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.  A1’ = A2 = 0.  Φ is continuous at x = 0.  Φ1(0) = Φ2(0).  A1 = A2’.
∂Φ2/∂x|x=0 = ∂Φ1/∂x|x=0 + (2mC/ħ2)Φ(0).
-ρA2' - (2mC/ħ2)A2' = ρA2',  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.