Introduction to the Theory of Superconductivity and

Super uidity

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 super uidity 4

A Landau criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

B Two- uid hydrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

C First and second sounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

D Vortices in a rotating super uid . . . . . . . . . . . . . . . . . . . . . . . 12

E Vortex near a wall. Feynman critical velocity . . . . . . . . . . . . . . . . 15

II 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 Meissner e�ect. Magnetic ux quantization . . . . . . . . . . . . . . . . . 41

IV Vortices in type II superconductors 44

A Transition into superconducting state in a magnetic �eld . . . . . . . . . . 45

B Single vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

C Vortex free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

V 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 super uids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Weak Links 93

A Aslamazov{Larkin model . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

2

1 D.C. Josephson E�ect . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

2 A.C. Josephson E�ect . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Highly layered case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

XI 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

Super uidity is an ability of a uid to ow without friction through narrow tubes. We

shall see why one needs a narrow tube in this de�nition. The most striking realizations of

super uidity are thermo-mechanical e�ects. 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 satis�ed if

jvj > �(p)

jpjor if at least

jvj > min

��(p)

p

�(1)

Excitations are created if this condition is satis�ed. If the minimum is nonzero

vc � min

��(p)

p

�6= 0 (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

super uidity.

4

FIGURES

T+∆T T

j jn s

FIG. 1. The fan rotates if one end of the closed tube is heated.

T T+∆T

∆p

superleak

FIG. 2. A pressure head is established between two vessels kept at di�erent 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 = �nvn + �svs (3)

The velocity vn is associated with the motion of excitations, �n is the coeÆcient of propor-

tionality 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 vs is associated with the motion without friction;

it is called the super uid velocity, while the proportionality coeÆcient �s is the super uid

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 super uid 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 super uid motion is a potential ow, characterized by the zero-vorticity

condition

curlvs = 0 (5)

Therefore, the full time derivative (the material time derivative) can be presented as a

gradient

dvsdt

=@vs@t

+ (vs � r)vs = �r�s (6)

where �s is the potential (per particle mass) that acts on the super uid component. There-

fore, �s has the meaning of the chemical potential of the super uid part of the liquid. Using

Eq. (5) we obtain

@vs@t

+r��s +

v2s2

�= 0 (7)

Consider a reference frame K0 that moves with the velocity vs with respect to the

laboratory frame K. We have in the laboratory frame (per unit volume)

6

j = �vs + j0 (8)

E =�v2s2

+ vsj0 + E0 (9)

Here E0 and j0 are the energy and current in the moving frame which depend on the uid

velocity vn � vs in the moving frame. The energy obeys the thermodynamic identity

dE0 = T dS + � d�+ (vn � vs) � dj0 (10)

Here � is the chemical potential (de�ned per particle mass), S is the entropy per unit volume.

The current density is

j0 = �n (vn � vs)

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

d ~E = T d ~S � p d ~V + ~�dN + v � d~J

The chemical potential per particle ~� is de�ned as

~�N = ~E � T ~S + p ~V � 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

d~� = �~S~V

~V

NdT +

~V

Ndp�

~V

N

~J~V� dv

= �S~V

NdT +

~V

Ndp�

~V

Nj � dv

Dividing this by the particle mass we get

7

d� = �S�dT + ��1 dp� ��1j � dv

where � = mN= ~V and � = ~�=m.

Therefore, in our case, the chemical potential (per unit particle mass) satis�es the ther-

modynamic relation

d� = �� dT + ��1 dp� ��1j0 � d (vn � vs) (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

super uid particles. Indeed, if we apply Eq. (11) to super uid component alone we should

put the entropy to zero since the super uid 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 super uid part. As a

result, Eq. (11) for super uid component gives

r�s = ��1s rp

At the same time, according to the Euler equation,

�sdvsdt

= �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

@vs@t

+r��+

v2s2

�= 0 (12)

In addition, the ow obeys the continuity equation

@�

@t+ div j = 0 (13)

Since the entropy of the super uid 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

@t+ div (Svn) =WS (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@t

+@�ik

@rk= 0 (15)

where �ik is the tensor of the momentum ow. In the absence of the normal viscosity, it is

�ik = �nvn ivnk + �svs ivs k + pÆik (16)

The two- uid hydrodynamic equations (12) { (15) are quite complicated because �n, �,

�, etc., depend on vn�vs 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@t

+r� = 0 (17)

@�

@t+ div j = 0 (18)

@(��)

@t+ (��) divvn = 0 (19)

@j

@t+rp = 0 (20)

We assume also that system is in thermal equilibrium with !� � 1 so that the dissipation

is absent.

Eqs. (18), (20) give

@2�

@t2= r2p (21)

9

Eq. (19) yields

@2(��)

@t2+ (��) div

@vn@t

= 0 (22)

while Eqs. (17), (20) result in

div@vs@t

+r2� = 0

�ndiv@vn@t

+ �sdiv@vs@t

+r2p = 0

which gives

div@vn@t

= (�s=�n)r2� � ��1n r2p

Inserting this into Eq. (22) we obtain

�n��

@2(��)

@t2+ �sr2��r2p = 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 + ��1r2p (24)

As a result,

@2�

@t2=�s�

2

�nr2T (25)

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

� =@�

@pÆp+

@�

@TÆT

� =@�

@pÆp+

@�

@TÆT

we obtain from Eqs. (21) and (25)

@�

@p

@2Æp

@t2+@�

@T

@2ÆT

@t2�r2Æp = 0 (26)

@�

@p

@2Æp

@t2+@�

@T

@2ÆT

@t2� �s�

2

�nr2ÆT = 0 (27)

We look for a solution in the form of a plane wave

Æp; ÆT / exp [�i! (t� x=u)] (28)

The equations yield �@�

@pu2 � 1

�Æp+

@�

@Tu2ÆT = 0 (29)

@�

@pu2Æp+

�@�

@Tu2 � �s�

2

�n

�ÆT = 0 (30)

These equations can be simpli�ed if we neglect a small thermal expansion and put

@�

@T=@�

@p= 0

Equations (29), (30) then have two solutions:

u1 =

s@p

@�while ÆT = 0 (31)

and

u2 =

s�2�s

�n(@�=@T )while Æp = 0 (32)

Equation (31) describes the usual (�rst) sound, i.e., variation of density and pressure at a

constant temperature. Note that we neglected the di�erence between adiabatic and isother-

mal 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 super uid

In a rotating container, the normal component rotates with the vessel such that its

velocity is vn = � r. Note that this ow �eld has curlvn = 2. Due to the potentiality of

super ow Eq. (5) the super uid 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 super uid 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 super uid component Eq. (5) implies that the super uid velocity can be presented as a

gradient of a ow potential

vs = r� (33)

This equations yields identically that curlvs = 0 everywhere in the uid, as it should be.

However, if it were correct absolutely at any point in the uid, the super uid 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 super uid 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,

curlvs = AÆ(2) (r? � r0?) (34)

where is the unit vector along the rotation axis and A is a coeÆcient to be determined.

The coordinate r0? determines thus a singular point on a plane where the curl-free condition

of super ow is violated. In three dimensions, we thus obtain a singular line.

According to the Stokes theorem, Eq. (5) can be written as

12

Ivs dr = 0

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

Ivs dr =

Zcurlvs dS = A

where the second integral is taken over the area con�ned within the contour. Again, the

contour around the singular line can be taken arbitrary because curlvs = 0 everywhere

outside the line. Multiplying the above equation by the particle mass m we have

Imvs dr = mA

According to the general principles of the quantum mechanics (Bohr{Sommerfeld quantiza-

tion) the above integral should be quantized. In other words, one should have mA = 2�~n

where n is an integer. Therefore,

Ivs dr =

2�~n

m� �0n (35)

or

curlvs = �0neL 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 super uid 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 �� = �0n 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

vs' =�0n

2�r(37)

13

or

vs =�0n ez � r

2�r2(38)

If there are several vortex lines, the super uid velocity is a superposition of velocities of

each vortex line. For straight parallel lines,

vs =Xi

�0niez � (r� ri)

2�jr� rij2 (39)

where ri is a (two-dimensional) position vector of the i-th vortex. Eq. (36) takes the form

curlvs =Xi

�0niÆ(2) (r� ri) (40)

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)

Ivs dr =

Z Xi

�Æ(2) (r� ri) dS = nL�R2�

where � = �0n. This gives

2�Rvs' = nL�R2�

or

vs' = nL�R=2 = R

The last equality holds if

nL = 2=� (41)

We see that the super uid 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 super uid 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�sv

2s

2dV =

�2�sL

4�

Z rmax

rmin

dr

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)

r20 ��

2�(42)

where we put nL = 1=�r20.

