=
~
d
(
~
k
)
·
~σ,
where
d
z
(
~
k
) =
m
,
d
x
(
~
k
) = 2
k
3
cos(3
θ
) = 2
k
x
(
k
2
x

3
k
2
y
)
,
d
y
(
~
k
) = 2
k
3
sin(3
θ
) = 2
k
y
(

k
2
y
+ 3
k
2
x
)
.
As in problem 4 of the previous problem set, we will argue below that the result can be
obtained by mapping to a sphere. If you are interested in seeing some of the math worked out,
we have included it at the end.
Let
ˆ
d
be a unit vector in the direction of
~
d
.
The Berry curvature
F
xy
= a unit area
on the sphere.
When
m >
0 and
k
= 0,
ˆ
d
= ˆ
z
.
As we increase
k
,
ˆ
d
spans the upper
hemisphere.
In particular, if we write
~
k
= (
k, θ
) and
~
d
=
d
(sin
θ
0
cos
φ,
sin
θ
0
sin
φ,
cos
θ
0
) =
(2
k
3
cos(3
θ
)
,
2
k
3
sin(3
θ
)
, m
), we see by analogy that
φ
= 3
θ
.
It is clear in general that for a
Hamiltonian with
k
n
, we will have
φ
=
nθ
.
If we map the Brillouin zone to the sphere, we
see that for a fixed sign of
m
, we cover the hemisphere
n
times. We thus see immediately (see
previous problem set) that
σ
xy
=
n
2
sign(
m
)
e
2
h
.
From here it follows that Δ
σ
xy
=
ne
2
/h
, and
we see by comparison with the previous question that in general, (Chern number) =

(Berry
Phase)
/
2
π
.
Below we show the integral done explicitly.
As in the previous problem set, we write:
F
xy
=
1
2

~
d
(
~
k
)

3
~
d
(
~
k
)
·
(
∂
k
x
~
d
(
~
k
)
×
∂
k
y
~
d
(
~
k
))
=
1
2

~
d

3
abc
d
a
∂
k
x
d
b
∂
k
y
d
c
,
3
Phys 268r Fall 2013
Problem Set 4  Part I  Solutions
B. Halperin
where

~
d

=
p
m
2
+ (2
k
3
)
2
. The nonzero contractions are (
a, b, c
) =
{
(
z, x, y
)
,
(
z, y, x
)
}
, which
gives:
F
xy
=
1
2(
m
2
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 Fall '13
 BertrandI.Halperin
 Physics, Resistance, Orthogonal matrix, Hilbert space, Unitary matrix