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In Griffiths' text on QM, I am trying to understand his logic as to why there can be no states of negative energy. He writes:

What if I apply the lowering operator repeatedly? Eventually I'm going to reach a state with energy less than zero, which (according to the general theorem of Problem 2.2) does not exist!

And here is Problem 2.2:

Show that $E$ must exceed the minimum value of $V(x)$ for every normalizable solution to the time-independent Shroedinger equation.

Taking this as a given, is the reason we can't have negative energy states for the quantum harmonic oscillator because, in this situation, the potential is always non-negative? Otherwise, I don't see how Problem 2.2 helps us.

More generally, if the potential of a system is negative, then it is possible to have negative energy states?

Qmechanic
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theQman
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1 Answers1

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Yes. It's precisely because the minimum value of $V(x) = \frac{1}{2}kx^2$ is zero that there cannot be states with negative energy.

More generally, if the potential is negative, there can be states with negative energy. For example, for hydrogen atom, it's a well-know result that the electron "orbiting" a hydrogen atom to have energy:

\begin{align*} E_n \approx -\frac{13.6 }{n^2} \text{eV} \end{align*}

where $n$ is the principle quantum number.

This is precisely because the electrostatic potential energy in the system is $-\frac{1}{4\pi \epsilon_0} \frac{e^2}{r}$, which is negative for all $r$

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