Fermi's golden rule

Problem:

Consider spinless non-relativistic free particles of mass M moving in a two dimensional square box.  Find an expression for the density of states ρ(E).
Remember ρ(E)dE is defined as the number of energy levels per unit volume between E and E + dE.

Solution:

• Concepts:
  The two-dimensional, infinite square well, the density of states
• Reasoning:
  We are asked to find the density of states for a two-dimensional, infinite square well
• Details of the calculation:
  The eigenfunctions and eigenvalues of the 2D infinite square well are
  Φ_nm(x,y) = (2/L)sin(k_xx)sin(k_yy),  with k_x = πn/L,  k_y = πm/L,  n, m = 1, 2, ...  .
  The associated eigenvalues are
  E_nm = (n² + m²)π²ħ²/(2ML²) = (k_x² + k_y²)ħ²/(2M) = k²ħ²/(2M)
  k_x = πn/L,  k_y = πm/L, Δk_x = Δk_y = π/L.

  
  Let N(k') be the number of states with |k| ≤ k'.
  The area of a circle of radius k' in 2D k-space is  A = πk^'2.
  The number of states in the circle is  N(k') = (A/4)/(π/L)² = k^'2L²/(4π).
  Only positive values of k_x and k_y are allowed.
  The density of states in k-space is  n(k') = dN(k')/dk' = k'L²/(2π).
  E = k²ħ²/(2M).  dE/dk = kħ²/M.
  n(E) = dN/dE = (dN/dk)(dk/dE) = ML²/(2πħ²).

Problem:

Consider a particle in the ground state of a one-dimensional square well of with a and depth U₀.
Assume that the well is very deep and
k/k₀ = (2m(E + U₀)/(2mU₀))^½ << 1
for the ground state, so that the ground state wave function is nearly identical to that of the infinite square well.  At t = 0, a time dependent perturbation W(t) = Wcosωt is turned on.
(a)  What is the minimum frequency necessary to free the particle from the well? 
(b)  For frequencies greater than this minimum frequency, use perturbation theory to find the transition rate.
You can assume that the free particles will be in a box of size L >> a.

Solution:

• Concepts:
  Fermi's golden rule, the infinite square well
• Reasoning:
  The initial wave function of the particle look very much like the wave function of a particle in an infinite square well.  The final state of the particle is a continuum state.
• Details of the calculation:
  (a)  Let H₀ be the Hamiltonian of the infinite square well.
  Eigenfunctions of H₀:  Φ_n(x) = (2/a)^½sin(nπx/a) with eigenvalues E_n = n²π²ħ²/(2ma²).
  The ground state energy is  E₁ = π²ħ²/(2ma²).
  In a very deep well the ground state energy is approximately equal to -V₀ + E₁ ~ -V₀.
  The minimum frequency required to free the particle from the well therefore is
  ω_min = V₀/ħ.
  
  (b)  To find the transition rate we use Fermi's golden rule.
  If W(t) = Wcos(ωt),  then the transition probability per unit time is given by
  w(i,E) = (π/(2ħ))ρ(E)|W_Ei|²δ_E-Ei,ħω, where W_Ei = <Φ_E|W|Φ_i>.
  Assume that the ejected particle is a free particle confined to a region of width L >> a.   Use periodic boundary conditions.  Then
  Φ_k(x) = L^-½exp(ikx), with k = 2πn/L,  n = 0, ±1, ±2, ... .   k² = 2mE/ħ² = 2m(ω - ω_min)/ħ.
  The number of states with wave vectors whose magnitudes lie between k and k + dk is
  dN = 2 dk/(2π/L)  = Ldk/π.  (The particle can move towards the left or to the right.)
  dN/dk = L/π.
  The density of states is dN/dE = (dN/dk)(dk/dE).
  With E = ħ²k²/(2m) we have ρ(E) = dN/dE = Lm/(πkħ²).
  W_Ei = (2/La)^½W∫₀^adx e^ikxsin(πx/a)
  = (2/La)^½W(e^ika+1)πa/(π² - k²a²)
  |W_Ei| = W²(2/La)cos²(ka/2)4π²a²/(π² - k²a²)².
  w(i,E) = (π/(2ħ))Lm/(πkħ²)W²(2/La)cos²(ka/2)4π²a²/(π² - k²a²)²
  = W²cos²(ka/2)4π²am/[kħ³(π² - k²a²)²].
  This is the transition probability per unit time.

