Problem 1:

Consider the one-dimensional potential step defined by U(x) = 0 , x < 0,  U(x) = U0, x > 0.

Suppose a wave incident from the left has energy E = 4U0.  What is the probability that the wave will be reflected?

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.  We have a potential that has a value of 0 for x < 0 and a value of U0 (barrier) for  x > 0.  We are asked to find the the reflectance R.
• Details of the calculation:
We have E > U0.
The most general solutions in regions 1, and 2 are
Φ1(x) = A1exp(ik1x) + A1'exp(-ik1x),
Φ2(x) = A2exp(ik2x),
for particles incident from the left.
Here k12 = (2m/ħ2)E and k22 = (2m/ħ2)(E - U0).
Φ is continuous at x = 0. Therefore A1 + A1' = A2.
(∂/∂x)Φ(x) is continuous at x = 0.  Therefore ik1A1 - ik1A1' = ik2A2 .
We can then solve for A1'/A1.  We find A1'/A1 = (k1 - k2)/(k1 + k2).
The reflectance R  = |A1'/A1|2  = 1 - 4k1k2/(k1 + k2)2.
k12 = 8mU02,  k22 = 6mU02,  k1k2 = (48)1/2(mU02),
R = 1 - 4(48)1/2/(√8 + √6)2 = 0.005

Problem 2:

You are given a one dimensional potential barrier of height U which extends from x = 0 to x = a.  A particle of mass m and energy E < U is incident from the left.
(a)  Derive expressions for the transmittance T and reflectance R for the barrier.
(b)  Show that T + R =  1.

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.  We have a potential that has a value of U from x = 0 to x = a and is zero otherwise.  We are asked to find the transmittance T and reflectance R.
• Details of the calculation:
Divide space into 3 regions;
region 1: x < 0, region 2: 0 < x < a, region 3: x > a.
(a)  potential barrier, E < U.
The most general solutions in regions 1, 2, and 3 are
Φ1(x) = A1exp(ikx) + A1'exp(-ikx),
Φ2(x) = B2exp(ρx) + B2'exp(-ρx),
Φ3(x) = A3exp(ikx).
Here k2 = (2m/ħ2)E and ρ2 = (2m/ħ2)(U - E).
The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.
x = 0:  A1 + A1' = B2 + B2', ikA1 - ikA1' = ρB2 - ρB2'.
x = a:  B2exp(ρa) + B2'exp(-ρa) = A3exp(ika),
ρB2exp(ik2a) - ρB2'exp(-ρa) = ikiA3exp(ika).
We need to solve these equations for A3 and A1' in terms of A1.

(i)  Solve for B2 and B2' in terms of A3.
B2 = ½exp((ik - ρ)a)(1 + ik/ρ )A3 = C A3.
B2' = ½exp((ik + ρ)a)(1 - ik/ρ)A3 = C' A3.

(ii)  Now solve for A1 in terms of A3 to find T.
2A1 = B2 + B2' + (ρ/ik)(B2 - B2') = (C + C' + (ρ/ik) (C - C'))A3.
A1 = ([(k2 - ρ2)/(2ikρ)]sinhρa + coshρa)exp(ika)A3.
T = |A3/A1|2 = 4k2ρ2/[(k2 - ρ2)2sinh2ρa + 4k2ρ2cosh2ρa]
= 4k2ρ2/[(k2 + ρ2)2sinh2ρa + 4k2ρ2].
[cosh2x - sinh2x = 1.]
T = 4E(U - E)/[U2sinh2[(2m(U - E)/ħ2)1/2a] + 4E(U - E)].

(iii)  Now solve for A1' in terms of A1 to find R.
A1 + A1' = B2 + B2' = (C + C')A3.
A1 - A1' = (ρ/(ik))(B2 - B2') = (ρ/(ik))(C - C')A3.
A1(C + C' - (ρ/(ik))(C - C')) = A1'(C + C' - (ρ/(ik))(C - C')).
C + C' - (ρ/(ik))(C - C') = [[(k2 - ρ2))/(ikρ)]sinhρa + coshρa]]exp(ika).
C + C' - (ρ/(ik))(C - C') = [(k2 + ρ2))/(ikρ)]sinhρa exp(ika).
R = |A1'/A1|2 = (k2 + ρ2)2sinh2ρa/[(k2 + ρ2)2sinh2ρa + 4k2ρ2],  or
R = U2sinh2[(2m(U - E)/ħ2)1/2a] /[U2sinh2[(2m(U - E)/ħ2)1/2a] + 4E(U - E)].

