Marder M.P. Condensed Matter Physics
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Problems 833
3.
Magnetic susceptibility of Fermi liquids: Extend the Stoner model to Fermi
liquids and find their magnetic susceptibility.
(a) Write the energy of a Fermi liquid in the presence of a magnetic field.
(b) Assume that all ôfa are occupied only for energies £° < £/r
—
A, while all
<5/j,
are occupied for £° < £/? + A. Assume that the density of states D(£)
can be taken constant and that the kk! dependence of the parameters
u-^
^, ,
can be ignored. Rewrite the total energy in terms of integrals over energy.
(c) Find the value of A by minimizing the total energy, and show that the sus-
ceptibility is
*=w-
(26l49)
4.
Diagonalizing spin waves: Show that Eq. (26.63) diagonalizes (26.62) using
Eq. (26.64).
5.
Proj ection operators :
(a) Find expressions analogous to Eq. (26.96) for
P\
and P^.
(b) Verify that
Pb
. . . /*? satisfy the requirement of all projection operators
P
2
= P. (26.150)
(c) Why is Eq. (26.99) true?
6. Local moment: Derive Eq. (26.120), and find an expression for Ä~,.
7.
Mean field treatment of magnetic moments: Consider the Anderson model
given in Eq. (26.95). An approximate solution can be obtained through the
mean field replacement
t/n
0
T«0| ~ U (n
0
|) "0| +Un
0
f {n
Ql
)-U (n
0
t) («o|) • (26.151)
The brackets mean that one finds the expectation value, or mean occupancy, of
the impurity states. Neglect the final term on the right-hand side of Eq.
(26.151
),
which leads to a constant energy shift, and define
e
0
a =
eo
+
U
(no,-*) • (26.152)
One has then to solve the Hamiltonian
l
k = ^'K
a
(26.153a)
a
where
834 Chapter 26. Quantum Mechanics of Interacting Magnetic Moments
(a) Define the Green function
G
a
(£) = [£- 'ka + 177] "
1
. r, is very small. (26.154)
Let |0) be the vacuum state, and let
|O)=4|0) and \k) =
cl\0).
(26.155)
Show that
[£-e
0a
+ ir
]
}(0\G
(7
\0)-^v
k
(k\G
(T
\0) = l (26.156a)
k
[E-e
i
+ ir
l
}{k\G
IT
\0)-vS{0\G
IT
\0)=0. (26.156b)
(b) Show that
<0|G
ff
|0) =
?
*
A-L
.
A
, (26.157a)
where
|2
5e =
E^:
and
A = 7r^|^|
2
5(£-e
J
).(26.157b)
*
6
* it
(c) Recall from Eq. (18.38) that
n
0
„(£.)d£ = --Im(0|G
a
(£)|0)d£ (26.158)
7T
gives the probability that the impurity state with spin a will be occupied in the
energy range [£, £ +d£]. Take the limit r/
—>
0, assume that
<5e
is negligibly
small, and assume that A can be treated as a constant. Find two self-consistent
expressions for («oj) and («oj.) by employing
) = f
F
d£ noa(£)- (26.159)
J —
oo
r&F
{no,
x
Assume eoo- > £f- The expressions should be in the form (wof) = /(("oi))
for some function /.
(d) Take £
F
- e
0
= U/2 and U/A = 1. Plot /((%)) and /
_1
(("oi)) versus
(noi )
on a graph where x and y axes range between 0 and 1. The crossings
of these two curves give possible values for the spin occupation of the impu-
rity. There is only one solution. What magnetic moment is predicted for the
impurity?
(e) Repeat the previous part of this problem for U/A = 5. There is more than
one solution. What magnetic moment is predicted for the impurity?
References 835
8. Renormalization of J:
(a) In multiplying (26.122a) and (26.122b), to be applied to
\ip\),
one can take
ê
^
â
l>a =
ô
w'-
wh
y
?
(b) Using identities such as
S
Z
S~ = --S~ or S~S
+
= --S
z
, (26.160)
2 2
verify Eq. (26.126).
9. The t-J model: Consider the Hubbard model at half filling:
]T -t c^c/v + 4
CT
c
/(J
+[/^CJT^T^U^I-
(26.161)
a
The number of electrons and the number of lattice sites are both equal to N.
(a) What is the ground state of the model when t = 0 and U > 0? What is the
degeneracy of the ground state? What are the energies of the excited states?
(b) Introduce the hopping term proportional to t as a perturbation. The task is to
find the new ground state. Show that to second order in perturbation theory
one must diagonalize an effective Hamiltonian of the form
JeffJ2 (srSi>-\)
,
(26.162)
and find the value of
i
e
ff-
This model is known as the t-J model.
