# Quantum mechanics: Finite square well problem

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.

• 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 9:00
• 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 9:11
• A negative potential is a potential $V(x)<V(\pm\infty)$ everywhere. – an offer can't refuse Oct 24 '14 at 9:12
• 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 9:26
• 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 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 $V_1 \lt V_0$.