Required math: calculus
Required physics: Schrödinger equation
Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.33.
Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.4, Exercise 5.4.2 (b).
We’ve analyzed the scattering problem in the finite square well, and we can use similar techniques to analyze a finite square barrier, which has the potential
where is a positive constant energy, and is a constant location on the axis.
There are three distinct cases here:
- Energy below the barrier:
- Energy exactly equal to the barrier:
- Energy greater than the barrier:
In all three cases, the wave function away from the barrier on either side has the same form; it is only within the barrier that the three cases differ. We’ll consider first the case where .
In this case, the Schrödinger equation within the barrier is
This has solution
Outside the barrier, the Schrödinger equation is
Outside the barrier on the left, the solution is the sum of travelling waves (assuming particles are incident from the left only), while to the right we have right propagating waves only. Thus
Since the potential is finite everywhere, both and are continuous everywhere, which gives us four boundary conditions.
At we have
We can solve these linear equations to get the other four constants in terms of . The results are
From here we can get the transmission coefficient as
The reciprocal of is then, substituting to get the physical quantities back:
The second case is where . In this case, the outer two solutions are the same as before, but in the barrier region we have
which has the solution
Applying the boundary conditions we have, at
At we have
Solving these equations we get
In this case the transmission coefficient is
Finally, for the Schrödinger equation within the barrier is
The solution within the barrier is thus
Boundary conditions give at
Solving these four equations gives
The transmission coefficient is
The reciprocal is