PHZ7427: Spring 2014
HOMEWORK ASSIGNMENT #4: Solutions
due Monday, 04/07
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
P1 Piezoelectric electron-phonon interaction
In addition to the deformation-potential type of the electron-phonon interaction which is found in all materials,
materials without inversion symmetry, e.g., GaAs, have another mechanism: piezoelectric interaction. In this
mechanism, electrons interact with the dipole moment produced by ions displaced from their equilibrium positions. If the material has inversion symmetry, the average dipole moment of a unit cell is equal to zero. However,
if inversion symmetry is absent, the net dipole moment is non-zero. (Piezolectricity, i.e., electric polarization
induced by stress, stems from the same mechanism, hence the name ”piezoelectric interaction.) The operator
of the electrostatic potential produced by a phonon mode with dispersion ωq can be written as1
1/2
β X
~
b̂†q eiq·r−iωq t + H.c.,
φ̂(r, t) = √
2ρωq
V q
where β is the piezoelectric modulus (in units of electric field).
Suppose that ωq = sq for 0 ≤ q ≤ qD . Find the temperature dependence of the electron quantum lifetime
and transport scattering time at temperatures both much higher and much lower than the Bloch-Grueneisen
temperature.
—bf Solution:
To estimate the quantum lifetime, we use formula for 1/τ on p. 22 of the notes ”Scattering mechanisms and
electronic transport in conductors” which, upon integrating over the angle, reduces to (restoring ~)
Z
1
2
=
dqqwq 2Nq + 1 − fεp −~ωq + fεp +~ωq ,
(1)
τ
π~vF
where
w=
e2 β 2 ~
.
2ρωq
(2)
At T the Bloch-Grueneisen temperature, the Bose function Nq ≈ T /~ωq 1, so that the Fermi functions,
which a bounded by 0 and 1 can be neglected. The q integral is determined primarily by its upper limit: q ∼ qD .
The T factor from Nq is the only source of the T dependence and 1/τ ∝ T . In this regime, scattering is of the
large-angle type and thus the quantum and transport scattering times coincide. The T dependence at higher T
does not depend on a particular mechanism of the electron-phonon interaction. At T the Bloch-Grueneisen
temperature, typical phonons have thermal wave numbers R(q ∼ qT = T /~s),
R q while both the Fermi and Bose
q
functions are on order one. Hence, 1/τ scales as 1/τ ∝ 0 T dqqwq ∝ 0 T dq ∝ T . However, scattering is
of the small-angle type in this regime; hence the transport rate differs from the quantum rate by a factor of
1 − cos θ ≈ q 2 /2kF2 ∼ qT /2kF2 ∝ T 2 and 1/τtr ∝ T 3 .
P2 Density of states in the BCS theory
In the BSC theory of superconductivity the excitation spectrum of the single-electron states is
q
2
Ek = ± (εk − εF ) + ∆2 ,
where εk is the energy spectrum of an electron in the normal metal. Assuming that ∆ does not depend on k
(s-wave pairing) and that εk = ~2 k 2 /2m, find the the density of states as a function of the excitation energy
ε = Ek .
2
Solution: As usual,
1 2 dk
k
.
π 2 dE
(3)
q
p
ε F + E 2 − ∆2 .
(4)
g(E)dE =
Solving for k in the upper band:
r
k=
2m
~2
Obviously, a solution exists only for |E| > ∆.
r
2m
E
1
dk
√
p
=
√
dE
~2 2 E 2 − ∆2 ε F + E 2 − ∆2
(5)
Then
1
g(E) =
2π 2
2m
~2
3/2 εF +
p
E 2 − ∆2
1/2
√
E
θ(|E| − ∆).
− ∆2
E2
(6)
For max ∆, E εF , the last expression can be simplified to
1
g(E) =
2π 2
2m
~2
3/2
√
εF √
E
θ(|E| − ∆).
− ∆2
E2
(7)
P3 Two-fluid model
The two-fluid model of a superconductor assumes that, at any finite temperature below Tc , there are two types
of electrons: normal (with number density nn and superconducting (with number density ns ) with equal masses.
Normal electrons obey the Drude equation
pn
dpn
= −eE −
,
dt
τ
while superconducting electrons obey the same equation with τ = ∞.2 Find the frequency-dependent complex
conductivity of a superconductor in the two-fluid model.
Solution:
The normal component behaves as the usual Fermi gas with conductivity given by
σn =
nn e2
i
.
m ω + iτ
(8)
The conductivity of the superconducting component can be obtained from the last formula by replacing nn by
ns and by letting s1/τ → 0+ . Then
σs =
ns e2
i
.
m ω + i0+
The total conductivity is the sum of the two partial components:
e2 i
1
1
nn
+ ns
.
σ = σn + σs =
m
ω + iτ
ω + i0+
(9)
(10)
The real part of σ is
Reσ =
ne2
1/τ
nn 2
+
πn
δ(ω)
s
m
ω + 1/τ 2
(11)
3
P4 The London equation
Find the spatial profile of the magnetic field inside an infinite superconducting plate of thickness d. The magnetic
field outside the plate is parallel to the plane of the plate and has magnitude B0 .
Solution:
The projection of the magnetic field on the plane of the plate field satisfies the London equation inside the plate
1
d2 B
= 2 B,
dx2
Λ
where x is along the normal to the plate. At x = ±d/2, B = B0 . A general solution inside the plate is
B = a exp(x/Λ) + b exp(−x/Λ).
(12)
(13)
The boundary conditions give
a=b=
B0
d
2 cosh 2Λ
(14)
and
B = B0
x
cosh Λ
d
cosh 2Λ
.
(15)
P5 Elements of the BCS Theory
– Recall the Bogoliubov transformation
bp↑ = up ap↑ − vp a†−p↓
bp↓ = up ap↓ + vp a†−p↑
(16)
Show that bp,σ satisfies the fermionic commutation rules, i.e., [bpσ , b†p,σ ]+ = 1 with [A, B]+ = AB + BA, if
|vp |2 + |up |2 =1.
– Prove Eqs. (7) and (8) in the lecture notes “BCS Theory” posted at
http://www.phys.ufl.edu/maslov/phz7427/lectures.html
Solution
h
i
h
i
bp↑ , b†p↓ = up ap↑ − vp a†−p↓ , u∗p a†p↑ − vp∗ a†−p↑ = |up |2 + |vp |2 = 1.
+
(17)
+
To invert the Bogoliubov transformation, leave the first equation intact while change p → −p in the second
equation and take a conjugate of it. This gives
bp↑ = up ap↑ − vp a†−p↓
b†−p↓ = up a†−p↓ + vp ap↑
(18)
Multiplying the first eq-n by u and the second by v and using the condition |up |2 + |vp |2 = 1, we find
ap↑ = up bp↑ + vp b†−p↓ .
(19)
ap↓ = up bp↓ − vp b†−p↑ .
(20)
Likewise, we obtain the second relation
Eqs. (7) and (8) follow from trivial algebra.
1
2
V. F. Gantmakher and Y. B. Levinson, Carrier Scattering in Metals and Semiconductors (North-Holland, Amsterdam, 1987).
Note that the condition τ = ∞ defines not really a superconductor but rather a perfect conductor. The important difference
between the two is that, in addition to zero resistance, a superconductor also exhibits the Meissner effect, whereas a perfect
metal has no Meissner effect. See A& M, pp. 737-738.