Marder M.P. Condensed Matter Physics
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Fermi Liquid Theory
513
Equation (17.168) provides one explanation of why Fermi liquid theory expands
the energy out to quadratic order in quasi-particle densities; the second derivatives
needed to find sound speeds could not be obtained correctly otherwise.
Because fi is the natural independent variable, the easiest quantity to calculate
in Eq. (17.168) is dN/d/i. At very low temperatures, where the sound speed is
being computed, one obtains
öf
i
= 6(ß-£.
I
)-9(£.
F
-£.
i
\
ß=£F
) (17.169)
=»-^
= <*(£
2
-M)(l-lT*) (17-170)
jx
is oscillating around some equilibrium value as the wave passes by, and the occupation numbers
Sf are given by subtracting from / the occupation number appropriate in equilibrium.
= *(£j-/i)[l-£
«g,^].
(17.171)
l'a'
Now define
x-^ dô ft,
k'a'
Equation (17.171) guarantees that k! lies on the Fermi surface, and because k also
lies on the Fermi surface, while u-^, depends only on the angle between k and k',
A is independent of
k.
Then using Eqs. (17.171) and (17.172), one obtains
A = j[dk'} „_,£(£-,-^(l-A) (17.173)
=
B(\-A),
where 5 = £ "«,<*(£* "A*) = *o (17-174)
k'a'
B FÂ
^A = = —2—. (17.175)
\+B 1+F
0
*
The total number of particles N in the system differentiated by ß is therefore
*=£^
<m76)
ka
= C(5(£j-/z)(l-A) (17.177)
ka
= VD(E
F
)-^— (17.178)
mD{£
F
)
[l+F*) (17.179)
/
tfl
■■V
F
\ (1+Fn). Use Eqs. (17.120), (17.128), and (17.154).
(17.180)
3m
514
Chapter 17.
Transport
Phenomena and
Fermi
Liquid Theory
Zero Sound. When
the
frequency
of
oscillation
is
very fast compared
to the
scattering time
of
quasi-particles, the population
of
quasi-particles has
no
time
to
arrive locally
at
the equilibrium described
by
Eq. (17.169). To describe traveling
waves in this limit, one can turn to the Boltzmann equation (17.10), interpreting
g
=
fa
as the occupation number
of
quasi-particle excitations and using Eq.
(
17.118)
as
the energy functional corresponding to Eq. (17.2).
It
is
enough
to
consider cases
in
which deviations
of
the distribution function
from equilibrium are small, and then linearize
in
the deviations. Letting
fi
0)
from
Eq. (17.119) denote the equilibrium distribution and then expanding
to
first order,
one finds immediately
0i
*
dfX'n
at, «J di
k
(17.181)
coll.
If
6f-ï
varies
in
space, so does
£-7
through
its
dependence
on ôf,
but the variation
of
£-7
with 7 is
a
small quantity.
The point
of
focusing upon time scales short compared with
the
scattering time
(near the Fermi surface), is that the scattering term becomes negligible. To proceed,
carry
out a
Fourier transform
in
time
and
space, going over
to
variables
UJ
and
q
respectively, to obtain
(uj-q-V-~
k
)Ôf~
k
-8(£,-~
k
-£.
F
)q-V-~
k
(Y^
U
ï
pSf
l
,)=0.
Because S/.""
/S£T.
=
-<5(£j-£
f
).
l'a'
(17.182)
All
of
the action
is at
the Fermi surface,
and
Of
behaves
as a
delta function near
there. To make this fact explicit, define
^5(£
F
-£
2
)
= %
(17.183)
Then
[u-q-v^-q-v^Y^
U
U'
(
H'
=0
-
(17.184)
l'a'
To simplify matters
at
this point, assume that D(Ep)
u-g,,
is
spherically symmetric
and can be replaced by
F
Q
S
'.
