PHZ7427: Spring 2014
Problem set # 2: Solutions
due Monday, 02/17
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Office hours: TR 3 pm-4 pm
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
...
P1 20 points Kohn anomaly in two dimensions
Following the discussion in class and the notes, find the singular part of the charge susceptibility near
the 2kF point in two dimensions. Determine the form of the Friedel oscillation at large distances from
the center.
Solution:
The only difference compared to the 3D case is the measure of angular integration. In 2D,
Z
Z 0
1
χc (q) = 2e2 g(εF )Re dθ
dξk
,
0
−2ξk + ~q vF + 2εF θ2 + iδ
−Λ
where q 0 = q − 2kF . As in 3D, we subtract off χ(q = 2kF ), which eliminates the need to keep the cutoff
Λ:
Z
Z 0
1
δχc (q) = χc (q) − χc (2kF ) = 2e2 g(εF )Re dθ
dξk
0 v + 2ε θ 2 + iδ
−2ξ
+
~q
k
F
F
−∞
1
−
−2ξk + 2εF θ2 + iδ
Z ∞
θ2
= e2 g(εF )Re
dθ ln q0
.
(1)
2
0
kF + θ + iδ
In contrast to the 3D case, the integral over θ is convergent if the upper limit is extended to infinity,
so there is no need of keep the cutoff here.
p
For q 0 > 0, the argument of the log is positive definite. Dropping the Re sign, we rescale θ by q 0 /kF
to obtain
p
δχc (q) = e2 g(εF ) q 0 /kF × C,
(2)
where
∞
Z
C=
ln
0
x2
x2 + 1
is a number. To calculate C, change the variable as y = 1 + 1/x2
Z
1 ∞
ln y
C=−
dy
2 1
(y − 1)3/2
and integrate by parts to obtain
Z
C=−
1
∞
1
1
y (y − 1)1/2
Changing the variable to z =
√
y − 1, we finally find
Z ∞
dz
C = −2
= −π.
1 + z2
0
Thus
p
p
δχc (q) = −πe2 g(ε) q 0 /F = −πe2 g(ε) (q − 2kF )/kF ; q > 2kF .
For q 0 < 0, we take the real part of the log to obtain
Z ∞
Z ∞
θ2
θ2
= 2e2 g(εF )
.
δχc (q) = e2 g(εF )
dθ ln q0
dθ ln
0
|q
| kF + θ2 |
| − kF| + θ2 |
0
0
Rescaling θ by
(3)
p
|q 0 |/kF , we arrive at the same form as in Eq. (2) but with C replaced by
C0 =
Z
∞
dx ln
0
x2
.
|x2 − 1|
Following the same steps as before, we find that C 0 = 0 and thus
δχc (q) = 0; q < 2kF .
It means that
√ the Kohn anomaly in 2D is asymmetric: χ√c is equal to its value at 2kF for q < 2kF and
falls off as q − 2kF for q > 2kF ; the derivative has a 1/ q − 2kF singularity. Although we cannot see
this from the present analysis which is limited to the vicinity of 2kF , it turns out χc (q) is independent
of q in the entire interval 0 ≤ q ≤ 2kF in 2D.
P2 20 points Kohn anomaly in one dimension
In the 1D case, one can find χc (q, 0) without any approximations. Do it and compare the 1D Kohn
anomaly to its 2D and 3D counterparts.1
Solution
In 1D and assuming that q > 0, we obtain
2
χc (q) = 4e Re
Z
fk
dk
= 4e2 Re
2π εk+q − εk + iδ
2e2 m vF q +
=
ln πq
vF q −
q2
2m
q2
2m
Z
kF
−kF
dk
2π
fk
k
mq
+
q2
2m
+ iδ
.
(4)
χc itself has a logarithmic singularity at q = 2kF .
P3 20 points Effective electron mass in two dimensions
Following the discussion for the 3D case, find the effective mass for 2D electrons interacting via a
screened Coulomb potential V (q) = 2πe2 /(q + κ) (q is in the plane of 2D electrons). As we did in the
3D case, simplify the finite result using a (necessary) assumption that the electron-electron interaction
is weak, i.e., that κ kF .
Solution
The last formula on p. 8 on notes ”Hartree-Fock Theory” for the 2D case reads (with ~ = 1)
Z ∞
Z π
1
q cos θ
q2
δεk − δεkF = ξk
1+
dqq
dθVsc (q)δ vF q cos θ +
(2π)2 0
2m
kF
−π
Integrating over θ and using Vsc = 2πe2 /(q + κ), we obtain
2
δεk − δεkF
e2
= ξk
πvF
2kF
Z
0
q
1 − 2k
2
dq
F
r
2 .
q+κ
q
1 − 2kF
Although the last integral can be solved exactly, there is really no need in doing so because we have to
impose the condition κ kF upon which the last factor in the integrand can be replaced by 1. The
upper limit must be chosen of order kF rather than equal to 2kF because we do not pretend to able to
calculate the number under the log. To leading order in κ/kF
Z kF
dq
e2
e2
kF
δεk − δεkF = ξk
= ξk
ln
πvF 0 q + κ
πvF
κ
The coefficient in front of ξk is the velocity renormalization factor. Thus
e2
kF
∗
m =m 1−
ln
πvF
κ
P4 20 points Ashcroft & Mermin, Ch. 17, Problem 4
Equation (17.19) (with ~ = 1) reads
εk =
k2
2e2 kF
−
F (k/kF )
2m
π
with
F (x) =
1 1 − x2 1 + x +
ln 2
4x
1 − x
Expanding logs to 3rd order in x, we obtain
F (x 1) = 1 −
x2
+ ...
3
Then
εk =
k2
2e2 kF
−
2m
π
1
1 − (k/kF )2
3
=−
2e2 kF
k2
+
π
2m
1+
4e2 m
3πkF
(5)
Using the definition of rs (here, I use dimensionless rs which is rs of AM divided by the Bohr radius
a0 = 1/me2 )
4π
(rs a0 )3 n = 1
3
and recalling that
n = kF3 /3π 2 ,
we find that
rs =
9π
4
1/3
me2
kF
or
k2
2e2 kF
+
εk = −
π
2m
with m∗ = m/(1 + 0.22rs ).
4
1+
3π
4
9π
!
1/3
rs
=−
2e2 kF
k2
+
π
2m∗
P5 20 points Prove Eqs. (17.20) and (17.22) from Ashcroft & Mermin See the notes ”Hartree-Fock theory”.
1
A stronger Kohn anomaly is one of the reasons for a breakdown of the Fermi-liquid theory in 1D.