Chemistry Reference
In-Depth Information
=
√
2
t
where
is the number of unit cells. In our linearization
scheme, in Eq. (4) the summation is over the two circular patches (Fig. 2).
For a finite range of
v
F
and
N
Im χ
0
(
ω,
q
) can be evaluated exactly
[
18
]
:
q
and
ω
,
ω
2
−
qv
F
)
2
q
3
/
2
√
ω − qv
F
1
16
2
(
1
16
√
2
Im χ
0
(
q
,ω
)=
qv
F
)
2
∼
,
v
F
v
F
ω
2
−
(
with a square root divergence at the edge of the particle-hole continuum
in (
q
) space. This expression has the same form as density of states
of a particle in 1
ω,
D
(with energy measured from
v
F
q
). Note that in fact
Im χ
0
(
q, ω
)=
πρ
q
(
ω
)
,
where
ρ
q
(
ω
) is free particle-hole pair
DOS
for a
fixed center of mass momentum
.
That is, the particle-hole pair has a
phase space for scattering which is effectively one dimensional
.Thuswe
have a particle-hole bound state in the spin triplet channel for arbitrarily
small U. However, we also have a prefactor
q
q
3
/
2
, that scales the density
of states. This together with the square root divergence of the density
of states at the bottom of the particle-hole continuum gives us a bound
state for every q as
αq
3
,as
shown below. The square root divergence has the following phase space
interpretation. The constant energy (
q →
0, with the binding energy vanishing as
ω
) contour of a particle-hole pair of
a given total momentum
q
defines an ellipse in k-space:
ω
=
v
F
(
|
k
+
q
|
+
). In our convention, the points on the ellipse denote the momentum
co-ordinates of the electron of the electron-hole pair. As the energy of the
particle-hole pair approaches the bottom of the continuum, i.e.,
|
k
|
p−h
→
v
F
q
, the minor axes of the ellipses become smaller and smaller and the
elliptic contours degenerate into parallel line segments of effective length
∼ q
3
2
. The asymptotic equi-spacing of these line segments leads to an
effective one-dimensionality and the resulting square root divergence.
The collective mode in
magnetic
channel is the solution of:
− Uχ
0
(
q
1
,ω
)=0
)=
U
. The asymptotic
Im χ
0
(
q
Re χ
0
(
q
or equivalently,
,ω
)=0 a
nd
,ω
Re χ
0
(
q
expression for
,ω
) is found to be
√
2
√
1
√
2
√
1
1
Re χ
0
(
q
,ω
)
≈
(
k
c
+
arctan (
))
π
2
v
F
4
− z
− z