The assumption
is
not reasonable:
Ff
must be included
as well, but the additional algebra is left
for
Problem 12. Proceeding as
if
FQ
were
enough, one obtains
\u>
- q
■ Vr](j>7
- q
■
VrFn
/
——
<f>r,
= 0.
The integral
is a
surface (17.185)
*
K
J 4-TT
integral over the Fermi
uj —
q-u^J 4TT
Equation (17.186) shows that
cfy
k
depends only
on the
angle between
q and
v-^
(which points along k) and can be written as
<?!>(cos
9).
Let
oo
(/>(C0S
6) = ^2
Pl(
C0fi
&)4>l Legendre polynomials, again. (17.187)
/=0
Fermi Liquid Theory 515
0(cos
6)
= - [l
CO
qvp cos 9
•l
F
°^°2
/,
d{C
°
S
e)
I
1
CJ
=
-Fed
1+fi?
l4-^ln(
+
2 gu/r a; + guf
1
+ 1-^-1
n(^^
2 gt>/r w + qvf
'
co —
qvp cos
to —
qvf,
0.
(17.188)
(17.189)
(17.190)
(17.191)
If
FQ
> 0, there is always a solution of Eq. (17.191) for some value of co/qvp. In
other words, there is a linear dispersion relation between
co
and q. Unfortunately
Eq. (17.191) does not compare well with experiment, because F[ does not turn out
to be small. Including Ff in the analysis gives instead (Problem 12)
j/tf(i
+!/?)
+ (
u
qv
F
)
2
FI
l + !-^ln(^^
2 qvp
co
+ qvp
'
+
1
+
0. (17.192)
A numerical solution of Eq. (17.192) gives the dispersion relation Co = ui/q for
zero sound.
17.5.7 Comparison with Experiment in
3
He
The Fermi liquid parameters that enter into Eq. (17.192) were measured in
3
He
by Greywall (1983) and are listed in Table 17.3. Abel et al. (1966) studied sound
induced by a quartz crystal at 15.4 and 45.5 MHz, and they measured sound speed
and attenuation as a function of temperature. Ideally the experiment would be
conducted as a function of frequency, showing that at low frequencies sound prop-
agates with the first sound velocity of Eq. (17.180) and that at higher frequencies
it propagates at a new velocity given by solving Eq. (17.192). In the absence of
tunable crystals to create sound of arbitrary frequency in the tens of megahertz
range, the experiment was conducted as a function of temperature. The reasoning
was that at sufficiently low temperature one should see zero sound, but at higher
temperatures additional scattering should bring about thermal equilibrium more
rapidly, and the velocity should cross over to first sound. The scattering time for
quasi-particles varies as T~
2
, and it was estimated that for T = 1 K the scattering
time is of order 10~
12
s. Therefore sound at 10 MHz should begin to travel as zero
sound when the temperature is on the order of 10 mK. Precisely such a transition
is demonstrated in Figure 17.8.
Table 17.3. Fermi liquid parameters for
3
He
/'(bar) R
o
l
n
o
pa
m*/m Vf(ms )
0 9.15 5.27 -0.700 -0.55 2.76 59.7
3 15.83 6.40 -0.725 -0.73 3.13 54.3
Source: Greywall (1983).
516
Chapter 17. Transport Phenomena and
Fermi
Liquid Theory
196
194
o
192 -
190
186
Predicted zero sound speed
° % '
O m
m
° H
0
►
o
►
o
o ■
°° ■
-
-°o°"
"
a
°oo
Predicted sound speed c
i
' ' '
CO
45.5 MHz
15.4 MHz
-
► S
°-i.'
-
10 100
T(mK)
Figure 17.8. Sound propagation velocity as a function of temperature in
3
He,
at
a
pressure
of 0.32 bar. [Source: Abel et
al.
(1966), p. 76.]
