Introduction to the Theory of Superconductivity and
Superuidity
N. B. Kopnin
Low Temperature Laboratory, Helsinki University of Technology, P.O. Box 2200, FIN-02015
HUT, Finland
Abstract
Kyl-0.104 (3 cr, L) 2+2. Post-graduate course. HUT. Fall, 2002
Contents
I
Two-uid description of superuidity
4
A Landau criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
B
Two-uid hydrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
C
First and second sounds . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
D Vortices in a rotating superuid . . . . . . . . . . . . . . . . . . . . . . . 12
E
II
Vortex near a wall. Feynman critical velocity . . . . . . . . . . . . . . . . 15
The Gross{Pitaevskii model
19
A GP equation and the coherence length . . . . . . . . . . . . . . . . . . . . 19
B
Quantized vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
C
Galilean invariance, Critical velocity and Excitations . . . . . . . . . . . . 23
III Ginzburg{Landau theory
29
A Ginzburg{Landau equations . . . . . . . . . . . . . . . . . . . . . . . . . . 33
B
Discussion of the GL equations . . . . . . . . . . . . . . . . . . . . . . . . 34
C
Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
1
D The Ginzburg{Landau parameter. Type I and type II superconductors . . 40
E
IV
Meissner eect. Magnetic ux quantization . . . . . . . . . . . . . . . . . 41
Vortices in type II superconductors
44
A Transition into superconducting state in a magnetic eld . . . . . . . . . . 45
V
B
Single vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
C
Vortex free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
London model
57
A Single vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
B
Bean{Livingston vortex barrier near the surface of a superconductor . . . 59
C
System of straight vortices . . . . . . . . . . . . . . . . . . . . . . . . . . 62
D Uncharged superuids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
VI
Time-dependent Ginzburg{Landau theory
68
A Microscopic values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
B
Discussion of TDGL equations . . . . . . . . . . . . . . . . . . . . . . . . 72
C
Energy balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
D Charge neutrality and a dc electric eld . . . . . . . . . . . . . . . . . . . 74
E
An a.c. electric eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
F
Critical current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
VII Motion of vortices
80
A
A moving vortex and the electric eld . . . . . . . . . . . . . . . . . . . . 81
B
Flux ow: Low vortex density . . . . . . . . . . . . . . . . . . . . . . . . 82
C
Flux ow: High vortex density . . . . . . . . . . . . . . . . . . . . . . . . 86
VIII Paraconductivity
89
IX
93
Weak Links
A Aslamazov{Larkin model . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
2
1 D.C. Josephson Eect . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
2 A.C. Josephson Eect . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
X
Layered superconductors.
98
A Lawrence{Doniach model . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
B
Anisotropic superconductors . . . . . . . . . . . . . . . . . . . . . . . . . 101
C
Upper critical eld for parallel orientation . . . . . . . . . . . . . . . . . . 102
1 Continuous limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2
XI
Highly layered case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Josephson junctions. Josephson vortices.
106
A Josephson junctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
B
Long Josephson junctions in the magnetic eld. . . . . . . . . . . . . . . . 106
3
I. TWO-FLUID DESCRIPTION OF SUPERFLUIDITY
Superuidity is an ability of a uid to ow without friction through narrow tubes. We
shall see why one needs a narrow tube in this denition. The most striking realizations of
superuidity are thermo-mechanical eects. Some of them are ilustrated in Figs. 1 and 2
A. Landau criterion
One can understand the absence of dissipation in a quantum liquid from the following
arguments. Assume that our uid is at zero temperature in a ground state. Since a quantum
system cannot change its energy continuously, the system has to create an excitation to
absorb the dissipated energy. Let the energy of an excitation in the stationary uid be (p)
with a momentum p. For a uid moving with a velocity v, the energy of the excitation in
the laboratory frame is
(p) + p v
It is favorable to create such an excitation if
(p) + p v < 0
This inequality can be satised if
or if at least
jvj > j(ppj)
jvj > min (pp)
(1)
Excitations are created if this condition is satised. If the minimum is nonzero
(p)
vc min
6= 0
p
(2)
no excitations can be created when the ow has a velocity below vc : The quantum uid
has no dissipation if it ows with a velocity v < vc . This is called the Landau criterion of
superuidity.
4
FIGURES
T+∆T
T
jn
js
FIG. 1. The fan rotates if one end of the closed tube is heated.
∆p
T
superleak
T+∆T
FIG. 2. A pressure head is established between two vessels kept at dierent temperatures and
connected by a superleak.
For a nite temperatures, there are already thermally created excitations which, if they
move, experience friction through their interactions with the container walls. At the same
time, for ows with v < vc there is a part of uid that moves without dissipation. In fact,
one can present the total mass current (momentum) per unit volume of the uid as a sum
of two parts
j= v
n
n
+ s vs
(3)
The velocity vn is associated with the motion of excitations, n is the coeÆcient of proportionality with the dimension of the mass density. The motion of excitations produces friction
as in a usual or \normal" uid. Thus vn and n are called the normal velocity and density
5
of the uid, respectively. The velocity
v
s
is associated with the motion without friction;
it is called the superuid velocity, while the proportionality coeÆcient s is the superuid
density. If the entire uid moves as a whole with a velocity vs = vn = v, the mass current
is
j = (
n
+ s ) v = v
where is the total density of the uid. Therefore
= n + s
(4)
B. Two-uid hydrodynamics
Since the superuid motion below vc does not create excitations, it does not transfer
momentum to and thus it does not exert a force on an external object immersed into uid.
This means that the superuid motion is a potential ow, characterized by the zero-vorticity
condition
curl vs = 0
(5)
Therefore, the full time derivative (the material time derivative) can be presented as a
gradient
dvs @ vs
=
+ (vs r) vs =
dt
@t
r
(6)
s
where s is the potential (per particle mass) that acts on the superuid component. Therefore, s has the meaning of the chemical potential of the superuid part of the liquid. Using
Eq. (5) we obtain
2
@ vs
v
s
+ r s +
=0
@t
2
Consider a reference frame K0 that moves with the velocity
(7)
v
s
with respect to the
laboratory frame K . We have in the laboratory frame (per unit volume)
6
j = v + j0
v2
E=
+v j
(8)
s
2
s
0 + E0
s
(9)
Here E0 and j0 are the energy and current in the moving frame which depend on the uid
velocity vn
v
s
in the moving frame. The energy obeys the thermodynamic identity
dE0 = T dS + d + (vn
v ) dj0
s
(10)
Here is the chemical potential (dened per particle mass), S is the entropy per unit volume.
The current density is
j0 = (v
n
n
v)
s
which, together with Eq. (8), indeed reproduces Eq. (3).
To nd the thermodynamic relation for the chemical potential we consider a system of
N particles with an energy E~ in a volume V~ and with a momentum J~ . Its energy obeys
dE~ = T dS~
p dV~ + ~ dN + v dJ~
The chemical potential per particle ~ is dened as
~N = E~
T S~ + pV~
v J~
Calculating the variation of the both sides of this equation and using the identity for the
energy we nd
Nd~ = S~ dT + V~ dp
J~ dv
or
S~ V~
V~
V~ J~
dT
+
dp
dv
N
N V~
V~ N
V~
V~
V~
= S dT + dp
j dv
N
N
N
d~ =
Dividing this by the particle mass we get
7
d =
S
dT + 1 dp
1 j dv
where = mN=V~ and = ~=m.
Therefore, in our case, the chemical potential (per unit particle mass) satises the thermodynamic relation
d = dT + 1 dp 1 j0 d (vn
v)
s
(11)
where = S= is the entropy per unit particle mass.
Eq. (11) helps to identify the function s in Eq. (7) as the chemical potential of the
superuid particles. Indeed, if we apply Eq. (11) to superuid component alone we should
put the entropy to zero since the superuid part of the uid is all in one (the ground) state.
Moreover, we also put j0 = 0 since there is no relative motion in the superuid part. As a
result, Eq. (11) for superuid component gives
r
s
= s 1 rp
At the same time, according to the Euler equation,
s
dvs
=
dt
rp
Comparing these two equations we arrive at Eq. (7).
In equilibrium, the chemical potentials of all the components of the uid should be equal,
therefore, s = , and
2
@ vs
v
s
+r +
=0
@t
2
(12)
In addition, the ow obeys the continuity equation
@
+ div j = 0
@t
(13)
Since the entropy of the superuid part is zero, the entropy of the uid is only associated
with the normal component (excitations). Therefore, the continuity equation for entropy is
8
@S
+ div (S vn ) = WS
@t
(14)
Here WS is the dissipation function that describes the entropy production due to viscosity
of the normal component.
The momentum conservation takes the form
@ji @ ik
+
=0
@t @rk
(15)
where ik is the tensor of the momentum ow. In the absence of the normal viscosity, it is
ik = n vn i vn k + s vs i vs k + pÆik
(16)
The two-uid hydrodynamic equations (12) { (15) are quite complicated because n , ,
, etc., depend on vn
v
s
and cannot be calculated without a microscopic theory. However,
there are situations where some progress can be made already on the basis of these general
equations.
C. First and second sounds
Assume that the uid is at rest on average, and consider small variations of its density
and velocity. Within the rst approximation in vs and vn we obtain from Eqs. (12) { (15)
@ vs
+ r = 0
@t
@
+ div j = 0
@t
@ ( )
+ ( ) div vn = 0
@t
@j
+ rp = 0
@t
We assume also that system is in thermal equilibrium with !
(17)
(18)
(19)
(20)
1 so that the dissipation
is absent.
Eqs. (18), (20) give
@2
= r2 p
@t2
9
(21)
Eq. (19) yields
@ 2 ( )
@v
+ ( ) div n = 0
2
@t
@t
(22)
while Eqs. (17), (20) result in
@ vs
+ r2 = 0
@t
@v
@v
n div n + s div s + r2 p = 0
@t
@t
div
which gives
div
@ vn
= (s =n ) r2 @t
n 1 r2 p
Inserting this into Eq. (22) we obtain
n @ 2 ( )
+ s r2 @t2
r2 p = 0
(23)
We express here @ 2 =@t2 through Eq. (21) writing to the rst approximation in small
variations
@ 2 ( )=@t2 = @ 2 =@t2 + @ 2 =@t2
We neglect the term (@=@t)(@=@t) which is of the second order in variations. Next we use
the thermodynamic relation Eq. (11) whence
r2 = r2T + 1 r2 p
(24)
@ 2 s 2 2
=
rT
@t2
n
(25)
As a result,
Equations (21) and (25) determine the behavior of thermodynamic variables in a sound
wave.
Denote Æp and ÆT variations of pressure and temperature in the sound wave. Since
10
@
@
Æp + ÆT
@p
@T
@
@
Æ = Æp + ÆT
@p
@T
Æ =
we obtain from Eqs. (21) and (25)
@ @ 2 Æp @ @ 2 ÆT
2 Æp = 0
+
r
2
2
@p @t
@T @t
@ @ 2 Æp @ @ 2 ÆT s 2 2
+
r ÆT = 0
@p @t2 @T @t2
n
(26)
(27)
We look for a solution in the form of a plane wave
Æp; ÆT
/ exp [ i! (t x=u)]
(28)
The equations yield
@ 2
u
@p
@ 2
u ÆT = 0
@T
s 2
ÆT = 0
n
1 Æp +
@ 2
@ 2
u Æp +
u
@p
@T
(29)
(30)
These equations can be simplied if we neglect a small thermal expansion and put
@ @
=
=0
@T @p
Equations (29), (30) then have two solutions:
s
@p
while ÆT = 0
@
(31)
2 s
while Æp = 0
n (@=@T )
(32)
u1 =
and
s
u2 =
Equation (31) describes the usual (rst) sound, i.e., variation of density and pressure at a
constant temperature. Note that we neglected the dierence between adiabatic and isothermal processes when we put the thermal expansion to zero. Equation (32) describes variations
of temperature and entropy. The pressure does not change. Since the thermal expansion is
absent, the density is also constant. This type of perturbation is called the second sound.
11
D. Vortices in a rotating superuid
In a rotating container, the normal component rotates with the vessel such that its
velocity is vn = r. Note that this ow eld has curl vn = 2
. Due to the potentiality of
superow Eq. (5) the superuid component cannot rotate with a rotating container; it should
be stationary in the laboratory frame. However, this fact contradicts to the experiment
which shows that the entire uid rotates with the same angular velocity . For example,
if the superuid did not rotate, the meniscus height would contain an extra factor n = as
compared to the normal uid (see Problem 3).
Consider the apparent contradiction in more detail. The condition of a potential ow of
the superuid component Eq. (5) implies that the superuid velocity can be presented as a
gradient of a ow potential
v
s
= r
(33)
This equations yields identically that curl vs = 0 everywhere in the uid, as it should be.
However, if it were correct absolutely at any point in the uid, the superuid could not
rotate. Is it possible to \spoil" Eq. (5) in such a way that the damage to it would be
minimal? One can argue as follows. Assume that Eq. (5) is violated only at separate points
in the uid while the rest of the superuid remains curl-free. To have a nonzero contribution
to the bulk of the uid we should assume a Æ -function-type of the violation. This Æ -function
should depend only on the coordinates r? in the plane perpendicular to the rotation axis
since vs should be uniform along the rotation axis. In particular,
^ Æ (2) (r?
curl vs = A
r0?)
(34)
^ is the unit vector along the rotation axis and A is a coeÆcient to be determined.
where The coordinate r0? determines thus a singular point on a plane where the curl-free condition
of superow is violated. In three dimensions, we thus obtain a singular line.
According to the Stokes theorem, Eq. (5) can be written as
12
I
v dr = 0
s
where the contour integral is taken along any contour in the uid. However, if now we take
the contour that encircles the above singular line, the result is
I
v dr =
s
Z
curl vs dS = A
where the second integral is taken over the area conned within the contour. Again, the
contour around the singular line can be taken arbitrary because curl vs = 0 everywhere
outside the line. Multiplying the above equation by the particle mass m we have
I
mvs dr = mA
According to the general principles of the quantum mechanics (Bohr{Sommerfeld quantization) the above integral should be quantized. In other words, one should have mA = 2 ~n
where n is an integer. Therefore,
I
v dr = 2m~n 0n
s
(35)
or
curl vs = 0 neL Æ (2) (r?
r0?)
(36)
where eL is the unit vector along the line.
This generalizes Eq. (5): For n = 0 one has the curl-free condition (5). Nevertheless
there can be singular lines such that the circulation of superuid velocity around them is
nonzero; it is quantized with the circulation quantum 0 = 2 ~=m. These singular lines are
called quantized vortex lines. Note that Eq. (33) introduces a function that is multivalued:
It changes by = 0 n after encircling each vortex line once.
For one straight vortex line along the z axis of a cylindrical coordinate frame (r; '; z ),
the velocity around it is from Eq. (35) vs = (0; vs' ; 0) where
n
vs' = 0
2r
13
(37)
or
v
ne r
= 0 z2
2r
s
(38)
If there are several vortex lines, the superuid velocity is a superposition of velocities of
each vortex line. For straight parallel lines,
v
s
=
e (r r )
2 jr r j2
X 0 ni
z
(39)
i
i
i
where ri is a (two-dimensional) position vector of the i-th vortex. Eq. (36) takes the form
curl vs =
X
0 ni Æ (2) (r
r)
(40)
i
i
Consider a large cylinder with a radius R. Assume that it is lled with vortex lines of
the same n with an uniform density nL . The velocity circulation around it is from Eq. (40)
I
v dr =
Z X
s
Æ (2) (r
r ) dS = n
i
L
R2 i
where = 0 n. This gives
2Rvs' = nL R2 or
vs' = nL R=2 = R
The last equality holds if
nL = 2
=
(41)
We see that the superuid component has on average the same velocity eld as the rotating
normal component provided the uid is lled by quantized vortex lines with the density
satisfying Eq. (41). Therefore, the entire liquid rotates on average with the angular velocity
. The superuid velocity has, of course, a highly non-uniform local distribution such that
the velocity is large near each vortex line while decreasing away from it.
14
The energy of a straight vortex line can be calculated from Eq. (37)
EL =
Z
Z
s vs2
2 s L rmax dr
dV =
2
4 rmin r
where L is the length of the line. The upper limit rmax is determined by the larger distance
at which the slow r 1 dependence of the velocity eld changes to a faster decay. In an array
of vortex lines, this happens at a distance of the order of the distance between vortices, r0 .
