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

Suppose a wave incident from the left has energy E = 4U_{0}.
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 U_{0}(barrier) for x > 0. We are asked to find the the reflectance R. - Details of the calculation:

We have E > U_{0}.

The most general solutions in regions 1, and 2 are

Φ_{1}(x) = A_{1}exp(ik_{1}x) + A_{1}'exp(-ik_{1}x),

Φ_{2}(x) = A_{2}exp(ik_{2}x),

for particles incident from the left.

Here k_{1}^{2}= (2m/ħ^{2})E and k_{2}^{2}= (2m/ħ^{2})(E - U_{0}).

Φ is continuous at x = 0. Therefore A_{1 }+ A_{1}' = A_{2}.

(∂/∂x)Φ(x) is continuous at x = 0. Therefore ik_{1}A_{1 }- ik_{1}A_{1}' = ik_{2}A_{2}.

We can then solve for A_{1}'/A_{1}. We find A_{1}'/A_{1}= (k_{1}- k_{2})/(k_{1}+ k_{2}).

The**reflectance**R = |A_{1}'/A_{1}|^{2}_{ }= 1 - 4k_{1}k_{2}/(k_{1}+ k_{2})^{2}.

k_{1}^{2}= 8mU_{0}/ħ^{2}, k_{2}^{2}= 6mU_{0}/ħ^{2}, k_{1}k_{2}= (48)^{1/2}(mU_{0}/ħ^{2}),

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

Φ_{1}(x) = A_{1}exp(ikx) + A_{1}'exp(-ikx),

Φ_{2}(x) = B_{2}exp(ρx) + B_{2}'exp(-ρx),

Φ_{3}(x) = A_{3}exp(ikx).

Here k^{2}= (2m/ħ^{2})E and ρ^{2}= (2m/ħ^{2})(U - E).

The boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous at x = 0 and x = a.

x = 0: A_{1}+ A_{1}' = B_{2 }+ B_{2}', ikA_{1 }- ikA_{1}' = ρB_{2 }- ρB_{2}'.

x = a: B_{2}exp(ρa) + B_{2}'exp(-ρa) = A_{3}exp(ika),

ρB_{2}exp(ik_{2}a) - ρB_{2}'exp(-ρa) = ik_{i}A_{3}exp(ika).

We need to solve these equations for A_{3}and A_{1}' in terms of A_{1}.

(i) Solve for B_{2}and B_{2}' in terms of A_{3}.

B_{2}= ½exp((ik - ρ)a)(1 + ik/ρ )A_{3}= C A_{3}.

B_{2}' = ½exp((ik + ρ)a)(1 - ik/ρ)A_{3}= C' A_{3}.

(ii) Now solve for A_{1}in terms of A_{3}to find T.

2A_{1}= B_{2}+ B_{2}' + (ρ/ik)(B_{2 }- B_{2}') = (C + C' + (ρ/ik) (C - C'))A_{3}.

A_{1}= ([(k^{2}- ρ^{2})/(2ikρ)]sinhρa + coshρa)exp(ika)A_{3}.

T = |A_{3}/A_{1}|^{2}= 4k^{2}ρ^{2}/[(k^{2}- ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}cosh^{2}ρa]

= 4k^{2}ρ^{2}/[(k^{2}+ ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}].

[cosh^{2}x - sinh^{2}x = 1.]

T = 4E(U - E)/[U^{2}sinh^{2}[(2m(U - E)/ħ^{2})^{1/2}a] + 4E(U - E)].

(iii) Now solve for A_{1}^{'}in terms of A_{1 }to find R.

A_{1}+ A_{1}' = B_{2 }+ B_{2}' = (C + C')A_{3}.

A_{1}- A_{1}' = (ρ/(ik))(B_{2 }- B_{2}') = (ρ/(ik))(C - C')A_{3}.

A_{1}(C + C' - (ρ/(ik))(C - C')) = A_{1}'(C + C' - (ρ/(ik))(C - C')).

C + C' - (ρ/(ik))(C - C') = [[(k^{2}- ρ^{2}))/(ikρ)]sinhρa + coshρa]]exp(ika).

C + C' - (ρ/(ik))(C - C') = [(k^{2}+ ρ^{2}))/(ikρ)]sinhρa exp(ika).

R = |A_{1}'/A_{1}|^{2}= (k^{2}+ ρ^{2})^{2}sinh^{2}ρa/[(k^{2}+ ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}], or

R = U^{2}sinh^{2}[(2m(U - E)/ħ^{2})^{1/2}a] /[U^{2}sinh^{2}[(2m(U - E)/ħ^{2})^{1/2}a] + 4E(U - E)].

(b) T + R = 4k^{2}ρ^{2}/[(k^{2}+ ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}] + (k^{2}+ ρ^{2})^{2}sinh^{2}ρa/[(k^{2}+ ρ^{2})^{2}sinh^{2}ρa + 4k^{2}ρ^{2}] = 1.

The one-dimensional square well
shown in the figure rises to infinity at x = 0 and has range a and depth U_{1}.

(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 k^{2}= (2m/ħ^{2})(E + U_{1}), ρ^{2}= (2m/ħ^{2})(-E), and k_{0}^{2}= (2m/ħ^{2})U_{1}.

Note: I am assuming that U_{1}is a positive number denoting the depth, the potential energy at the bottom of the well is -U_{1}.

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) = B exp(-ρa)

kA cos(ka) = -ρB exp(-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.

For only two solution to exist we need the following condition for the slope 1/k_{0}.

2a/(5π) < 1/k_{0}< 2a/(3π), 3π/(2a) < k_{0}< 5π/2a, 3π/(2a) < ((2m/ħ^{2})U_{1})^{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/k_{0}, cot(ka) < 0, and ρ is determined once k is determined.

Consider the scattering of a particle of mass m and total
energy E = ħ^{2}k^{2}/(2m) under the influence of a
localized one-dimensional potential.

(a) Let the potential be a delta function potential well, U(x) = -aU_{0}δ(x)
with a > 0 and U_{0 }= ħ^{2}k_{0}^{2}/(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) = -aU_{0}δ(x) - aU_{0}δ(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.

Φ_{1}(x) = A_{1}exp(ikx) + A_{1}'exp(-ikx) for x < 0. k^{2}= 2mE/ħ^{2}.

Φ_{2}(x) = A_{2}exp(ikx) for x > 0.

(b) Φ is continuous at x = 0. Φ_{1}(0) = Φ_{2}(0). A_{1 }+ A_{1}’ = A_{2}.

∂Φ/∂x has a finite discontinuity at x = 0.

∂^{2}Φ(x)/∂x^{2}+ (2m(E - U)/ħ^{2})Φ(x) = 0.

Let us evaluate this equation at x = ε and at x = -ε and write down a difference equation.

∂Φ_{ε}(x_{1 }+ ε)/∂x + ∂Φ_{ε}(x_{1 }- ε)/∂x = -(2m/ħ^{2})∫_{x1-ε}^{x1+ε }(E + aU_{0}δ(x)) Φ(x) dx

= -(2maU_{0}/ħ^{2})Φ(0).

iA_{1}k – iA_{1}’k = iA_{2}k + A_{2}2maU_{0}/ħ^{2}, A_{1}– A_{1}’ = A_{2}+ A_{2}2maU_{0}/(ikħ^{2}) = [1 + 2maU_{0}/(ikħ^{2})]A_{2}.

Eliminate A_{1}’:

2A_{1}= [2 + 2maU_{0}/(ikħ^{2}) ]A_{2}, A_{2}/A_{1}= 1/[1 + maU_{0}/(ikħ^{2})] = ikħ^{2}/(ikħ^{2}+ maU_{0}).

T(E) = (k|A_{2}|^{2})/(k|A_{1}|^{2}) = ħ^{4}k^{2}/(ħ^{4}k^{2}+ m^{2}a^{2}U_{0}^{2}) = E/(E + ma^{2}U_{0}^{2}/(2ħ^{2})).

Eliminate A_{2}:

A_{1}– A_{1}’ = A_{2}- A_{2}2maU_{0}/(ikħ^{2}) = [1 - 2maU_{0}/(ikħ^{2})](A_{1}+ A_{1}’).

A_{1}'/A_{1}= 2maU_{0}/(ikħ^{2})/ [2 - 2maU_{0}/(ikħ^{2})] = -maU_{0}/(ikħ^{2}- maU_{0}).

R(E) = (k|A_{1}'|^{2})/(k|A_{1}|^{2}) = ma^{2}U_{0}^{2}/(ħ^{4}k^{2}+ m^{2}a^{2}U_{0}^{2}).

T + R = 1

(c) T(E) = E/(E + ma^{2}U_{0}^{2}/(2ħ^{2})).

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.