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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?

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  • $\begingroup$ Hint: the bound states of hydrogen have negative energy, the unbound (ionized) states have positive energy. $\endgroup$ May 3 '16 at 0:16
  • $\begingroup$ @AlfredCentauri I guess that the answer to this is pretty obvious but... in what case would the state of hydrogen have zero energy, then? $\endgroup$
    – Tendero
    May 3 '16 at 0:20
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    $\begingroup$ M.S., it seems that it would be the '$n = \infty$' state, that state that is 'on the fence' between the countably infinite bound states and the continuum of unbound states. Image $\endgroup$ May 3 '16 at 0:26
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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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  • $\begingroup$ Just to see if I got you right: you say that one example of this would be $V(x)=\frac12 m\omega^2x^2-\frac{\hbar \omega}{2}$? $\endgroup$
    – Tendero
    May 3 '16 at 0:16
  • $\begingroup$ Yup, that's right. $\endgroup$
    – knzhou
    May 3 '16 at 0:18
  • $\begingroup$ Nice answer, thank you very much. Could you please dwell on the "consequence of the uncertainty principle" thing? I don't see the relationship and I think that it is important $\endgroup$
    – Tendero
    May 3 '16 at 0:28
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    $\begingroup$ Consider the harmonic oscillator with $V(x) = m \omega^2 x^2 / 2$. Classically there's a state just sits at the bottom of the well. This can't happen in quantum mechanics because then the particle would have a definite position (zero) and momentum (zer0). So the particle needs to be spread around the minimum a little, giving it a little extra energy. $\endgroup$
    – knzhou
    May 3 '16 at 0:32
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    $\begingroup$ That's true for a classical particle, because the particle could be sitting at the minimum ($x = 0$) and not moving ($p = 0$). For a quantum particle, it's not possible at all. $\endgroup$
    – knzhou
    May 3 '16 at 0:50

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