# What happens on a finite negative well when the particle has smaller energie than the potential well? [duplicate]

This is a common exercise in the classroom. They always check the cases when $$E>0$$ and $$V for a finite negative well $$V<0$$.

But what about the case $$E. What happens here?

• There are no states in this energy range. Commented Apr 25, 2023 at 17:53
• @Who In that case, particle is not constrained by potential energy of quantum well. You can ask similar question for classical physics as well. For example, what will be if you jump into a hole with potential energy $-mgh_0$ and suddenly out of nowhere you have potential energy $-mgh_1$, where $h_1 \lt h_0$ ? Whatever it will be,- you are not "describable" by previous potential well. And what will happen to you - nobody knows. Maybe you will be captured by a deeper other potential well, or maybe you are tunneling to somewhere else, in quantum case analogy. Anyway, well has no meaning here. Commented Apr 25, 2023 at 18:30
• Possible duplicates: physics.stackexchange.com/q/391446/2451 and links therein. Commented Apr 26, 2023 at 3:23
• It is also possible to show from analysing the Schrödinger's equation that assuming an energy that is too low leads to impossibility to normalise the resulting wavefunction. Commented Apr 26, 2023 at 6:19

For any state $$|\psi\rangle$$ we have that $$\langle H \rangle = \frac{1}{2m}\langle p^2\rangle + \langle V \rangle\;.$$
As it is a square $$\langle p^2 \rangle \ge 0$$.
Now if we have that, for all $$x$$, $$V(x) \ge V_{min}$$, we can say $$\langle V \rangle = \int dx\; \psi^*(x) V(x) \psi(x) \ge V_{min} \int dx |\psi(x)|^2 = V_{min}\;.$$ Putting this together we have that $$\langle H \rangle \ge 0 + V_{min}\;.$$ Since all states have an energy expectation greater than or equal to $$V_{min}$$, there can be no eigenstates of $$H$$ with energy less than $$V_{min}$$
• It's certainly provable without much effort that the square of a self-adjoint operator must have non-negative expected value in any state, but given the level of the question perhaps it would be valuable to suggest doing so on their own (rather than simply saying that $\langle p^2\rangle>0$, which is not as trivial as it appears). After all, $\langle - \nabla^2 \rangle > 0$ is perhaps less obvious to a newer student. Commented Apr 25, 2023 at 18:23