I've read a little bit about zero-energy states, but I just don't get it. I'm just starting to study quantum mechanics and, at least for all the potentials I've seen until now (the most popular ones, like the infinite well or the harmonic oscillator), there is a certain minimum value $E_0\neq0$.
So the question is: given the time-independent Schrodinger's equation $$-\frac{\hbar}{2m}\frac{\partial^2\Psi(x)}{\partial x^2}+V(x)\Psi(x)=E\Psi(x)$$
is it possible, for any potential $V(x)$, that $E=0$? If it is, What would an example of that potential be?
Answer
Yes. For example, consider the harmonic oscillator potential $V(x)$, where the ground state has energy $\hbar \omega / 2$. Then the ground state of the potential $V(x) - \hbar \omega / 2$ has zero energy.
This works because in quantum mechanics, like in classical mechanics, absolute energies don't matter. You can always add or subtract constants.
However, there's a more interesting question I think you wanted to ask instead, which is, is it possible to have a state with energy lower than the minimum of $V(x)$? The answer is no, because in such a case, $\partial^2 \Psi / \partial x^2$ will always be positive, and you can check that there is no normalizable $\Psi(x)$ that satisfies this. So particles always sit a little higher than the bottom of a well. This can also be thought of as a consequence of the uncertainty principle.
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