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:
  (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:

Among the values of L and S obtained from the general rules for addition of angular momenta are those which correspond to states forbidden by the Pauli principle.  Assume you have an atom with two electrons with the same principal quantum number n in d states (nd²) outside closed shells.
Find the allowed terms ^2S+1L_J.

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:
  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 nd 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 therefore are
  ¹G₄, ³F_4,3,2, ¹D₂, ³P_2,1,0, ¹S₀.

Problem 3:

(a)  A spin ½ particle can have S_z = ±½ħ.  Construct states for two such particles that are symmetric or anti-symmetric under interchange and find their total spin S.
(b)  Assuming that ³H and ³He consist of protons and neutrons in L = 0 states, find their total angular momentum.

Solution:

• Concepts:
  Two spin ½ particles, identical particles
• Reasoning:
  The eigenfunctions of S² and S_z for two spin ½ particles are the triplet states
  |11> = |++>,  |10> = ½^½(|+-> + |-+>),  |1-1> = |-->,
  and the singlet state  |00> = ½^½(|+-> - |-+>).
• Details of the calculation:
  (a)  The triplet states are symmetric under interchange and the singlet state is anti-symmetric under interchange of the two particles.
  For the triplet states we have S² = 2ħ²,  S_z = m_sħ (m_s = -1, 0, 1),
  and for the singlet state we have S² = S_z = 0.
  (b)  The L = 0 ground state wave function of the two neutrons in ³H or the two protons in ³He is symmetric under interchange of the two particles.  So the spin state of those two particles is the anti-symmetric singlet state.  The total angular momentum is therefore the spin angular momentum of the third particle.
  For the total angular momentum we have j = ½, J² = ¾ħ².