Download SOLID-STATE PHYSICS 3, Winter 2008 O. Entin-Wohlman

SOLID-STATE PHYSICS 3, Winter 2008
O. Entin-Wohlman
1.
ELECTRONIC TRANSPORT PROPERTIES
2.
WEAK LOCALIZATION
3.
ANDERSON LOCALIZATION, THE METAL-INSULATOR TRANSITION
4.
TRANSPORT BY LOCALIZED ELECTRONS: PHONON-ASSISTED HOP-
PING
5.
LINEAR RESPONSE AND THE FLUCTUATION-DISSIPATION THEOREM
6.
LINEAR RESPONSE–THE KUBO FORMULA
Second-quantization of single-particle operators. Here we consider particles whose
number is conserved, e.g., electrons, and ignore the spin index (for the time-being). Let
us imagine a system of many particles which do not interact with one another. The Hamiltonian is
H=
X
Hi ,
(6.1)
i
where i runs over all particles, and
Hi ≡ H(ri ) = −
~2 2
∇ + U (ri ) .
2m i
(6.2)
(It is assumed that all particles are identical, e.g., they have the same mass, m.) Since the
particles do not interact with each other, we may consider just H(ri ), and omit the particle
index i. Using the field operators Ψ and Ψ† , the Hamiltonian is
Z
H = drΨ† (r)H(r)Ψ(r) .
(6.3)
Let us denote the eigen functions and the eigen values of the single-particle Hamiltonian by
φα and ²α , respectively (where α includes all quantum numbers),
H(r)φα (r) = ²α φα (r) .
1
(6.4)
For example, when the particle has just the kinetic energy, i.e. U = 0, then φα (r) ≡ φk (r) =
p
1/Ω) exp(ik · r), and ²α ≡ ²k = ~2 k 2 /2m. We now write the field operator in terms of the
basis φα ,
Ψ(r, t) =
X
φα (r)cα (t) , Ψ† (r, t) =
α
X
φ∗α (r)c†α (t) .
(6.5)
α
The operator cα destroys a particle in the state α, and similarly, the operator c†α creates a
particle in the state α. For fermions, they obey anti-commutation relations
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c†α , cα0 ≡ c†α cα0 + cα0 c†α = δαα0 , cα , cα0 ≡ cα cα0 + cα0 cα = 0 .
(6.6)
For bosons, they obey commutation relations,
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aα , a†α0 ≡ aα a†α0 − a†α0 aα = δαα0 , aα , aα0 ≡ aα aα0 − aα0 aα = 0 .
(6.7)
The Schrödinger equation is
i~
X
X
∂Ψ X
∂c
=
φα (r)i~ α = HΨ =
Hφα (r)cα =
²α φα (r)cα
∂t
∂t
α
α
α
⇓
i~
∂cα
=²α cα ,
∂t
(6.8)
(since φα form a complete set) and similarly
−i~
∂c†α
= ²α c†α .
∂t
(6.9)
The Hamiltonian itself, Eq. (6.3), is now
Z
¡X † ∗ ¢
¡X
¢ X †
H = dr
cα φα (r) H(r)
φα0 (r)cα0 =
²α cα cα .
α0
α
(6.10)
α
For example, the free electron gas (without any interactions), in terms of the wave vector
basis, is just the kinetic energy
Hfree =
X ~2 k 2
k
2m
c†k ck .
(6.11)
Example: The density operator. In terms of the field operators, the density operator,
ρ(r) is ρ(r) = Ψ† (r)Ψ(r). Using the expansion (6.5),
ρ(r) =
X
φ∗α (r)φα0 (r)c†α cα0 .
αα0
2
(6.12)
For example, the density operator of the free electron gas (namely, when there no interactions
at all) is
ρ(r) =
1 X i(k0 −k)·r †
e
ck ck0 ,
Ω kk0
(6.13)
and in Fourier space
Z
Z
X †
X †
1 X i(k0 −k)·r †
iq·r
ck+q/2 ck−q/2 .
ck+q ck =
e
ck ck0 =
ρq = dre ρ(r) = dreiq·r
Ω kk0
k
k
(6.14)
Example: The current density operator. The current density operator is
´
¡
¢
~ ³ †
Ψ (r)∇Ψ(r) − ∇Ψ† (r) Ψ(r) .
2im
(6.15)
´
~ X † ³ ∗
cα cα0 φα (r)∇φα0 (r) − φα0 (r)∇φ∗α (r) .
2im αα0
(6.16)
j(r) =
Using the expansion (6.5),
j(r) =
The Fourier transform of the current density of the free electron gas is then
Z
Z
³
´
~ X † 1
~ X
q
iq·r
i(q+k0 −k)·r
0
jq = dre j(r) =
dre
ik + ik =
(k − )c†k ck−q
ck ck0
2im kk0
Ω
m k
2
~ X †
kck+q/2 ck−q/2 .
(6.17)
=
m k
♣Exercise. Verify the (operator) continuity equation for fermions, ∂ρ/∂t + ∇ · j = 0, (using
the above two examples).
Example: The thermal average of the current of free fermions. We have found
above that the Hamiltonian of free fermions, in second quantization, reads (ignoring spin
components for simplicity)
Hfree =
X ~2 k 2
k
2m
c†k ck ,
(6.18)
and that the (Fourier component of the) current density operator is
jq =
~ X †
kck+q/2 ck−q/2 .
m k
3
(6.19)
Let us find the average current of the free fermion system (obviously, we must obtain that
there is no current flowing in the system). The expression for the thermal average [written
for the Hamiltonian (6.18)] can be put in the form
¡
¢
Tr e−βHfree jq
~ X
¡ −βH ¢ =
hjq i =
khc†k+q/2 ck−q/2 i .
free
m k
Tr e
(6.20)
Since the Hamiltonian (6.18) is diagonal in k−states, then only the q−component of jq may
give a non-vanishing contribution to average current. Note that this means that if there is
an average current, it will be uniform in space. The q = 0 component of hjq i is
~ X
hjq=0 i =
khc†k ck i .
m k
(6.21)
The thermal average of the number operator c†k ck with the free Hamiltonian (6.18) is just
the Fermi distribution, and hence
1
~ X
~2 k 2
k β²
, ²k =
.
m k e k +1
2m
hjq=0 i =
(6.22)
Since ²k depends solely on |k|, the sum over k here vanishes (each k−contribution is cancelled
by the contribution of −k) and consequently there is no average current in the system
described by the free Hamiltonian.
♣Exercise. Find the thermal average of the density in a system described by the free Hamiltonian (6.18).
We wish to study the way in which a system responds to a certain perturbation applied to
it, which breaks the symmetry between k and −k. When the perturbation is sufficiently
weak, one may expect the response to be linear in the perturbation (hence the name ‘linear
response’). Here we will calculate the current generated in an electronic system which is
placed in an electric field, and obtain the Kubo formula for the electrical conductivity.
Electrical conductivity. Let us denote the Hamiltonian of the system itself (without the
electric field) by H0 . Then, in the presence of the field, the Hamiltonian of the system is
H = H0 + Hext .
(6.23)
It is convenient to describe the electric field by introducing a vector potential, A(r, t), such
that the electric field, E(r, t), is
E(r, t) = −
4
1 ∂A(r, t)
.
c ∂t
(6.24)
In order to find the explicit form of Hext , we remember that in the presence of a vector
potential, the kinetic energy of the electron, in terms of field operators, is changed as follows
Z
Z
¡
¡
~2
~2
e ¢
e ¢
†
dr∇Ψ · ∇Ψ ⇒
dr ∇ + i A Ψ† · ∇ − i A Ψ ,
(6.25)
2m
2m
~c
~c
and the current density operator is (here we add the electron charge, −e, to our definition)
J=
¢
e
e
−e~ ³ †
e2
Ψ (∇ − i A)Ψ − [(∇ + i A)Ψ† ]Ψ = j +
An ,
2mi
~c
~c
mc
(6.26)
where n ≡ Ψ† Ψ is the density operator, and j is the current density operator without the
vector potential. In deriving this expression we have used the gauge choice ∇ · A = 0,
namely, the vector potential is a transverse field (since in Fourier space it is normal to the
wave vector, i.e., q · A(q) = 0).
Returning now to the kinetic energy, we see that it can be written in the form
Z
³ ~2
´
1
dr
∇Ψ† · ∇Ψ + j · A + O(A2 ) .
2m
c
(6.27)
The quadratic term in the vector potential is not needed, since we are interested in linear
response only. It therefore follows that the perturbation term in the Hamiltonian is
Z
1
drj · A .
(6.28)
Hext =
c
We now wish to find the average current density induced in the system. (Note: in the
following, in order to shorten the notations, we will use units in which ~ = 1, namely,
the dimension of time is 1/energy.) Suppose that the field is switched-on at t → −∞.
Formally, this means that we have to multiply Hext by eηt , with η → 0+ , so that at t → −∞
the external perturbation is zero. In the following we will omit this factor, but in actual
calculations it must be inserted. Just before the field is switched-on, the system was in one
of the eigen states of H0 . Let us denote that state by φα . Since after switching on the field
that state is no more an eigen state, it evolves in time according to
i
∂φα (t)
= (H0 + Hext )φα (t) .
∂t
(6.29)
As we know that for Hext = 0 the solution for this equation is e−iH0 t φα , we write the solution
in the presence of Hext in the form
φα (t) = e−iH0 t U (t)φα .
5
(6.30)
(Note that for any operator O, eiOt commutes with O.) Inserting the form (6.30) into the
Schrödinger equation (6.29), we obtain an equation for the operator U ,
dU (t)
= Hext (t)U (t) ,
dt
(6.31)
Hext (t) = eiH0 t Hext e−iH0 t ,
(6.32)
i
where
gives the time-dependent external Hamiltonian in the interaction picture.
The formal solution of Eq. (6.31), with the initial condition U (−∞) = 1, may be found by
iterations,
Z
Z
t
0
U (t) = 1 − i
−∞
dt Hext (t ) −
Z
t
0
t0
0
dt
−∞
−∞
dt00 Hext (t0 )Hext (t00 ) + . . . .
(6.33)
The full formal solution is usually written in the form of a time-ordered product, T , such
that
U (t) = T e−i
Rt
−∞
dt0 Hext (t0 )
.
(6.34)
The time-ordering operator T arranges the operators according to their time arguments,
such that those of later times appear to the left.
However, for the purpose of the present discussion we need just terms linear in the perturbation Hamiltonian Hext , so we write
Z
t
U (t) = 1 − i
−∞
dt0 Hext (t0 ) ,
(6.35)
and consequently
−iH0 t
φα (t) = e
Z
³
t
1−i
−∞
´
dt0 Hext (t0 ) φα .
(6.36)
The expectation value of the current density, Eq. (6.26), in the state φα is hence
Z t
Z t
0
0
iH0 t
−iH0 t
hα(t)|J|α(t)i = hα|(1 + i
dt Hext (t ))e Je
(1 − i
dt00 Hext (t00 ))|αi
−∞
−∞
Z t
Z t
= hα|(1 + i
dt0 Hext (t0 ))J(t)(1 − i
dt00 Hext (t00 ))|αi , J(t) = eiH0 t Je−iH0 t . (6.37)
−∞
−∞
Since our calculation is valid only to first order in Hext ∝ A, then [using Eq. (6.26)]
Z t
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e2
Ahα|n|αi + i
dt0 hα| Hext (t0 ), j(t) |αi .
(6.38)
hα(t)|J|α(t)i = hα|j|αi +
mc
−∞
6
(Note the appearance of the commutator of two operators at different times.)
The expression (6.38) has been derived for a certain eigen function of the Hamiltonian H0 .
Let us now sum over all eigen states, giving each one of them its statistical (Boltzmann)
weight. The result will be the thermodynamic average of the current density operator, i.e.,
Z t
£
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e2
hJ(t)i = hji +
Ahni − i
dt0 h j(t), Hext (t0 ) i ,
(6.39)
mc
−∞
P −βE
P −βE
α , where E
α hα|O|αi/
where for any operator O, hOi =
α are the eigen
αe
αe
energies of the system Hamiltonian, H0 . Inspecting Eq. (6.39), we see that the first term is
the average current in the absence of the field. Obviously, this quantity is zero. The second
term includes just the density of the electrons, and this quantity obviously does not depend
on the temperature (for metals). The last term gives the real current induced by the electric
field.
In order to see clearly that it is the third term of Eq. (6.39) that gives the actual current,
we need to introduce the perturbation Hamiltonian into Eq. (6.39). We will simplify a
bit the derivation by assuming that the vector potential depends on time, but not on the
P
coordinates. (This assumption can be easily relaxed.) Then Hext = (1/c) ν jν Aν , where
the index ν runs over the coordinates, ν = x, y or z. Equation (6.39) now takes the form
Z
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iX t
e2
Aν (t)hni −
dt0 Aν 0 (t0 )h jν (t), jν 0 (t0 ) i .
(6.40)
hJν (t)i =
mc
c ν 0 −∞
(The vector potential can be taken out of the commutator since it is not an operator.) It is
now convenient to define the current-current response function,
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Rνν 0 (t − t0 ) = −iΘ(t − t0 )h jν (t), jν 0 (t0 ) i .
Here, Θ is the Heavyside function,

1 , x>0
Θ(x) =
0 , x<0

,
.
(6.41)
(6.42)
Note that the response function (6.41) depends only on the time difference. This is because
the average cannot depend separately on the time arguments (homogeneity of time). Since
the Heavyside function takes care of the fact that in the integration t0 is always smaller than
t, we can extend the integration upper bound to ∞ and write
Z
e2
1X ∞ 0
hJν (t)i =
Aν (t)hni +
dt Aν 0 (t0 )Rνν 0 (t − t0 ) .
mc
c ν 0 −∞
7
(6.43)
Going over to Fourier space, this result reads
hJν (ω)i =
e2
1X
Aν (ω)hni +
A 0 (ω)Rνν 0 (ω) .
mc
c ν0 ν
(6.44)
Remembering now that the Maxwell equation gives [see Eq. (6.24)]
E(ω) = −
iω
A(ω) ,
c
(6.45)
we may write the result (6.44) in the form of Ohm’s law,
hJν (ω)i =
X
σνν 0 (ω)Eν 0 (ω) ,
(6.46)
ν0
in which the frequency-dependent (i.e., ac) conductivity tensor is
σνν 0 (ω) = i
e2 hni
i
δνν 0 + Rνν 0 (ω) .
mω
ω
(6.47)
This is the Kubo formula for the electrical conductivity. The dc conductivity is obtained
upon taking the limit ω → 0. A remarkable property of this formula is that the response
function (6.41) is calculated from the eigen spectrum of the Hamiltonian of the system in
the absence of the field, H0 .
The Kubo formula for the susceptibility tensor. The spin operator of the electron is
customarily described using the three Pauli matrices,
1
S= σ ,
2
(6.48)
σ = x̂σx + ŷσy + ẑσz ,
(6.49)
where
with

σx = 

0 1
1 0

 , σy = 

0 −i
i 0

 , σz = 

1 0
0 −1
 .
(6.50)
The Pauli matrices satisfy
σ`2 = 1 , σ` σm = iε`mn σn ,
(6.51)
where here 1 is the unit matrix and ε`mn is the full antisymmetric (Levy-Chivita) tensor.
Thus, S 2 = 3/4 as it should, since for the electron S = 1/2.
8
The spin-density operator of the electrons, using field operators, reads
1X † ¡ ¢
S(r) =
Ψ (r) σ σσ0 Ψσ0 (r) .
2 σσ0 σ
(6.52)
Therefore, in second quantization (in terms of creation and destruction operators in
k−states), its Fourier component is
S(q) =
1 XX †
c
σ 0c
0 .
2 σσ0 k k+q/2,σ σσ k−q/2,σ
(6.53)
[Note the similarity with the (Fourier component of) the density operator, Eq. (6.14).] In
particular, the z− component of the spin-density operator is
´
1 X³ †
1 XX †
z
S z (q) =
ck+q/2,σ σσσ
ck+q/2,↑ ck−q/2,↑ − c†k+q/2,↓ ck−q/2,↓ . (6.54)
0 ck−q/2,σ 0 =
2 σσ0 k
2 k
The transverse components are conveniently written as
S + (q) = S x (q) + iS y (q)
X †
¡ x
¢
1 XX †
y
ck+q/2,↑ ck−q/2,↓ ,
ck+q/2,σ σσσ
=
0 + iσσσ 0 ck−q/2,σ 0 =
2 σσ0 k
k
(6.55)
and
S − (q) = S x (q) − iS y (q)
X †
¡ x
¢
1 XX †
y
ck+q/2,↓ ck−q/2,↑ .
ck+q/2,σ σσσ
=
0 − iσσσ 0 ck−q/2,σ 0 =
2 σσ0 k
k
(6.56)
It is seen that the operators in S + (q) are such that a down-spin is destroyed and an up-spin
is created, while in S − (q) it is the other way around. We also see that S z counts the number
of up-spins minus the number of down-spins.
Let us now consider a system of electrons (described by an Hamiltonian H0 which is not
specified at the moment) that is paramagnetic, namely, there is no net magnetization at
equilibrium. In other words, hSi vanishes. However, when one applies a magnetic field H
on this system, the situation is changed. We will consider here only the coupling of the
magnetic field to the spins of the electrons, and will neglect the coupling of the field to the
orbital motion (i.e., we will disregard the Lorentz force).
The interaction of an electric field with the electron spins is the Zeeman interaction. It adds
to the system Hamiltonian the term
HZ = −
Z
drH(r, t) · S(r) .
9
(6.57)
[We absorb into the definition of the magnetic field the Bohr magneton µB .] One can now
repeat the linear response formalism to calculate the thermal average of the spin variables.
This gives us the linear relation between the thermal average of the α− component of the
pins, hSα (r, t)i (where α = x, y or z) and the γ− component of the magnetic field, Hγ ,
(where γ = x, y, or z) such that
hSα (r, t)i =
XZ
Z
dt
0
dr0 χαγ (r − r0 , t − t0 )Hγ (r0 , t0 ) ,
(6.58)
γ
where the susceptibility tensor is
χαγ (r − r0 , t − t0 ) = iΘ(t − t0 )h[Sα (r, t), Sγ (r0 , t0 )]i .
(6.59)
[It depends only on the difference r − r0 and on each of them separately, since we assume
the system to be spatially homogeneous.]
♣Exercise. Derive the expression for the electronic spin-susceptibility, Eq. (6.59).
In an isotropic system, or when the system has a cubic symmetry, one expects that the
susceptibility tensor has only diagonal matrix elements, and that they are equal, namely,
χxx = χyy = χzz and χxy = χyz = χzx = 0. It turns out that it most convenient to calculate
χ−+ ,
χ−+ (r − r0 , t − t0 ) = iΘ(t − t0 )h[S− (r, t), S+ (r0 , t0 )]i .
(6.60)
Since the susceptibility depends only on t − t0 , we may also put t0 = 0. The susceptibility is
then [see Eqs. (6.55) and (6.56)]
χ−+ (r − r0 , t) =
X
£
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0
e−iq·(r−r ) iΘ(t)h c†k+q↓ (t)ck↑ (t), c†k0 −q↑ ck0 ↓ i .
(6.61)
kk0 q
The q−Fourier component of the susceptibility is
χ−+ (q, t) = iΘ(t)
X£ †
¤
h ck+q↓ (t)ck↑ (t), c†k0 −q↑ ck0 ↓ i .
(6.62)
kk0
Adding the Fourier transform with respect to the time t we have
XZ ∞
£
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χ−+ (q, ω) = i
dtei(ω+iη)t h c†k+q↓ (t)ck↑ (t), c†k0 −q↑ ck0 ↓ i ,
kk0
0
where at the end of the calculation we have to take the limit η → 0+ .
10
(6.63)
Pauli susceptibility–the spin-susceptibility of free electrons. Let us find the spin
P
susceptibility of a system of free electrons, whose Hamiltonian is just Hfree = kσ ²k c†kσ ckσ ,
†
with ²k = k 2 /2m. In that case c†kσ (t) = ei²k t ckσ
and Eq. (6.63) becomes
XZ ∞
£
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Pauli
χ−+ (q, ω) = i
dtei(ω+iη)t ei(²k+q −²k )t h c†k+q↓ ck↑ , c†k0 −q↑ ck0 ↓ i .
kk0
(6.64)
0
Since now the operators in the commutator do not depend on time, we can calculate the
commutator explicitly, to find
χPauli
−+ (q, ω)
=i
XZ
k
∞
0
dtei(ω+iη)t ei(²k+q −²k )t hc†k+q↓ ck+q↓ − c†k↑ ck↑ i .
(6.65)
The thermal average of the number operator for free electrons does not depend on the spin
index, and is just the Fermi function, hc†k ck i = f (²k ), where f (²k ) = 1/(eβ²k + 1) (we take
the chemical potential as zero). Performing also the time-integration in Eq. (6.65), we find
χPauli
−+ (q, ω) =
X f (²k+q ) − f (²k )
.
²k − ²k+q − ω − iη
k
(6.66)
We first consider the static limit, ω = 0. In that case, we can send η to zero, and then
χPauli
−+ (q, ω = 0) =
X f (²k+q ) − f (²k )
k
²k − ²k+q
.
(6.67)
In the particular case q → 0 as well, an expansion [f (²k+q ) = f (²k ) + (²k+q − ²k )∂f (²k )/∂²k ]
gives the simple result
χPauli
−+ (q
X³ ∂f (² ) ´ Z
∂f (²) ´
k
−
= d²N (²)(−
= 0, ω = 0) =
' N (0) ,
∂²k
∂²
k
(6.68)
where N (²) is the density of states at energy ², and N (0) is the density of states at the
Fermi level. The final step in Eq. (6.68) holds at temperatures lower than the Fermi energy
(which practically always holds for metals). The density of states (per spin) at the Fermi
level for a three-dimensional free electron gas, is kF m/(2π 2 ), where kF is the Fermi wave
vector.
Example: The zero-temperature static spin susceptibility at finite wave vectors. We first
write Eq. (6.67) in the form
χPauli
−+ (q, ω = 0) =
X
k
³
f (²k )
´
X
1
1
1
.
−
= 4m
f (²k ) 2
²k−q − ²k ²k − ²k+q
q
+
2k
·
q
k
11
(6.69)
At zero temperature, f (²k ) = 1 for k < kF , and is zero otherwise. Hence
χPauli
−+ (q, ω
2π
= 0) = 4m
(2π)3
Z
kF
Z
dk k
π
2
0
dθ sin θ
0
q2
1
.
+ 2kq cos θ
(6.70)
Performing the angular integration,
χPauli
−+ (q, ω
2π
= 0) = 4m
(2π)3
Z
kF
0
dk k 2
¯ q + 2k ¯
1
¯
¯
ln ¯
¯.
2kq
q − 2k
(6.71)
It is convenient now to write q = 2kF x and k = kF y to obtain [denoting the density of states
at the Fermi level by N (0) = mkF /(2π 2 )]
χPauli
−+ (q, ω
Z 1
¯x + y ¯
1
¯
¯
= 0) = N (0)
dyy ln ¯
¯
2x 0
x−y
1³
1 − x2 ¯¯ 1 + x ¯¯´
= N (0) 1 +
ln ¯
¯ .
2
2x
1−x
(6.72)
Note that as x ≡ q/(2kF ) tends to zero, χPauli
−+ tends to N (0), while when x becomes very
large, it tends to N (0)/x2 . Thus, for a parabolic energy band, the Pauli spin susceptibility
is largest as q → 0. This is very important when interaction effects are included.
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