Download Lecture Notes: Condensed Matter Theory I (TKM1)

Lecture Notes: Condensed Matter Theory I (TKM1)
J. Schmalian
(Dated: December 1, 2011)
1
I.
INTRODUCTION
Condensed matter physics is concerned with the behavior of large aggregates of atoms or
molecules in liquid or solid form. It is one of the largest branches of physics, with a wide
variety of di¤erent systems, approaches, challenges and concepts. Often, it is subdivided
in soft condensed matter physics and hard condensed matter physics. While the transition between the two branches is gradual, one way to distinguish them is by the role of
quantum mechanics for the elementary excitations of the systems. Soft condensed matter
physics (the physics of polymers, liquid crystals, the statistical mechanics of bio-molecules
etc.) is frequently termed "~ = 0"-physics, stressing that classical dynamics su¢ ces for
an understanding of the motion and aggregation of these systems. In distinction, for hard
condensed matter physics, i.e. "~ = 1"-physics, the motion of electrons, lattice vibrations
etc. is determined by Schrödinger’s equation.
Physics is a basic science and its ultimate purpose is the accumulation of new knowledge.
In addition, condensed matter physics is closely connected to materials science as well as
mechanical, chemical, and electric engineering that focus on the design of novel materials
and devices, ranging from better batteries, thermoelectric devices for waste heat conversion
to superconductors, magnets, all the way to better agents in drug delivery. This applied
aspect of condensed matter physics is exciting and important. Still, we should not forget
that the …eld also contributes to the accumulation of fundamental knowledge and to major
philosophical issues of our times. The value of Planck’s quantum, ~, or of the electron
charge, e, are de…ned via solid state e¤ects in semiconductors and metals (the quantum
Hall e¤ect and the Josephson e¤ect). It is quite amazing that these fundamental constants
of nature are not determined in an experiment at a particle accelerator but rather in a
solid state laboratory. Other frequently cited examples about the fundamental importance
of condensed matter physics are that the famous Higgs particle was …rst proposed in the
context of superconducting phase transitions, that asymptotic freedom (important for our
current understanding of hadronization of quarks) occurs in case of the Kondo e¤ect of a
magnetic impurity in a metal, that fractional charges emerge naturally in the context of the
fractional Quantum Hall e¤ect etc. etc. The beauty of condensed matter physics is that it
combines hands-on applications with the development of fundamentally new concepts, often
even in the same material!
2
Lets discuss one epistemological issue that is heavily debated these days: Particle and
string theorists search to …nd a better way to formulate the fundamental laws of physics.
This is very exciting research. But suppose for a moment, we knew the fundamental "theory
of everything" (TOE). Does this mean that the physics would stop existing as a basic science?
Well, in condensed matter physics we have a TOE since 1927, yet major discoveries continue
to take place. The TOE of condensed matter physics is the many particle Schrödinger
equation
i~
@
@t
(1)
=H
with Hamiltonian
H = Te + Ti + Vee + Vii + Vei
(2)
with individual terms:
Te =
Vee =
Ne
X
~2 2
rj
2m
j=1
Ne
X
j;j 0 =1
e2
jrj
Ne ;Ni
Vei =
X
j;l=1
rj 0 j
and Ti =
and Vii =
Ni
X
Ni
X
~2 2
rl
2M
l
l=1
e2 Zl Zl0
jRl Rl0 j
l;l0 =1
e2 Zl
:
jrj Rl j
(3)
Here rj (Rl ) refers to the coordinates of the Ne (Ni ) electrons (nuclei) with mass m (Ml ).
Zl is the corresponding nuclear charge. The wave function depends in …rst quantization on
all coordinates
(r1 ;
; rNe ; R1 ;
; RNi ) =
(frj g ; fRl g) :
(4)
With the exception of radiation e¤ects and spin-orbit interaction (both can easily be included into the formalism) can all phenomena of condensed matter physics be described by
this Hamiltonian and the corresponding equation of motion. At least there is no experiment (or Gedankenexperiment) that is in con‡ict with this assertion. Thus, the "theory of
everything" of condensed matter physics is well known and established. It would however be
foolish to believe that phenomena like superconductivity, the fractional quantum Hall e¤ect,
electron localization etc. could be derived from Eqs.1,2. Instead a combination of experimental ingenuity, symmetry based reasoning, and a clever analysis of the relevant time and
length scales of a given problem (formalized in terms of the renormalization group theory)
ultimately allows for such conclusions and lead to conceptually new insights. The message
3
is that the knowledge of a fundamental "theory of everything" has very little bearing on the
fascinating possibilities that emerge when many particles interact with each other and organize into new states of matter. There is no reason, other than habit and accepted custom, to
believe that the situation is much di¤erent in other areas of physics, such as particle physics
or quantum gravity. These issues have been lucidly discussed by by Phillip W. Anderson,
Science 177, 393–396 (1972) and, more recently, by Robert B. Laughlin and David Pines
(Proceedings of the National Academy of Sciences 97, 28-31 (2000).
II.
QUANTUM THEORY OF SOLIDS
A.
The Born-Oppenheimer approximation
We start our analysis of Eqs.1,2. Great progress in our understanding can be made if
one recognizes that the two kinetic energy parts of the Hamiltonian are very di¤erent. The
ratio of the masses of the electrons and nuclei is small.
m=Ml ' 10 3 =Zl
1:
(5)
Thus, the motion of the nuclei is much slower compared to that of the electron and we can
decouple their dynamics. This decoupling goes back to Max Born and Robert Oppenheimner
(1927). One assumes that on the time scale of the electrons the nuclei are frozen. To this
end we assume that the positions of the nuclei are …xed and play for the electronic wave
function,
(r1 ;
; rNe ; R1 ;
; RNi ) =
(frj g ; fRl g)
(6)
solely the role of given parameters. This leads to the simpli…ed Schrödinger equation
i~
@
@t
= Hel
(7)
with electronic Hamiltonian
Hel = Te + Vee + Vei :
(8)
From the perspective of the electrons the electron-ion Coulomb interaction plays a role of
an "external" potential:
Vei =
Ne
X
j=1
4
U (rj )
(9)
where U (rj ) is the single particle potential that originates from the nuclei:
Ni
X
U (rj ) =
l=1
e2 Zl
:
jrj Rl j
(10)
The time independent Schrödinger equation
Hel
n
= Eel;n
(11)
n
determines the eigenvalues of the electronic system that depend parametrically on the ion
positions
Eel;n = Eel;n (R1 ;
(12)
; RNi ) :
Next we check whether this approximate treatment is indeed appropriate for m=Ml
We use the fact that, no matter whether our decoupling is correct or not, the
n
1.
form
a complete set of states for the electronic variables. This allows to expand the full wave
function
(frj g ; fRl g) =
Formally, the
n
X
n
n
(frj g ; fRl g)
n
(fRl g)
(13)
(fRl g) are the coe¢ cients of this expansion into a complete set of states.
Physically, they correspond to the amplitude of the ions to be found at positions fRl g if the
electrons are in the state
H
n.
We insert this ansatz into the full Schrödinger equation
X
= (Te + Ti + Vee + Vii + Vei )
=
X
n
n
n
n
n
(Te + Ti + Vee + Vii + Vei )
n
=
X
(Eel;n + Ti + Vii )
n
(14)
n
n
It holds immediately:
Vii
n
n
Ni
X
~2 2
r
2Ml l
l=1
n
n
=
n Vii
(15)
n:
In addition we have
Ti
n
n
=
=
=
Ni
X
~2
rl ((rl
2M
l
l=1
Ni
X
~2
2Ml
l=1
r2l
n)
n
5
n
n
+
n rl
+ 2 (rl
n)
n ) rl
n
+
2
n rl
n
(16)
Suppose we are allowed to ignore the …rst two terms. Then follows:
(Eel;n + Ti + Vii )
n
n
'
n
(Eel;n + Ti + Vii )
n:
(17)
Thus, after the solution of the electronic problem is accomplished, we have to solve the
purely ionic Schrödinger equation
Ti + Viief f
n
=E
(18)
n
with e¤ective ion-ion interaction Viief f = Vii + Eel;n . This determines the ionic wave function
and …nally the total energy eigenvalue E. A coupled problem of ions and electrons is therefore
simpli…ed into two separate problems of the two subsystems.
To justify this approach it must obviously hold that the two terms
~2
2Ml
are negligible compared to
functions
n
and
n
~2
2Ml
r2l
2
n rl
n.
n
n
+ 2 (rl
n ) rl
(19)
n
This must be checked a posteriori, after the wave
are determined. At this point it is a little bit to early to do this analysis.
As the course proceeds, we will be able to perform this calculation and we will …nd:
~2
2M
2
n rl
~2
(rl n ) rl
M
~2
r2l n
2Ml
For m=M
n
n
n
m 1=2
"F
M
m 3=4
'
"F
M
m
"F n n ;
'
M
'
n
n;
n
n;
1 follows that terms that contain a derivative rl
to the leading term
~2
2M
2
n rl
(20)
n
can be neglected relative
n.
We conclude, that the Born-Oppenheimer approximation, valid in the limit m=M
1,
justi…es the investigation of a purely electronic Hamiltonian
Hel =
Ne
Ne
Ne
X
X
X
~2 2
e2
rj +
U (rj ) +
:
2m
jrj rj 0 j
j=1
j=1
j;j 0 =1
(21)
Still, the problem is only de…ned if we know the positions of the nuclei. In case of crystalline
solids, where the nuclei are arranged on a periodic lattice, the potential U (r) is periodic
with respect to the discrete translations of the crystal, which simpli…es this aspect of the
problem signi…cantly.
6
After we established Hel as e¤ective theory of the electronic system only, we study the
impact of the potential U (r) in case of periodic lattices. To this end we …rst ignore the
rj 0 j 1 . This seems at …rst glance a foolish thing to do,
electron-electron interaction e2 jrj
as the Coulomb interaction between electrons is neither small nor irrelevant. At this point
we make this assumption without further justi…cation and come back to it later. We will see
that there are numerous situation where the behavior of the electrons is e¤ectively described
by non-interacting fermions, while in other cases (quasi-one dimensional systems, systems
where plasma excitations of the electrons matter, systems that order magnetically etc.) the
neglect of the Coulomb interaction cannot be justi…ed.
Once electrons are noninteracting the Hamiltonian is a sum over single particle Hamiltonians
Hel =
Ne
X
j=1
with
H (p; r) =
H (pj ; rj )
~2 2
r + U (r) :
2m
Since we assumed that the electrons are non-interacting, their thermodynamic behavior is
that of an ideal Fermi gas. The remaining task, needed to analyze this Fermi gas, is to
determine the single particle eigenfunctions 'n (r) and the corresponding eigenvalues "n of
H. Before we discuss these issues we repeat the statistical mechanics of quantum gases and
the formalism of second quantization.
III.
QUANTUM STATISTIC OF IDEA GASES
We consider consequences of indistinguishability in quantum statistics. The quantity of
interest in statistical mechanics is the partition function
Z = tre
H
=
P
exp (
(22)
E )
where the sum is over all many body states j i with energy E and
=
1
.
kB T
Once we
know Z we can determine all thermodynamic quantities of a given equilibrium system. For
example, the Boltzmann probability
p =
1
exp (
Z
7
E )
P
is normalized ( p = 1) because of the denominator Z. The mean energy is
U = hHi =
P
1P
E e
Z
E p =
@
log Z
@
=
E
(23)
For the entropy holds
S =
=
kB
P
kB P
e
Z
p log p =
1
U + kB log Z
T
From thermodynamics we know that F = U
E
(
E
log Z)
(24)
T S is the free energy. Thus, it follows:
F (T; V; N ) =
(25)
kB T log Z:
In some situations one can also include variable particle numbers and one has to obtain the
grand canonical partition function
Zg = tre
H
=
P
exp (
(E
(26)
N ))
where N is the number of particles in the state j i. In this gas holds for the grand potential
(or Gibbs free energy)
(T; V; ) =
(27)
kB T log Zg :
and the mean particle number is
hN i =
1 P
N exp (
Zg
(E
N )) =
@
:
@
(28)
For indistinguishable particles it would be necessary to introduce the symmetrized or
antisymmetrized states in order to perform the above sum, which is technically very complicated. A way out of this situation is the so called second quantization, which simply respects
the fact that labeling particles was a stupid thing to begin with and that one should characterize a quantum many particle system di¤erently. If the label of a particle has no meaning,
a quantum state is completely determined if one knows which states of the system are occupied by particles and which not. The states of an ideal quantum gas are obviously the
momenta since the momentum operator
bl =
p
8
~
rl
i
(29)
commutes with the Hamiltonian of an ideal quantum system
N
X
b 2l
p
H=
:
2m
l=1
(30)
In case of interacting systems the set of allowed momenta do not form the eigenstates of
the system, but at least a complete basis the eigenstates can be expressed in. Thus, we
characterize a quantum state by the set of numbers
n1 ; n2 ; :::nM
(31)
which determine how many particles occupy a given quantum state with momentum p1 , p2 ,
...pM . In a one dimensional system of size L those momentum states are
~2 l
L
pl =
(32)
which guarantee a periodic wave function. For a three dimensional system we have
plx ;ly ;lz =
~2 (lx ex + ly ey + lz ez )
:
L
(33)
A convenient way to label the occupation numbers is therefore np which determined the occupation of particles with momentum eigenvalue p. Obviously, the total number of particles
is:
N=
X
np
(34)
np " (p)
(35)
p
whereas the energy of the system is
E=
X
p
If we now perform the summation over all states we can just write
!
X
X
Z=
exp
np " (p) N;Pp np
fnp g
where the Kronecker symbol
N;
number are taken into account.
P
p
(36)
p
np ensures
that only con…gurations with correct particle
Returning to our earlier problem of noninteracting quantum gases we therefore …nd
!
