Assume we have a (single) particle in a potential well of the following shape:

For $x \leq 0$, $V = \infty$ (Region **I**)

For $x \geq L$, $V = 0$ (Region **III**)

For the interval $x > 0$ to $x < L$, $V = -V\_0\frac{L-x}{L}$ (Region **II**).

The potential geometry is reminiscent of the potential energy function of a diatomic molecule (with $x$ the intra-nuclear distance).
In Region II the potential energy is a field with (positive) gradient $\frac{V\_0}{L}$.

**A few observations:**

In Region II, $V(x)$ is non-symmetric, so we can expect eigenfunctions without definite parity.

In Region II we can expect $\psi(0) = 0$.

We can also expect $\psi(\infty) = 0$, so the wave functions should be normalisable.

A quick analytic look at the Schrödinger equation in Region II using Wolfram Alpha’s DSolve facility shows the solutions involve the Airy Functions $A\_i$ and $B\_i$.

For $\frac{V\_0}{L} = 0$, the problem is reduced to an infinite potential *wall* (not a well). Incoming particles from Region III would simply be reflected by the wall at $x = 0, V = \infty$. There would be no bound states.

**And this raises an interesting question:** for which value of $\frac{V\_0}{L}$ is there at least one bound state and approximately at which value of the Hamiltonian $E$?

I have a feeling this can be related to the Uncertainty Principle because aren’t the confinement energies of bound particles in 1 D wells inversely proportional to $L^2$? If so would calculating a $\sigma\_x$ not allow calculating a $\langle p^2 \rangle$ and thus a minimum $E$ for a bound state?