Problem:

A hypothetical particle of mass 2m can decay into two different final states under the action of a perturbation H'.
(a)  2 m --> m + γ,
(b)  2 m --> 1.5 m + γ.
Here m and 1.5 m represent particles of mass m and 1.5 m respectively.  If the matrix elements of H' are equal, i.e.  <2 m|H'|m> = <2 m|H'|1.5 m>, what will be the relative likelihood of decay to m as opposed to 1.5 m?  You may assume the recoil fragment (m or 1.5 m) receives negligible kinetic energy. 

Solution:

• Concepts:
  Fermi's golden rule,
• Reasoning:
  The transition probability per unit time is
  w(i,βE) = (2π/ħ)ρ(β,E)|W_Ei|²δ_E-Ei,ħω;
  ρ(β,E)dE is the number of final states in the interval dE characterized by some discrete index b.
  Here ρ(β,E) is the density of photon states for photons of energy ħω = (m_initial - m_final)c².
• Details of the calculation:
  The number of photon states ρ(β,E)dΩdE, i.e.  the number of photon states in an energy interval between E and E + dE with wave vector k in the solid angle dΩ is proportional to d³k.
  ρ(E)dΩdE ∝ d³k ∝ p²dpdΩ.
  Therefore ρ(E) ∝ p²dp/dE.
  p = E/c,  ρ(E) ∝ E²/c³ ∝ ω²/c³.
  So if the matrix elements are equal, the likelihood of decay is proportional to (m_initial - m_final)².
  P_(a)/P_(b) = (m/0.5 m)² = 4.

Problem:

A uniform periodic electric field acts upon a hydrogen atom, which at t = 0 is in its ground state.  Determine the minimum frequency of the field necessary to ionize the atom and use perturbation theory to evaluate the probability for ionization per unit time.  For the sake of simplicity, assume the electron in the final state to be free.

Solution:

• Concepts:
  Time dependent perturbation theory, Fermi's golden rule
• Reasoning:
  Fermi's golden rule gives the probability per unit time for a transition from a state of the discrete spectrum to a state corresponding to an infinitesimal interval in the continuous spectrum.
• Details of the calculation:
  w(i,βE) = (π/2ħ)ρ(β,E)|W_Ei|²d_E-Ei,ħω for W(t) = Wcos(ωt).
  ρ(β,E)dE is the number of final states in the interval dE characterized by some discrete index β.
  Here W(t) = W_DE = er∙E = eE₀zcos(ωt).
  The minimum frequency necessary to ionize the hydrogen is ω₀ = E_I/ħ = μe⁴/(2ħ³).
  Assume the ejected electron is free.
  The eigenfunctions and eigenvalues of a free electron confined to a cubical 3D infinite square well with periodic boundary conditions are
  ψ_nml(x,y,z) = (1/L)^3/2exp(ik∙r)= (1/L)^3/2exp(ik_xx)exp(ik_yy)exp(ik_zz), 
  with k_x = 2πn/L,  k_y = 2πq/L,  k_z = 2πl/L, n, q, l = ±1, ±2, ... .
  The associated eigenvalues are
  E_nql = (n² + q² + l²)4π²ħ²/(2mL²) = (k_x² + k_y² + k_z²)ħ²/(2m) = ħ²k²/(2m).
  [H = p_x²/(2m) + p_y²/(2m) + p_z²/(2m), if x, y , z < L.
  ψ(x,y,z) = Φ(x)χ(y)γ(z).
  (-ħ²/(2m))(∂²/∂x²)Φ(x) = E_xΦ(x).
  Φ(x) = (1/L)^½exp(ik_xx),  Φ(x) = Φ(x + L) --> k_x = 2πn/L.
  E_x = ħ²k_x²/(2m) = 4π²n²ħ²/(2mL²) .]
  Δk_x = Δk_y = Δk_z = 2π/L.
    

