Assignment 6

Problem 1:

(a)  Determine the energy levels and normalized wave functions ψ(r) of a particle with zero angular momentum in a spherical "potential well" U(r) = 0, (r < a), U(r) = ∞, (r > a).
(b)  If the particle is initially in the ground state of this well and the U(r) is suddenly changed so that U(r) = 0, (r < 3a), U(r) = ∞, (r > 3a), what is the probability that after that change an energy measurement will find that the energy of the particle has changed?

Solution:

• Concepts:
  Three-dimensional "square" potentials, the sudden approximation
• Reasoning:
  We are asked to determine the energy levels and normalized wave functions ψ(r) of a particle in a three-dimensional square potential.  The sudden approximation can be used to calculate transition probabilities when the Hamiltonian changes rapidly.  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:
  (a)  The Hamiltonian is H = -(ħ²/(2m)(1/r)(∂²/∂r²)r + L²/(2mr²) + U(r).
  The wave function ψ_klm(r,θ,φ) = R_kl(r)Y_lm(θ,φ) = [u_kl(r)/r]Y_lm(θ,φ) is a product of a radial function R_kl(r) and the spherical harmonic Y_lm(θ,φ).
  The differential equation for u_kl(r) is
  [-(ħ²/(2m)(∂²/∂r²) + ħ²l(l + 1)/(2mr²) + U(r)]u_kl(r) = E_klu_kl(r).
  For l = 0 we have [-(ħ²/(2m)(∂²/∂r²) + U(r)]u_k0(r) = E_k0u_k0(r).
  With U(r) = 0, (r < a), U(r) = ∞, (r > a) the solutions are
  u_k0(r) = A sinkr, ka = nπ, k = nπ/a.
  We therefore label u_k0(r) with the quantum number n as u_n0(r).
  E_n0 =  n²π²ħ²/(2ma²) are the energy levels when l = 0.
  ∫|ψ_n00(r,θ,φ)|²d³r = 1, A²∫₀^a sin²(nπr/a)dr = 1,  A = (2/a)^½.
  ψ_n00(r,θ,φ) = (1/(2πa)^½)sin(nπr/a)/r.
  
  (b)  Before U is changed the wave function of the particle is
  ψ₁₀₀(r,θ,φ)_i = Y₀₀ (2/a)^½ sin(πr/a)/r, and its energy is E₀₀ = π²ħ²/(2ma²).
  The probability of finding the particle in the energy eigenstate ψ^*_nlm(r,θ,φ)_f of the changed well is
  |∫d³r ψ^*_nlm(r,θ,φ)_f ψ₁₀₀(r,θ,φ)_i|².
  ∫d³r ψ^*_nlm(r,θ,φ)_f ψ₁₀₀(r,θ,φ)_i = 0 unless l = m = 0 because of the orthogonality of the spherical harmonics.
  After U is changed the eigenfunction of the new H with E = π²ħ²/(2ma²) is
  ψ₃₀₀(r,θ,φ)_f = Y₀₀ (2/(3a))^½ sin(πr/a)/r.
  ∫d³r ψ^*₃₀₀(r,θ,φ)_f ψ₁₀₀(r,θ,φ)_i = (2/(a√3))  ∫₀^adr sin²(πr/a) = 1/√3.
  |<ψ_1f|ψ_0i>|² = 1/3.
  The probability that after that change an energy measurement will find that the energy of the particle has not changed is 1/3.
  Therefore, the probability that the energy of the particle has changed is 2/3.

Problem 2:

The Lα line of the characteristic X-ray spectra of heavy atoms consists of several components of different frequencies corresponding to the various allowed transitions from levels with n = 3 to levels with n = 2.  Predict the number of different frequencies to be observed, on the basis of the selection rules  Δl = ±1, Δj = 0, ±1 (except j_i = j_f = 0).

Solution:

• Concepts:
  Selection rules for "allowed" atomic transitions
• Reasoning:
  Neutral heavy atoms have filled n = 2 and n = 3 shells.
• Details of the calculation:
  If we have a hole in the n = 2 shell, the shell angular momentum quantum numbers are l = 0 or l = 1 and j = l ± ½.
  We can have the following energy levels: 2s_1/2,  2p_1/2,  2p_3/2.
  If an electron from the n = 3 shell fills that hole we are left with a hole in the n = 3 shell.
  We then can have the following energy levels: 3s_1/2,  3p_1/2,  3p_3/2,  3d_3/2,  3d_5/2.  
  The allowed transitions are indicated by a + while the forbidden transitions are indicated by a x sign.

  		n = 3	l	0	1	1	2	2
  		l	j	1/2	1/2	3/2	3/2	5/2
  n = 2
  l	j
  0	1/2			x	+	+	x	x
  1	1/2			+	x	x	+	x
  1	3/2			+	x	x	+	+

  There are seven allowed transitions.

Problem 3:

A hydrogen atom with Hamiltonian H₀(r) is placed in a time-dependent electric field E = E(t) k. The perturbed Hamiltonian is H(r,t) = H₀(r) + H'(r,t).
(a)  Show that H'(r,t) = q_eE(t) r cos(θ).
(b)  Assuming the electron is initially in the ground state, and recalling that the first excited state of hydrogen is quadruply degenerate, to which state of the quadruply degenerate first excited states is a dipole transition from the ground state possible?  Prove this.
(c)  If the electron is in the ground state at t = 0, find the probability (to first order in perturbation theory) that at time t the electron will have made the transition to the state determined in (b), as a function of E(t).