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

�L =�2�s4�

ln

�r0rc

�(43)

The vortex energy is proportional to square of circulation. Therefore, the energy of N

singly quantized vortices with � = �0 is N�1 where �1 is the energy Eq. (43) for n = 1.

However, the energy of one vortex with N quanta, i.e., � = N�0 is N2�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

P

vsFM

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 in ate the loop. Indeed, the energy of the half-loop is from Eq. (43)

E =�2�sR

4lnR

rc

The momentum of the loop is (see Problem 4 to Sect. I)

P = ��snS (44)

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 super ow vs. For a half-circle, S = �R2=2.

The energy of the loop in the ow is then

E + P � vs = �2�sR

4lnR

rc� ��s�R

2

2

The energy is plotted in Fig. 4. As a function of the loop radius, the energy has a maximum

at

Rc =�

4�vslnR

rc(45)

16

This maximum is a potential barrier

E0 =�3�s16�vs

ln2�

�

4�vsrc

�(46)

that separates the state where the loop vanishes at the wall R = 0, E = 0 and the state

where the loop is in�nitely 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

E

R Rc

0

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

vc =�

4�Rln

�R

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 in ate 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 super uidity! This is why the velocity Eq. (47) is called the critical

velocity of super uidity. 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-of-

magnitude 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 super uid 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 super uids 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 satis�es the

Schr�odinger equation

i~@

@t= �

�~2

2mr2 + �

� +

Zj (r0)j2 U(r � r0) d3r0:

Here � is the chemical potential and U > 0 is the interaction (repulsive) interaction. As-

suming that varies slowly at distances of an atomic scale, one can take it out from the

integral denoting

U0 �ZU(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 Schr�odinger

equation takes the form

i~@

@t= � ~

2

2mr2 � 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 = j jei�. In a spatially uniform

situation j j = pN 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~2[ �r � r �] (49)

It appears when the phase � varies in space. Equation (49) suggests that P = ~r� is the

momentum of a condensate particle, while

vs =~

mr� (50)

is its velocity. The super ow is thus potential; Eq. (33) yields

� =~�

m

As a result, the particle current becomes j = �svs where �s = Nm = mj j2 the mass densityof 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@t

+r��s +

v2s2

�= 0 (51)

where

�s = � ~

m

@�

@t� v2s

2(52)

is the chemical potential of the super uid component. The GP equation (48) provides an

expression for �s. Let us put = j jei�. The complex equation (48) yields two real equations

j j ~m

@�

@t= �j jv

2s

2+

~2

2mr2j j+ j j� �1� j j2=N� ; (53)

@j j@t

= � ~

2m[rj jr�+r(j jr�)] (54)

Equation (53) gives the expression for the chemical potential

�s = ��j j2=N � 1

�� ~2

2mj j�1r2j j (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 j and �nd

20

@(mN)

@t+ div j = 0 (56)

According to Eq. (48) the characteristic length of spatial variations of the wave function

is

� =~p

2mU0N

Using the expression for a, we �nd

� � N�1=3��1=2 � N�1=3

where � = aN1=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 �N1=3 � �=d� 1

that allows one to construct a tractable model of super uidity.

f

x

1

1 3

j/j

v /v

max

maxs

FIG. 5. The condensate density f , velocity v and current near the vortex axis for a sin-

gle-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

=pNein'f (r=�) (57)

21

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 2�n on going around the z axis. As a result, the super uid velocity has a

nonzero circulation

Ivs � dr = ~

m

Ir� � dr = 2�~n

m� n�0

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 super uid is quantized. The ansatz Eq. (57)

corresponds to an n-quantum vortex.

The amplitude function f(x) satis�es the equation

1

x

d

dx

�xdf

dx

�� n2f

x2+ f � f 3 = 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)

vs =n~e'mr

=n�0e'2�r

=n�0ez � r

2�r2

where e' is the unit vector in the azimuthal direction while ez is the unit vector in the

direction of the z axis. This agrees with Eq. (38). The particle current is

j =n~�f 2

mre'

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

~

mv2t

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

de�ne a critical velocity above which super uidity is destroyed, in contradiction to the

Landau criterion2 of super uidity. Consider this in more detail.

On one hand, we can look for a time independent solution

=pNfeik�r (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 � k2�2)

This expression has a maximum at k� = 1=p3 with

23

jmax =2~�

3p3m�

; vmax =~p3m�

(63)

These quantities de�ne the maximum supercurrent and the maximum super uid velocity

which can exist in the super uid in a time-independent state. The time-independent state

becomes unstable when the super ow exceeds vmax.

It is interesting to compare the maximum velocity vmax with the sound velocity in a non-

ideal 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)

! =

"U0N

mk2 +

�~k2

2m

�2#1=2

(64)

For long wave lengths or small wave vectors k one obtains the sound-like dispersion

! = uk

with the sound velocity

u =

rU0N

m=

~p2m�

(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 de�ned

according to the Landau criterion as

24

vc = min

��(p)

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 de�nes the

critical velocity equal to the sound velocity. The spectrum of excitations in a real 4He 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

p p(a) (b)

vcvc

FIG. 6. The excitation spectrum in the GP model (a) and in the real 4He (b). The Landau

critical velocity is given by the slope of the corresponding dashed line.

Equation (66) de�nes the velocity limit above which the excitations (phonons in our

case) are produced by the moving super uid. We can see from Eq. (65) that vmax is of the

same order as the Landau critical velocity but it is by a factor ofp3=2 larger than vc:

vmax =

r3

2vc =

r3

2u: (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 time-

independent 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

~

mv2t

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 e�ects 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 signi�cance. 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 e�ects 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 super uids.

The main limitation of the GP equation is the lack of an explicit mechanism that es-

tablishes equilibrium towards a particular state. In principle, a source of dissipation can

be identi�ed within the GP model: it is associated with an emission of phonons. These

phonons either escape to in�nity or are absorbed at the walls. The efective relaxation mech-

anism 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~�@

@t= � ~

2

2mr2 � 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 e�ective 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 super uid helium. The resistivity behavior as a function of temperature is shown

in Fig. 7.

ρ

TTc

ρn

ρ = 0

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 pa-

rameter, for example, of the system temperature (or magnetic �eld, etc.). The well known

example is the spontaneous magnetization of a ferromagnet. For superconductors (or super-

uids) 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 + b

2jj4

�: (69)

where a and b are the expansion parameters and V is the volume of the system. At temper-

atures 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 Fsnmin = �V jaj2=2b

Near the transition temperature, one can thus put the coeÆcient a = a0(T � Tc) while

b = const.

F

Re Ψ Im ΨχFIG. 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 demon-

strated 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 speci�cally its modulus is chosen to be the

energy gap in the electronic spectrum which opens after transition into superconducting

state

30

Ep =q(�p � EF )2 + j�j2

where �p is the spectrum in the normal state. The order parameter � is a complex function

� = j�jei� 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:

Fsn =

Z "�j�j2 + �

2j�j4 +

������i~r� 2e

cA

��

����2#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 homoge-

neous state. The total energy consists of the normal state energy Fn, the superconducting

free energy Fsn, and the magnetic energy:

F = Fn + Fsn +

Zh2

8�dV (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�0h

2

2dV

where �0 is permeability of vacuum. We omit the (constant) free energy of the normal state

Fn in what follows.

The gradient term in Eq. (70) is the momentum operator in presence of the magnetic

�eld

P = �i~r� 2e

cA (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 +~c

2erf

The free energy expression is supplemented with the Maxwell equation

curlh =4�

cj (73)

where j is now the electric current and the microscopic �eld is

h = curlA (74)

In the SI units we have instead of Eq. (73)

curlh = 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 � T

Tc; � =

7�(3)�

8�2T 2c

(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

8~Tcy(�Tc=~) (76)

where D = v2F �=3 is the di�usion coeÆcient, and

y(x) =8

�2

1Xn=1

1

(2n+ 1)2[(2n+ 1)2�x+ 1](77)

This function is y = 1 for a dirty limit �Tc=~� 1, and it is

32

y(�Tc=~) =7�(3)~

2�3�Tc

for a clean case Tc�=~� 1. Therefore

=

8><>:��D=8Tc~ ; dirty

7�(3)�v2F=48�2T 2

c ; clean(78)

so that dirty= clean � (�Tc=~)� 1.

A. Ginzburg{Landau equations

Variation of F with respect to �, �� and A gives

ÆF =

Z ( "��+ �j�j2�+

��i~r� 2e

cA

�2

�

#�� + c:c:

!

+

�curl curlA

4�� 2e

c

���

��i~r� 2e

cA

��+ c:c:

��ÆA

+ div

�~

���

�~r� 2ie

cA

��+ c:c:

�+

1

4�ÆA� curlA

��dV

The requirement of extremum of the free energy gives the GL equation

��+ �j�j2�+

��i~r� 2e

cA

�2

� = 0 (79)

together with the de�nition of the current

j = 2e

���

��i~r� 2e

cA

��+�

�i~r� 2e

cA

���

�(80)

which follow from the Maxwell equation (73) and Eq. (74). In addition, the surface term

requires, in particular,

n ��~r� 2ie

cA

�� = 0 (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 con-

ductors, 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 =pj�j=� =

�8�2

7�(3)

� 1

2

Tc (1� T=Tc)1

2 (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 = �j�j2=2� (83)

It is called the condensation energy. It de�nes the thermodynamic critical magnetic �eld

H2c = 4�j�j2=� (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

H2c = j�j2=�0�

The thermodynamic critical �eld is linear in temperature

Hc(T ) � Hc(0)

�1� T

Tc

�

34

where Hc(0) is formally de�ned 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

H �B4�

In the superconducting state B = h = 0, and the Gibbs free energy density is Gs = Fsn =

Fc = �j�j2=2�. In the normal state, Fsn = 0, B = h = H so that Gn = �H2=8�. The

energy in the normal state becomes smaller than Gs for H > Hc.

Eq. (79) de�nes the length

�(T ) =p ~2=j�j / (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

�(T ) =

�7�(3)

12

� 1

2

�0 (1� T=Tc)� 1

2 (86)

where

�0 = ~vF=(2�Tc) (87)

is the \zero-temperature" coherence length. In the dirty case

�(T ) =�p�0`

2p3

(1� T=Tc)� 1

2 (88)

where ` = vF � is the electron mean free path. The impurity parameter can be expressed

through the ratio of �0 and `:

Tc�=~ = `=2��0 (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�r� 2ie

~cA

�2

�+���j�j2=�2GL = 0 (90)

which contains only one parameter �.

Let us put � = j�jei�. The complex equation (90) gives two real equations. One is

�2�r2j�j � 4e2

~2c2Q2j�j

�+ j�j � j�j3=�2

GL = 0 (91)

Here we introduce the gauge invariant vector potential

Q = A� ~c

2er� (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 de�nition of the momentum

operator Eq.(72) we introduce the superconducting velocity operator

2mvs = P (94)

for a Cooper pair with the mass 2m. Now the current becomes

j = �e2Ns

mc

�A� ~c

2er��= Nsevs (95)

where

vs =~

2m

�r�� 2e

~cA

�= � e

mcQ (96)

and

Ns = 8m j�j2 (97)

is the density of \superconducting electrons". For low currents or magnetic �elds, the order

parameter does not depend on the current, j�j = �GL. In the clean case

36

Ns = 8m �2GL = 8m j�j=� =

8�EF

3

�1� T

Tc

�= 2N

�1� T

Tc

�(98)

where N = p3F=3�2 is the total number of electrons. Equation (98) is the same as Ns =

N�1� T 2

T 2c

�. The last equality in Eq. (98) holds for a simple metal with a parabolic spectrum

�p = p2=2m. For a dirty case

Ns =16�3�EF

21�(3)

TC�

~

�1� T

Tc

�=

4�3

7�(3)

Tc�

~

�1� T

Tc

�N: (99)

It is much smaller than in the clean case: the scattering on impurities impedes the super-

current which e�ectively 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

curl curlA = �4�Nse2

mc2

�A� ~c

2er��

(100)

or

curl curlA = ���2L�A� ~c

2er��

(101)

where we de�ne the characteristic length

�L =

�mc2

4�Nse2

� 1

2

(102)

In SI units,

�L =

�m

�0Nse2

� 1

2

For low currents

�L =

�c2�

32�e2 j�j� 1

2

/�1� T

Tc

�� 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 = � c

4��2L

�A� ~c

2er��

(104)

37

We de�ne the Ginzburg-Landau current

jGL = 4e~ �2GL=� = c2=(8�e�2�) (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

F =

Z ���2j�j4 + h2

8�

�dV + ~

2

ZdS��

�r� 2ie

~cA

��

The surface term vanishes because of the boundary conditions Eq. (81). We get

F =

Z ���2j�j4 + h2

8�

�dV (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

Fsn =

Z "a jj2 + b

2jj4 + 1

2m

������i~r� 2e

cA

�

����2#dV: (107)

The constants a and b satisfy

jajb

=mc2

16�e2�2L

and determine the new order parameter magnitude jGLj2 = jaj =b which is

jGLj2 = 2m �2GL

in terms of the previous de�nition 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

small, i.e., when 1� T=Tc � 1. However, being a mean-�eld theory, the GL theory cannot

be valid very close to the transition temperature because of an increasing magnitude of

uctuations. Let us calculate the average j�j2 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, j�j = �GL + Æj�j, is

ÆF =1

2

�Æ2FÆj�j2

�(Æj�j)2 = 4�3j�j (Æj�j)2

The average can be calculated with the probability of uctuations

P (Æj�j) = C exp f�ÆF=Tg

where the normalization constant C is found fromZC exp f�ÆF=Tg d (Æj�j) = 1 (108)

We obtain

(Æj�j)2� =

ZC exp f�ÆF=Tg (Æj�j)2 d (Æj�j)

=

ZC exp

(�4�3j�j (Æj�j)2

T

)(Æj�j)2 d (Æj�j)

= 2T=j�j�3

This uctuation should be smaller than �2GL, i.e., 2T=j�j�3 � j�j=� or

2T

�3� j�j2

�

This can be written as ����1� T

Tc

������

TcH2

c (0)�30

�2� Gi (109)

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

Gi � T 4c

E4F

� ~4

(�0pF )4(110)

We can see that the Ginzburg number is very small for superconductors where �0pF=~ �102 � 103. Therefore, Eq. (109) can be easily ful�lled for superconductors: In superconduc-

tors, 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 exis-

tence of the large parameter �0pF=~ � �(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

� =�L(T )

�(T )=

��c2

32�~2e2 2

� 1

2

(111)

It is independent of T and is determined by the material characteristics.

For clean superconductors it is

� =

�9�4

14�(3)

� 1

2��a0pF

2�~

��e2=a0EF

�� 1

2 ~c

e2TcEF

(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 � 100K and EF � 1000K. The �ne structure

constant e2=~c = 1=137.

40

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

superconductors. Those with � < 1=p2 belong to the type I, while those with � > 1=

p2

are type II superconductors.

E. Meissner e�ect. Magnetic ux quantization

Eq. (101) describes the Meissner e�ect, 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

@2hy@x2

� ��2L 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

h

hy

x0

S

λ L

FIG. 9. The Meissner e�ect: Magnetic �eld penetrates into a superconductor only over dis-

tances shorter than �L.

B

l

FIG. 10. Magnetic ux through the hole in a superconductor is quantized.

Therefore,

B = H + 4�M = 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 e�ect 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 e�ect 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

2er��dl =

ZS

curlA dS � ~c

2e�� = �� ~c

2e2�n (115)

Here � is the magnetic ux through the contour. The phase change along the closed contour

is �� = 2�n where n is an integer because the order parameter is a single valued function.

Since j = 0 in the bulk, we obtain � = �0n where

�0 =�~c

e� 2:07 � 10�7 Oe � cm2 (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 in ate a vortex loop, Eq. (46), in a superconductor for a maximum

possible super uid velocity vs � ~=m�. Compare it with the temperature Tc. What is the

probability of such a uctuation?

Problem 2

Find the jump of the speci�c heat at the superconducting transition.

Problem 3

Find the behavior of the order parameter near a contact with the normal region. Mag-

netic �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 super uids.

In superconductors, each vortex carries exactly one magnetic- ux quantum. Being magnet-

ically 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 e�ect: 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 e�ect.

In super uids, vortices appear in a container with helium rotating at an angular ve-

locity above a critical value which is practically not high and can easily be reached in

experiment7. Vortices are also created if a super uid 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 super uid and normal

components and produces a \mutual friction" between them. As a result, the super ow

is no longer persistent. Remember the de�nition of super uidity: a liquid is super uid 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 super uidity. This was

realized by Feynman in the early days of super uidity19.

44

A. Transition into superconducting state in a magnetic �eld

Vortices play an important role also in bringing about the super uid 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).

z

yx

H

FIG. 11.

We can linearize the GL equations in a small �:

�2�r� 2ie

~cA

�2

�+� = 0 (117)

The vector potential can be taken in the Landau gaugeA = (0; Hx; 0). The order parameter

depends on x and y. Now we have

@2�

@x2+

�@

@y� 2ieHx

~c

�2

�+ ��2� = 0 (118)

This is the well known Schr�odinger equation for a charge in a magnetic �eld. We put

� = eikyf(x)

and obtain the oscillator equation

@2f

@x2��k � 2eHx

~c

�2

f + ��2f = 0 (119)

with

45

!0 =2eH

mc; E =

~2

2m�2

The energy spectrum E = ~!0(n + 1=2) is

~

2m�2=

2eH

mc

�n+

1

2

�

The highest H = Hc2 is for n = 0:

Hc2 =~c

2e�2=

�0

2��2/ 1� T

Tc: (120)

In SI units,

Hc2 =�0

2��0�2

It is the upper critical magnetic �eld below which the transition into superconducting state

occurs. Comparing it with Hc we observe that

Hc2 =p2�Hc =

p8����2

GL

For type II superconductors with � > 1=p2, 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

f = C exp

"� 1

2�2

�x� ~ck

2eHc2

�2#

(121)

The solution Eq. (121) is centered at x = ~ck=2eH. Actually, the full solution is a linear

combination of these solutions for di�erent k. One can construct a periodic solution in a

form

� =Xn

Cneiqny exp

"� 1

2�2

�x� ~cqn

2eHc2

�2#

(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

�

�x+

p~cq

2eHc2

; y

�= eipqy�(x; y)

One sees that j�j is periodic with the period

X0 =~cpq

2eHc2

The simplest case is realized when all the coeÆcients C are equal, p = 1. The array

forms a rectangular lattice. The periods are

X0 =~cq

2eHc2; Y0 =

2�

q:

The unit cell area is

X0Y0 = �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

X0Y0 = �0=Hc2 = 2��2

which again corresponds to exactly one ux quantum per unit cell. If X0=Y0 = 2=p3 one

obtains a hexagonal lattice (see Fig. 12).

47

0.2

0.10.30.5 0.9

0.95

0.40.6 0.7

0.8 0.9

FIG. 12. Left panel: Lines of constant j�j in a square lattice according to Ref.20. Right panel:

The same for a hexagonal lattice according to Ref.21.

The j�j-pattern has zeroes at the points x = X0=2 +X0n, y = Y0=2 + Y0m, surrounded

by current lines. Indeed, the current is

jx = � ~c2

16��2Le�2GL

�i��@�

@x� i�

@��

@x

�

jy = � ~c2

16��2Le�2GL

���

�i@�

@y+2eHc2x

~c�

���

�i@��

@y� 2eHc2x

~c�

��

To transform this further we use the identity which holds for the function of the type of Eq.

(122):

@�

@x=

��i @@y� 2eHc2x

~c

�� (123)

With help of Eq. (123) we get

jx = � ~c2

16��2Le�2GL

@j�j2@y

(124)

jy =~c2

16��2Le�2GL

@j�j2@x

(125)

These expressions suggest that j�(x; y)j2 is a stream function, i.e., that the current ows

along the lines of constant j�j. If we place the node of j�j 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 j�j 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 belowHc2 such thatHc2�H � Hc2. The solution

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

4�

@hz@y

; jy = � c

4�

@hz@x

we obtain from Eqs. (124,125)

Æhz = � ~cj�0j24�2Le�

2GL

(126)

Therefore, the vector potential becomes A = A0 +A1 where A0 = (0; Hc2x; 0),

49

A1 = (0; (H �Hc2)x; 0) + ÆA

and ÆA is due to the correction to the magnetic �eld induced by the supercurrent such that

Æh = curl ÆA. As a result, h1z = H �Hc2 + Æhz and hz = H + Æhz.

Since the non-disturbed function of Eq.(122) satis�es the linearized GL equation with

A = A0, we obtain for the correction �1 �r� 2ie

~cA

�2

�1 + ��2�1

�2ie

~c

�r� 2ie

~cA

�(A1�0)� 2ie

~cA1

�r� 2ie

~cA

��0

���2j�0j2�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

Zj�0j2(H �Hc2 + Æh) dx dy +��2

GL��2

Zj�0j4 dx dy = 0 (127)

We introduce the average

hj�j2i = S�1Zj�j2 dx dy

Using Eq. (126) we obtain

�2GLhj�0j2i

�1� H

Hc2

�=�1� 1

2�2

�hj�0j4i (128)

It is convenient to introduce the ratio

�A =hj�j4ihj�j2i2 � 1 (129)

The value �A is determined by the structure of the vortex lattice. Now we get

hj�0j2i = 2�2�2GL

(2�2 � 1)�A

�1� H

Hc2

�(130)

Eq. (130) shows that j�j2 has a small magnitude proportional to 1�H=Hc2 if � > 1=p2.

For � < 1=p2 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

Type II superconductors have � > 1=p2 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 = hhzi = H + hÆhzi. Then

Mz =B �H

4�=hÆhzi4�

= � Hc2 �H

4��A(2�2 � 1)(131)

The applied �eld can be expressed through the magnetic induction

H = Hc2 � �A(2�2 � 1)(Hc2 � B)

1 + �A(2�2 � 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 + (2�2 � 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

con�guration: it corresponds to an extremum rather than to a minimum of the free energy.

The stable con�guration 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

quantum, �0 = SB the intervortex distance is r0=� �pHc2=B. The limit of isolated

vortices corresponds to B � Hc2 which can be realized if �� 1.

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

� = �GLf(r)ei'

The vector potential has only a '-component: A = (0; A'; 0). We have for f

�2�@2

@r2+1

r

@

@r� 4e2Q2

~2c2

�f + f � f 3 = 0 (133)

The gauge invariant vector potential in our case is Q = (0; A' � ~c=2er; 0).

Equation (101) becomes

�2Lcurl curlA+ f 2Q = 0 (134)

For r 6= 0 it is

�2Lr2Q� f 2Q = 0 (135)

or

@2Q

@r2+1

r

@Q

@r� Q

r2� f 2Q

�2L= 0 (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

2e�K1(r=�L)

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:

K1(z) =� 2

�z

� 1

2

exp(�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 = curlzQ =1

r

@(rQ)

@r=

~c

2e�2LK0(r=�L)

52

Here K0(z) is the Bessel function of zero order. For z � 1 it is K0(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 � �02��2L

ln�Lr

The logarithm is cut o� at r � � for r� �.

For r � �, Q = �~c=2er and the order parameter satis�es the equation

�2�@2

@r2+1

r

@

@r� 1

r2

�f + f � f 3 = 0 (137)

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

Æf =2e2�2

~2c2Q2 =

�2

2�2LK2

1 (r=�L)

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; j�j 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 λ rLFIG. 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 di�erence between the energy

Eq. (70) in the presence of a vortex and the energy without a vortex and the magnetic �eld.

53

We have

FL =

Z ���j�j2 � j�GLj2

�+�

2

�j�j4 � j�GLj4�

+

������i~r� 2e

cA

��

����2

+h2

8�

#dV (138)

The kinetic energy term contains (4e2=c2)Q2j�j2. Since Q / 1=r for � � r � �L this

gives a logarithmically large contribution at distances r � �L. Simple arguments suggest

that we can thus put j�j = �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

FL =4e2

c2

ZQ2j�j2 d2r = ~ �2

GL ln

��L�

�

=�20

16�2�2Lln

��L�

�=

��~

m

�2mNs

4�ln

��L�

�(139)

The possibility to put j�j = �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 j�j2��2GL and j�j4��4

GL in Eq. (138).

One can check however, that these two terms cancel each other because 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

FL =n2�2

0

16�2�2Lln

��L�

�: (140)

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 = nLFL = (B=�0)FL. The proper thermodynamic

potential in an external �eld H is the Gibbs free energy G = F �HB=4�

54

G =BFL

�0

� BH

4�=

B�0

16�2�2Lln

��L�

�� BH

4�

It becomes negative for H > Hc1 where

Hc1 =�0

4��2Lln

��L�

�: (141)

Therefore, vortices appear for H > Hc1.

We note that

Hc1 = Hc2ln�

2�2= Hc

ln�p2�

(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.

Meissner

VortexNormal

H

T T

HH

H

c

c

c1

c2

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 � � placed in a magnetic �eld

tilted by an angle � from the normal to the �lm, Fig. 16

z

yx

HΘ

FIG. 16.

56

V. LONDON MODEL

The GL equation for the vector potential Eq. (100)

curl curlA+4�Nse

2

mc2

�A� ~c

2er��= 0 (143)

can be extended to a more general case if currents are small and j�j 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 = Nsevs

whence vs = j=Nse and write the free energy as a sum of the kinetic energy of supercon-

ducting electrons and the magnetic energy

F =

Z �Nsmv

2s

2+h2

8�

�dV

Using the Maxwell equation j = (c=4�)curlh, we transform this to the following form [F.

London and H. London, 1935]

F =1

8�

Zd3r

�h2 + �2L (curlh)

2� (144)

Here �L is

�L =

�mc2

4�Nse2

� 1

2

(145)

is the London penetration depth. Variation with respect to h gives the London equation:

h+ �2Lcurl curlh = 0 (146)

this is just a curl of Eq. (143).

57

A. Single vortex

Equation (146) holds if there are no vortices. However, the London equation should be

modi�ed if there is a vortex. For a vortex we have

curlr� = 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 + �2Lcurl curlh = �0Æ(2)(r) (147)

where �0 is the vector along the vortex axis with the magnitude of one ux quantum.

For a system of vortices

h + �2Lcurl curlh = �0

Xn

Æ(2)(r� rn) (148)

where the sum is over all the vortex positions rn.

If h = (0; 0; hz(r)), Eq. (147) gives for � 6= 0:

�2L

�@2h

@r2+1

r

@h

@r

�� h = 0

Which gives h = aK0(r=�L). The constant can be found from Eq. (147) by integrating it

over d2r over a small area with r� �L:

�2L

Zcurlh d l = �0

so that we obtain a = �0=2��2L. Thus

h =�0

2��2LK0

�r

�L

�

It is

h(0) =�0

2��2Lln

��Lr

�

near the vortex axis r� �L.

58

The free energy of the vortex per unit length is

F =1

8�

Z �h�h+ �2Lcurl curlh

�+ div[h� curlh]

�d2r

=1

8�

Zhz�0Æ

(2)(r) d2r +1

8�

Z[h� curlh] dl (149)

The last integral is taken along a remote contour and vanishes. The �rst integral gives

FL =�20

16�2�2Lln� (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 su-

perconductor 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. Supercon-

ductor occupies the half-space x > 0.

The magnetic �eld satis�es the London equation (147), i.e.,

h + �2curl curlh = z�0Æ(2)(r� rL)

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; [curlh]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 = hv + him

The image vortex is the vortex at the point (�xL; 0) having the circulation along the �zdirection. 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 =�0

4�

�He�xL=� +

1

2h2(xL)�H

�(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 =�0

4�

�He�x=� +

�0

4��2ln�

�� 1

2hv(2xL)�H

�

=�0

4�

�He�x=� � 1

2hv(2xL) +Hc1 �H

�(152)

It is shown in Fig. 17 as a function of the vortex position.

60

x

x

x

G

G

G

(a)

(b)

(c)

H - Hc1

c1H - H

c1H - H

λ

λ0

0

0

FIG. 17. The free energy of the vortex as a function of its distance from the surface. (a)

H < Hc1. The energy far from the surface is larger. (b) Hs > H > Hc1. The energy far from the

surface is smaller but there is an energy barrier. (c) H > Hs. The barrier disappears.

Using

hv(2xL) =�0

2��2K0

�2xL�

�

we can calculate the force on the vortex

Fx = � @G@xL

=�0

4��

�He�x=� +

1

2

@hv(2xL)

@xL

�

=�0

4��

�He�x=� � �0

2��2K1

�2xL�

��

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

4��

�H � �0

4��xL

�(153)

61

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���� Hc

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 =

Zeiq�rhq

d2q

(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,

rn = (ka; mb). Equation (148) gives

hq =�0

1 + �2Lq2

Xn

e�iq�rn (154)

The free energy becomes

F =1

8�

Z �1 + �2Lq

2�hqh�q

d2q

(2�)2=

�20

8�

Z1

(1 + �2Lq2)

Xn;m

eiq(rn�rm) d2q

(2�)2(155)

We note that

Xn

eiqrn = (2�)2nLÆ(2)(q�Q) (156)

where Q is any vector of the reciprocal lattice. They are de�ned (see Fig. 18) in such a way

that Q = Q1K +Q2M where K and M are integer and

62

Q1 =2�[b� z]

([a� b] � 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

reciprocal lattice is rn � Q = 2�N 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 S0A0 = (2�)2.

a

b

Q

Q

2

1

FIG. 18. The vectors of the direct (full lines) and reciprocal (dashed lines) lattices.

Using Eq.(156) we can write

Xn

Zeiqrnf(q)

d2q

(2�)2=XQ

nLf(Q) (158)

Therefore, the free energy becomes

F =SnL�

20

8�

Z1

(1 + �2Lq2)

Xn�m

eiqRn�md2q

(2�)2=Sn2L�

20

8�

XQ

1

1 + �2LQ2

(159)

where S is the area of the sample. The energy per unit volume becomes

F =B2

8�

XQ

1

1 + �2LQ2=B2

8�

1 +

XQ6=0

1

1 + �2LQ2

!(160)

To simplify calculation of the sum, we replace it with the integral according to the rule

XQ

) S0

Zd2Q

(2�)2

where S0 = �0=B is the area of the unit cell of the real lattice.

Now we have

63

F =B2

8�

�1 + S0

Z Qmax

Q>Q0

d2Q

(2�)21

1 + �2LQ2

�=B2

8�+

B�0

32�2�2Lln

��2

1 + �2LQ20

�(161)

Here Q0 is the �rst reciprocal-lattice vector in the approximation of a round cell: S0�Q20 =

4�2, i.e., Q20 = 4�B=�0. We putQmax = 1=� since the maximumwave 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

H = B +�0

8��2Lln

��2

1 + �2LQ20

�� B

2(1 + �2LQ20)

(162)

Magnetization 4�M = B �H becomes a logarithmic function

M = � �0

32�2�2Lln

��2

�2LQ20

�= � �0

32�2�2LlnHc2

B(163)

for �LQ0 � 1. On the other hand,

@B

@H!1 (164)

for B ! 0 as it follows from Eq. (162). The magnetization of a type II superconductor is

shown in Fig. 19.

− 4π

M

H Hc1 c2HHc

FIG. 19. Full line: Magnetization of a type II superconductor. The linear part at low �elds

corresponds to the full Meissner e�ect Eq. (114). The vertical slope at H = Hc1 is determined

by Eq. (164). The linear part near Hc2 follows from Eq. (131). The magnetization for �elds

Hc1 < H < Hc2 is logarithmic as a function of H, Eq. (163). Dashed line: Magnetization of a type

I superconductor. The Meissner e�ect persists up to the thermodynamic critical �eld Hc.

64

D. Uncharged super uids

There is only one known uncharged Fermi super uid, namely, 3He. 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 super uid 3He.

In this section we discuss only some selected topics. We note �rst that the mean-�eld GL

expansion for 3He 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 super uids. In an uncharged super uid,

the London screening length �L = 1. Vortices in uncharged super uids are created by

rotating the container with a super uid. 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 super uid. One �nds

F � L � = F � [� r] � ÆFÆvs

= F (vs �� r)� m

2

Z[� r]2 jj2 dV

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 super uidity 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 super uids, where pairs of particles are super uid, 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

cA ! 2m � r

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 de�nition of superconducting velocity vs used for superconductors is equiv-

alent to the super uid velocity measured in the rotating frame

vs $ ~

2mr� � � r

The results obtained for vortices in superconductors can be, for some cases, generalized

to uncharged super uids (more speci�cally, to super uid 3He) if one takes the limit e ! 0

which implies �L ! 1. Thus, an uncharged super uid 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 de�nes the rotation velocity

c2 =~

4m�2

for which the vortex cores overlap.

The last expression in Eq. (139) contains the circulation quantum of super uid velocity

around a vortex �0 = (�~=m) and the super uid mass density mNs. Note that this expres-

sion 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 super uids

with vortices. For an uncharged super uid, �L ! 1, therefore, the logarithm is cut o� at

intervortex distances

66

FL =�20mNs

4�ln

�r0�

�(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 =~

2mR2ln�R�

�

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 time-

dependent 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 ful�lled 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 di�erently 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 satis�ed in a very narrow vicinity of Tc. We will not go into microscopic details here.

A justi�cation 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:

ÆFsn

��= 0 ;

ÆFsn

ÆA+ c�1j = 0 (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

��@�@t

=ÆFsn

��(167)

where � 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+~c

2erf ; �! �� ~

2e

@f

@t(168)

without changing the electromagnetic �elds

E = �1

c

@A

@t�r�; h = curlA (169)

In superconductors, the gauge transformation Eq. (168) is accompanied by the transforma-

tion 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

���@�

@t+2ie�

~�

�=ÆFsn

��(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� jn which accounts for thefact that a part of current can be produced by normal electrons in presence of nonequilibrium

electric �eld:

1

c(j� jn) = �ÆFsn

ÆA(172)

where the normal current is jn = �nE. The electric �eld is

E =

��1

c

@A

@t�r�

�

=

��1

c

@Q

@t�r�

�(173)

We introduce here the scalar gauge invariant potential � in addition to the vector potential

Q de�ned by Eq. (92) so that the full set of gauge invariant potentials becomes

69

Q = A� ~c

2er� ; � = �+

~

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 e�ect later.

