More Problems

Problem 1:

(a)  Consider three bosons, without spin, in a one-dimensional infinite square well with energy levels
E_n = n²E₁, n = 1, 2, 3, ... .  What is the energy of the ground state?
(b)  Repeat for three fermions, assumed without spin.

Solution:

• Concepts:
  The Pauli exclusion principle for fermions
• Reasoning:
  The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers. 
• Details of the calculation
  (a)  For bosons, place the 3 particles in the n = 1 state, E_total = 3E₁.
  (b)  For fermions place one particle in each the n = 1, n = 2, and n = 3 state.
  E_total = (1 + 4 + 9)E₁ = 14E₁.

Problem 2:

For the Titanium atom (Z = 22) in its ground state find the allowed terms ^2S+1L_J in the L-S (Russell-Sanders) coupling scheme and use Hund's rule to find the ground state term.

Solution:

• Concepts:
  The Pauli exclusion principle, energy levels of multi-electron atoms
• Reasoning:
  The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers.  In the LS coupling scheme, the energy level of a multi-electron atom in the absence of external fields is characterized by the quantum numbers n, L, S, and J.  In the LS coupling scheme the notation for the angular momentum is the term ^2S+1L_J.  Because of the Pauli exclusion principle not all terms one obtains by simply adding the angular momentum and spin of all the electrons are allowed.
• Details of the calculation:
  The ground state configuration of titanium is is (1s)², (2s)², (2p)⁶, (3s)², (3p)⁶, (4s)², (3d)².  We have two l = 2 electrons outside of closed shells.  All closed shells have L = 0, S = 0.
  The total orbital angular momentum quantum number L is the orbital angular momentum quantum number of the two 3d electrons.
  L = 4, 3, 2, 1, and 0 terms are possible.  For two equivalent electrons, ψ_space or χ_spin, the "stretched case" is always symmetric under exchange of the two particles. The symmetry then alternates, every time L or S decreases by one down to zero.  Here the "stretched case for the orbital angular momentum is L = 4 and for the spin angular momentum it is S = 1.
  The possible spectroscopic terms ^2S+1L_J for Titanium therefore are
  ¹G₄, ³F_4,3,2, ¹D₂, ³P_2,1,0, ¹S₀.
  Hund's rule:
  The level with the largest multiplicity has the lowest energy.  (³F_4,3,2 and ³P_2,1,0)
  For a given multiplicity, the level with the largest value of L has the lowest energy. (³F_4,3,2)
  For less than half-filled shells:
  The component with the smallest value of J has the lowest energy (³F₂).
  The ground state tem is ³F₂.

Problem 3:

Two identical spin-1 particles obeying Bose-Einstein statistics are placed in a 3D isotropic harmonic potential.
(a)  If the particles are non-interacting, give the energy and degeneracy of the ground state of the two-particle system.
(b)  Now assume that the particles have a magnetic moment and interact through a term in the Hamiltonian of the form AS₁∙S₂.  How are the energies and degeneracies of the states in (a) changed by this interaction?

Solution:

• Concepts:
  The 3D harmonic oscillator, identical particles
• Reasoning:
  We are given two Bosons in a 3D harmonic oscillator potential.
• Details of the calculation:
  (a)  The ground-state energy of a single particle in a the 3D harmonic oscillator potential is E = (3/2 )ħω.
  For two identical non-interacting Bosons the lowest possible energy is
  E_ground = 3ħω.
  The state vector has to be symmetric under exchange of the two bosons.  The ground-state wave function in coordinate space is symmetric, so the spin function must be symmetric as well.
  The possible values for the total spin are S = 2, 1, 0.
  |S = 2, m_s> is symmetric, there are 5 different values for m_s.
  |S = 1, m_s> is anti-symmetric.
  |S = 0, m_s = 0> is symmetric.
  The ground state is therefore 6-fold degenerate.
  
  (b)  S₁∙S₂ = (S² - S₁² -  S₂²)/2 = ħ²(S(S + 1) - 1(1 + 1) - 1(1 + 1))/2
  = ħ²(S(S + 1) - 2)/2.
  S = 2: AS₁∙S₂ = 2Aħ².
  S = 0: AS₁∙S₂ = -Aħ².
  The degeneracy of the states in (a) is partially removed.  Assuming A is a positive constant, the ground state is now non-degenerate.