1. No category

Solution manual - Chapter 2
Jens Zamanian
March 9, 2014
2.1 The Free and Independent Electron Gas in Two Dimensions
2
2
a) In two dimensions
each state takes up an
see Fig. 1. The two dimensional
Ground-Stat<'
uf area
lhl' 4⇡ /L in k-space,35
!<,
imeo""'nal k-sp:u:c of the fo rm J. , =
L Noh: th:ot the::
per point is Jl"l
t he n:•lwn.: r<:r point is t:! n L r .
•
•
•
•
•
•
•
•
k,
•
•
•
•
•
•
•
f k. That latter volume (see Figure 2.2) is j ust t2n L) 3 • We therefore
Figure 1: Two dimensional k-space. Each state takes up an area (2⇡/L)2 .
region of k-space tlf \'Olumc: n will contain
nv
n
equivalent of the Fermi sphere is a circular area with radius kF . This area must contain all the
(2. 17)
electrons.
The= number
(1rr, L):J
lfn3 of states in the Fermi circle is the area of the circle multiplied by the the
number of states per area multiplied by a factor 2 to account for spin. We hence have
f k.
equivalently. that the number of allowed k-valucs
per unit
L2
=
N
lalso known itS the k-sp:.1cc density
lcvd sl2⇡k
isF2just
4⇡ 2
and hence Jl
(1)
r
(2.18)
p
2⇡N
=
2⇡n,
(2)
L2
where,regions
in 2D, it makes
sense(to talk
of electrons
per area instead of per volume and
all deal with k-sp;H:e
so large
I about
ponumber
ints) and
so regular
2
we
hence
use
n
=
N/L
.
) that to all intents and purposes (2.17) and (2.1 S) can be regarded
b) In 3D rs is the radius of a sphere which has the same volume as the volume per conduction
l begin to applyelectron.
these important
formulas
In 2D it mustcounting
be defined as
the radius shortly.
of a circle which has the same area as the area
sume the electrons
non interacting
Gill build
upL the
N-clcctron
perare
conduction
electron. Forwe
N electrons
in an
⇥ L solid
we have
g1t3 .
kF =
on
placing electro ns into the allowed one-electronL2le,•els
we have j ust
1
=
=
⇡rs2(as it does
exclusion principle plays a vital role in thts construction
N
n
e states many dcctron atoms} : we may place at most one electron
ctron level. The one-electron levels are specified by the wave vectors
ection of the electron's spin along an arbitrary axis, 1which can take
values f• / 2 or - lr 2. Therefore associated w1th each allowed wave
electronic levels, one for each direction of tl1e electron's spin.
ng up the N-electron ground sla te we begin by placing two electrons
(3)
which gives
1
rs = p .
⇡n
(4)
c) In two dimensions one is often interested in calculating the sum of a function over all states.
We may write
X
2L2 X
2
F (k) =
F (k) k
(5)
4⇡ 2
k
k
where the first factor 2 is due to the possible spin states and where k = 4⇡ 2 /L2 . We then get
that
Z
X
2 X
k
dk
F
(k)
=
2
F
(k)
!
F (k) 2 ,
(6)
L2
4⇡ 2
2⇡
k
k
in the limit L ! 1. Often one is interested in functions F that only depends on k through the
energy E(k) = ~2 k 2 /(2m), i.e. F (k) = F (E(k)) = F (k). Switching to cylindrical coordinates we
have
Z
Z 1
Z 2⇡
Z
dk
1
1 1
I=
F
(E(k))
=
dkkF
(E(k))
d
=
dkkF (E(k)) .
(7)
4⇡ 2
2⇡ 2 0
⇡ 0
0
Now we wish to switch variables to E = ~2 k 2 /(2m). The di↵erential is found as
dE =
So that
I=
m
⇡~2
Z
~2
kdk
m
1
)
kdk =
dEF (E) =
0
where the density of states is
⇢
Z
1
m
dE.
~2
(8)
dEg(E)F (E),
(9)
1
m
⇡~2
E >0
(10)
0
E < 0.
R1
d) One is often interested in integrals like 1 dEH(E)f (E) where f (E) is the Fermi-Dirac
distribution. In the book they derive the Sommerfeld expansion which is
g(E) =
Z
1
dEH(E)f (E) =
1
Z
µ
dEH(E) +
1
1  2j
X
d
j
1
dE 2j
H
1
✏=µ
Z
The integral to calculate n in two dimensions is given by
Z 1
n=
dEg(E)f (E),
1
1
dE
(E
µ)2j
(2j)!
✓
@f (E)
@E
◆
.
(11)
(12)
1
where g(E) is the density of state as above. Since this is a constant only the first term in the
expansion survives. We then have
Z µ
Z µ
m
µm
n=
dEg(E) =
dE 2 =
,
(13)
2
⇡~
⇡~
1
0
which yields
n⇡~2
k 2 ~2 ⇡
k 2 ~2
= F
= F ,
(14)
m
2⇡m
2m
where we have used (2) to express n in terms of the Fermi wave vector kF . The last term is
identified as the Fermi energy so we hence have
µ=
µ = EF ,
2
(15)
independently of temperature! Although this is very close to the exact result, there is a small
correction to this and the Sommerfeld expansion does not give the exact result in this case.
e) If we instead directly use (12) to calculate n we have
Z 1
Z 1
m
1
m
1
n=
dE (E µ)/k T
=
dE (E µ)
,
(16)
B
⇡~2 0
⇡~2 0
e
+1
e
+1
where we have defined = 1/(kB T ) for convenience. This integral can easily be solved using
for example tables, but as we are physicist we do it the real way. First we switch variables to
x = e(E µ) which yields dE = dx/( x) 1 The integral is then
Z
1 1
1
dx
.
(17)
x(x
+ 1)
µ
e
This is now solved by the standard method of separating the fraction. If we try to write this as
two separate brackets we get
1
A
B
= +
,
(18)
x(x + 1)
x
x+1
which gives A + Ax + Bx = 1 which gives us A = 1 and B = A = 1. The integral is now easily
solved

