Assignment 13, solutions

Problem 1:

Solve the Schroedinger equation for a particle 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/(2ph)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 µ n5/3.

Solution:

Concepts, principles, relations that apply to the problem:
The three-dimensional, infinite square well, the Pauli exclusion principle, the density of states
Why do they apply?
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.
How do they apply?
(a) The eigenfunctions and eigenvalues of the 3D infinite square well with periodic boundary conditions are
ynml(x,y,z) = (1/L)3/2exp(ikxx)exp(ikyy)exp(ikzz), 
with kx = 2pn/L,  ky = 2pm/L,  kz = 2pl/L, n, m, l = 1, 2, ... .
The associated eigenvalues are
Enml = (n2 + m2 + l2)4p2h2/(2ML2) = (kx2 + ky2 + kz2)h2/(2M)
[H = px2/(2m) + py2/(2m) + pz2/(2m), if x, y , z < L.
y(x,y,z) = f(x)c(y)g(z).
(-h2/(2m))(2/x2)f(x) = Exf(x).
f(x) = (1/L)1/2exp(ikxx),  f(x) = f(x + L) --> kx = 2pn/L.
Ex = h2kx2/(2m) = 4p2n2h2/(2mL2) .]
 Dkx = Dky = Dkz = 2p/L.

To each allowed knml there corresponds a wavefunction ynml(x,y,z).
The tips of the vectors knml divide k-space into elementary cubes of edge length 2p/L.  We have one vector per (2p/L)3 volume of k-space.
The number of vectors in a volume d3k of k-space therefore is dN = d3k/(2p/L)3.
d3k = k2dkdW = (1/h3)p2dpdW = (1/h3)d3p.
dN = L3d3p/(2ph)3.
The number of vectors in a spherical volume of of radius k in k-space is
N = V/(2p/L)3 = (4p/3)k3L3/(2p)3

(b)  For a spin 1/2 particle the number of states with p between p' and p' + dp' is
dNparticle = 2*4p(L3/(2ph)3)p2dp.
dNparticle/dp = 8pL3p2/(2ph)3).
E = p2/(2m), dE = pdp/m.
The number of states with E between E' and E' + dE' is 
dNparticle = 21/2m3/28pL3E1/2dE/(2ph)3.
dNparticle/dE = AL3E1/2,  A = 21/2m3/28p/(2ph)3.
If all the lowest energy levels are filled, then the energy of the highest occupied state is
Emax = h2k2/(2m).
Using  Nparticle/2 = (4p/3)k3L3/(2p)3 we have
Emax = (h2/(2m))[3Nparticlep2/L3]2/3 µ (Nparticle/L3)2/3.

The total energy of the system is 
Et = ò0EmaxE(dNparticle/dE)dE = AL3ò0EmaxE3/2dE = (2/5)AL3Emax5/2.
Et µ L3Emax5/2.
E/L3 = u µ Emax5/2 = (Nparticle/L3)5/3 = n5/3.

Details of the calculation:
None

Problem 2:

The Hamiltonian for two interacting spin ½ identical fermions in one dimension is
H = p12/(2m) + p22/(2m) + (mw2/2)(x2 - x1)2.
What is the energy spectrum and what are the corresponding eigenfunctions?

Solution:

Concepts, principles, relations that apply to the problem:
Two interacting indistinguishable particles, center of mass and relative motion
Why do they apply?
To solve the problem we separate the total Hamiltonian into a Hamiltonian for motion of the center of mass and a Hamiltonian for the motion about the center of mass.  The total state vector must be antisymmetric under the exchange of the particles.
How do they apply?
(a)  Let R = ½(x1+x2) and r = (x2-x1).
T = ½m[ (dx1/dt)2 +  (dx2/dt)2]
= (1/4)m[(dx1/dt + dx2/dt)2 + (dx1/dt - dx2/dt)2]
= m(dR/dt)2 + (1/4)m(dr/dt)2 = mV2 + (1/4)mv2.
U = (mw2/2)(x2-x1)2 = (mw2/2)r2 = ½Br2.
L = T - U

