### 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 (ψα(x), ψβ(x), ψγ(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 = ψa(x1) ψb(x2) ψc(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 = (nx + ny + nz + 3/2)ħω, nx, ny, nz = 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 (nx, ny, nz = 0).
E1 = (1 + 3/2)ħω, 6 undistinguishable spin ½ particles can occupy this state (nx or ny or nz = 1 with the other ni = 0).
E2 = (2 + 3/2)ħω, 12 undistinguishable spin ½ particles can occupy this state (nx or ny or nz = 2 with the other ni = 0, or two of the ni = 1 and one of the ni = 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(ωx2x2 + ωy2y2 + ωz2z2).  Assume that ωx = ωy = 2ωz.
(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 Lx, Ly, or Lz 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 = (nx + ½)ħωx + (ny + ½)ħωy + (nz + ½)ħωz = (N + 5/2)ħωz.
Here N = 2nx + 2ny + nz.  nx, ny, nz, = 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 nx, ny, and nz 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.
[Lx, H] ≠ 0, [Ly, H] ≠ 0, [Lz, H] = 0.
Lx and Ly do not have good eigenvalues, Lz has good eigenvalues.