Phys 268r Fall 2013
Problem Set 3 Solutions
B. Halperin
Solutions to Problem Set 3
Problem 1.
We start in a reference frame moving with the lattice at a velocity
~v
=
d
~
R/dt
, (so that the
problem looks stationary), and mark it as the primed frame. In this frame, the current will be
the standard quantized Hall current:
~
j
0
=
ν
e
2
h
~
E
0
×
ˆ
z.
where
~
E
0
is the electric field in the moving frame.
This may be obtained from the standard
Lorentz Transformations (letting
~
β
=
~v/c
):
~
E
0
=
γ
~
E
+
~
β
×
~
B
~
B
0
=
γ
~
B

~
β
×
~
E
.
We have already discarded higher order terms in
β
, which we assume small, so we approxi
mate
γ
= 1 Further, the assumption that
v
D
is small implies
E/B
small. Thus
~
β
×
~
E
is small,
and
~
B
0
=
~
B
. Consequently, we may assume that
ν
is the same in the moving frame as it would
have been for the system at rest. We assume here that
~
B
is oriented along the zaxis, and we
define
ρ
B
=
B/
Φ
0
=
eB/hc
.
We thus find:
~
j
0
=
eνρ
B
c
~
E
0
×
~
B
B
2
(1.1)
=
eνρ
B
c
~
E
+
~
β
×
~
B
×
~
B
B
2
(1.2)
=
eνρ
B
~v
D
+
eνρ
B
~v
×
~
B
×
~
B
B
2
(1.3)
=
eνρ
B
~v
D

eνρ
B
~v
(1.4)
where in the last step, we used the vector triple product identity, and the fact that
~
B
·
~v
= 0.
Finally, we transform back to the lab frame. To do this, we note that:
~
j
=
~
j
0
+
eρ
el
~v
Combining this with our previous result, and using
ρ
el
=
νρ
B
+
ηρ
V
we see that:
~
j
=
eνρ
B
~v
D
+
eηρ
V
~v
as claimed.
1
Phys 268r Fall 2013
Problem Set 3 Solutions
B. Halperin
Problem 2.
We proceed similar to the derivation of
σ
xy
in class. We first calculate expectation value of the
spaceintegrated current
< J >
, induced by a small timedependent perturbation
δH
due to
δX
(
t
). To first order in the perturbation due to
δX
, we may write our states as:

ψ
(
t
)
>
=

0
>
+

δψ >
t
where

δψ >
t
=

i
X
N
6
=0

N >
Z
t
∞
< N

∂H
0
∂X

0
> δX
(
t
0
)
e
i
(
E
N

E
0
)
t
0
dt
0
.
We have:
< J >
=
<
0

J

0
>
+
<
0

J

δψ >
+
< δψ

J

0
>
+
....
The first term on RHS is 0 as discussed in class, since there is no electric field in the unperturbed
Hamiltonian.
We next calculate the second term, noting that the third is just its complex
conjugate (which we will add in later).
We also note that the
J
operator can be written as
∂H/∂k
.
Let us now assume that
δX
(
t
) = Re[
λe
iωt
+
γt
], where
γ
is a positive infinitesimal, and
λ
is
small.
We thus have:
<
0

J

δψ >
=

ie
X
N
<
0

∂H
∂k

N >< N

∂H
∂X

0
>
Z
t
∞
δX
(
t
0
)
e
i
(
E
N

E
0
)
t
0
dt
0
=

ie
X
N
<
0

∂H
∂k

N >< N

∂H
∂X

0
>
Z
t
∞
λe

iωt
0
+
λ
*
e
iωt
0
2
e
i
(
E
N

E
0
)
t
0
dt
0
=

ie
X
N
<
0

∂H
∂k

N >< N

∂H
∂X

0
>
λe

iωt
2
i
(
ω

E
N
+
E
0
)
+
λ
*
e
iωt
2
i
(
ω
+
E
N

E
0
)
=
e
dX
dt
X
N
<
0

∂H
∂k

N >< N

∂H
∂X

0
>
i
(
E
N

E
0
)
2
,
Using the fact that
h
δψ

J

0
i
=
h
0

J

δψ
i
*
, we obtain an expression relating
h
J
i
to
λ
and
λ
*
.
In the limit of
ω
→
0, we can simplify that expression by noting that
λe

iωt
2
i
(
ω

E
N
+
E
0
)
+
λ
*
e
iωt
2
i
(
ω
+
E
N

E
0
)

c.c.
→
iωλe

iωt
(
E
N

E
0
)
2

c.c.
as was noted in class. We also note that
iωλe

iωt

c.c.
=

2
i dX/dt.
2
Phys 268r Fall 2013
Problem Set 3 Solutions
B. Halperin
We thus find
< J >
=
ie
dX
dt
X
N
6
=0
<
0

∂H
∂k

N >< N

∂H
∂X

0
>

(
k
↔
X
)
(
E
N

E
0
)
2
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 Fall '13
 BertrandI.Halperin
 Physics, Trigraph, Chern class, B. Halperin