Z
⇣
⌘
1 1
1
1
1
1
dx
= [ln x ln(x + 1)]e µ = µ + kB T ln 1 + e µ/(kB T ) .
(19)
x x+1
e µ
Inserting this into Eq. (16) we get
⇣
µ + kB T ln 1 + e
µ/(kB T )
⌘
= EF .
(20)
f ) Typically we have kB T ⌧ EF and then we must also have µ
kB T (try the limit µ
and see that this lead to nonsens (EF ! 0). For µ ⇠ EF
kB T we get
⌘ k T
EF µ
kB T ⇣
B
=
ln 1 + e µ/(kB T ) ⇡
e µ/(kB T ) ,
µ
µ
µ
kB T
(21)
which is very small.
The reason why the Sommerfeld expansion did not work perfectly is because the function g(E)
has a discontinuity at E = 0 where it jumps down to zero. This means that the function is not
analytical and cannot be Taylor expanded exactly with no amount of coefficients.
2.3 The Classical Limit of Fermi-Dirac Statistics
a) In the classical limit
Furthermore, we know that
The density is now
n
=
=
=
1A
f (E) ⇡ e
(E µ)/(kB T )
.
(22)
1
4⇡rs3
=
.
n
3
calculated from the distribution function as
p
Z 1
Z
m3/2 2eµ/(kB T ) 1 p
dE Ee E/(kB T )
dEg(E)f (E) =
3 ⇡2
~
1
0
p

Z
m3/2 2eµ/(kB T ) (kB T )3/2 1 p
y = E/(kB T )
=
dy ye
dE = kB T dy
~3 ⇡ 2
0
p
p
m3/2 2eµ/(kB T ) (kB T )3/2 ⇡
m3/2 eµ/(kB T ) (kB T )3/2
=
,
~3 ⇡ 2
2
21/2 ~3 ⇡ 3/2
slightly shorter solution is obtained by instead using the variable substitution z = e
3
(E
(23)
y2
=
(24)
µ)
.
where we have looked up the integral using a table (it is actually related to the -function if you
are interested). Inserting this into Eq. (23) we get
rs3 =
3~3 ⇡ 1/2
23/2 m3/2 eµ/(kB T ) (k
or
rs =
b) Note that this shows that if e
µ/(3kB T ) 1/3 1/6
e
3 ⇡
(2kB T )1/2
µ/(kB T )
✓
rs
3/2
(25)
.
(26)
BT )
~
1 we also have
~2
2mkB T
◆1/2
.
(27)
To see the significance of this we define the thermal velocity vT by
vT2 =
We then have
✓
rs
~2
m2 vT2
2
mvT
2
= kB T or
2kB T
.
m
◆1/2
(28)
=
~
,
mvT
(29)
i.e. this says that the typical distance between the electrons rs much be much larger than the
(thermal) de Broglie wavelength of the particle.
c) In terms of the Bohr radius a0 = ~2 /(me2 ) [in SI-units it is a0 = 4⇡✏0 ~2 /(me2 )] we have
✓
rs
a0
me4
2~2 kB T
◆1/2
.
(30)
If we now insert the number in CGS units (SI-units in brackets, but not that you need to use a0
in SI-units in that case)
m
~
=
=
kB
=
e
=
9.1095 ⇥ 10
28
1.3807 ⇥ 10
16
1.05459 ⇥ 10
4.80324 ⇥ 10
(9.1095 ⇥ 10
g
27
erg · s
erg/K
10
esu
31
kg)
(1.05459 ⇥ 10
(1.3807 ⇥ 10
(1.60219 ⇥ 10
(31)
34
Js)
(32)
J/K)
(33)
C),
(34)
23
19
where 1erg = 1gcm2 /s2 is what energy is measured in in the CGS-system and 1esu = 1g1/2 cm3/2 /s
is the unit of charge. We then get
rs
a0
✓
157888 K
T
◆1/2
⇡
✓
105 K
T
◆1/2
.
(35)
Note that in metals rs /a0 ⇠ 1 and we have Fermi-Dirac statistics up to quite high temperatures.
d) To solve this (which involves a bit of algebra) we use
n
=
kF3
3⇡ 2
EF
=
k B TF =
(36)
~2 kF2
2m
(37)
Now I’ll give you the start
m3
m3 n
m3 3⇡ 2 n
=
=
= use kF = (2mkB TF )1/2 /~ . . .
4⇡ 3 ~3
4⇡ 3 ~3 n
4⇡ 3 ~3 kF3
4
(38)