Problem 1:
Two identical spin1 particles
obeying BoseEinstein statistics are placed in a 3D isotropic harmonic
potential.
(a) If the particles are noninteracting, give the energy and degeneracy of the
ground state of the twoparticle system.
(b) Now assume that the particles have a magnetic moment and interact through a
term in the Hamiltonian of the form AS_{1}∙S_{2}.
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 groundstate energy of a single particle in a the 3D
harmonic oscillator potential is E = (3/2 )ħω.
For two identical noninteracting Bosons the lowest possible energy is
E_{ground} = 3ħω.
The state vector has to be symmetric under exchange of the two bosons. The
groundstate 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 antisymmetric.
S = 0, m_{s} = 0> is symmetric.
The ground state is therefore 6fold degenerate.
(b) S_{1}∙S_{2} =
(S^{2}  S_{1}^{2}  S_{2}^{2})/2 = ħ^{2}(S(S + 1)
 1(1 + 1)  1(1 + 1))/2
= ħ^{2}(S(S +
1)/2  2).
S = 2: AS_{1}∙S_{2}
= Aħ^{2}.
S = 0: AS_{1}∙S_{2}
= 2Aħ^{2}.
The degeneracy of the states in (a) is partially
removed. Assuming A is a positive constant, the ground state is now
nondegenerate.
Problem 2:
Solve the Schroedinger equation for a particle of mass M in a cubical box of volume
L^{3}. Assume periodic boundary conditions.
(a) Show that in the limit L > ∞ the number of
states with momentum p in the range d^{3}p = dp_{x}dp_{y}dp_{z}
(that is p_{x} between p_{x} and p_{x }+ dp_{x}
etc.) is L^{3}d^{3}p/(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 ∝ n^{5/3}.
Solution:
 Concepts:
The threedimensional, infinite
square well, the Pauli exclusion principle, the density of states
 Reasoning:
We are asked to find the
eigenfunctions and eigenvalues of the threedimensional, infinite square
well. When we consider a system of N noninteracting Fermions in that well,
we must take into account that each state can only be occupied by one spin up
and one spin down particle.
 Details of the calculation:
(a) The eigenfunctions and
eigenvalues of the 3D infinite square well with periodic boundary conditions
are
Φ_{nml}(x,y,z)
= (1/L)^{3/2}exp(ik_{x}x)exp(ik_{y}y)exp(ik_{z}z),
with k_{x} = 2πn/L, k_{y} = 2πm/L, k_{z} = 2πl/L,
n, m, l = 1, 2, ... .
The associated eigenvalues are
E_{nml} = (n^{2} + m^{2} + l^{2})4π^{2}ħ^{2}/(2ML^{2})
= (k_{x}^{2} + k_{y}^{2} + k_{z}^{2})ħ^{2}/(2M).
[H = p_{x}^{2}/(2M) + p_{y}^{2}/(2M)
+ p_{z}^{2}/(2M), if x, y , z < L.
Φ(x,y,z)
= f(x)c(y)g(z).
(ħ^{2}/(2M))(∂^{2}/∂x^{2})f(x)
= E_{x}f(x).
f(x) = (1/L)^{½}exp(ik_{x}x),
f(x) = f(x + L) > k_{x} = 2πn/L.
E_{x} = ħ^{2}k_{x}^{2}/(2M) = 4π^{2}n^{2}ħ^{2}/(2ML^{2})
.]
∆k_{x} = ∆k_{y} = ∆k_{y }
= 2π/L.To each allowed k_{nml}
there corresponds a wavefunction Φ_{nml}(x,y,z).
The tips of the vectors k_{nml}
divide kspace into elementary cubes of edge length 2π/L.
We have one vector per (2π/L)^{3} volume of
kspace.
The number of vectors in a volume d^{3}k of kspace therefore is dN =
d^{3}k/(2π/L)^{3}.
d^{3}k = k^{2}dkdΩ = (1/ħ^{3})p^{2}dpdΩ = (1/ħ^{3})d^{3}p.
dN = L^{3}d^{3}p/(2πħ)^{3}.
(b) For a spin ½ particle the number of states with p between p' and
p' + dp' is
dN_{particle} = 2*4π p'^{2}dp' L^{3}/(2πħ)^{3}.
dN_{particle}/dp' = 8πL^{3}p'^{2}/(2πħ)^{3}.
E = p^{2}/(2M), dE = pdp/M.
