mathematical physics - Solution for inverse square potential in $d=3$ spatial dimensions in quantum mechanics

Can a particle in an inverse square potential $$V(r)=-1/r^{2}$$ in $d=3$ spatial dimensions be solved exactly? Also please explain me the physical significance of this potential in comparison with Coulomb potential? That problem was talking about positive repulsive potential and what I am looking for is an attractive potential.

Answer

One may show that the inverse square potential

$$V(r)~=~\frac{\hbar^2}{2m}\frac{c}{r^2}~\propto~\frac{1}{r^2}, \tag{1}$$

in $d$ spatial dimensions mathematically speaking has three alternatives, depending on the dimensionless proportionality constant $$c_d~\equiv~\frac{(d-1)(d-3)}{4}+c~\equiv~\varkappa^2-\frac{1}{4}~\in~\mathbb{R}.\tag{2}$$

1. 
   

   Case $c_d < -\frac{1}{4}$: The spectrum is unbounded from below, i.e. the system is unstable.

   
   



2. 
   

   Case $c_d\geq \frac{3}{4}$: There are no bound states at all.

   
   


3. 
   

   Case $-\frac{1}{4} \leq c_d < \frac{3}{4}$: It is possible to define asymptotic boundary conditions (ABCs) at $r=0$ / selfadjoint extensions of the Hamiltonian, such that the spectrum is bounded from below. Some of these extensions have bound states, others have not.

   
   

From a physics point of view, it may seem unnatural that conclusions depend on adjustable ABCs imposed at $r=0$. The usual physics lore is that an inverse square potential is only an effective description, and that presumably new physics should emerge to resolve the singularity at $r=0$. From now on in this answer, we only discuss the purely mathematical problem at hand, even if it is somewhat academic.

First of all, angular excitations only push the energy up, never down, so it is enough to analyze spherically symmetric $s$-waves $$ \psi(r) ~\equiv~r^{\frac{1-d}{2}}u(r) , \qquad u(r)~\equiv~\sqrt{r}~ v(r) ,\tag{3}$$ and hence the energy functional simplifies to

$$\begin{align}\frac{2m}{\hbar^2} \frac{\langle \psi | \hat{H} \psi \rangle}{{\rm Vol}(S^{d-1})} &~=~ \int_{\mathbb{R}_+} \!r^{d-1}~ dr~ \psi(r)^{\ast} \left(- \frac{\partial^2}{\partial r^2}-\frac{d-1}{r}\frac{\partial}{\partial r} +\frac{c}{r^2}\right) \psi(r) \cr &~=~ \int_{\mathbb{R}_+} \! dr~ u(r)^{\ast} \left(- \frac{\partial^2}{\partial r^2}+\frac{c_d}{r^2}\right) u(r)\cr &~=~ \int_{\mathbb{R}_+} \! dr~ v(r)^{\ast} \left(- \frac{\partial}{\partial r}r\frac{\partial}{\partial r}+\frac{\varkappa^2}{r}\right) v(r). \tag{4}\end{align} $$

1) Sketched proof of case $c_d< -\frac{1}{4}$: Let us study a trial/test function

$$ u(r)~=~ r^{p}e^{-r/2}, \qquad \frac{1}{2}~

in the limit

$$p ~\searrow~ \frac{1}{2}. \tag{6}$$

The square norm

$$\frac{||\psi||^2}{{\rm Vol}(S^{d-1})}~=~ \int_{\mathbb{R}_+} \! dr~ |u(r)|^2~=~ (2p)! ~\searrow~ 1 \quad\text{for}\quad p ~\searrow~ \frac{1}{2} \tag{7}$$

is finite. The energy functional (4) simplifies to (after integration by parts)