The intervortex distance is determined by Eq. (41)
r02 2 (42)
where we put nL = 1=r02.
The lower limit is determined by the microscopic thickness of the vortex line. On the
atomic scale, the vortex line has a structure characterized by the so called vortex core with
the radius rc . Finally, the energy of a vortex line per unit length, i.e., the linear tension, is
2 r
L = s ln 0
4
rc
(43)
The vortex energy is proportional to square of circulation. Therefore, the energy of N
singly quantized vortices with = 0 is N1 where 1 is the energy Eq. (43) for n = 1.
However, the energy of one vortex with N quanta, i.e., = N0 is N 2 1 . At the same
time, to create a velocity eld corresponding to a given average rotation velocity one
needs either certain number N of singly quantized vortices or one vortex with N quanta.
Therefore, it is more energetically favorable to have singly quantized vortices with = 0 .
E. Vortex near a wall. Feynman critical velocity
Consider a vortex near a at wall of a container. Due to boundary conditions of vanishing
of the normal component of ow at the wall, the vortex ends should be perpendicular to the
wall. Energy minimum suggests that the vortex should have the form of a half-loop with its
ends terminating at the wall (see Fig. 3).
15
vs
FM
P
FIG. 3. The vortex half-loop attached to the wall.
The linear tension tries to contract this loop. However, if there is a ow parallel to the
wall, it will try to inate the loop. Indeed, the energy of the half-loop is from Eq. (43)
2 s R R
ln
E=
4
rc
The momentum of the loop is (see Problem 4 to Sect. I)
P = nS
(44)
s
where S is the area of the loop, and n is the unit vector perpendicular to the plane of the
loop; its positive direction is determined by the right screw rule with the vortex circulation.
Assume that n is anti-parallel to the external superow vs . For a half-circle, S = R2 =2.
The energy of the loop in the ow is then
E + P vs =
2 s R R
ln
4
rc
s R2
2
The energy is plotted in Fig. 4. As a function of the loop radius, the energy has a maximum
at
Rc =
R
ln
4vs rc
16
(45)
This maximum is a potential barrier
3 s 2
E0 =
ln
16vs
4vs rc
(46)
that separates the state where the loop vanishes at the wall R = 0, E = 0 and the state
where the loop is innitely large (is far from the wall) R =
1, E = 1.
A loop with a
radius smaller than Rc in a ow vs will shrink while that with R > Rc will grow.
E
E0
R
Rc
FIG. 4. The energy of the half-loop as a function of its radius.
The velocity that corresponds to an unstable equilibrium at the barrier is
R
ln
vc =
4R
rc
(47)
This is the same as the result of Problem 5 to Sect. I which also determines the equilibrium
velocity of the loop. If, instead of a at wall, we have a ow channel of a dimension R
the ow with a velocity from Eq. (47) will inate loops with the maximum possible radius
which is of the order of R. These loops will be torn away from the channel walls and will
come into motion in the uid. The vortex motion, in fact, gives rise to a dissipation due to
interaction of vortices with the normal component which is called the mutual friction. As a
result, there will be no superuidity! This is why the velocity Eq. (47) is called the critical
velocity of superuidity. It was rst introduced by Feynman. The exact magnitude of the
critical velocity depends on the ow geometry, this is why Eq. (47) gives only an -order-ofmagnitude estimate. The Feynman critical velocity is much smaller that the Landau critical
velocity for creation of rotons (or phonons); it decreases for wider channels.
17
Problems
Problem 1
Find the fountain pressure head in two vessels connected by a superleak if one is heated
with respect to another (see Fig 2).
Problem 2
Find the velocity of the fourth sound, when the normal component is clamped in a narrow
tube due to a large viscosity.
Problem 3
Find the meniscus of a rotating superuid in a vortex-free state.
Problem 4
Find the momentum of a vortex loop with a radius R.
Problem 5
Find the energy of the vortex loop and its velocity in the uid.
Problem 6
Find the rotation velocity above which the rst vortex becomes energetically favorable
in a container with a radius R.
18
II. THE GROSS{PITAEVSKII MODEL
A. GP equation and the coherence length
The simplest theory pretending to describe kinetic processes in superuids is known to
be the Gross{Pitaevskii theory designed for a weakly non-ideal Bose gas (with repulsion
between particles) at zero temperature. It is assumed that almost all particles are in the
condensed state, i.e., the wave function is = + 0 where the non-condensate part of the
wave function 0 is small compared to the condensate wave function . It thus satises the
Schrodinger equation
@
i~ =
@t
~2
2m
r2 + +
Z
j (r0)j2 U (r r0 ) d3r0:
Here is the chemical potential and U > 0 is the interaction (repulsive) interaction. Assuming that
varies slowly at distances of an atomic scale, one can take it out from the
integral denoting
U0
Z
U (r) d3r
The potential U0 determines the scattering length a = mU0 =4 ~2 . For the Bose condensate
of an ideal gas, = 0. In an interacting gas, the chemical potential is known to be2
= NU0
where N is the density of number of atoms in a non-perturbed liquid. We shall see that this
expression gives the correct normalization of the condensate wave function. The Schrodinger
equation takes the form
i~
@
~2 2
=
r
@t
2m
U0 N
j j2 :
(48)
The r.h.s. of this equation is nothing but the energy of the system.
This equation was obtained by Pitaevskii and Gross in 1961. The macroscopic wave
function of the condensate atoms is generally complex
situation j
=
j je
i
. In a spatially uniform
p
j = N while the energy is independent of the phase .
19
The particle ow is determined by the usual quantum-mechanical expression for the uid
momentum per unit volume
j=
i~ [ r
2
r
]
(49)
It appears when the phase varies in space. Equation (49) suggests that P = ~r is the
momentum of a condensate particle, while
v
s
=
~
m
r
(50)
is its velocity. The superow is thus potential; Eq. (33) yields
=
~
m
As a result, the particle current becomes j = s vs where s = Nm = mj j2 the mass density
of condensate atoms. For a weakly interacting gas, s coincides with .
We can write Eq. (50) in the form of Eq. (7) used in the two-uid description:
@ vs
vs2
+ r s +
=0
@t
2
(51)
where
s =
~ @
m @t
vs2
2
(52)
is the chemical potential of the superuid component. The GP equation (48) provides an
expression for s . Let us put = j jei. The complex equation (48) yields two real equations
v 2 ~2
j j m~ @
= j j s + r2 j j + j j 1 j j2 =N ;
(53)
@t
2 2m
@j j
~
=
[rj jr + r(j jr)]
(54)
@t
2m
Equation (53) gives the expression for the chemical potential
~2
s = j j2 =N 1
j j 1r2j j
2m
(55)
The chemical potential is zero in a spatially uniform state with j j2 = N . Equation (54) is
nothing but the continuity equation. To see this we multiply Eq. (54) by 2mj
20
j and nd
@ (mN )
+ div j = 0
@t
(56)
According to Eq. (48) the characteristic length of spatial variations of the wave function
is
=
~
p2mU
N
0
Using the expression for a, we nd
N 1=3 1=2 N 1=3
where = aN 1=3 a=d is the \gas" parameter and d is the distance between atoms. Since
it is assumed that 1 the coherence length is indeed much longer that the interatomic
distance.
This example shows that it is indeed the existence of a large parameter N 1=3 =d 1
that allows one to construct a tractable model of superuidity.
vs /vmax
1
f
j/j max
3
1
x
FIG. 5. The condensate density f , velocity v and current near the vortex axis for a single-quantum vortex n = 1. vmax and jmax are determined by Eq. (63) below.
B. Quantized vortex
Equation has a time-independent vortex solution. It corresponds to a wave function in
the form
p
= Nein' f (r= )
21
(57)
where ' is the azimuthal angle in the cylindrical frame (r; '; z ), and n is an integer. With
this choice of the ow potential, the function becomes non-single valued: The phase
= n' varies by 2n on going around the z axis. As a result, the superuid velocity has a
nonzero circulation
I
I
v dr = m r dr = 2m~n n0
~
s
where
0 = 2 ~=m
(58)
is the quantum of circulation. It is the same as in Eq. (35) derived using semi-classical
arguments in Sect I. Velocity circulation in a superuid is quantized. The ansatz Eq. (57)
corresponds to an n-quantum vortex.
The amplitude function f (x) satises the equation
1 d
df
x
x dx dx
n2 f
+f
x2
f3 = 0
(59)
For a singly quantized vortex n = 1, the function f (x) vanishes as f
/ x at the vortex axis
x ! 0. It saturates at f = 1 for x ! 1. The region r near the vortex axis is called the
vortex core.
The condensate velocity near the vortex is from Eq. (50)
v
s
where
e
'
=
n~e' n0 e' n0 ez r
=
=
mr
2r
2r2
is the unit vector in the azimuthal direction while
e
z
is the unit vector in the
direction of the z axis. This agrees with Eq. (38). The particle current is
2
f
e
j = n~mr
'
It vanishes at r = 0 and decays as 1=r for r ! 1.
More extended discussion of vortices will be given later on the basis of the Ginzburg{
Landau theory.
22
C. Galilean invariance, Critical velocity and Excitations
Note that Eq. (48) is Galilean invariant. If 0 (r) is a solution to Eq. (48) then the
function
= 0 (r
vt) exp
i
~
mv r
i mv 2 t
~ 2
(60)
is also a solution to that equation carrying a particle current
j = j0 + v
where j0 is the particle ow associated with the wave function 0 .
This fact has a particular meaning in terms of vortex motion. Indeed, if the system is in
a state with quantized vortices in the absence of a net ow, the vortices will move together
with the whole liquid with a velocity
v = j= if a current j is driven through the system.
This agrees with the Helmholtz theorem of conservation of vorticity in an ideal (non-viscous)
uid.
Taken more seriously, the Galilean invariance of the GP equation shows that one cannot
dene a critical velocity above which superuidity is destroyed, in contradiction to the
Landau criterion2 of superuidity. Consider this in more detail.
On one hand, we can look for a time independent solution
p
= Nfeikr
(61)
where wave vector k is related to the supercurrent through Eq. (49): j = (~=m)kf 2 .
Equation (48) yields
f 2 = 1 k2 2
(62)
This results in the current
j = (~=m)k(1
p
This expression has a maximum at k = 1= 3 with
23
k2 2 )
jmax =
p2~ ; vmax = p ~
3 3m
3m
(63)
These quantities dene the maximum supercurrent and the maximum superuid velocity
which can exist in the superuid in a time-independent state. The time-independent state
becomes unstable when the superow exceeds vmax.
It is interesting to compare the maximum velocity vmax with the sound velocity in a nonideal Bose gas. It is known that the GP equation allows excitations which have the form of
phonons at long wave lengths2 . The excitations are perturbations of the wave function in
the form of an oscillatory wave
= 0 + A cos (!t
k r + )
where 0 is a stationary wave function, A and are the amplitude and phase of oscillations.
The excitation spectrum which can be found from Eq. (48) has the form (see Problem 3 to
Sect. II)
2 2 #1=2
U0 N 2
~k
!=
k +
m
2m
"
(64)
For long wave lengths or small wave vectors k one obtains the sound-like dispersion
! = uk
with the sound velocity
r
u=
~
U0 N
=p
m
2m
(65)
For large wave vectors ~k mu, the spectrum is particle-like
=
p2
2m
where = ~! and p = ~k.
The sound velocity u given by Eq. (65) coincides with the Landau critical velocity dened
according to the Landau criterion as
24
(p)
vc = min
p
(66)
that indeed gives the phonon velocity u for the excitation spectrum (p) = ~! with ! from
Eq. (64).
The Landau criterion Eq. (66) has a simple interpretation on a plot of as a function of
p. Indeed, calculating the minimum of =p we nd that it is a solution of
d =
dp p
On the plot, it means that the line = Cp where C is a constant is tangent to the curve
(p). The constant is then C = vc . The spectrum of excitations in the GP model is shown
in Fig. 6 (a). The tangent coincides with the initial part of the spectrum and denes the
critical velocity equal to the sound velocity. The spectrum of excitations in a real 4 He is
shown in Fig. 6 (b). It has a minimum which is historically called the \roton minimum".
The critical velocity is smaller than the sound velocity.
ε
ε
vc
vc
p
p
(a)
(b)
FIG. 6. The excitation spectrum in the GP model (a) and in the real 4 He (b). The Landau
critical velocity is given by the slope of the corresponding dashed line.
Equation (66) denes the velocity limit above which the excitations (phonons in our
case) are produced by the moving superuid. We can see from Eq. (65) that vmax is of the
same order as the Landau critical velocity but it is by a factor of
r
vmax =
3
v =
2 c
r
3
u:
2
p
3=2 larger than vc :
(67)
The question is what velocity limit should be used? Is it vc or vmax that sets the limit of
existence of a time-independent state? Or, maybe, the GP model itself breaks down at vc
25
since creation of excitations violates the basic assumption of vanishing of non-condensate
part of the wave function?
This consideration, however, contradicts to the general Galilean invariance of the GP
equation. Doing the calculations of the maximum velocity, we restricted ourselves by timeindependent solutions. However, the Galilean invariance tells us that, in the presence of a
ow, a solution of the form of Eq. (60) also exists. If the initial state is uniform in space, the
state of the system in the presence of a ow has the same (velocity-independent) amplitude
and an additional phase factor
exp
i
~
mv r
i mv 2 t
~ 2
This holds for any ow no matter how large its velocity is.
The apparent contradiction cannot be resolved within Eq. (48). The problem is that
Eq. (48) does not include any interaction with the environment. In particular, the walls
of the container are not involved. In real physical situation at a nite temperature, the
environment plays a very important role since it provides a source of excitations and couples
strongly to the non-condensate part of the uid. Taken this into account we realize, that
at nite temperatures, as long as the container itself does not take part in the motion, the
Galilean invariance is not applicable to the entire system which also comprises eects of
container walls. A zero-temperature limit has to be taken with caution, keeping in mind the
relative importance of the environment and internal processes in the uid. The issue of the
proper choice of the reference frame is here of a crucial signicance. Taking the laboratory
frame where the container is at rest, we assume that creation of excitations dominates over
the intrinsic processes far from container walls. This situation is relevant to the phenomenon
of the critical velocity. One can expect the following behavior. Below the Landau critical
velocity vc there are no excitations created in the uid. The condensate part is Galilean
invariant, and the state Eq. (60) is realized. Note that it is due to the fact that the entire
system is not Galilean invariant that the critical limit vc does exist. Above vc excitations
are created, and the validity of the GP model has to be investigated separately for each
26
particular problem. The opposite limit of a fully Galilean-invariant behavior of the entire
system assumes that eects of the environment on the bulk properties are small. Both of
these pictures are only valid during some transient period whose duration is determined
by properties of the uid and its interaction with the environment. In this sense, the GP
equation itself does not provide a comprehensive description of superuids.
The main limitation of the GP equation is the lack of an explicit mechanism that establishes equilibrium towards a particular state. In principle, a source of dissipation can
be identied within the GP model: it is associated with an emission of phonons. These
phonons either escape to innity or are absorbed at the walls. The efective relaxation mechanism could then be included into the GP equation in the form of a complex-valued factor
= 0 + i 00 instead of = 1 in front of the time derivative:
i~
@
~2 2
=
r
@t
2m
U0 N
j j2 :
(68)
A purely imaginary factor corresponds to the so called time-dependent Ginzburg{Landau
model which reasonably well describes non-stationary behavior of superconductors in some
simple situations. However, the problem here is that the rate of phonon creation depends on
the particular state such that it seems unlikely to construct a GP-like equation that includes
an universal eective relaxation parameter .
27
Problems
Problem 1
Find the behavior of the wave function magnitude near the vortex axis for an n-quantum
vortex.
Problem 2
Show that Eq. (60) is indeed a solution of Eq. (48) if 0 is.
Problem 3
Find the spectrum of excitations described by Eq. (48).
28
III. GINZBURG{LANDAU THEORY
Superconductivity manifests itself mainly as an absence of resistivity below some critical
temperature. It is easy to measure resistivity, much easier than to measure quantities
relevant to superuid helium. The resistivity behavior as a function of temperature is shown
in Fig. 7.
ρ
ρn
ρ=0
Tc
T
FIG. 7. Below the transition temperature, the resistivity drops to zero.
The Ginzburg-Landau theory of superconductivity created by Ginzburg and Landau in
19509 is based the Landau theory of second-order phase transitions2. The basic notion is the
order parameter which describes a new property of the system that rst leads to breaking
of certain symmetry and then continuously develops under changing of some external parameter, for example, of the system temperature (or magnetic eld, etc.). The well known
example is the spontaneous magnetization of a ferromagnet. For superconductors (or superuids) one can suggest the density of \superconducting electrons" as an order parameter.
It appeared more productive, however, to introduce the probability amplitude or the\wave
function" of superconducting electrons to play the role of the order parameter.
In these lecture notes we only consider s-wave superconductors in isotropic media. If
is the wave function of superconducting electrons the free energy of the system near the
transition into the superconducting state can be written as an expansion in terms of a small
number of superconducting electrons. If the state is spatially uniform and the magnetic eld
is absent, the superconducting free energy measured from the normal state has the form
29
Fsn = V a jj2 + 2b jj4 :
(69)
where a and b are the expansion parameters and V is the volume of the system. At temperatures above the phase transition temperature Tc , the parameter a should be positive. This
ensures = 0 to be a local minimum of the free energy. Below the transition temperature,
however, the coeÆcient A should become negative. This leads to shifting the minimum to a
nonzero . The free energy will have a minimum if the coeÆcient b is positive:
min = jaj=b
F
sn
min = V jaj2 =2b
Near the transition temperature, one can thus put the coeÆcient a = a0 (T
Tc ) while
b = const.
F
Re Ψ
Im Ψ
χ
FIG. 8. Below the transition temperature, the free energy Eq. (69) has a minimum at a
nonzero order parameter magnitude. The minimum energy is degenerate with respect to the order
parameter phase .
The development of the microscopic theory of superconductivity14 has further demonstrated that the order parameter is actually a wave function of \superconducting pairs" of
electrons which make a \Bose condensate". The most convenient normalization of the wave
function is such that its modulus is related to a certain characteristics of the quasiparticle
energy spectrum in the superconductor. More specically its modulus is chosen to be the
energy gap in the electronic spectrum which opens after transition into superconducting
state
30
Ep =
q
EF )2 + jj2
(p
where p is the spectrum in the normal state. The order parameter is a complex function
= jjei where is the same for all condensate particles if there is no current. In the
presence of current and magnetic eld, both the magnitude and the phase vary in space.
The superconducting free energy is expanded now in terms of and its gradients:
2 #
Z "
2
e
Fsn = jj2 + 2 jj4 + i~r c A dV
(70)
Here c is speed of light. We use the gaussian units for electromagnetic quantities. These
units are commonly used in the physics of superconductivity. We will give the conversion to
SI units in some cases. For example, in the SI units one has to put c = 1 in Eq. (70).
The coeÆcient should be positive to ensure a minimum energy for a spatially homogeneous state. The total energy consists of the normal state energy Fn , the superconducting
free energy Fsn, and the magnetic energy:
F =F
n + Fsn +
Z
h2
dV
8
(71)
Here h is the microscopic magnetic eld. The average of h gives the magnetic induction B.
In the SI units, the magnetic energy is
Z
0 h2
dV
2
where 0 is permeability of vacuum. We omit the (constant) free energy of the normal state
F
n
in what follows.
The gradient term in Eq. (70) is the momentum operator in presence of the magnetic
eld
P^ =
i~r
2e
A
c
(72)
Here the charge 2e accounts for the charge of a Cooper pair (e is the electronic charge). It
implies that there is a gauge invariance: the free energy of the system and the magnetic eld
do not change if one makes a simultaneous transformation
31
! + f (r) ;
A ! A + ~2ec rf
The free energy expression is supplemented with the Maxwell equation
curl h =
4
j
c
(73)
where j is now the electric current and the microscopic eld is
h = curl A
(74)
In the SI units we have instead of Eq. (73)
curl h = j:
At the transition temperature, T = Tc , the coeÆcient changes its sign and becomes
negative for T < Tc , while and are positive constants. Microscopic theory gives15;16
= Tc
Tc
T
; =
7 (3)
8 2 Tc2
(75)
where is the single-spin density of states at the Fermi level, (3) 1:202, and we use the
units with kB = 1. Equation (75) demonstrates that the expansion in Eq. (70) goes in the
parameter =Tc . One has thus to assume that the order parameter is small compared to T .
The coeÆcient depends on purity of the sample. The purity is characterized by the
parameter Tc =~, where is the electronic mean free time due to the scattering by impurities.
Superconductors are called clean when this parameter is large, and they are dirty in the
opposite case. One has
=
D
y (Tc=~)
8~Tc
(76)
where D = vF2 =3 is the diusion coeÆcient, and
1
1
8 X
y (x) = 2
2
n=1 (2n + 1) [(2n + 1)2x + 1]
This function is y = 1 for a dirty limit Tc =~ 1, and it is
32
(77)
y (Tc =~) =
7 (3)~
2 3 Tc
for a clean case Tc =~ 1. Therefore
=
8
>
< D=8Tc ~ ;
dirty
>
: 7 (3)vF2 =48 2 Tc2
;
(78)
clean
so that dirty =clean (Tc =~) 1.
A. Ginzburg{Landau equations
Variation of F with respect to , and A gives
Z ( "
!
2 #
2
e
ÆF =
+ jj2 + i~r
A Æ + c:c:
c
2e
curl curlA 2e i~r
A + c:c: ÆA
+
4
c
c
2ie
1
+ div ~ Æ ~r
A + c:c: + 4 ÆA curlA dV
c
The requirement of extremum of the free energy gives the GL equation
2e 2
2
A =0
+ jj + i~r
c
(79)
together with the denition of the current
j = 2e i~r
2e
A + i~r
c
2e
A c
(80)
which follow from the Maxwell equation (73) and Eq. (74). In addition, the surface term
requires, in particular,
n ~r
2ie
A =0
c
(81)
where n is the unit vector along the normal to the surface. This is the so called \natural"
boundary condition. In particular, it tells that the current through the surface vanishes.
It only applies at the boundary with vacuum. In other cases such that contacts with conductors, etc., one has to include also the energy of interaction between the superconductor
33
and contacting media, which will change the boundary conditions. For the same reason,
the vector-potential term can not be put to zero independently from the corresponding
contribution of external elds.
The rst equation (79) is very similar to the time-independent version of the Gross{
Pitaevskii equation (48) for a noncharged system, e = 0. The particle current j=e from Eq.
(80) also coincides with Eq. (49) in the absence of charge.
B. Discussion of the GL equations
Consider rst the equation for the order parameter Eq. (79). In a homogeneous case
without a current and a magnetic eld, it gives
= GL =
p
1
8 2 2
jj= = 7 (3) Tc (1 T=Tc) 12
(82)
As we already know, the ratio GL =Tc should be small. This implies that the GL theory
works for temperatures close to Tc such that 1 T=Tc 1.
The free energy density in a homogeneous case is
Fc =
jj2=2
(83)
It is called the condensation energy. It denes the thermodynamic critical magnetic eld
Hc2 = 4 jj2=
(84)
when the magnetic energy is equal to the condensation energy, i.e., to the energy in the
absence of magnetic elds in the bulk superconductor (the so called Meissner state, see a
discussion later). In the SI units
Hc2 = jj2 =0 The thermodynamic critical eld is linear in temperature
Hc(T ) Hc(0) 1
34
T
Tc
where Hc(0) is formally dened by Eq. (84) where we put T = 0 in the coeÆcient .
Above the eld Hc , the superconducting state without currents has a larger energy than
the normal state. Indeed, the proper thermodynamic potential in an applied eld is the
Gibbs free energy
G=F
Z
HB
4
In the superconducting state B = h = 0, and the Gibbs free energy density is Gs = Fsn =
Fc = jj2 =2 . In the normal state, Fsn = 0, B = h = H so that Gn = H 2 =8 . The
energy in the normal state becomes smaller than Gs for H > Hc .
Eq. (79) denes the length
(T ) =
p
~2 =jj / (1 T=Tc )
1
2
(85)
which is a characteristic scale of variations of the order parameter. It is called the coherence
length. In the clean case Tc =~ 1 the coherence length is
7 (3)
(T ) =
12
1
2
0 (1 T=Tc)
1
2
(86)
where
0 = ~vF =(2Tc )
(87)
is the \zero-temperature" coherence length. In the dirty case
p
`
(T ) = p 0 (1 T=Tc)
2 3
1
2
(88)
where ` = vF is the electron mean free path. The impurity parameter can be expressed
through the ratio of 0 and `:
Tc =~ = `=20
(89)
so that a dirty limit corresponds to ` 0 while a clean limit is for ` 0 . Using (T ) and
0 one can write Eq. (79) in the form
35
2
2ie 2
A + jj2 =2GL = 0
~c
r
(90)
which contains only one parameter .
Let us put = jjei. The complex equation (90) gives two real equations. One is
2
r2jj
4e2 2
Q jj + jj
~2 c2
jj3=2GL = 0
(91)
Here we introduce the gauge invariant vector potential
Q=A
~c
2e
r
(92)
The other equation is the conservation of supercurrent
div js = 0
(93)
which agrees with the Maxwell equation (73).
Consider now the expression for current Eq.(80). Using the denition of the momentum
operator Eq.(72) we introduce the superconducting velocity operator
^
2mv^ s = P
(94)
for a Cooper pair with the mass 2m. Now the current becomes
e2 Ns
j = mc A
~c
= Nsevs
(95)
2e
e
A
=
Q
~c
mc
(96)
2e
r
where
v
s
=
~
2m
r
and
Ns = 8m jj2
(97)
is the density of \superconducting electrons". For low currents or magnetic elds, the order
parameter does not depend on the current, jj = GL . In the clean case
36
T
= 2N 1
Tc
8EF
1
Ns = 8m 2GL = 8m jj= =
3
T
Tc
(98)
where N = p3F =3 2 is the total number of electrons. Equation (98) is the same as Ns =
N 1
2
2
T
Tc
. The last equality in Eq. (98) holds for a simple metal with a parabolic spectrum
p = p2 =2m. For a dirty case
T
4 3 Tc =
1
Tc
7 (3) ~
16 3 EF TC Ns =
1
21 (3) ~
T
N:
Tc
(99)
It is much smaller than in the clean case: the scattering on impurities impedes the supercurrent which eectively leads to a reduction in the superconducting density. Note that the
critical temperature of a s-wave superconductor Tc is itself insensitive to the impurities.
The Maxwell equation Eq.(73) combined with Eq. (95) gives
4Ns e2
curl curl A =
mc2
~c
A
2e
r
(100)
or
curl curl A = L 2
~c
A
2e
r
(101)
where we dene the characteristic length
1
mc2 2
L =
4Nse2
(102)
In SI units,
m
L =
0 Ns e2
1
2
For low currents
c2 L =
32e2 jj
1
2
/
T
Tc
1
1
2
(103)
The length L is called the London penetration length. It determines the characteristic scale
of variations of the magnetic eld. Now the current can be written as
j = 4c 2
L
A
37
~c
2e
r
(104)
We dene the Ginzburg-Landau current
jGL = 4e~ 2GL = = c2 =(8e2 )
(105)
It sets the order of magnitude of the largest current which the superconductor can sustain.
The maximum current is called also the pair-breaking current since superconductivity is
destroyed by larger currents. The pair-breaking current corresponds to the critical current,
Eq. (63), with the largest possible gradient of the order parameter phase r 1= .
With the Ginzburg{Landau equation (79) we can transform the free energy expression
Eq. (70) to another form. Let us perform integration by parts in the gradient term and
substitute the kinetic energy term using Eq. (79). We obtain
Z Z
4 h2
2
F=
jj + 8 dV + ~ dS r
2
2ie
A ~c
The surface term vanishes because of the boundary conditions Eq. (81). We get
Z 4 h2
F=
jj + 8 dV
2
(106)
Sometimes, it is convenient to use the normalization of the order parameter such that it
has the form of the wave function of superconducting electrons. The free energy becomes
2 #
Z "
b 4 1 2e
2
Fsn = a jj + 2 jj + 2m i~r c A dV:
(107)
The constants a and b satisfy
jaj = mc2
b
16e2 2L
and determine the new order parameter magnitude jGL j2 = jaj =b which is
jGLj2 = 2m 2GL
in terms of the previous denition of . The coherence length is now expressed through the
electronic mass
2 =
~2
2mjaj
:
More discussion of the GL equations can be found in Ref.4 .
38
C. Fluctuations
We have seen that the free energy expansion Eq. (70) holds when the parameter =Tc is
T=Tc 1. However, being a mean-eld theory, the GL theory cannot
small, i.e., when 1
be valid very close to the transition temperature because of an increasing magnitude of
uctuations. Let us calculate the average jj2 due to uctuations at temperatures slightly
below Tc . The variation of free energy in a volume of the order of 3 near the equilibrium
value, jj = GL + Æ jj, is
1 Æ2F
ÆF =
(Æ jj)2 = 4 3jj (Æ jj)2
2
2 Æ jj
The average can be calculated with the probability of uctuations
P (Æ jj) = C exp f Æ F =T g
where the normalization constant C is found from
Z
C exp f Æ F =T g d (Æ jj) = 1
(108)
We obtain
(Æ jj)2 =
Z
C exp f Æ F =T g (Æ jj)2 d (Æ jj)
(
)
Z
4 3 jj (Æ jj)2
= C exp
(Æ jj)2 d (Æ jj)
T
= 2T=jj 3
This uctuation should be smaller than 2GL , i.e., 2T=jj 3 jj= or
2T
jj2
3
This can be written as
1
2
T Tc
Gi
(109)
Tc Hc2(0)03
Equation (109) is the Ginzburg criterion10;11 ; the dimensionless quantity Gi is called the
Ginzburg number. It is the second power of the ratio of the critical temperature and the
39
zero-temperature condensation energy in the volume of a cube with the size of the coherence
length. Using the microscopic values for the GL parameters, we obtain
T4
~4
Gi c4 EF (0 pF )4
(110)
We can see that the Ginzburg number is very small for superconductors where 0 pF =~ 102 103 . Therefore, Eq. (109) can be easily fullled for superconductors: In superconductors, there exists a broad region near Tc where uctuations are not important.
We observe again that the applicability of the GL theory depends crucially on the existence of the large parameter 0 pF =~ (T )=d. For helium II, as we know, such parameter
does not exist, thus one cannot directly apply a GL-type description17 (see18 ) to helium
II except for the GP model for weakly non-ideal Bose gas, discussed in Section II, which
fortunately does have such a parameter.
D. The Ginzburg{Landau parameter. Type I and type II superconductors
The ratio of the two characteristic lengths is called the Ginzburg-Landau parameter
21
L (T )
c2
(111)
=
=
(T )
32 ~2e2 2
It is independent of T and is determined by the material characteristics.
For clean superconductors it is
1 1
9 4 2 a0 pF e2 =a0 2 ~c Tc
=
14 (3)
2 ~
EF
e2 EF
(112)
Here a0 is the interatomic distance. Usually, it is of the order of 1=2 ~pF . The ratio
of the Coulomb interaction energy of conducting electrons, e2 =a0 , to the Fermi energy is
of the order of unity for good metals, but it may become larger for systems with strong
correlations between the electrons. The last factor in Eq. (112) is usually small: Tc =EF
is of the order of 10 3 for usual superconductors, but it is of the order of 10 1 10 2 for
high temperature superconductors with Tc
constant e2 =~c = 1=137.
100K and E 1000K .
F
40
The ne structure
We see that for usual clean superconductors the Ginzburg-Landau parameter is normally
small 1, though, in some cases it may be of the order of 1. On the contrary, for high
temperature superconductors, which have a tendency to be strongly correlated systems with
a not very low ratio of Tc =EF , the parameter is usually very large. The Ginzburg-Landau
parameter increases for dirty superconductors:
dirty clean (~=Tc )
(113)
Therefore, dirty alloys normally have a large .
The magnitude of divides all superconductors between two types: type I and type II
p
p
superconductors. Those with < 1= 2 belong to the type I, while those with > 1= 2
are type II superconductors.
E. Meissner eect. Magnetic ux quantization
Eq. (101) describes the Meissner eect, i.e., an exponential decay of weak magnetic elds
and supercurrents in a superconductor. The characteristic length over which the magnetic
eld decreases is just L . Consider a superconductor which occupies the half-space x > 0.
A magnetic eld hy is applied parallel to its surface (Fig. 9). Taking curl of Eq. (101) we
obtain
@ 2 hy
@x2
L 2 hy = 0
which gives hy = hy (0) exp( x=). The eld decays in a superconductor such that there is
no eld in the bulk. The supercurrent also decays and vanishes in the bulk acording to Eq.
