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

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})^{½}e^{-iθ}/(ρ^{2}+ k^{2})^{½}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**.

The one-dimensional square well shown in the figure rises to infinity at x =
0 and has a range “a” and a depth U_{0}.
Find the bound state solutions.

Solution:

- Concepts:

Square potentials - Reasoning:

We want to solve for the bound states in a square well. - Details of the calculation:

Let region 1 extend from x = 0 to x = a and region 2 from x = a to infinity.

For bound states we have E < 0.

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

Note: I am assuming that U_{0}is a positive number denoting the depth, the potential at the bottom of the well is -U_{0}.

In region 1 we have Φ_{1}(x) = A sin(kx), since Φ_{1}(0) = 0.

In region 2 we have Φ_{2}(x) = Bexp(-ρx), since Φ_{2}(∞) = 0.

At x = a we need hat Φ(a) and (∂/∂x)Φ(x)|_{a}are continuous.

A sin(ka) = Bexp(-ρa).

kA cos(ka) = -ρBexp(-ρa).

Therefore cot(ka) = -ρ/k.

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

We can find a graphical solution by plotting |sin(ka)| and k/k_{0}versus k. The intersections of the two plots in regions where cot(ka) < 0 gives the values of k for which a solution exist.

It is possible to have no solutions, when π/(2k_{0}a) > 1.

Consider a particle with mass m in a potential well in one
dimension as shown. The potential energy U is zero between x = 0 and x =
a, U_{0}
between x = a and x = 2a, and infinite otherwise.

(a) Assume E = U_{0}. Find the smallest value of U_{0} for which that state exists.

(b) Sketch the wave function and the probability density
in the whole well for this state.

Solution:

- Concepts:

This is a "square potential" problem. We solve Hψ(x) = Eψ(x) in regions where U(x) is constant and apply boundary conditions. - Reasoning:

Assume U(x) = U = constant in certain regions of space.

In such a region the Schroedinger equation yields (∂^{2}/∂x^{2})ψ(x) + (2m/ħ^{2})(E - U)ψ(x) = 0.

(i) Let E > U: (∂^{2}/∂x^{2})ψ(x) + k^{2}ψ(x) = 0. E - U = ħ^{2}k^{2}/(2m).

The most general solution is ψ(x) = Aexp(ikx) + A'exp(-ikx), with A and A' complex constants.

(ii) Let E < U: (∂^{2}/∂x^{2})ψ(x) - α^{2}ψ(x) = 0. U - E = ħ^{2}α^{2}/(2m).

The most general solution is ψ(x) = Bexp(αx) + B'exp(-αx), with B and B' complex constants.

(iii) Let E = U: (∂^{2}/∂x^{2})ψ(x) = 0. ψ(x) = Cx + C', with C and C' complex constants.

(Note: A solution exists in the classically forbidden region.)

At a finite step the boundary conditions are that ψ(x) and (∂/∂x)ψ(x) are continuous. - Details of the calculation:

(a) Let 0 < x < a define region 1, a < x < 2a define region 2, and let E = U_{0}= ħ^{2}k^{2}/(2m). Then, to satisfy the boundary conditions, we need

ψ_{1}(x) = Asin(kx), ψ_{2}(x) = C(x - 2a), A sin(ka) = -Ca, kAcos(ka) = C, tan(ka) = -ka.

We have π/2 < ka < π for the smallest non-zero value of ka, ka = (π/2)(1 + ε).

Use a calculator to find ε ~ 0.3.

k = (π/(2a))(1 + ε), U_{0}= ħ^{2}k^{2}/(2m).(b) In region 1 ψ(x) = A sin(kx). The wavelength λ = 2π/k = 4a/(1 + ε) is somewhat shorter than 4a. In region 2 the wave function decreases linearly to zero.

Wave function ψ(x): Probability density: