Recently I read a paper about solving radial Schrodinger equation with inverse power law potential.
Consider the radial Schrodinger equation(simply set $\mu=\hbar=1$):
$$\left(-\frac{1}{2}\Delta+V(\mathbf{r})\right)\psi(\mathbf{r})=E\psi(\mathbf{r}).$$
A well-known substitution gives a one-dimension equation:
$$-\frac{1}{2}D^2\phi(r)+\left(V(r)+\frac{1}{2}\frac{l(l+1)}{r^2}\right)\phi(r)=E\phi(r),$$
where $D=\dfrac{d}{dr}$, and $l$ is the azimuthal quantum number.
If we only consider the ground state, then $l=0$, so
$$-\frac{1}{2}D^2 \phi(r)+V(r)\phi(r)=E\phi(r).$$
We want to find eigenvalue $E$ such that $\phi(0)=\phi(+\infty)=0$.
The central potential discussed in the paper is of this form:
$$V(r)=\alpha r^{-\beta}.$$
It states (see page 4) that if $\beta>2$ then the potential is repulsive (i.e. $\alpha>0$).
My questions are:
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Is this conclusion (i.e. If $\beta>2$ then we must have $\alpha>0$) generally valid in physics?
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What would happen if $\alpha<0$ and $\beta>2$? Is there 'ground state' in this condition?
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What about the condition $\alpha>0$ and $\beta=1,2$? I have tried to solve the equation numerically with $\alpha=1,\beta=1,2$, and the ground state energy in this two conditions seem to be $0$, is my result correct?
P.S. When I try to find ground state when $\alpha=-1,\beta=2$, the energy seem to be $-\infty$, which is qualitatively different from $\alpha=-1,\beta=1$.
Best Answer
What they show in the paper is that, for $\beta>2$, there are no solutions with the given asymptotic form as $r\to 0$ unless we assume $\alpha>0$. I think you can go further and show that there are no nonsingular solutions for $\beta>2,\alpha<0$, but I'm not sure.
What does this mean physically? Well, when we have a potential with a singularity, usually we think of it as just an approximation that breaks down sufficiently close to the singularity. For instance, we model the hydrogen atom with a potential $V\propto r^{-1}$, but really the potential near $r=0$ doesn't go to infinity, due to the nonzero size of the nucleus. We get away with using the singular potential because the "bad" behavior at $r=0$ doesn't qualitatively change the solutions. (And of course people do correct for nonzero-nuclear-size effects in atomic physics.)
If it's true that the ground state of the Schrodinger equation is singular for potentials of the given form ($\alpha<0,\beta> 2$), what that means is that this procedure doesn't work. To be specific, suppose that you solved the Schrodinger equation for a potential that looks like the given one down to some "cutoff" $r_0$, and is constant for smaller $r$. What you'd find is that the solution doesn't tend to some limit as $r_0\to 0$ -- the solution depends qualitatively on the size of that cutoff, no matter how small it is.
To answer your last question, for any repulsive potential ($\alpha,\beta>0$), you expect to find only continuum (unbound) states. Those states have $E>0$, and all positive values of $E$ are allowed. So if you try to solve numerically for the ground state, I'm not surprised that you seem to get zero.
Addition: After the discussion in the comments, it occurs to me that we can see why the case $\beta=2$ behaves the way it does. The Schrodinger equation in that case is $$ -{1\over 2}\nabla^2\psi + {\alpha\over r^2}\psi=E\psi. $$ Suppose that you'd found a bound-state solution $\psi_0$ corresponding to some energy $E_0<0$. Define a new solution by simply scaling the radial coordinate: $$ \psi_1(r)=\psi_0(cr) $$ for any $c>0$. Then $\psi_1$ is also a solution to the Schrodinger equation, with energy $E_1=c^2E_0$. In particular, for $c>1$ this corresponds to squeezing the wavefunction into a smaller space and making the energy more negative. If you try to find the lowest-energy solution, you'll end up with the $c\to\infty$ case -- an infinitely concentrated wavefunction, with energy $-\infty$.
If you try the procedure I suggest in my last comment (cutting off the singularity in the potential at some $r_0$ and then varying $r_0$), something similar occurs. The ground state solution for all positive $r_0$'s look the same, with radial coordinates scaled by the value of $r_0$, and the ground state energy goes like $r_0^2$. As $r_0\to 0$, the ground-state energy approaches $-\infty$, and the wavefunction becomes infinitely concentrated at $r=0$.
This only works for the case $\beta=2$, because for this value of $\beta$ both the kinetic and potential terms on the left side of the Schrodinger equation scale in the same way when you rescale your coordinates (i.e., both go like $c^2$). Another way to put it: only in the case $\beta=2$ is the constant $\alpha$ dimensionless. For any other $\beta$, the value of $\alpha$ determines a length scale (so that you can't just rescale one solution to get a new one), but when $\beta=2$ the problem is scale-invariant.