X
X
Zg =
exp
np (" (p)
)
(37)
p
fnp g
9
for the grand partition function. This can be rewritten as
Zg =
XX
:::
np1 np2
Y
np ("(p)
e
)
=
p
YX
p
np ("(p)
e
)
(38)
np
Fermions: In case of fermions np = 0; 1 such that
ZgFD =
Y
("(p)
1+e
)
(39)
p
which gives (FD stands for Fermi-Dirac)
FD
=
X
kB T
("(p)
log 1 + e
)
(40)
p
Bosons: In case of bosons np can take any value from zero to in…nity and we obtain
X
e
np ("(p)
)
=
X
) np
("(p)
e
=
np
np
1
1
e
("(p)
(41)
)
which gives (BE stands for Bose-Einstein)
ZgBE =
Y
1
X
log 1
1
("(p)
)
e
("(p)
)
log 1 + e
("(p)
)
e
(42)
p
as well as
BE
= kB T
(43)
:
p
A.
Analysis of the ideal fermi gas
We start from
=
kB T
X
(44)
p
which gives
hN i =
@
@
=
X
p
e ("(p) )
1 + e ("(p)
)
=
X
p
1
e
("(p)
)
+1
=
X
p
hnp i
(45)
i.e. we obtain the averaged occupation number of a given quantum state
hnp i =
Often one uses the symbol f (" (p)
1
e
("(p)
)
+1
(46)
) = hnp i. The function
f (!) =
10
1
e +1
!
(47)
is called Fermi distribution function. For T = 0 this simpli…es to
8
< 1 " (p) <
hnp i =
: 0 " (p) >
States below the energy
are empty.
(48)
are singly occupied (due to Pauli principle) and states above
(T = 0) = EF is also called the Fermi energy.
In many cases will we have to do sums of the type
Z
X
V
I=
f (" (p)) = 3 d3 pf (" (p))
h
p
(49)
these three dimensional integrals can be simpli…ed by introducing the density of states
Z 3
dp
(! " (p))
(50)
(!) = V
h3
such that
I=
Z
(51)
d! (!) f (!)
We can determine (!) by simply performing a substitution of variables ! = " (p) if " (p) =
" (p) only depends on the magnitude jpj = p of the momentum
Z
Z
V4
V4
dp
2
I=
p dpf (" (p)) = 3
d! p2 (!) f (!)
3
h
h
d!
such that
(!) = V
with A0 =
4
h3
p
p
4 m p
2m!
=
V
A
!
0
h3
(52)
(53)
2m3=2 . Often it is more useful to work with the density of states per particle
0
(!) =
p
(!)
V
=
A0 !.
hN i
hN i
(54)
We use this approach to determine the chemical potential as function of hN i for T = 0.
Z
Z EF
Z EF
2
3=2
hN i = hN i
! 1=2 d! = V A0 EF
(55)
0 (!) n (!) d! = hN i
0 (!) d! = V A0
3
0
0
which gives
h2
EF =
2m
If V = d3 N it holds that EF
h2
d 2.
2m
0 (EF ) =
6
2
hN i
V
2=3
(56)
Furthermore it holds that
V 2m
hN i 4 2 h2
6
11
2
hN i
V
1=3
=
3 1
2 EF
(57)
Equally we can analyze the internal energy
U =
=
@
log Zg =
@
X
p
such that
U=
X
p
" (p)
e
("(p)
" (p) hnp i =
)
@ X
log 1 + e
@ p
("(p)
)
(58)
+1
Z
(!) !n (!
) d! =
3
hN i EF
5
(59)
At …nite temperatures, the evaluation of the integrals is a bit more subtle. The details,
which are only technical, will be discussed in a separate handout. Here we will concentrate
on qualitative results. At …nite but small temperatures the Fermi function only changes in
a regime
kB T around the Fermi energy. In case of metals for example the Fermi energy
with d ' 1
10A leads to EF ' 1:::10eV i.e. EF =kB ' 104 :::105 K which is huge compared
to room temperature. Thus, metals are essentially always in the quantum regime whereas
low density systems like doped semiconductors behave more classically.
If we want to estimate the change in internal energy at a small but …nite temperature
one can argue that there will only be changes of electrons close to the Fermi level. Their
excitation energy is
Due to
0
(EF )
1
EF
kB T whereas the relative number of excited states is only
it follows in metals
U'
0
(EF ) kB T
3
hN i EF + hN i
5
0
0
(EF ) kB T .
1. We therefore estimate
(EF ) (kB T )2 + :::
(60)
at lowest temperature. This leads then to a speci…c heat at constant volume
cV =
CV
1 @U
=
hN i
hN i @T
2kB2
0
(EF ) T = T
(61)
which is linear, with a coe¢ cient determined by the density of states at the Fermi level. The
correct result (see handout3 and homework 5) is
2
=
3
kB2
0
(EF )
(62)
which is almost identical to the one we found here. Note, this result does not depend on the
speci…c form of the density of states and is much more general than the free electron case
with a square root density of states.
12
Similarly one can analyze the magnetic susceptibility of a metal. Here the energy of the
up ad down spins is di¤erent once a magnetic …eld is applied, such that a magnetization
M =
=
B
(hN" i
B hN i
For small …eld we can expand
Z
hN# i)
EF
0
(! +
B B)
0
(!
B B)
d!
(63)
0
0
(! +
B B)
'
M = 2
2
B
hN i B
= 2
2
B
hN i B
Z
EF
@
0
0
@
(!) +
0
0 (!)
@!
BB
which gives
(!)
d!
@!
0
(64)
(EF )
This gives for the susceptibility
=
@M
=2
@B
2
B
hN i
0
(65)
(EF ) :
Thus, one can test the assumption to describe electrons in metals by considering the ratio
of
and CV which are both proportional to the density of states at the Fermi level.
B.
The ideal Bose gas
Even without calculation is it obvious that ideal Bose gases behave very di¤erently at low
temperatures. In case of Fermions, the Pauli principle enforced the occupation of all states
up to the Fermi energy. Thus, even at T = 0 are states with rather high energy involved.
The ground state of a Bose gas is clearly di¤erent. At T = 0 all bosons occupy the state
with lowest energy, which is in our case p = 0. An interesting question is then whether this
macroscopic occupation of one single state remains at small but …nite temperatures. Here,
a macroscopic occupation of a single state implies
hnp i
> 0.
hN i!1 hN i
(66)
lim
We start from the partition function
BE
= kB T
X
log 1
e
X
("(p)
("(p)
)
(67)
:
(68)
p
which gives for the particle number
hN i =
@
@
=
p
13
1
e
)
1
Thus, we obtain the averaged occupation of a given state
1
hnp i =
e
("(p)
)
1
(69)
:
Remember that Eq.68 is an implicit equation to determine (hN i). We rewrite this as
Z
1
hN i = d! (!) (! )
:
(70)
e
1
The integral diverges if
> 0 since then for ! '
Z
(!)
hN i d!
!1
(!
)
(71)
if ( ) 6= 0. Since (!) = 0 if ! < 0 it follows
(72)
0:
The case
= 0 need special consideration. At least for
convergent and we should not exclude
(!)
! 1=2 , the above integral is
= 0.
Lets proceed by using
p
(!) = V A0 !
with A0 =
4
h3
p
(73)
2m3=2 . Then follows
hN i
= A0
V
< A0
Z
1
Z0 1
d!
d!
0
p
e
e
= A0 (kB T )3=2
!
(!
p
)
!
!
Z
1
1
1
dx
0
It holds
Z
0
We introduce
1
dx
x1=2
ex
1
=
kB T0 = a0
p
2
h2
m
&
3
2
hN i
V
x1=2
ex
1
' 2:32
(74)
(75)
2=3
(76)
with
a0 =
2
&
3 2=3
2
' 3:31:
(77)
The above inequality is then simply:
T0 < T:
14
(78)
Our approach clearly is inconsistent for temperatures below T0 (Note, except for prefactors,
kB T0 is a similar energy scale than the Fermi energy in ideal fermi systems). Another way
to write this is that
3=2
T
:
(79)
hN i < hN i
T0
Note, the right hand side of this equation does not depend on hN i. It re‡ects that we could
not obtain all particle states below T0 .
The origin of this failure is just the macroscopic occupation of the state with p = 0.
It has zero energy but has been ignored in the density of states since
(! = 0) = 0. By
introducing the density of states we assumed that no single state is relevant (continuum
limit). This is obviously incorrect for p = 0. We can easily repair this if we take the state
p = 0 explicitly into account.
hN i =
X
p>0
1
e
("(p)
)
1
+
1
e
1
for all …nite momenta we can again introduce the density of states and it follows
Z
1
1
hN i = d! (!) (! )
+
e
1 e
1
(80)
(81)
The contribution of the last term
N0 =
1
e
(82)
1
is only relevant if
N0
> 0:
hN i!1 hN i
(83)
lim
If
< 0, N0 is …nite and limhN i!1
N0
hN i
= 0. Thus, below the temperature T = T0 the
chemical potential must vanish in order to avoid the above inconsistency. For T < T0
follows therefore
3=2
T
+ N0
(84)
T0
which gives us the temperature dependence of the occupation of the p = 0 state: If T < T0
!
3=2
T
N0 = hN i 1
:
(85)
T0
hN i = hN i
and N0 = 0 for T > T0 . Then
< 0.
For the internal energy follows
U=
Z
d! (!) !
15
1
e
(!
)
1
(86)
which has no contribution from the "condensate" which has ! = 0. The way the existence
of the condensate is visible in the energy is via
U = V A0
It holds again
R1
0
3=2
dx e x x
Z
=
1
d!
3p
4
! 3=2
e
!
1
(T < T0 ) = 0 such that for T < T0
= V A0 (kB T )
5=2
Z
1
dx
0
x3=2
e
x
1
(87)
& (5=2) ' 1:78. This gives
U = 0:77 hN i kB T
T
T0
3=2
(88)
leading to a speci…c heat (use U = T 5=2 )
C=
This gives
S=
Z
0
which leads to
T
@U
5 3=2 5 U
=
T
=
@T
2
2T
c (T 0 ) 0 5
dT =
T0
2
=U
Z
T 3=2 .
T
T 01=2 dT 0 =
0
TS
5
3
T 3=2 =
2
U
3
N=
(89)
5U
3T
(90)
(91)
The pressure below T0 is
p=
@
5 @U
m3=2
=
= 0:08 3 (kB T )5=2
@V
3 @V
h
(92)
which is independent of V . This determines the phase boundary
(93)
pc = pc (vc )
with speci…c volume v =
V
hN i
at the transition:
h2
pc = 1:59 v
m
IV.
A.
5=3
:
(94)
SECOND QUANTIZATION
The harmonic oscillator: raising and lowering operators
Lets …rst reanalyze the harmonic oscillator with potential
V (x) =
16
m! 2 2
x
2
(95)
where ! is the frequency of the oscillator. One of the numerous approaches we use to
solve this problem is based on the following representation of the momentum and position
operators:
r
~
b
ay + b
a
2m!
r
m~! y
pb = i
b
a
b
a :
2
x
b =
(96)
From the canonical commutation relation
[b
x; pb] = i~
follows
(97)
b
a; b
ay = 1
[b
a; b
a] = b
ay ; b
ay = 0:
(98)
Inverting the above expression yields
r
m!
2~
r
m!
b
ay =
2~
b
a =
x
b+
x
b
i
pb
m!
i
pb
m!
(99)
demonstrating that b
ay is indeed the operator adjoined to b
a. We also de…ned the operator
b =b
N
ay b
a
(100)
which is Hermitian and thus represents a physical observable. It holds
We therefore obtain
i
i
b = m! x
N
b
pb
x
b+
pb
2~
m!
m!
m! 2
1
i
=
x
b +
pb2
[b
p; x
b]
2~
2m~!
2~
1
pb2
m! 2 2
1
=
+
x
b
:
~! 2m
2
2
b = ~! N
b+1
H
2
:
(101)
(102)
b are given as En = ~! n + 1 we conclude that the eigenvalues of
Since the eigenvalues of H
2
b
the operator N are the integers n that determine the eigenstates of the harmonic oscillator.
b jni = n jni :
N
17
(103)
Using the above commutation relation b
a; b
ay = 1 we were able to show that
b
a jni =
b
ay jni =
p
p
n jn
1i
n + 1 jn + 1i
(104)
The operator b
ay and b
a raise and lower the quantum number (i.e. the number of quanta).
For these reasons, these operators are called creation and annihilation operators.
B.
second quantization of noninteracting bosons
While the above results were derived for the special case of the harmonic oscillator there
is a similarity between the result
1
2
(105)
"p np
(106)
En = ~! n +
for the oscillator and our expression
Efnp g =
X
p
for the energy of a many body system, consisting of non-interacting indistinguishable particles. While n in case of the oscillator is the quantum number label, we may alternatively
argue that it is the number of oscillator quanta in the oscillator. Similarly we can consider
the many body system as a collection of a set of harmonic oscillators labelled by the single
particle quantum number p (more generally by p and the spin). The state of the many
body system was characterized by the set fnp g of occupation numbers of the states (the
number of particles in this single particle state). We the generalize the wave function jni to
the many body case
jfnp gi = jn1 ; n2 ; :::; np ; :::i
(107)
p
np jn1 ; n2 ; :::; np 1; :::i
p
b
ayp jn1 ; n2 ; :::; np ; :::i = np + 1 jn1 ; n2 ; :::; np + 1; :::i
(108)
and introduce operators
b
ap jn1 ; n2 ; :::; np ; :::i =
That obey
h
i
b
ap ; b
ayp0 =
18
p;p0 :
(109)
It is obvious that these operators commute if p 6= p0 . For p = p0 follows
p
b
ap b
ayp jn1 ; n2 ; :::; np ; :::i =
np + 1b
ap jn1 ; n2 ; :::; np + 1; :::i
= (np + 1) jn1 ; n2 ; :::; np ; :::i
(110)
and
b
aypb
ap jn1 ; n2 ; :::; np ; :::i =
p
np b
ayp jn1 ; n2 ; :::; np
1; :::i
= np jn1 ; n2 ; :::; np ; :::i
which gives b
ap b
ayp
(111)
b
aypb
ap = 1. Thus the commutation relation follow even if the operators
are not linear combinations of position and momentum. It also follows
n
bp = b
aypb
ap
(112)
for the operator of the number of particles with single particle quantum number p. The
b =P b
total number operator is N
aypb
ap . Similarly, the Hamiltonian in this representation is
p
given as
X
b=
H
p
which gives the correct matrix elements.
"p b
aypb
ap
(113)
We generalize the problem and analyze a many body system of particles with single
particle Hamiltonian
b2
p
b
h=
+ U (b
r)
2m
(114)
which is characterized by the single particle eigenstates
b
hj
i=" j
i:
(115)
is the label of the single particle quantum number. We can then introduce the occupation
number representation with
jn1 ; n2 ; :::; n ; :::i
h
i
y
and corresponding creation and destruction operators b
a ;b
a 0 =
(116)
;
0
. We can then perform
a unitary transformation among the states
j i=
P
U
j i=
19
P
j ih j i
(117)
The states j i are in general not the eigenstates of the single particle Hamiltonian (they
only are if U
= h j i =
). We can nevertheless introduce creation and destruction
operators of these states, that are most naturally de…ned as:
b
a =
and the corresponding adjoined equation
b
ay =
P
P
h j ib
a
(118)
h j i b
ay :
(119)
This transformation preserves the commutation relation (see below for an example).
We can for example chose the basis
second quantization
b=
U
and we can transform the result as
b =
U
=
It holds of course h jU (r)j
0
X
; ;
X
;
0
i=
R
as the eigenbasis of the potential. Then holds in
X
h jU (r)j i ay a
h j i h jU (r)j i h j 0 i b
ay b
a
0
h jU (r)j
d3 r
0
ib
ay b
a
(r) U (r)
(120)
0
0
(r).
0
In particular, we can chose j i = jri such that h j i = hrj i =
use the notation b
ar = b (r) and our unitary transformations are
b (r) = P
The commutation relation is then
h
(122)
h
i
y
a ;b
a 0
0 (r ) b
P
(r0 ) = hrj i h jr0 i
(123)
y
(124)
(r) b
ay
0
i
b (r) ; b (r0 ) = P
; 0
P
=
0
(r)
(r)
= hrjr0 i = (r
and it follows
b=
U
Z
r0 )
d3 rU (r) b (r) b (r)
20
(r). In this case we
(r) b
a
b y (r) = P
;
(121)
Similarly holds for the kinetic energy
Z
y
~2
b
d3 rd3 r0 r r2 r0 b (r) b (r0 )
T =
2m
Z
y
~2
=
d3 rd3 r0 b (r) r2 (r r0 ) b (r0 )
2m
Z
y
~2
=
d3 r b (r) r2 b (r)
2m
(125)
Thus we …nd
H =
=
X
Z
a
" b
ay b
~2 r2
+ U (r) b (r)
2m
y
d3 r b (r)
(126)
y
With the help of the …eld operators b (r) and b (r) is it possible to bring the many body
Hamiltonian in occupation number representation into the same form as the Hamiltonian of
a single particle.
C.
Example 1: a single particle
We consider the most general wave function of a single spinless boson:
Z
y
j i = d3 r (r) b (r) j0i
(127)
where j0i is the completely empty system. Let the Hamiltonian be
Z
~2 r2r
3 by
H = d r (r)
+ U (r) b (r)
2m
It follows
Hj
Z
Z
~2 r2r
+ U (r)
2m
Z
Z
y
~2 r2r
3
=
d r d3 r0 b (r)
+ U (r)
2m
Z
Z
~2 r2r
3
3 0 by
+ d r d r (r)
+ U (r)
2m
i =
3
dr
y
d3 r0 b (r)
(128)
y
(r0 ) b (r) b (r0 ) j0i
y
(r0 ) b (r0 ) b (r) j0i
(r0 ) (r
r0 ) j0i
(129)
The …rst term disappears since b (r) j0i = 0 for the empty state. Performing the integration
over r0 gives
Hj
i =
=
Z
Z
y
d3 r b (r)
d3 r
~2 r2r
+ U (r)
2m
~2 r2r
+ U (r)
2m
21
(r) j0i
y
(r) b (r) j0i
(130)
Thus, we need to …nd the eigenvalue of and eigenfunction of
~2 r2
+ U (r)
2m
to obtain
Hj
i="
Z
(r) = "
(131)
(r)
y
(r) b (r) j0i = " j
d3 r
i:
(132)
Thus, for a single particle problem we recover the original formulation of the "…rst quantization". The function
(r) in Eq.127 is therefore the wave function of the single particle
problem.
P
y
Using b (r) =
is nothing but
ay =
(r) b
ay follows b
R
y
d3 r
i=b
ay j0i
j
(r) b (r) and our above wave function
(133)
Applying the Hamiltonian to the wave function in this basis is obviously giving the same
answer.
Hj
D.
i=
X
0
" 0b
ay 0 b
a 0b
ay j0i = "a j
i
(134)
Second quantization of interacting bosons
Next we analyze the formulation of particle-particle interactions within the second quantization. We consider a two body interaction Vb that has, by de…nition, matrix elements that
depend on the states of two particles. Thus the expression for a single particle where
b=
U
X
;
0
h jU j
0
ib
ay b
a
will be determined by a matrix elements of the kind:
Z
0 0
h jV j
i = d3 rd3 r0 (r) (r0 ) V (r; r0 )
In general there will be a two particle basis j
Vb j
(135)
0
0
(r0 )
0
(r) :
(136)
i where the interaction is diagonal
i=V
j
i
(137)
b,
i. In this basis we can proceed just like for the interaction U
P
where the operator was given by
h jU j i b
ay b
a . In case of a two particle interaction we
where V
= h
jV j
22
have contributions if there are two particles, one in state
interaction must be
where Pb
the other in state . This the
1X
V Pb
Vb =
2
(138)
is the operator which counts the number of pairs of particles in the states j i
and j i. The prefactor
1
2
takes into account that each pair is considered only once.
If j i = j i, the number of pairs is n (n
1), while for j i =
6 j i it is n n , where the
n are the occupation numbers of those states. It follows
Pb
and we …nd
= n
b n
b
n
b
= ay ay a a = ay ay a a
1X
1X
Vb =
V ay ay a a =
h
2
2
jV j
(139)
i ay ay a a
Transforming this expression into an arbitrary basis j i =
operators in the new basis
b
ay =
b
a =
and it follows
P
P
1 X
h j ih j ih
Vb =
2 ;
which simpli…es to:
If for example
1X
Vb =
h
2
j i h j i ; we insert the
h j ib
ay :
h j ib
a
jV j
jV j
hr; r0 jV j r00 r000 i = v (r
P
(141)
i h j i h j i ay ay a a
i ay ay a a
r0 ) (r00
(140)
r0 ) (r000
(142)
(143)
r)
(144)
for an interaction that only depends on the distance between the two particles, it follows
Z
y
y
1
b
V =
d3 rd3 r0 v (r r0 ) b (r) b (r0 ) b (r0 ) b (r) :
(145)
2
23
E.
1.
Second quantization of noninteracting fermions
The fermionic "harmonic oscillator"
When we introduced the second quantized representation for bosons we took advantage
of the fact that the eigenstates of a free bose system
E=
X
(146)
" n
could be expressed in terms of the set fn g of occupation numbers. In case of bosons these
occupation numbers were allowed to take all integer values na = 0; 1;
; 1, reminiscent of
the quantum number of the harmonic oscillator. The latter then led to the introduction of
creation and annihilation operators of the bosons, where the occupation number operator
of a given state was n
b =b
ay b
aa . The Hamiltonian was then written as
b=
H
X
" n
b
(147)
Obviously, this approach cannot be used to describe fermions where n = 0 or 1. In case of
fermions, the single particle quantum state always includes the spin, for example
= (k; ).
We need to …nd the fermion analog to the harmonic oscillator, i.e. a state that only
allows for the two occupations n = 0 or 1. We want to express the Hamiltonian for a single
quantum state as
This is easily done with the help of a (2
b
h = "b
n
(148)
2) matrix representation (note, these matrices
have nothing to do with the spin of the system). If we introduce
0 1
0 1
1
0
j0i = @ A and j1i = @ A
0
1
(149)
for the empty and occupied state, it holds
0
n
b=@
0 0
0 1
1
A
(150)
We can equally introduce lowering and raising operators
b
a j0i = 0 and b
a j1i = j0i
24
(151)
as well as
b
ay j1i = 0 and b
ay j0i = j1i :
(152)
It follows easily that this is accomplished by
0
1
0
1
0 1
0
0
A and b
A .
b
a=@
ay = @
0 0
1 0
(153)
As in case of bosons, b
ay is the adjoined operator of b
a.
The action of these operators of a state with arbitrary occupation is then
b
a jni = n jn
b
ay jni = (1
1i = n j1
ni
n) jn + 1i = (1
n) j1
ni
If we now determine ay a it follows
0
10
1 0
1
0 1
0 1
0 0
A@
A=@
A
b
ay b
a=@
1 0
0 0
0 1
(154)
(155)
and we have, just as for bosons,
n
b=b
ay b
a:
(156)
b
ab
ay + b
ay b
a=1
(157)
b
ay b
ay = b
ab
a=0
(158)
However, an important di¤erence is of course that now holds
in addition we immediately see
Fermionic creation and annihilation operators do not commute, they anticommute:
b
a; b
ay
b
ay ; b
ay
+
+
= 1
= [b
a; b
a]+ = 0:
Note, we could have introduced equally
0
1
0 1
A and b
b
a= @
ay =
0 0
with the only change that now b
a j1i =
unchanged.
0
@
j0i and b
ay j0i =
25
(159)
0 0
1 0
1
A .
(160)
j1i and all other results remain
2.
Many fermi states
To generalize the single fermi result to many fermions we the any body wave function in
occupation number representation
jn1 ; n2 ; : : : ; n ; : : :i
(161)
aa for the individual
We then need to analyze the creation and annihilation operators b
ay and b
states, respectively.
At …rst glance it is natural to introduce(1
n) jn + 1i
b
aa jn1 ; n2 ; : : : ; n ; : : :i = n jn1 ; n2 ; : : : ; n
b
aya jn1 ; n2 ; : : : ; n ; : : :i = (1
1; : : :i
n ) jn1 ; n2 ; : : : ; n + 1; : : :i
(162)
(Note, these equations will turn out to be incorrect!)
This implies however that
while for di¤erent states
6=
0
b
a ;b
ay
+
(163)
=1
follows
h
i
b
a ;b
ay 0 = 2b
a b
ay 0
(164)
X
(165)
+
h
i
y
a result that follows from b
a ;b
a 0 = 0 for
basis to another, with
jli =
6=
0
. If we now want to transform from one
j i h jli
Just like in case of bosons the new operators should be linear combinations of the old ones,
which yields
b
al =
and the corresponding adjoined equation
b
ayl =
We now require
P
P
hlj i b
a
(166)
hlj i b
ay :
(167)
h
i
b
al ; b
ayl = 1
(168)
+
which leads to
1=
P
;
0
h
i
hlj i h 0 jli b
a ;b
ay 0
+
26
(169)
For a complete set of states hlj i this is only possible if
h
i
b
a ;b
ay 0 =
;
+
i.e. for
6=
0
(170)
0
the anticommutator and not the commutator must vanish. We conclude that
Eq.162 cannot be correct.
Jordan and Wigner realized that a small change in the de…nition of these operators can
…x the problem. To proceed we need to order the quantum numbers in some arbitrary but
…xed way. We then introduce
X1
=
(171)
n
0 =1
as the number of occupied states that precede the -th state. We can then introduce
1
0
1
0
0 0
0 1
A and b
A
ay = ( 1) @
(172)
b
a = ( 1) @
0 0
1 0
The matrix acts on the occupation of the -th state. As shown above, a prefactor
1 in the
de…nition of these operators causes no problem. It then follows
b
aa jn1 ; n2 ; : : : ; n ; : : :i = ( 1)
n jn1 ; n2 ; : : : ; n
1; : : :i
n jn1 ; n2 ; : : : ; 1
n ; : : :i
b
aya jn1 ; n2 ; : : : ; n ; : : :i = ( 1)
(1
n ) jn1 ; n2 ; : : : ; n + 1; : : :i
(1
n ) jn1 ; n2 ; : : : ; 1
= ( 1)
= ( 1)
n ; : : :i
(173)
It obviously holds that
We next analyze (assume
0
prior to )
b
a ;b
ay
b
aa0 b
aya jn1 ; : : : ; n 0 ; : : : ; n ; : : :i = ( 1)
+
n )b
aa0 jn1 ; : : : ; n 0 ; : : : ; 1
(1
+
= ( 1)
(174)
=1
0
n 0 (1
n ) jn1 ; : : : ; 1
n ; : : :i
n 0; : : : ; 1
n ; : : :i
On the other hand:
b
ayab
aa0 jn1 ; : : : ; n 0 ; : : : ; n ; : : :i = ( 1)
= ( 1)
0
n 0b
aya jn1 ; : : : ; 1
+
0
1
27
n 0 (1
n 0 ; : : : ; n ; : : :i
n ) jn1 ; : : : ; 1
n 0; : : : ; 1
n ; : : :i
It then follows
+
b
aa0 b
aya + b
ayab
aa0 = ( 1)
The same holds of course if we assume
Thus, we …nd
0
0
(n 0 (1
n )
n 0 (1
n )) = 0
(175)
to occur after .
h
i
b
a ;b
ay 0 =
(176)
0
;
+
as desired, yielding after a unitary transformation
h
i
h
i
P
b
al ; b
ayl0 =
hlj i h 0 jl0 i b
a ;b
ay 0 =
+
;
+
0
l;l0
(177)
i.e. the anticommutation relation of fermionic operators is independent on the speci…c
representation.
The Hamiltonian of noninteracting fermions is then
H=
X
" b
ay b
a
(178)
"k b
ayk; b
ak;
(179)
which in case of free particles reads
H=
X
k;
where k goes over all momentum values and
1
.
2
=
To incorporate interaction e¤ects is now rather similar to the case of bosons. We start
from the two particle basis j
Here V
=h
jV j
i where the interaction is diagonal
Vb j
i=V
j
i:
(180)
i. In this basis we can proceed just like for bosons. In case of a two
particle interaction we have contributions if there are two particles, one in state
the other
in state . This the interaction must be
where Pb
1X
Vb =
V Pb
2
(181)
is the operator which counts the number of pairs of particles in the states j i and
j i. The prefactor
1
2
takes into account that each pair is considered only once. It follows
again
28
Pb
= n
b n
b
n
b
= ay a ay a
=
ay a
ay ay a a + ay
ay a
a
= ay ay a a
and we …nd
(182)
1X
Vb =
V ay ay a a
2
(183)
just as in case of bosons. Transforming this expression into an arbitrary basis j i =
P
P
j i h j i where b
a = h j ib
a it holds
If for example
1X
Vb =
h
2
jV j
hr; r0 jV j r00 r000 i = v (r
i ay ay a a
r0 ) (r00
r0 ) (r000
(184)
r)
(185)
for an interaction that only depends on the distance between the two particles, it follows
Z
y
y
1
b
d3 rd3 r0 v (r r0 ) b (r) b (r0 ) b (r0 ) b (r) :
(186)
V =
2
3.
Example 1: free electron gas
We want to derive the ground state wave function of the free electron gas. The Hamiltonian of an individual electron is
h=
~2 r2
2m
(187)
~2 k 2
:
2m
(188)
which leads to the single particle eigenvalues
"k =
The Hamiltonian of the many fermion system is then
H=
X
k;
"k b
ayk; b
ak; :
29
(189)
The ground state wave function is the state where all single particle states with energy
(190)
"k < EF
are occupied and all states above the Fermi energy are empty. ,
~2
EF =
2m
3
2=3
2
hN i
V
(191)
was determined earlier. The ground state wave function is then
j
0i
Q
=
k; ("k <EF )
This state is normalized:
h
0j
0i
=
=
*
Q
0
*
k; ("k <EF )
Q
0
=
b
ayk; j0i
b
ak; b
ayk;
1
k; ("k <EF )
Q
h0j0i = 1
k; ("k <EF )
(192)
+
0
b
ayk; b
ak;
+
0
(193)
and it is indeed the eigenstate of the problem
Hj
0i =
X
k;
It follows immediately that
b
ayk; b
ak;
Q
q; ("q <EF )
"k b
ayk; b
ak; j
b
ayq; j0i = (EF
0i
"k ) j
(194)
0i :
(195)
Either k is among the states below the Fermi surface or it isn’t. This yields
Hj
with
E0 =
0i
X
= E0 j
(EF
0i
"k ) "k
(196)
(197)
k;
The states that contribute to the ground state energy are all those with an energy below
EF . Since "k =
~2 k2
2m
is implies that the magnitude of the momentum must be smaller than
a given value
~2 k 2
~2 kF2
<
= EF
2m
2m
30
(198)
Here kF is the so called Fermi momentum. All momentum states inside the sphere of radius
kF are occupied. Those outside are empty. Our above result for the Fermi energy yields
kF =
where
F.
3
1=3
2
(199)
= hN i =V is the electron density.
Example 2: two particles
The natural state of two noninteracting particles is
;
0
Applying the Hamiltonian
H=
to this wave function gives
H
;
0
=
=
=
X
=b
ay b
ay 0 j0i
(200)
" b
ay b
a
(201)
X
" b
ay b
a b
ay b
ay 0 j0i
X
" b
ay b
ay b
a b
ay 0 j0i + " b
ay b
ay 0 j0i
" 0b
ay 0 b
ay j0i + " b
ay b
ay 0 j0i
= (" + " 0 )
The eigenvalue is E
;
0
;
0
= " + " 0 . The wave function can also be written as
Z
y
y
1
d3 rd3 r0 (r) 0 (r0 ) b (r) b (r0 ) j0i
; 0 = p
2
(202)
(203)
Since a labelling of the particles is not necessary within the second quantization, there is no
need to symmetrize
(r)
0
(r0 ) in this formulation.
To determine the wave function in real space we analyze
0
(r; r0 ) = r; r0 j
;
0
(204)
It holds
y
y
jr; r0 i = b (r) b (r0 ) j0i
31
(205)
such that
hr; r0 j = h0j b (r0 ) b (r)
and we can analyze
0
r; r j
0
;
1
= p
2
Z
d3 r00 d3 r000
(206)
(r00 )
y
0
(r000 )
(207)
y
h0j b (r0 ) b (r) b (r00 ) b (r000 ) j0i
It holds
y
y
y
h0j b (r0 ) b (r) b (r00 ) b (r000 ) j0i =
+ (r
=
(r
(208)
y
h0j b (r0 ) b (r00 ) b (r) b (r000 ) j0i
y
r00 ) h0j b (r0 ) b (r000 ) j0i
y
r000 ) h0j b (r0 ) b (r00 ) j0i
y
r00 ) h0j b (r0 ) b (r000 ) j0i
+ (r
=
r000 ) (r0 r00 ) + (r
(r
r00 ) (r0 r000 )
Inserting this yields
0
1
(r; r0 ) = p (
2
(r)
0
(r0 )
(r0 )
0
(r))
(209)
This is of course the correct result for the wave function of two indistinguishable fermions.
V.
PERIODIC STRUCTURES AND BLOCH THEOREM
Following our Born-Oppenheimer decoupling we consider the ionic problem of the coupled
electron-ion system, governed by the Schrödinger equation:
Ti + Viief f
n
=E
n
(210)
with e¤ective ion-ion interaction Viief f = Vii + Eel;n . Here, the Eel;n are determined by the
electronic Schrödinger equation
Hel =
Ne
Ne
Ne
X
X
~2 2 X
e2
r +
U (rj ) +
:
2m j j=1
jrj rj 0 j
j=1
j;j 0 =1
The atoms in a solid tend to order in a crystalline form, i.e. most systems arrange
their atomic constituents at su¢ ciently low temperatures in a periodic lattice. here are a
32
number of exceptions, such as glasses (systems believed to be kinetically unable to reach
the crystalline ground state), quasicrystals (possibly metastable structures that arrange in
highly ordered, yet non-crystalline form), or He (forming in equilibrium a quantum ‡uid
without periodic order, related to its large zero-point ‡uctuations of the crystalline state).
The reason for the periodic arrangement is ultimately the dominance of the potential energy
term Viief f over the kinetic energy Ti .
Thus, it holds
n
(Ri ) = ei
n
(Ri + t)
where t is a vector that characterizes the periodic structure.
We can write
t =l1 a1 + l2 a2 + l3 a3
where the ai are the primitive translation vectors and the li are integers. The three primitive
translation vectors span a parallelepiped with volume
= a1 (a2
a3 ) :
This parallelepiped is also called the primitive unit cell of the crystalline solid. Inside this
unit cell one might …nd a number of distinct atoms, at locations rl relative to the origin
of the primitive cell. The set of the atoms within the unit cell and their corresponding
locations are refereed to as the basis of the unit cell. The choice of the primitive translation
vectors and thus, of the primitive unit cell is not unique. A very convenient choice is the
Wigner-Seitz cell. The Wigner–Seitz cell around a lattice point is de…ned as the locus of
points in space that are closer to that lattice point than to any of the other lattice points.
A Wigner–Seitz cell is a primitive cell spanning the entire Bravais lattice without leaving
any gaps or holes. Its volume is given by
de…ned above. The cell may be chosen by …rst
picking a lattice point. Then, lines are drawn to all nearby (closest) lattice points. At the
midpoint of each line, another line is drawn normal to each of the …rst set of lines. In the
case of a three-dimensional lattice, a perpendicular plane is drawn at the midpoint of the
lines between the lattice points. By using this method, the smallest area (or volume) is
enclosed in this way and is called the Wigner–Seitz primitive cell. All area (or space) within
the lattice will be …lled by this type of primitive cell and will leave no gaps.
Consider now a set of points t constituting a Bravais lattice, and a plane wave de…ned
33
by:
eir K
If this plane wave has the same periodicity as the Bravais lattice, then it satis…es the equation:
ei(t+r) K = eir K
such that
eit K
the reciprocal lattice as the set of all vectors K that satisfy the above identity for all lattice
point position vectors t. Thus
t K =2 m
with integer m. This reciprocal lattice is itself a Bravais lattice, and the reciprocal of
the reciprocal lattice is the original lattice.
The reciprocal lattice can be determined by
generating its three reciprocal primitive vectors, through
b1 =
b2 =
b3 =
2
2
2
a2
a3 ;
a3
a1 ;
a1
a2 :
where
K =m1 b1 + m2 b2 + m3 b3
with integers mi . The Brillouin zone is a primitive unit cell of the reciprocal lattice. In
other words, the Wigner–Seitz cell in the reciprocal lattice is the …rst Brillouin zone.
The simple cubic Bravais lattice, with cubic primitive cell of side a, has for its reciprocal a
simple cubic lattice with a cubic primitive cell of side 2 =a. The cubic lattice is therefore said
to be self-dual, having the same symmetry in reciprocal space as in real space. The reciprocal
lattice to an face-centered cubic (FCC) lattice is the body-centered cubic (BCC) lattice and
vice versa. The reciprocal to a simple hexagonal Bravais lattice with lattice constants c and
p
a is another simple hexagonal lattice with lattice constants 2 =c and 4 = 3a rotated
through 30 about the c axis with respect to the direct lattice. The reciprocal lattice of the
reciprocal lattice is the original, or direct lattice.
34
Consider a lattice periodic function, such as the electron density
(r).
Due to its
periodicity, we can expand it in a Fourier series
X
(r) =
iK r
;
Ke
K
where the periodicity implies that K is a vector of the reciprocal lattice. In order to determine the Fourier coe¢ cients, we write multiply
(r) with e
iK r
and integrate over the
volume
Z
iK r
3
d r (r) e
X
=
V
K0
K0
X
=
Z
0
d3 reiK r e
iK r
V
K0 V
=V
K0 ;K
K
K0
which yields
K
1
=
V
Z
d3 r (r) e
Z
d3 r (r) e
iK r
:
V
When we evaluate this integral it is, given the periodicity, su¢ cient to integrate only over
the unit cell with volume V0
K
1
=
V0
iK r
:
V0
Suppose the masses are distributed in point form
(r) = M
X
(r
t)
t
it follows
K
Z
MX
=
d3 r (r
V0 t V0
=
M
e
V0
iK t0
=
t) e
iK r
M
V0
where t0 is the one Bravais lattice point inside the unit cell. We obtain
(r) =
M X iK r
e :
V0 K
In particular, we …nd the identity
X
eiK r = V0
X
t
K
35
(r
t) :
Scattering measurements can determine the Fourier coe¢ cients
K.
Our above results
implies that all Fourier coe¢ cients are exactly the same. This is indeed only true in case of
a system with only one point-like atom per unit cell. Lets consider two pointlike atoms per
unit cell, one at the origin and one at a1 =2. It follows
(r) = M1
X
(r
t) + M2
t
X
(r
t
a1 =2) :
t
If we determine the Fourier coe¢ cients of this expression we …nd
K
M1 M2
+
e
V0
V0
M1 M2
=
+
e
V0
V0
M1 M2
=
+
e
V0
V0
iK a1 =2
=
:
im1 b1 a1 =2
i m1
=
M1
M2
+ ( 1)m1
V0
V0
The structure inside the unit cell is therefore re‡ected in a modulation of the density components in the reciprocal lattice. This is a general property that is being used to determine
the structure of complex unit cells.
A.
Bloch Theorem
Potential of ions are periodic with periods being the vectors of the Bravais lattice. We
are looking for the eigenfunctions of the Hamiltonian
~2 2
r + U (r)
2m
H=
where U (r) = U (r + t). According to the Bloch Theorem, the eigenstates have the
following form:
nk
(r) = eik r unk (r)
where unk (r) is periodic, i.e., unk (r) = unk (r + t) and k is Element of the …rst Brillouin
zone.
To prove the theorem we de…ne translation operator Tt such that Tt f (r) = f (r + t). Tt
is unitary (unitary operators satisfy U
1
= U y ). We have obviously
Tt
1
=T
36
t
(211)
To obtain Tty we note the following
h 1 j Tt j 2 i =
=
=
=
Thus Tty = T
t
Z
Z
Z
Z
d3 r
1 (r)Tt 2 (r)
d3 r
1 (r) 2 (r
d3 r
1 (r
d3 r(T
+ t)
t) 2 (r)
t 1 (r))
2 (r)
(212)
= Tt 1 .
Furthermore, Tt commutes with H, [Tt ; H] = 0.
Tt H
= H(r + t) (r + t) = H(r) (r + t) = HTt
(213)
All operators Tt commute with each other.
Tt1 Tt2
= Tt2 Tt1
= (r + t1 + tr )
(214)
which implies
Tt1 Tt2 = Tt2 Tt1 = Tt1 +t2 :
(215)
This means that the set of operators H, Tt (all of them) have common eigenstates (a full
set of them).
H
= E
(216)
Tt
= ct
(217)
From unitarity follows jct j = 1. From commutativity of Tt : ct1 ct2 = ct1 +t2 .
t are the vectors of Bravais lattice. Thus t = n1 a1 + n2 a2 + n3 a3 . This gives
ct = (ca1 )n1 (ca2 )n2 (ca3 )n3
We de…ne caj = e2
ixj
(218)
. Then
ct = e2
i(n1 x1 +n2 x2 +n3 x3 )
37
(219)
Now we start using the reciprocal lattice. We de…ne k =
P
xj bj where bj are the
elementary vectors of the reciprocal lattice. Then we can rewrite as follows
ct = eik t
Indeed, k t =
P
jl xj nl bj al = 2
Thus we obtain
P
j
(220)
xj nj (for reciprocal lattice we have bj al = 2
Tt = eik t
;
jl ).
(221)
i.e., each eigenvector is characterized by a vector k. Thus we have
= eik r u(r) ;
where u(r + t) = u(r). (We can de…ne u as e
ik r
(222)
).
Thus all eigenstates are split into families, characterized by di¤erent vectors k. Only
k belonging to the …rst Brillouin zone (the Wigner-Seitz unit of the reciprocal lattice) or
any other primitive unite of the reciprocal lattice give di¤erent families. This follows from
ei(k+K) t = eik t . Indeed, if k is outside the …rst Brillouin zone, then we can …nd K in the
reciprocal lattice so that q = k
K is in the …rst Brillouin zone. Then we use
= eik r u = eik r e
where u~
e
iK r
iK r
u = eiq r u~ ;
(223)
u and u~(r + R) = u~(r).
In each family we introduce the index n counting the states of the family. The functions
u depend on k 2 …rst B.Z. and on n. Thus, …nally we obtain:
nk
B.
(r) = eik r unk (r) :
(224)
Born-von Karmann boundary conditions
The B-v-K conditions read:
(r + Nj aj ) = (r) ;
for j = 1; 2; 3 and N1 ; N2 ; N3
(225)
1. The total number of primitive cells is then N1 N2 N3 .
P
This limits the possible values of k. Namely we must have eiNj k aj = 1. With k = xj bj
where bj are the elementary vectors of the reciprocal lattice we obtain xj = mj =Nj .
38
Although it would be better to chose all allowed values of k within the …rst Brillouin
zone it is simpler here to use a di¤erent primitive cell in the reciprocal lattice. Namely we
can chose mj = 0; 1; :::; Nj
1. This gives
k=
X mj
Nj
j
for mj = 0; 1; :::; Nj
(226)
bj ;
1. There are N = N1 N2 N3 allowed vectors k.
The volume in the reciprocal lattice per one vector k:
k1 k2 k3 =
k1 ( k2
k3 ) =
b1
N1
b2
N2
b3
N3
1 (2 )3
;
=
N
where
V
N
= a1 (a2
(227)
a3 ).
To calculate a sum over the whole primitive cell (1-st B.Z.) we use in the limit of large
N:
X
=
k
1.
Z
d3 k
N
!
k1 k2 k3
(2 )3
Z
V
dk=
(2 )3
3
Z
d3 k
(228)
Schrödinger equation with B-K boundary conditions
We expand both the wave function and the potential in the basis of momentum states,
i.e., plane waves. Thus:
(r) =
X
cq eiq r :
(229)
q
The boundary conditions, e.g., those of Born-von Karmann, make the set of k-vectors discrete:
k=
X mj
j
Nj
(230)
bj ;
where mj 2 Z. The sum is not limited to the …rst Brillouin zone.
The potential energy is a periodic function (Bravais-lattice). Thus it can be expanded as
U (r) =
X
UQ eiQ r ;
(231)
Q
where Q runs over the reciprocal lattice. We have
Z
1
UQ =
d3 r U (r)e
P:U:
39
iQ r
;
(232)
where the integration is over a primitive unit of the Bravais lattice and
the primitive unit. Since U is real (Hermitian) we have U
Q
is the volume of
= UQ .
The Schrödinger equation now reads
E
= E
X
~2 2
r +U
2m
cq eiq r =
q
=
X ~2 q 2
2m
q
=
X ~2 q 2
2m
q
cq eiq r +
X
UQ cq ei(Q+q) r
Q;q
cq eiq r +
where in the last line we substituted q ! q
X
UQ cq
Q
eiq r ;
(233)
Q;q
Q. The coe¢ cients in front of each harmonics
must satisfy this equation. Thus
~2 q 2
2m
E
cq =
X
UQ cq
Q
(234)
:
Q
We see that only q’s related by a vector of the reciprocal lattice in‡uence each other. Each
such family can be characterized by a vector in the 1-st Brillouin zone. Thus, in each family
we introduce k and all the q’s in the family are given by k + K, where K runs over the
reciprocal lattice. This gives
E
~2 (k + K)2
2m
ck+K =
X
UQ ck+K
Q
:
(235)
Q
The number of equations for each k 2 1-st B.Z. is in…nite as K runs over the whole reciprocal
lattice.
We will use the index n to count the solutions of Eq. (235). The solution number n is a
set cn;k+K for all vectors K 2 reciprocal lattice. Since Eq. (235) is a Schrödinger equation
and the sets cn;k+K are the wave functions, they are orthonormal, i.e.,
X
cn1 ;k+K cn2 ;k+K =
n1 ;n2
;
(236)
:
(237)
K
and complete
X
cn;k+K1 cn;k+K2 =
K1 ;K2
n
(Note that K serves here as coordinate of the wave function.)
The eigenstates in the coordinate representation then read
n;k (r)
=
X
cn;k+K ei(k+K) r = eik r
K
X
K
40
cn;k+K eiK r = eik r un;k (r) ;
(238)
where
X
un;k (~r)
cn;k+K eiK r :
(239)
K
Now, if we slightly change k, only the LHS of the equation (235) changes slightly. One
can expect that in each family n the states and the eigen-energies change smoothly. We
obtain bands.
2.
Properties of the Bloch states
Bloch states are orthonormal.
We obtain
Z
d3 r
n1 ;k1 (r)
n2 ;k2 (r)
=
X
Z
d3 r ei(k2 +K2
k 1 K1 )
cn1 ;k1 +K1 cn2 ;k2 +K2
k2 +K2 ;k1 +K1
;
cn1 ;k1 +K1 cn2 ;k2 +K2
K1 ;K2
= V
X
(240)
K1 ;K2
Since k1 and k2 both are in the 1-st B.Z. we have k2 +K2 ;k1 +K1 = k1 ;k2
Z
X
d3 r n1 ;k1 (r) n2 ;k2 (r) = k1 ;k2 V
cn1 ;k1 +K1 cn2 ;k1 +K1
K1 ;K2 .
Thus
K1
= V
k1 ;k2
In the thermodynamic limit V ! 1 we have V
n1 ;n2
k1 ;k2
:
! (2 )3 (k1
(241)
k2 ).
Basis of Bloch states is complete.
X X
n
=
X X
n
=
k21:B:Z
n;k (r1 )
X
n;k (r2 )
cn;k+K1 cn;k+K2 e
iK1 r1 iK2 r2 ik (r2 r1 )
e
e
k21:B:Z K1 ;K2
X X
ei(k+K) (r2
r1 )
k21:B:Z K
Momentum and crystal momentum
41
= V (r2
r1 ) :
(242)
The vector ~k is not the momentum and the Bloch states are not eigenstates of the
momentum operator. Indeed
b
p
n;k
=
i~r
n;k
= ~k
n;k
+ eik r run;k :
(243)
The vector ~k is called "crystal momentum".
In order to determine the expectation value of the momentum operator
Z
hnk jb
pj nki = i~
d3 re ik r unk (r) reik r unk (r)
Z
Z V
2
3
i~
d3 runk (r) runk (r)
= ~k
d r junk (r)j
V
V
Z
= ~k i~
d3 runk (r) runk (r) :
V
We insert
nk
(r) = eik r unk (r) into the Schrödinger equation
~2 r2
+U
2m
i
~2
~2 k 2
k r Enk +
m
2m
unk (r) = 0
We change k ! k+ k which yields Enk ! Enk + Enk = Enk + (@Enk =@k)
k as well as
unk ! unk + unk and it follows
~2 r2
+U
2m
i
~2
~2 k 2
k r Enk +
m
2m
unk = i
~2
k runk + Enk unk
m
~2 k k
unk
m
This is an inhomogeneous di¤erential equation for unk with inhomogeneity on the right
hand side. In order for a unique solution to exist, the solution of the homogeneous equation,
i.e. unk , must be orthogonal to the inhomogeneity:
Z
~2
3
k runk + Enk unk
d runk (r) i
m
V
~2 k k
unk
m
Inserting Enk yields
Z
~2
k
d3 runk (r) i runk + (@Enk =@k) unk
m
V
Since this should be true for an arbitrary k we obtain
Z
~2
d3 runk (r) i runk + (@Enk =@k) unk
m
V
which yields
i~
Z
V
d3 runk (r) runk = m~
42
1
~2 k
unk
m
~2 k
unk
m
(@Enk =@k)
~k
= 0:
=0
=0
such that
1
hnk jb
pj nki = m~
(@Enk =@k) = mvnk :
This, essential for the determination of the momentum is the energy dispersion Enk . Here
1
we also introduced the velocity vnk = ~
(@Enk =@k).
Discreetness of states indexed by n.
The Schrödinger equation for a given k
~2 (k + K)2
2m
E
ck+K =
X
UQ ck+K
Q
(244)
Q
can be rewritten for the function
X
uk (r)
ck+K eiK r :
(245)
K
as
~2 (k
ir)2
uk (r) = U (r)uk (r) ;
(246)
2m
accompanied by the periodic boundary conditions uk (r + t) = uk (r). The problem
E
thus must be solved in one primitive unit of the Bravais lattice and can give only
discreet spectrum.
VI.
ALMOST FREE ELECTRONS.
We start from the Schrödinger equation
~2 (k + K)2
2m
En;k
cn;k+K =
X
UQ cn;k+K
Q
(247)
Q
for the coe¢ cients of the function
X
un;k (~r)
cn;k+K eiK r :
(248)
K
Renaming K1
K and K2
En;k
K
Q we obtain
~2 (k + K1 )2
2m
cn;k+K1 =
X
UK1
K2 cn;k+K2
(249)
K2
We start from the limit of free electrons U = 0. The solutions of (249) are trivial: for
each n there is Kn such that
En;k =
~2 (k + Kn )2
2m
(0)
n;k
43
(250)
n and cn;k+Kl =
n;l .
Now consider U 6= 0. First, UQ=0 gives a total shift of energy. Thus, we take it into
account and put UQ=0 = 0. There are two possibilities:
1) For a given k there are no other vectors of the reciprocal lattice Kl such that
(0)
l;k
(0)
n;k
(more precisely the di¤erence of the two energies of order or smaller than U ). Then we are
in the situation of the non-degenerate perturbation theory. This gives for l 6= n
cn;k+Kl =
UKl Kn
(0)
(0)
n;k
l;k
+ O(U 2 )
(251)
and for the band energy we obtain
En;k =
(0)
n;k
+
X UK
Kl UKl Kn
(0)
(0)
n;k
l;k
n
l6=n
+ O(U 3 )
(252)
The bands repel each other.
2) There are some (at least one in addition to Kn ) vectors Kl 6= Kn such that
(0)
l;k
(0)
n;k .
We denote all m such vectors (including Kn ) by Kl with l = 1; : : : ; m. The degenerate
perturbation theory tells us to solve the following system of m equations (j = 1; : : : ; m):
Ek
~2 (k + Kj )2
2m
ck+Kj =
m
X
UKj
Ki ck+Ki
(253)
i=1
Consider a special, but most important case when the degeneracy is between two energies
corresponding to vectors K1 and K2 . First we note that the condition on k for this to
happen coincides with the one for the Bragg scattering of the X-rays. Namely, the condition
of degeneracy reads jk + K1 j = jk + K2 j = jk + K1
and K
K1
(K1
K2 )j. Introducing q
k + K1
K2 (K 2 reciprocal lattice) we see that the relation between the wave vectors
in the expanded band picture q = k + K1 and q
K = k + K2 is like between the wave
vectors of the incident and the re‡ected waves in the Bragg scattering. Both have to end
at the so called "Bragg plane" as depicted in Fig. 1. In particular the condition on q reads
jq Kj = 21 jKj.
The eigenvalues are determined as zeros of the determinant of the following matrix
0
1
(0)
Ek
U
K
1;k
@
A
(254)
(0)
U K Ek
2;k
The solutions read
E~k =
(0)
1;k
+
2
(0)
2;k
v
u
u
t
44
(0)
1;k
(0)
2;k
2
!2
+ jUK j2
(255)
In particular, the splitting exactly at the Bragg plain, where
E2;k
(0)
1;k
=
(0)
2;k
is given by
E1;k = 2jUK j.
Example in 1D:
Extended zone and Bragg scattering at
the BZ boundary in one dimension.
VII.
A.
TIGHT BINDING APPROXIMATION
Wannier functions
One can show that the Bloch states can be presented in a di¤erent form:
n;k (r)
=
X
eik R wn (r
(256)
R) ;
R
where R is a vector of the Bravais lattice and
wn (k) =
1
N
V
=
N
X
n;k
(r)
k21: B:Z:
Z
1: B:Z:
d3 k
(2 )3
n;k
(257)
(r) :
By operation of translation we obtain
wn (r
R) =
1
N
X
n;k (r
k21: B:Z:
45
R) =
1
N
X
k21: B:Z:
e
ik R
n;k
(r) :
(258)
Indeed, substituting Eq.(258) into Eq.(256) we obtain
X
X
ik R 1
e ik R n;k (r)
(r)
=
e
n;k
N
q21: B:Z:
R
X
=
r) = n;k (~r) :
k;q n;q (~
(259)
q21: B:Z:
Wannier functions of di¤erent bands n are orthogonal. Also orthogonal are the Wannier
functions of the same band but shifted to di¤erent R’s.
B.
Schrödinger equation for Wannier functions
Assume the total potential is a sum of atomic ones (for a simple Bravais lattice with one
atom per unit):
U (r) =
X
Ua (r
(260)
R) :
R
Then from
H
n;k
=
we obtain
En;k
X
eik R wn (r
X
~2
+
Ua (r
2m
R
R)
X
~2
+
Ua (r
2m
R
R) =
~
R
!
n;k
= En;k
!
R1 )
1
(261)
n;k
eik R wn (r
R) :
(262)
In the r.h.s. we separate the terms with R1 = R from those where R1 6= R:
X
X
~2
eik R wn (r R) =
En;k
+ Ua (r R) eik R wn (r R)
2m
R
~
R
X X
+
Ua (r R1 ) eik R wn (r R)
R R1 6=R
=
X
R
+
where
C.
U (r; R)
X
~2
+ Ua (r
2m
R)
U (r; R)eik R wn (r
eik R wn (r
R)
(263)
R) ;
~
R
P
R1 6=R
Ua (r
R1 ) = U (r)
Ua (r
R).
Linear Combination of Atomic Orbitals (LCAO)
Simplest approximation for the Wannier function w =
orbitals, such that Ha
m
= Ea;m
m.
P
m bm m
where
m
are the atomic
This can be, e.g., a multiplet of the orbital momentum
46
L with 2L + 1 degenerate states (we omit the band index n). This gives
X
bm (Ek
Ea;m )
m
X
eik R
m (r
R) =
X
bm
m
~
R
X
U (r; R)eik R
m (r
R) ;
~
R
We have restricted our Hilbert space to linear combinations of atomic orbitals
m (r
R)
shifted to all vectors of the Bravais lattice. While we cannot guarantee that Eq. (??) holds
exactly (in the whole Hilbert space) we can choose the coe¢ cients bm and the energy E~k so
that Eq. (??) holds in our restricted space. That is we demand that Eq. (??) projected on
all
m (r
R) holds. Due to the periodicity of the l.h.s. and the r.h.s. of Eq. (??) it is
su¢ cient to project only on
Projecting on
l (r)
m (r)
.
we obtain
(Ek
Ea;l )bl +
X
X
X
bm (Ek
Ea;m )
m
=
bm
m
eik R
R
Z
Introducing
Sl;m (R)
Z
tl;m (R)
we obtain
=
(E~k
Ea;l )bl +
X
X
m
bm
l (r)
Z
d3 r
X
U (r; R)
d3 r
ik R
m (r
l (r) m (r
l (r)
bm (Ek
m
e
e
ik R
R6=0
d3 r
and
X
U (r; R)
Ea;m )
Z
d3 r
l (r) m (r
R)
R) ;
(264)
R)
(265)
m (r
X
R)
(266)
eik R Sl;m (R)
R6=0
tl;m (R) ;
(267)
R
This is a homogeneous matrix equation on coe¢ cients bm . To have solutions one has to
demand that the determinant of the matrix vanishes, This determines the band energies
En;k . The number of bands is equal to the number of states in the multiplet.
D.
Single orbital (s states), one band
We obtain
P ik R
e
t (R)
PR
Ek = Ea +
;
ik
1 + R6=0 e R S(R)
47
(268)
Assume that only nearest neighbors matrix elements, do not vanish (and also t(0)). It is
important to note that S(R)
1. Thus
Ek
Ea + t(0) +
X
eik R t (R) ;
(269)
R;nn
Then, for di¤erent Bravais lattices with one ion per primitive cell we obtain:
1) 1-D lattice with step a.
(270)
Ek = Ea + t(0) + 2W cos(ak) ;
where W = t(a).
2) sc-lattice, a1 = ax; a2 = ay; a3 = az, (r) is rotational symmetric:
(271)
Ek = Ea + h(0) + 2t (cos(akx ) + cos(aky ) + cos(akz )) ;
where W = t (a).
3) bcc-lattice. One of the possible choices of the primitive basis is: a1 = ax; a2 =
ay; a3 = 21 a(x + y + z), however the nearest neighbors are at R = a2 ( x
p
8 neighbors each at distance 3a=2. We obtain
y
z). Altogether
E~k = Ea + t(0) + 8W cos(akx =2) cos(aky =2) cos(akz =2) ;
(272)
p
where W = t( 3a=2). (Interesting exercise: show that the reciprocal lattice in fcc).
E.
Alternative formulation of tight-binding method
Each primitive cell is characterized by states jR; mi. Index m can count either states of
the same atom or states of di¤erent atoms in the cell. For example in graphen we would
have m = A; B, where A and B denote sub-lattices. The overlaps of di¤erent states vanish:
hR1 ; m1 j jR2 ; m2 i =
R1 ;R2 m1 ;m2 .
H=
One postulates a tunneling Hamiltonian
X X
tm1 ;m2 (R1
R2 )ayR2 ;m2 aR1 ;m1
(273)
R1 ;m1 R2 ;m2
The Hamiltonian is Hermitian, i.e., tm1 ;m2 (R) = tm2 ;m1 ( R).
The Bloch states:
j
Ri
=
X
eik R
X
m
R
48
bm ayR;m j0i :
(274)
The Wannier w.f.: jwi =
P
y
m bm aR2 ;m2
j0i.
The energies and the coe¢ cients bm are determined by substituting the Bloch wave function into the Schrödinger equation: H j
Ri
= Ek j
R i.
We obtain
Hj
Ri =
X X
R2 )ayR2 ;m2 aR1 ;m1
tm1 ;m2 (R1
R1 ;m1 R2 ;m2
X X
=
tm1 ;m2 (R1
R2 )ayR2 ;m2
R1 ;m1 R2 ;m2
+
X X
R1 ;m1 R2 ;m2
= Ek
X
eik R2
R2
X
X
eik R
R
eik R
R
tm1 ;m2 (R1
X
m2
ik R1
R2 )e
bm1 ayR2 ;m2
X
m
X
m
bm ayR;m j0i
bm ayR;m aR1 ;m1 j0i
j0i
bm2 ayR2 ;m2 j0i
(275)
Comparing coe¢ cients in front of ayR2 ;m2 j0i we obtain
X
tm1 ;m2 (R1
R2 )eik R1 bm1 = Ek eik R2 bm2 :
(276)
R1 ;m1
With R
R1
R2
X
eik R bm1 tm1 ;m2 (R) = Ek bm2 :
(277)
R;m1
We again have reduced the problem to a matrix equation.
1.
Example: Copper-Oxide High temperature superconductors
The copper oxide high Tc superconductors are attracted the attention of the condensed
matter physics community for a number of decades now. Here we want to develop a simple
model to describe the electronic states in the vicinity of the Fermi level. Consider for example
the system La2 x Srx CuO4 . Electronically interesting states are usually those that are not
associated with closed electronic shells. It is known that elements like La or Sr tend to
donate electrons to obtain a closed shell con…guration. From the periodic table follows that
La3+ and Sr2+ are the corresponding valence states. Generally, oxygen tends to attract two
electrons to achieve its closed shell con…guration, i.e. one can assume (for a more detailed
discussion, see below) an O2 valence state. Charge neutrality then determines the valence
vCu of copper. Consider …rst x = 0 and we obtain:
2
3 + vCu
49
4
2=0
(278)
yielding vCu = 2. The electronic con…guration of Copper is Ar3d10 4s1 , where Ar stands
for the closed shell con…guration of argon. Thus, we expect for a Cu2+ state an electronic
con…guration Ar3d9 , i.e. with one hole per 3d-shell. In a spherical environment all 3d orbitals
are degenerate Already a cubic environment leads to a so called crystal …eld splitting of
the 3dxy , 3dxz , 3dyz orbitals from the 3dx2
y2
and 3dz2
orbitals. These two subsets of
3r2
orbitals are still degenerate among each other. However, the crystal structure of the cuprates
is tetragonal which leads to a further splitting of the 3dx2
energetically highest orbital is the 3dx2
y2
y2
3r2
orbitals. The
orbital. Thus, to achieve the 3d9 con…guration,
the electron will be transferred from the 3dx2
a model of a single hole in the 3dx2
and 3dz2
y2
y2
to the oxygen states. This leaves us with
orbital of copper. For x > 0, i.e. substituting Sr for
La, charge neutrality implies
(2
x)
3+x
2 + vCu
4
(279)
2=0
which leads to
(280)
vCu = 2 + x:
Substitution of La by Sr yields therefore additional hole doping.
Now, the Cu-sites are arranged on a square lattice, with weak hybridization perpendicular
to those planes. In between two copper sites is an oxygen site. The key orbital of those
oxygen states are the 2px and the 2py orbitals, depending of whether they connect two
copper sites along the x- or y-axis. Consider the overlap integral
Z
tx2 y2 ;x (ax) = d3 r x2 y2 (r) U (r; ax) x (r ax)
and
tx2
The potential
y 2 ;x (
ax) =
Z
d3 r
x2 y 2 (r)
U (r; R) is given by:
X
U (r; R) =
Ua (r
U (r; ax)
R1 ) = U (r)
x (r
Ua (r
(281)
+ ax)
(282)
R)
(283)
R1 6=R
Thus, in case of inversion symmetry (which exists in case of the copper-oxides) we have:
U (r; R) =
Since
x2 y 2 (r)
=
x2 y 2 (
r) and
tpd
x (r)
tx2
=
x(
y 2 ;x (ax)
=
50
(284)
U ( r; R)
r), follows
tx2
y 2 ;x (
ax)
(285)
The same is true for the y-direction.
Thus, if we con…ne ourselves to the overlap between those nearest neighbor copper and
oxygen states it follows:
Ek bm2 =
X
tm1 ;m2 (R)eik R bm1
(286)
R;m1
performing the sum explicitly yields:
Ek bd = "d bd + tpd 1
e
ikx a
bpx + tpd 1
Ek bpx = "p bpx + tpd 1
eikx a bd
Ek bpy = "p bpy + tpd 1
eiky a bd
iky a
e
bp y
(287)
which implies that we determine E from the eigenvalues of
0
It holds
det b
a = ("p
"d
B
B
b
a = B tpd 1
@
tpd 1
E
e
ikx a
e
"p
eiky a
4t2pd
E) "p "d
tpd 1
ikx a
tpd 1
E
0
e
iky a
0
"p
H
1
C
C
C
A
("p + "d ) E + E 2 + 2t2pd (cos kx a + cos ky a)
(288)
(289)
and we obtain
1
E =
2
"p + "d
and
q
("p
"d )2 + 16t2pd (k)
E 0 = "p
(290)
(291)
where
(k) = 1
1
(cos (kx a) + cos (ky a))
2
(292)
Thus, we obtain a bonding a non-bonding and an antibonding band. The bonding and
antibonding bands are mixtures of copper and oxygen states that hybridize. Charge neuality
demands that the number of holes in the lowest (bonding) band is 1 + x.
This model is not yet su¢ cient to describe the physics of the copperoxides. A prediction
of this model would be that the material for x = 0 would be a metal, as the Fermi level is
51
in the middle of the band. A widely accepted model that includes electron correlations is
the Hubbard model with
H = Htight
binding
+U
X
ndR" ndR#
(293)
R
where ndR
= dyR dR is the occupation number on the copper site. The strong local
Coulomb repulsion strongly suppresses charge ‡uctuations and for x = 0 and for su¢ ciently
large values of U does indeed lead to an insulating state. This state is called a Mott insulator.
VIII.
A.
ELECTRON-ELECTRON INTERACTIONS
The role of long range Coulomb interactions
Until now we have ignored the electron-electron Coulomb interaction. To get a …rst
qualitative sense for the role of this important interaction, we consider a simpli…ed model
where the electronic band structure is given by a parabolic band with Ek = ~k 2 = (2m).
Here m could also stand for the e¤ective mass that results from the inclusion of the periodic
lattice. Thus we analyze the Hamiltonian
H=
1 e2 X 1
~2 X 2
:
rri +
2m i
2 4 " i6=j jri rj j
(294)
In order to get a qualitative understanding of the role of the interaction term it is always
useful to introduce a dimensionless measure. To this end we consider a system with electron
density n. The resulting characteristic length scale of the problem is the mean electron
distance
d = n1=d
(295)
where d is the dimension of space. This is useful in particular due to the fact that a
realization of the above Hamiltonian is achieved in semiconductor heterostructures, where
in a narrow inversion layer an e¤ective two dimensional system has been realized, i.e. d = 2.
If we now introduce dimensionless positions
xi = ri =d
52
(296)
we can rewrite the above Hamiltonian as
!
X
X
~2
1
H=
r2xi +
2md2
4 i6=j jxi xj j
i
(297)
where we introduced a dimensionless strength of the interaction
=
where aB =
~2 "
me2
d
e2 md
=
2
~"
aB
(298)
is the Bohr radius known from the Hydrogen atom problem. Here the
dielectric constant " of the solid (or in d = 2, the dielectric constant of the substrate) can
lead to larger values compared to the canonical value in vacuum aB;0 =
10
10
~2
me2
= 0:529177249
m.
These considerations imply that systems with high electron density (which for the above
energy dispersion would be metallic) have small values of d, i.e. small values of the electronelectron interaction. This somewhat surprising result is a consequence of the fact that the
kinetic energy only dominates once the crystal momenta are large, which is the case in dense
systems. In this high-density limit we will …nd that the electron-electron interaction can
frequently be ignores or treated in a simpli…es perturbative way as small corrections to the
kinetic energy. The main e¤ect will be shown to be the emergence of plasma-oscillations
(collective density oscillations). On the other hand, in dilute systems, where the electronelectron interaction becomes large compared to aB , interactions are expected to be dominant. Here the kinetic energy can often be considered a small perturbation compared to
the electron-electron interaction. The ground state is then a result of the optimal Coulomb
interaction, At …xed density this leads to the formation of an electron crystal, referred to
as the Wigner crystal. The kinetic energy leads solely to vibrations of the electrons in the
crystal, similar to lattice vibrations of atoms in solids.
The above conclusions are a direct consequence of the assumed parabolic spectrum. Another, system of current interest in graphene, where electrons near isolated point in the BZ
have a linear spectrum. In this case we would start from the Hamiltonian
X
1 e2 X 1
;
H = i~v
rri +
2
4
"
jr
r
j
i
j
i
i6=j
(299)
the two atoms per unit cell. In terms of dimensionless length scales, follows
!
~v X
1X
1
H= i
rxi +
;
d
4 2 i6=j jxi xj j
i
(300)
where
=(
x;
y)
is the two dimensional vector of Pauli matrices, acting in the space of
53
where
=
e2
~v"
(301)
is a density-independent coupling strength. This is a result of the fact that both, the kinetic
and the potential energy, are proportional to inverse length-scales. Inserting the value of
the electron velocity of graphene yields
' 2:2=" and can be tuned by varying the dielectric
constant of the substrate. The key aspect of these considerations is that even the same
interaction leads to quite di¤erent behavior, depending on the nature of the band-structure
of a system. For example, the value of the velocity of graphene which implies
< 2:2,
excludes the option of a Wigner crystal in this system.
B.
Hartree-Fock approximation
The Hartree-Fock approximation is among the simplest methods to treat interactions in
many body systems. Here the dynamic problem of interacting electrons is replaced by an
e¤ective one-electron problem, where electrons are moving in a static …eld caused by the
other electrons.
We consider in second quantization the Hamiltonian of the many electron problem
XZ
~2 r2r
+ U (r)
(r)
d3 r y (r)
H =
2m
Z
1X
+
d3 rd3 r0 v (r r0 ) y (r) y 0 (r0 ) 0 (r0 ) (r) :
(302)
2 ; 0
The main spirit of the Hartree-Fock approximation is based on the identity
AB = (A
hAi) (B
hBi) + A hBi + B hAi
hAi hBi :
(303)
The …rst term is the joint deviation of the operators A and B from their expectation values.
This …rst term is called the correlation of the two operators AB and is being ignored in the
Hartree-Fock approach:
AB ' A hBi + B hAi
hAi hBi :
(304)
On the level of operators in the Hamiltonian, constants (like the last term hAi hBi) are
frequently being ignored. In case of the electron-electron interaction the corresponding
54
approximation is
y
(r)
y
0
0
(r )
0
0
(r )
y
(r) '
(r)
y
y
(r)
(r)
0
0
0
(r )
(r0 )
y
y
(r)
0
0
0
(r ) +
(r0 )
(r)
D
Thus, if we introduce
0
it follows for the Hamiltonian
XZ
HHF =
d3 r
XZ
;
(r; r0 ) =
y
0
y
D
0
0
(r )
y
0
y
d3 rd3 r0
E
E
(r)
y
(r)
y
(r)
(r)
0
(r0 )
r0 )
(r) v (r
(r )
(r0 )
~2 r2r
+ U (r) + UH (r)
2m
(r)
0
0
0
(r0 ; r)
(r)
0
(r0 )
(305)
0
where
UH (r) =
XZ
0
e2
=
4 "
d3 r0 v (r
Z
with electron density
d3 r0
0 0
(r0 ; r0 )
(r0 )
jr r0 j
X
(r) =
r0 )
0 0
(306)
(r0 ; r0 ) :
(307)
0
The last term in HHF is the Fock term It cannot be expressed in terms of the electron density
alone. Still HHF is a one particle Hamiltonian albeit with a non-local e¤ective potential.
We expand
(r) =
X
(308)
' (r) c
is ' (r) = hrj i is some complete set of states (for example a = k,n in case of Bloch
electrons). Then we obtain
HHF =
XX
;
with matrix elements
H
0
= H0
0
+
XX
00
V
H
0
cy c
(309)
0
0
D
cy 00 c
00
E
0
X
V
determined by the non-interacting matrix elements
Z
~2 r2r
0
+ U (r) ' (r)
H = d3 r' (r)
2m
55
D
cy 0 c
E
(310)
(311)
(r0 ) :
as well as the two-particle matrix elements:
Z
V
= d3 rd3 r0 ' (r) ' (r0 ) v (r
r0 ) ' (r0 ) ' (r) :
(312)
To solve the Hartree-Fock equations one starts from a trial ansatz for the matrix elements
D
E
y
c 0 c . Given a basis set, the V
are given and only need be determined once. The
next step is to evaluate all matrix elements of H
0
. Diagonalization of H
0
leads to the
eigenvalues of a problem that is formally noninteracting, i.e. the occupations of this oneD
E
particle problem can be analyzed explicitly. This leads to cy 0 c
and we can compare it
with our initial guess. If it agrees, the Hartree Fock equations are solved. Otherwise, we
D
E
y
may take those recalculated expectation values c 0 c
as starting point for the subsequent
evaluation of the H
C.
0
until the procedure converges.
Stoner Theory of Ferromagnetism
The Stoner theory of ferromagnetism is a famous application of the Hartree-Fock approach. We …rst make a simplifying assumption for the relevant matrix elements V
. We
consider only one band and take a single particle basis ' (r) = hrj i as Wannier functions,
i.e.
= i corresponds to a lattice site. Then we only include the matrix element with same
sites (i.e.
=
=
U
=
= i)
Viiii =
Z
d3 rd3 r0 'i (r) 'i (r0 ) v (r
r0 ) 'i (r0 ) 'i (r)
(313)
Then follows with the additional assumption that there is a well de…ned spin-quantization
D
E
0 hni i
axis cyi 0 ci =
Hij
0
=
Hij0 +
X
00
U hni 00 i
= Hij0 + U hni i
Here we introduced the notation that
!
ij
U hni i
0
(314)
0
is the spin opposite to . The resulting single-particle
Hamiltonian is therefore
HHF =
X
ij;
Hij0 + U hni i
56
ij
cyi cj :
(315)
This result could have been obtained from the above approximation for AB with A = ni"
and B = ni# if we start from the Hubbard Hamiltonian
HHF =
X
Hij0 cyi cj + U
ij;
X
ni" ni# :
(316)
i
Thus, we perform the decoupling
ni" ni# ' hni" i ni# + hni# i ni"
hni# i hni" i :
(317)
Ignoring the constant shift in the overall energy, this leads to exactly the same single-particle
Hamiltonian HHF . To solve this problem we assume that lattice translation invariance in
unbroken, i.e. hni" i = hn" i which allows Fourier transformation to Block states
HHF =
X
+ U hn i) cyk ck
("k
k;
(318)
which allows immediately to determine the occupations of momentum states
D
E
y
hnk i = ck ck
=
exp ( ("k
1
+ U hn i)) + 1
(319)
The occupation in real space follows as
hni i =
1 X
hnk i eik Ri :
N k2BZ
(320)
Given the assumed translation invariance we can evaluate this sum at any lattice site i and
take for convenience Ri = 0:
hn i =
1 X
N k2BZ exp ( ("k
1
+ U hn i)) + 1
(321)
This is a coupled equation for the occupations of the two spin states. Introducing
n = hn" i + hn# i
(322)
m = hn" i
(323)
for the total charge density and
hn# i
for the spin polarization (which will determine the macroscopic magnetization
M =N
57
Bm
(324)
with Bohr magneton
B.
It follows
hn i =
and we …nd
n=
and
1
(n + m)
2
(325)
1 X X
N k2BZ
e ("k
(n
+U
2
m))
1 X X
N k2BZ
e ("k
+U
(n
2
m))
m=
1
;
(326)
:
(327)
+1
+1
The …rst equation can be used to determine the chemical potential
of the system. It is
convenient to absorb the spin independent shift U2 n into a rede…nition of
!
r
=
U n=2.
Here we concentrate on the second term. that becomes
m =
=
1 X
N k2BZ
1 X
N
k2BZ
1
e ("k
1
U
2
+ 1 e ("k
sinh ( U m=2)
cosh ( U m=2) + cosh ( ("k
r
m)
U
r+ 2
m)
+1
(328)
r ))
Obviously, one solution of this nonlinear equation is always m = 0. To check whether there
are other solutions we write the sum over momenta in terms of an integral over the density
of states
Z
sinh ( U m=2)
d! (!)
cosh ( U m=2) + cosh ( (!
r ))
R
which is normalized to one d! (!) = 1. This follows from the de…nition
m=
(!) =
1 X
(!
N k2BZ
"k )
along with the fact that there are exactly N crystal momenta in the BZ:
Z
Z
1 X
d! (!) =
d! (! "k )
N k2BZ
1 X
=
1 = 1:
N k2BZ
At high temperatures follows with sinh (x
1) ' x and cosh (x
Z
U
U
m=m
d! (!) = m
:
4kB T
4kB T
58
(329)
(330)
(331)
1) ' 1
(332)
Since by assumption kB T
U it follows that only m = 0 is an allowed solution. Suppose
we are looking for very small, but …nite m as solutions, which justi…es a Taylor expansion
on the right hand side:
m = am
bm3 +
(333)
with coe¢ cients:
Z
U
1
d! (!)
a =
2
1 + cosh ( (!
r ))
3 3 Z
U
cosh ( (!
2
r ))
b =
d! (!)
2
48
(1 + cosh ( (!
r )))
(334)
Nontrivial solutions with small m can only exist if a > 1 and b > 0. Thus, at the transition
from ordered state to disordered state yields
Z
1
U c
d! (!)
2
1 + cosh ( c (!
r ))
=1
(335)
An interesting limit is Tc ! 0 which addresses the issue of when does the ferromagnetic
ground state is allowed to exist. Here we use that
(x) = lim
!1
such that Tc ! 0 corresponds to
cosh ( x)
U
( r ) = 1:
2
(336)
(337)
Thus, at T = 0 a ferromagnetic ground state is expected for U ( r ) =2 > 1. This condition
is called the Stoner criterion.
In case where b < 0 the transition cannot have an in…nitesimally small solution. However
a detailed analysis yields that a discontinuous jump to a …nite value is possible. This
behavior corresponds to a …rst order phase transition.
D.
Landau Theory of Fermi-liquids
The key concept underlying Fermi liquid theory is adiabacity, i.e. the assumption that
the low energy excitations of an interacting Fermi system are in one-to-one correspondence
to the excitations of a non-interacting Fermi gas. The theory was originally developed for
3
He, which at low temperatures is a structureless fermion due to the net spin 1=2 in the
59
nucleus. The proton charge is compensated by the two electrons that forma a singlet state
and therefore don’t contribute to the total spin of the system. One simplifying aspect of
3
He is the absence of an underlying crystalline lattice. The bare dispersion is then given in
form of the free particle dispersion
~k 2
:
(338)
2m
The quantum numbers of the excited states of a free fermi gas are the occupations nk of
=
"free
k
single particle states, the corresponding single particle states are characterized by momentum
and spin: jk i.
Momentum and spin of excitations can only serve as quantum numbers if an excitation
with jk i doesn’t immediately decay into other states. To demonstrate that the decay rate
of fermions with excitation energies
kB T is indeed small, we use Fermi’s golden rule for
the decay of one particle with energy " above the Fermi surface. Assume further that all
other states below the Fermi surface are …lled and those outside the Fermi surface are empty.
Then we would consider a collision of the particle with energy " with an occupied state of
energy "2 leading to …nal states with energy "3 and "4 outside the Fermi sea. Measuring all
energies relative to the Fermi surface we obtain for the decay rate
Z
Z
Z
2
= V
d"2 ("2 ) d"3 ("3 ) d"4 ("4 )
f ("2 ) (1
f ("3 )) (1
f ("4 )) (" + "2
"3
"4 )
(339)
Here, the integration should go over all energies with …nite density of states. Since the
integral is convergent and dominated by small energies, we can without problem extend the
integration limit to
1. First we integrate over "2 :
Z
Z
2 3
= V F d"3 d"4 f ("3 + "4 ") (1
f ("3 )) (1
f ("4 )) :
(340)
where we also approximated all densities of states by their value at the Fermi level. Next we
take into account that for T = 0, "3 ; "4 > 0 and "3 +"4 < ", which determines the integration
limits
Z
"
Z
" "3
1
d"4 = V 2 3F "2
2
0
0
Similarly we can perform the integration at " = 0 and for …nite temperatures. Now
Z
Z
1
1
1
2 3
(T ) = V F d"3 d"4 ("3 +"4 )
"
"
3
e
+1e
+1e 4 +1
Z
2
"3
1
2 3
= V 2 3F d"3 "3
=
V
(kB T )2 :
F
"
3
e
1e
+1
4
(") = V
2 3
F
d"3
60
(341)
(342)
Thus we …nd that for su¢ ciently small energy is
max ("; kB T ). In case of a wave
function, where damping leads to
(t) / exp
i
"t
~
1
t
~
(343)
it implies that the coherent oscillations go on for many periods until the damping leads
to a suppression of the oscillating amplitude. Thus, we demonstrated that single particle
excitations above the Fermi energy do not immediately radiate and decay.
In the ground state we continue to assume that the system is characterized by a …lled
Fermi sea with
(0)
nk = (kF
(344)
k)
where kF is the same as for an ideal Fermi gas with same density:
Z
d3 k (0)
N
= 2
nk
V
(2 )3
Z kF
1
kF3
2
= 2
k dk = 2
3
0
which yields kF = (3 2 N=V )
1=3
(345)
. Excitations are now characterized by changes nk of the
occupations, i.e.
(0)
nk = nk + n k :
(346)
The corresponding change in energy is then given by
X
"k n k :
E=
(347)
k;
Since the labelling of quantum numbers is the same compared to the free fermi system, purely
statistical aspects, like the entropy, should also only be determined by the corresponding
ideal fermi gas expressions. This implies
X
S = kB
(nk log nk + (1
nk ) log (1
(348)
nk ))
k;
Maximizing this expression with the condition that E =
yields
k;
"k nk and N =
P
k;
nk ,
:
(349)
e "k + 1
is measured relative to the Fermi energy. Near EF we make
nk =
where the excitation energy "k
1
P
the assumption
"k ' "0k = v (k
61
kF )
(350)
where the parameter v = kF =m is often expressed in terms of the e¤ective mass m . This
immediately leads to the density of states
F
m kF
2 ~2
=
(351)
that is enhanced by the factor m =m compared to the free fermion expression. An immediate
consequence of this modi…ed density of states emerges or the heat capacity
Z 1
X
d
1
d
nk = F
d""
C =
"k
"=(k
T) + 1
B
dT
dT e
1
k;
=
(352)
T
where
2 2
kB
m
3
m free
is the heat capacity of non-interacting fermions.
=
where
free
F
(353)
=
A key additional aspect of the Landau theory is that changes in the occupations nk
will lead to changes "k of the quasi-particle energies. Thus one writes generally
"k = "0k + "k
where
1 X
fk
N k0 ; 0
"k =
where fk
;k0
0
;k0
0
(354)
n k0
(355)
0
is a phenomenological interaction parameter that determines the extend to
which the energy of state jk; i is a¤ected by a change in population nk0
0
of jk0 ; 0 i. If we
do not want to have a preference of one spin direction over the other (in the absence of an
external magnetic …eld) we write
fk
;k0
0
s
= fk;k
0 +
0 a
fk;k0 :
(356)
In addition we assume that all relevant momenta are on the Fermi surface, i.e. k =kF ek
and k0 =kF ek0 where e2k = e2k0 = 1 are unit vectors. In an isotropic system like 3 He, with no
s;a
preferred direction, one expects that fk;k
0 only depend on the angle
on cos
k;k0
between k and k0 , i.e.
= ek ek0 :
s;a
s;a
fk;k
cos
0 = f
k;k0
:
(357)
Since f s;a are of dimension energy, dimensionless quantities follow via
F s;a =
62
Ff
s;a
:
(358)
We expand F s;a cos
in Legendre polynomials
k;k0
F
s;a
cos
k;k0
=
1
X
(2l + 1) Fls;a Pl cos
(359)
k;k0
l=0
and use the usual representation in terms of spherical harmonics
F
s;a
cos
k;k
0
1 X
l
X
=4
Fls;a Ylm (ek ) Ylm (ek0 ) :
(360)
l=0 m= l
The orthogonality of the Legendre polynomials
Z
1 1
l;l0
d cos Pl (cos ) Pl0 (cos ) =
2 1
2l + 1
(361)
allows for the representation
Z
1 1
s;a
d cos F s;a (cos ) Pl0 (cos )
Fl =
2 1
Z 2
Z
1
=
d'
sin d F s;a (cos ) Pl0 (cos )
4 0
Z 2
Z0
Z
1
d'
sin d
d"F s;a (cos ) Pl0 (cos ) (")
=
4 0
0
2 1 X s;a
=
Fk;k0 Pl cos k;k0 ("k ) :
F N k0
(362)
Suppose we are at T = 0 we add an external perturbation of the type
"0k = vl Ylm (ek )
(363)
where the spherical harmonic determines the directional dependence on the momentum
and vl is the amplitude of the perturbation. In case of an external magnetic …eld holds
"0k =
B B,
i.e. we have m = l = 0 and, due to Y00 = 4 , v0 =
amounts to "0k =
in the chemical potential
B B=4
. A change
, i.e. m = l = 0 and v0 =
=4 .
This allows us to determine physical observables like the magnetic susceptibility
s
with magnetization M =
B
P
k;
=
1 @M
V @B
(364)
B=0
nk . Another option is the charge susceptibility
c
=
1 @N
V @( )
63
(365)
=0
with particle number N =
P
k;
nk .
c
is closely related to the compressibility
1 @V
V @p
=
In case where the free energy of a system can be written as
(366)
F (V; N ) = N f (n)
where n = N=V is the particle density, it follows
p=
@F
@V
= n2
N
@f (n)
@n
(367)
which allows us to write
@p
@P
=n
@V N
@n
On the other hand it holds for the chemical potential
1
=
=
@F
@N
(368)
V
= f (n) + n
N
@f (n)
@n
(369)
The change of the pressure and chemical with density are then:
@p
@f (n)
@ 2 f (n)
= 2n
+ n2
@n
@n
@n2
2
@f (n)
@ f (n)
@
= 2
+n
@n
@n
@n2
(370)
which implies
@p
@
=n
@n
@n
(371)
and we obtain
=n
2 @n
@
(372)
:
In case of a free Fermi gas follows
0
s
=
2 0
B F
s
=
0
F:
(373)
where the mass is m.
Since fermions of a Fermi liquid are interacting it there is no reason that an external …eld
or chemical potential change will change the quasiparticle energies "k in the exact same
fashion. Thus, we assume
"k = "0k +
1 X
fk
N k0 ; 0
= tl Ylm (ek )
64
;k0
0
nk0
0
(374)
where in general tl 6= vl . Suppose there is such an energy shift, then we can determine the
associated particle density shift from Eq.349
1
nk =
e (
= (kF
"0k
)+1
+ "k
"0k
k)
(375)
"k
which yields
"0k
nk =
(376)
"k :
This result can now be inserted into Eq.374 which yields
1 X
fk
N k0 ; 0
"k = "0k
;k0
0
"0k0
;k0
0
" k0
0
(377)
0
or equivalently
tl Ylm (ek ) = vl Ylm (ek )
1 X
fk
N k0 ; 0
Now we can use the above expansion of fk
tl Ylm (ek ) = vl Ylm (ek )
= vl Ylm (ek )
= vl Ylm (ek )
F
2
F
;k0
0
Z
Z
2
Z
1
d
2
"0k0
0
tl Ylm (ek0 )
0
in spherical harmonics
d
k0
X
fk
;k0
0
0
tl Ylm (ek0 )
0
d
4
k0
k0
X
fk
;k0
0
0
tl Ylm (ek0 )
0
1
l0
XX
X
0
(Fls0 +
Yl0 m0 (ek ) Yl0 m0 (ek0 ) tl Ylm (ek0 )
We use the orthogonality of the spherical harmonics
Z
d k0 Yl0 m0 (ek0 ) Ylm (ek0 ) =
tl = vl
0
Fla0 )
l0 =0 m0 = l0
0
and obtain
(378)
1X 0 s
t (Fl +
2 0 l
ll0 mm0
0
Fla )
If we consider for example a change in the chemical potential with, v0 =
(379)
(380)
(381)
=4 , it follows
for t0 = t0 spin independent, that
t0 = v0
65
t0 F0s
(382)
which leads to
t0 =
v0
:
1 + F0s
(383)
It is now straightforward to determine the charge susceptibility via
c
To determine
4
1 1 @N
V @t0
1 @N
1 1 @N
=
V @( )
4 V @v0
1 1 @N @t0
=
4 V @t0 @v0
1 1 @N
1
:
=
s
1 + F0 4 V @t0
=
(384)
we use
1
nk =
kF
m
e
The derivative of N =
P
k;
(k kF )+4 t0
+1
nk with respect to t0 can be performed, e.g. by resorting to
our above result for the charge susceptibility of a free electron gas (with the di¤erence that
we need to consider the e¤ective mass, not the bare mass). It follows
4
1 1 @N
V @t0
=
F
and we
obtain
c
=
F
1 + F0s
Thus, the charge susceptibility is di¤erent from the free fermi gas value in two ways. First,
in the density of states the mass m is replaced by the e¤ective mass m . In addition the
interactions lead to an overall coe¢ cient (1 + F0s ) 1 . The theory is therefore stable as long
as F0s <
1. If F0s !
1 the system will undergo a transition to a regime where di¤erent
densities phase segregate. An analogous analysis can be performed for the spin susceptibility
(homework).
IX.
A.
DYNAMICS OF BLOCH ELECTRONS
Semi-classical equation of motion of Bloch electrons
We want to describe the evolution of electron’s wave function when a weak and slowly
changing external …eld is added. That is the Hamiltonian now reads
H=
i~r + ec A(r)
2m
66
2
+ U (r)
e (r) ;
(385)
where (the signs are consistent with negative charge, that is e = jej > 0, but the charge of
the electron is
e < 0.) The potential U (r) is periodic while A and
change little on the
scale of primitive cell of the Bravais lattice (slow …elds).
Our aim to prove that the electrons in the band n with energy En (k) are governed by
the following e¤ective Hamiltonian
He
B.
;n
= En
ir +
e
A
~c
(386)
e :
Wave packet argument
We localize the electron of a certain band n into a wave packet:
Z
Z
3
(r) = d k g(k) n;k (k) = d3 k g(k)un;k (k)eik r :
(387)
The function g(k) is centered around a certain crystal momentum k0 and has a width
such that the width of the wave packet in the real space
related as
k r
r is small enough. The two are
1.
The time evolution of the wave packet is given by
Z
(r; t) = d3 k g(k)un;k (r)eik r
We expand around k0 and
whole interval of
k
n;k0 .
un;k0 (~r)e
n;k t=~
(388)
:
We assume one can approximate un;k (r)
k. Then
(r; t)
i
ik0 r i
n;k0 t=~
Z
i k r
d3 k g(k)e
@ n;k
t=~
@k
:
un;k0 in the
(389)
Thus we conclude that the wave packet propagates with the velocity
v=
@r
1 @ n;k
=
:
@t
~ @k
(390)
Assume now the electron is in‡uenced by an electric …eld E. The work done by the …eld
pro unit of time is
eE v. This work is "used" to change the energy of the electron. Thus
we obtain
@
@ n;k dk
dk
=
= ~v
=
@t
@k dt
dt
eE v :
(391)
Thus we obtain
dk
= eE :
(392)
dt
The quasi-momentum ~k satis…es the same equation as the usual momentum for free elec~
trons!
67
C.
Proof for potential perturbation (not for vector potential)
We consider the following problem
H=
( i~r)2
+ U (r) + Uext (r) = H0 + Uext (r) ;
2m
Here U is the periodic lattice potential and Uext =
(393)
e is the external and weak potential.
More precisely what has to be weak is the external electric …eld, i.e.,
rUext .
We want to solve the time-dependent Schrödinger equation:
i~
@
=H
@t
(394)
We expand (t) in basis of Wannier functions
(t) =
X
R)
(395)
R) :
(396)
an;R (t)wn (r
n;R
Recall the representation of a Bloch wave function
r) =
n;k (~
X
eik R wn (r
R
In this case an;R = eik R . Wannier functions are given by
wn (r) =
1
N
and
wn (r
R) =
1
N
X
n;k (r)
(397)
:
k21: B:Z:
X
e
X
e
ik R
H0
n;k (r)n
e
ik R
"n;k
n;k (r)
ik R
n;k (r)
(398)
:
k21: B:Z:
First we investigate how H0 acts on the (shifted) Wannier functions using the fact
H0
n;~k
=
n;k
n;k .
H0 wn (r
R) =
1
N
1
=
N
k21: B:Z:
X
(399)
:
k21: B:Z:
We use now the Wannier expansion (396) and obtain
H0 wn (r
X 1 X
"n;k eik (R1 R) wn (r
N
R1
k21: B:Z:
X
=
"n (R1 R)wn (r R1 ) ;
R) =
R1
68
R1 )
(400)
where
1
N
"n (R)
X
"n;k eik R :
(401)
k21: B:Z:
The Schrödinger equation now reads:
X
@
= i~
a_ n;R (t)wn (r R)
@t
n;R
X
= H = (H0 + Uext ) =
an2 ;R2 (t)(H0 + Uext )wn2 (r
i~
X
=
an2 ;R2 (t)
n2 ;R2
+
X
X
R2 )
n2 ;R2
"n2 (R1
R2 )wn2 (r
R1 )
R1
an2 ;R2 (t)Uext wn2 (R
(402)
R2 ) :
n2 ;R2
The Wannier functions form a complete orthonormal basis. Thus we just compare the
coe¢ cients:
i~a_ n;R =
X
an;R2 "n2 (R
R2
+
X
an2 ;R2
n2 ;R2
Z
R2 ))
d3 r wn (r
R)Uext (r)wn2 (r
(403)
R2 ) :
The …rst term in the r.h.s. of (403) is rewritten as follows
X
an;R2 "n (R
R2 ) =
R2
= "n (k !
X
"n (R1 )an;R
R1
ir)an;R :
Here we have used an;R
R1
=e
iR1 ( ir)
R1
=
X
"n (R1 )e
iR1 ( ir)
an;R
~1
R
(404)
an;R . That is already here we consider an;R as a
well behaved function in all the space, i.e., an;r .
The second term of the r.h.s. of (403) is approximated as
Z
X
an2 ;R2 d3 r wn (r R)Uext (r)wn2 (r R2 )
Uext (R)an;R :
(405)
n2 ;R2
That is only diagonal matrix elements of Uext are left. Since Uext is slowly changing in space,
i.e,. it changes very little on the scale of primitive cell, while the Wannier functions are
localized on the scale of a cell this approximation is justi…ed.
Thus, the SE for the "envelope" wave function an;R~ reads
i~a_ n;R = [ n ( ir) + Uext (R)] an;R :
69
(406)
~ and de…ne it in the
If we now "forget" that an;R~ is de…ned only in the locations R
whole space, an;r we obtain a Schrödinger equation with the e¤ective Hamiltonian He
;n
=
"n ( ir) + Uext (r). In presence of vector potential it becomes (with no proof given here)
He
;n
= "n
ir +
70
e
A
~c
e :
(407)