**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 -U_{0 }< E < 0. The wave function must remain finite at x = ± ∞. We therefore have:

Φ_{1}(x) = B_{1}e^{iρx}, Φ_{2}(x) = A_{2}e^{ikx }+ A_{2}'e^{-ikx}, Φ_{3}(x) = B_{3}'e^{-iρx},

with k^{2}= (2m/ħ^{2})(E + U_{0}) and ρ^{2}= (2m/ħ^{2})(-E).

Φ and ∂Φ/∂x are continuous at x = ± a/2 This implies:

B_{1}e^{-ρa/2}= A_{2}e^{-ika/2 }+ A_{2}'e^{ika/2}, B_{3}'e^{-ρa/2}= A_{2}e^{ika/2 }+ A_{2}'e^{-ika/2}.

ρB_{1}e^{-ρa/2}= ikA_{2}e^{-ika/2 }- ikA_{2}'e^{ika/2}, -ρB_{3}'e^{-ρa/2}= ikA_{2}e^{ika/2 }- ikA_{2}'e^{-ika/2}.

Solving for A_{2}and A_{2}^{’}in terms of B_{1}we obtain

A_{2}= e^{(-ρ + ik)(a/2)}(ρ + ik)/(2ik)B_{1}, -A_{2}' = e^{-(ρ + ik)(a/2)}(ρ - ik)/(2ik)B_{1}.

Solving for A_{2}and A_{2}^{’}in terms of B_{3}^{’}we obtain

-A_{2}= e^{-(ρ + ik)(a/2)}(ρ - ik)/(2ik)B_{3}', A_{2}' = e^{(-ρ + ik)(a/2)}(ρ + ik)/(2ik)B_{3}'.

This yields two equations for B_{3}^{’}in terms of B_{1},

-B_{3}' = [(ρ + ik)/(ρ - ik)]e^{ika}B_{1}and -B_{3}' = [(ρ - ik)/(ρ + ik)]e^{-ika}B_{1},

which can only simultaneously be satisfied if

[(ρ - ik)/(ρ + ik)]^{2}= e^{2ika}.

There are two possible solutions.

**Solution 1:**

[(ρ - ik)/(ρ + ik)] = -e^{ika}

But we also have

[(ρ - ik)/(ρ + ik)] = [(ρ^{2}+ k^{2})^{1/2}e^{-iθ}/(ρ^{2}+ k^{2})^{1/2}e^{iθ}] = e^{-i2θ}, with cotθ = ρ/k.

This implies

e^{-i2θ}= -e^{ika}, -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 k_{0}^{2}= 2mU_{0}/ħ^{2}= k^{2}+ ρ^{2}.

Then 1/cos^{2}(ka/2) = 1 + tan^{2}(ka/2) = (k^{2}+ ρ^{2})/k^{2}= k_{0}^{2}/k^{2},

or |cos(ka/2)| = k/k_{0}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 k_{0}in the graph. As we increase k_{0}more solutions become possible. For every k_{0}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)] = +e^{ika}

This implies e^{-i2θ}= e^{ika}, -2θ = ka, cotθ = cot(-ka/2) = -cot(ka/2) = ρ/k.

Then 1/sin^{2}(ka/2) = 1 + cot^{2}(ka/2) = (k^{2}+ ρ^{2})/k^{2}= k_{0}^{2}/k^{2},

or |sin(ka/2)| = k/k_{0}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)] = ±e

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

if [(ρ - ik)/(ρ + ik)] = -e^{ika }then A_{2 }= A_{2}', B_{1 }= B_{3}', Φ(-x) = Φ(x), the solutions are even;

if [(ρ - ik)/(ρ + ik)] = +e^{ika }then A_{2 }= -A_{2}', B_{1}= -B_{3}', Φ(-x) = -Φ(x), the solutions are odd.For k

_{0}≤ π/a only one (even) solution exists.

If π/a ≤ k_{0}≤ 2π/a the first odd solution becomes possible.

If k_{0}is very large, then the slope of the straight line is very small and solutions appear near every k = nπ/a (n = integer). Consequently

E_{n}= n^{2}π^{2}ħ^{2}/(2ma^{2}) - U_{0}.

These are the energy levels of the**infinite square well**.

We find the eigenfunction of H by solving HΦ

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 > U
_{min}. 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}.)

- There exists an eigenfunction for every E > U
- 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 can solve for the bound states in a square-well potential using a graphical solution.

- There exists an eigenfunction for every E > E

**Confinement leads to energy quantization.**