The experiment was conducted at a pressure P = 0.32 bar, somewhat different
from any for which the Fermi liquid parameters were measured directly. Using lin-
ear interpolation to estimate the values of the various parameters at this pressure,
one finds that for ordinary sound from Eq. (17.180) c = 188 m s
-1
, while for zero
sound, Eq. (17.192) has a solution at ui/qvp = 3.3 which predicts a zero sound ve-
locity co = 195 m s
_1
. These values compare well with the measurements in Figure
17.8.
The test is not trivial, because Fermi liquid theory has predicted the existence
of a new high-frequency mode, whose properties are to be found solely on the ba-
sis of specific heat and low-frequency sound measurements. Abel et al. (1966) also
performed another test of the theory by measuring sound attenuation. Sound atten-
uation can be calculated within the context of the relaxation time approximation to
the Boltzmann equation, using the relaxation time r as an additional phenomeno-
logical parameter. Attenuation should increase as T
2
for zero sound, then decay as
T~
2
for regular sound. This behavior is seen as well, with the crossover occurring
at the same temperature as the crossover in sound speed.
Problems
1.
Boltzmann statistics: Calculate the analog of Eq. (17.60), proceeding from
Eq. (17.56), assuming that the probability / of occupying a state is given
by Boltzmann rather than Fermi statistics, and verify that the final answer is
unchanged.
2.
Semiclassical equations: Use Eq. (17.1) to derive Eq. (17.3b). Remember
Problems
517
that the vector potential A may be time dependent and contribute to the electric
field.
3.
Boltzmann equation with anomalous velocity:
(a) Demonstrate that Eq. (17.22) follows from Eqs. (17.16) and (17.17).
(b) Verify Eq. (17.24)
4.
Onsager relations:
(a) Derive Eq. (17.51). Use the fact that TS is linear in the forces, and also use
the fact that v^ = — v_^.
(b) Show that Eq. (17.51) leads to the symmetry Eq. (17.50).
5.
AC conductivity: Use the Boltzmann equation to find the response of a free
Fermi electron gas to an electric field which oscillates in time at frequency io.
6. Current driven by thermal gradient: Consider a metal subject to a constant
temperature gradient dT/dx. Assume energy near the Fermi surface to be
isotropic and of the form £^ = m*vl/2. Show that the electric current j
x
is
nonzero and is proportional to
1
m*
dT
dx '
(17.193)
where cy is the specific heat. Find the dimensionless constant of proportion-
ality.
7.
Thermoelectric figure of merit:
Cold,
T\
Cold, T,
Hot,
T
2
Figure 17.9.
A
thermoelectric refrigerator
operates by passing current through a first
element p with positive Peltier coefficient,
and into a second n with negative Peltier
coefficient, causing transport of heat from
OtoL.
Consider the operation of the thermoelectric refrigerator shown in Figure 17.9.
518
Chapter 17.
Transport
Phenomena and Fermi Liquid Theory
(a) Argue that the thermal currents
JQ
and
JQ
flowing along x in the p and n
elements are
J
p
0
=
n
p
j-A
P
K
P
—
(17.194a)
dT
J
Q
= -Ii
n
J-A„K
n
—
,
(17.194b)
where J is the electrical current, K
P
and the thermal conductivities of
the two elements, U
p
and H
n
are the Peltier coefficients of the two elements,
and A
p
and A
n
are the cross-sectional areas.
(b) The temperature gradients are not uniform, because the current flow pro-
duces Joule heating per unit length
J
2
R
P:n
/L,
where R
p
and R
n
are the resis-
tances of the two elements. Assume that the Peltier coefficients are indepen-
dent of
temperature.
Solve for the temperature field in each element, and show
that the flow of heat at x = 0 is
J
p
Q
=
Ii
p
J-A
p
K
p
T
^-
J
^-
(17.195a)
J
n
Q
=
-H
n
J-A
n
K
n
T
^^-
J
-^.