As we know, the vector potential Q is related to the superconducting velocity

Q = �mcevs (175)

The physical meaning of the scalar potential � can be seen from Eq. (173). According to

it,

@vs@t

=e

m(E+r�) (176)

Since the electric �eld already includes the gradient of the scalar potential �, the quantity

�(e=m)� = � ~

2m

@�

@t� (e=m)�

is the di�erence 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 super uids

if we put A = 0 for an uncharged super uid and neglect small terms of the second order

70

in vs. The term v2s is indeed small in superconductors: Its maximum possible magnitude

is of the order ~2=m2�2 � v2max � �2=p2F while (~=m)@�=@t � ~j�j=m�. We shall see that

~j�j=� � �2=Tc. Therefore, the ratio of these two terms is � Tc=EF � 1.

Finally, the TDGL equations take the form

���@�

@t+2ie�

~�

�= �j�j�+ �j�j2�� ~2

�r� 2ie

~cA

�2

�; (178)

and

j = jn + js (179)

where, as before, the supercurrent is (compare with Eq. (80))

js = �4e2

c

���

�A+

i~c

2er��+�

�A� i~c

2er���

�= �8e2

cj�j2Q (180)

Note that it is the total current j that enters the Maxwell equation (73).

A. Microscopic values

Microscopic theory justi�es 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 = v2F �=3 is the di�usion coeÆcient. Therefore, the relaxation time for current is

�j =��n

8e2 j�j =7�(3)~

4�3Tc

�1� T

Tc

��1(182)

For strong inelastic scattering, the microscopic theory24;25;23 gives

� = ��~=8Tc (183)

71

B. Discussion of TDGL equations

Equation (178) looks like a GP equation with one important di�erence: 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) de�ne two characteristic relaxation times. Separat-

ing real and imaginary parts of Eq. (178) we obtain for the real part

��@j�j@t

= �j�j + �j�j3 � ~2�r2j�j � 4e2

~2c2Q2j�j

�

or

��@j�j@t

= j�j � j�j3

�20

+ �2�r2 � 4e2

~2c2Q2

�j�j (184)

where we introduce the order parameter relaxation time

�� =�

j�j /�1� T

Tc

��1

(185)

The equation for the total current

j = ���1

c

@Q

@t+r�

�� 8e2 j�j2Q

c(186)

determines the relaxation time for the vector potential (or current)

�j =��n

8e2 j�j =�n

8e2 �2GL

/�1� T

Tc

��1(187)

One can de�ne the ratio of these two relaxation times

u = ��=�j (188)

which is independent of T .

For the microscopic value of �, the order-parameter relaxation time is

72

�� =�~

8Tc

�1� T

Tc

��1

Ratio of the two times is u = �4=14�(3) � 5:79.

The imaginary part of Eq. (178) gives

c��j�j2� + �2r �Qj�j2� = 0 (189)

Eq. (189) can also be obtained by calculating the variation of the free energy with respect

to the order parameter phase

2ÆFsn

�� i

��ÆFsn

Æ���� ÆFsn

��

�

= �i~ "�

�r +

2ie

~cA

�2

�� ���

�r� 2ie

~cA

�2

�

#

= i~2 div

���

�r� 2ie

~cA

����

�r +

2ie

~cA

���

�

= � ~

2ediv js (190)

With help of Eq. (171) we obtain from Eq. (190)

ÆFsn

�= ��j�j2

�@�

@t+2e�

~

�= � ~

4ediv 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 deriva-

tive of the total free energy, Fsn plus the magnetic energy we obtain the energy balance

equation

@Ftot

@t+

Zdiv jF = �

ZW dV (192)

where the free energy current is

jF = � ~2��

@�

@t+2ie��

~

��r +

2ie

~cA

��� + c:c:

�+

c

4�[E� h] (193)

73

and the dissipation function is

W = 2�

�����@�

@t+2ie��

~

�����2

+ �nE2 (194)

It is a positively de�nite function consisting of two contributions. First comes from the en-

ergy 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

divE = �r2� = 4�e(Ne � 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

� = �e� such that

ÆN = Ne � Ni =@Ne

@�� = �2�e�

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

r2� = 8�e2��

It demonstrates that the potential � decays at distances of the order of the Debye screening

length

�D =�8�e2�

��1=2which 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)

@t+ div j = 0

Eq. (191) now gives

2e�j�j2~

� = � ~

4ediv jn =

~�n4e

�r2� +

1

cr@Q@t

�(196)

or

l2E

�r2� +

1

cr@Q@t

�=j�j2�2

0

� (197)

This equation de�nes 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

de�ned 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 super uidity.

75

I II +Is sn n

SN

I I

I

I

n

n

s

tot

x0FIG. 20. A boundary between normal metal and superconductor. Normal current converges

into supercurrent at distances lE while the total current is constant.

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�2E �

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 superconduc-

tor. Normal current

jn = �nE 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 divE 6= 0. An oscillating, transverse electric �eld, divE � 0, behaves di�erently. 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 = ���1

c

@Q

@t+r�

�� c

4��2LQ

Eq. (197) can be satis�ed with � = 0. The electric �eld decays into the bulk at distances

of the order of (see Problem 1 to Sect. VII)

1

�2eff=

�1

�2L+

1

�2skin

�

where

�skin = (1 + i)cp8��!

is the skin depth in the normal state.

z

yx

E

FIG. 21. A radio-frequency wave incident on the superconductor.

If !�� � 1 one has �eff = �L, and the �eld is expelled due to the Meissner e�ect. On the

contrary, if !�� � 1 one has �eff = �skin. The penetration is determined the normal-state

skin e�ect.

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

� = �0feikx=�

The supercurrent takes the form

j = jGLf2k (201)

where jGL is de�ned by Eq. (105). The order parameter equation (184) becomes

��@f

@t= f � f 3 � j2

j2GL

1

f 3(202)

The time independent solution is

j

jGL= f 2

p1� 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

maximum jmax=jGL = 2=(3p3) at f =

p2=3 or k = 1=

p3 which coincides with Eq. (63).

Let us study the stability of this solution. We put f = f0 + f1 where f0 satis�es Eq.

(203). We �nd for the deviation f1 by linearizing Eq. (202)

��@f1@t

= 2�2� 3f 20

�f1 = �2f1 @ (j=jGL)

@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 >p2=3 or 1 < k <

p1=3. Equivalently, the

solution is stable when

78

@j

@k> 0

(see Fig. 23). The solution in the range 0 < f0 <p2=3 or

p1=3 < k < 1 is unstable.

With increasing current, the order parameter decreases from f = 1 until the minimum

value f =p2=3 is reached. With further increase in the current, the stability is lost, and

superconductivity is destroyed.

j/jGL

k13−1/20

2/33/2

j/jGL

f1(2/3)1/20

2/33/2

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 di�erent 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

FL =�0

c

hjtr � h

i(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 e�ect.

F

F

F

env

E

FL

v L

j tr

FIG. 24. Forces exerted on a moving vortex.

80

We already mentioned in Section II that vortices in ideal super uids 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� vLt) + �1 ; A(r) = Ast(r� vLt) +A1; (206)

where �st andAst are the values for a stationary vortex, and �1 andA1 are small corrections.

Therefore, within the main terms in the vortex velocity vL, we can write

@�

@t= �(vL � r)�st ;

@A

@t= �(vL � r)Ast: (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 = �1

c

@A

@t�r� = �1

c

@Q

@t�r�� ~

2e

�@

@tr��r@�

@t

�: (208)

The last term here is nonzero for a moving vortex. Indeed,

� ~

2e

�@

@tr��r@�

@t

�=

~

2e[(vL � r)r��r(vL � r)�]

= � ~

2e[vL � curlr�]

= ��0

c[vL � z]sgn(e)Æ(2)(r): (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

e

@vs@t

= E+r� +�0

c[vL � z]sgn(e)Æ(2)(r) (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

@t= �(vL � r)Qst

the average of this term vanishes because the gauge-invariant vector potential Qst 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

hEiL =1

cB� vL (211)

which is the Faraday's law. The average is de�ned as

h: : :iL = S�10

ZS0

(: : :) dS (212)

where S0 = �0=B is the area occupied by each vortex.

B. Flux ow: Low vortex density

We consider here the limit � � 1. In the limit of individual vortices for � � 1 and

H � Hc2 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

ZW d2r = 2�

Z �����@�

@t+2ie��

~

�����2

d2r +

Z�nE

2 d2r

= 2�

Z "�@j�j@t

�2

+4e2

~2j�j2�2

#d2r +

Z�nE

2 d2r (213)

82

Putting E = �r� we rewrite the second integral in the form

Z�nE

2 d2r = �Z�nEr� d2r =

Z�n� divE

= �2�Z

4e2

~2j�j2�� d2r

Here we omit the surface contribution and make use of Eq. (196). Inserting this into Eq.

(213) we obtain

ZW d2r = 2�

Z "�@j�j@t

�2

+4e2

~2j�j2� (�� �)

#d2r

= 2�

Z "�@j�j@t

�2

+2e

~j�j2�@�

@t

#d2r

= 2�

Z �[(vL � r) j�j]2 � 2e

~j�j2� (vL � r)�

�d2r (214)

The two terms under the integral in Eq. (214) represent two di�erent dissipation mech-

anisms 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 j�j = �GLf(r) for a static vortex satis�es Eq. (137).

The gradient of j�j has only a component along the radius-vector in the cylindrical frame

(r; '; z):

vL � rj�j = �GLvLrdf

dr

The gauge invariant potential � obeys Eq. (197) whence

r2� � l�2E f 2� = 0

The scalar potential � should be �nite. Therefore, we require

�! � ~

2e(vL � r)� (215)

83

for r ! 0. We put r = �~r and

� = �~vL'�

2e�(216)

where ' is the azimuthal angle. The function �(~r) satis�es the equation�d2�

d~r2+1

~r

d�

d~r� �

~r2

�� uf 2� = 0: (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 vL.

The integral over the angle ' in Eq. (214) can be calculated using

vLr = vL cos' ; vL' = �vL sin'

Equation (214) gives ZW d2r = 2���2

GLav2L (218)

where

a =

Z "�df

d~r

�2

+f 2�

~r

#~r d~r: (219)

Equation (218) can be written in terms of the vortex viscosity �ZW d2r = �v2L (220)

where

� = 2���2GLa

The moving vortex experiences a friction force

Ffrict = ��vL (221)

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

�0

c

hjtr � h

i= �vL (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 = �f hEiL (223)

where the ux ow conductivity is

�f =2�c2��2

GLa

B�0

: (224)

With the microscopic values of � and �GL taken for dirty superconductors, we �nd

�f =ua

2�n

Hc2

B: (225)

The same result can be obtained in a di�erent way. The dissipation in a volume occupied

by one vortex can be written asZW d2r = �f hEi2L S0 = �f hEi2L �0=B = �fB�0v

2L=c

2

where we use

hEi2L = B2v2L=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�nHc2

B: (226)

It is close to the expression �f = �nHc2=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 =�n1:45

B

Hc2� 0:69�n

B

Hc2

The ux- ow resistivity is linear in B.

85

C. Flux ow: High vortex density

Consider now high �elds, H ! Hc2. We put A = (0; Hx; 0), and choose the static

potential in the form � = �Exx, 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 satis�es the linearized GL equation

@2�

@x2+

�@

@y� 2ieHx

~c

�2

�+ ��2� = 0 (227)

(see Section IV). It has the form Eq. (122):

� =Xn

Cneiqny exp

"� 1

2�2

�x� ~cqn

2eH

�2#: (228)

We calculate the average dissipative function Eq. (194):

hW iL = 2�

��@�

@t� 2ieExx

~�

��@��

@t+2ieExx

~��

��L

+ �nE2x

= 2�v2y

��@�

@y� 2ieHx

~c�

��@��

@y+2ieHx

~c��

��L

+ �nE2x (229)

To calculate the average we make use of the identities

@�

@x=

��i@�

@y� 2eHx

~c�

�;@��

@x=

�i@��

@y� 2eHx

~c��

�; (230)

derived for the order parameter Eq. (228).

We have ��@�

@y� 2ieHx

~c�

��@��

@y+2ieHx

~c��

��L

=

�@�

@x

@��

@x

�L

:

Therefore ��@�

@y� 2ieHx

~c�

��@��

@y+2ieHx

~c��

��L

= �1

2

*�

"�

�@

@y� 2ieHx

~c

�2

�+@2�

@x2

#+L

=1

2�2j�j2�

L: (231)

We use here Eq. (227). We obtain from Eq. (229)

86

hW iL = ���2v2yj�j2�

L+ �nE

2x

=

��n +

�c2

�2H2

j�j2�L

�E2x = �fE

2x (232)

where the ux ow conductivity is

�f = �n + 4e2~�2��2j�j2�

L: (233)

The average magnitude of the order parameter is from Eq. (130)

j�j2�L=

2�2�2GL

(2�2 � 1)�A

�1� B

Hc2

�:

Therefore

�f = �n +4e2~�2��2�2

GL

�A[1� 1=(2�2)]

�1� B

Hc2

�

= �n

�1 +

u

2

(1� B=Hc2)

�A[1� (1=2�2)]

�: (234)

The ux- ow resistivity is

�f = �n

�1� u

2

(1�B=Hc2)

�A[1� (1=2�2)]

�

It is linear in B. The slope is

@�f@B

����B=Hc2

=�nHc2

u

2�A[1� (1=2�2)]� 2:5

[1� (1=2�2)]

�nHc2

The ux- ow resistivity is shown in Fig. as a function of the magnetic induction.

ρ /ρ

B/H

f n

c2

1

1

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 = A0e�i!t.

Problem 2

Find the solution of linearized TDGL equation in a uniform electric �led with a potential

� = �Exx 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 or-

der 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) =Xp

�peip�r~

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

Xp

� (L1L2L3)�1

Xnx;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

Xp

(3D)= (L1L2L3)

�1X

nx;ny;nz

= (L1L2L3)�1

ZL1L2L3

(2�~)3dpx dpy dpz =

Zd3p

(2�~)3

In a �lm with a thickness L1 = d � �, the number nx can only have one value nx = 0.

Therefore,

Xp

(2D)= (dL2L3)

�1Xny;nz

= (dL2L3)�1

ZL2L3

(2�~)2dpy dpz = d�1

Zd2p

(2�~)2

89

Similarly, in the one dimensional case, i.e., for a wire with the cross-section area S � �2

Xp

(1D)= S�1

Zdp

2�~

In the absence of the electric and magnetic �elds, the free energy of the superconductor

is, within the leading terms in small j�j

Fsn =

Z ��j�j2 + ~2 jr�j2� dV

=Xp

�� + p2

� j�pj2 (235)

Note that � > 0 above Tc. The probability of uctuations is

Pf�g = C exp (�Fsn=T )

= C exp

�T�1

Xp

��+ p2

� j�pj2!

= CYp

exp��T�1

�� + p2

� j�pj2�

(236)

For each p we have

Pf�pg = Cp exp��T�1

�� + p2

� j�pj2�

(237)

and C =Q

pCp. The normalization constant is

C�1p =

Z 1

0

exp��T�1

�� + p2

� j�pj2�d j�pj

The average uctuation is

j�pj2�= Cp

Z 1

0

j�pj2 exp��T�1

�� + p2

� j�pj2�d j�pj

=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

���@�

@t+2ie�

~�

�= ��� ~2r2� (239)

90

with � = �E � r. We put � = �(0)p + �

(1)p where �

(0)p is the order parameter without the

electric �eld, and �(1)p is a correction proportional to E. For a time-independent state, it

satis�es the equation

��+ p2

��(1)p =

2ie�

~(E � r)�(0)

p (240)

where the operator r in the Fourier representation is r = i~@=@p. Therefore,

�(1)p = �2e� �� + p2

��1E � @

@p�(0)p (241)

The uctuation-induced supercurrent is

hjsi = 2e h[�� (�i~r�) +� (i~r��)]i

= 4e Xp

pj�pj2

�= 4e

Xp

pD���(0)

p

��2 +��(0)p �(1)

p +�(0)p ��(1)

p

E(242)

The sum over momenta of the �rst term disappears. The correction gives

hjsi = �8e2 �Xp

p

�+ p2

���(0)p E � @

@p�(0)p +�(0)

p E � @@p

��(0)p

�

= �8e2 �Xp

p

�+ p2

�E � @

@p

D���(0)p

��2E� (243)

Using Eq. (238) we �nd

hjsi = 8Te2 2�Xp

p (p �E)(�+ p2)3

=�e2Tc

~ (T � Tc)

Xp

p (p �E) �4=~2(1 + �2p2=~2)3

(244)

Here we use the microscopic value

� =�~�

8Tc

and the de�nition of the coherence length � = ~p =j�j.

