Assignment 12, solutions

Problem 1:

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 AS1S2.  How are the energies and degeneracies of the states in (a) changed by this interaction?

Solution:

Problem 2:

Solve the Schroedinger equation for a particle of mass M in a cubical box of volume L3.  Assume periodic boundary conditions.
(a)  Show that in the limit L --> ∞ the number of states with momentum p in the range  d3p = dpxdpydpz (that is px between px and px + dpx etc.) is L3d3p/(2πħ)3.
(b)  Assume that the lowest energy levels in the box are filled with N electrons, taking due account of the Pauli exclusion principle.  Show that the energy per unit volume, u, is related to the number of particles per unit volume n by u ∝ n5/3.

Solution:

Problem 3

Consider a system of two non-interacting particles (1, 2) and two orthonormal energy eigenstates (α, β) with energies Eα and Eβ = 3Eα.
Determine the factor by which the probability for finding both particles in the same state for bosons exceeds that for classical particles.

Solution:

Problem 4:

Determine the energies and the degeneracies of the two lowest levels of a system composed of three particles with equal masses m, where the particles are
(a)  distinguishable non-interacting spinless particles in a 3-dimensional simple harmonic potential with spring constants kx = ky = kz = k.
(b)  distinguishable non-interacting spinless particles in a 3-dimensional Coulomb potential
V(x,y,z) = -Ze2/r, where r = (x2 + y2 + z2)½.
(c)  indistinguishable non-interacting spin ½ particles in a 3-dimensional cubic box (with impenetrable walls) of dimensions Lx × Ly × Lz, where Lx = Ly = Lz = L.

Solution:

Problem 5:

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

Solution: