# Why do we always require the potentials to be bounded from below?

Why do we always require the potentials $V(x)$ (both in classical and quantum mechanics) to be bounded from below i.e., require that $V(x)$ should not go to $-\infty$ as $x\rightarrow \pm \infty$?

For example, a potential like $V_1(x)=x^2+x^3$ is often considered "pathological" because it goes to $-\infty$ as $x\rightarrow-\infty$ unlike the potential $V_2(x)=x^2$. It is argued that for $V_1(x)$, the potential has a metastable/local minima, and under small (classical/quantum) perturbation, the particle may roll down to $x\rightarrow -\infty$ in the direction of ever-decreasing potential energy, releasing energy in the form of kinetic energy (if the particle is not electrically charged).

When the particle rolls down to $x\rightarrow -\infty$, the potential energy decreases continuously, and kinetic energy increases from energy conservation. So it doesn't violate the conservation of energy.

Why is this a problem then?

• The title is not true. The Coulomb potential in the Hydrogen atom is not bounded from below. – Qmechanic Dec 23 '16 at 21:44

## 1 Answer

In classical physics, it makes sense to deal with a potential that is not bounded below. You can solve the equations of motion for a system as the potential energy becomes ever more negative and the kinetic energy increases. This is often not a terribly interesting scenario even in Newtonian mechanics, because there is no stable equilibrium state. (But there are interesting potentials that are unbounded below, such as the gravitational potential.) Moreover, if you want to think realistically about what happens as the particle energy gets very large, you will eventually need to take relativity into account. And in relativity, the notion of potential energy becomes problematic.

In quantum mechanics, the problem can be much more severe, although there are still interesting potentials that are unbounded below (again, the $$1/r$$ potential being the most important example). The issue is that if the potential becomes infinitely negative too quickly, there is no normalizable wave function; the wave function solutions diverge as the potential does, making them not square integrable. In the operator formulation, you would want to construct the states as excited states above the lowest-energy state, but there is no lowest-energy state, making this impossible. So if there is too much negative divergence in the potential, the quantum theory simply fails to be well defined.

• But the fact that gravitational potential at $r=0$ is singular implies that the theory is sick at $r=0$ because it leads to infinite force. But such a problem is not there for the potential $V(x)=x$, for example. What is the problem with this potential if $x\rightarrow-\infty$ is accessible for the particle.? – SRS Dec 23 '16 at 20:24
• @SRS Quantum mechanically, the energy eigenstate solutions of the Schroedinger equation (Airy functions) are never normalizable. – Buzz Dec 23 '16 at 20:40
• Great! But why does the energy eigenstates do not all become non-normalizable in H-atom too the potential is unbounded? Is it that somehow we are avoiding $r=0$ value as unphysical or inaccessible for the electron? We also do have a ground state and excited states above it. @Buzz – SRS Nov 5 '18 at 19:08