Problem 1:
Two identical spin zero bosons are placed in a one-dimensional infinite square
well, U(x) = ∞, x < 0 and x > a, U(x) = 0, 0 < x < a. The bosons interact
weakly with one another via the potential energy function U(x1,x2)
= -aU0δ(x1 - x2), where U0 is a
constant with the dimensions of energy.
(a) First, ignoring the interaction between the particles, find the ground
state and first excited state wave functions and the associated energies.
(b) Use first-order perturbation theory to estimate the effect of the
particle-particle interaction on the energies of the ground state and the first
excited state.
Solution:
-
Concepts:
Identical particles, stationary perturbation theory
-
Reasoning:
For identical bosons, the state must be symmetric under the exchange of the two
particles.
- Details of the calculation:
(a) ψg(x1,x2) = ψ1(x1) ψ1(x2).
ψex(x1,x2) = 2-½(ψ1(x1)
ψ2(x2) + ψ1(x2) ψ2(x1)),
Eg = 2E1, Eex = E1 + E2,
where ψn(x) = (2/a)½sin(nπx/a). En = n2π2ħ2/(2ma2)
(b) The unperturbed ground state energy is Eg = 2E1, the
unperturbed energy of the first excited state is Eex = E1
+ E2. Both energy levels are non-degenerate.
The first order energy correction to the ground state is
Eg1 = -aU0∫dx1∫dx2 ψ1(x1)
ψ1(x2) δ(x1 - x2) ψ1*(x1)
ψ1*(x2) = -aU0∫dx|ψ1(x)|4
= -(4U0/a)∫0adx sin4(πx/a).
Eg1 = -(3/2)U0.
(∫0 π sin4(x) dx = 3π/8)
To first order, the ground state energy is lowered by Eg1
to Eg = 2π2ħ2/(2ma2) - (3/2)U0.
The first order energy correction to the first excited state is
Eex1 = -½aU0∫dx1∫dx2 (ψ1(x1)
ψ2(x2) + ψ1(x2) ψ2(x1))
δ(x1 - x2) (ψ1*(x1) ψ2*(x2)
+ ψ1*(x2) ψ2*(x1))
= -½aU0∫dx1 (ψ1(x1) ψ2(x1)
+ ψ1(x1) ψ2(x1)) (ψ1*(x1)
ψ2*(x1) + ψ1*(x1) ψ2*(x1))
= -2aU0∫dx|ψ1(x)|2|ψ2(x)|2
= -(4U0/a)∫0adx sin2(πx/a)sin2(2πx/a).
Eex1 = -4U0.
(∫0 π sin2(x)sin2(2x) dx = π/4)
To first order, the excited state energy is lowered by Eex1
to Ex = 5π2ħ2/(2ma2) - 4U0.
Problem 2:
Write down the electronic configuration for the Magnesium atom (Z =
12) in its ground state. Then enumerate the allowed term symbols 2S+1LJ for
the ground state from the point of view of angular momentum alone.
Solution:
- Concepts:
The energy levels of multi-electron atoms
- Reasoning:
We use the Pauli exclusion
principle and the rules for adding angular momentum to find the allowed
terms 2S+1L.
- Details of the calculation:
The electronic configuration is (1s)2, (2s)2,
(2p)6 (3s)2.
All shells are full and have L = 0, S = 0, J = 0.
There is only one allowed term, 1S0 .
Problem 3:
Titanium (Ti) is element No. 22 in the Periodic Table.
(a) Write down the electronic shell configuration of Ti (i.e., 1s2,
2s2, 2p6, ... ).
(b) Find the total orbital angular momentum quantum number L, total spin
angular momentum quantum number S, and the total angular momentum quantum number
J for the ground state of the Ti atom.
(c) Write down the corresponding spectroscopic term symbol 2S+1XJ
for the ground state, where X represents the orbital angular momentum.
(d) In a photoemission experiment, an electron is ejected from the 2p
shell. In this process, we will assume that the angular momentum states of the
valence electrons remain frozen. Find the possible values of S, L, and J
for the ionized 2p level, ignoring all other levels in the atom.
