Problem:

Consider the square potential well shown in the figure below.

(a)  Find the most general solution Φ(x) of the eigenvalue equation HΦ(x) = EΦ(x), (E < 0), in regions 1, 2, and 3 and apply boundary conditions.
(b)  Solve the equation that results from part (a) graphically, and find the conditions under which even and odd solutions exist.

Solution:

• Concepts:
This is a "square potential" problem.  We solve HΦ(x) = EΦ(x) in regions where U(x) is constant and apply boundary conditions.
• Reasoning:
We are given a piecewise constant potential and are asked to find bound-state solutions.
• Details of the calculation:
Let -U0 < E < 0.  The wave function must remain finite at x = ± ∞.  We therefore have:
Φ1(x) = B1eiρx,   Φ2(x) = A2eikx + A2'e-ikx,  Φ3(x) = B3'e-iρx,
with k2 = (2m/ħ2)(E + U0) and ρ2 = (2m/ħ2)(-E).
Φ and ∂Φ/∂x  are continuous at x = ± a/2  This implies:
B1e-ρa/2 = A2e-ika/2 + A2'eika/2,  B3'e-ρa/2 = A2eika/2 + A2'e-ika/2.
ρB1e-ρa/2 = ikA2e-ika/2 - ikA2'eika/2,  -ρB3'e-ρa/2 = ikA2eika/2 - ikA2'e-ika/2.
Solving for A2 and A2 in terms of B1 we obtain
A2 = e(-ρ + ik)(a/2)(ρ + ik)/(2ik)B1,  -A2' = e-(ρ + ik)(a/2)(ρ - ik)/(2ik)B1.
Solving for A2 and A2 in terms of B3 we obtain
-A2 = e-(ρ + ik)(a/2)(ρ - ik)/(2ik)B3',  A2' = e(-ρ + ik)(a/2)(ρ + ik)/(2ik)B3'.
This yields two equations for B3 in terms of B1,
-B3' = [(ρ + ik)/(ρ - ik)]eikaB1  and  -B3' = [(ρ - ik)/(ρ + ik)]e-ikaB1,
which can only simultaneously be satisfied if
[(ρ - ik)/(ρ + ik)]2 = e2ika.
There are two possible solutions.

Solution 1:
[(ρ - ik)/(ρ + ik)] = -eika
But we also have
[(ρ - ik)/(ρ + ik)] = [(ρ2 + k2)1/2e-iθ/(ρ2 + k2)1/2e] = e-i2θ, with cotθ = ρ/k.
This implies
e-i2θ = -eika,  -2θ = ka - π,  cotθ = cot(π/2 - ka/2) = tan(ka/2) = ρ/k.
How do you solve an equation like this for k?
We can try a graphical solution.
Define  k02 = 2mU02 = k2 + ρ2.
Then 1/cos2(ka/2) = 1 + tan2(ka/2) = (k2 + ρ2)/k2 = k02/k2,
or |cos(ka/2)| = k/k0  for all k for which tan(ka/2) ≥ 0.
Note: We changed from a tangent to a cosine function for easier graphing.

In regions (1), (2), (3,) ...  tan(ka/2) ≥ 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.

Solution 2:
[(ρ - ik)/(ρ + ik)] = +eika
This implies e-i2θ = eika,  -2θ = ka,  cotθ = cot(-ka/2) = -cot(ka/2) = ρ/k.
Then 1/sin2(ka/2) = 1 + cot2(ka/2) = (k2 + ρ2)/k2 = k02/k2,
or |sin(ka/2)| = k/k0 for all k for which cot(ka/2) ≤ 0.
We construct a similar graph to get more solutions.

After we have found our solutions for k we can substitute  [(ρ - ik)/(ρ + ik)] = ±eika  back into the equations giving the relations between the A’s and B’s.  We find:
if [(ρ - ik)/(ρ + ik)] = -eika  then A2 = A2', B1 = B3', Φ(-x) = Φ(x), the solutions are even;
if  [(ρ - ik)/(ρ + ik)] = +eika  then A2 = -A2', B1 = -B3', Φ(-x) = -Φ(x), the solutions are odd.

For k0 ≤ π/a only one (even) solution exists.
If π/a ≤ k0 ≤ 2π/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 = nπ/a (n = integer).  Consequently
En = n2π2ħ2/(2ma2) - U0.
These are the energy levels of the infinite square well.

Summary
The operator representing the energy of a system is H.  The eigenvalues of H are E.  If the potential U(x) is independent of time, then separation of variables is possible, and we can write ψ(x,t) = Φ(x)Χ(t).  If the wave function is of this form, then Φ(x) = ΦE(x) is an eigenfunction of the operator H, and the energy of the system is certain.
We find the eigenfunction of H by solving HΦE(x) = EΦE(x).

We can solve this equation in regions of piecewise constant potentials.

• Regions that do not contain a well
• There exists an eigenfunction for every E > Umin.  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 = k|Φ(x)|2.)

• Regions that do contain a well
• There exists an eigenfunction for every E > Erim.  These eigenfunctions are not square integrable.
• For E < Erim eigenfunctions exist only for selected eigenvalues.  We can solve for the bound states in a square-well potential using a graphical solution.