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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 ??

Jose Javier Garcia
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1 Answers1

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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.

Counterexample:

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

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.

II)

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.

III)

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.

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