(73).
41
S
hy
h
x
λL
0
FIG. 9. The Meissner eect: Magnetic eld penetrates into a superconductor only over distances shorter than .
L
B
l
FIG. 10. Magnetic ux through the hole in a superconductor is quantized.
Therefore,
B = H + 4M = 0
in a bulk superconductor. The magnetization and susceptibility are
M = H=4 ; = @M
@H
=
1
4
(114)
as for an ideal diamagnetic. The Meissner eect in type I superconductors persists up
to the eld H = Hc1 above which superconductivity is destroyed, see Fig. 19. Type II
superconductors display the Meissner eect up to much lower elds, after which vortices
appear (see the following section).
42
Let us consider an non-singly-connected superconductor with dimensions larger than L
placed in a magnetic eld (Fig. 10). We choose a contour which goes all the way inside the
superconductor around the hole and calculate the contour integral
I A
~c
r dl =
2e
Z
S
curl A dS
~c
2e
= ~c
2e
2n
(115)
Here is the magnetic ux through the contour. The phase change along the closed contour
is = 2n where n is an integer because the order parameter is a single valued function.
Since j = 0 in the bulk, we obtain = 0 n where
0 =
~c
2:07 10 7 Oe cm2
e
(116)
is the quantum of magnetic ux. In SI units, 0 = ~=e.
Problems
Problem 1
Estimate, in terms of the microscopic parameters, the hight of the energy barrier one
needs to overcome to inate a vortex loop, Eq. (46), in a superconductor for a maximum
possible superuid velocity vs
~=m .
Compare it with the temperature Tc . What is the
probability of such a uctuation?
Problem 2
Find the jump of the specic heat at the superconducting transition.
Problem 3
Find the behavior of the order parameter near a contact with the normal region. Magnetic eld and currents are absent.
Problem 4
Find the surface energy of the boundary with the normal region. Magnetic eld and
currents are absent.
43
IV. VORTICES IN TYPE II SUPERCONDUCTORS
Vortices are the objects which play a very special role in superconductors and superuids.
In superconductors, each vortex carries exactly one magnetic-ux quantum. Being magnetically active, vortices determine the magnetic properties of superconductors. In addition,
they are mobile if the material is homogeneous and there are no defects which can attract
vortices and \pin" them somewhere in the superconductor. In fact, a superconductor in the
vortex state is no longer superconducting in a usual sense. Indeed, there is no complete
Meissner eect: some magnetic eld penetrates into the superconductor via vortices. In
addition, regions with the normal phase appear: since the order parameter turns to zero at
the vortex axis and is suppressed around each vortex axis within a vortex core with a radius
of the order of the coherence length, there are regions with a nite low-energy density of
states. Moreover, mobile vortices come into motion in the presence of an average (transport)
current. We shall see that there is a nite resistivity (the so-called ux ow resistivity): a
superconductor is no longer \superconducting"! This is certainly an important eect.
In superuids, vortices appear in a container with helium rotating at an angular velocity above a critical value which is practically not high and can easily be reached in
experiment7 . Vortices are also created if a superuid ows in a tube with a suÆciently high
velocity. The driving force that pushes vortices is now the Magnus force. Vortices move
and experience reaction from the normal component; this couples the superuid and normal
components and produces a \mutual friction" between them. As a result, the superow
is no longer persistent. Remember the denition of superuidity: a liquid is superuid if
it can ow without friction through a narrow tube. One needs to restrict the uid to a
narrow tube because it is a narrow tube that inhibits vortex motion thus switching o the
dissipation. As we can see, vortex dynamics is the very heart of superuidity. This was
realized by Feynman in the early days of superuidity19 .
44
A. Transition into superconducting state in a magnetic eld
Vortices play an important role also in bringing about the superuid phase transition
itself. Let us put a sample of a type-II superconductor at a temperature below Tc into a
high magnetic eld and start to decrease the applied eld H . At some eld magnitude, the
sample will become superconducting. We shall see that this transition is of the second order.
Close to the transition point thus GL . Let the magnetic eld be along the z axis (see
Fig. 11).
H
z
y
x
FIG. 11.
We can linearize the GL equations in a small :
2ie 2
2
r
A +=0
~c
(117)
The vector potential can be taken in the Landau gauge A = (0; Hx; 0). The order parameter
depends on x and y . Now we have
@2
@
+
2
@x
@y
2ieHx 2
+ 2 = 0
~c
(118)
This is the well known Schrodinger equation for a charge in a magnetic eld. We put
= eiky f (x)
and obtain the oscillator equation
@2f
k
@x2
2eHx 2
f + 2f = 0
~c
with
45
(119)
2eH
~2
!0 =
;E=
mc
2m 2
The energy spectrum E = ~!0 (n + 1=2) is
2eH
1
=
n+
2
2m
mc
2
~
The highest H = Hc2 is for n = 0:
Hc2 =
~c
= 02
2
2e
2
/ 1 TT :
c
(120)
In SI units,
Hc2 =
0
20 2
It is the upper critical magnetic eld below which the transition into superconducting state
occurs. Comparing it with Hc we observe that
p
Hc2 = 2Hc =
p
82GL
p
For type II superconductors with > 1= 2, the upper critical eld Hc2 > Hc: transition
occurs to the state which, as we will see soon, has persistent currents in the superconducting
bulk.
The solution for the lowest energy level is a Gaussian function
"
#
~ck 2
1
x
f = C exp
2 2
2eHc2
(121)
The solution Eq. (121) is centered at x = ~ck=2eH . Actually, the full solution is a linear
combination of these solutions for dierent k. One can construct a periodic solution in a
form
=
X
n
"
Cn e
iqny
exp
1
x
2 2
#
~cqn 2
2eHc2
(122)
It is periodic in y with the period 2=q . It would be periodic in x as well if the coeÆcients
Cn satisfy periodicity condition Cn+p = Cn . Then,
46
p~cq
; y = eipqy (x; y )
x+
2eHc2
One sees that jj is periodic with the period
~cpq
X0 =
2eHc2
The simplest case is realized when all the coeÆcients C are equal, p = 1. The array
forms a rectangular lattice. The periods are
~cq
2
X0 =
; Y0 = :
2eHc2
q
The unit cell area is
X0 Y0 = 0 =Hc2 = 2 2;
which corresponds to exactly one ux quantum per unit cell. If q is chosen in such a way
that X0 = Y0 , we obtain a square lattice.
However, the period 2=q is not the least possible period in y for p 6= 1. The analysis
shows that if p = 2 and
C0 = iC1
the period in y is
Y0 =
q
The unit cell area is
X0 Y0 = 0 =Hc2 = 2 2
p
which again corresponds to exactly one ux quantum per unit cell. If X0 =Y0 = 2= 3 one
obtains a hexagonal lattice (see Fig. 12).
47
0.8
0.9
0.95
0.1
0.6 0.2
0.5
0.4 0.7
0.3 0.9
FIG. 12. Left panel: Lines of constant jj in a square lattice according to Ref.20 . Right panel:
The same for a hexagonal lattice according to Ref.21 .
The jj-pattern has zeroes at the points x = X0 =2 + X0 n, y = Y0 =2 + Y0 m, surrounded
by current lines. Indeed, the current is
~c2
@
@ i
i
jx =
162L e2GL
@x
@x
2
~c
@ 2eHc2 x
i +
jy =
162L e2GL
@y
~c
@ i
@y
2eHc2 x
~c
To transform this further we use the identity which holds for the function of the type of Eq.
(122):
@
=
@x
@
i
@y
2eHc2 x
~c
(123)
With help of Eq. (123) we get
@ jj2
(124)
162L e2GL @y
~c2
@ jj2
jy =
(125)
162L e2GL @x
These expressions suggest that j(x; y )j2 is a stream function, i.e., that the current ows
jx =
~c2
along the lines of constant jj. If we place the node of jj in the middle of the Bravais
unit cell, then the current along the boundary of a unit cell is zero: due to periodicity, the
lines of constant jj are perpendicular to the boundary (see Fig. 13). We now calculate the
48
contour integral of A (~c=2e)r = 0 and obtain = 2 since the ux through the unit
cell is equal to one ux quantum. The phase of the order parameter acquires the increment
of 2 after encircling the point where the order parameter is zero.
FIG. 13. The Bravais unit cell for a square lattice. Arrows show the ow pattern of the
supercurrent; the current lines are perpendicular to the unit-cell boundary.
Here we come to a vortex: A quantized vortex is a linear (in three dimensions) object
which is characterized by a quantized circulation of the order parameter phase around this
line. We see that transition into a superconducting state in a magnetic eld below Hc2 gives
rise to formation of vortices. Vortices in superconductors were theoretically predicted by
Abrikosov20 in 1957.
Let us consider a magnetic eld slightly below Hc2 such that Hc2 H
H 2. The solution
c
of the GL equation is = 0 + 1 where 0 is the solution Eq. (122) of the linearized
equation and 1 is a small correction 1 . This correction is caused by (i) nonlinear term
in the GL equation, (ii) variations in
A due to the supercurrent Eqs.
(124,125), and (iii)
deviation of H from Hc2 .
Using the Maxwell equation
jx =
c @hz
c @hz
; jy =
4 @y
4 @x
we obtain from Eqs. (124,125)
Æhz =
~cj0 j2
42L e2GL
Therefore, the vector potential becomes A = A0 + A1 where A0 = (0; Hc2 x; 0),
49
(126)
A1 = (0; (H
Hc2 ) x; 0) + Æ A
and Æ A is due to the correction to the magnetic eld induced by the supercurrent such that
Æ h = curl Æ A. As a result, h1z = H
Hc2 + Æhz and hz = H + Æhz .
Since the non-disturbed function of Eq.(122) satises the linearized GL equation with
A = A0, we obtain for the correction 1
2ie
~c
r
2ie 2
r ~c A 1 + 21
2ie
2ie
2ie
A (A10 ) ~c A1 r ~c A 0
~c
2 j0 j2 0 =2GL = 0
Now we multiply this equation by 0 from Eq. (122) and integrate it over dx dy . After
integration by parts using Eqs. (123), (126) we obtain
2e
~c
Z
j0 j2(H
Hc2 + Æh) dx dy + GL2 2
Z
j0j4 dx dy = 0
(127)
We introduce the average
hjj2i = S 1
Z
jj2 dx dy
Using Eq. (126) we obtain
2GL hj0 j2 i 1
H = 1
Hc2
1 hj0j4i
22
(128)
It is convenient to introduce the ratio
A =
hjj4i 1
hjj2i2
(129)
The value A is determined by the structure of the vortex lattice. Now we get
22 2GL
H
2
hj0j i = (22 1) 1 H
(130)
A
c2
p
Eq. (130) shows that jj2 has a small magnitude proportional to 1 H=Hc2 if > 1= 2.
p
For < 1= 2 the order parameter for H close to Hc2 is no longer small: it jumps to a
nite value making the transition a rst-order transition.
50
p
Type II superconductors have > 1= 2 and experience a second-order transition into
superconducting state below the upper critical eld with formation of quantized vortices.
Using Eq. (126) we can calculate the magnetization of the superconductor. The magnetic
induction is B = hhz i = H + hÆhz i. Then
Mz =
B
4
H
=
hÆh i =
z
4
Hc2 H
4A (22 1)
(131)
The applied eld can be expressed through the magnetic induction
H = Hc2
A (22 1)(Hc2 B )
1 + A (22 1)
Using Eqs. (129), (130), and (126) one can calculate the free energy density Eq. (106).
As a function of the magnetic induction it becomes
F=
B2
8
(Hc2 B )2
8 [1 + (22 1)A]
(132)
For a given B , the free energy decreases with decreasing A . For a square lattice A =
1:18 while for a hexagonal lattice A = 1:16. Actually, a square lattice is an unstable
conguration: it corresponds to an extremum rather than to a minimum of the free energy.
The stable conguration is a hexagonal lattice (Fig. 12).
B. Single vortex
The previous case corresponds to the situation where vortices are closely packed together:
the distance between them is of the order of the coherence length. Let us now consider an
example of an isolated single-quantum vortex when the distance to the neighbor vortex is
much larger than . Since each vortex unit cell with an area S carries one magnetic ux
p
H 2=B .
vortices corresponds to B H 2 which can be realized if 1.
quantum, 0 = SB the intervortex distance is r0 =
c
The limit of isolated
c
An isolated vortex is axially symmetric. It has a phase which changes by 2 after rotation
around its axis which we choose as the z axis. We take = ' where ' is the azimuthal angle
51
in the cylindrical frame (r; '; z ). We thus assumed a cylindrical symmetry of the vortex
and can look for a solution in the form
= GL f (r)ei'
The vector potential has only a '-component:
2
A = (0; A
'
4e2 Q2
f +f
~2 c2
@2 1 @
+
@r2 r @r
; 0). We have for f
f3 = 0
The gauge invariant vector potential in our case is Q = (0; A'
(133)
~c=2er; 0).
Equation (101) becomes
2L curl curl A + f 2 Q = 0
(134)
2L r2 Q f 2 Q = 0
(135)
For r 6= 0 it is
or
@ 2 Q 1 @Q
+
@r2 r @r
Q
r2
f 2Q
=0
2L
(136)
This equation can be solved in the limit 1. Here L and one can put f = 1 for
r . Eq. (136) gives
Q=
~c
K (r=L )
2e 1
Here K1 (z ) is the Bessel function of rst order of an imaginary argument. For z
1 the
function K1 (z ) = 1=z , and it decreases exponentially for large z :
2 1
2
K1 (z ) =
exp( z )
z
The constant at K1 is chosen in such a way that A = Q + (~c=2er) does not diverge for
r . The magnetic eld is
hz = curlz Q =
1 @ (rQ)
~c
=
K (r=L)
r @r
2e2L 0
52
1 it is K0 (z) =
Here K0 (z ) is the Bessel function of zero order. For z
ln z , and it
decreases exponentially for large z in the same way as K1 . The magnetic eld produced by
a single vortex decays exponentially for r and it is
hz 0 2 ln L
2L r
The logarithm is cut o at r for r .
For r , Q = ~c=2er and the order parameter satises the equation
2
@
1 @
1
2
+
f + f f3 = 0
(137)
2
2
@r r @r r
This equation coincides exactly with the vortex GP equation (59) for n = 1. Its solution
saturates at the equilibrium value f = 1 for r , and decreases as f
/ r for r ! 0.
The
behavior at large distances can be found from Eq. (133). Putting f = 1 Æf we nd
2e2 2
2
Æf = 2 2 Q2 = 2 K12 (r=L )
~c
2L
For r we have Æf = 2 =2r2 .
A vortex has a core, i.e., the region near its axis with the size of the order of where
the order parameter is decreased from its value in the bulk; jj vanishes at the vortex axis.
The vortex core is surrounded by persistent currents which decay away from the vortex core
at distances of the order of L (Fig. 14).
H
∆
ξ
λL
r
FIG. 14. Structure of a single vortex. The core region with the radius is surrounded by
currents. Together with the magnetic eld, they decay at distances of the order of .
L
C. Vortex free energy
Let us calculate the free energy of a single vortex, i.e., the dierence between the energy
Eq. (70) in the presence of a vortex and the energy without a vortex and the magnetic eld.
53
We have
Z j
j4 jGL j4
2
#
2
2
2
e
h
+ i~r
A + 8 dV
c
The kinetic energy term contains (4e2 =c2 )Q2 jj2 . Since Q / 1=r for F
L
=
jj2
jGL j2
+
gives a logarithmically large contribution at distances r
L
(138)
r
L
this
. Simple arguments suggest
that we can thus put jj = GL everywhere at large distances from the core. The rst
two terms then vanish, together with the gradients of . The magnetic eld gives also a
non-logarithmic contribution. As a result, we obtain per unit length of the vortex
Z
4e2 2
2
2
2
FL = c2 Q jj d r = ~ GL ln L
2
L
~ mNs
20
ln
=
ln L
=
2
2
16 L
m
4
(139)
The possibility to put jj = GL at large distances in Eq. (138) is, however, not so obvious
because the function f approaches unity not fast enough, only proportionally to r 2 . This
results in a logarithmically divergence of the terms jj2 2GL and jj4
One can check however, that these two terms cancel each other because
4GL in Eq. (138).
the extra factor 12
in front of the second term is compensated by a two times larger power of as compared
to the rst term.
For an n-quantum vortex we will obtain
n2 20
(140)
FL = 1622 ln L :
L
The energy is proportional to n2 . Therefore, vortices with n > 1 are not favorable: The
energy of n single-quantum vortices is proportional to the rst power of n and is thus smaller
than the energy of one n-quantum vortex.
Equation (139) allows to nd the lower critical magnetic eld, i.e., the eld H above
which the rst vortex appears. The free energy of a unit volume of a superconductor with
a set of single-quantum vortices is FL = nL FL = (B=0 )FL. The proper thermodynamic
potential in an external eld H is the Gibbs free energy G = F
54
HB=4
B FL
G=
0
BH
B 0
=
ln L
2
2
4
16 L
BH
4
It becomes negative for H > Hc1 where
Hc1 = 0 2 ln L :
4L
(141)
Therefore, vortices appear for H > Hc1 .
We note that
Hc1 = Hc2
ln ln = Hc p
2
2
2
(142)
i.e., for superconductors with a large , the critical eld Hc1 is considerably lower than Hc2.
The phase diagram of a type II superconductor is shown in Fig. 15.
Normal
H
Vortex
Hc2
Hc
H c1
Meissner
T
Tc
FIG. 15. Phase diagram of a type II superconductor
For more reading on vortices in type II superconductors see Refs.5;6 .
55
Problems
Problem 1
Derive Eq. (132).
Problem 2
The lm with thickness d has the same upper critical eld as a bulk superconductor.
Find the upper critical eld for a lm with a thickness d
tilted by an angle from the normal to the lm, Fig. 16
z Θ
H
y
x
FIG. 16.
56
placed in a magnetic eld
V. LONDON MODEL
The GL equation for the vector potential Eq. (100)
4Nse2
curl curl A +
A
mc2
~c
2e
r
=0
(143)
can be extended to a more general case if currents are small and jj is constant. The
temperature needs not to be close to Tc . This approximation is good for distances r .
The latter condition implies that we are restricted to the case of 1.
We assume that the current has a form
j = N ev
s
whence
v
s
s
= j=Nse and write the free energy as a sum of the kinetic energy of supercon-
ducting electrons and the magnetic energy
F=
Z Ns mvs2 h2
+
dV
2
8
Using the Maxwell equation j = (c=4 )curl h, we transform this to the following form [F.
London and H. London, 1935]
F
1
=
8
Z
d3 r h2 + 2L (curl h)2
(144)
Here L is
1
mc2 2
L =
4Nse2
(145)
is the London penetration depth. Variation with respect to h gives the London equation:
h + 2 curl curl h = 0
L
this is just a curl of Eq. (143).
57
(146)
A. Single vortex
Equation (146) holds if there are no vortices. However, the London equation should be
modied if there is a vortex. For a vortex we have
curl r = 2 ^zÆ (2) (r)
where ^z is the unit vector in the direction of the vortex axis. Therefore, the London equation
for a vortex becomes
h + 2 curl curl h = 0Æ(2) (r)
(147)
L
where 0 is the vector along the vortex axis with the magnitude of one ux quantum.
For a system of vortices
h + 2 curl curl h = 0
X
L
Æ (2) (r
r)
n
(148)
n
where the sum is over all the vortex positions rn.
If h = (0; 0; hz (r)), Eq. (147) gives for 6= 0:
@ 2 h 1 @h
2L
+
@r2 r @r
h=0
Which gives h = aK0 (r=L ). The constant can be found from Eq. (147) by integrating it
over d2 r over a small area with r L :
2
L
Z
curl h d l = 0
so that we obtain a = 0 =22L. Thus
r
h = 0 2 K0
2L
L
It is
h(0) = 0 2 ln L
2L
r
near the vortex axis r L .
58
The free energy of the vortex per unit length is
F
Z
1 =
h
h
+ 2L curl curl h + div[h curl h] d2 r
8 Z
Z
1
1
(2)
2
hz 0 Æ (r) d r +
[h curl h] dl
=
8
8
(149)
The last integral is taken along a remote contour and vanishes. The rst integral gives
20
FL = 1622 ln L
(150)
which is the same as Eq. (140).
B. Bean{Livingston vortex barrier near the surface of a superconductor
Consider a straight vortex placed near the surface of a superconductor in an applied
magnetic eld H parallel to the surface. It experiences a force from the applied led that
pushes it into the superconductor since the magnetic eld wants to penetrate into the superconductor through this vortex. Another force originates from its interaction with the
surface. It is created by an image vortex and attracts our vortex to the surface in the same
way as it was for the vortex loop considered earlier in Sect. I.
Let the surface be the (y; z ) plane and the magnetic eld is along the z axis. Superconductor occupies the half-space x > 0.
The magnetic eld satises the London equation (147), i.e.,
h + 2curl curl h = ^z0 Æ(2) (r r
L
)
here r is two-dimensional coordinate in the (x; y ) plane, rL = (xL ; 0) is the coordinate of the
vortex, and z^ is the unit vector along the z axis. The boundary conditions at the surface
are
h = H; [curl h]
x
= (4=c)jx = 0
The magnetic eld has the form
59
h = h1 + h2
Here h1 =
H exp(
x=) is due to the decaying external eld. The eld h2 is due to the
vortex placed near the surface. It is composed by the vortex eld and the eld created by
the image vortex.
h2 = h
v
+ him
The image vortex is the vortex at the point ( xL ; 0) having the circulation along the z^
direction. Its currents ow in the opposite way as compared to the real vortex. As a result,
these two vortices produce zero current and
h2 = 0
at the surface (in the middle between the two vortices) thus satisfying both the boundary
conditions jx = 0 and h = h1 = H at the surface together with the London equation in the
half-space x > 0.
The Gibbs free energy has the form (see Problem 1)
G = 40 He
xL =
1
+ h2 (xL ) H
2
(151)
It vanishes when vortex approaches the surface xL = 0 since h2 (0) = 0. For xL
, the
free energy can also be written as (see Problem 1)
G = 40 He
0
=
He
4
1
+ 0 2 ln
h (2x ) H
4 2 v L
1
x=
h (2x ) + Hc1 H
2 v L
x=
It is shown in Fig. 17 as a function of the vortex position.
60
(152)
G
(a)
0
G
(b) 0
H c1 - H
λ
x
λ
x
H - H c1
G
(c) 0
x
H - H c1
FIG. 17. The free energy of the vortex as a function of its distance from the surface. (a)
H < H 1. The energy far from the surface is larger. (b) H > H > H 1 . The energy far from the
surface is smaller but there is an energy barrier. (c) H > H . The barrier disappears.
c
s
c
s
Using
2xL
hv (2xL ) = 0 2 K0
2
we can calculate the force on the vortex
@G
1 @hv (2xL )
= 0 He x= +
Fx =
@xL 4
2 @xL
0
0
2xL
x=
=
He
K
4
22 1 At large distances x
the function K1 /
exp( 2xL =) decays faster than the rst
exponent. The force due to the external eld dominates: it tends to push the vortex into
the sample. However, the free energy of the vortex in the bulk is lower than the energy at the
surface (i.e., zero) only for H > Hc1 [see Eq. (152)]. This is because the vortex experiences
an attraction to the surface at short distances. Indeed, at small distances x
,
the
function K1 (2xL =) = =2xL and the force
Fx = 0 H
4
0
4xL
61
(153)
is dominated by the image vortex: it attracts the vortex to the surface. Therefore, the
vortex energy has a barrier (see Fig. 17).
However, Eq.(153) is only valid when x . The force from the image vortex saturates
when x . Therefore, if the applied eld H > Hs where
Hs =
0
4
H
c
the force is always positive, barrier vanishes and the vortex can freely penetrate into the
superconductor.
C. System of straight vortices
If the vortices make a regular lattice, one can solve Eq.(148) by means of the Fourier
transformation. We introduce
hz =
Z
eiqr hq
d2 q
(2 )2
Here rn form a regular lattice with the unit cell vectors a and b so that every rn = ka + mb
where k and m are integer numbers, see Fig. 18. In a simple case of a rectangular lattice,
r
n
= (ka; mb). Equation (148) gives
hq =
0 X iqrn
e
1 + 2L q 2 n
(154)
The free energy becomes
F
1
=
8
Z
1 + 2 q 2 hq h
L
Z
X
d2 q
20
1
=
eiq(rn
q
(2 )2 8 (1 + 2L q 2 ) n;m
rm )
d2 q
(2 )2
(155)
We note that
X
eiqrn = (2 )2 nL Æ (2) (q
Q)
(156)
n
where Q is any vector of the reciprocal lattice. They are dened (see Fig. 18) in such a way
that Q = Q1 K + Q2 M where K and M are integer and
62
Q1 = ([2a[bb] z^^z] ) ; Q2 =
2 [a ^z]
([a b] z^)
(157)
It can be seen that a dot product of any vector of the real lattice and any vector of the
r Q = 2N where N is an integer. The area of the unit cell of the
reciprocal lattice A0 = [Q1 Q2 ]^z is connected to the area of the unit cell of the real lattice
S0 = [a b]^z by S0 A0 = (2 )2 .
reciprocal lattice is
n
Q1
b
a
Q2
FIG. 18. The vectors of the direct (full lines) and reciprocal (dashed lines) lattices.
Using Eq.(156) we can write
XZ
n
eiqrn f (q)
X
d2 q
=
nL f (Q)
(2 )2
Q
Therefore, the free energy becomes
Z
X
1
SnL 20
eiqRn
F = 8
(1 + 2L q 2 ) n m
m
d2 q
Sn2L 20 X
1
=
2
(2 )
8 Q 1 + 2L Q2
where S is the area of the sample. The energy per unit volume becomes
!
X
B2 X
1
B2
1
F=
=
1+
8 Q 1 + 2L Q2 8
1 + 2L Q2
Q6=0
(158)
(159)
(160)
To simplify calculation of the sum, we replace it with the integral according to the rule
Z
2
X
) S0 (2d Q)2
Q
where S0 = 0 =B is the area of the unit cell of the real lattice.
Now we have
63
B2
F=
1 + S0
8
Z
Qmax
0
Q>Q
d2 Q
1
B2
B 0
2
=
+
ln
(2 )2 1 + 2L Q2
8 32 2 2L
1 + 2L Q20
(161)
Here Q0 is the rst reciprocal-lattice vector in the approximation of a round cell: S0 Q20 =
4 2 , i.e., Q20 = 4B=0 . We put Qmax = 1= since the maximum wave vectors are determined
by the shortest distance which is in the London model.
The magnetic induction B = 0 =S0 is determined by the minimum of the Gibbs free
energy F
HB=4 which gives
0
2
H =B+
ln
82L
1 + 2L Q20
Magnetization 4M = B
B
2(1 + 2L Q20 )
(162)
H becomes a logarithmic function
2 0
H
0
M=
ln 2 2 =
ln c2
2
2
2
2
32 L
L Q0
32 L B
(163)
for L Q0 1. On the other hand,
@B
@H
for B
! 0 as it follows from Eq.
!1
(164)
(162). The magnetization of a type II superconductor is
− 4πM
shown in Fig. 19.
H c1
Hc
H
H c2
FIG. 19. Full line: Magnetization of a type II superconductor. The linear part at low elds
corresponds to the full Meissner eect Eq. (114). The vertical slope at H = H 1 is determined
by Eq. (164). The linear part near H 2 follows from Eq. (131). The magnetization for elds
H 1 < H < H 2 is logarithmic as a function of H , Eq. (163). Dashed line: Magnetization of a type
I superconductor. The Meissner eect persists up to the thermodynamic critical eld H .
c
c
c
c
c
64
D. Uncharged superuids
There is only one known uncharged Fermi superuid, namely, 3 He. Its order parameter
has a p-wave pairing symmetry. Therefore, the above GL model does not directly apply.
However, it is possible to construct the corresponding Ginzburg{Landau energy functional
and derive the GL equations which are quite similar to those discussed above. We refer the
reader to Ref.22 for more detail on the GL theory for superuid 3 He.
In this section we discuss only some selected topics. We note rst that the mean-eld GL
expansion for 3 He has a wide range of applicability in the same way as for superconductors,
see Section III C. Here again it relies on the existence of a large parameter pF 0 =~ 102 103
of the same order of magnitude as in superconductors. According to the Ginzburg criterion
Eq. (109) the uctuation region near Tc is thus very narrow; it is practically out of range of
the present-day experiment.
Let us consider some aspects relevant to rotating superuids. In an uncharged superuid,
the London screening length L =
1.
Vortices in uncharged superuids are created by
rotating the container with a superuid. According to the general thermodynamic principles,
for a rotating uid one has to minimize the combination F
L where L = ÆF =Æ
is the
angular momentum of the superuid. One nds
F L
=F
ÆF
Æ vs Z
r) m2 [
r]2 jj2 dV
[
r] = F (vs
The last term should be combined with the contribution from the normal component after
which it becomes a constant solid-body rotation energy.
The rst term suggests that, in a Bose system where superuidity is carried out by
individual atoms, the momentum operator in the corresponding order parameter equation,
like in Eq. (48), should be replaced with
i~r
m
r
65
For Fermi superuids, where pairs of particles are superuid, the comparison of the rst term
to the GL free energy Eq. (70), (107) with the help of the expression for the superconducting
velocity Eq. (96) yields that the vector potential is equivalent to the rotation velocity with
the substitution rule
2e
A ! 2m r
c
A magnetic eld B (which is uniform in space since = L =
1) is thus equivalent to a
rotation velocity = eB=2mc, which agrees with the Larmor theorem.
Moreover, the denition of superconducting velocity vs used for superconductors is equivalent to the superuid velocity measured in the rotating frame
v $ 2~m r r
s
The results obtained for vortices in superconductors can be, for some cases, generalized
to uncharged superuids (more specically, to superuid 3 He) if one takes the limit e ! 0
which implies L
! 1.
Thus, an uncharged superuid to some extent is similar to an
extreme type II superconductor with a very large . With e ! 1, the electric current, of
course, disappears but the particle current j=e remains nite.
For example, the upper critical eld is still meaningful: it denes the rotation velocity
c2 =
~
4m 2
for which the vortex cores overlap.
The last expression in Eq. (139) contains the circulation quantum of superuid velocity
around a vortex 0 = ( ~=m) and the superuid mass density mNs . Note that this expression for 0 is two times smaller than the expression Eq. (58) for a Bose gas. The reason
is that one should insert the mass 2m of a Cooper pair for the mass m in Eq. (58); with
this in mind the two expressions for 0 are equivalent. Eq. (139) is suitable for superuids
with vortices. For an uncharged superuid, L ! 1, therefore, the logarithm is cut o at
intervortex distances
66
2 s
r
FL = 04mN
ln 0
(165)
The lower critical eld, Hc1 , however, corresponds to vanishing rotation velocity: it
formally requires the limit L ! 1. However, in a nite-size container, the container radius
R has to be taken instead of L . The lower critical eld Eq. (141) corresponds now to
the critical rotation velocity for which a rst vortex becomes favorable in the center of the
container. It takes the form7 (see Problem 6 to Sect. I)
c1 =
~
2mR2
R
ln
where R is the container radius.
Problems
Problem 1
The vortex line is placed parallel to the plane surface of the superconductor. Calculate
its energy in an applied magnetic eld parallel to the surface (and to the vortex).
67
VI. TIME-DEPENDENT GINZBURG{LANDAU THEORY
The simplest theory which describes dynamics of superconductors is the so called timedependent Ginzburg-Landau theory (TDGL). It generalizes the usual GL theory to include
relaxation processes for nonequilibrium superconductivity. Unlike for its static counterpart,
the validity of the TDGL theory in a strict sense is much more limited. It is not enough
just to have temperatures close to the critical temperature. One also needs that relaxation
processes are fast. For real superconductors it is, in general, a very strong limitation which
can be fullled only for the so called gapless superconductivity. The latter corresponds to
a situation where mechanisms working to destroy Cooper pairs are almost successful: the
energy gap in the spectrum disappears, but the order parameter and supercurrent still exist.
These mechanisms are: interaction with magnetic impurities which act dierently on the
electrons with opposite spins in a Cooper pair, inelastic (not conserving energy) interactions
with phonons, etc. For example, inelastic scattering by phonons is characterized by mean
free time . The condition for applicability of the TDGL theory is ~. Due to a
comparatively large magnitude of (for example, 10 9 s for Al), this condition is
only satised in a very narrow vicinity of Tc . We will not go into microscopic details here.