P = L/V = 2mV, p = L/v = (1/2)mv.
H = P2/(2M) + p2/(2m) + ½Br2.
Here M = 2m, m = m/2, and B = mw2.
To make the transition to Quantum Mechanics we let the canonical variables R and P and the canonical variables r and p become operators.
Hy(R,r) = [(-h2/(2M) )(2/R2) + (-h2/(2m))(2/r2) + ½Br2]y(R,r) = Ey(R,r)
is the eigenvalue equation for H.
Separation of variables is possible.
Let y(R,r) = c(R)f(r).
Then
(-h2/2M )(2/R2)c(R) = ERc(R), c(R) = exp(ikR),  ER = h2k2/(2M).
The solutions are plane wave solutions.  There exists a solution for every k.
[(-h2/2m)(2/r2) + ½Br2]f(r) = Erf(r), Er = (nr + ½)hwr, wr2 = B/m = 2w2.
The solutions are the solutions of the 1D harmonic oscillator.
E = ER + Er, nr = 0, 1, 2, ... .
E =  h2k2/(2M) + (nr + ½)21/2hw.
Details of the calculation:
Under exchange R --> R, r --> -r.  Interchanging the two particles leaves R unchanged, but changes the sign of r.
fnr(r) is symmetric under exchange if nr = even and fnr(r) is antisymmetric if nr = odd.
For identical fermions the total wavefunction must be antisymmetric under the exchange of the two particles.
Therefore, two spin 1/2 particles in an orbital state with nr = even must be in the singlet (S = 0) spin state, and two spin 1/2 particles in an orbital state with nr = odd must be in the triplet (S = 1) spin state.

Problem 3:

Scandium has a ground state configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d.  Consider a transition from the ground state 4s2 3d to the excited state 3d 4s 4p.
(a) Assuming LS coupling find L, S, and J values for the terms derived from the 4s2 3d and 3d 4s 4p configurations.  Write down each term in standard notation.
(b) Write down dipole selection rules for LS coupling and indicate the allowed transitions.

Solution:

Concepts, principles, relations that apply to the problem:
Atomic spectra, addition of angular momentum, selection rules
Why do they apply?
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+1LJ.  We then can use the dipole selection rules to find the allowed transitions.

How do they apply?
(a) For the initial state we have:

  4s2 3d
L 0 2
S 0 1/2

LS coupling: L = 2; S = 1/2
Possible terms: 2D5/2, 2D3/2
 (The Pauli exclusion principle requires that L = S = 0 for closed shells.)

 For the final state we have:

  3d 4s 4p
L 2 0 1
S 1/2 1/2 1/2

LS coupling: L = 3, 2, 1; S = 3/2,1/2
Possible terms: 4F9/2,7/2,5/2,3/2; 2F7/2,5/2,; 4D7/2,5/2,3/2,1/2; 2D5/2,3/2; 4P5/2,3/2,1/2; 2P3/2,1/2
(No two electrons have the same quantum numbers n and l.)

(b) LS coupling selection rules for allowed transitions:
DL = ±1, DJ = 0,±1, DMJ = 0,±1, DS = 0.

Ground state ß   à Excited state
2D5/2   2F7/2,5/2,  2P3/2
2D3/2   2P3/2,1/2,  2F5/2,
Details of the calculation:
None

Problem 4:

For the Nitrogen atom find the three terms in the LS coupling scheme which lie lowest in energy.

Solution:

Concepts, principles, relations that apply to the problem:
The energy levels of multi-electron atoms, angular momentum coupling, Hund's rules
Why do they apply?
We use the Pauli exclusion principle and the rules for adding angular momentum to find the allowed terms 2S+1L.
How do they apply?
The Nitrogen atom has 7 electrons.
Electronic configuration, 1s2 2s2 2p3 = [He] 2s2 2p3
The 1s and 2s shells are filled and have L = 0, S = 0.
The 2p shell is half filled.
The possible terms 2S+1L for 3 equivalent p electrons (np)3 are 4S, 2P, and 2D.
The p -electron can have:
 
ml = 1 0 -1 1 0 -1
ms = 1/2 1/2 1/2 -1/2 -1/2 -1/2
label state as 1 2 3 4 5 6

To satisfy the Pauli exclusion principle a configuration must not contain identical electrons.
Let us write down all possible configurations with non-negative ML = Sml and MS = Sms.
 

State ML MS
123 0 3/2
124 2 1/2
125 1 1/2
126 0 1/2
134 1 1/2
135 0 1/2
234 0 1/2

Pick the states starting with the largest ML and the largest MS.
ML = 2, Ms = 1/2.  The presence of the ML = 2, Ms = 1/2 term shows that a 2D term is among the possible terms.  To this term we must further assign states with ML = 1,0, MS = 1/2.
What is left?  ML = 1, Ms = 1/2.  The presence of the ML = 1, Ms = 1/2 term shows that a 2P term is among the possible terms.  To this term we must further assign states with ML = 0, MS = 1/2.
What is left?  ML = 0, Ms = 3/2.  The presence of the ML = 0, Ms = 3/2 term shows that a 4S term is among the possible terms.  To this term we must further assign states with ML = 0, MS = 1/2.

Details of the calculation:
Hund's rule:
The level with the largest multiplicity has the lowest energy.
For a given multiplicity, the level with the largest value of L has the lowest energy.
When a shell is half filled, there is no multiplet splitting.
E(4S) < E(2D) < E(2P).

Problem 5:

Two neutrons are confined two a cubical box whose sides are 1Å in length.  The particles interact strongly, whenever the distance between them is less than 10-15m.  Approximate the interaction potential by V(r1-r2) = -(4pR3V0/3)d(r1-r2), with R = 10-15m, and use perturbation theory to calculate the ground state energy to first order and ground state wave function to zeroth order.

Solution:

Concepts, principles, relations that apply to the problem:
A particle in a 3D square well, stationary perturbation theory, identical fermions
Why do they apply?
We have two neutrons in 3D square well.  The Hamiltonian is perturbed.  We must use properly antisymmetrized eigenfunctions of H0 when we calculate correction terms using perturbation theory.
How do they apply?
The eigenfunctions and eigenvalues of the 3D infinite square well with side length L are
fnml(x,y,z) = (2/L)3/2sin(kxx)sin(kyy)sin(kzz),
with kx = pn/L,  ky = pm/L,  kz = pl/L,   n, m, l = 1, 2, ... .
The associated eigenvalues are
Enml = (n2 + m2+ l2)p2h2/(2ML2).
The ground state has n = m = l = 1 and
f111(x,y,z) = (2/L)3/2sin(px/L)sin(py/L)sin(pz/L),
E0 = 3p2h2/(2ML2) = 5.5*10-2 eV
For two non-interacting particles we write
y = f(r1,r2)cS(1,2).
The total wavefunction must be antisymmetric under the exchange of the two neutrons.  The spin function cS(1,2) is symmetric for S = 1 (triplet state) and antisymmetric for S = 0 (singlet state).
The space function f(r1,r2) therefore must be antisymmetric for S = 1 and symmetric for S = 0.
For the space eigenfunctions of the unperturbed Hamiltonian we therefore have in general
f(r1,r2) = (2)-1/2(fnml(x1,y1,z1)fn'm'l'(x2,y2,z2) + fnml(x2,y2,z2)fn'm'l'(x1,y1,z1))
if S = 0 and
f(r1,r2) = (2)-1/2(fnml(x1,y1,z1)fn'm'l'(x2,y2,z2) - fnml(x2,y2,z2)fn'm'l'(x1,y1,z1))
if S = 1.
In general, a state characterized by the quantum numbers n,l,m, n', l', m' is two-fold degenerate.  The ground state, however, is not degenerate. 
The ground state wavefunction to zeroth order is
 f0(r1,r2) = f111(x1,y1,z1)f111(x2,y2,z2), and S = 0.
The first order correction to the energy is given by
 DE = -(4pR3V0/3)òòf*0(r1,r2)d(r1-r2)f0(r1,r2)d3r1d3r2
= -(4pR3V0/3)ò|f0(r1,r1)|4d3r1 = -(4pR3V0/3)ò|f111(x,y,z)|4dxdydz
= -(9pR3V0/(2L3)).
R/L ~ 10-5, (R/L)3 ~ 10-15.
Ground state energy:  E0 = 2*3p2h2/(2ML2) - 10-15(9pV0/2).
Details of the calculation:
None

