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What will happen if the potential is less than 0, for instance $V(x)=-10eV$. Is this means there will be no bound states? Since solution to the time independent Schrodinger equation (those discrete energies) must be greater than 0.

Qmechanic
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blue sky
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  • You are talking nonsense. There is always at least one bound state for the negative potential – an offer can't refuse Oct 24 '14 at 09:00
  • Can you explain in more details? what is the difference between negative and positive potential? Sorry, I'm quite new to this area... – blue sky Oct 24 '14 at 09:11
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    A negative potential is a potential $V(x)<V(\pm\infty)$ everywhere. – an offer can't refuse Oct 24 '14 at 09:12
  • absolute potential is not defined, only potential differences are in play, as luming's comment makes explicit. plus for negative potential energy there are bound states. – Nikos M. Oct 24 '14 at 09:26
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    Actually, we require that the energy eigenvalue be greater than the minimum of the potential energy (otherwise the solution is not normalizable). In your example, the energy eigenstate will be a bound state if the associated energy eigenvalue is greater than -10eV and less than the potential energy at infinity. – Alfred Centauri Oct 24 '14 at 11:26
  • "There is always at least one bound state for the negative potential in 1D", I should correct my statement, see http://physics.stackexchange.com/questions/143630/why-the-statement-there-exist-at-least-one-bound-state-for-negative-potential – an offer can't refuse Oct 29 '14 at 08:23

1 Answers1

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Your finite square well potential looks like:

Square well

where $V_0$ is the potential energy outside the well and $V_1$ is the potential inside the well. The depth of the well is $\Delta V = V_0 - V_1$.

We normally take $V_0$ to be zero, in which case $V_1$ is negative (like your $-10$eV) and $\Delta V = V_1$. However you can add any constant value to the potential energies without changing the physics. That's because the wavefunction only depends on the well depth $\Delta V$, and adding the same constant term to $V_0$ and $V_1$ doesn't change $\Delta V$. We say the potential energy has a global gauge symmetry.

It's not clear from your question what you intended to have the value $-10$eV. You could set $V_0 = -10$eV and particles would still be bound in the well as long as $V_1$ was less than $-10$eV, that is $V_1 \lt V_0$.

John Rennie
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