Problem 1:

Consider a particle in a 1D box of length L, i.e. there are hard walls at x = 0 and at x = L.  Assume that the particle is in the ground state.  At t = 0 the wall at x = L is suddenly moved to x = 2L.  What is the likelihood to find the particle in the ground state of the larger box?

Solution:

• Concepts:
  A particle in an infinite square well, 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:
  For t < 0, ψ_1i(x) = (2/L)^½sin(πx/L),  E_1i = π²ħ²/(2mL²).
  For t > 0, ψ_1f(x) = (1/L)^½sin(πx/(2L)),E_1f = π²ħ²/(8mL²).  
  P = |<ψ_f,1|ψ_i,1>|² = |(1/L)^½(2/L)^½∫₀^Ldx sin(πx/(2L))sin(πx/L)|²
   = (2/L²)[(2L/π)]∫₀^π/2dx' sin(x')sin(2x')|² = (8/π²)[(sin(x)/2 - sin(3x)/6))|₀^π/2]²
   = (8/π²)(4/6)² = 0.36.
   The probability of finding the particle in the ground state of the larger box is 0.36.

Problem 2:

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

Problem 3:

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).