A justication of using TDGL is that it is simple, gives a generally reasonable picture of
superconducting dynamics, and in some cases it can even be derived from the microscopic
theory23 .
Time-dependent Ginzburg-landau theory is constructed in the following way24;25 . In
equilibrium, the full energy of a superconductor Eq. (71) has a minimum with respect to and A:
Æ Fsn
ÆF
= 0 ; sn + c 1 j = 0
Æ
ÆA
(166)
where the superconducting energy is determined by Eq. (70)
If the system is out of equilibrium, the order parameter relaxes to its equilibrium value.
The rate of relaxation depends on the deviation from equilibrium:
68
@ Æ Fsn
= @t
Æ
where
(167)
is a positive constant. This equation is, however, not gauge invariant. Remember
that according to the gauge invariance, the electromagnetic potentials can be changed as
A ! A + ~2ec rf ; ! ~ @f
(168)
2e @t
without changing the electromagnetic elds
1 @A
c @t
E=
r; h = curl A
(169)
In superconductors, the gauge transformation Eq. (168) is accompanied by the transformation of the order-parameter phase
!+f
(170)
To preserve the gauge invariant property of equation (167) we have to add there the scalar
potential in the form
@ 2ie
ÆF
+
= sn
@t
~
Æ
(171)
The second equilibrium condition Eq. (166) is generalized in the following way. Instead
of total current in Eq. (166) we put there the supercurrent js = j
j
n
which accounts for the
fact that a part of current can be produced by normal electrons in presence of nonequilibrium
electric eld:
1
(j
c
j ) = ÆÆFA
sn
n
(172)
where the normal current is jn = n E. The electric eld is
E=
=
1 @A
c @t
1 @Q
c @t
r
r
(173)
We introduce here the scalar gauge invariant potential in addition to the vector potential
Q dened by Eq. (92) so that the full set of gauge invariant potentials becomes
69
Q=A
~c
r ;
2e
=+
~ @
2e @t
(174)
The second line in Eq. (173) was obtained by a gauge transformation using the order
parameter phase as a generating function. This can be done if there are no moving vortices
in the superconductor: Otherwise, the time and spatial derivatives of the phase do not
commute. We shall discuss this eect later.
As we know, the vector potential Q is related to the superconducting velocity
Q=
mc
v
e s
(175)
The physical meaning of the scalar potential can be seen from Eq. (173). According to
it,
@ vs e
= (E + r)
@t m
(176)
Since the electric eld already includes the gradient of the scalar potential , the quantity
(e=m) =
~ @
2m @t
(e=m)
is the dierence between the chemical potential per unit particle mass
s =
~ @
2m @t
(177)
of the superconducting particles (i.e., of each of two electrons in a Cooper pair) and the
chemical potential
n = (e=m)
of the normal particles. Note that these chemical potentials are both measured from a
constant chemical potential of the normal state without the electric eld, i.e., from the
Fermi energy EF per particle mass.
Note that Eq. (176) coincides with Eq. (7) used in the two-uid description of superuids
if we put
A = 0 for an uncharged superuid and neglect small terms of the second order
70
in vs . The term vs2 is indeed small in superconductors: Its maximum possible magnitude
2 2 =p2 while (~=m)@=@t ~jj=m . We shall see that
is of the order ~2 =m2 2 vmax
F
~jj= 2 =Tc . Therefore, the ratio of these two terms is Tc =EF 1.
Finally, the TDGL equations take the form
@ 2ie
+
=
@t
~
jj + jj2
~2
r
2ie 2
A ;
~c
(178)
and
j=j
n
+ js
(179)
where, as before, the supercurrent is (compare with Eq. (80))
4e2 i~c
js = c A + 2e r + A
i~c
8e2 2
r
=
jj Q
2e
c
(180)
Note that it is the total current j that enters the Maxwell equation (73).
A. Microscopic values
Microscopic theory justies the TDGL approach only for dirty gapless superconductors.
Therefore, to compare our results with microscopic calculations we use the value for the GL
coeÆcient as for a dirty superconductor. We remind that
= D=8~Tc
(181)
where D = vF2 =3 is the diusion coeÆcient. Therefore, the relaxation time for current is
n
7 (3)~
T 1
j = 2
=
1
(182)
8e jj 4 3 Tc
Tc
For strong inelastic scattering, the microscopic theory24;25;23 gives
= ~=8Tc
71
(183)
B. Discussion of TDGL equations
Equation (178) looks like a GP equation with one important dierence: The coeÆcient
in front of the time derivative of the wave function is real in contrast to Eq. (48). This
is a crucial distinction: Eq. (178) describes a relaxation dynamics of superconductors. For
example, it does not poses a Galilean invariance. The reason is that the heat bath is assumed
to be in equilibrium, i.e., at rest in the laboratory frame. It makes a preferable reference
frame violating the Galilean invariance.
Equations (178) and (173, 179, 180) dene two characteristic relaxation times. Separating real and imaginary parts of Eq. (178) we obtain for the real part
@ jj
= jj + jj3
@t
~2
r2 jj
4e2 2
Q jj
~2 c2
or
@ jj
= jj
@t
jj3 + 2 r2
20
4e2 2
Q
~2 c2
jj
where we introduce the order parameter relaxation time
T 1
=
jj / 1 Tc
(184)
(185)
The equation for the total current
j = 1c @@tQ + r
8e2 jj2 Q
c
determines the relaxation time for the vector potential (or current)
n
n
T 1
j = 2
=
/ 1 T
8e jj 8e2 2GL
c
(186)
(187)
One can dene the ratio of these two relaxation times
u = =j
which is independent of T .
For the microscopic value of , the order-parameter relaxation time is
72
(188)
~
=
1
8Tc
T
Tc
1
Ratio of the two times is u = 4 =14 (3) 5:79.
The imaginary part of Eq. (178) gives
Qjj2 = 0
c jj2 + 2r
(189)
Eq. (189) can also be obtained by calculating the variation of the free energy with respect
to the order parameter phase
ÆF
2 sn
Æ
i ÆÆFsn
"
ÆF
sn
Æ
#
2ie 2 2ie 2
= i~ r + A r
A ~c
~c
2ie
2ie
2
A r + ~c A = i~ div r
~c
=
~
2e
div js
(190)
With help of Eq. (171) we obtain from Eq. (190)
Æ Fsn
=
Æ
jj2 @ + 2e =
@t
~
~
4e
div js
(191)
Comparing Eqs. (180) and (191) we obtain Eq. (189).
C. Energy balance
The TDGL equations support the energy balance equation24 . Calculate the time derivative of the total free energy,
F
sn
plus the magnetic energy we obtain the energy balance
equation
@ Ftot
+
@t
Z
div jF =
Z
W dV
(192)
where the free energy current is
jF = ~2
@ 2ie
+
@t
~
r + 2~iec A + c:c: + 4c [E h]
73
(193)
and the dissipation function is
W =2
@
@t
+
2ie 2
~
+ n E2
(194)
It is a positively denite function consisting of two contributions. First comes from the energy loss due to relaxation of the order parameter while the second contribution is associated
with the Joule heating by the normal currents.
D. Charge neutrality and a dc electric eld
In metals, all bulk charges are screened. It is seen from the Poisson equation
div E =
r2 = 4e(N
e
Ni )
where Ne and Ni are densities of electrons and ions, respectively. In equilibrium, Ne = Ni .
Introduction of an additional electronic charge results in a shift of the chemical potential
Æ = eÆ such that
ÆN = Ne
Ni =
@Ne
Æ = 2eÆ
@
Here 2 is the density of states, i.e., the number of states within the unit energy interval.
The factor 2 accounts for two spin projections of an electron. The Poisson equation becomes
r2Æ = 8e2 Æ
It demonstrates that the potential decays at distances of the order of the Debye screening
length
D = 8e2 1=2
which is of the order of interatomic distance in good metals. This result means that variations
in the electronic charge density are practically zero at distances of the order of the coherence
length, and we have to put the constraint
74
ÆNe = 0
or
div j = 0
(195)
which follows from the continuity equation
@ (eNe )
+ div j = 0
@t
Eq. (191) now gives
2e jj2
~
~
=
4e
div jn =
~n
4e
r2 + 1 r @ Q
c
@t
(196)
or
l2
E
1 @Q
j2
2
r + c r @t = j
20 (197)
This equation denes the new characteristic length
lE =
~n
8e2 20
1
2
= j
1
2
(198)
It is the electric-eld penetration length, i.e., the distance over which an d.c. electric eld
decays into a superconductor.
In the bulk of a superconductor, in case of equilibrium, we always should have
= +
~ @
2e @t
=0
(199)
which is the famous Josephson relation. It shows that the superconducting chemical s
dened by Eq. (177) is equal to the normal chemical potential n = (e=m) in equilibrium.
Remember that we have used this condition in the two-uid description of superuidity.
75
N
S
In+Is
In
In
Is
I tot
Is
In
0
x
FIG. 20. A boundary between normal metal and superconductor. Normal current converges
into supercurrent at distances l while the total current is constant.
E
Consider an interface between a normal metal and a superconductor, and assume that
the superconductor occupies the space x > 0 (Fig. 20). If there is a current along the x axis
owing from the normal metal into the superconductor, the electric eld E = j=n exists
in the normal metal. However, deep in the superconductor, a d.c. electric eld can not
exist: It would produce continuous acceleration of superconducting electrons and destroy
superconductivity. Indeed, Eq. (197) describes relaxation of E . For a constant-in-time
electric eld and for = GL
r2 = l
E
2
whence
= ElE exp( x=lE )
(200)
The constant is chosen to satisfy continuity of current through the interface. According
to this equation, electric eld together with normal current decay into the superconductor,
while supercurrent increases until js takes all the current which ows into the superconductor. Normal current
jn = n E exp( x=lE )
76
converges into supercurrent over the distance lE .
E. An a.c. electric eld
This consideration refers to an electric eld constant in time, the so called longitudinal
eld div E 6= 0. An oscillating, transverse electric eld, div E 0, behaves dierently. An
example is a radio-frequency electromagnetic wave with a frequency ! incident on a surface
of a superconductor (see Fig. 21). In this case, the current in the superconductor is
j = 1c @@tQ + r
c
Q
42L
Eq. (197) can be satised with = 0. The electric eld decays into the bulk at distances
of the order of (see Problem 1 to Sect. VII)
1
1
1
= 2 + 2
2
ef f
L skin
where
c
8!
skin = (1 + i) p
is the skin depth in the normal state.
z
y
E
x
FIG. 21. A radio-frequency wave incident on the superconductor.
If ! 1 one has ef f = L , and the eld is expelled due to the Meissner eect. On the
contrary, if ! 1 one has ef f = skin. The penetration is determined the normal-state
skin eect.
77
F. Critical current
Consider a thin wire with a radius smaller than (see Fig. 22) and nd the maximum
current which can ow without dissipation. This problem is exactly the same as the one
considered earlier for the GP equation.
I
FIG. 22. Thin current-carrying superconducting wire
We assume the order parameter in the form
= 0 feikx=
The supercurrent takes the form
j = jGL f 2 k
(201)
where jGL is dened by Eq. (105). The order parameter equation (184) becomes
@f
j2 1
= f f 3 2 3
@t
jGL f
(202)
The time independent solution is
j
jGL
p
= f 2 1 f 2 = k 1 k2
(203)
In the last part of this equation, Eq. (201) is used. As a function of f , the current has a
p
maximum jmax =jGL = 2=(3 3) at f =
p
p
2=3 or k = 1= 3 which coincides with Eq. (63).
Let us study the stability of this solution. We put f = f0 + f1 where f0 satises Eq.
(203). We nd for the deviation f1 by linearizing Eq. (202)
@ (j=jGL )
@f
1 = 2 2 3f02 f1 = 2f1
@t
@k
(204)
where we express the function f0 in terms of k using Eq. (203). We can see that the time
independent solution Eq. (203) is stable for f0 >
solution is stable when
78
p
2=3 or 1 < k <
p
1=3. Equivalently, the
@j
>0
@k
(see Fig. 23). The solution in the range 0 < f0 <
p
2=3 or
p
1=3 < k < 1 is unstable.
With increasing current, the order parameter decreases from f = 1 until the minimum
value f =
p
2=3 is reached. With further increase in the current, the stability is lost, and
superconductivity is destroyed.
j/jGL
j/jGL
2/33/2
2/33/2
0
3−1/2
1
0
k
(2/3)1/2
1
f
FIG. 23. The supercurrent as a function of k (left panel) or f (right panel). The dashed part
of each curve is unstable.
Recall that the problem of the critical current was ambiguous for the GP theory because
it did not contain the proper reference frame. For the TDGL theory, such ambiguity does
not arise since the reference frame is xed to be the one where an equilibrium is established
in time independent conditions. The situation is quite dierent here because, in contrast
to the GP theory, superconductivity is created on the background of thermal equilibrium
at high temperatures with a very eÆcient relaxation of the order parameter. There is no
Galilean invariance because the major part of the uid consists of the normal component
xed to the laboratory frame. The Landau criterion cannot be used either since the normal
excitations do already exist in a stationary system.
Problems
Problem 1
Derive Eq. (192).
Problem 2
Find the resistance of the SN interface for low currents assuming lE .
79
VII. MOTION OF VORTICES
TDGL model can be employed to consider the vortex dynamics. Assume that there is
an average (transport) current passing through a superconductor with vortices. It exerts a
Lorentz force
F
i
0 h
^
=
j
h
L
c tr
(205)
on vortices and pushes them in a direction perpendicular to the current (here h^ is the unit
vector along the magnetic eld). The magnitude and direction of the vortex velocity is
determined by a balance of the Lorentz force and the forces Fenv acting on a moving vortex
from the environment, Fig. 24. In the absence of pinning these forces include friction
(longitudinal with respect to the vortex velocity) and gyroscopic (transverse) forces. As a
result, vortices move at an angle to the transport current. An electric eld perpendicular
to the vortex velocity is generated by the moving ux, and a voltage appears across the
superconductor with the electric eld components both parallel and perpendicular to the
current. The longitudinal component produces dissipation in a superconductor while the
transverse one is responsible for the Hall eect.
Fenv
E
F
F
j tr
vL
FL
FIG. 24. Forces exerted on a moving vortex.
80
We already mentioned in Section II that vortices in ideal superuids at T = 0 move
with the ow. We shall see now that, in the TDGL model that describes highly dissipative
systems, vortices move perpendicular to the ow and generate an electric led parallel to it,
as in a usual dissipative conductor.
A. A moving vortex and the electric eld
The time derivatives @ =@t; @ =@t and @ A=@ t; and the scalar potential can be easily
calculated if the vortex velocity vL is small. We note that, for a small vortex velocity,
(r) = st (r
v
L
t) + 1 ;
A(r) = A (r v
st
L
t) + A1 ;
(206)
where st and Ast are the values for a stationary vortex, and 1 and A1 are small corrections.
Therefore, within the main terms in the vortex velocity vL , we can write
@
@A
= (vL r)st ;
= (vL r)Ast :
@t
@t
(207)
The scalar potential ' is to be found from Eq. (189). For small vL it contains the order
parameter of the stationary vortex and the time derivative of the order parameter phase.
Let us calculate the average electric eld. The local electric eld is
E = 1c @@tA
r = 1c @@tQ
r
@
r
2e @t
~
r @
:
@t
(208)
The last term here is nonzero for a moving vortex. Indeed,
@
r
2e @t
~
r @
@t
=
~
2e
[(vL r)r
r(v r)]
L
~
[v curl r]
2e L
0
=
[v z^]sgn(e)Æ (2) (r):
c L
=
(209)
Here ^z is the unit vector along the vortex axis in the positive direction of its circulation. The
vortex circulation is determined by the sense of the phase increment, it coincides with the
magnetic eld direction for positive charge of carriers and is antiparallel to it for negative
carriers. Equation (208 can be written as
81
m @ vs
= E + r + 0 [vL ^z]sgn(e)Æ (2) (r)
e @t
c
(210)
This equation generalizes Eq. (176) for the case of moving vortices carrying a magnetic ux.
