Physics 112 Fall 2010
Professor William Holzapfel
Homework 8 Solutions
Problem 1, Kittel 7.1: Density of Orbitals in one and two dimensions
In one dimension the orbitals are of the form
ψ
n
(
x
) =
A
sin(
nπx/L
) where
n
is a positive integer.
The energy is
n
=
2
2
m
πn
L
2
(1)
and the density in
n
space is simply 2
dn
, since there is one state per unit interval (the 2 comes from
the spin degeneracy). We can invert (1) to get
n
( )
=
L
π
√
2
m
=
⇒
dn
=
L
2
π
2
m
d .
Since the density of states
D
( ) is defined by
D
( )
d
≡
2
dn
, we have
D
( ) =
L
π
2
m
.
(b)
Now the orbitals are of the form
ψ
n
(
x
) =
A
sin(
n
x
πx/L
) sin(
n
y
πy/L
), where
n
x
, n
y
>
0 and
the energy has the same form (1), but with
n
=
n
2
x
+
n
2
y
. The density of states in
n
space is
1
4
·
2
·
2
πn dn
=
πn dn
(2)
where the 1/4 is because we’re only interested in the first quadrant, the 1/2 is for spin and the 2
πndn
is just the area element in 2D
n
space in polar coordinates. As above, we have
D
( )
d
≡
πn dn
=
π
L
π
√
2
m
·
1
2
L
π
2
m
d
=
mL
2
π
2
d
=
⇒
D
( )
=
mL
2
π
2
.
Problem 2, Kittel 7.2: Energy of a Relativistic Fermi Gas
(a)
For a relativistic gas in 3D the orbitals are still of the form
ψ
n
(
x, y, z
) =
A
sin(
n
x
πx/L
) sin(
n
y
πy/L
) sin(
n
z
πz/L
)
but now the energy of the orbitals takes the form
n
=

p

c
=
π c
L
n
(3)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
where
n
=
n
2
x
+
n
2
y
+
n
2
z
. We can still, however, use Kittel’s result that in
n
space the occupied modes
fill up a sphere of radius
n
F
=
3
N
π
1
/
3
(4)
since this result is independent of the dispersion relation (3). The Fermi energy is then just
F
≡
n
F
=
π c
L
n
F
=
πc
3
n
π
1
/
3
.
(b)
Now using (3) and (4) the ground state energy is easily found:
U
0
=
2
×
1
8
n
F
0
4
πn
2
n
dn
=
π
2
c
L
n
F
0
n
3
dn
=
π
2
c
4
L
3
N
π
4
/
3
=
3
4
N
F
.
Problem 3, Kittel 7.5: Liquid
3
He
as a Fermi Gas
(a)
The Fermi velocity
v
F
is defined in terms of the Fermi energy:
F
≡
1
2
mv
2
F
. To find the Fermi
energy, we need to know the number density; we can find this from the given mass density of
3
He
:
n
=
ρ
m
M
3
He
=
0
.
081 g
/
cm
3
3 amu
=
0
.
081 g
/
cm
3
3
·
1
.
66
×
10

24
g
= 1
.
63
×
10
22
1
cm
3
The Fermi velocity is then
v
F
=
2
f
m
=
v
F
=
2
2
2
m
(3
π
2
n
)
2
/
3
m
=
m
(
3
π
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '06
 StevenG.Louie
 Physics, Energy, Potential Energy, Work, Statistical Mechanics, Fermi gas

Click to edit the document details