Problem 1:
Suppose you have two non-interacting, spin-less particles
in a one-dimensional potential, each in one of two single-particle, orthonormal
eigenstates ψ1(x) and ψ2(x)
of that potential. Now suppose you measure the position of one of the
particles in an ensemble of identically prepared experiments. What is the
expectation value of position if the particles are
(a) distinguishable particles,
(b) identical bosons,
(c) identical fermions?
In
any one of these cases, if there is more than one answer to the question,
provide all of them.
Solution:
- Concepts:
Indistinguishable particles
- Reasoning:
For distinguishable particles, we do not have to worry about symmetrization
requirements.
The two-particle wave function of indistinguishable bosons must be symmetric
under the exchange of the two particles, and the two-particle wave function
of indistinguishable fermions must be anti-symmetric under the exchange of the
two particles.
-
Details of the calculation:
(a) Since the particles are distinguishable, we assume we know which
particle is in state ψ1 and which particle is in state ψ2,
We then have ψ12 = ψ1(x1)ψ2(x2)
or ψ12 = ψ1(x2)ψ2(x1)
depending on the information we have about the particles.
If ψ12
= ψ1(x1)ψ2(x2),
then the expectation value for the position of particle 1 is
∫-∞∞dx ψ1*(x)
x ψ1(x) = <ψ1|x|ψ1>, and the expectation value
for the position of particle 2 is
∫-∞∞dx ψ2*(x) x ψ2(x) =
<ψ2|x|ψ2>.
If ψ12 = ψ1(x2)ψ2(x1),
then the expectation value for the position of particle 1 is <ψ2|x|ψ2>,
and the expectation value for the position of particle 2 is <ψ1|x|ψ1>.
(b) ψ12 = 2-½(ψ1(x1)ψ2(x2)
+ ψ1(x2)ψ2(x1)).
The expectation value for x1 is
<x1> = ½∫-∞∞(ψ1*(x1)ψ2*(x2)
+ ψ1*(x2)ψ2*(x1))
x1 (ψ1(x1)ψ2(x2) + ψ1(x2)ψ
2(x1)) dx1 dx2.
<x2> = <x1> since we cannot distinguish between the
particles.
<x1> = ½∫-∞∞(ψ1*(x1)ψ2*(x2)
+ ψ1*(x2)ψ2*(x1))
x1 (ψ1(x1)ψ
2(x2) + ψ1(x2)ψ
2(x1)) dx1 dx2
= ½∫-∞∞ψ1*(x1)
x1 ψ1(x1) dx1 ∫-∞∞ψ2*(x2)
ψ2(x2) dx2
+ ½∫-∞∞ψ1*(x1)
x1 ψ2(x1) dx1 ∫-∞∞ψ2*(x2)
ψ1(x2) dx2
+ ½∫-∞∞ψ2*(x1)
x1 ψ1(x1)
dx1 ∫-∞∞ψ1*(x2)
ψ2(x2) dx2
+ ½∫-∞∞ψ2*(x1)
x1 ψ2(x1)
dx1 ∫-∞∞ψ1*(x2)
ψ1(x2) dx2.
∫-∞∞ψ2*(x)
ψ2(x) dx = ∫-∞∞ψ1*(x)
ψ1(x) dx = 1.
∫-∞∞ψ1*(x)
ψ2(x) dx
= ∫-∞∞ψ2*(x)
ψ1(x) dx
= 0.
<x1> = ½[<ψ1|x|ψ1> + <ψ2|x|ψ2>]
since ψ1(x) and ψ2(x) are
orthonormal and the cross terms are zero.
The expectation value of position is ½[<ψ1|x|ψ1> + <ψ2|x|ψ2>].
(c) ψ12 = 2-½(ψ1(x1)ψ2(x2)
- ψ1(x2)ψ 2(x1)).
The expectation value of position is ½[<ψ1|x|ψ1> + <ψ2|x|ψ2>],
the same as in (b), since ψ1(x) and ψ2(x)
are orthonormal and the cross terms are zero.
Problem 2:
(a) If the spin of the electron were (3/2)ħ,what spectroscopic terms
2S+1LJ would exist for the electron configuration 2p, 3p?
(b)
If the spin of the electron were (3/2)ħ,what spectroscopic terms
2S+1LJ
would exist for the electron configuration (3p)2?
Solution:
- Concepts:
The Pauli exclusion principle, addition of angular momenta
- 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:
(a) Since the particles have a different quantum number n, all values
of L and S obtained from the general rules for addition of angular momenta
are allowed by the Pauli exclusion principle.
Possible L quantum numbers: 2, 1, 0.
Possible S quantum numbers: 3, 2, 1, 0.
Allowed terms 2S+1LJ: 7D5,4,3,2,1, 5D4,3,2,1,0, 3D3,2,1,
1D2,
7P4,3,2, 5P3,2,1, 3P2,1,0, 1P1,
7S3, 5S2, 3S1, 1S0.
(b) Since the particles have the same quantum number n, some of
the values of L and S obtained from the general rules for addition of
angular momenta can correspond to states forbidden by the Pauli exclusion
principle.
Since the particles are fermions, the total state vector has to be
anti-symmetric under exchange of the particles.
The symmetry of the space part of the state vector is determined by L, the
symmetry of the spin part of the state vector is determined by S,
When adding the angular momenta of two particles with the same quantum
numbers l or s, the "stretched" case Lmax or Smax is
always symmetric or even (e) under exchange. The symmetry alternates
between even (e) and odd (o) as we count down to L = 0 or S = 0 in integer
steps.
Possible L quantum numbers: 2(e), 1(o), 0(e).
Possible S quantum numbers: 3(e), 2(o), 1(e), 0(o).
Allowed terms 2S+1LJ: 5D4,3,2,1,0,
1D2, 7P4,3,2, 3P2,1,0, 5S2,1S0.
Problem 3:
Consider N >> 1 non-interacting
spin-½ 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.
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
Φnmp(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
Enmp = (n2 + m2 + p2)π2ħ2/(2ML2)
= (kx2 + ky2 + kz2)ħ2/(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 4:
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.
Useful integrals:
∫0
π sin4(x) dx = 3π/8
∫0 π sin2(x)sin2(2x) dx = π/4
Solution:
- Concepts:
Identical particles
- Reasoning:
Identical particles
-
Details of the calculation:
Identical particles
For identical bosons, the state must be symmetric under the exchange of the two
particles.
(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.
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
= -(8U0/a)∫0adx
sin2(πx/a)sin2(2πx/a).
Eex1
= -2U0.
To first order, the excited state energy is lowered by Eex1
to Ex = 5π2ħ2/(2ma2)
- 2U0.