Continuity for Double Delta Function Potential

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.
  ∂Φ_ε(x₁ + ε)/∂x - ∂Φ_ε(x₁ - ε)/∂x = (2m/ħ²)∫_x1-ε ^x1+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_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.

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 = ħ²k₀ 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) = -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.
  ∂Φ_ε(x₁ + ε)/∂x + ∂Φ_ε(x₁ - ε)/∂x = (2m/ħ²)∫_x1-ε ^x1+ε (-aU₀δ(x) - E) Φ(x) dx
  = (2maU₀/ħ²)Φ(0).
  iA₁k - iA₁'k = iA₂k - A₂2maU₀/ħ²,  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 + m²a²U₀ ²) = E/(E + ma²U₀ ²/(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) =  ma²U₀ ²/(ħ⁴k² + m²a²U₀ ²).
  T + R = 1
  (c)  T(E) = E/(E + ma²U₀ ²/(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.

  Additional information:
  Because U(x) = U(x + a) we also have Φ(x) = exp(ikx)u(x).  Why?
  Let T_a describe an operation called a translation.   Every part of the system is displaced by the amount a by this operation.   Let U(T_a) be the operator that maps the wave function before the translation onto the wave function after the translation.
  ψ'(x) = U(T_a)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(T_Δx) = (1 - Δx(i/ħ)p).  (p denotes the momentum operator.)
  U(T_{x + Δx}) = U(T_x)U(T_Δx) = U(T_Δx)U(T_x) = (1 - Δx(i/ħ)p)U(T_x).
  U(T_{x + Δx}) - U(T_x) = -Δx(i/ħ)p)U(T_x).  dU(T_x)/dx = -(i/ħ)p)U(T_x),  U(T_x) = exp(-ikx), with p = ħk.
  Therefore U(T_a) = exp(-ika).
  Note:  U(T_a) 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(T_a) are of the form Φ(x) = exp(ikx)u(x) with u(x + a) = u(x).
  U(T_a)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(T_a) commutes with H, and therefore U(T_a) 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).

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Source: http://electron6.phys.utk.edu/PhysicsProblems/QM/2-one-dimensional%20eigenvalue/delta.html

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