$$\begin{align}\frac{2m}{\hbar^2} \frac{\langle \psi | \hat{H}| \psi \rangle}{{\rm Vol}(S^{d-1})} &~\stackrel{(4)}{=}~ \int_{\mathbb{R}_+} \! dr \left( |u^{\prime}(r)|^2 +\frac{c_d}{r^2} |u(r)|^2\right) \cr &~=~ \int_{\mathbb{R}_+} \! dr \left( \left(p-\frac{r}{2}\right)^2+c_d \right)r^{2p-2}e^{-r}\cr &~=~ \underbrace{(p^2+c_d)}_{<0}\underbrace{\int_{\mathbb{R}_+} \! dr ~r^{2p-2}e^{-r}}_{=(2p-2)!}+\text{finite terms}\cr &~\to~ -\infty\quad\text{for}\quad p ~\searrow~ \frac{1}{2},\tag{8} \end{align}$$

and is unbounded from below. $\Box$

In light of the previous section, let us assume that $\varkappa^2~\equiv~c_d+\frac{1}{4}\geq 0$ from now on. The radial TISE

$$\frac{\hbar^2}{2m}\left(-\frac{1}{r} \frac{\partial}{\partial r}r\frac{\partial}{\partial r}+\frac{\varkappa^2}{r^2}\right) v(r) ~=~ -|E|v(r), \qquad v(r\!=\!\infty)~=~0, \tag{9}$$

for bound states becomes the modified Bessel equation. The solution

$$ \begin{align}v(r)~~\propto~&~ K_{\varkappa}(\rho) ~~\propto~~ -\frac{\sin (\varkappa\pi)}{\pi/2}K_{\varkappa}(\rho) ~=~I_{\varkappa}(\rho)-I_{-\varkappa}(\rho) \cr ~~\stackrel{\rho\ll 1}{\sim}&~~ \frac{(\rho/2)^{\varkappa}}{\Gamma(1+\varkappa)} -\frac{(2/\rho)^{\varkappa}}{\Gamma(1-\varkappa)} , \end{align} \tag{10}$$ $$ \rho ~\equiv~\frac{\sqrt{2m|E|}}{\hbar}r, \tag{11}$$

is a modified Bessel function of the second kind. The problem (9) has a scale symmetry, so that the energy spectrum becomes unbounded from below unless we impose a pertinent ABC at $r=0$.

2) Sketched proof of case $c_d\geq \frac{3}{4}\Leftrightarrow \varkappa\geq 1$: The $\psi$-wavefunction (10) is not square integrable at $r=0$. In other words, pertinent ABC at $r=0$ prohibits the $\psi$-wavefunction (10). $\Box$

3) Sketched proof of case $-\frac{1}{4} \leq c_d < \frac{3}{4}\Leftrightarrow 0\leq \varkappa<1$: We now impose ABC (12) at $r=0$:

$$ v(r)~~\propto~~(k_0r)^{\varkappa} + \lambda (k_0r)^{-\varkappa} + O(r), \qquad k_0r ~\ll~1, \qquad \lambda~\in~\mathbb{R}, \tag{12}$$

where

$$k_0~\equiv~\frac{mc}{\hbar}\tag{13}$$

is the Compton wavenumber, i.e. the reciprocal reduced Compton wavelength. Here $\lambda\in\mathbb{R}$ is a fixed dimensionless parameter. Comparing eqs. (10) & (11), in the case $0< \varkappa<1$, this leads to a single bound state

$$ E~=~- 2mc^2 \left|\lambda\frac{\Gamma(1-\varkappa)}{\Gamma(1+\varkappa)} \right|^{-1/\varkappa} \quad \text{for}\quad \lambda~<~0,\tag{14}$$

and no bound states if $\lambda\geq 0$. The case $\varkappa=0$ has similar conclusions. See Ref. 3 for details. $\Box$

References:

1. 
   

   A.M. Essin & D.J. Griffiths, Quantum mechanics of the $1/x^2$ potential, Am. J. Phys. 74 (2006) 109.

   
   


2. 
   

   L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S$35.

   
   
   


3. 
   

   D.M. Gitman, I.V. Tyutin & B.L. Voronov, arXiv:0903.5277. (Hat tip: JamalS.)