  

  To each allowed k_nml there corresponds a wave function ψ_nml(x,y,z).
  The tips of the vectors k_nml divide k-space into elementary cubes of edge length 2π/L.  We have one vector per (2π/L)³ volume of k-space.
  The number of vectors in a volume d³k of k-space therefore is dN = d³k/(2π/L)³.
  d³k = k²dkdΩ, 
  dN = k²dkdΩ/(2π/L)³ = k²dkVdΩ/(2π)³ = number of state with a wave vector of magnitude between k and k + dk in a solid angle dΩ.
  dN/dk = k²dΩ/(2π/L)³ = k²VdΩ/(2π)³. (V = L³.)
  The density of states for the electron is dN/dE = (dN/dk)(dk/dE), with E = ħ²k²/(2m).
  dN/dE = ρ(β,E) = 2^½m^3/2VE^½dΩ/(2πħ)³.
  (We do not multiply by 2 since the interaction cannot flip the spin.)
  
  The ground state wave function of hydrogen is (πa₀²)^-½exp(-r/a₀).
  We therefore have W_E1 = (Vπa₀²)^-½∫exp(-ik∙r-r/a₀)(-er∙E)d³r.
  To evaluate this integral let k point into the z-direction and use spherical coordinates.
  r = r(r,θ,φ),  E = E(r₀,θ₀,φ₀).
  Then E∙r = E₀r[cosθcosθ₀ + sinθsinθ₀cos(φ-φ₀)] and
  W_E1 = -(Vπa₀²)^-½eE₀∫∫∫exp(-ik∙r-r/a₀)[cosθcosθ₀ + sinθsinθ₀cos(φ-φ₀)]r³drsinθdθdφ
  = 2πcosθ₀(Vπa₀²)^-½eE₀∫_-1¹dcosθ[∫₀^∞exp(-ikrcosθ - r/a₀)r³dr]cosθ.
  W_E1 = -2πcosθ₀(Vπa₀²)^-½eE₀∫_-1¹dx (3! x)/(a₀^-1 + ikx)⁴ = -i2πcosθ₀(Vπa₀²)^-½eE₀ 16ka₀⁵/(1 + k²a₀²)³.
  
  Notation:
  We have E_I = ħω₀ = μe⁴/(2ħ²), a₀ = ħ²/(me²), so we may write a₀² = ħ⁴/(m²e⁴) = ħ/(2mω₀).
  Energy conservation requires that E = ħω - ħω₀.
  Therefore k² = 2mE/ħ² = 2m(ω - ω₀)/ħ, k²a₀² = (ω - ω₀)/ω₀, 1 + k²a₀² = ω/ω₀.
  The probability that an electron is ejected into a solid angle dΩ with energy E per unit time therefore is
  w(1,dΩ E) = ( π/2ħ)2^½m^3/2VE^½dΩ/(2πħ))³|-i2πcosθ₀(Vπa₀²)^-½eE₀ 16ka₀⁵/(ω/ω₀)³|²
  = E₀²(cosθ₀)²(ω/ω₀)⁶[(ω - ω₀)/ω₀]^3/2a₀³dΩ 2⁸/(4πħ).

  
  To find the probability per unit time that an electron is ejected with energy E we must integrate over all possible angles between the direction of the field and the direction of k.
  ∫₀^2π∫_-1¹(cosθ₀)²dcosθ₀dφ₀ = 4π/3.
  w(1,dΩ E) = (256/3)(a₀³/ħ) E₀²(ω/ω₀)⁶[(ω - ω₀)/ω₀]^3/2.
  Near the threshold for ionization the probability increases from zero as (ω - ω₀)^3/2.  If ω >> ω₀ it decreases as ω^-9/2.  It has a maximum for ω = 4ω₀/3.