(e) Couple the ground state orbital angular momentum found in (b) with the one
you found in (d). Do the same for the spin angular momenta in (b) and (d).
What possible values of L and S do you get?
(f ) Find the six possible spectroscopic term symbols for the ionized Ti atom
(or final state of the photoemission process). Just leave the value of J blank,
i.e., write them as 2S+1X.
(g) Find the corresponding multiplicities.
Solution:
- Concepts:
The Pauli exclusion principle, energy levels of multi-electron atoms, Hund's
rules
- 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:
(a) 1s2, 2s2, 2p6, 3s2, 3p6,
4s2, 3d2
(b)
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 to 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 therefore are
1G4, 3F4,3,2, 1D2,
3P2,1,0, 1S0.
Hund's rules (largest multiplicity, largest multiplicity, smallest J) tell
us that the ground state is the 3F2 state.
(c) Term symbol 3F2.
(d) |∑ms|= S =1/2, |∑ml| = L = 1 , J = L+S,
L+S-1,...|L-S|= 3/2,1/2 .
(e) L = Lb + Ld, Lb + Ld -
1,......|Lb - Ld|
S = Sb + Sd, Sb + Sd - 1,......|Sb
- Sd |
For Lb = 3 and Ld = 1, we have L = 4, 3, 2.
For Sb =1 and Sd
= 1/2, we have S = 3/2,1/2.
(We labeled the angular momentum quantum numbers with subscripts b and d,
according to the answers found in problems b and d.)
(f) 4G, 2G, 4F, 2F, 4D,
2D
(g) Multiplicity = (2S+1)×(2L+1) = 36, 18, 28, 14, 20, 10.
Problem 4:
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 5:
The Hamiltonian for two interacting spin ½ identical fermions in one
dimension is
H = p12/(2m) + p22/(2m) + (mω2/2)(x2
- x1)2.
What is the energy spectrum and what are the corresponding eigenfunctions?
Consider both the motion of the center of mass and the motion about the center
of mass.
Solution:
- Concepts:
Two interacting indistinguishable particles, center of mass and relative
motion
- Reasoning:
To solve the problem we separate the total Hamiltonian into a Hamiltonian
for the 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.
- Details of the calculation:
(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 + ¼mv2.
U = (mω2/2)(x2 - x1)2
= (mω2/2)r2
= ½Br2.
L = T - U.
P = ∂L/∂V
= 2mV, p = ∂L/∂v
= ½mv.
H = P2/(2M) + p2/(2μ) +
½Br2.
Here M = 2m, μ = m/2, and B = μω2.
To make the transition to Quantum Mechanics we let the canonical variables R
and P and the canonical variables r and p become operators.
Hψ(R,r) = [(-ħ2/(2M))(∂2/∂R2) + (-ħ2/(2m))(∂2/∂r2) + ½Br2]ψ(R,r)
= Eψ(R,r).
is the eigenvalue equation for H.
Separation of variables is possible.
Let ψ(R,r)
= χ(R)Φ(r).
Then
(-ħ2/2M )(∂2/∂R2)χ(R)
= ERχ(R),
χ(R)
= exp(ikR), ER
= ħ2k2/(2M).
The solutions are plane wave solutions. There exists a solution for
every k.
[(-ħ2/2m)(∂2/∂r2) + ½Br2]Φ(r)
= ErΦ(r),
Er = (n + ½)ħω, ω2
= B/m.
The solutions are the solutions of the 1D harmonic oscillator.
E = ER + Er, n = 0, 1, 2, ... .
E = ħ2k2/(2M)
+ (n + ½)ħω.
Under exchange R --> R, r --> -r.
Interchanging the two particles leaves R unchanged, but changes the sign of
r.
Φn(r)
is symmetric under exchange if n = even and
Φn(r)
is antisymmetric if n = odd.
For identical fermions the total wavefunction must be antisymmetric under the
exchange of the two particles.
Therefore, two spin ½ particles in an orbital state with
n = even must
be in the singlet (S = 0) spin state, and
two spin ½ particles in an orbital state with
n = odd must be
in the triplet (S = 1) spin state.