Problem 6:

Two identical spin-one-half particles are bound to one another to form a composite particle by a potential

V = ½mw2r2

where r = |r1 – r2|.
(a)  Find the allowed energy levels and the degeneracy of the three lowest lying allowed states.  Hint. You may assume that for the oscillator Ekl = hw(2k + l + 3/2),  k, l = 0, 1, 2, …,  where l and k are the angular momentum and radial quantum number, respectively.
(b)  How do these change with the addition of the perturbation

U = CL · S,

where C is a constant?  What are possible spin and orbital angular momentum eigenvalues?

Solution:

Concepts, principles, relations that apply to the problem:
Two interacting indistinguishable particles, addition of angular momentum
Why do they apply?
The total state vector must be antisymmetric under the exchange of the particles.  This determines the allowed combinations of l, s, and j.
How do they apply?
(a)  Ground state: n = 0, l = 0, s = 0.  E00 = (3/2)hw.
The symmetry of the space function is (-1)l.  It is determined from the symmetry properties of the spherical harmonics.  The total wave function must be anti-symmetric under exchange of the two fermions, so the particles must be in the singlet state.  The ground state is not degenerate.

The next allowed energy level has n = 0, l = 1, s = 1.  E01 = (5/2)hw.
The space function is anti-symmetric and the spin function is symmetric. The magnetic quantum numbers m and ms can both take on the values -1, 0, and +1.  The energy level is 9-fold degenerate.

The next allowed energy level can have n = 1, l = 0, s = 0, or n = 0, l = 2, s = 0.
E10 = E02 = (7/2)hw.
The level is six-fold degenerate.

Details of the calculation:
(b)  L · S = ½ (J2 – L2 – S2)
The eigenvalues of L2 are l(l+1)h2, the eigenvalues of S2 are s(s+1)h2, and the eigenvalues of J2 are j(j+1)h2. 

For the ground state l = s = j = 0, DE = 0.  There is no first order energy correction.
For the first excited energy level we have l = 1, s = 1, j = 0, 1, 2.
The perturbation splits the level into three sublevels:
j = 0: no degeneracy, first order energy correction DE = -2Ch2.
j = 1: 3-fold degenerate, first order energy correction DE = -Ch2.
j = 2: 5-fold degenerate, first order energy correction DE = Ch2.

For the next excited energy level we have l = 0, s = 0, j = 0, or l = 2, s = 0, j = 2.
We have:
j = 0: no degeneracy, first order energy correction DE = 0.
j = 2: 5-fold degenerate, first order energy correction DE = 0.
The level is still six-fold degenerate in first order.

Problem 7:

Consider N non-interacting bosons of mass m in a 3D rigid cubical box with volume a3.  What is the ground state energy in terms of h, N, m and a?

Solution:

Concepts, principles, relations that apply to the problem:
The infinite square well in 3D.
Why do they apply?
The bosons are confined to an infinite square well in 3D.  They are allowed to all occupy the lowest possible energy state of this potential.
How do they apply?
The energy eigenfunctions of a particle in the three-dimensional, infinite square well are
fnx,ny,nz(x,y,z) = (2/a)3/2sin(nxpx/a)sin(nypy/a)sin(nzpz/a)
with eigenvalues
Enx,ny,nz = (nx2 + ny2 +nz2)p2h2/(2ma2),  nx, ny, nz = 1, 2, 3, ... .
In the ground state the quantum numbers of all bosons are nx = ny = nz = 1.
The ground-state energy of the N bosons therefore is E = 3Np2h2/(2ma2).
Details of the calculation:
None