Hydrogen atom energy eigenfunctions:

Φ₁₀₀(r,θ,φ)  =  π^-½a₀^-3/2 exp(-r/a₀).
Φ₂₀₀(r,θ,φ)  = (4π)^-½ (2a₀)^-3/2 (2 - r/a₀) exp(-r/(2a₀)),
Φ₂₁₁(r,θ,φ)  = (8π)^-½ (2a₀)^-3/2 (r/a₀) exp(-r/(2a₀)) sinθ e^iφ,
Φ₂₁₀(r,θ,φ)  = (4π)^-½ (2a₀)^-3/2 (r/a₀) exp(-r/(2a₀)) cosθ,
Φ_21-1(r,θ,φ) = (8π)^-½ (2a₀)^-3/2 (r/a₀) exp(-r/(2a₀)) sinθ e^-iφ.

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:
  (a)  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).
  
  (b)  For a dipole transition to be allowed, we need <Φ_f|H'(r,t)|Φ_i> to be nonzero.
  Φ_nlm(r,θ,φ) = R_nl(r)Y_lm(θ,φ).
  The ground state wave function is Φ₁₀₀(r,θ,φ)  =  π^-½a₀^-3/2 exp(-r/a₀).
  The wave functions of the quadruply degenerate first excited states are
  Φ₂₀₀(r,θ,φ)  = (4π)^-½ (2a₀)^-3/2 (2 - r/a₀) exp(-r/(2a₀)),
  Φ₂₁₁(r,θ,φ)  = (8π)^-½ (2a₀)^-3/2 (r/a₀) exp(-r/(2a₀)) sinθ e^iφ,
  Φ₂₁₀(r,θ,φ)  = (4π)^-½ (2a₀)^-3/2 (r/a₀) exp(-r/(2a₀)) cosθ,
  Φ_21-1(r,θ,φ) = (8π)^-½ (2a₀)^-3/2 (r/a₀) exp(-r/(2a₀)) sinθ e^-iφ.
  
  <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).
  
  (c)  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).
  Let Φ_i = Φ₁₀₀ and Φ_f = Φ₂₁₀.  Then ω_fi = ω₂₁ = (E₂ - E₁)/ħ = (10.2 eV)/ħ and
  W_fi(t) = <Φ₂₁₀|H'(r,t)|Φ₀₀₀> = [q_eE(t) /(2^3/2a₀⁴)] ∫₀^∞r⁴dr exp(-3r/(2a₀))∫₀^πsinθ dθ cos²θ
  = q_eE(t) 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 = 0.555 q_e²a₀²|∫₀^texp(iω₂₁t')E(t')dt'|².

Problem 4:

Consider spinless non-relativistic free particles of mass M moving in a three dimensional cubical box of side length L, with L very large.  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 three-dimensional, infinite square well, the density of states
• Reasoning:
  We are asked to find the density of states for a three-dimensional, infinite square well
• Details of the calculation:
  The eigenfunctions and eigenvalues of a free particle 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 4πk²dk of k-space therefore is 4πk²dk/(2π/L)³.
  dN = 4πk²dk/(2π/L)³. = number of state with a wave vector of magnitude between k and k + dk.
  dN/dk = 4πk²/(2π/L)³.
  The density of states for the paeticle is dN/dE = (dN/dk)(dk/dE), with E = ħ²k²/(2M).
  dE/dk = ħ²k/M.
  dN/dE = ρ(E) = 4πkL³M/((2π)³ħ²) =  2^½M^3/24πL³E'^½dE/(2πħ)³.

Problem 5:

A one-dimensional harmonic oscillator is in its ground state for t < 0.
The unperturbed Hamiltonian is H₀ = p²/(2m) + ½mω₀²x².
For  t > 0 it is subjected to a time-dependent but spatially uniform force in the x-direction, 
F = F₀cos(ωt), ω << ω₀.
Using time-dependent perturbation theory to first order, obtain the probability of finding the oscillator in an excited state for t > 0, t << 2π/ω.

Solution:

• Concepts:
  The harmonic oscillator, time dependent perturbation theory
• Reasoning:
  We are asked to find the transition probability from the ground state to an excited state for a perturbed harmonic oscillator.
• Details of the calculation:
  Time dependent perturbation theory:
  P_n0(t) = (1/ħ²)|∫₀^texp(iω_n0t')W_n0(t')dt'|²
  W_n0 = <n|W(t)|0>, where W(t) = -F₀xcos(ωt), and {|n>} are the normalized eigenstates of the harmonic oscillator Hamiltonian.
  <n|W(t)|0> = -F₀cos(ωt)<n|x|0> = CF₀cos(ωt)<n|(a + a^T)|0> = CF₀cos(ωt)δ_1n.
  Here C = -(ħ/(2mω₀))^½.  Only transitions to the first excited states occur.
  For the harmonic oscillator with the potential energy function ½mω₀²x² we have
  E_n = (n + ½)ħω₀.  Therefore ω_n0 = nω₀.
  P₁₀(t) = (1/ħ²)|∫₀^texp(iω₁₀t')W_n0(t')dt'|² = (CF₀/ħ)²|∫₀^texp(iω₀t')cos(ωt)dt'|².
  Since ω << ω₀, we have an approximately constant perturbation for t << 2π/ω..
  ∫₀^texp(iω₀t')cos(ωt)dt' ≈∫₀^texp(iω₀t')dt' = (2/ω₀)exp(iω₀t/2)sin(ω₀t/2).
  
  P₁₀(t) =  (2CF₀/(ω₀ħ))²sin²(ω₀t/2).
  The probability of finding the oscillator in an excited state for t > 0 is  P(t) = (2F₀/(mω₀³ħ))sin²(ω₀t/2).