(17.195b)
(c) Find for any given temperature difference
T2 —
T\
the value of
the
current J
m
producing the greatest total heat flux JQ = JQ+JQ.
(d) Show that the maximum temperature difference
T2 —
T\
that the refrigerator
can sustain is
fn„-m
2
(17.196)
{l\
p
-H
n
)
2
.
2RK
along the way, define R and
K.
It is conventional to define a figure of merit
Z==(n
^""
)2
'
r
= (
r
>+^
2
-
(17
-
197)
which has dimensions of inverse kelvin. Sometimes the dimensionless ZT is
also reported.
8. Thomson coefficient: The Thomson effect is the name given to the fact that
heat dissipation is different in a wire where electric currents flow with a tem-
perature gradient than it is in the same wire when the direction of the current
is reversed. Derive the effect first by noting that the flow of entropy may be
given by
dS - -, -, -,
T— = -V-J
0
+ G-J. (17.198)
dt
Use the general equations of linear transport theory to eliminate G everywhere
in favor off. As a result, find an equation for entropy generation just in terms
Problems
519
of electron current and temperature gradients. From this expression, isolate
the term that will change when the relative direction of heat and temperature
flow is altered. Relate this term to the temperature derivative of the ther-
mopower, and describe an experiment that would allow one to measure this
term.
The simple result of
Eq.
(17.93) is only correct if
L
21
and L
11
commute.
9. Hall effect—elementary argument: A simple treatment of the Hall effect
begins by assuming that electrons obey the equation of motion
mv = -e lE + -xB) . (17.199)
(a) Assume that current is not permitted to pass the y boundaries in Figure 17.3,
so that charge builds up there and creates an electric field E
y
in the y direction.
Assume that when the electron current reaches steady state, electrons drift in
the x direction at a constant
rate.
Furthermore, in steady state the y component
of all electron velocities must vanish, or else more charge would build up
on the boundaries, and the electric field along y would continue to change.
Rewrite Eq. (17.199) so as to take into account these two observations.
(b) Solve the resulting equations to find the Hall coefficient, defined to be
RH
= ^, (17.200)
Bjx
where j
x
is the current in the x direction.
10.
Hall effect—Boltzmann equation: In order to find the Hall coefficient in
the context of
the
Boltzmann equation, one may proceed in the following way:
Begin with
dg _ d j9 g-f
Q
t
—
V
V) Q-fS
K
pjt
S
~
T
' Combining Eqs. (17.9) and (17.26).
In steady state, dg/dt vanishes, so one has
(17.201)
Q J, 0 a-f
v(f)—g + k^g = -^^-. (17.202)
or dk T
When a magnetic induction B is present, one obtains
hk = -e(E + vxB/c). (17.203)
(a) Guess a solution of this equation in the form
- - df
g = f+[D-k-e\^-. (17.204)
Find expressions for
T>
and C, assuming that E lies in the plane of the metal,
and B is perpendicular to it. Neglect a term involving d
2
f/d/j?.
520
Chapter 17.
Transport
Phenomena and
Fermi
Liquid Theory
(b) Compute currents along x and y for arbitrary values of B and È, and find the
electric field strength E
y
necessary to guarantee that J
y
, the current in the y
direction vanishes.
(c) As a result, find the Hall coefficient
RH
= ^-- (17.205)
11.
Temperature dependence of
/J:
In the context of Fermi liquid theory, find the
leading temperature dependence of
/x
at low temperatures. Begin by writing
the temperature derivative of the total number of particles, and by requiring it
to be zero, find
dfi/dT.
12.
Higher-order expression for zero sound: Verify that if all Fermi liquid pa-
rameters except for F( and
FQ
vanish, then the dispersion relation for zero
sound is
U (l
+
\rf)
+
(—)
2
Fi\
h +
i—
in
£^)1+
1 +
^=0.
[ V 3 V \qvFJ J
L
2qvF u + qv
F
'i 3
(17.206)
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