Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications
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Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
98 3 Free Electron Theory of Metals
Substitute this into (3.100) to obtain w
x
; the result is 335
w
x
=
1
Θ
∂Θ
∂x
K
2
K
1
K
2
−
1
Θ
∂Θ
∂x
K
3
=
K
3
K
1
−K
2
2
K
1
Θ
−
∂Θ
∂x
. (3.105)
Thus the thermal conductivity κ
T
= k
B
w
x
−
∂Θ
∂x
−1
is 336
κ
T
= k
B
K
3
K
1
−K
2
2
K
1
Θ
. (3.106)
If we evaluate K
n
as a function of Θ using (3.101) and the result 337
ζ
ζ
0
l
1 −
π
2
12
Θ
ζ
0
2
l, (3.107)
we find (for τ independent of energy)
338
K
1
n
0
τ
m
,
K
2
n
0
τζ
0
m
1+
5
12
π
2
Θ
ζ
0
2
, (3.108)
K
3
n
0
τζ
2
0
m
1+
7
6
π
2
Θ
ζ
0
2
.
Now, substitute these results into (3.106) to obtain
339
κ
T
= k
B
π
2
3
n
0
τ
m
Θ.
Thus the Sommerfeld expression for κ
T
can be written 340
κ
T
=
π
2
3
k
2
B
n
0
τ
m
T. (3.109)
The Lorenz ratio for the Sommerfeld model L
S
is given by 341
L
S
=
κ
T
σT
=
π
2
3
k
B
e
2
2.71 × 10
−13
esu. (3.110)
Recall that for the Drude model
342
L
D
=
3
2
k
B
e
2
1.24 × 10
−13
esu, (3.111)
and the average experimental result is L≈2.72 × 10
−13
esu. 343
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
3.15 Magnetoconductivity 99
3.14 Critique of Sommerfeld Model 344
The main achievements of the Sommerfeld model were as follows: 345
1. It explained the specific heat dilemma by showing that C
v
for the electrons 346
was very small. 347
2. It showed that even though there was one free electron per atom, the Pauli 348
principle made only those in an energy range k
B
T about ζ
0
effectively free. 349
3. It not only explained the Wiedemann–Franz law but it gave a very accurate 350
value for the Lorenz ratio. 351
4. It correctly predicted the Pauli spin paramagnetism of metals. 352
5. It predicted Fermi energies that agreed with observed X-ray band widths. 353
The main shortcomings of the Sommerfeld model were: 354
1. It said nothing about the relaxation time τ (ε), τ appeared only as a 355
phenomenological parameter. To agree with observed conductivities, the 356
mean free path l = v
F
τ had to be of the order of a 100 atomic spac- 357
ings (∼5 ×10
−6
cm), and had to vary as T
−1
at room temperature. These 358
requirements were difficult to understand in 1928. 359
2. The model ignored the interaction of the free electrons with the fixed ions 360
and with one another. These interactions were surely large. How could 361
one achieve such excellent agreement with experiment when they were 362
ignored. Furthermore, attempts to include these interaction ran into great 363
difficulties. 364
3.15 Magnetoconductivity 365
In the presence of a large dc magnetic field B, the conductivity of a metal 366
displays some new effects. These can be understood very simply using the 367
Drude model (the Sommerfeld model gives exactly the same result but involves 368
much more mathematics). 369
In the presence of an electric field E and a dc magnetic field B, the Drude 370
model would predict a drift velocity v
D
which was a solution of the equation 371
m
dv
D
dt
+
v
D
τ
= −eE −
e
c
v
D
× B. (3.112)
Let us choose the z-axis along B and assume that E is spatially uniform but
372
varies in time as e
iωt
. Then (3.112) can be rewritten (we drop the subscript 373
D of v
D
in the rest of this section) as follows: 374
(1 + iωτ)v
x
= −
eτ
m
E
x
−
eτ
mc
Bv
y
,
(1 + iωτ)v
y
= −
eτ
m
E
y
+
eτ
mc
Bv
x
, (3.113)
(1 + iωτ)v
z
= −
eτ
m
E
z
.
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
100 3 Free Electron Theory of Metals
Define the cyclotron frequency ω
c
=
eB
mc
and solve for v. The result is 375
v
x
= −
eτ
m
[(1 + iωτ)E
x
− ω
c
τE
y
]
(1 + iωτ)
2
+(ω
c
τ)
2
,
v
y
= −
eτ
m
[ω
c
τE
x
+(1+iωτ)E
y
]
(1 + iωτ)
2
+(ω
c
τ)
2
, (3.114)
v
z
= −
eτ
m
E
z
1+iωτ
.
The current density is given by j = − en
0
v. This can be written j = σ · E, 376
where σ is called the magnetoconductivity tensor. Its components are σ
xz
= 377
σ
zx
= σ
yz
= σ
zy
=0,and 378
σ
xx
= σ
yy
=
σ
0
(1 + iωτ)
(1 + iωτ)
2
+(ω
c
τ)
2
,
σ
xy
= −σ
yx
=
σ
0
(−ω
c
τ)
(1 + iωτ)
2
+(ω
c
τ)
2
, (3.115)
σ
zz
=
σ
0
1+iωτ
.
379
Here σ
0
=
n
0
e
2
τ
m
is just the Drude’s dc conductivity. 380
3.16 Hall Effect and Magnetoresistance 381
If we apply an electric field E in the x-direction, the Lorentz force, −
e
c
v × B 382
causes a drift velocity in the y-direction. If ω = 0 charge will accumulate on 383
the surfaces normal to the y-direction until a field E
y
builds up that exactly 384
cancels the Lorentz force. (See Fig. 3.5.) The condition j
y
=0gives 385
j
y
= σ
xx
E
y
− σ
xy
E
x
=0,
E
E
B
v
B
Fig. 3.5. Schematics of the Hall effect experiment. The initial drift of the negatively
charged electron is illustrated
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
3.17 Dielectric Function 101
or 386
E
y
=
σ
xy
σ
xx
E
x
.
The Hall coefficient R is defined as the ratio of E
y
to j
x
B: 387
R =
E
y
j
x
B
. (3.116)
But j
x
= σ
xx
E
x
+ σ
xy
E
y
=(σ
2
xx
+ σ
2
xy
)/σ
xx
E
x
. If we substitute it into the 388
expression for the Hall coefficient, we find 389
R =
E
y
j
x
B
=
(σ
xy
/σ
xx
) E
x
σ
2
xx
+ σ
2
xy
/σ
xx
E
x
B
=
σ
xy
σ
2
xx
+ σ
2
xy
1
B
.
Making use of (3.115) in the limit ω −→ 0gives
390
R =
1
n
0
(−e)c
. (3.117)
391
Because R depends on the carrier concentration, the Hall effect is often used 392
to measure n
0
. 393
Magnetoresistance 394
When B =0,j
x
= σ
0
E
x
. In the presence of the magnetic field B,wehave 395
j
x
=
σ
2
xx
+ σ
2
xy
σ
xx
E
x
. (3.118)
For the free electron model
σ
2
xx
+σ
2
xy
σ
xx
= σ
0
(one can check this relation as 396
an exercise). Therefore even in the presence of the B-field j
x
= σ
0
E
x
.The 397
magnetic field causes no change in the ratio
E
x
j
x
= ρ,theresistivity,andwe 398
would say that 399
Δρ = ρ(B) − ρ(0) = 0,
or that the magnetoresistance vanishes. This does not occur in more general
400
cases than the simple free electron model as we shall see later. 401
3.17 Dielectric Function 402
The electrical current density j can be thought of as the time rate of change 403
of the polarization P. Assume D, P,andE vary as e
iωt
.Thenj =
˙
P =iωP 404
and D = εE = E +4πP where ε is the dielectric function. This gives us the 405
relation 406
ε(ω)=1−
4πi
ω
σ(ω), (3.119)
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
102 3 Free Electron Theory of Metals
where σ(ω) is the frequency dependent Drude conductivity. In the presence 407
of a dc magnetic field, the dielectric function and conductivity become tensor 408
quantities: σ(ω)andε(ω), whose off-diagonal components result from the 409
Lorentz force. 410
Thedielectricfunctionε(ω) or conductivity σ(ω) appear in Maxwell’s 411
equation for ∇×B: 412
∇×B =
1
c
˙
E +
4π
c
j =
iω
c
1 −
4πi
ω
σ
(ω)
·E =
1
c
ε
(ω) ·
˙
E. (3.120)
In the Drude model
413
ε(ω)=1−
4πi
ω
σ(ω),
=1−
4πi
ω
n
0
e
2
τ/m
1+iωτ
.
Define the plasma frequency ω
p
by ω
2
p
=
4πn
0
e
2
m
;thenwehave 414
ε(ω)=1−
ω
2
p
ω(ω − i/τ)
. (3.121)
ε(ω) has real and imaginary parts, ε
1
and ε
2
, respectively, as 415
ε
1
(ω)=1−
ω
2
p
ω
2
+1/τ
2
, (3.122)
ε
2
(ω)=−
ω
2
p
/ωτ
ω
2
+1/τ
2
.
416
In general, we can ask how electromagnetic waves propagate in a medium 417
described by a dielectric tentor ε(ω). The wave equation can be obtained 418
from the two Maxwell equations: 419
∇×E = −
1
c
˙
B, (3.123)
∇×B =
1
c
ε
·
˙
E.
Assume E and B vary as e
iωt−iq·r
. The two Maxwell equations can then 420
be combined, by eliminating B,togive 421
q(q · E) − q
2
E +
ω
2
c
2
ε · E =0. (3.124)
422
This can be applied to a case in which a dc magnetic field B
0
is present and 423
oriented in the z-direction. Then without loss of generality we can choose 424
q =(0,q
y
,q
z
) and write (3.124) as 425
⎛
⎜
⎝
ω
2
c
2
ε
xx
− q
2
ω
2
c
2
ε
xy
0
−
ω
2
c
2
ε
xy
ω
2
c
2
ε
xx
− q
2
z
q
y
q
z
0 q
y
q
z
ω
2
c
2
ε
zz
− q
2
y
⎞
⎟
⎠
⎛
⎜
⎝
E
x
E
y
E
z
⎞
⎟
⎠
=0. (3.125)
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
3.17 Dielectric Function 103
The description of the bulk modes are given by setting the determinant of the 426
matrix multiplying the column vector equal to zero. 427
For surface modes at a metal–dielectric interface, think of ω and q
y
(wave 428
vector along the surface) as given and determine the allowed values of q
z
in 429
the solid (with a given ε(ω)) and in the dielectric. One can get modes from 430
applying standard boundary conditions. 431
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
104 3 Free Electron Theory of Metals
Problems 432
3.1. A two-dimensional electron gas is contained within a square box of side L. 433
(a) Apply periodic boundary conditions and determine E(k
x
,k
y
)and434
Ψ
k
(x, y) for the free electron Hamiltonian H =
1
2m
p
2
x
+ p
2
y
.
435
(b) Determine the Fermi wave number k
F
in terms of the density n
0
=
N
L
2
. 436
(c) Evaluate G(ε)andg(ε) for this system. 437
(d) Use n
0
=
"
∞
0
dεg(ε)f
0
(ε) together with the Fermi function integration 438
formula to determine how the chemical potential ζ depends on T . 439
3.2. In the simple Drude model it is assumed that the probability that an 440
electron has a collision in a time interval dt is given by
dt
τ
. 441
(a) Show that an electron picked at random at a given instant had no 442
collision during the past t seconds with probability e
−t/τ
.Demonstrate 443
that it will have no collision during the next t seconds with the same 444
probability, 445
(b) Show that the probability that the time interval between two successive 446
collisions falls in the range t to t +dt is
dt
τ
e
−t/τ
, 447
(c) as a consequence of (a), the mean time (at any given instant) back to 448
the last collision (or up to the next collision) averaged over all electrons 449
is τ, 450
(d) and, as a consequence of (b), the mean time between collisions is τ. 451
3.3. Think about the following argument, and use physical intuition (not 452
formal mathematics) to say why it its wrong. 453
Argument: Both (c) and (d) in Problem 2 cannot be simultaneously correct 454
for the following reason. If I catch a bus each morning to go to work, and the 455
buses arrive at random times except that the average time between buses is 456
15 min, then (c) would imply that, on the average, I have to wait 15 min. 457
However, if a friend got off each bus at my stop, and then got on the next bus 458
with me, then the average time waited by my friend boarding the bus with 459
me would be 15 min also. Clearly the friend boarding with me is waiting at 460
the bus stop when I arrive. How can it be that we wait the same amount of 461
time? 462
3.4. A metal is described by the conductivity tensor σ
xx
=σ
yy
=
σ
0
(1+iωτ )
(1+iωτ )
2
+(ω
c
τ)
2
, 463
σ
xy
=−σ
yx
=
σ
0
(−ω
c
τ)
(1+iωτ )
2
+(ω
c
τ)
2
,andσ
zz
=
σ
0
1+iωτ
in the presence of a dc magnetic 464
field B = Bˆz. Consider the propagation of an electromagnetic wave E
±
= 465
(E
x
± iE
y
)e
iωt
parallel to the z-axis. Use Maxwell’s equations to obtain the 466
wave equation, and show that c
2
k
2
= ω
2
ε
±
(ω), where ε = 1 −
4πi
ω
σ(ω). 467
Consider the cases ω
c
τ 1andω
c
ω and show that ω =
c
2
k
2
ω
c
ω
2
p
for one 468
circular polarization. 469
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
3.17 Dielectric Function 105
DIELECTRIC
METAL
D
Fig. 3.6. Interface between a dielectric and a metal
3.5. Let us consider the interface between a dielectric of dielectric constant 470
ε
D
and a metal of dielectric function ε(ω)=1−
ω
2
p
ω
2
,whereω
2
p
=
4πne
2
mω
2
.Itis 471
illustrated in Fig. 3.6. 472
If the normal to the surface is in the z direction and the wave vector 473
q =(0,q
y
,q
z
), consider the region of ω − q
y
space in which q
z
is imaginary 474
(i.e. q
2
z
< 0) both in the dielectric and in the metal. Impose the appropriate 475
boundary conditions at z =0andat| z |−→ ∞, and determine the dispersion 476
relation (ω as a function of q
y
) for these surface plasma modes. 477
3.6. At a temperature T a semiconductor contains n
e
electrons and n
h
holes 478
per unit volume in parabolic energy bands (i.e. E ∝ k
2
). The mass, charge, 479
and collision time of the electrons and holes are m
e
, −e, τ
e
,andm
h
, e, τ
h
, 480
respectively. 481
(a) Use the equations of motion of charged particles in the presence of a 482
dc magnetic field B = Bˆz and an ac electric field Ee
iωt
to determine 483
σ
e
(ω)andσ
h
(ω), the electron and hole contributions to the frequency 484
dependent magnetoconductivity tensor. 485
(b) Consider ω
ce
=
eB
m
e
c
and ω
ch
=
eB
m
h
c
to be large compared to τ
−1
e
and 486
τ
−1
h
, respectively. Determine the Hall coefficient for ω =0. 487
(c) Under the conditions of part (b), determine the magnetoresistance. 488
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
106 3 Free Electron Theory of Metals
Summary 489
In this chapter, first we have briefly reviewed classical kinetic theories of an 490
electron gas both by Drude and by Lorentz as simple models of metals. Then 491
Sommerfeld’s elementary quantum mechanical theory of metals is discussed. 492
In the Drude model, the electrical conductivity σ =
n
0
e
2
τ
m
is determined 493
by the Newton’s law of motion given by 494
m
dv
D
dt
+
v
D
τ
= −eE.
Here, n
0
=
N
V
and −e are the electron concentration and the charge on an 495
electron. The thermal conductivity is given by 496
κ =
w
−∂T/∂x
=
1
3
n
0
v
2
T
τ
dE
dT
=
1
3
v
2
T
τC
v
,
where C
v
= n
0
dE
dT
is the electronic specific heat. 497
The electrical current density j and thermal current density w are given, 498
in terms of distribution function f ,by 499
j(r,t)=
(−e)v f (r, v,t)d
3
v and w(r,t)=
εv f (r, v,t)d
3
v.
In the Sommerfeld model, states are labeled by {k,σ} =(k
x
,k
y
,k
z
)andσ, 500
where σ is a spin index. The Fermi energy ε
F
(≡ Θ
F
) , Fermi velocity v
F
,and 501
Fermi temperature T
F
=
Θ
F
k
B
are defined, respectively, by
502
ε
F
=
¯h
2
k
2
F
2m
=
1
2
mv
2
F
=Θ
F
,
where the Fermi wave number k
F
is related to the carrier concentration n
0
by 503
k
3
F
=3π
2
n
0
. The density of states of an electron gas is 504
g(ε)=
1
2π
2
2m
¯h
2
3/2
ε
1/2
.
For electrons moving in a periodic potential, g(ε) does not have such a simple
505
form. At a finite temperature, the chemical potential ζ is determined from 506
N = V
∞
0
g(ε)f
0
(ε)dε.
The internal energy U is given by
507
U
V
= u =
∞
0
dεεg(ε)f
0
(ε).
Uncorrected Proof
BookID 160928 ChapID 03 Proof# 1 - 29/07/09
3.17 Dielectric Function 107
These integrals are of the form I =
"
∞
0
dε f
0
(ε)
dF (ε)
dε
. At low temperatures, 508
we have, to order Θ
2
, 509
I = F (ζ)+
π
2
6
Θ
2
F
(ζ).
The electronic heat capacity C
v
=
∂U
∂T
V
is given, at low temperature, by 510
C
v
= γT, where γ =
π
2
k
2
B
2ζ
0
N for free electrons. 511
The electrical and thermal current densities j
x
and w
x
are, respectively, 512
written as 513
j
x
=
e
2
E + eΘ
∂
∂x
ζ
Θ
K
1
+
e
Θ
∂Θ
∂x
K
2
and 514
w
x
= −
eE +Θ
∂
∂x
ζ
Θ
K
2
−
1
Θ
∂Θ
∂x
K
3
.
where
515
K
n
=
n
0
mζ
3/2
0
∞
0
dε
−
∂f
0
∂ε
ε
n+1/2
τ.
The function K
n
is given by 516
K
n
=
n
0
mζ
3/2
0
ζ
n+1/2
τ(ζ)+
π
2
6
Θ
2
d
2
dε
2
ε
n+1/2
τ(ε)
|
ε=ζ
.
The electrical and thermal conductivities are given, in terms of K
1
,byσ = eK
1
517
and κ
T
= k
B
K
3
K
1
−K
2
2
K
1
Θ
. The Sommerfeld expression for κ
T
is κ
T
=
π
2
3
k
2
B
n
0
τ
m
T. 518
In the presence of an electric field E and a dc magnetic field B, the mag- 519
netoconductivity tensor has nonzero components, for the case B along the 520
z-axis, as follows: σ
xx
= σ
yy
=
σ
0
(1+iωτ)
(1+iωτ)
2
+(ω
c
τ )
2
, σ
xy
= −σ
yx
=
σ
0
(−ω
c
τ )
(1+iωτ)
2
+(ω
c
τ )
2
, 521
σ
zz
=
σ
0
1+iωτ
.Hereω
c
=
eB
mc
and σ
0
=
n
0
e
2
τ
m
is just the Drude’s dc conductivity. 522
The electrical current density j canbethoughtofasthetimerateofchange 523
of the polarization P,thatis,j =
˙
P =iωP,whereP =
ε−1
4π
E and D, P,and 524
E are assumed to vary as e
iωt
. Hence, we have the relation 525
ε(ω)=1−
4πi
ω
σ(ω).
ε(ω) has real and imaginary parts, ε
1
and ε
2
, respectively, and in the Drude 526
model, we have ε
1
(ω)=1−
ω
2
p
ω
2
+1/τ
2
and ε
2
(ω)=−
ω
2
p
/ωτ
ω
2
+1/τ
2
. The two Maxwell 527
equations ∇×E = −
1
c
˙
B and ∇×B =
1
c
ε ·
˙
E can be combined to obtain the 528
wave equation 529
q(q · E) − q
2
E +
ω
2
c
2
ε · E =0.