### Delta function potentials

#### Problem:

A one-dimensional potential well is given in the form of a delta function at x = 0, U(x) = Cδ(x), C < 0.  A stream of non-relativistic particles of mass m and energy E approaches the origin from one side.
(a)  Derive an expression for the reflectance R(E).
(b)  Can you express R(E) in terms of sin2(δ), where δ is the phase shift of the transmitted wave?

Solution:

• Concepts:
This is a "square potential" problem.  We solve HΦ(x) = EΦ(x) in regions where V(x) is constant and apply boundary conditions.
• Reasoning:
U(x) = 0 everywhere except at x = 0.
• Details of the calculation:
(a)  Φ1(x) = A1 exp(ikx) + A1'exp(-ikx) for x < 0.  k2 = 2mE/ħ2.
Φ2(x) = A2 exp(ikx)  for x > 0.
Φ is continuous at x = 0.  Φ1(0) = Φ2(0).  A1 + A1' = A2.
2Φ(x)/∂x2 + (2m(E - U(x))/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 (Cδ(x) - E) Φ(x) dx
= (2mC/ħ2)Φ(0).
If U does not remain finite at the step, then ∂Φ/∂x has a finite discontinuity at the step.
iA1k - iA1'k = iA2k - A22mC/ħ2,  A1 - A1' = A2 - A22mC/(ikħ2)  = [1 - 2mC/(ikħ2)]A2.
Eliminate A1':
2A1 = [2 - 2mC/(ikħ2)]A2,  A1 = [1 + imC/(kħ2)]A2
T(E) = (k|A2|2)/(k|A1|2) = ħ4k2/(ħ4k2 + m2C2) = E/(E + mC2/(2ħ2)).
R(E) = 1 - T(E) = [mC2/(2ħ2)]/(E + mC2/(2ħ2)) = m2C2/(ħ4k2 + m2C2).

(b)  A2 = A1/ [1 + imC/(kħ2)] =  A1(1 - imC/(kħ2))/[1 + m2C2/(k2ħ4)]  = |A1|exp(iδ).
tanδ = -mC/(kħ2).
sinδ = [-mC/(kħ2)]/[1 + m2C2/(k2ħ4)]1/2.
sin2δ = [m2C2/(k2ħ4)]/[1 + m2C2/(k2ħ4)] = R(E).

#### Problem:

A one-dimensional potential well is given in the form of a delta function at x = 0,
U(x) = Cδ(x), C < 0.
(a)  A non-relativistic particle of mass m and energy E is incident from one side of the well.
Derive an expression for the coefficient of transmission T(E).
(b)  Since a bound state can exist with the attractive potential, find the binding energy of the ground state of the system.

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.
• Details of the calculation:
(a)  E > 0.
Φ1(x) = A1 exp(ikx) + A1'exp(-ikx) for x < 0.  k2 = 2mE/ħ2.
Φ2(x) = A2 exp(ikx)  for x > 0.
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.
∂Φ(x1 + ε)/∂x - ∂Φ(x1 - ε)/∂x = -(2m/ħ2)∫x1-εx1+ε (E - Cδ(x)) Φ(x) dx
= (2mC/ħ2)Φ(0).
iA1k - iA1'k = iA2k - A22mC/ħ2,  A1 - A1' = A2 - A22mC/(ikħ2)  = [1 - 2mC/(ikħ2)]A2.
Eliminate A1':
2A1 = [2 - 2mC/(ikħ2) ]A2,  A2/A1 = 1/[1 - mC/(ikħ2)] = ikħ2/(ikħ2 - mC).
T(E) = (k|A2|2)/(k|A1|2) = ħ4k2/(ħ4k2 + m2C2) = E/(E + mC2/(2ħ2)).

(b)  E < 0.
Φ1(x) = A1 exp(ρx) + A1'exp(-ρx) for x < 0.  ρ2 = -2mE/ħ2.
Φ2(x) = A2 exp(ρx) + A2'exp(-ρx) for x > 0.
Φ is finite at infinity.  If we choose ρ > 0, then A1' = A2 = 0.  Φ is continuous at x = 0.  Φ1(0) = Φ2(0).  A1 = A2' = A.
∂Φ2/∂x|x=ε = ∂Φ1/∂x|x=-ε + (2mC/ħ2)Φ(0), as ε --> 0
-ρA - (2mC/ħ2)A = ρA,  ρ = -mC/ħ2.
m2C24 = -2mE/ħ2,  E = -mC2/(2ħ2).
Only one bound state exists.

#### Problem:

Consider the non-relativistic motion in one dimension of a particle outside an infinite barrier at x ≤ 0 with an additional delta function potential at x = a, i.e. U(x) = ∞ for  x ≤ 0,   U(x) = Fδ(x - a) for x > 0, where F is a positive constant.  Derive an analytical expression for the phase shift δ(k) for a particle approaching the origin from x = +∞ with momentum ħk.

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 for x > 0 except at x = a.  U(x) = ∞ for x ≤ 0.
• Details of the calculation:
The most general solution of the" time-independent" Schroedinger equation in region 1 is Φ1(x) = A sin(kx) because Φ1(x) = 0 due to the boundary condition at x = 0.
The most general solution in region 2 is Φ2(x) = B sin(kx + δ(k)).  The boundary conditions at x = a are Φ1(a) = Φ2(a),  ∂Φ1/∂x|a = ∂Φ2/∂x|a - (2mF/ħ2)Φ(a).
A sin(ka) = B sin(ka + δ(k)),  kA cos(ka) = kB cos(ka + δ(k)) - (2mF/ħ2)A sin(ka).
cot(ka) + (2mF/ħ2) = cot(ka + δ(k)),
δ(k) = cot-1(cot(ka) + (2mF/ħ2)) - ka.