The number of states with E between E' and E' + dE' is
dN_{particle} = 2^{½}M^{3/2}8πL^{3}E'^{½}dE'/(2πħ)^{3}.
dN_{particle}/dE' = AL^{3}E'^{½}, A = 2^{½}M^{3/2}8π/(2πħ)^{3}.
If all the lowest energy levels are filled, then the energy of the highest
occupied state is
E_{max} = ħ^{2}k_{max}^{2}/(2M).
The
volume of a sphere in 3D kspace is V = (4/3)πk^{3}.
The number of particle states in the sphere is N_{particle} = 2V/(2π/L)^{3}
= k_{max}^{3}L^{3}/(3π^{2}).
E_{max} = ħ^{2}[3N_{particle}π^{2}/L^{3}]^{2/3}/(2M)
∝ (N_{particle}/L^{3})^{2/3}.
The total energy of the system is
E_{t }= ∫_{0}^{Emax} E(dN_{particle}/dE)dE
= AL^{3}∫_{0}^{Emax}E^{3/2}dE
= (2/5)AL^{3}E_{max}^{5/2}.
E_{t} ∝ L^{3}E_{max}^{5/2}.
E/L^{3} = u ∝ E_{max}^{5/2}
= (N_{particle}/L^{3})^{5/3} = n^{5/3}.
Problem 3
Consider a system of two noninteracting particles (1, 2) and two orthonormal
energy eigenstates (α, β) with energies E_{α} and E_{β}
= 3E_{α}.
Determine the factor by which the probability for finding both particles in the
same state for bosons
exceeds that for classical particles.
Solution:
 Concepts:
Indistinguishable particles, identical bosons, the Boltzmann factor
 Reasoning:
Bosons and classical particles obey different probability laws.
 Details of the calculation:
For classical particles the probability of finding both particles in the
same state is
P_{c} = [exp(2E_{α}/(kT)) + exp(2E_{β}/(kT))]/[exp(2E_{α}/(kT))
+ exp(2E_{β}/(kT)) + 2 exp((E_{α }+ E_{β})/(kT))]
= [exp(2) + exp(6]/[exp(2) + exp(6) + 2 exp(4)].
For bosons the probability of finding both particles in the same state is
P_{b} = [exp(2E_{α}/(kT)) + exp(2E_{β}/(kT))]/[exp(2E_{α}/(kT))
+ exp(2E_{β}/(kT)) + exp((E_{α}+E_{β})/(kT))]
= [exp(2) + exp(6]/[exp(2) + exp(6) + exp(4)].
P_{b}/P_{c} = [exp(2) + exp(6) + 2 exp(4)]/[exp(2) +
exp(6) + exp(4)] = 1.117.
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
noninteracting spinless particles in a 3dimensional simple harmonic potential
with spring constants k_{x} = k_{y} = k_{z} = k.
(b) distinguishable
noninteracting spinless particles in a 3dimensional Coulomb potential
V(x,y,z) = Ze^{2}/r, where r = (x^{2 }+ y^{2 }+ z^{2})^{½}.
(c) indistinguishable
noninteracting spin ½ particles in a 3dimensional cubic box (with impenetrable
walls) of dimensions L_{x} × L_{y} × L_{z}, where L_{x}
= L_{y} = L_{z} = L.
Solution:
 Concepts:
Degeneracy, indistinguishable particles
 Reasoning:
The degeneracy of an energy level of a multiparticle
system is different for distinguishable and indistinguishable particles.
 Details of the calculation:
(a) For each particle, E = (n_{x} + ½)ħω + (n_{y} + ½)ħω + (n_{z}
+ ½)ħω = (n_{x} + n_{y} + n_{z} + 3/2)ħω,
where ω^{2} = k/m and n_{x}, n_{y}, n_{z}^{
}= 0, 1, 2, 3, ... .
The lowest level for the three
particles has energy E^{(1)} = 3*(3/2)ħω. The degeneracy is 1 because
there is only one possibility. Each particle must have n_{x} = n_{y}
= n_{z}^{ }= 0.
The second lowest single particle level has
energy E = (5/2)ħω and the second lowest three particle level has
energy E^{(2)} = (11/2)ħω. The degeneracy is 9 because two particles
must have all n_{i} = 0 and one particle (3
possibilities) must have one n_{i} = 1 (3
possibilities) and the other two n_{i} = 0.
(b) In a 3D Coulomb potential V(x,y,z) = Ze^{2}/r the energy for each
particle is Z^{2}*13.6 eV/n^{2}, n^{ }= 1, 2, 3, ... .
