Assignment 11

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) = -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 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 + ½)ħω + (n_y + ½)ħω + (n_z + ½)ħω = (n_x + n_y + n_z + 3/2)ħω,
where ω² = 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 3*3 = 9, because one particle (3 particles - 3 possibilities) must have one n_i = 1 (3 n_i - 3 possibilities) and the other two n_i = 0. The other two particles must have n_x = n_y = n_z= 0 (1 possibility).

(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 must have n_i = 2
(4 possibilities), and the other two n_i = 1 (3 possibilities).

(c) For a single particle in a cubic box
E_n = [π²ħ²/(2mL²)] (n_x² + n_y² + n_z²), 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 E₁ = 3π²ħ²/(2mL²).
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
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*1, 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*(3*4 + 3) = 30.
For two particles each contributing energy E₂, there are 6 possibilities for the spatial quantum numbers, (n_1x, n_1y, n_1z) = (1, 1, 2), (1, 2, 1), or (2, 1, 1) and (n_2x, n_2y, n_2z) = (1, 1, 2), (1, 2, 1), or (2, 1, 1), they can be the same (3 possibilities), or they can be different (3 possibilities).
When n_1i= n_2i for all i (3 of the 3 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 3*4 = 12 possible two-particle states.
For the one particle contributing energy E₁ we have n_x = n_y = n_z= 1, and the spin can be up or down (2 possible one-particle states).

Problem 2:

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/|r₁- 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/|r₁- 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 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 e²/|r₁- r₂| term in your qualitative discussion.

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 = H₀ + H', H₀ = P₁²/(2m) + P₂²/(2m) - 2e²/r₁ - 2e²/r₂, H' = e²/|r₁- r₂|.
Let us make a central field approximation.
(a) Assume both electrons are in the n = 1 state.
Then ψ_space = ψ₁₀₀(r₁)ψ₁₀₀(r₂) = 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 = ψ₁₀₀(r₁)ψ₁₀₀(r₂)χ_2m or ψ = ψ₁₀₀(r₁)ψ₁₀₀(r₂)χ₀₀.
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^-½(ψ₁₀₀(r₁)ψ₂₀₀(r₂) + ψ₁₀₀(r₂)ψ₂₀₀(r₁)) = ψ_ex(even),
or
ψ_space = 2^-½(ψ₁₀₀(r₁)ψ₂₀₀(r₂) - ψ₁₀₀(r₂)ψ₂₀₀(r₁)) = ψ_ex(odd).
We need ψ = ψ_ex(even)*χ(even) or ψ = ψ_ex(odd)*χ(odd).
ψ = ψ_ex(e)*χ_2m or ψ = ψ_ex(e)*χ₀₀ or ψ = ψ_ex(o)χ_1m.
The the states with the antisymmetric ψ_space have smaller <H'> = <e²/|r₁- r₂|> since they vanish at r₁ = r₂.
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)χ₀₀    (degeneracy: 5 + 1)
excited state -------
                               ------- ψ_ex(o)_χ1m                         (degeneracy: 3)

ground state   ------- ψ_gχ_2m or ψ_gχ₀₀                          (degeneracy: 5 + 1)

Problem 3:

Consider an excited Nitrogen atom with a an electronic configuration (1s)², (2s)², (2p)², (3s).
Find the spectroscopic terms ^2S+1L_J that characterize the system.
Solution:

Concepts:
The Pauli exclusion principle, energy levels of multi-electron atoms, addition of angular momentum
Reasoning:
The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers. We use it to find the allowed terms ^2S+1L.

Details of the calculation:
For identical fermions the total wave function must be antisymmetric under exchange of any two particles.
Let (o) denote antisymmetric or odd, (e) denote symmetric or even.
First add the angular momentum of the two 2p electrons.

(3p)²		Possible terms
L₁₂	S₁₂
0 (e)	0 (o)	¹S
1 (o)	1 (e)	³P
2 (e)	0 (o)	¹D

Now add to this the angular momentum of the 3s electron.

		(3p)²(3s)				Possible terms
L₁₂	S₁₂	l	s	L, (L = L₁₂ + l)	S, (S = S₁₂ + s)
0	0	0	½	0	½	²S
1	1	0	½	1	½, 3/2	²P, ⁴P
2	0	0	½	2	½	²D

Spectroscopic terms:

^2S+1L_J

²S_1/2

²P_{3/2,
1/2}

⁴P_{5/2,
3/2, 1/2}

²D_{5/2,
3/2}

Problem 4:

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 = dp_xdp_ydp_z (that is p_x between p_x and p_x+ dp_x 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 ∝ n^5/3.

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
Φ_nml(x,y,z) = (1/L)^3/2exp(ik_xx)exp(ik_yy)exp(ik_zz),
with k_x = 2πn/L, k_y = 2πm/L, k_z = 2πl/L, n, m, l = ±1, ±2, ... .
The associated eigenvalues are
E_nml = (n² + m² + l²)4π²ħ²/(2ML²) = (k_x² + k_y² + k_z²)ħ²/(2M).
[H = p_x²/(2M) + p_y²/(2M) + p_z²/(2M), if x, y, z < L.
Φ(x,y,z) = f(x)c(y)g(z).
(-ħ²/(2M))(∂²/∂x²)f(x) = E_xf(x).
f(x) = (1/L)^½exp(ik_xx), f(x) = f(x + L) --> k_x = 2πn/L.
E_x = ħ²k_x²/(2M) = 4π²n²ħ²/(2ML²) .]
∆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 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
dN_particle = 2*4π p'²dp' L³/(2πħ)³.
dN_particle/dp' = 8πL³p'²/(2πħ)³.

E = p²/(2M), dE = pdp/M.
The number of states with E between E' and E' + dE' is
dN_particle = 2^½M^3/28πL³E'^½dE'/(2πħ)³.
dN_particle/dE' = AL³E'^½, A = 2^½M^3/28π/(2πħ)³.
If all the lowest energy levels are filled, then the energy of the highest occupied state is
E_max = ħ²k_max²/(2M).
The volume of a sphere in 3D k-space is V = (4/3)πk³.
The number of particle states in the sphere is N_particle = 2V/(2π/L)³ = k_max³L³/(3π²).
E_max = ħ²[3N_particleπ²/L³]^2/3/(2M) ∝ (N_particle/L³)^2/3.

The total energy of the system is
E_t= ∫₀^Emax E(dN_particle/dE)dE = AL³∫₀^EmaxE^3/2dE = (2/5)AL³E_max^5/2.
E_t ∝ L³E_max^5/2.
E/L³ = u ∝ E_max^5/2 = (N_particle/L³)^5/3 = n^5/3.