(b)  T + R = 4k2ρ2/[(k2 + ρ2)2sinh2ρa + 4k2ρ2] + (k2+ ρ2)2sinh2ρa/[(k2 + ρ2)2sinh2ρa + 4k2ρ2] = 1.
Problem 3:

The one-dimensional square well shown in the figure rises to infinity at x = 0 and has range a and depth U1.

(a)  Derive the condition for a spinless particle of mass m to have two and only two bound states in the well.
(b)  Sketch the wave function of these two states inside and outside the well and give their analytic expressions.  These expressions may involve undetermined constants.

Solution:

• Concepts:
Square potentials
• Reasoning:
We want to solve for the bound states in a square well.
• Details of the calculation:
(a)  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 + U1), ρ2 = (2m/ħ2)(-E), and k02 = (2m/ħ2)U1.
Note: I am assuming that U1 is a positive number denoting the depth, the potential energy at the bottom of the well is -U1.
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) = B exp(-ρa)
kA cos(ka) = -ρB exp(-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.
For only two solution to exist we need the following condition for the slope 1/k0.
2a/(5π) < 1/k0 < 2a/(3π),  3π/(2a)  < k0 < 5π/2a,  3π/(2a)  < ((2m/ħ2)U1)1/2 < 5π/2a.

(b)  Inside the well Φ(x) = A sin(kx), outside the well for positive x > a, Φ(x) = Bexp(-ρx) = A sin(ka)exp(ρa)exp(-ρx), where k is a solution of |sin(ka)| = k/k0, cot(ka) < 0, and ρ is determined once k is determined.
Problem 4:

Consider the scattering of a particle of mass m and total energy  E = ħ2k2/(2m) under the influence of a localized one-dimensional potential.
(a)  Let the potential be a delta function potential well, U(x) = -aU0δ(x) with a > 0 and U0 = ħ2k02/(2m).  What are the asymptotic boundary conditions at x = ∞ and the matching conditions at x = 0 for the wave function?
(b)  Define the transmission coefficient T and the reflection coefficient R and find the relationship between T and R.
(c)  How does the transmission coefficient depend on E?
(d)  Now the potential is replaced by a double delta function potential well.  The delta functions are a distance b apart, i.e. U(x) = -aU0δ(x) - aU0δ(x - b).  By inspecting the matching conditions without solving the algebra equation, explain intuitively the limiting behavior of the transmission coefficient T for E --> 0 and E --> ∞.

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:
U(x) = 0 everywhere except at x = 0 (and x = b in part d).
• Details of the calculation:
(a) Φ is continuous at x = 0.  ∂Φ/∂x has a finite discontinuity at x = 0.
Assume the particle is incident from the left.
Φ1(x) = A1 exp(ikx) + A1'exp(-ikx) for x < 0.  k2 = 2mE/ħ2.
Φ2(x) = A2 exp(ikx)  for x > 0.
(b)  Φ is continuous at x = 0.  Φ1(0) = Φ2(0).  A1 + A1’ = A2.
∂Φ/∂x has a finite discontinuity at x = 0.
2Φ(x)/∂x2 + (2m(E - U)/ħ2)Φ(x) = 0.
Let us evaluate this equation at x = ε and at x = -ε and write down a difference equation.
∂Φε(x1 + ε)/∂x + ∂Φε(x1 - ε)/∂x = -(2m/ħ2)∫x1-εx1+ε (E + aU0δ(x)) Φ(x) dx
= -(2maU02)Φ(0).
iA1k – iA1’k = iA2k + A22maU02,  A1 – A1’ = A2 + A22maU0/(ikħ2)  = [1 + 2maU0/(ikħ2)]A2.
Eliminate A1’:
2A1 = [2 + 2maU0/(ikħ2) ]A2,  A2/A1 = 1/[1 + maU0/(ikħ2)] = ikħ2/(ikħ2 + maU0).
T(E) = (k|A2|2)/(k|A1|2) = ħ4k2/(ħ4k2 + m2a2U02) = E/(E + ma2U02/(2ħ2)).
Eliminate A2:
A1 – A1’ = A2 - A22maU0/(ikħ2)  = [1 - 2maU0/(ikħ2)](A1 + A1’).
A1'/A1 =  2maU0/(ikħ2)/ [2 - 2maU0/(ikħ2)] = -maU0/(ikħ2 - maU0).
R(E) = (k|A1'|2)/(k|A1|2) =  ma2U02/(ħ4k2 + m2a2U02).
T + R = 1
(c)  T(E) = E/(E + ma2U02/(2ħ2)).
As E --> 0, T --> 0,   R --> 1.
As E --> ∞, T --> 1,   R --> 0.
(d)  As E --> 0, nothing is transmitted past the first delta function.
As E --> ∞,everything is transmitted across both delta functions.