Exercise 10 - institute for theoretical physics cologne

Institute of Theoretical Physics
University of Cologne
PD Dr. T. Quella
Dr. D. Bagrets, Dr. V. Osipov
Quantum Field Theory II
Exercise sheet 10
22.01.2014
17. Kubo formula for the Hall conductivity σxy
This exercise is devoted to the derivation of the the Smrˇcka-Stˇreda formula for the off-diagonal
conductivity σxy (ω = 0). We consider a two-dimensional gas of non-interacting electrons in
the presence of the perpendicular magnetic field. The system is described by the Hamiltonian
ˆ = 1 (−i∇ − A)2 + Vˆ , where Vˆ is a one-particle potential (disorder and/or boundary potential)
H
2m
and A = B2 r × e3 is the vector potential of the external field.
To evaluate the off-diagonal conductivity, we start from the real-space expression
1
σxy (ω) = −
ωm L2
Z
2
2 0
Z
d xd x
0
β
dτ eiωm τ hˆjx (x, τ )ˆjy (x0 , 0)i
iωm →ω+i0
,
where ˆji is the i-th component of the current operator.
a) Using the Wick’s theorem to compute the expectation value of the current operators
and making the analytical continuation from the Matsubara to real frequency, reduce
σxy (ω) to the form
e2
σxy (ω) = −
ω
where vˆi =
Z
e
m (−i∇i
dE
R
A
A
R
R
A
f
(E)Tr
v
ˆ
(G
−
G
)ˆ
v
G
+
v
ˆ
G
v
ˆ
(G
−
G
)
, (1)
x
y
x
y
E
E
E−ω
E+ω
E
E
2πL2
− Ai ) denotes the velocity operator.
b) Our next task is to evaluate the zero-frequency limit of Eq. (1). To start with, prove
that
Tr vˆx GR
ˆy GR
ˆx GA
ˆy GA
(2)
Ev
E = Tr v
Ev
E = 0,
with the use of the following commutation relations
vˆi = −i[ˆ
xi , H0 ] = i[ˆ
xi , GR
−1
] = i[ˆ
xi , GA
−1
],
(3)
which can be easily checked.
I and σ II , where the 1st contribution includes R − A
c) Decompose Eq. (1) in two terms, σxy
xy
terms, while the 2nd one comprises R − R and A − A terms. Show that in the limit
I comes from the electron states concentrated
ω → 0 and T = 0 the contribution to σxy
on the Fermi surface only and reads
I
σxy
=
e2 A Tr vˆx GR
v
ˆ
G
y
E
E 2π
E=µ
where µ is the chemical potential.
1
(4)
II in the limit ω → 0. Show that
d) Let us now analyse the 2nd term, σxy
Z
n
o
dE
II
2
R
A 0
σxy = e
f
(E)
Tr
i(ˆ
y
v
ˆ
−
x
ˆ
v
ˆ
)
(G
−
G
)
.
x
y
E
E
4πL2
(5)
where 0 denotes the derivative with respect to energy E. To derive this representation
II you can once again employ the commutation relations (??) together with the
for σxy
relation
0
2
GR,A
= −(GR,A
(6)
E
E ) .
e) Show that in the symmetric gauge, Ai = −Bij xj /2, the following relation holds
∂H0
−e
=
ij x
ˆi vˆj ,
∂B
2c
(7)
II in the form
where ij the antisymmetric unity tensor. Using this relation rewrite σxy
II
σxy
= −ec
∂n
∂B
(8)
µ
Here n(B) is a concentration of the electron gas at fixed chemical potential. Think of
the physical meaning of this result.
2

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