We now average Eq. (208) over the vortex array. Since
@Q
= (vL r)Qst
@t
the average of this term vanishes because the gauge-invariant vector potential
Q
st
is a
bounded oscillating function (periodic for a periodic vortex array). The average of
r
vanishes, too, since is also a bounded oscillating function periodic for a regular vortex
lattice. The only nonzero term in Eq. (208) is thus the one that follows from Eq. (209). It
gives
hEi
L
=
1
B vL
c
(211)
which is the Faraday's law. The average is dened as
h: : :iL = S0 1
Z
0
S
(: : :) dS
(212)
where S0 = 0 =B is the area occupied by each vortex.
B. Flux ow: Low vortex density
We consider here the limit H
H2
c
1.
In the limit of individual vortices for 1 and
we can neglect the magnetic eld and the vector potential compared to the
gradient of the order parameter.
Let us calculate the dissipative function W in Eq. (194) in a volume occupied by one
vortex. We have per unit vortex length
Z
W d2 r = 2
=2
2
Z @
2
ie
@t +
~
"
2
Z 2
@ jj
@t
d2 r +
#
Z
n E2 d2 r
4e
+ 2 jj2 2 d2 r +
~
82
Z
n E2 d2 r
(213)
Putting E =
r we rewrite the second integral in the form
Z
n E2 d2 r =
Z
= 2
n Er d2r =
Z
Z
n div E
4e2 2
jj d2r
~2
Here we omit the surface contribution and make use of Eq. (196). Inserting this into Eq.
(213) we obtain
Z
W d2 r = 2
=2
=2
Z "
Z
Z
#
@ jj 2 4e2 2
+ 2 jj ( ) d2 r
@t
~
"
#
2
@ jj
2e
@ 2
+ jj2 dr
@t
~
@t
2e 2
2
jj (vL r) d2r
[(vL r) jj]
~
(214)
The two terms under the integral in Eq. (214) represent two dierent dissipation mechanisms working during the vortex motion. The rst term is Tinkham's mechanism8 of
relaxation of the order parameter: Due to the motion of a vortex, the order parameter at
a given point varies in time which produces a relaxation accompanied by dissipation. The
second is the so-called Bardeen and Stephen26 contribution. It accounts for normal currents
owing through the vortex core.
The order parameter magnitude jj = GL f (r) for a static vortex satises Eq. (137).
The gradient of jj has only a component along the radius-vector in the cylindrical frame
(r; '; z ):
v rjj = GL v
L
Lr
df
dr
The gauge invariant potential obeys Eq. (197) whence
r2 l 2f 2 = 0
E
The scalar potential should be nite. Therefore, we require
!
~
2e
(vL r) 83
(215)
for r ! 0. We put r = r~ and
=
~vL ' (216)
2e
where ' is the azimuthal angle. The function (~r) satises the equation
2
d 1 d +
uf 2 = 0:
2
2
dr~ r~ dr~ r~
(217)
The condition Eq. (215) requires ! 1=r~ for r~ ! 0. We see from Eqs. (216), (217) that a
moving vortex induces a dipole-like scalar potential proportional to the vortex velocity.
We choose the x axis of the Cartesian coordinate frame along the vortex velocity
v
L
.
The integral over the angle ' in Eq. (214) can be calculated using
vL r = vL cos ' ; vL ' = vL sin '
Equation (214) gives
Z
where
a=
W d2 r = 2 2GL avL2
Z "
(218)
#
df 2 f 2 +
r~ dr~:
dr~
r~
(219)
Equation (218) can be written in terms of the vortex viscosity Z
W d2 r = vL2
(220)
where
= 2 2GL a
The moving vortex experiences a friction force
Ffrict = v
(221)
L
This force is the only (longitudinal) component Fk
Ffrict
of the force Fenv exerted by
the environment, see Fig. 24. It counteracts the Lorentz force Eq. (205) produced by the
transport current, i.e.,
84
i
0 h
^
j h = vL
c tr
(222)
We see that the vortex moves perpendicular to the current.
Expressing the vortex velocity through the electric eld using Eq. (211) we nd
jtr = hEi
f
(223)
L
where the ux ow conductivity is
f =
With the microscopic values of
2c2 2GL a
:
B 0
(224)
and GL taken for dirty superconductors, we nd
f =
ua Hc2
:
2 n B
(225)
The same result can be obtained in a dierent way. The dissipation in a volume occupied
by one vortex can be written as
Z
W d2 r = f hEi2L S0 = f hEi2L 0 =B = f B 0 vL2 =c2
where we use
hEi2
L
= B 2 vL2 =c2
Comparing it with Eq. (218) we again arrive at Eq. (224).
To calculate a we need rst to solve Eq. (217). For u = 5:79, one nds a 0:502. The
ux ow conductivity becomes27
f
1:45 HB2 :
(226)
c
n
It is close to the expression f = n Hc2 =B known as the Bardeen{Stephen model26. This
simple prediction implies that the ux ow resistivity is just a normal-state resistivity times
the fraction of space occupied by vortex cores.
f =
n B
1:45 Hc2
0:69 HB
The ux-ow resistivity is linear in B .
85
n
2
c
C. Flux ow: High vortex density
Consider now high elds, H
! H 2.
We put
c
A = (0; Hx; 0), and choose the static
potential in the form = Ex x, where the electric eld Ex = (vy =c)H is homogeneous
in the leading approximation and has only an x component. The vortex velocity has only
a y component vL = (0; vy ; 0). Recall that the static order parameter in the vicinity of the
upper critical eld satises the linearized GL equation
@2
@ 2ieHx 2
+
+ 2 = 0
@x2
@y
~c
(see Section IV). It has the form Eq. (122):
=
X
"
Cn eiqny exp
n
~cqn
1
x
2 2
2eH
2 #
(227)
:
(228)
We calculate the average dissipative function Eq. (194):
hW i
L
@ 2ieEx x
@ 2ieEx x =2
+
+ n Ex2
@t
~
@t
~
L
2ieHx @
2
ieHx
@
= 2 vy2
+
+ n Ex2
@y
~c
@y
~c
L
(229)
To calculate the average we make use of the identities
@
=
@x
@
i
@y
2eHx
@ @ ;
= i
~c
@x
@y
2eHx ;
~c
(230)
derived for the order parameter Eq. (228).
We have
@
@y
2ieHx
~c
@ 2ieHx +
@y
~c
@ @ =
@x @x
L
L
:
Therefore
2ieHx
@ 2ieHx +
~c
@y
~c
* " #+ L
2
2
1
@ 2ieHx
@
1 =
+ 2
= 2 jj2 L :
2
@y
~c
@x
2
L
@
@y
We use here Eq. (227). We obtain from Eq. (229)
86
(231)
hW i
L
= 2 vy2 jj2 L + n Ex2
c2 2 = n + 2 2 jj L Ex2 = f Ex2
H
(232)
where the ux ow conductivity is
f = n + 4e2 ~ 2 2 jj2 L :
(233)
The average magnitude of the order parameter is from Eq. (130)
jj2 L =
22 2GL
1
(22 1) A
B
:
Hc2
Therefore
4e2 ~ 2 2 2GL
B
f = n +
1
A [1 1=(22 )]
Hc2
u (1 B=Hc2 )
= n 1 +
:
2 A [1 (1=22 )]
(234)
The ux-ow resistivity is
f = n 1
u (1 B=Hc2 )
2 A [1 (1=22 )]
It is linear in B . The slope is
@f u
= n
@B B=Hc2 Hc2 2A [1 (1=22 )]
[1
2:5
n
(1=22 )] Hc2
The ux-ow resistivity is shown in Fig. as a function of the magnetic induction.
ρf /ρn
1
1
B/H c2
FIG. 25. Flux ow resistivity within the TDGL model as a function of the magnetic eld.
87
Problems
Problem 1: additional problem to Sect. VI
Find the penetration depth for a magnetic eld oscillating with a small frequency ! such
that A = A0 e
i!t
.
Problem 2
Find the solution of linearized TDGL equation in a uniform electric led with a potential
= Ex x that describes a moving vortex lattice for H close to Hc2 .
Problem 3
Calculate the current for a moving vortex lattice from the previous problem.
88
VIII. PARACONDUCTIVITY
A small superconducting order parameter can appear above the transition temperature
due to uctuations. In the presence of an electric eld, existence of a nonzero uctuating order parameter will give rise to a nonzero supercurrent which would enhance the conductivity
of the superconductor. This uctuation-induced additional conductivity above Tc is called
the paraconductivity. Also, a nonzero uctuation magnetization and other superconducting
properties can appear already at temperatures above Tc . Here we calculate the uctuation
conductivity using the TDGL model. Of course, the applicability of this result is limited
by the validity of the TDGL model itself. This result was rst obtained by Aslamazov and
Larkin in 1968.
Consider the uctuation-induced order parameter above Tc . In the Fourier representation
(r) =
X
p
p eipr~
The sum over momenta is the sum over the states in a unit volume. In a system with
dimensions L1 , L2 and L3 the momentum components assume the values px = 2 ~nx =L1 ,
py = 2 ~ny =L2 , and pz = 2 ~nz =L3 where nx ; ny ; nz are integer numbers. The sum is
X
p
(L1L2 L3 )
1
X
nx ;ny ;nz
Since the order parameter varies over distances of the order of , a characteristic scale for
p is ~= . In a three dimensional case, where all L1 ; L2 ; L3 , the numbers nx ; ny ; nz can
assume large values. Therefore, in three dimensions, the sum is
X(3D)
p
= (L1 L2 L3 ) 1
X
nx ;ny ;nz
In a lm with a thickness L1 = d
= (L1 L2 L3 ) 1
Z
L1 L2 L3
dp dp dp =
(2 ~)3 x y z
, the number n
x
Z
d3 p
(2 ~)3
can only have one value nx = 0.
Therefore,
X(2D)
p
= (dL2 L3 ) 1
X
ny ;nz
= (dL2 L3 ) 1
89
Z
L2 L3
dpy dpz = d 1
2
(2 ~)
Z
d2 p
(2 ~)2
Similarly, in the one dimensional case, i.e., for a wire with the cross-section area S 2
X(1D)
p
=S 1
Z
dp
2 ~
In the absence of the electric and magnetic elds, the free energy of the superconductor
is, within the leading terms in small jj
F
sn
Z
jj2 + ~2 jrj2 dV
X
=
+ p2 jp j2
=
(235)
p
Note that > 0 above Tc . The probability of uctuations is
P fg = C exp (
F =T )
= C exp
=C
Y
p
exp
sn
T 1
X
p
2
+ p
jp j2
!
T 1 + p2 jp j2
(236)
For each p we have
P fpg = Cp exp
and C =
Q
p Cp .
T 1 + p2 jp j2
(237)
The normalization constant is
Z 1
Cp 1 =
0
exp
T 1 + p2 jp j2 d jp j
The average uctuation is
jpj2 = Cp
=
Z 1
0
jp j2 exp T
1 + p2 jp j2 d jp j
T
2 ( + p2 )
(238)
Consider the TDGL equation above Tc so that > 0. The order parameter appears due
to uctuations and is thus small. The linearized TDGL equation has the form
@ 2ie
+
= ~2 r2 @t
~
90
(239)
(1)
(0)
We put = (0)
p + p where p is the order parameter without the
electric eld, and (1)
p is a correction proportional to E . For a time-independent state, it
with =
E r.
satises the equation
2ie
+ p2 (1)
(E ^r) (0)
p =
p
(240)
~
where the operator ^r in the Fourier representation is ^r = i~@=@ p. Therefore,
(1)
p = 2e
@
+ p2 1 E (0)
@p p
(241)
The uctuation-induced supercurrent is
hj i = 2e h[ ( i~r) + (i~r )]i
X = 4e p jp j2
s
p
= 4e
X
p
p
D
(0) 2 + (0) (1)
p
(0) (1)
p + p p
p
E
The sum over momenta of the rst term disappears. The correction gives
hj i
s
= 8e2 X
p
= 8e2 X
p
@
@ (0)
p(0) E (0)
+ (0)
E
p
p
2
@p
@p p
p + p
p
Using Eq. (238) we nd
@ D (0) 2 E
E
p
+ p2
@p
e2 Tc X p (p E) 4 =~2
=
s
2 3 ~ (T Tc )
2 2 23
p ( + p )
p (1 + p =~ )
Here we use the microscopic value
hj i = 8T e2 2
X
p (p E)
=
(242)
(243)
(244)
~
8Tc
p
and the denition of the coherence length = ~ =jj.
Let us calculate the uctuation supercurrent for a lm. In the 2D case, Eq. (244) gives
Z
8T e 2 2
p (p E) d2p
h js i = d
( + p2 )3 (2 ~)2
Z 1
2T e2 2
p3 dp
=E
~2 d 0 ( + p2 )3
= 0 E
(245)
91
where we use that d2 p = pdp d'p and
Z
2
0
pi pk
d'p p2
= Æik
2
2
The excess uctuation conductivity is
Z 1
Tc e2
p3 dp
x3 dx
=
3 ~2 d
0 ( + p2 )
0 (1 + x2 )3
e2
Tc
1
Tc
T e2
=
= c2 =
2 ~ d 16~d T Tc 8RQ d T Tc
2T e2 2
0 =
~2 d
Z 1
(246)
Remarkably, the uctuation conductivity in a lm does not depend on the material parameters of the superconductor. We denote here
2~
= RQ
e2
6 k
It is called the quantum resistance.
In three and one dimensions, the uctuation conductivity is
0 =
8
>
< e2 T=[32~(T
Tc ) ]
>
: e2 Tc =[16S ~ (T
in 3D
Tc )]
in 1D
p
(247)
The uctuation conductivity diverges as 1= 1 T=Tc in the three dimensional case, while
it diverges as (1 T=Tc ) 3=2 in the one-dimensional case.
Problems
Problem 1
Calculate the uctuation conductivity in the three-dimensional case.
Problem 2
The same for the one-dimensional case.
92
IX. WEAK LINKS
A. Aslamazov{Larkin model
1. D.C. Josephson Eect
Consider two superconductors S1 and S2 separated by an insulating layer with a hole
in it with an area A 2 . The thickness of the insulating layer is L . This contact
is equivalent to a constriction between two superconductors, the so called superconducting
bridge (see Fig. 26). The both congurations belong to the so called superconducting weak
links.
S2
S1
S1
S2
L
a
z
χ1
χ1
χ2
χ2
L
(a)
(b)
FIG. 26. Examples of superconducting weak links. (a) A hole in an insulating layer between two
bulk superconductors S1 and S2. (b) Superconducting bridge between two bulk superconductors.
It is possible to derive a simple equation for the supercurrent through such weak links.
The GL equation (90) is
2
r
2ie 2
A +
~c
jj2 =2GL = 0
(248)
For distances of the order of L, the gradient terms are of the order (=L)2 , i.e., they are
much larger than all other terms. We obtain the Laplace equation
r2 = 0
93
Let us take the order parameter in the form
= 1 ei1 [1 f (r)] + 2 ei2 f (r)
The function f (r) also satises the Laplace equation
r2 f = 0
with the boundary conditions f = 0 in the superconductor 1, and f = 1 in the superconductor 2.
The supercurrent density Eq. (80) becomes
j
s
= 2e ~ [ i r + ir ]
= 2ie ~1 2 ei(2 1 ) e i(2
1)
rf
1 ) (rf )
= 4e ~1 2 sin (2
(249)
Integrating over the cross section of the hole we get the full current along the z axis
Is = Ic sin (2
1 )
(250)
where
Ic = 4e ~1 2
Z
(rz f ) dS
(251)
Note that the function f increases from region 1 into the region 2, therefore, the value Ic is
positive. By the order of magnitude,
Z
(rz f ) dS
A=L
If the two superconductors are connected by a one-dimensional channel (wire) of a length L
and a cross section area A, such that the ends correspond to z = 0 and z = L the function
is f = z=L and we get exactly
Z
(rz f ) dS = A=L
94
Equation (250) describes the d.c. (stationary) Josephson eect: the supercurrent through
a weak link is proportional to a sine of the phase dierence. Eq. (250) is called also the
current-phase relation. It is sinusoidal for a short weak link with L . We shall discuss
other realizations of the Josephson eect in the following sections.
2. A.C. Josephson Eect
Using the same arguments one can also calculate the normal current. For a short contact,
the gradient term is the largest. Therefore,
j
n
= n r
and also
div jn = 0
The potential thus satises the Laplace equation
r2 = 0
We can put
= V [1 f (r)]
where V is the voltage across the weak link. The function f satises the Laplace equation
with the same boundary conditions as before; it is therefore the same function as was
introduced above. The normal current is
j
n
= n V rf (r)
The full normal current becomes
In = V=Rn
where
95
Z
1
= n (rz f ) dS n A=L
(252)
Rn
is the total resistance of the hole in the normal state. For a one-dimensional wire one has
exactly Rn = n 1 L=A.
