I am studying bound states for different types of potential, and got confused by a particular circumstance: suppose we have a potential $V({\bf r})$ that goes to zero when the position $r$ goes to $\infty$, but is positive for every finite value of the position:

$$ \qquad V({\bf r}) \ge 0 \quad \text{ but } \quad V({\bf r}) \to 0 \,\text{ for }\, r\to\infty. $$

Is it possible to find a bound state within such a potential region?

  • $\begingroup$ to be clear, are you thinking of a potential which is just positive everywhere, or for a monotonically decreasing one? $\endgroup$ – glS Oct 17 '16 at 19:15
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    $\begingroup$ I edited the post trying to make question clearer. If you think the current version does not reflect appropriately what you meant to ask feel free to revert the edit $\endgroup$ – glS Oct 17 '16 at 19:35
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    $\begingroup$ One could have an infinite potential well with infinite high walls of finite thickness. $\endgroup$ – Qmechanic Oct 17 '16 at 19:48
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    $\begingroup$ This seems to be a rather mathematical question. You would need some further conditions on your potential $V$ to ensure that the Hamiltonian operator is ess. self adjoint on a dense subspace of the square-integrable functions, which is needed to make the problem well-posed. Assuming these conditions in addition to your requirements, I would be surprised if such a system can have any bound states. Actually proving this might not be easy though. (I'm assuming $V(x)$ is a function, which is finite everywhere) $\endgroup$ – Adomas Baliuka Oct 17 '16 at 19:57
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    $\begingroup$ Look at the potentials in schoolphysics.co.uk/age16-19/Thermal%20physics/Heat%20energy/… and recall that potentials are defined up to an arbitrary constant. Molecules exist which means that atoms in them are in a bound state. Also, in most cases, the "position" $\mathbb r$ is to be considered as relative position, e.g. as the distance between atoms in a diatomic molecule (the main argument is symmetry w.r.t. spatial translations). $\endgroup$ – Phoenix87 Oct 17 '16 at 21:12

Suppose the Hamiltonian with the given potential $V(x)$ for the one-dimensional system is well-defined and $V(x)\geq 0$ falls off at $\pm\infty$. Suppose $0\neq \psi\in L^2(\mathbb{R})$ is a bound state, i.e. there is an energy $E$ such that $$\psi''=2m(V-E)\psi.$$

For $\psi$ to be normalizable, we certainly need $E>0$. Then looking at the limit towards $\infty$, we can argue that $$\psi''(x)\approx -2mE\psi(x)\qquad (x\to\infty).$$ This is solved by sine and cosine waves and I think that is enaugh to argue that $\psi$ cannot be normalized. This would lead to the conclusion that the system has no bound states. I hope someone knows how to make this argument precise in terms of mathematics and generalize to $3$ spacial dimensions.

  • $\begingroup$ Don't just accept it, this is far from a correct proof of the statement... Was hoping someone would know the correct argument, maybe I'll try to think of how $\endgroup$ – Adomas Baliuka Nov 8 '16 at 15:59

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