The Square Well Potential

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Let -V₀<E<0. The wave function must remain finite at x = ± ¥. We therefore have:

and

f and ¶f/¶x are continuous at x = ± a/2 This implies:

Solving for A₂ and A₂^’ in terms of B₁ we obtain

Solving for A₂ and A₂^’ in terms of B₃^’ we obtain

This yields two equations for B₃^’ in terms of B₁,

which can only simultaneously be satisfied if

There are two possible solutions.

But we also have

This implies

We have

How do you solve an equation like this for k?
We can try a graphical solution. Define

Then

for all k for which tan(ka/2) ³ 0.
Note: We changed from a tangent to a cosine function for easier graphing.
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In regions (1), (2), (3,) ... tan(ka/2) ³ 0. Three solutions exist for the given k₀ in the graph. As we increase k₀ more solutions become possible. For every k₀ at least one solution is possible. There exists at least one bound state. But we have not yet found the complete set of solutions.

This implies

We have

for all k for which cot(ka/2) £ 0.
We construct a similar graph to get more solutions.

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After we have found our solutions for k we can substitute

back into the equations giving the relations between the A’s and B’s. We find:

	if then A₂=A₂', B₁=B₃', f(-x)=f(x), the solutions are even;
	if then A₂=-A₂', B₁=-B₃', f(-x)=-f(x), the solutions are odd.

For k₀ £ p/a only one (even) solution exists.
If p/a £ k₀ £ 2p/a the first odd solution becomes possible.
If k₀ is very large, then the slope of the straight line is very small and solutions appear near every k=np/a (n = integer). Consequently

These are the energy levels of the infinite square well.

Link:

The infinitely deep square well

Summary

The operator representing the energy of a system is H. The eigenvalues of H are E. If the potential V(x) is independent of time, then separation of variables is possible, we can write y(r,t)=f(r)c(t). If the wave function is of this form, then f(r)=f_E(r) is an eigenfunction of the operator H, and the energy of the system is certain. We find the eigenfunction of H by solving Hf_E(r)=Ef_E(r).

We have solved this equation in regions of piece-wise constant potentials.

Regions that do not contain a well

there exists an eigenfunction for every E>V. These eigenfunctions, however, are plane waves and are not square integrable. They cannot represent a single particle, but can represent a constant flux of particles. We calculate transmission and reflection coefficients by comparing fluxes. (Flux µ k|f(x)|².)

Regions that do contain a well

	There exists an eigenfunction for every E>E_rim. These eigenfunctions are not square integrable.
	For E<E_rim eigenfunctions exist only for selected eigenvalues. We have solved for the bound states in a square-well potential using a graphical solution. The program below demonstrates how to find a numerical solution.

Write your own program

This link shows you how to find a numerical solution of the time-independent Schroedinger equation in one dimension to find the bound states of a particle in a square well potential and presents you with an example program. The link introduces you to the Numerov method.