With the general expression Eq. (252) we can write the critical supercurrent Eq. (251)
in the form
4e ~1 2 1 2
=
(253)
n Rn
4TceRn
where we use the expression = D=8Tc~ for dirty superconductors. We see that Ic is
Ic =
now expressed through the resistance of the link in the normal state.
The total current takes the form
V
+ I sin '
(254)
Rn c
where the phase dierence is ' = 2 1 . Deep in the superconducting region, the Josephson
equation (199) holds
~ @
= +
=0
(255)
2e @t
Since the full current through the small hole is also small, the current density far from
I=
the hole is vanishingly small. Thus, the phases 1 and 2 do not vary far from the hole,
1;2 = const. As a result, the dierence of the phases at the both sides from the hole obeys
the Josephson relation
@'
= 2eV
(256)
@t
This equation describes the so called resistively shunted Josephson junction (RSJ) model
~
(see Fig. 27).
I
V
R
J
FIG. 27. The resistively shunted Josephson junction.
96
The full equation for the current is
I=
~ @'
+ Ic sin '
2eRn @t
(257)
If I < Ic, the phase is stationary:
' = arcsin(I=Ic)
and voltage is zero. the phase dierence reaches =2 for I = Ic. If I > Ic , the phase starts to
grow with time. Denote t0 the time it takes to increase from =2 to =2 + 2 . The average
voltage is
(2e=~)V = 2=t0
Calculating t0 (see Problem 1) we nd the current{voltage curve dependence
p
V = Rn I 2
Ic2
(258)
It is shown in Fig. 28.
V
Ic
I
FIG. 28. The current{voltage curve for resistively shunted Josephson junction.
Problems
Problem 1
Derive Eq. (258).
97
X. LAYERED SUPERCONDUCTORS.
A. Lawrence{Doniach model
Consider a stack of thin superconducting lms with a thickness smaller than separated
by insulating layers of a thickness d. Let the z axis be perpendicular to the lms (see Fig.
29).
z
FIG. 29. The model of a layered superconductor.
The free energy can be written as
F
sn
=d
XZ
n
"
jn (r)j2 + jn (r)j4 + k 2
i~r
#
2
2e
A n(r) + Fint(n) d2r (259)
c
Here the sum is over all the layers; n (r) is the value of the order parameter at the layer n,
A and r are the two-dimensional vectors in the plane of layers.
The term Fint is the density of the interaction energy between the layers. One could
write it in the following form
~2
Fint (n) = ?2 jn+1
d
n j2
where ? is a constant that describes the strength of the interaction. For d ! 0, the sum is
equivalent to
d
X
=
n
Therefore,
98
Z
dz
d
X
Fint (n) =
n
X
n
Z
@ n 2
@ n 2
2
2
= ? ~
d?~ dz @z @z which gives the missing z -component of the gradient energy in Eq. (259).
However, Eq. (259) is not gauge invariant. To preserve the gauge invariance one needs
to introduce the phase factors containing the vector potential
~2 Fint (n) = ?2 n+1 e
d
( +1;n)
iA n
2
n (260)
where
2e
A (n + 1; n) =
~c
Z zn+1
zn
Az dz
(261)
For d ! 0, this expression will then go over into the correct expression
2
@
2
ie
2
Az Fint (n) = ?~ @z ~c
Finally, the free energy of the system of layers is
XZ
"
Fsn = d
jn (rj2 + jn(rj4 + k i~r
2
n
2
?~2 iA (n+1;n)
n d2 r
+ 2 n+1 e
d
2
2e
A
n (r)
c
(262)
This energy is the basis for the Lawrence{Doniach model for layered superconductors.
Variation of the free energy with respect to the order parameter gives
Æ Fsn
2ie 2
2
2
= n + jnj n k ~ r
A n(r)
Æ n
~c
?~2
n+1 e iA (n+1;n) + n 1 e+iA (n;n 1) 2n = 0
2
d
(263)
The second line is the discrete analogue of the second derivative with respect to z .
Let us calculate the variation of the free energy with respect to ÆAz . We have
Z
Z
zn+1
2ie? ~ X
Æ Fsn =
d2 r n+1 e iA (n+1;n) n+1 eiA (n+1;n) n c:c
ÆAz dz
dc n
zn
Z
Z
zn+1
2ie?~ X
2
iA (n+1;n) iA (n+1;n)
=
d r n+1 e
n n+1 e
n
ÆAz dz
dc n
zn
99
Since
Z zn+1
zn
d
ÆAz dz [ÆAz (n + 1) + ÆAz (n)]
2
we have
Z
ie ~ X 2 Æ Fsn = ? d
d r n+1 e
dc
n
[ÆAz (n + 1) + ÆAz (n)]
( +1;n) n
n+1 eiA (n+1;n) n
iA n
Changing the summation index n for n 1 in the term with A(n + 1) we get
Z
ie ~ X 2
Æ Fsn = ? d
d rÆAz (n) n+1 e iA (n+1;n) n
dc
n
iA (n;n 1) + n e
n 1 n eiA (n;n 1) n 1
n+1 eiA (n+1;n) n
Using
jz = c
ÆF
ÆAz
we nd
jz =
ie? ~ n+1 e iA (n+1;n) n n+1 eiA (n+1;n) n
d
+ n e iA (n;n 1) n 1 n eiA (n;n 1) n 1
Putting
n = jn jein
we nally obtain
2e ~
jz = ? jn+1 n j sin n+1
d
+ jn n 1 j sin n
n 1
100
n
2e
~c
2e
~c
Z zn
zn
1
Z zn+1
zn
Az dz
Az dz
(264)
B. Anisotropic superconductors
If the distance between the layers is small d ! 0, one has for the free energy
2
Z "
2e
2
4
Fsn = j(r; z)j + 2 j(r; z)j + k i~r c A (r; z)
2 #
@
2
e
+ ? i~
Az d2 r dz
@z
c
(265)
This expression is similar to Eq. (70); the distinction is that the constants k and ? are now
dierent. These constants determine two coherence lengthes: one in the symmetry plane,
k , and another perpendicular to the plane, ?,
q
p
k = ~ k =jj; ? = ~ ?=jj
(266)
Eq. (265) describes an anisotropic superconductor with uniaxial anisotropy.
If the order parameter is normalized in such a way that it is the wave function of superconducting electrons , the free energy has the form
F
sn
=
Z "
b
1 a jj2 + jj4 +
2
2mk 1 +
2m? @
i~
@z
i~r
#
2
2e
Az dV:
c
2
2e
A
c
(267)
The constants a and b determine the magnetic eld penetration length k
jaj = mk c2
b
16e2 2k
for currents owing in the symmetry plane and also
jaj = m?c2
b
16e2 2?
for currents along the z axis. The coherence lengthes along the plane and perpendicular to
the plane are now also expressed through the eective masses
k2 =
~2
~2
; ?2 =
2mk jaj
2m? jaj
101
C. Upper critical eld for parallel orientation
Let the magnetic eld be parallel to the layers. We nd a second-order phase transition
into the superconducting state with lowering the eld. We choose the y axis along the eld,
with the gauge A = (0; 0; Hx) and look for a solution in the form of
(n; r) = Ceiqnd fq (x)
The equation for f is obtained from Eq. (263). Using the denitions of k and ? from Eq.
(266) we arrive at the Mathieu equation:
@ 2 f 2?2
2eHdx
2
k 2
1 cos qd +
f +f =0
@x
d2
~c
(268)
There are two limiting cases: (1) Weakly layered (nearly continuous) limit, and (2)
Highly layered case.
1. Continuous limit
The continuous limit ?
d coincides with the anisotropic case.
We shall see in a
moment (see also Problem 2) that
Hc2 =
0
2k ?
(269)
Since x k we nd that the argument of the cosine function in Eq. (268) is
dx
k ?
d
?
The continuous limit corresponds to
d ?
In this case, the cosine function in Eq. (268) can be expanded in small argument and the
equation becomes
@f
k2 2
@x
2eHx
?2 q +
~c
102
2
f +f = 0
The lowest-level solution is
"
#
1
~cq 2
x+
2k2
2eH
f = exp
(270)
with H = Hc2 from Eq. (269).
2. Highly layered case
As we shall see, this limit corresponds to a high upper critical eld and is realized for a
certain relation between d and ?. We introduce the parameter
s=
2eHd2k
~c?
(271)
For a continuous case,
d2
s= 2
?
1
On the contrary, a highly layered case is realized for s 1 when H ! 1.
For this limit, a solution of Eq. (268) can be obtained by expansion in powers of a small
1=s. The lowest-level periodic solution corresponds to an even Mathieu function
2
2eHdx
1
4eHdx
f = 1 + 2 cos
+ 4 cos
+ :::
s
~c
2s
~c
under the condition
s2 =
1
(d2 =2?2 )
1
This condition determines the upper critical eld
Hc2 = 0 2 ?
2d k
1
1 (d2 =2?2 )
p
p
This solution is valid for s 1, or ?(T ) ! d= 2.
In a situation such that ?(0)
d,
(272)
a weakly layered case with ?(T )
d at high
temperatures can transform into a highly layered limit for lower temperatures. One can say
103
that an anisotropic three-dimensional superconductor experiences a transition into eectively
p
a two-dimensional superconductor at the temperature when ?(T ) = d= 2.
p
We observe that the upper critical eld diverges as ?(T ) approaches d= 2 with lowering
the temperature. In this limit, a vortex cores t in-between the superconducting layers, and
the supercurrents do not destroy superconductivity.
Divergence of the upper critical eld is similar to the Little and Parks eect for a nonsingly-connected superconductor (Fig. 10) where the critical temperature is a periodic function of the magnetic eld instead of vanishing with an increasing eld. Suppose we have
a cylinder of radius R made of a thin lm with thickness d such that d
.
The cylin-
der is placed in a magnetic eld parallel to its axis. Because of the small thickness, the
superconducting velocity inside the lm is not fully screened:
v
s
=
~
2e
A =6 0
~c
r
2m
Its circulation around the cylinder is
Z
v dl = 2Rv
s
s
~
=
2m
2n
2e
~c
since the circulation of the phase can only be 2n where n is an integer. Here is the ux
inside the cylinder. Note that is not quantized, because d L . Thus the superconducting
velocity is
vs =
~
2mR
0
n
We see that the superconducting velocity is a periodic function of the magnetic ux, the
number of quanta n being adjusted in such a way that vs is limited by
vs max =
~
4mR
Superconductivity is not destroyed if this maximum value is smaller than the pair-breaking
velocity vpb
~=m .
Instead, the critical temperature will oscillate: It is maximum when
the ux is exactly integer number of ux quanta. The condition for the Little and Parks
eect is thus R .
104
Problems
Problem 1
Derive Eq. (263)
Problem 2
Find the upper critical eld for an anisotropic superconductor in a magnetic eld tilted
by an angle with respect to the anisotropy axis (see Fig. 16).
105
XI. JOSEPHSON JUNCTIONS. JOSEPHSON VORTICES.
A. Josephson junctions
Assume that there are only two layers, n = 1 and n = 2, Fig. 30. The current between
them will be
jz = jc sin 2
1
2ed
A
~c z
(273)
where the quantity
jc =
4e?~
j1 2 j
d
is called the critical Josephson current. Equation (273) has the same form as the current
through the short weak link.
χ2
χ1
FIG. 30. The Josephson junction between two superconductors.
Equation (273) describes the d.c. Josephson eect: The supercurrent can ow through
the insulating layer provided there is an interaction between the superconducting regions. In
practice this interaction is realized via a quantum-mechanical tunneling of electrons through
the insulating barrier. In this case, ?
/ exp( Ap d=~), where A is a constant of order
F
unity.
B. Long Josephson junctions in the magnetic eld.
Consider two large superconductors 1 and 2 separated by a thin insulating layer with a
thickness d. The (x; y ) plane is in the middle of the insulating layer. The superconductor
106
1 is at z < d=2, the superconductor 2 occupies the region z > d=2. The magnetic eld is
applied parallel to the insulating layer along the x axis, Fig. 31. We choose A = (0; Ay ; 0).
Therefore,
hx =
@Ay
@z
z
y
x
H
FIG. 31. A long Josephson junction in a magnetic eld.
Deep in the superconductors
@ 2e
= A
@y ~c y
due to the Meissner screening. The eld decays as
hx (z ) =
8
>
< He
(z
2)=2
d=
>
: He(z +d=2)=1
z > d=2
z < d=2
We denote H the eld in between the two superconductors at z = 0; 1;2 is the London
penetration length in the superconductor 1 or 2, respectively. The phase is essentially
independent of z because the current through the junction is small. In the bulk it is thus
the same as at the boundary of the contact.
The vector potential deep in the superconductor 2 is
Ay (2) =
Z 1
0
hx dz =
Z
2
d=
0
hx dz
Z 1
2
d=
hx dz = H
d
2
H2
Here we put Ay (0) = 0 and assume that H is independent of z at 0 < z < d=2. Similarly,
107
Ay (1) =
Z
1
0
hx dz =
Z
0
d=
2
Z
hx dz +
2
d
hx dz = H
+ 1
2
1
d=
As a result
H=
~c
@1
~c
@2
=
2e(1 + d=2) @y
2e(2 + d=2) @y
so that
@'
2e(1 + 2 + d)H
=
@y
~c
(274)
where
' = 2
1
The magnetic eld H is generally a function of the coordinate y into the junction.
Using the Maxwell equation
c
curl h = jc sin ' or
4 z
c @H
= jc sin '
4 @y
we obtain
@2'
= jc sin '
8e(2 + 1 + d) @y 2
~c2
or
@2'
2J 2 = sin '
@y
where
s
J =
~c2
8ejc (2 + 1 + d)
(275)
(276)
is called the Josephson length. We can write Eq. (274) in the form
H=
4jc 2J @'
c @y
(277)
Equation (275) is similar to equation of motion for a pendulum. Indeed, the latter has
the form
108
g
d2
=
sin 2
dt
l
where the angle is measured from the bottom. Equation (275) is obtained from it by
replacing = ' and putting 2 = g=l. This means that the pendulum angle is
J
measured from the top, Fig. 32.
φ
θ
FIG. 32. A pendulum moves in time similar to variations of along the y axis.
Consider rst a low eld H applied outside the junction at y = 0. In this case ' is also
small. Expanding Eq. (275) we nd
2J
@2'
='
@y 2
whence (see Fig. 33, curve 1)
' = '0 e
y=J
The magnetic eld decays as
H (y ) = H (0)e
y=J
where we nd from Eq. (277)
H (0) =
4jc J '0
c
(278)
Magnetic eld decays into the junction in a way similar to the Meissner eect. The penetration length is J . This length is larger than L because the screening current cannot exceed
the Josephson critical current jc . Such behavior exists for elds smaller than 4jc J =c.
109
φ
2π
π
0
1
Y0
2
2Y0
3Y0
y
−π
FIG. 33. The phase dierence ' as a function of the distance into the junction measured from
the left edge. Curve 1: small magnetic elds. Curve 2: large elds, phase runs from ' = at
the edge through 2n values making Josephson vortices. The curves correspond to H (0) < 0.
To nd a solution of Eq. (275) for larger elds, we multiply it by @'=@y and obtain
2J @' 2
+ cos ' = C
(279)
2 @y
where C is a constant. Eq. (279) gives
Z
'
pJ pC d'cos ' = y
(280)
2 '0
As discussed above, the stable solution for y ! 1 corresponds to ' = 0 and @'=@y = 0.
Eq. (279) results in C = 1. The applied led is then
4jc 2j @'(0) 8jcj '0
H (0) =
=
sin
c
@y
c
2
The sign is chosen to agree with Eq. (278) for small elds. Increasing eld leads to an
increase in '0 until it reaches . This threshold corresponds to the led
H1 =
8jcJ
c
Above this eld, the constant C > 1, and the phase ' can vary within unlimited range.
Consider for example the case H (0) < 0. The phase runs indenitely from at the edge
through the values 2n producing the so called solitons (Fig. 33, curve 2). The phase
solitons are also called the Josephson vortices: the phase dierence across the junction
varies by 2 each time as we go past one Josephson vortex. The distance between vortices
is L = 2Y0 J .
110
Problems
Problem 1
Find the distance between Josephson vortices for H close to H1 .
111
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