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Consider e.g. the finite square well: $V = -V_o$ between $x=-a$ and $x=a$, $V=0$ elsewhere

Now for scattering states, $E$ must be $> 0$. For normalizable bound states, $E$ must be $< 0$ and $> V_{\rm min}$ (=$-V_o$ in example).

But if a particle in a lab has Energy which $< 0$ and $<V_{\rm min}$, is it bound or will it scatter?

And I don't know why I haven't thought about this before but what does it mean for a photon to have negative energy?

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If $E< V(x) $ everywhere, and if we assume that the kinetic energy operator $T=\frac{p^{\dagger}p}{2m}$ is a (semi)positive operator, then the TISE implies

$$ 0 ~\leq~ \langle \psi | T | \psi \rangle ~=~ \langle \psi | (E-V) | \psi \rangle~<~ 0, $$

which is impossible. Here $H=T+V$ is the Hamiltonian operator.

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This particle with have an unphysical wave function which blows up (as can be quite easily derived). Therefore, in quantum mechanics, we do not have any particles with $E<V_\text{min}$.

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  • $\begingroup$ Disregarding the mathematical blowing up of the wavefunction, practically we create a particle with the energy we want and then we pass it through a particular potential function(the particle doesn't "see" the potential Vmin until it comes across it). So I can't see why we can't create a particle with an energy value less than Vmin and then pass it through the potential. Unless I am missing something. $\endgroup$
    – user120404
    Commented Apr 7, 2014 at 12:20
  • $\begingroup$ 'Disregarding the mathematical blowing up' is simply discarding the physical reasonableness of your theory. A nonrenormalizable wave function makes no sense in terms of probabilities. $\endgroup$
    – Danu
    Commented Apr 7, 2014 at 12:21
  • $\begingroup$ ok my comment didn't come off like I intended. What I meant is that sometimes real physics happens where current mathematical models can't go. And in particular, Normalizability shouldn't be an indicator as to whether a particle with a certain energy can exist or not; particles with scattering state energies don't have normalizable wavefunctions, for instance. $\endgroup$
    – user120404
    Commented Apr 7, 2014 at 15:11
  • $\begingroup$ I assume you're talking about momentum eigenstates of, for instance, a free particle? In that case, that's exactly why those don't actually exist: We must have a superposition of momentum eigenstates (i.e. a wave packet) for things to work out. $\endgroup$
    – Danu
    Commented Apr 7, 2014 at 16:35

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