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_2ε(x₁ + ε)/∂x - ∂Φ_1ε(x₁ - ε)/∂x = -(2m/ħ²)∫_x1-ε^x1+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.
  m²C²/ħ⁴ = -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.