Particle in a finite well: Potential energy function

I'm a chemist, not a physicist, but am taking a quantum chemistry course right now and I'm having difficulty grappling with the following:

For the particle in a finite potential well, so long as no forces act on it inside the well its potential will be constant and we set it to $0$. That makes sense, what I don't understand is why the particles potential energy must be $U_0$ outside the well. I understand it obviously cant be less than $U_0$, and it makes sense that it could be exactly $U_0$, but why cant it be greater than $U_0$?

If its greater than $U_0$ wouldn't it still be found outside the walls of the box?

I've tried looking for answers but I cant find a satisfactory one, which makes me think I'm thinking about the problem incorrectly. Does it have something to do with the energy being less than the potential? Any help on rectifying this?

• Only in an infinite potential well is the probability of finding the particle zero outside the well. The wavefunction goes like $exp(-kx)$ outside of the well. Im pretty sure this is a repeat question, can anyone else find an earlier post? – Haru Fujimura Oct 1 '16 at 21:17
• Wave functions and energy eigenvalues of bound states ($E<U_0$): en.wikipedia.org/wiki/… – Gert Oct 1 '16 at 23:05
• I don't understand the question. The energy of the particle can be greater than $U_0$, but the potential energy is fixed. The amount that the energy exceeds the potential energy is found in the particle's kinetic energy, and the particle can be found outside the box. But these are classical arguments. There's nothing specific to quantum mechanics about that. I also don't know what you are trying to say in the last two sentences. But I might be misreading. – garyp Oct 1 '16 at 23:30
• I wasn't grasping the fixed potential outside the well. I thought the potential was only constant inside the well, and was just imagining a particle with arbitrarily high energy rocketing out of the well before eventually stopping at some height far above U (0) where all of its energy would be potential. – Brian Oct 1 '16 at 23:37

The potential well has by definition this potential energy structure. The potential is zero inside and outside it has a constant value denoted by U(0), which should not be confused with the potential inside.

• So its purely a definition thats set that way to simplify the resulting equations? – Brian Oct 1 '16 at 21:22
• You are correct. The zero of potential energy can be chosen arbitrarily and it is most convenient to take the bottom of the well for it. – freecharly Oct 1 '16 at 21:28
• The choice of the constant potential outside the well is, of course, arbitrary. Depending on the total energy of the particle, the higher the potential step is, the lower is the probability to find the particle outside the well. – freecharly Oct 2 '16 at 1:08

What's going on here is the potential energy is constant inside the well, and a different constant outside. What you're looking for is a bound state of the well. In order to be bound the total energy of the particle has to be less than the potential energy outside of the well, otherwise it would have enough energy to escape. Because of the choice that the energy inside of the well is $0$, the total energy of the particle will be entirely kinetic energy.

The quantum mechanical nature of the wave function means that the particle will penetrate the walls, what is known as the "classically forbidden region." That penetration will have an exponentially damped length scale that depends on how far below $U_0$ the total energy is.

• I think I understand it a bit better. Would then, if we were attempting to solve the scattered state of the well, consider E> U(0) wherein the particle could be arbitrarily high above and far away from the well's walls? – Brian Oct 1 '16 at 21:45
• Yes, scattering states have $E > U_0$. – Sean E. Lake Oct 1 '16 at 21:46
• Is the potential energy function defined at the x positions 0 and L? It looks as if it could take on any value from 0 to U(0) at those two points the way the well is drawn but that doesnt seem to make intuitive sense. – Brian Oct 1 '16 at 23:41
• It doesn't matter. The wave function can only respond to one of the following: a difference in potential that takes up non-zero volume/distance in space, or a difference in potential that has infinite size (a delta function). Anything else, i.e. point-like finite discontinuities, don't produce a difference in the inner product $$\langle \phi| H | \psi\rangle$$ for any wave functions, $\phi$ and $\psi$, that have finite energy. – Sean E. Lake Oct 2 '16 at 0:14
• Another way to think about it is that $\psi$ responds to the local average potential: $$\langle V(x) \rangle \equiv \lim_{\delta\rightarrow 0} \frac{1}{2\delta} \int_{x-\delta}^{x+\delta} V(u) \operatorname{d}u.$$ – Sean E. Lake Oct 2 '16 at 0:15