(c) Describe physically why the ground state of the t
—
J model, and therefore
of the Hubbard model for large U, are antiferromagnetic at half filling.
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838 Chapter 26. Quantum Mechanics of Interacting Magnetic Moments
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27.
Superconductivity
27.1 Introduction
Kamerlingh Onnes (1911) found that mercury loses all electrical resistance below
a critical temperature of 4 K. In copper, a very good conductor at temperatures
of 77 K, electrical currents decay with a characteristic time of 2
•
10
-13
s. But a
superconductor is qualitatively better. Currents in a superconductor have been seen
to persist for over a year. Since Kamerlingh Onnes' initial discovery, most metals
and alloys have been found to exhibit superconductivity at some temperature, so
the phenomenon is quite universal.
The phenomenon of superconductivity posed for many years a serious chal-
lenge to quantum mechanics. Perhaps the dense combination of many electrons
really could not be described by the same equations that had been deduced from ex-
periments on dilute systems; perhaps quantum mechanics was just an approxima-
tion to more complicated many-body equations, and superconductivity was the key
to finding the true form. Sommerfeld and Bethe (1933) wrote that "[d]espite failure
up to now, we may assert that superconductivity will be solved on the basis of our
present quantum mechanical understanding." This assertion remained a matter of
faith for another 25 years as one microscopic theory after another collapsed until
Bardeen, Cooper, and Schrieffer (1957) created a compelling quantum-mechanical
model.
An important idea to grasp in understanding superconductivity is that it really
is a macroscopic phenomenon, and corresponds to a definite macroscopic state
of matter. The important feature of a superconductor is not how it responds to
external electrical fields, but how it responds to external magnetic fields. Meissner
and Ochsenfeld (1933) found that magnetic flux is completely expelled from a
superconductor. Suppose one has a loop with a current flowing in it. Magnetic flux
lines must thread the loop, as in Figure
27.1.
In order for the current to diminish,
Figure 27.1. The flux lines thread-
ing a current loop have to pass
through the current if the current is
to diminish. But for a superconduc-
tor, that is impossible, and the cur-
rent persists.
839
Condensed Matter
Physics,
Second Edition
by Michael P. Marder
Copyright © 2010 John Wiley & Sons, Inc.
840
Chapter 27. Superconductivity
the total magnetic field it creates would have to be reduced as well. In the course
of the reduction, a flux line would have to pass through the current loop, but for
a superconductor that is nearly impossible. It is for this reason that currents in
superconductors are so persistent. The impurity scattering sites in the metal and
the collisions of electrons with phonons do not go away, but whenever an electron
is slowed down by such a collision, a flux line heads toward the metal surface. The
metal repels it, and in so doing pushes the tardy electron back up to speed.
27.2 Phenomenology of Superconductivity
Perfect Diamagnets. When a superconductor is placed in an external field H,
all the flux lines are expelled from it so B is zero in its interior. The magnetic
permeability \i described by Eq. (24.6) is zero, and the superconductor is a perfect
diamagnet. This perfect diamagnetism can be explained phenomenologically by
supposing superconductors to be solids in which electrons accelerate in the pres-
ence of electric fields without displaying any damping. Like magnets and dielectric
media, they are captured by proposing a relation between electrical field and cur-
rent, which is
(27.1)
(27.2)
Using
Eq. (24.5) and restricting (27.3)
attention
to slowly varying currents so
that
dD/dt can be ignored.
Using
again Eq. (20.5b) and integrating (27.4)
in
time,
obtaining a constant of
integration
Bo-
This line of argument cannot determine the constant field Bo. Because of Meissner
and Ochsenfeld's observation that all magnetic fields are expelled from a super-
conductor, London
( 1961 )
proposed that superconductivity could be explained by
setting
BQ
to zero to obtain
ß + A£VxVxß = 0 (27.5)
where the London penetration depth is
/ 2 The assumption is that any features of mag-
\, =
*
/ . netic response not captured explicitly by j are (27.6)
Vf
Airline
2
captured by a coefficient jx.
Equation (27.5) is called the London equation. The length \i gets its name from
the solutions of
Eq.
(27.5). Consider a superconductor which occupies all of space
for z > 0. Look for solutions of Eq. (27.5) which are independent of x and y.
Rewrite Eq. (27.5) as
B +
X
2
L
(v(V-ß)-V
2
ß) =0. (27.7)
mv =
—eE
dj ne
2
s
ot m
d-■ -> B 4iTne
2
-L -
-VxVx - = Vx£
ot ß mc
- - B 4nne
2
.-,
=> Vx Vx- = (B-BO).
