**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 k_{x} = k_{y} = k_{z} = k.

(b) distinguishable
non-interacting spinless particles in a 3-dimensional Coulomb potential

V(x,y,z) = –Ze^{2}/r, where r = (x^{2 }+ y^{2 }+ z^{2})^{˝}.

(c) indistinguishable
non-interacting spin ˝ particles in a 3-dimensional 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 multi-particle system is different for distinguishable and indistinguishable particles. | |

Details of the calculation: (a) For each particle, E = (n _{x} + ˝)ħω + (n_{y} + ˝)ħω + (n_{z}
+ ˝)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)} = (1˝)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 1-particle energy levels aren = 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 = –Z ^{2}*13.6 eV(1/n_{1}^{2}
+ 1/n_{2}^{2} +1/n_{3}^{2}).Hence the lowest level for the 3-particle 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 3-particle 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 ˝ 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)} = (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 two-particle 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 one-particle
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 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 one-particle states).
For the two other 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 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) ^{3}He atoms in liquid ^{3}He, which has an atomic volume
of about 0.05 nm^{3} per atom.

Solution:

Concepts: The Pauli exclusion principle, the Fermi energy | |

Reasoning: 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. | |

Details of the calculation: (a) The eigenfunctions and eigenvalues of the 3D infinite cubic well are Φ _{nm}(x,y,z) = (2/L)^{3/2}sin(k_{x}x)sin(k_{y}y)sin(k_{z}z),with k _{x} = πn/L, k_{y} = πm/L, k_{z} = πp/L, n, m,
p = 1, 2, ... .The associated eigenvalues are E _{nm} = (n^{2} + m^{2} + p^{2})π^{2}ħ^{2}/(2ML^{2})
= (k_{x}^{2} + k_{y}^{2} + k_{z}^{2})ħ^{2}/(2M).The spacing between successive allowed values of k _{i} is ∆k_{x}
= ∆k_{y} = ∆k_{s} = π/L. Only positive values of k_{x},
k_{y}, and k_{z} 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)k ^{3}/(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 E _{F} = ħ^{2}k^{2}/(2M),
with k^{2} = (3Nπ^{2}/V)^{2/3}.E _{F} = ħ^{2}(3Nπ^{2}/V)^{2/3}/(2M)
= ħ^{2}(3nπ^{2})^{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 }m^{3}).
The electron mass is 9.1*10^{-31} kg, so E_{F} ~ 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} m^{3}. The mass of a
nucleon is 1.67*10^{-27} kg, so E_{F} ~
40 MeV.(iii) The density is ~5*10 ^{25}/m^{3}, and the mass is
~3*1.67*10^{-27} kg, so E_{F} ~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 = P_{1}^{2}/(2m) + P_{2}^{2}/(2m) - 2e^{2}/r_{1}
- 2e^{2}/r_{2} + e^{2}/|**r**_{1 }-** r**_{2}|.

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
e^{2}/|**r**_{1 }-** r**_{2}| 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 ψ_{nlm}
and spin wave functions χ_{s,ms}. Show the splitting qualitatively,
and state the degeneracy of each level. Don’t forget to include the effect of the
e^{2}/|**r**_{1 }-** r**_{2}| term.

Solution:

Concepts: Indistinguishable particles | |

Reasoning: Spin 1 particles are bosons. For identical bosons, the state must be symmetric under the exchange of the two particles. | |

Details of the calculation: H = H _{0} + H', H_{0} = P_{1}^{2}/(2m) + P_{2}^{2}/(2m) - 2e^{2}/r_{1}
- 2e^{2}/r_{2}, H' = e^{2}/|r_{1 }- r_{2}|.Let us make a central field approximation. (a) Assume both electrons are in the n = 1 state. Then ψ _{space} = ψ_{100}(r_{1})ψ_{100}(r_{2})
= ψ_{g}. 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 = ψ _{100}(r_{1})ψ_{100}(r_{2})χ_{2m}
or ψ = ψ_{100}(r_{1})ψ_{100}(r_{2})χ_{00}.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 ψ _{space} can be symmetric or antisymmetric.ψ _{space} = 2^{-˝}(ψ_{100}(r_{1})ψ_{200}(r_{2})
+ ψ_{100}(r_{2})ψ_{200}(r_{2}))
= ψ_{ex}(even)or ψ _{space} = 2^{-˝}(ψ_{100}(r_{1})ψ_{200}(r_{2})
- ψ_{100}(r_{2})ψ_{200}(r_{2}))
= ψ_{ex}(odd).We need ψ = ψ _{ex}(even)χ(even) or ψ = ψ_{ex}(odd)χ(odd).ψ = ψ _{ex}(even)χ_{2m} or ψ = ψ_{ex}(even)χ_{00}
or ψ = ψ_{ex}(odd)χ_{1m}.The the states with the antisymmetric ψ _{space} have smaller
<H'> = <e^{2}/|r_{1 }- r_{2}|>
since they vanish at r_{1} = r_{2}.They therefore lie lower in energy than the states with the symmetric ψ _{space}.So for the energy level diagram we have ------- ψ _{ex}(e)_{χ2m} or ψ_{exe}(e)χ_{00}
(degeneracy: 5 + 1)excited state ------- ------- ψ _{ex}(o)_{χ1m}
(degeneracy: 3)ground state ------- ψ _{g}χ_{2m}
or ψ_{g}χ_{00}
(degeneracy: 5 + 1) |

**Problem 4:**

For the Titanium atom (Z = 22) in its ground state find the allowed terms
^{2S+1}L_{J} 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+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 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, ψ _{space} or χ_{spin} 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+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 half-filled shells: The component with the smallest value of J has the lowest energy ( ^{3}F_{2}).The ground state tem is ^{3}F_{2}. |

**Problem 5:**

Consider a system of 4 identical particles. Each particle has 3 possible
eigenvalues, E_{1}, E_{2}, E_{3}, of some observable.

Write down the normalized symmetric wave function in which two of the particles
have eigenvalue E_{1}, one has eigenvalue E_{2}, and one has
eigenvalue E_{3}.

12 possible states

Solution:

Concepts: Identical particles | |

Reasoning: For identical particles we have symmetrization requirements. For identical bosons, the state must be symmetric under the exchange of the two particles. | |

Details 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 ^{-˝}[|E_{1}, E_{1}, E_{2}, E_{3}>
+ |E_{1}, E_{2}, E_{1}, E_{3}> + |E_{1},
E_{3}, E_{2}, E_{1}> + |E_{2}, E_{1},
E_{1}, E_{3}> + |E_{2}, E_{1}, E_{3},
E_{1}> + |E_{2}, E_{3}, E_{1}, E_{1}>+ |E _{1},
E_{1}, E_{3}, E_{2}> + |E_{1}, E_{3},
E_{1}, E_{2}> + |E_{1}, E_{2}, E_{3},
E_{1}> + |E_{3}, E_{1}, E_{1}, E_{3}>
+ |E_{3}, E_{1}, E_{2}, E_{1}> + |E_{3},
E_{2}, E_{1}, E_{1}>] |