Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.5.

In a one-dimensional scattering problem, if we have an incident wave coming in from the left and scattering off a potential that is non-zero in the region and infinite for , then the wave effectively hits a brick wall at and is fully reflected back in the direction. Since the total probability of finding the particle can’t change (it won’t partially or fully disappear), the amplitude of the incoming wave function must equal the amplitude of the reflected wave function heading in the direction. However, interaction with the potential in the region could change the phase of the wave function, since the phase doesn’t affect the amplitude. That is, the wave function in the region has the form

where is the total phase change due to the reflection through the potential (the 2 is there since the wave function undergoes one phase shift of as it travels towards the origin and another phase shift of as it travels back towards , so is the total phase shift). The quantity as usual.

To find we must, as usual, solve the Schrödinger equation in the region and apply boundary conditions.

ExampleSuppose the potential is a half-finite square well of formwhere is a positive constant. Then the solution of the Schrödinger equation inside the well is

Because of the infinite barrier at , we must have so and

Since the potential is finite at , both and must be continuous there. This gives, from 1 and 6

Dividing the second equation by the first gives

We can solve for by multiplying the LHS by , so we get

Plugging this into 1 we see that the reflected wave is

The amplitude of the reflected wave is

So the reflected wave has the same amplitude as the incident wave, as required.

For a very deep well, so that , so from 11