Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let us suppose i can calculate the asymptotic of any potential $ V(x) $ in one dimension , and that i manage to prove that $ V(x) \ge 0 $ as $ x \rightarrow \infty $

could i conclude taht if or big 'x' the potential is POSITIVE that for big 'n' the enegies will be also POSITIVE $ E_{n} \ge 0 $ ?? as $ n \rightarrow \infty $

the idea is that if we define a turning point $ E=V(a) $ and 'a' is a turning point then if a is very big for big energy E then V(a) will be also positive but this is just in the WKB approximation isn't it ??

share|cite|improve this question
up vote 1 down vote accepted

I) It seems that the question(v1) is essentially asking the following:

If a 1D potential $V$ is a non-negative function on the real line $\mathbb{R}$, except for a compact interval $[a,b]$ where it is allowed to be negative, then would the number of (bounded) negative energy eigenstates for the corresponding 1D Hamiltonian $$H(x,p)~=~\frac{p^2}{2m}+V(x)$$ always be finite?

The answer is in general No.


$$V(x) ~=~ V_0 - \frac{A}{|x|^p},$$

where $V_0>0$ and $A>0$ are positive constants, and the power $p>2$. It is possible to prove that the spectrum is unbounded from below with infinitely many negative energy eigenstates in a very similar to e.g. methods used in this Phys.SE answer.


What if one additionally assumes that the potential $V$ is bounded from below?

Then the answer is Yes, because there must exist positive constants $V_0,L>0$ such that $$V(x)~\geq~ -V_0 ~\theta(L\!-\!|x|) .$$ In other words, one can find a finite potential well with lower energy levels, but we know that the spectrum of a finite potential well satisfies the sought-for statement.


What if one instead only additionally assumes that the spectrum (but not necessarily the potential $V$) is bounded from below, i.e. that the system has a stable ground state?

Then the answer is Yes, semiclassically (WKB), and I believe it is Yes non-semiclassically as well, but I don't (yet) have a rigorous proof.

share|cite|improve this answer
thanks Qmechanic what would happen if the potential $ V(x)=x^{2}- \frac{A}{|x|^{p}}$ for big 'x' would the energies be the ones of the harmonic oscillator ?? – Jose Javier Garcia Oct 3 '12 at 21:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.