JM Densmore
P613 HW # 2 Solutions
1. The Free and Independent Electron Gas in Two Dimensions
(a) In a 2D system with periodic BC you can solve the SE with momentum kx = 2º/Lnx and ky = 2º/Lny . The number of allowed values
of k space inside a volume ≠ will be
≠V
4º 2
The volume in k space is the area of a circle with radius kf . Taking
into account the number of spins for the electron
n
=
1
n
=
rs
=
kf2
2º
(1)
(b)
(2)
ºrs2
p
2
kf
(3)
(c) In 2D the density of particles in range k + dk is 2ºkdk
Z
2ºkdk
n =
f (≤k )
4º 2
Changing to an integral over energy we find
Z
m
n =
f (≤)d≤
ºh̄2
g=
g=
m
ºh̄2
0
(4)
(5)
(6)
≤>0
≤<0
(d) Since g is constant g0 = 0 all the terms in the BS expansion are equal
to zero. So we find
Z ≤f
m
m
n =
(7)
2 dE + (µ ° ≤f )
ºh̄
ºh̄2
0
m
n = n(µ ° ≤f ) 2
(8)
ºh̄
µ = ≤f
(9)
(e) From 2.67 we have
m
n=
ºh̄2
1
Z
1
eØ(≤°µ) + 1
d≤
This integral can be looked up in a table. Using the electron density
and the fermi energy it is easy to show
=
≤f
µ+
1
ln(1 + e°Øµ )
Ø
(10)
(f) µ(T ) does not have an analytic expansion about Tº 0, which is why
the BS expansion does not work in 2D.
2. Entropy of a free electron gas
(a) We’ll start with the partition function, Z, for a free electron gas
X
Z =
e°Ø(ni ≤i °ni µ)
(11)
ni
=
X
n1
e°Ø(n1 ≤1 °n1 µ) §
X
e°Ø(nN ≤N °nN µ)
(12)
nN
For Fermi-Dirac statistics n can be either 0 or 1. The sums are easily
evaluated.
= ¶(1 + e°Ø(≤i °µ) )
Z
(13)
The entropy S is
@≠
= ° @T
S
= kB (ln Z + Ø≤)
(14)
@
ln Z
@Ø
X (≤i ° µ)e°Ø(≤i °µ)
1 + e°Ø(≤i °µ)
(15)
With the energy defined as
≤ =
=
and
ln(fi ) = ln
=
°
√
e°Ø(≤i °µ)
1 + e°Ø(≤i °µ)
!
°Ø(≤i ° µ) ° ln(1 + e°Ø(≤i °µ) )
from (4), (6), & (8) it is easy to show that
X
S = °kb
[fi ln fi + (1 ° fi ) ln(1 ° fi )]
k
(b) We can use integration by parts to integrate
Z
s = °kB dEg(E)(f ln(f ) + (1 ° f ) ln(1 ° f ))
2
(16)
(17)
(18)