Problem:

Nuclei sometimes decay from excited states to the ground state by internal conversion, a process in which an atomic electron is emitted instead of a photon.  Let the initial and final nuclear states have wave functions Ψ_i(r₁,r₂,..,r_z) and Ψ_f(r₁,r₂,..,r_z), respectively, where r_i describes the protons.  The perturbation giving rise to the transition is the proton-electron interaction,

W = -∑_i=1^z e²/|r - r_i|,

where r is the electron coordinate.
(a)  Write down the matrix element for the process in lowest-order perturbation theory, assuming that the electron is initially in a state characterized by the quantum numbers nlm, and that its energy, after it has been emitted, is large enough so that its final state may be described by a plane wave.  Neglect spin.
(b)  Write down an expression for the internal conversion rate.
(c)  For light nuclei, the nuclear radius is much smaller than the Bohr radius for the given Z, and we can use the expansion

1/|r - r_i| ≈ 1/r + r∙r_i/r³.

Use this approximation to express the transition rate in terms of the dipole matrix element

d = <Ψ_f|∑_i=1^z r_i|Ψ_i>.

Solution:

• Concepts:
  Fermi's golden rule, a constant perturbation
• Reasoning:
  We find the internal conversion rate using Fermi's golden rule,
  w(0-->1) = (2π/ħ)ρ(E)|W₁₀|²δ_E1,E0,
  because the final state of the electron is a continuum state.
• Details of the calculation:
  (a)  The initial state is Φ₀ = ψ_lmn(r)Ψ_i(r₁,r₂,..,r_z),
  the final state is Φ₁ = V^-½ exp(ik∙r) Ψ_f(r₁,r₂,..,r_z).
  The final state of the electron is a plane wave.  We confine the plane wave to a volume V = L³ and use periodic boundary conditions.  The matrix element needed to calculate the internal conversion rate is 
  <Φ₁|W|Φ₀> = V^-½ ∫d³r exp(ik∙r)<Ψ_f(r_i)|∑_ie²/|r-r_i||Ψ_i(r_i)>ψ_lmn(r).
  The energy eigenstates of an electron confined to a cubical box with periodic boundary conditions are
  Φ_nx,ny,nz(x,y,z) = L^-3exp(i2π(n_xx+n_yy+n_zz)/L),
  with n_x, n_y, n_z = 0, ±1, ±2, ...  .
  We have k_x = 2πn_x/L,  k_y = 2πn_y/L,  k_z = 2πn_z/L.
  If k is large, then the number of states with wave vectors whose magnitudes lie between k and k + dk is
  dN = 4πk²dk/(2π/L)³  = 4πVk²dk/(2π)³.
  dN/dk = 4πVk²/(2π)³.
  The density of states is dN/dE = (dN/dk)(dk/dE).
  With E = ħ²k²/(2m) we have dN/dE = 2mVk/(4π²ħ²) = (2m)^3/2VE^½/(4π²ħ³).
  We are instructed to neglect spin.  If we do not neglect spin we multiply the density by 2.
  We therefore have for the internal conversion rate
  w(0-->1) = (½πħ)(2m/ħ²)^3/2E^½|∫d³r exp(ik∙r)<Ψ_f|∑_ie²/|r-r_i||Ψ_i>ψ_lmn(r)|²δ(E₁ - E₀),
  (c)  We use 1/|r - r_i| ≈ 1/r + r∙r_i/r³.
  <Ψ_f|∑_ie²/|r-r_i||Ψ_i> = (e²/r³)r∙<Ψ_f|∑r_i||Ψ_i> = (e²/r³)r∙d, since Ψ_i and Ψ_f depend only on the nuclear coordinates r_i and not on the electron coordinate r.
  d is independent of r.  We can therefore write
  w(0-->1) = (e⁴/2πħ)(2m/ħ²)^3/2E^½|d∙∫d³r exp(ik∙r) (r/r³) ψ_lmn(r)|²δ_E1,E0,