Assignment 6

Problem 1:

Consider a spin ½ particle whose Hamiltonian is H₀ = (A/ħ²)S² + (B/ħ)S_z.  For t < 0 the particle is in the |+> eigenstate of H₀.
At t = 0 the Hamiltonian changes to H = H₀ + (C/ħ)S_y.  Let A, B, C be positive constants and let C << B.
Use time-dependent perturbation theory to find the probability that the particle is found in the |-> eigenstate of H₀ states for times t > 0.

Solution:

• Concepts:
  Time-dependent perturbation theory
• Reasoning:
  A small perturbation is introduced and we are asked to calculate the probability or a transition between two eigenstates of the unperturbed Hamiltonian.
• Details of the calculation:
  The eigenstates of H₀ are denoted by |+>  and |->, the eigenstates of S² and S^z.
  Time dependent perturbation theory yields
  P_if(t) = (1/ħ²)|∫₀^texp(iω_fit')W_fi(t')dt'|², when H = H₀ + W(t),
  with  ω_fi = (E_f - E_i)/ħ and W_fi(t) = <Φ_f|W(t)|Φ_i>.
  Here W(t) = -(C/ħ)S_y, |Φ_i> = |+>, |Φ_f> = |->
  E_i = 3A/4 + B/2,  E_f = 3A/4 - B/2, 
  W_fi = (C/ħ)<-|S_y|+>, independent of t.
  In the {|+>, |->} basis, the matrix of S_y is
  .
  <-|S_y|+> = iħ/2.
  W_fi(t') = i(C/2).
  ω_fi = -B/ħ = -B'.
  P_if(t) = (C²/4)|∫₀^texp(-iB't')dt'|².
  = (C²/(4ħ²))|i(exp(-iB't - 1)/B'|²
  = (C²/(4ħ²)) sin²(B't/2)/(B'/2)² = (C²/B²)sin²(Bt/(2ħ)).

Problem 2:

A particle of mass m is in the ground state of an infinite square well (U = 0 for 0 < x < a and U = ∞ otherwise).
At time t = 0, the right "wall" (i.e. at x = a) shifts to x = 3a.  A measurement of the energy of the particle is made just after the wall shifts. (Assume that all this happens so quickly so the spatial wave function of the particle does not change).  What is the probability that the energy measurement yields a value EXACTLY the same as the energy of the ground state of original well?

Solution:

• Concepts:
  A particle in an infinite square well, the sudden approximation
• Reasoning:
  The reaction time is so short that the transition amplitude <β|U(t₂,t₁)|a> is simply given by the overlap <β|α>.  The transition probability is |<β|α>|².  Here |α> is the eigenstate of the Hamiltonian before the transition and |β> is the eigenstate of the Hamiltonian after the transition.
• Details of the calculation:
  For t < 0, the particle is in the n = 1 ground state.
  ψ_1i(x) = (2/a)^½sin(πx/a),  E_1i = π²ħ²/(2ma²).
  For t > 0, the wave function of the n = 3 excited state is ψ_2f(x) = (2/(3a))^½sin(πx/a).
  E_nf = n²π²ħ²/(18ma²).   E_3f = E_1i.  If we find the particle in state ψ_3f, its energy has not changed.
  The wave function just after the wall is removed is ψ_1i(x).
  The probability of finding the particle in the state ψ_3f(x) of the new well is |<ψ_3f|ψ_1i |².
  |<ψ_3f|ψ_1i>|² = (4/(3a²))| ∫₀^asin²(πx/a)dx|² = 1/3.

Problem 3:

In one dimension, consider a spinless particle trapped in a delta-function potential U(x) = -Cδ(x), C > 0.
At t = 0, a time dependent perturbation W(t) = Wcosωt is turned on.
Assume ω >> mC²/(2ħ³) so that the particle can be ejected from the trap.  Use perturbation theory to find the transition rate.
You can assume that the free particles will be in a box of size L,  L = very large.

Solution:

• Concepts:
  Fermi's golden rule, the delta-function potential
• Reasoning:
  Only one bound state exist in the delta-function potential.  This is the initial state.  The final state of the particle is a continuum state.
• Details of the calculation:
  (a)  Let H₀ = p²/(2m) - Cδ(x).
  Normalized eigenfunction of H₀:  Φ(x) = ρ^½ exp(-ρ|x|), 
  ρ² = -2mE_i/ħ²,  E_i = -mC²/(2ħ²),  ρ = mC/ħ².
  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>.
  Since ħω >> -mC²/(2ħ²), we can assume that the ejected particle is a nearly free particle with energy E = ħω + E_i and k² = 2mE/ħ².
  We assume that the particle is confined to a region of width L >> a and use periodic boundary conditions. 
  Then Φ_k(x) = L^-½exp(ikx), with k = 2πn/L,  n = 0, ±1, ±2, ...  is the final state wave function.
  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 = (ρ/L)^½W[∫₀^∞dx e^ikxe^-ρx  + ∫_-∞⁰dx e^ikxe^ρx]
  = -(ρ/L)^½W 2ρ/(ρ² + k²).
  
  w(i,E) = (π/(2ħ))(Lm/(πkħ²))(ρ/L)W² 4ρ²/(ρ² + k²)²
  = (2mW²/ħ³) ρ³/[k(ρ² + k²)²].
  We can rewrite this in terms of the given quantities W, C and ω.
  (ρ² + k²)² = 4m²ω²/ħ², ρ³/[(ρ² + k²)²] = mC³/(4ħ⁴ω²). 
  w(i,E) = (1/k)(2mW²/ħ³)mC³/(4ħ⁴ω²).
  w(i,E) = (2mE)^-½ 2m²W²C³/(4ω²ħ⁶), with E = ħω - mC²/(2ħ²).
  This is the transition probability per unit time.