The energy is independent of l and m. The possible quantum numbers for
the two lowest 1particle energy levels are
n = 1, l = 0, m = 0 (1 possibility),
and n = 2, with l = 0, m = 0 and l = 1 , m = 0, ±1 (4
possibilities).
The possible energy levels for
the 3particle system are E = Z^{2}*13.6 eV(1/n_{1}^{2}
+ 1/n_{2}^{2} +1/n_{3}^{2}).
Hence the lowest level for the
3particle system is E^{(1)} = 3Z^{2}*13.6 eV, n_{1}
= n_{2} = n_{2}^{ }= 1.
The degeneracy is 1.
In the second lowest level for the 3particle system, one of the three particles
has n = 2,
so E^{(2)} = (9/4)Z^{2}*13.6 eV.
The degeneracy is 3*4 = 12, because any one of the distinguishable particles
could have n = 2
(3 possibilities), and there are 4 possible n = 2
states (4 possibilities).
(c) For a single particle in a
cubic box
E_{n} = [π^{2}ħ^{2}/(2mL^{2})] (n_{x}^{2}
+ n_{y}^{2} + n_{z}^{2}), n_{x}, n_{y},
n_{z}^{ }= 1, 2, ... .
For a single particle, the
lowest level has n_{x} = n_{y} = n_{z}^{ }= 1
and energy E_{1} = 3π^{2}ħ^{2}/(2mL^{2}).
For a single particle,
the second lowest level has (n_{x}, n_{y}, n_{z})
= (1, 1, 2), (1, 2, 1), or (2, 1, 1) and
E_{2} = 6π^{2}ħ^{2}/(2mL^{2}).
For the system of three
indistinguishable spin ½ particles, we can put either one or two particles in
each level. When there are two particles in a given level, there is only one
possibility  one of the spins must be up and the other one down. When there is
only one particle in a given level, there are two spin possibilities, the spin
can be either up or down.
The lowest level of the 3particle system has energy
E^{(1)} = (E_{1} + E_{1} + E_{2}) = 6π^{2}ħ^{2}/(mL^{2}).
The degeneracy is 6, since two of the particles are in level E_{1} with
paired spins (1 possible twoparticle state) and
there are 3 possible sets of quantum numbers for the third particle in level E_{2},
each with 2 possible spins. (6 possible oneparticle
states).
The second lowest level of the
system has energy E^{(2)} = (E_{1} + E_{2} + E_{2})
= 15π^{2}ħ^{2}/(2mL^{2}).
The degeneracy = 2*(6*4 + 3) = 54.
For two particles each contributing energy E_{2}, there
are 9 possibilities for the spatial quantum numbers, (n_{1x}, n_{1y},
n_{1z}) = (1, 1, 2), (1, 2, 1), or (2, 1, 1) (3 possibilities) times
(n_{2x}, n_{2y}, n_{2z}) = (1, 1, 2), (1, 2, 1), or (2,
1, 1) (3 possibilities).
When n_{1i }= n_{2i}
for all i (3 of the 9 possibilities) the two particles are in the same spatial
state so the particles must be in the singlet state (3 possible
twoparticle states). For the other 6 of the 9 possible sets of spatial
quantum numbers, the two particles are in different spatial states, so each
particle can have spin up or spin down, hence 4 four possible spin states and
6*4 = 24 possible twoparticle states.
For the one particle contributing energy E_{1} we have n_{x} = n_{y}
= n_{z}^{ }= 1, and the spin can be up or down (2 possible oneparticle states).
Problem 5:
For the Titanium atom (Z = 22) in its ground state find the allowed terms
^{2S+1}L_{J} in the LS (RussellSanders) coupling scheme and
use Hund's rule to find the ground state term.
Solution:
 Concepts:
The Pauli exclusion principle, energy levels of multielectron 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 multielectron 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+1}L_{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}, (3s)^{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 two 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+1}L_{J} for Titanium
therefore are
^{1}G_{4}, ^{3}F_{4,3,2}, ^{1}D_{2},
^{3}P_{2,1,0}, ^{1}S_{0}.
Hund's rule:
The level with the largest multiplicity has the lowest energy. (^{3}F_{4,3,2}
and ^{3}P_{2,1,0})
For a given multiplicity, the level with the largest value of L has the
lowest energy. (^{3}F_{4,3,2})
For less than halffilled shells:
The component with the smallest value of J has the lowest energy (^{3}F_{2}).
The ground state tem is ^{3}F_{2}.