Counting statistics

Problem:

Assume that two particles are placed at random into one of two boxes.  What are the probabilities of finding the following distributions if the particles are classical particles, identical bosons, or identical fermions?

Solution:

• Concepts:
  Identical particles
• Reasoning:
  Fermions, bosons, and classical particles obey different probability laws. 

• Details of the calculation:
  Classical particles are distinguishable.  The state that has particle 1 in box 1 and particle 2 in box 2 differs from the state that has particle 2 in box 1 and particle 1 in box 2.  Bosons and Fermions are indistinguishable.  There is only one state with one of the indistinguishable particles in box 1 and the other in box 2.  Fermions obey the Pauli exclusion principle.  No two fermions can be in the same box.

  Classical	¼	½	¼
  Boson	(1/3)	(1/3)	(1/3)
  Fermion	0	1	0

Problem:

Assume three particles (1, 2, 3) and three distinct one particle states (ψ_α(x), ψ_β(x), ψ_γ(x)).
Describe in detail the possible three particle states can that be constructed if:

(a)  The three particles are distinguishable,
(b)  The three particles are identical fermions, and
(c)  The three particles are identical bosons.

Solution:

• Concepts:
  Indistinguishable particles
• Reasoning:
  The wave functions of indistinguishable particles must obey symmetrization requirements.
• Details of the calculation:
  (a)  If the particles are distinguishable, 27 possible linearly independent states can be constructed.
  (There are 3 possible state for each particle, or 3³ possible states.)
  The possible wave functions are ψ = ψ_a(x₁) ψ_b(x₂) ψ_c(x₃),  a, b, c = α, β, γ.  There is no symmetrization requirement.

  (b) If the particles are identical fermions only one state can be constructed.
  The wave function must be anti-symmetric under exchange.
  Notation |a, b, c>:  Let a denote the state of particle 1, b the state of particle 2, and c the state of particle 3.
  |ψ> = (1/6)^½[|α,β,γ> + |γ,α,β> + |β,γ,α> - |α,γ,β> - |β,α,γ> - |γ,β,α>]
  is antisymmetric under exchange of any two particles.

  (c)  If the particles are identical bosons, 10 linearly independent states can be constructed.
  The wave function must be symmetric under exchange.
  The following states satisfy this requirement.
  (i)  All three particles can be in the same state. (3 possibilities)
  |ψ> = |α,α,α>  or  |ψ> = | β,β,β>  or |ψ> = | γ,γ,γ>.
  (ii)  Each particle occupies a different state. (1 possibility)
  |ψ> = (1/6)^½[|α,β,γ> + |γ,α,β> + |β,γ,α> + |α,γ,β> + |β,α,γ> + |γ,β,α>]
  is symmetric under exchange of any two particles.
  (iii)  Two particles occupy the same state and one particle occupies one of the remaining two states.  (3 possibilities times two possibilities = 6 possibilities)
  Example: |ψ> = (1/3)^½[|α,α,γ> + |γ,α,α > + | α,γ,α>]

Problem:

Protons and neutrons making up a light nucleus move in an average potential that resembles that of a harmonic oscillator.
U(r) = -U₀ + Mω²r²/2
Find numbers of protons (or neutrons) corresponding to the first three closed shells (magic numbers).

Solution:

• Concepts:
  The energy levels of the 3D harmonic oscillator, the Pauli exclusion principle.
• Reasoning:
  The energy levels of the 3D harmonic oscillator are degenerate.
• Details of the calculation:
  We have a 3D harmonic oscillator.  For convenience, choose the zero of the potential energy such that U₀ = 0.  The energy levels of the isotropic, 3D harmonic oscillator are
  E = (n_x + n_y + n_z + 3/2)ħω, n_x, n_y, n_z = 0, 1, 2, ..., (Cartesian coordinate)
  or
  E = (2n + l + 3/2)ħω, n, l = 0, 1, 2, ..., (spherical coordinates). 

  The energy levels are degenerate.
  E₀ = (3/2)ħω, two undistinguishable spin ½ particles can occupy this state (n_x, n_y, n_z = 0).
  E₁ = (1 + 3/2)ħω, 6 undistinguishable spin ½ particles can occupy this state (n_x or n_y or n_z = 1 with the other n_i = 0).
  E₂ = (2 + 3/2)ħω, 12 undistinguishable spin ½ particles can occupy this state (n_x or n_y or n_z = 2 with the other n_i = 0, or two of the n_i = 1 and one of the n_i = 0).

  Magic numbers: 2, 2 + 6 = 8, 8 + 12 = 20.

Problem:

A nucleus can be considered as fermions moving in a three-dimensional harmonic oscillator potential.  Each nucleon has potential energy  U = ½ m(ω_x²x² + ω_y²y² + ω_z²z²).  Assume that ω_x = ω_y = 2ω_z.
(a)  Find the first five magic numbers for protons and neutrons.
(b)  Will the magnitude of the total angular momentum have good eigenvalues for this potential?
(c)  Will L_x, L_y, or L_z have good eigenvalues?

Solution:

• Concepts:
  Identical particles, good eigenvalues
• Reasoning:
  Identical fermions obey the Pauli exclusion principle.  Operators that commute with the Hamiltonian have "good" eigenvalues.
• Details of the calculation:
  (a)  The energy eigenvalues of a nucleon are
  E = (n_x + ½)ħω_x + (n_y + ½)ħω_y + (n_z + ½)ħω_z = (N + 5/2)ħω_z.
  Here N = 2n_x + 2n_y + n_z.  n_x, n_y, n_z, = 0, 1, 2, ... .
  The 5 lowest energy levels have N = 0, 1, 2, 3, and 4.
  To find the magic numbers we have to find the degeneracy of these energy levels.
  Each possible combination of n_x, n_y, and n_z yields two states when we account for the possible orientations of the spin.

  N	nx	ny	nz	partitions	states	magic number
  0	0	0	0	1	2	2
  1	0	0	1	1	2	4
  2	0	1	0
  	1	0	0
  	0	0	2	3	6	10
  3	0	1	1
  	1	0	1
  	0	0	3	3	6	16
  4	2	0	0
  	0	2	0
  	1	0	2
  	0	1	2
  	1	1	0
  	0	0	4	6	12	28

  (b)  The potential is not spherically symmetric, [L², U] ≠ 0, [L², H] ≠ 0, the stationary states are not eigenstates of L².  The magnitude of the total angular momentum does not have good eigenvalues.
  The potential is not invariant under rotations about the x-axis and y-axis, but is invariant under rotations about the z-axis.
  [L_x, H] ≠ 0, [L_y, H] ≠ 0, [L_z, H] = 0.
  L_x and L_y do not have good eigenvalues, L_z has good eigenvalues.