## 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:

• 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 ground-state energy of a single particle in a the 3D harmonic oscillator potential is E = (3/2 )ħω.
For two identical non-interacting Bosons the lowest possible energy is
Eground = 3ħω.
The state vector has to be symmetric under exchange of the two bosons.  The ground-state 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, ms> is symmetric, there are 5 different values for ms.
|S = 1, ms> is anti-symmetric.
|S = 0, ms = 0> is symmetric.
The ground state is therefore 6-fold degenerate.
(b)  S1S2 = (S2 - S12 -  S22)/2 = ħ2(S(S + 1) - 1(1 + 1) - 1(1 + 1))/2
= ħ2(S(S + 1)/2 - 2).
S = 2: AS1∙S2  = Aħ2.
S = 0: AS1S2  = -2Aħ2.
The degeneracy of the states in (a) is partially removed.  Assuming A is a positive constant, the ground state is now non-degenerate.

#### 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:

• Concepts:
The three-dimensional, infinite square well, the Pauli exclusion principle, the density of states
• Reasoning:
We are asked to find the eigenfunctions and eigenvalues of the three-dimensional, infinite square well.  When we consider a system of N non-interacting 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
Φnml(x,y,z) = (1/L)3/2exp(ikxx)exp(ikyy)exp(ikzz),
with kx = 2πn/L,  ky = 2πm/L,  kz = 2πl/L, n, m, l = 1, 2, ... .
The associated eigenvalues are
Enml = (n2 + m2 + l2)4π2ħ2/(2ML2) = (kx2 + ky2 + kz22/(2M).
[H = px2/(2M) + py2/(2M) + pz2/(2M), if x, y , z < L.
Φ(x,y,z) = f(x)c(y)g(z).
(-ħ2/(2M))(∂2/∂x2)f(x) = Exf(x).
f(x) = (1/L)½exp(ikxx),  f(x) = f(x + L) --> kx = 2πn/L.
Ex = ħ2kx2/(2M) = 4π2n2ħ2/(2ML2) .]
∆kx = ∆ky = ∆ky = 2π/L.

To each allowed knml there corresponds a wavefunction Φnml(x,y,z).
The tips of the vectors knml divide k-space into elementary cubes of edge length 2π/L.  We have one vector per (2π/L)3 volume of k-space.
The number of vectors in a volume d3k of k-space therefore is dN = d3k/(2π/L)3.
d3k = k2dkdΩ = (1/ħ3)p2dpdΩ = (1/ħ3)d3p.
dN = L3d3p/(2πħ)3.

(b)  For a spin ½ particle the number of states with p between p' and p' + dp' is
dNparticle = 2*4π p'2dp' L3/(2πħ)3.
dNparticle/dp' = 8πL3p'2/(2πħ)3.

E = p2/(2M), dE = pdp/M.
The number of states with E between E' and E' + dE' is
dNparticle = 2½M3/28πL3E'½dE'/(2πħ)3.
dNparticle/dE' = AL3E'½,  A = 2½M3/28π/(2πħ)3.
If all the lowest energy levels are filled, then the energy of the highest occupied state is
Emax = ħ2kmax2/(2M).

The volume of a sphere in 3D k-space is  V = (4/3)πk3.
The number of particle states in the sphere is  Nparticle = 2V/(2π/L)3 = kmax3L3/(3π2).
Emax = ħ2[3Nparticleπ2/L3]2/3/(2M) ∝ (Nparticle/L3)2/3.

The total energy of the system is
Et = ∫0Emax E(dNparticle/dE)dE = AL30EmaxE3/2dE = (2/5)AL3Emax5/2.
Et ∝ L3Emax5/2.
E/L3 = u ∝ Emax5/2 = (Nparticle/L3)5/3 = n5/3.

#### 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:

• 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
Pc = [exp(-2Eα/(kT)) + exp(-2Eβ/(kT))]/[exp(-2Eα/(kT)) + exp(-2Eβ/(kT)) + 2 exp(-(Eα + Eβ)/(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
Pb = [exp(-2Eα/(kT)) + exp(-2Eβ/(kT))]/[exp(-2Eα/(kT)) + exp(-2Eβ/(kT)) + exp(-(Eα+Eβ)/(kT))]
= [exp(-2) + exp(-6]/[exp(-2) + exp(-6) + exp(-4)].
Pb/Pc = [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 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:

• Concepts:
Degeneracy, indistinguishable particles
• Reasoning:
The degeneracy of an energy level of a multi-particle system is different for distinguishable and indistinguishable particles.
• Details of the calculation:
(a)  For each particle, E = (nx + ½)ħω + (ny + ½)ħω + (nz + ½)ħω = (nx + ny + nz + 3/2)ħω,
where ω2 = k/m and nx, ny, nz = 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 nx = ny = nz = 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 ni = 0 and one particle (3 possibilities) must have one ni = 1 (3 possibilities) and the other two ni = 0.

(b)  In a 3D Coulomb potential V(x,y,z) = -Ze2/r the energy for each particle is -Z2*13.6 eV/n2, n = 1, 2, 3, ... .  The energy is independent of l and m.  The possible quantum numbers for the two lowest 1-particle 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 3-particle system are E = -Z2*13.6 eV(1/n12 + 1/n22 +1/n32).
Hence the lowest level for the 3-particle system is E(1) = -3Z2*13.6 eV, n1 = n2 = n2 = 1.
The degeneracy is 1.
In the second lowest level for the 3-particle system, one of the three particles has n = 2,
so E(2) = -(9/4)Z2*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
En = [π2ħ2/(2mL2)] (nx2 + ny2 + nz2), nx, ny, nz = 1, 2, ... .
For a single particle, the lowest level has nx = ny = nz = 1 and energy E1 = 3π2ħ2/(2mL2).
For a single particle, the second lowest level has (nx, ny, nz) = (1, 1, 2), (1, 2, 1), or (2, 1, 1) and
E2 = 6π2ħ2/(2mL2).
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 3-particle system has energy
E(1) = (E1 + E1 + E2) =  6π2ħ2/(mL2).
The degeneracy is 6, since two of the particles are in level E1 with paired spins (1 possible two-particle state) and there are 3 possible sets of quantum numbers for the third particle in level E2, each with 2 possible spins. (6 possible one-particle states).

The second lowest level of the system has energy E(2) = (E1 + E2 + E2) =  15π2ħ2/(2mL2).
The degeneracy = 2*(6*4 + 3) = 54.
For two particles each contributing energy E2, there are 9 possibilities for the spatial quantum numbers, (n1x, n1y, n1z) = (1, 1, 2), (1, 2, 1), or (2, 1, 1) (3 possibilities) times
(n2x, n2y, n2z) = (1, 1, 2), (1, 2, 1), or (2, 1, 1) (3 possibilities).
When n1i = n2i 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 two-particle 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 two-particle states.
For the one particle contributing energy E1 we have nx = ny = nz = 1, and the spin can be up or down (2 possible one-particle states).

#### 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:

• 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+1LJ.  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, ψspace or χspin, 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+1LJ for Titanium therefore are
1G4, 3F4,3,2, 1D2, 3P2,1,0, 1S0.
Hund's rule:
The level with the largest multiplicity has the lowest energy.  (3F4,3,2 and 3P2,1,0)
For a given multiplicity, the level with the largest value of L has the lowest energy. (3F4,3,2)
For less than half-filled shells:
The component with the smallest value of J has the lowest energy (3F2).
The ground state tem is 3F2.