Assignment 12, solutions

Problem 1:

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)˝.
(c)  indistinguishable non-interacting spin ˝ particles in a 3-dimensional cubic box (with impenetrable walls) of dimensions Lx × Ly × Lz, where Lx = Ly = Lz = L.


Degeneracy, indistinguishable particles
The degeneracy of an energy level of a multi-particle system is different for distinguishable and indistinguishable particles.
bulletDetails of the calculation:
(a)  For each particle, E = (nx + ˝)ħω + (ny + ˝)ħω + (nz + ˝)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) = (1˝)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 ˝ 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 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).
For the two other 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 spins must be paired (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.

Problem 2:

Consider N >> 1 non-interacting spin-˝ particles with mass M confined to a cubical box of volume V.
(a)  Derive an expression for the Fermi energy.
(b)  Make some numerical estimates, assuming in each case that the particles in question are non-interacting, for the Fermi energy of
(i)  electrons in a typical metal,
(ii)  nucleons in a large nucleus,
(iii)  3He atoms in liquid 3He, which has an atomic volume of about 0.05 nm3 per atom.


The Pauli exclusion principle, the Fermi energy
The Fermi energy is the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at T = 0.
bulletDetails of the calculation:
(a)  The eigenfunctions and eigenvalues of the 3D infinite cubic well are
Φnm(x,y,z) = (2/L)3/2sin(kxx)sin(kyy)sin(kzz),
with kx = πn/L,  ky = πm/L,  kz = πp/L,   n, m, p = 1, 2, ...  .
The associated eigenvalues are
Enm = (n2 + m2 + p22ħ2/(2ML2) = (kx2 + ky2 + kz22/(2M).
The spacing between successive allowed values of ki is  ∆kx = ∆ky = ∆ks = π/L.  Only positive values of kx, ky, and kz are allowed.
If k is large, then the number of states with wave vectors whose magnitudes is less or equal to k is N = (1/8)(4π/3)k3/(π/L)3.
Each state can be occupied by two particles, one with spin up and one with spin down.
The energy of the highest occupied state is EF = ħ2k2/(2M), with k2 = (3Nπ2/V)2/3.
EF = ħ2(3Nπ2/V)2/3/(2M) = ħ2(3nπ2)2/3/(2M), where n is the number of particles per unit volume.

(b)  (i)  The spacing between metal atoms is a few Angstrom and each atom contributes on the order of 1 conduction electron, so we have n ~  1/(10-29 m3).  The electron mass is 9.1*10-31 kg, so EF ~ 10 eV.
(ii)  The “radius” of a nucleon is ~1 fm, so the density of protons or neutrons is on the order of 1/10-44 m3.  The mass of a nucleon is 1.67*10-27 kg, so EF ~ 40 MeV.
(iii)  The density is  ~5*1025/m3, and the mass is ~3*1.67*10-27 kg, so EF ~10-5 eV.

Problem 3:

Consider a helium atom where both electrons are replaced by identical charged particles of spin quantum number s = 1.  Ignoring the motion of the nucleus and the spin-orbit interaction, the Hamiltonian is given by
H = P12/(2m) + P22/(2m) - 2e2/r1 - 2e2/r2 + e2/|r- r2|.
Construct an energy level diagram (qualitatively) for this "atom", when

(a) both particles are in the n = 1 state, and when
(b) one particle is in the n = 1 state and the other is in the state (n l m) = (2 0 0). 

Do this by treating the e2/|r- r2| term in the Hamiltonian as a perturbation.  Write out the space and spin wave functions for each level in terms of the single particle hydrogenic wave functions ψnlm and spin wave functions χs,ms.  Show the splitting qualitatively, and state the degeneracy of each level.  Don’t forget to include the effect of the e2/|r- r2| term.


Indistinguishable particles
Spin 1 particles are bosons.  For identical bosons, the state must be symmetric under the exchange of the two particles.
bulletDetails of the calculation:
H = H0 + H',  H0 = P12/(2m) + P22/(2m) - 2e2/r1 - 2e2/r2,  H' = e2/|r- r2|.
Let us make a central field approximation.
(a)  Assume both electrons are in the n = 1 state. 
Then ψspace = ψ100(r1100(r2) = ψg.  The orbital part of the wave function is symmetric under exchange of the two electrons.
Possible spin quantum numbers are S = 2, 1, 0,  The S = 2 and 0 spin states are even under the exchange of the particles and the S = 1 state is odd.
The total ψ must be symmetric under exchange, so we need S = 2 or 0.
ψ = ψ100(r1100(r22m  or ψ = ψ100(r1100(r200.
The ground state is 5 +1 = 6 fold degenerate.
(b)  Assume one electron is in the |100> and the other in the |200> state.
Then ψspace  can be symmetric or antisymmetric.
ψspace = 2100(r1200(r2) + ψ100(r2200(r2)) =  ψex(even)
or ψspace = 2100(r1200(r2) - ψ100(r2200(r2)) = ψex(odd).
We need ψ = ψex(even)χ(even) or ψ = ψex(odd)χ(odd).
ψ = ψex(even)χ2m or ψ = ψex(even)χ00 or ψ = ψex(odd)χ1m.
The the states with the antisymmetric ψspace have smaller <H'> = <e2/|r- r2|> since they vanish at r1 = r2.
They therefore lie lower in energy than  the states with the symmetric ψspace.
So for the energy level diagram we have
                               ------- ψex(e)χ2m or ψexe(e)χ00    (degeneracy: 5 + 1)
excited state -------
                               ------- ψex(o)χ1m                         (degeneracy: 3)

ground state   -------  ψgχ2m  or ψgχ00                          (degeneracy:  5 + 1)

Problem 4:

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.


The Pauli exclusion principle, energy levels of multi-electron atoms
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.
bulletDetails 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 he 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 3 3d electrons.
L = 4, 3, 2, 1, and 0 terms are possible.  For two equivalent electrons, ψspace or χspin for 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.

Problem 5:

Consider a system of 4 identical particles.  Each particle has 3 possible eigenvalues, E1, E2, E3, of some observable.
Write down the normalized symmetric wave function in which two of the particles have eigenvalue E1, one has eigenvalue E2, and one has eigenvalue E3.

12 possible states


Identical particles
For identical particles we have symmetrization requirements.
For identical bosons, the state must be symmetric under the exchange of the two particles.
bulletDetails of the calculation:
We have 3 states and 4 particles.  Let us use the following notation.
|a, b, c, d>:  Let a denote the state of particle 1, b the state of particle 2, and c the state of particle 3 and d the state of particle 4.
There are 12 possible states.
ψ = 12[|E1, E1, E2, E3> + |E1, E2, E1, E3> + |E1, E3, E2, E1> + |E2, E1, E1, E3> + |E2, E1, E3, E1> + |E2, E3, E1, E1>
             + |E1, E1, E3, E2> + |E1, E3, E1, E2> + |E1, E2, E3, E1> + |E3, E1, E1, E3> + |E3, E1, E2, E1> + |E3, E2, E1, E1>]