The other branch, which is singular, has been discarded since the wave function should be normalizable, and the kinetic energy finite. For the $d=3$ case, see e.g. this Phys.SE post and links therein.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.