### Square potentials, bound states

#### 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) = B1eρx,   Φ2(x) = A2eikx + A2'e-ikx,  Φ3(x) = B3'e-ρx,
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)½e-iθ/(ρ2 + k2)½e] = 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.

#### Problem:

The one-dimensional square well shown in the figure rises to infinity at x = 0 and has a range “a” and a depth U0.  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 k2 = (2m/ħ2)(E + U0), ρ2 = (2m/ħ2)(-E), and k02 = (2m/ħ2)U0.
Note: I am assuming that U0 is a positive number denoting the depth, the potential at the bottom of the well is -U0.
In region 1 we have Φ1(x) = A sin(kx), since  Φ1(0) = 0.
In region 2 we have Φ2(x) = Bexp(-ρx), since Φ2(∞) = 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/sin2(ka) = 1 + cot2(ka) = (k2 + ρ2)/k2  = k02/k2.
We can find a graphical solution by plotting |sin(ka)| and k/k0 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/(2k0a) > 1.

#### Problem:

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, U0 between x = a and x = 2a, and infinite otherwise.
(a)  Assume E = U0.  Find the smallest value of U0 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/∂x2)ψ(x) + (2m/ħ2)(E - U)ψ(x) = 0.
(i)  Let E > U:  (∂2/∂x2)ψ(x) + k2ψ(x) = 0.   E - U = ħ2k2/(2m).
The most general solution is ψ(x) = Aexp(ikx) + A'exp(-ikx), with A and A' complex constants.
(ii)  Let E < U:  (∂2/∂x2)ψ(x) - α2ψ(x) = 0.   U - E = ħ2α2/(2m).
The most general solution is ψ(x) = Bexp(αx) + B'exp(-αx), with B and B' complex constants.
(iii)  Let E = U: (∂2/∂x2)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 = U0 = ħ2k2/(2m).  Then, to satisfy the boundary conditions, we need
ψ1(x) = Asin(kx),  ψ2(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), U0 = ħ2k2/(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: