Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications
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Uncorrected Proof
BookID 160928 ChapID 11 Proof# 1 - 29/07/09
354 11 Many Body Interactions – Introduction
But recall that Ω
2
p
=4π
n
0
z
(ze)
2
M
and k
2
s
=
4k
F
πa
0
. Substituting into (11.245) 746
gives the result given by (11.240). However, q need not be small compared to 747
k
F
, even though ω s
l
q will still be small compared to qv
F
and ω
p
. Then, we 748
must keep F (z)andwrite 749
ω
2
s
2
l
q
2
F (z)+
πa
0
4k
F
q
2
. (11.246)
750
Because F (z) has an infinite first derivative at q =2k
F
(or z = 1), the phonon 751
dispersion relation will show a small anomaly at q =2k
F
that is called the 752
Kohn anomaly
8
. 753
8
W. Kohn, Phys. Rev. Lett. 2, 393 (1959).
Uncorrected Proof
BookID 160928 ChapID 11 Proof# 1 - 29/07/09
11.8 Screening 355
Problems 754
11.1. Let us consider the paramagnetic state of a degenerate electron gas, in 755
which n
kσ
=1forε
kσ
<ε
F
and zero otherwise. 756
(a) Show that the exchange contribution to the energy of wave vector k 757
and spin σ is 758
Σ
Xσ
(k)=−
1
Ω
k
n
k
σ
4πe
2
|k
2
− k
|
2
.
(b) Convert the sum over k
to an integral and perform the integral to 759
obtain 760
Σ
Xσ
(k)=−
e
2
k
F
π
1+
1 − x
2
2x
ln
1+x
1 − x
,
where x = k/k
F
. 761
(c) Plot Σ
Xσ
(k) as a function of
k
k
F
. 762
(d) Show that the total energy (kinetic plus exchange) for the N particle 763
paramagnetic state in the Hartree–Fock approximation is 764
E
P
=
kσ
n
kσ
¯h
2
k
2
2m
+Σ
Xσ
(k)
= N
3
5
¯h
2
k
2
F
2m
−
3e
2
k
F
4π
.
11.2. Consider the ferromagnetic state of a degenerate electron gas, in which
765
n
k↑
=1fork<k
F↑
and n
k↓
=0forallk. 766
(a) Determine the Hartree–Fock energy ε
kσ
=
¯h
2
k
2
2m
+Σ
Xσ
(k). 767
(b) Determine the value of k
F
(Fermi wave vector of the nonmagnetic state) 768
for which the ferromagnetic state is a valid Hartree–Fock solution. 769
(c) Determine the value of k
F
for which E
F
=
k
ε
k↑
has lower energy than 770
E
P
obtained in Problem 11.1. 771
11.3. Evaluate I
xx
(q, ω)inthesamewayasweevaluatedI
zz
(q, ω), which is 772
given by (11.166). 773
11.4. The longitudinal dielectric function is written as 774
ε
(l)
(q, ω)=1−
ω
2
p
ω
2
[1 + I
zz
(q, ω)].
Use ln(x+iy)=
1
2
ln(x
2
+ y
2
)+iarctan
y
x
to evaluate ε
(l)
2
(z,u), the imaginary 775
part of ε
(l)
(q, ω), where z = q/2k
F
and u = ω/qv
F
. 776
Uncorrected Proof
BookID 160928 ChapID 11 Proof# 1 - 29/07/09
356 11 Many Body Interactions – Introduction
11.5. Let us consider the static dielectric function written as 777
ε
(l)
(q, 0) = 1 +
3ω
2
p
q
2
v
2
F
F (z),
where z = q/2k
F
and F (z)=
1
2
+
1
4
1
z
− z
ln
1+z
1−z
.
778
(a) Expand F (z)inpowerofz for z 1. Repeat it in power of 1/z for 779
z 1. 780
(b) Determine the expressions of the static dielectric function ε
(l)
(q, 0) in 781
the corresponding limits. 782
11.6. In the absence of a dc magnetic field, we see that |ν = |k
x
,k
y
,k
z
≡|k, 783
the free electron states. 784
(a) Show that k
|V
q
|k =
¯h
m
(k +
q
2
)δ
k
,k+q
, where V
q
= v
0
e
iq·r
+e
iq·r
v
0
. 785
(b) Derive the Lindhard form of the conductivity tensor given by 786
σ
(q,ω)=
ω
2
p
4πiω
1
+
m
N
k
f
0
(E
k+q
) − f
0
(E
k
)
E
k+q
− E
k
− ¯hω
¯h
m
2
k +
q
2
k +
q
2
.
(c) Show that the Lindhard form of the dielectric tensor is written by 787
ε
(q,ω)=
1 −
ω
2
p
ω
2
1
−
mω
2
p
Nω
2
k
f
0
(E
k+q
) − f
0
(E
k
)
E
k+q
− E
k
− ¯hω
¯h
m
2
k +
q
2
k +
q
2
.
11.7. Suppose that a system has a strong and sharp absorption line at a 788
frequency ω
A
and that ε
2
(ω) can be approximated by 789
ε
2
(ω)=Aδ(ω − ω
A
)forω>0.
(a) Evaluate ε
1
(ω) by using the Kronig–Kramers relation. 790
(b) Sketch ε
1
(ω) as a function of ω. 791
11.8. The equation of motion of a charge (−e)ofmassm harmonically bound 792
to a lattice point R
n
is given by 793
m(
¨
x + γ
˙
x + ω
2
0
x)=−eEe
iωt
.
Here, x = r −R, ω
0
is the oscillator frequency, and the electric field E = Eˆx. 794
(a) Solve the equation of motion for x(t)=X(ω)e
iωt
. 795
(b) Let us consider the polarization P (ω)=−en
0
X(ω), where n
0
is the 796
number of oscillators per unit volume. Write P (ω)=α(ω)E(ω)and 797
determine α(ω). 798
(c) Plot α
1
(ω)andα
2
(ω)vs.ω,whereα = α
1
+ α
2
. 799
(d) Show that α(ω) satisfies the Kronig–Kramers relation. 800
11.9. Take H =
1
2m
p +
e
c
A
2
− eφ and H
=
1
2m
p +
e
c
A
2
− eφ
where 801
A
= A + ∇χ and φ
= φ −
1
c
˙χ. 802
Uncorrected Proof
BookID 160928 ChapID 11 Proof# 1 - 29/07/09
11.8 Screening 357
(a) Show that H
−
e
c
˙χ =e
−
ieχ
¯hc
He
ieχ
¯hc
. 803
(b) Show that ρ
=e
−
ieχ
¯hc
ρe
ieχ
¯hc
satisfies the same equation of motion, viz. 804
(c)
∂ρ
∂t
+
i
¯h
[H
,ρ
]
−
=0asρ does. 805
11.10. Let us take ˜ρ
0
(H, η)=
exp(
H−η
Θ
)+1
−1
as the local thermal equi- 806
librium distribution function (or local equilibrium density matrix). Here 807
η(r,t)=ζ + ζ
1
(r,t) is the local value of the chemical potential at posi- 808
tion r and time t, while ζ is the overall equilibrium chemical potential. 809
Remember that the total Hamiltonian H is written as H = H
0
+ H
1
.Write 810
˜ρ
0
(H, η)=ρ
0
(H
0
,ζ)+ρ
2
and show that, to terms linear in the self-consistent 811
field, 812
k|ρ
2
|k
=
f
0
(ε
k
) − f
0
(ε
k
)
ε
k
− ε
k
k|H
1
− ζ
1
|k
.
11.11. Longitudinal sound waves in a simple metal like Na or K can be
813
represented by the relation ω
2
=
Ω
2
p
ε
(l)
(q,ω)
,whereε
(l)
(q, ω) is the Lindhard 814
dielectric function. We know that, for finite ω, ε
(l)
(q, ω) can be written as 815
ε
(l)
(q, ω)=ε
1
(q, ω)+iε
2
(q, ω). This gives rise to ω = ω
1
+iω
2
,andω
2
is pro- 816
portional to the attenuation of the sound wave via excitation of conduction 817
electrons. Estimate ω
2
(q) for the case ω
2
1
q
2
Ω
2
p
k
2
s
ω
2
2
. 818
Uncorrected Proof
BookID 160928 ChapID 11 Proof# 1 - 29/07/09
358 11 Many Body Interactions – Introduction
Summary 819
In this chapter, we briefly introduced method of second quantization and 820
Hartree–Fock approximation to describe the ferromagnetism of a degener- 821
ate electron gas and spin density wave states in solids. Equation of motion 822
method is considered for density matrix to describe gauge invariant theory of 823
linear responses in the presence of the most general electromagnetic distur- 824
bance. Behavior of Lindhard dielectric functions and static screening effects 825
are examined in detail. Oscillatory behavior of the induced electron density in 826
the presence of point charge impurity and an anomaly in the phonon dispersion 827
relation are also discussed. 828
In the second quantization or occupation number representation, the 829
Hamiltonian of a many particle system with two body interactions can be 830
written as 831
H =
k
ε
k
c
†
k
c
k
+
1
2
kk
ll
k
l
|V |klc
†
k
c
†
l
c
l
c
k
,
where c
k
and c
†
k
satisfy commutation (anticommutation) relation for Bosons 832
(Fermions). 833
The Hartree–Fock Hamiltonian is given by H =
i
E
i
c
†
i
c
i
, where 834
E
i
= ε
i
+
j
n
j
[ij|V |ij−ij|V |ji] .
The Hartree–Fock ground state energy of a degenerate electron gas in the
835
paramagnetic phase is given by E
ks
=
¯h
2
k
2
2m
−
e
2
k
F
2π
2+
k
2
F
−k
2
kk
F
ln
k
F
+k
k
F
−k
.
836
The total energy of the paramagnetic state is 837
E
P
= N
3
5
¯h
2
k
2
F
2m
−
3
4π
e
2
k
F
.
If only states of spin ↑ are occupied, we have
838
E
k↑
=
¯h
2
k
2
2m
−
2
1/3
e
2
k
F
2π
2+
2
2/3
k
2
F
− k
2
2
1/3
k
F
k
ln
2
1/3
k
F
+ k
2
1/3
k
F
− k
; E
k↓
=
¯h
2
k
2
2m
.
The total energy in the ferromagnetic phase is
839
E
F
=
E
k↑
= N
2
2/3
3
5
¯h
2
k
2
F
2m
− 2
1/3
3
4π
e
2
k
F
.
The exchange energy prefers parallel spin orientation, but the cost in kinetic
840
energy is high for a ferromagnetic spin arrangement. In a spin density wave 841
state, the (negative) exchange energy is enhanced with no costing as much in 842
kinetic energy. The Hartree-Fock ground state of a spiral spin density wave 843
can be written as |φ
k
=cosθ
k
|k ↑ +sinθ
k
|k + Q ↓. 844
Uncorrected Proof
BookID 160928 ChapID 11 Proof# 1 - 29/07/09
11.8 Screening 359
In the presence of the self-consistent (Hartree) field {φ, A}, the Hamilto- 845
nian is written as H = H
0
+H
1
,whereH
0
is the Hamiltonian in the absence of 846
the self-consistent field and H
1
=
e
2c
(v
0
·A + A · v
0
) −eφ, up to terms linear 847
in {φ, A}. Here, v
0
=
p
m
and the equation of motion of ρ is
∂ρ
∂t
+
i
¯h
[H, ρ]
−
=0. 848
The current and charge densities at (r
0
,t) are given, respectively, by 849
j(r
0
,t)=Tr
−e
1
2
vδ(r − r
0
)+
1
2
δ(r − r
0
)v
ˆρ
; n(r
0
,t)=Tr[−eδ(r − r
0
)ˆρ] .
Here −e
1
2
vδ(r −r
0
)+
1
2
δ(r − r
0
)v
is the operator for the current density at 850
position r
0
, while −eδ(r−r
0
) is the charge density operator. Fourier transform 851
of j(r
0
,t)gives 852
j(q,ω)=σ(q,ω) · E(q,ω)
where the conductivity tensor is given by σ
(q,ω)=
ω
2
p
4πiω
[1 + I(q,ω)] . Here 853
I(q,ω)=
m
N
k,k
f
0
(ε
k
) − f
0
(ε
k
)
ε
k
− ε
k
− ¯hω
k
|V
q
|kk
|V
q
|k
∗
and the operator V
q
is defined by V
q
=
1
2
v
0
e
iq·r
+
1
2
e
iq·r
v
0
. 854
The longitudinal and transverse dielectric functions are written as 855
ε
(l)
(q, ω)=1−
ω
2
p
ω
2
[1 + I
zz
(q, ω)]; ε
(tr)
(q, ω)=1−
ω
2
p
ω
2
[1 + I
xx
(q, ω)].
Real part (ε
1
) and imaginary part (ε
2
) of the dielectric function satisfy the 856
relation 857
ε
1
(ω)=1+
2
π
P
∞
0
ω
ε
2
(ω
)
ω
2
− ω
2
dω
.
The power dissipation per unit volume is then written P(q,ω)=
ω
2π
ε
2
(q, ω) | 858
E
0
|
2
. 859
Due to collisions of electrons with lattice imperfections, the conductivity 860
of a normal metal is not infinite at zero frequency. In the presence of collisions, 861
the equation of motion of the density matrix becomes, in a relaxation time 862
approximation, 863
∂ρ
∂t
+
i
¯h
[H, ρ]
−
= −
ρ − ˜ρ
0
τ
.
Here, ˜ρ
0
is a local equilibrium density matrix. Including the effect of collisions, 864
the induced current density becomes 865
j(q,ω)=
ω
2
p
4πiω
#
1
+ I −
iωτ
1+iωτ
(K
1
− K
2
)(K
1
− K
2
)
L
1
+iωτL
2
$
· E.
In the static limit, the dielectric function reduces to
866
ε
(l)
(q, 0) = 1 +
3ω
2
p
q
2
v
2
F
F (z),
Uncorrected Proof
BookID 160928 ChapID 11 Proof# 1 - 29/07/09
360 11 Many Body Interactions – Introduction
where F (z)=
1
2
+
1
4
1
z
− z
ln
1+z
1−z
and z = q/2k
F
. The self-consistent 867
screened potential is written as 868
φ(q)=
4πe
q
2
+ k
2
s
F (q/2k
F
)
.
where k
s
=
2
4k
F
πa
0
. For high density limit (πa
0
k
F
1) and large distances 869
from the point charge impurity, the induced electron density is given by 870
n
1
(r)=
6n
0
a
0
k
F
cos 2k
F
r
(2k
F
r)
3
.
Electronic screening of the charge fluctuations in the ion density modifies the
871
dispersion relation of phonons, for example, 872
ω
2
s
2
l
q
2
F (z)+
πa
0
4k
F
q
2
showing a small anomaly at q =2k
F
. 873
Uncorrected Proof
BookID 160928 ChapID 12 Proof# 1 - 29/07/09
12 1
Many Body Interactions: Green’s Function 2
Method 3
12.1 Formulation 4
Let us assume that there is a complete orthogonal set of single particle states 5
φ
i
(ξ), where ξ = r,σ. By this we mean that 6
φ
i
| φ
j
= δ
ij
and
i
| φ
i
φ
i
|= 1. (12.1)
We can define particle field operators ψ and ψ
†
by 7
ψ(ξ)=
i
φ
i
(ξ)a
i
and ψ
†
(ξ)=
i
φ
∗
i
(ξ)a
†
i
, (12.2)
where a
i
(a
†
i
) is an annihilation (creation) operator for a particle in state i. 8
From the commutation relations (or anticommutation relations) satisfied by 9
a
i
and a
†
j
, we can easily show that 10
[ψ(ξ),ψ(ξ
)] =
ψ
†
(ξ),ψ
†
(ξ
)
=0,
ψ(ξ),ψ
†
(ξ
)
= δ(ξ − ξ
). (12.3)
The Hamiltonian of a many particle system can be written. (Here we set
11
¯h =1.) 12
H =
"
d
3
r
#
1
2m
∇ψ
†
α
(r) ·∇ψ
α
(r)+U
(1)
(r)ψ
†
α
(r)ψ
α
(r)
$
+
1
2
"
d
3
r d
3
r
ψ
†
α
(r)ψ
†
β
(r
)U
(2)
(r, r
)ψ
β
(r
)ψ
α
(r).
(12.4)
Summation over spin indices α and β is understood in (12.4). For the moment, 13
let us omit spin to simplify the notation. Then 14
ψ(r)=
i
φ
i
(r)a
i
. (12.5)
Uncorrected Proof
BookID 160928 ChapID 12 Proof# 1 - 29/07/09
362 12 Many Body Interactions: Green’s Function Method
We can write the density at a position r
0
as 15
n(r
0
)=
d
3
rψ
†
(r)ψ(r)δ(r − r
0
)=ψ
†
(r
0
)ψ(r
0
). (12.6)
16
The total particle number N is simply the integral of the density 17
N =
d
3
rn(r)=
d
3
rψ
†
(r)ψ(r). (12.7)
If we substitute (12.5) in (12.7), we obtain
18
N =
d
3
r
9
i
φ
∗
i
(r)a
i
:
⎛
⎝
j
φ
j
(r)a
j
⎞
⎠
=
ij
φ
i
| φ
j
a
†
i
a
j
. (12.8)
By φ
i
| φ
j
= δ
ij
, this reduces to 19
N =
i
a
†
i
a
i
, (12.9)
so that ˆn
i
= a
†
i
a
i
is the number operator for the state i and
ˆ
N =
i
ˆn
i
is the 20
total number operator. It simply counts the number of particles. 21
12.1.1 Schr¨odinger Equation 22
The Schr¨odinger equation of the many particle wave function Ψ (1, 2,...,N) 23
is (¯h ≡ 1) 24
i¯h
∂
∂t
Ψ = HΨ. (12.10)
25
We can write the time dependent solution Ψ(t)as 26
Ψ(t)=e
−iHt
Ψ
H
, (12.11)
27
where Ψ
H
is time independent. If F is some operator whose matrix element 28
between two states Ψ
n
(t)andΨ
m
(t) is defined as 29
F
nm
(t)=Ψ
n
(t)|F |Ψ
m
(t), (12.12)
we can write |Ψ
m
(t) =e
−iHt
|Ψ
Hm
and Ψ
n
(t)| = Ψ
Hn
|e
iHt
.ThenF
nm
(t) 30
can be rewritten 31
F
nm
(t)=Ψ
Hn
|F (t)|Ψ
Hm
, (12.13)
where F
H
(t)=e
iHt
F e
−iHt
. The process of going from (12.12) to (12.13) is 32
a transformation from the Schr¨odinger picture (where the state vector Ψ (t) 33
depends on time but the operator F does not) to the Heisenberg picture (where 34
Ψ
H
is a time-independent state vector but F
H
(t) is a time-dependent opera- 35
tor). The transformation from (to) Schr¨odinger picture to (from) Heisenberg 36
picture can be summarized by 37
Uncorrected Proof
BookID 160928 ChapID 12 Proof# 1 - 29/07/09
12.1 Formulation 363
Ψ
S
(t)=e
−iHt
Ψ
H
and F
H
(t)=e
iHt
F
S
e
−iHt
. (12.14)
From these equations and (12.10), it is clear that
38
∂F
H
(t)
∂t
=i[H, F
H
] . (12.15)
39
12.1.2 Interaction Representation 40
Suppose that the Hamiltonian H can be divided into two parts H
0
and H
, 41
where H
represents the interparticle interactions. We can define the state 42
vector Ψ
I
(t)intheinteraction representation as 43
Ψ
I
(t)=e
iH
0
t
Ψ
S
(t). (12.16)
Operate i∂/∂t on Ψ
I
(t) and make use of the fact that Ψ
S
(t) satisfies the 44
Schr¨odinger equation. This gives 45
i
∂Ψ
I
(t)
∂t
= H
I
(t)Ψ
I
(t), (12.17)
where
46
H
I
(t)=e
iH
0
t
H
e
−iH
0
t
. (12.18)
From (12.12) and the fact that Ψ
S
(t)=e
−iH
0
t
Ψ
I
(t)itisapparentthat 47
F
I
(t)=e
iH
0
t
F
S
e
−iH
0
t
. (12.19)
By explicit evaluation of
∂F
I
∂t
from (12.19), it is clear that 48
∂F
I
∂t
=i[H
0
,F
I
(t)] . (12.20)
49
The interaction representation has a number of advantages for interacting 50
systems; among them are: 51
1. All operators have the form of Heisenberg operators of the noninteracting 52
system, i.e., (12.19). 53
2. Wave functions satisfy the Schr¨odinger equation with Hamiltonian H
I
(t), 54
i.e., (12.17). 55
Because operators satisfy commutation relations only for equal times, H
I
(t
1
) 56
and H
I
(t
2
)donotcommuteift
1
= t
2
. Because of this, we cannot simply 57
integrate the Schr¨odinger equation, (12.17), to obtain 58
Ψ
I
(t) ∝ e
−i
"
t
H
I
(t
)dt
. (12.21)
Instead, we do the following:
59
1. Assume that Ψ(t)isknownatt = t
0
. 60
2. Integrate the differential equation from t
0
to t. 61