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Consider the reduced radial Schrodinger equation:

$$-\frac{1}{2}\frac{\text{d}^2}{\text{d}r^2}\phi(r)+V(r)\phi(r)=E\phi(r).$$

We try to find a bound state (i.e. $\phi(0)=\phi(+\infty)=0$).

Here $V(r)=1/r$ or $1/r^2,1/r^3$ etc.

A numerical approach showed that the energy $E$ tends to zero when $a$ (i.e. the upper bound for $r$, use it to represent $+\infty$ in the numerical method) increases.

table1

(In this table $V(r)=1/r^{\alpha}$. $E_1$ is the calculated energy.)

My question is, how to explain this result physically?

Is that the real $E$ should be $0$, or allowed $E$s actually are continuous?

I think it is because of the repulsive potential $V(r)$, that $E$ should be $0$.

(P.S. I'm interested in the "ground state", if it exists. My numerical procedure tries to find the minimum energy. )

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There may be many bound states, each having its own E. Which one are you interested in? And how does you numerical procedure make sure that it selects the the eigenvalue that you are interested in? As for continuous spectrum, it got eliminated once you switched to the reduced equation (specific l has been selected, which means the motion is bounded). –  Slaviks Aug 19 '11 at 14:42
    
I'm interested in the "ground state", if it exists. My numerical procedure tries to find the minimum energy. –  NGY Aug 19 '11 at 14:48
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1 Answer 1

up vote 4 down vote accepted

There are no bound states for this potential, for any positive $\alpha$. All energies $E>0$ are allowed, and all of those are unbound states, in which $\phi\sim e^{ikr}$ for large $r$, with $E=k^2/2$.

The easiest way to prove that there are no states with $E<0$ is to show that the expectation values of both kinetic energy and potential energy, $\langle T\rangle$ and $\langle V\rangle$, are positive for any valid wavefunction. For an energy eigenfucntion, $E=\langle E\rangle=\langle T\rangle+\langle V\rangle>0$.

The fact that these aren't bound states -- that is, that they go as $e^{ikr}$ rather than as a decaying function of $r$ at large distances -- is most easily seen from the fact that, at sufficiently large $r$, the potential term is always negligible in comparison to the right-hand side. So at large $r$ we have $-\phi''/2=E\phi$.

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I think your answer is clear and convincing. And two more questions(maybe silly), does "all $E$ are allowed" imply that $E$ can be very large(even larger than $mc^2$)? And as my numerical result shows that $E$ tends to $0$ when I try to find a bound state, is it correct to say that bound state exists for $E=0$? –  NGY Aug 19 '11 at 15:16
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All positive $E$ are allowed as solutions to this equation. The Schrodinger equation is nonrelativistic, so that solutions with very large $E$ (of order $mc^2$ or more) would not be physically relevant -- you'd want to use a relativistic theory in this case. And no, I would not say that there is an $E=0$ bound state. Unbound solutions exist for $E>0$; there is no solution for $E=0$. –  Ted Bunn Aug 19 '11 at 15:29
    
Thanks, but how to see there is no solution for $E=0$ more clearly? Actually I have thought that my numerical solution tends to a solution with $E=0$. –  NGY Aug 19 '11 at 16:19
    
The argument I gave above proves that there is no such solution. For any solution, $E=\langle E\rangle=\langle T\rangle+\langle V\rangle$. Both of the latter two expectation values are strictly positive. –  Ted Bunn Aug 19 '11 at 16:51
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