Square Potentials

We study "square" potentials, or piece-wise constant potentials, because they may crudely approximate real potentials. Let us concentrate on one-dimensional "square" potentials. Assume V(x)=V=constant in certain regions of space. In such regions the Schroedinger equation yields

Let E>V:

, .

The most general solution is f(x)=Ae^ikx+A'e^-ikx, with A and A’ complex constants.

Let E<V:

, .

The most general solution is f(x)=Be^irx+B'e^-irx, with B and B’ complex constants.
Note: A solution exists in the classically forbidden region.

Let E=V:

The most general solution is f(x)=Cx+C', with C and C’ complex constants.

How does the wave function behave at a point x=x₁ where V is discontinuous, i.e. at a step?

Assume V has a step at x=x₁. Assume V¹±¥ on either side of the step.

Approximate V(x) by V_e(x), which is equal to V(x) everywhere except in a small region x₁±e, where it does not have the step but varies continuously. We can write

as a difference equation;

As e ® 0, V_e(x) remains finite, and therefore the integral remains finite and decreases as e decreases. For e=0 the integral is zero. Therefore ¶f_e/¶x |_{e ® 0} is continuous. At a finite step the boundary conditions are that f(x) and ¶f(x)/¶x are continuous.

Assume V(x) does not remain finite on one side of the step. Approximate V(x) by V_e(x) such that V_e(x) has a step DV over a small interval e=c/DV. As e ® 0, DV ® ¥, and

as e goes to zero. At an infinite step ¶f_e/¶x |_{e ® 0} is discontinuous, but it has a finite discontinuity. Therefore f_e(x) remains continuous as e ® 0.
In a region of finite width Dx, where |V| is infinite, the wave function must be zero at all times. Since the wave function is continuous, it must be zero also at the boundary in the adjacent region. If |V| is infinite in a region with width Dx=0 (a d-function, for example), the wave function can be finite there. It will have the same value at the boundaries of the two adjacent regions and its derivative with respect to x will be discontinuous.