More Problems

Problem 1:

Consider a free particle of mass M in two dimensions.  Since continuum states are not traditionally normalizable, we assume that the particle is confined to a cubical 2D box with periodic boundary conditions. Its eigenfunctions are
ψ_nq(x,y) = (1/L)exp(ik_xx)exp(ik_yy),
with k_x = 2πn/L,  k_y = 2πq/L, n, q = 1, 2, ... .
Derive and expression for the density of states dN/dE = ρ(E).

Solution:

• Concepts:
  The density of states
• Reasoning:
  It is customary to use periodic boundary conditions for continuum states.
• Details of the calculation:
  The energy eigenvalue associated with the eigenstate
  ψ_nq(x,y) = (1/L)exp(ik_xx)exp(ik_yy) is
  E_nq = (n² + q²)4π²ħ²/(2ML²) = (k_x² + k_y²)ħ²/(2M) = k²ħ²/(2M)
  k_x = 2πn/L,  k_y = 2πq/L, Δk_x = Δk_y = 2π/L.

  If k is large, then the number of states with wave vectors whose magnitudes lie between k and k + dk is
  dN = 2πkdk/(2π/L)²  = L²kdk/(2π).
  dN/dk = L²k/(2π).
  The density of states is dN/dE = (dN/dk)(dk/dE).
  With E = ħ²k²/(2M) we have ρ(E) = dN/dE = ML²/(2πħ²).

Problem 2:

In a one-dimensional structure a particle of mass m is in the ground state of a potential energy function U(x) = ½kx².  A phase transition occurs, and the effective spring constant suddenly becomes k' = k/2.  What is the probability that the particle will be found in an exited state after the phase transition?  Give a numerical answer.

Solution:

• Concepts:
  The QM harmonic oscillator, the sudden approximation
• Reasoning:
  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)|α> 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:
  The normalized ground-state wave function of the 1D harmonic oscillator for t < 0 is
  Φ_0i(x) = (mω/(πħ))^¼exp(-½mωx²/ħ),  ω = (k/m)^½.
  The normalized ground-state wave function for t > 0 is
  Φ_0f(x) = (mω/(πħ√2))^¼exp(-½mωx²/(ħ√2)).
  The probability that the oscillator remains in the ground state is
  P = |<Φ_0i|Φ_0f>|² =  |∫_-∞^∞dx Φ_0i*(x) Φ_0f(x)|²
  = 2^5/4/(1 + √2) = 0.985.
  The probability that the particle will be found in an exited state after the phase transition is
  1 - 0.985 = 0.015.

Problem 3:

An electron in a hydrogen atom is in one of the degenerate |n,l,m>  = |4,2,m> states.
For all possible |4,2,m> states, which radiative transitions to a lower energy state are allowed by the dipole selection rules?
Neglect spin.

Solution:

• Concepts:
  Selection rules for "allowed" atomic transitions
• Reasoning:
  The dipole selection rules are Δl = ±1, Δm = 0, ±1.
• Details of the calculation:
  Δl = ±1 implies that the final l value must be 3 or 1. There are no lower energy states with l = 3, so the final l value will be 1.
  Allowed transitions will go to states with n = 3, l = 1 or n = 2, l = 1.
  For both n_f = 3 and n_f = 2 we can make the same table.
  

  	mf	1	0	-1
  mi
  2		y
  1		y	y
  0		y	y	y
  -1			y	y
  -2				y