Problem 4:

A hydrogen atom in its ground state [(n, l ,m) = (1, 0, 0)] is placed between the plates of a capacitor.  A time dependent but spatially uniform electric field (not potential!) is applied as follows:
E = 0 for t < 0,
E = E₀exp(-t/τ) for t > 0.  E₀ = E₀ k.
Using first-order time-dependent perturbation theory compute the probability for the atom to be found at t >> τ in each of the three p-states n = 2, l = 1, m = ±1 or 0).

Useful information: <210|z|100> = (128*√2/243)a₀ and E₂ - E₁ = 3e²/8a₀.

Hydrogen atom wave functions:
R₁₀(r) = 2 a₀^-3/2 exp(-r/a₀), 
R₂₀(r) = (2a₀)^-3/2 (2 - r/a₀) exp(-r/(2a₀)),
R₂₁(r) = 3^-½(2a₀)^-3/2(r/a₀) exp(r/(2a₀)),
Y₀₀ = (4π)^-½,  Y_1±1 = ∓(3/(8π))^½sinθ exp(±iφ),  Y₁₀ = (3/(4π))^½cosθ.

Solution:

• Concepts:
  Time dependent perturbation theory
• Reasoning:
  We consider the interactions of the atomic electrons with the electric field perturbation to the atomic Hamiltonian.
• Details of the calculation:
  The uniform electric field changes the energy of the proton and the electron, and therefore perturbs the Hamiltonian.
  The energy of the proton in the field is -∫₀^zpq_eE(t)dz = -q_eE(t)z_p.
  The energy of the electron in the field is ∫₀^zeq_e E(t)dz = q_eE(t)z_e.
  The potential energy of both particles is
  q_eE(t)(z_e - z_p) = q_eE(t)z = q_eE(t) r cos(θ).
  H = H₀ + H'.
  The eigenfunctions of H₀ are the stationary states of the hydrogen atom, Φ_nlm(r,θ,φ) = R_nl(r)Y_lm(θ,φ).
  Time dependent perturbation theory:  P_if(t) = (1/ħ²)|∫₀^texp(iω_fit')W_fi(t')dt'|²,
  with  ω_fi = (E_f - E_i)/ħ and W_fi(t) = < Φ_f|H'(r,t)|Φ_i>, with H'(r,t)  = q_eE(t) r cos(θ).
  
  < Φ₁₀₀|H'(r,t)|Φ_2lm> ∝ ∫₀^∞r³dr R₀₀^*(r)R_2l(r)  ∫₀^πsinθ dθ∫₀^2πdφ Y₀₀^*(θ,φ) cosθ Y_lm(θ,φ)
  ∝ ∫₀^πsinθ dθ∫₀^2πdφ Y₀₀^*(θ,φ) Y₁₀(θ,φ) Y_lm(θ,φ).
  We can invoke the properties if the spherical harmonics.
  Or we can show explicitly
  <Φ₁₀₀|H'(r,t)|Φ₂₀₀> ∝ ∫₀^πsinθ cosθ dθ = 0,   <Φ₁₀₀|H'(r,t)|Φ₂₁₁> ∝ ∫₀^2πdφ e^iφ = 0,
  <Φ₁₀₀|H'(r,t)|Φ_21-1> ∝ ∫₀^2πdφ e^-iφ = 0,   <Φ₁₀₀|H'(r,t)|Φ₂₁₀> ∝ ∫₀^πsinθ dθ cos²θ ≠ 0.
  The only transition from the ground state to an n = 2 state that is allowed is the transition to the n = 2, l = 1, m = 0 state, Φ₂₁₀(r,θ,φ).
  
  Let Φ_i = Φ₁₀₀ and Φ_f = Φ₂₁₀.  Then ω_fi = ω₂₁ = (E₂ - E₁)/ħ = 3e²/(8a₀ħ) = (10.2 eV)/ħ and
  W_fi(t) = <Φ₂₁₀|H'(r,t)|Φ₀₀₀>
  = [q_eE₀exp(-t/τ)/(2^3/2a₀⁴)] ∫₀^∞r⁴dr exp(-3r/(2a₀))∫₀^πsinθ dθ cos²θ
  = q_eE₀exp(-t/τ) 2^½ 2⁷ a₀/3⁵.
  Therefore the probability that at time t the electron will have made the transition is
  P_if(t) = (1/ħ²)|∫₀^texp(iω_fit')W_fi(t')dt'|² = (1/ħ²)0.555 q_e²a₀²E₀²|∫₀^texp(iω₂₁t' - t'/τ)dt'|².
  |∫₀^texp(iω₂₁t' - t'/τ)dt'|² = [τ²/(ω₂₁²τ² + 1)][1 + exp(-2t/τ) - 2exp(-t/τ)cos(ω₂₁t)].
  As t -->infinity, |∫₀^texp(iω₂₁t') e^-t'/τdt'|² = [τ²/(ω²τ² + 1)].
  P_if(t) = (1/ħ²)|∫₀^texp(iω_fit')W_fi(t')dt'|² = (1/ħ²)0.555 q_e²a₀²E₀²[τ²/(ω₂₁²τ² + 1)].
  P_if(t) = 0.555 4πε₀e²a₀²E₀²[τ²/((3e²/(8a₀))²τ² + ħ²)].