ß mc
1
Phenomenology of Superconductivity 841
Taking the dot product of
Eq.
(27.7) with z gives
B
z
= 0. Because B is independent of x and y, all the x (27.8)
and y derivatives vanish.
The first conclusion is then that the normal component of the magnetic induction
always vanishes at the surface of a superconductor. Taking next a dot product with
respect to x gives
B
x
= \
2
L
<y^
=>
B
x
<xe-
Z
/
XL
. (27.9)
oz
The second conclusion is that a component of the induction parallel to a supercon-
ducting surface dies off exponentially over length A^. This decay is the origin of
the Meissner effect. Because the current within the superconductor is related to the
induction by Eq. (24.3), one obtains two additional important conclusions:
1.
The superconductor sustains currents in equilibrium.
2.
These currents inhabit a strip of width A/, near the superconductor's surface,
and they vanish everywhere else.
27.2.1 Phenomenological Free Energy
Trying to explain superconductivity with Eq. (27.2) as the starting point creates
two difficulties. First, the constant time-independent magnetic field
BQ
needs to
be eliminated without explanation in the midst of the analysis. Second, it is not
the most general relation leading to superconductivity. Both difficulties can be
counteracted by recasting the phenomenological theory in terms of a free energy,
rather than a linear relation between current and field.
Consider a material without macroscopic magnetic fields in equilibrium. Wein-
berg (1986) emphasizes the economy of beginning with a free energy whose mag-
netic-field-dependent part takes the form
?
=
Jdrdr' Y
J
\A
a
{7)G
aß
{r-r')A
0
{r')
+
5{r-^)^B{r)-B{r'). (27.10)
aß
G is an arbitrary phenomenological function. Because it is a function of
7 —
7',
it describes a macroscopically homogeneous medium.
Equation (27.10) differs from the previous magnetic free energy (24.25) in two
respects. It is less general, because Eq. (27.10) is only quadratic in
A,
and therefore
only suitable for small deviations from equilibrium. On the other hand, this form
puts forward A as the primary variable rather than B. A technical point that should
be mentioned is that Eq. (27.10) is therefore not gauge-invariant. This deficiency
can be remedied by replacing A wherever it appears by A
—
V</>,
where a gauge
change A
—>
A + Vx sends
4>
to
(fi
+
X-
The final results will all be expressed in
terms of B = V x A, and because V x V0 = 0 is a vector identity there is no need
to carry
cf>
along.
842
Chapter 27. Superconductivity
The minima
of
this
free energy describe equilibrium states. Requiring the func-
tional derivative
of the
free energy
to be
zero with respect
to the
vector potential
gives
[VxVx^]
ft
= - f df
J2G
aß
(r-r')A
ß
(r') (27.11)
J
ß
Write
B = VxA
before taking derivatives
of
Eq. (27.10).
^ja(7)
=
^[VxB]
a
= -c f df
J2G
a
ß(7-7
,
)A
ß
(r'). (27.12)
J
ί
In Fourier space,
Eq.
(27.11) becomes
E {G
aß
(k)
+
^(k
2
5
aß
-k
a
k
ß
)
]
jA
ß
= 0.
(27.13)
A system displays paramagnetic
or
diamagnetic behavior with permeability
/i
if in
the
limit
as k
—>
0,
G
aß
—>
( 1) k
2
S
aß
— k
a
k
ß
.
Not obvious—obtained
by
working backwards (27.14)
\n
/ 4TT L J
from
the
desired result.
To
see
why, note that
if
G
is
taken
to
have this form, then
the
term involving
G in
the free energy
is
-^ drd? S(r-r')A-V
x\7 x A
mite
J2
ß
lk
2
6
a
0-k
a
k
ß
]A
0
(27.15)
= —\kxk
xA]
a
.
—
—— /
drBCr)
2
.
Use the
identity
in Eq.
(24.21)
to
integrate
by
(27.16)
oVT
J
parts
and
turn
the
leftmost
A
into
B.
Adding Eq. (27.16)
to the
second term
in
Eq. (27.10) gives
a
free energy
s-è-J"*®-
(27
'
17)
which describes
a
conventional permeable magnetic substance.
A superconductor
is
produced
by
adding
any
term
to G
aß
that destroys
the
existence
of
solutions with spatially uniform magnetic fields. For example, suppose
that
lim
G
aß
=
-r-Tjoaß- (27.18)
k-^O
47TA£
In this event,
the
free energy takes
the
form
-
1
-
f
dr~A
2
(r)
+
\VxÂ\
2
, (27.19)
Φ7T
J
\
L
which
has an
extremum when
-jA
+
VxVxA^O. (27.20)