#### Problem:

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(ε)/∂x + ∂Φε(-ε)/∂x = -(2m/ħ2)∫(-aU0δ(x) - E) 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.

#### Problem:

Consider an electron trapped in a one-dimensional periodic potential with period a.
The electron's potential energy is given by
U(x) = ∑-∞+∞Aδ(x - na).
where A is a constant.

The eigenfunctions of the Hamiltonian have the form Φ(x) = exp(ikx)u(x), where u(x) is a periodic function with period a, u(x + a) = u(x) (Bloch's theorem).  Find the allowed energy eigenvalues of the electron.

Solution:

• Concepts:
Piecewise constant potentials
• Reasoning:
The potential energy is zero except at the location of the delta functions.
• Details of the calculation:
In the region from 0 to a, the most general solution of the time-independent Schroedinger equation is Φ(x) = B'exp(iαx) + B exp(-iαx), with α = (2mE/ħ2)½.
Because U(x) = U(x + a) we also have Φ(x) = exp(ikx)u(x).
Therefore u(x) = B'exp(i(α - k)x) + B exp(-i(α + k)x).
At x = a, Φ(x) is continuous, but (∂/∂x)Φ(x)|a+ε - (∂/∂x)Φ(x)|a-ε = (2mA/ħ2) Φ(a).
When applying the boundary conditions at x = a use Φ(a) = exp(ikx)u(a) for the x = a + ε side and the most general solution for the x = a - ε side of the boundary.
Φ(a + ε)ε-->0 = exp(ika)u(0) = exp(ika)(B' + B).
Φ(a - ε)ε-->0 = B'exp(iαa) + Bexp(-iαa).
Therefore B'(exp(iαa) - exp(ika)) = B(exp(ika) - exp(-iαa)).
As ε-->0 we have
(∂/∂x)Φ(x)|a+ε = (∂/∂x) exp(ikx)u(x)|a+ε = ik exp(ika)u(0) + exp(ika)du/dx|0
= ik exp(ika)(B' + B) + exp(ika)(i(α-k)B' - i(α+k)B),
(∂/∂x)Φ(x)|a-ε = iαB'exp(iαa) - iαBexp(-iαa).
Therefore
ik exp(ika)(B' + B) + exp(ika)(i(α - k)B' - i(α + k)B) - iαB'exp(iαa) + iαBexp(-iαa)
= (2mA/h2) Φ(a), or
iαB'(exp(ika) - exp(iαa)) -iαB(exp(ika) - exp(-iαa) = (2mA/ħ2) Φ(a).
Insert from above and eliminate B.
2iαB'(exp(ika) - exp(iαa)) = (2mA/ħ2)exp(ika)(B' + B)
= Cexp(ika)[B' + B' (exp(iαa) - exp(ika))/(exp(ika) - exp(-iαa))].
Here C = 2mA/ħ2.
Recast the expression in terms of real functions.
2iαexp(ika)(exp(ika) - exp(-iαa)) - 2iαexp(iαa)(exp(ika) - exp(-iαa))
= Cexp(ika) [(exp(ika) - exp(-iαa) + exp(iαa) - exp(ika)].
2iα(exp(ika) - exp(-iαa)) - exp(iαa) + exp(-ika)) = iC sin(αa).
iα(cos(ka) - cos(αa)) = C [(exp(iαa) - exp(-iαa).
(C/(2α))sin(αa) + cos(αa) = cos(ka).
Let αa = x.  Then (Ca/(2x))sinx + cosx = cos(ka).
All energies for which -1 < [(mAa/(ħ2x))sinx + cosx] < 1 are allowed.
Plot (Ca/(2x))sinx + cosx versus x.  For example, let Ca/(2) = 5.

We have a set of forbidden and allowed bands.

Because U(x) = U(x + a) we also have Φ(x) = exp(ikx)u(x).  Why?
Let Ta describe an operation called a translation.   Every part of the system is displaced by the amount a by this operation.   Let U(Ta) be the operator that maps the wave function before the translation onto the wave function after the translation.
ψ'(x) = U(Ta)ψ(x) = ψ(x - a).
For an infinitesimal translation (Δx --> 0)  ψ(x - Δx) = ψ(x) - Δx(dψ(x)/dx) = (1 - Δx(i/ħ)p)ψ(x)
Therefore U(TΔx) = (1 - Δx(i/ħ)p).  (p denotes the momentum operator.)
U(Tx + Δx) = U(Tx)U(TΔx) = U(TΔx)U(Tx) = (1 - Δx(i/ħ)p)U(Tx).
U(Tx + Δx) - U(Tx) = -Δx(i/ħ)p)U(Tx).  dU(Tx)/dx = -(i/ħ)p)U(Tx),  U(Tx) = exp(-ikx), with p = ħk.
Therefore U(Ta) = exp(-ika).
Note:  U(Ta) is a unitary operator, it is not a Hermitian operator.
It has complex eigenvalues of magnitude 1.
For a unitary operator there also exist a basis of orthonormal eigenfunctions.
The eigenfunctions of U(Ta) are of the form Φ(x) = exp(ikx)u(x) with u(x + a) = u(x).
U(Ta)Φ(x) = Φ(x - a) = exp(ikx)exp(-ika)u(x - a) = exp(-ika)Φ(x) .
The eigenvalue is exp(-ika).
For the given potential U(Ta) commutes with H, and therefore U(Ta) and H have common eigenfunctions.
The eigenfunctions of H are therefore also of the form Φ(x) = exp(ikx)u(x) with u(x + a) = u(x).