Download PHY 140A: Solid State Physics Solution to Homework #7

PHY 140A: Solid State Physics
Solution to Homework #7
Xun Jia1
December 5, 2006
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Fall 2006
c Xun Jia (December 5, 2006)
°
Physics 140A
Problem #1
Static magnetoconductivity tensor. For the drift velocity theory of (51), show
that the static current density can be written in matrix form as:





1 −ωc τ
0
Ex
jx
σ0
  Ey 
 jy  =
 ωc τ
1
0
(1)
1 + (ωc τ )2
2
0
0
1 + (ωc τ )
Ez
jz
In the high magnetic field limit of ωc τ À 1, show that
σyx = nec/B = −σxy
(2)
In this limit σxx = 0, to order 1/ωc τ . The quantity σyx is called the Hall conductivity.
Solution:
From Eqn. (51) in Kittel, in the static case, the velocity is not varying with
time, so dvi /dt = 0 for i = x, y, z, therefore, in the matrix form, we have:





m/τ eB/c
0
vx
Ex
 −eB/c m/τ
0   vy  = −e  Ey 
(3)
0
0
m/τ
vz
Ez
solve this equation by inverting the coefficient matrix, and let ωc = eB/mc, we get:





Ex
vx
1 −ωc τ
0
−eτ
  Ey 
 vy  =
 ωc τ
1
0
(4)
m[1 + (ωc τ )2 ]
2
Ez
vz
0
0
1 + (ωc τ )
since j = −nev and σ0 = ne2 τ /m, it follows that:




jx
vx
 jy  = −ne  vy 
jz
vz



1 −ωc τ
0
Ex
σ0
  Ey 
 ωc τ
1
0
=
1 + (ωc τ )2
2
Ez
0
0
1 + (ωc τ )
(5)
In the high magnetic field limit, ωc τ À 1, then:
σyx = −σxy =
and:
σxx
σ0 ωc τ
nec
σ0 ωc τ
σ0
=
≈
=
2
2
1 + (ωc τ )
(ωc τ )
ωc τ
B
σ0
σ0
=
≈
=0·
2
1 + (ωc τ )
(ωc τ )2
thus σxx = 0, to order 1/ωc τ .
1
µ
1
ωc τ
¶
+
σ0
(ωc τ )2
(6)
(7)
Fall 2006
c Xun Jia (December 5, 2006)
°
Physics 140A
Problem #2
Kinetic energy of electron gas. Show that the kinetic energy of a three dimensional gas of N free electrons at 0K is
3
U0 = N ²F
5
(8)
Solution:
For the Fermi-Dirac distribution at 0K, we have:
¯
½
¯
1
1
:
¯
f (²) = (²−² )/k T
=
¯
0
:
e F B + 1 T =0
² < ²F
² > ²F
(9)
and the density of states of electrons in 3-D is :
V
dk
V
D(²) =
4πk 2
= 2
3
(2π)
d²
4π
µ
2m
h̄2
¶3/2
²1/2
(10)
since for free electrons, the kinetic energy equals to the total energy ², the average
kinetic energy per electron is simply:
R∞
²̄ =
R²F
d²D(²)f (²)²
0
R∞
=
d²D(²)f (²)
0
0
R²F
d²D(²)²
d²D(²)
3
= ²F
5
(11)
0
therefore the kinetic energy of total N electrons is:
3
U0 = N ²̄ = N ²F
5
(12)
Problem #3
Chemical potential in two dimensions. Show that the chemical potential of a
Fermi gas in two dimensions is given by:
µ(T ) = kB T ln[exp(πnh̄2 /mkB T ) − 1]
(13)
for n electrons per unit area. Note: the density of orbitals of a free electron gas in
two dimensions is independent of energy: D(²) = m/πh̄2 , per unit area of specimen.
Solution:
At finite temperature T , the chemical potential is determined by the condition:
Z∞
n=
Z∞
d²D(²)f (²) =
0
d²
0
2
m
1
2 (²−µ)/kB T
+1
πh̄ e
(14)
Fall 2006
c Xun Jia (December 5, 2006)
°
Physics 140A
the right hand side can be computed explicitly as:
Z∞
Z∞
m
1
m
e−(²−µ)/kB T
d² 2 (²−µ)/k T
=
d²
B
+1
1 + e−(²−µ)/kB T
πh̄ e
πh̄2
0
0
Z∞
d(e−(²−µ)/kB T )
1 + e−(²−µ)/kB T
0
¶¸¯∞
·
µ
mkB T
² − µ ¯¯
=−
ln 1 + exp −
¯
kB T
πh̄2
²=0
¶¸
·
µ
mkB T
µ
=
2 ln 1 + exp
kB T
πh̄
mkB T
=−
πh̄2
so from above two equations, we can solve for µ as:
·
µ
¶
¸
πnh̄2
µ(T ) = kB T ln exp
−1
mkB T
(15)
(16)
Problem #4
Fermi gases in astrophysics.
(a). Given M¯ = 2 × 1033 g for the mass of the Sun, estimate the number of electrons
in the Sun. In a white dwarf star this number of electrons may be ionized and
contained in a sphere of radius 2 × 109 cm; find the Fermi energy of the electrons
in electron volts.
(b). The energy of an electron in the relativistic limit ² À mc2 is related to the
wavevector as ² ' pc = h̄kc. Show that the Fermi energy in this limit is
²F ≈ h̄c(N/V )1/3 , roughly.
(c). If the above number of electrons were contained within a pulsar of radius 10km,
show that the Fermi energy would be ≈ 108 eV . This explains why pulsars are
believed to be composed largely of neutrons rather than of protons and electrons,
for the energy release in the reaction n → p + e− is only 0.8 × 106 eV , which is
not large enough to enable many electrons to form a Fermi sea. The neutron
decay proceeds only until the electron concentration builds up enough to create
a Fermi level of 0.8 × 106 eV , at which point the neutron, proton, and electron
concentrations are in equilibrium.
Solution:
(a). The Sun is mainly composed by hydrogen atoms, whose mass is approximately
equal to the mass of proton mp . Since each hydrogen atom contains one electron,
the number of electrons in the Sun is:
M¯
N=
= 1.2 × 1057
(17)
mp
3
Fall 2006
c Xun Jia (December 5, 2006)
°
Physics 140A
if this number of electrons is contained in a white dwarf star with a radius
Rwd = 2 × 109 cm, the Fermi energy is:
2/3

µ
¶2/3

h̄2
h̄2 
2N
3π 2 N 
²F =
3π
=
3 
(18)
2m
V
2m 
4πRwd
3
−15
4
= 6.3 × 10 J = 4 × 10 eV
(b). In the relativistic case with a dispersion relation of ² ' pc = h̄kc, the density
of states of electrons is:
D(²) = 2
V ²2
V
2 dk
4πk
=
(2π)3
d²
π 2 h̄3 c3
(19)
then the Fermi energy is determined by the condition:
Z²F
²F =
D(²)d² =
0
which gives:
µ
²F = h̄c 3π
2N
V
V ²3F
3π 2 h̄3 c3
¶1/3
µ
∼ h̄c
N
V
(20)
¶1/3
(21)
(c). If above number of electrons is contained in a pulsar of radius Rp = 10km, the
Fermi would be:
1/3

µ ¶1/3
 N 
N
−11
8

(22)
²F = h̄c
= h̄c 
 4πR3  = 2.1 × 10 J = 1.3 × 10 eV
V
p
3
Problem #5
Liquid He3 . The atom He3 has spin 12 and is a fermion. The density of liquid
He3 is 0.081g/cm3 near absolute zero. Calculate the Fermi energy ²F and the Fermi
temperature TF .
Solution:
The mass of a single He3 is about:
mHe = 3a.m.u. = 3 × 1.67 × 10−27 kg = 5.01 × 10−27 kg
(23)
therefore the number density of He3 is
n=
ρ
= 1.62 × 1028 m−3
mHe
4
(24)
Fall 2006
c Xun Jia (December 5, 2006)
°
Physics 140A
so the Fermi energy is:
µ
¶2/3
h̄2
h̄2
2N
²F =
3π
= ²F =
(3π 2 n)2/3
2mHe
V
2mHe
= 6.74 × 10−23 J = 4.2 × 10−4 eV
(25)
and the Fermi temperature is:
TF =
²F
= 4.88K
kB
5
(26)