Let us calculate the uctuation supercurrent for a �lm. In the 2D case, Eq. (244) gives

hjsi = 8Te2 2�

d

Zp (p �E)(�+ p2)3

d2p

(2�~)2

= E2Te2 2�

�~2d

Z 1

0

p3 dp

(� + p2)3

= �0E (245)

91

where we use that d2p = pdp d'p andZ 2�

0

pipkd'p2�

=p2

2Æik

The excess uctuation conductivity is

�0 =2Te2 2�

�~2d

Z 1

0

p3 dp

(�+ p2)3=

Tce2�

�~2d�

Z 1

0

x3 dx

(1 + x2)3

=Tce

2�

2�~2d�=

e2

16~d

TcT � Tc

=1

8RQd

TcT � Tc

(246)

Remarkably, the uctuation conductivity in a �lm does not depend on the material param-

eters of the superconductor. We denote here

2~

e2= RQ � 6 k

It is called the quantum resistance.

In three and one dimensions, the uctuation conductivity is

�0 =

8><>:e2T=[32~(T � Tc)�] in 3D

�e2Tc�=[16S~ (T � Tc)] in 1D(247)

The uctuation conductivity diverges as 1=p1� 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 E�ect

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 con�gurations belong to the so called superconducting weak

links.

χ1 χ2

a

L

z

L

χ1 χ2

(a) (b)

S1 S2 S S21

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

~cA

�2

�+���j�j2=�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

� = �1ei�1 [1� f(r)] + �2e

i�2f(r)

The function f(r) also satis�es the Laplace equation

r2f = 0

with the boundary conditions f = 0 in the superconductor 1, and f = 1 in the supercon-

ductor 2.

The supercurrent density Eq. (80) becomes

js = 2e ~ [�i��r�+ i�r��]

= �2ie ~�1�2

�ei(�2��1) � e�i(�2��1)

�rf= 4e ~�1�2 sin (�2 � �1) (rf) (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(rzf) 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(rzf) 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(rzf) dS = A=L

94

Equation (250) describes the d.c. (stationary) Josephson e�ect: the supercurrent through

a weak link is proportional to a sine of the phase di�erence. 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 e�ect in the following sections.

2. A.C. Josephson E�ect

Using the same arguments one can also calculate the normal current. For a short contact,

the gradient term is the largest. Therefore,

jn = ��nr�

and also

div jn = 0

The potential thus satis�es the Laplace equation

r2� = 0

We can put

� = V [1� f(r)]

where V is the voltage across the weak link. The function f satis�es the Laplace equation

with the same boundary conditions as before; it is therefore the same function as was

introduced above. The normal current is

jn = �nVrf(r)

The full normal current becomes

In = V=Rn

where

95

1

Rn= �n

Z(rzf) dS � �nA=L (252)

is the total resistance of the hole in the normal state. For a one-dimensional wire one has

exactly Rn = ��1n L=A.

With the general expression Eq. (252) we can write the critical supercurrent Eq. (251)

in the form

Ic =4e ~�1�2

�nRn=��1�2

4TceRn(253)

where we use the expression = ��D=8Tc~ for dirty superconductors. We see that Ic is

now expressed through the resistance of the link in the normal state.

The total current takes the form

I =V

Rn+ Ic sin' (254)

where the phase di�erence is ' = �2��1. Deep in the superconducting region, the Josephsonequation (199) holds

� = �+~

2e

@�

@t= 0 (255)

Since the full current through the small hole is also small, the current density far from

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 di�erence of the phases at the both sides from the hole obeys

the Josephson relation

~@'

@t= 2eV (256)

This equation describes the so called resistively shunted Josephson junction (RSJ) model

(see Fig. 27).

R JV

I

FIG. 27. The resistively shunted Josephson junction.

96

The full equation for the current is

I =~

2eRn

@'

@t+ Ic sin' (257)

If I < Ic, the phase is stationary:

' = arcsin(I=Ic)

and voltage is zero. the phase di�erence 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

V = Rn

pI2 � I2c (258)

It is shown in Fig. 28.

II

V

c

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

Fsn = dXn

Z "�j�n(r)j2 + �

2j�n(r)j4 + k

������i~r� 2e

cA

��n(r)

����2

+ Fint(n)

#d2r (259)

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

Fint(n) = ?~

2

d2j�n+1 ��nj2

where ? is a constant that describes the strength of the interaction. For d! 0, the sum is

equivalent to

dXn

=

Zdz

Therefore,

98

dXn

Fint(n) =Xn

d ?~2

����@�n

@z

����2

= ?~2

Zdz

����@�n

@z

����2

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

Fint(n) = ?~

2

d2���n+1e

�i�A(n+1;n) ��n

��2 (260)

where

�A(n+ 1; n) =2e

~c

Z zn+1

zn

Az dz (261)

For d! 0, this expression will then go over into the correct expression

Fint(n) = ?~2

�����@�

@z� 2ie

~cAz

�����2

Finally, the free energy of the system of layers is

Fsn = dXn

Z "�j�n(rj2 + �

2j�n(rj4 + k

������i~r� 2e

cA

��n(r)

����2

+ ?~

2

d2���n+1e

�i�A(n+1;n) ��n

��2� d2r (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

ÆFsn

��n

= ��n + �j�nj2�n � k~2

�r� 2ie

~cA

�2

�n(r)

� ?~2

d2��n+1e

�i�A(n+1;n) +�n�1e+i�A(n;n�1) � 2�n

�= 0 (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

ÆFsn = �2ie ?~

dc

Xn

Zd2r��n+1e

�i�A(n+1;n)���

n+1ei�A(n+1;n) ���

n

�� c:c� Z zn+1

zn

ÆAz dz

=2ie ?~

dc

Xn

Zd2r��n+1e

�i�A(n+1;n)��n ���

n+1ei�A(n+1;n)�n

� Z zn+1

zn

ÆAz dz

99

Since

Z zn+1

zn

ÆAz dz � d

2[ÆAz(n+ 1) + ÆAz(n)]

we have

ÆFsn =ie ?~

dcdXn

Zd2r��n+1e

�i�A(n+1;n)��n ���

n+1ei�A(n+1;n)�n

�� [ÆAz(n + 1) + ÆAz(n)]

Changing the summation index n for n� 1 in the term with A(n + 1) we get

ÆFsn =ie ?~

dcdXn

Zd2rÆAz(n)

���n+1e

�i�A(n+1;n)��n ���

n+1ei�A(n+1;n)�n

�+��ne

�i�A(n;n�1)��n�1 ���

nei�A(n;n�1)�n�1

��

Using

jz = �c ÆFÆAz

we �nd

jz = � ie ?~d

���n+1e

�i�A(n+1;n)��n ���

n+1ei�A(n+1;n)�n

�+��ne

�i�A(n;n�1)��n�1 ���

nei�A(n;n�1)�n�1

��

Putting

�n = j�njei�n

we �nally obtain

jz =2e ?~

d

�j�n+1�nj sin

��n+1 � �n � 2e

~c

Z zn+1

zn

Az dz

�

+ j�n�n�1j sin��n � �n�1 � 2e

~c

Z zn

zn�1

Az dz

��(264)

100

B. Anisotropic superconductors

If the distance between the layers is small d! 0, one has for the free energy

Fsn =

Z "�j�(r; z)j2 + �

2j�(r; z)j4 + k

������i~r� 2e

cA

��(r; z)

����2

+ ?

������i~@�

@z� 2e

cAz

�����2#d2r dz (265)

This expression is similar to Eq. (70); the distinction is that the constants k and ? are now

di�erent. These constants determine two coherence lengthes: one in the symmetry plane,

�k, and another perpendicular to the plane, �?,

�k = ~

q k=j�j; �? = ~

p ?=j�j (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 super-

conducting electrons , the free energy has the form

Fsn =

Z "a jj2 + b

2jj4 + 1

2mk

������i~r� 2e

cA

�

����2

+1

2m?

������i~ @

@z� 2e

cAz

�

����2#dV: (267)

The constants a and b determine the magnetic �eld penetration length �k

jajb

=mkc

2

16�e2�2k

for currents owing in the symmetry plane and also

jajb

=m?c

2

16�e2�2?

for currents along the z axis. The coherence lengthes along the plane and perpendicular to

the plane are now also expressed through the e�ective masses

�2k =~2

2mkjaj ; �2? =

~2

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) = Ceiqndfq(x)

The equation for f is obtained from Eq. (263). Using the de�nitions of �k and �? from Eq.

(266) we arrive at the Mathieu equation:

�2k@2f

@x2� 2�2?

d2

�1� cos

�qd+

2eHdx

~c

��f + f = 0 (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

2��k�?(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

�2k@f

@x2� �2?

�q +

2eHx

~c

�2

f + f = 0

102

The lowest-level solution is

f = exp

"� 1

2�2k

�x+

~cq

2eH

�2#

(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 =2eHd2�k~c�?

(271)

For a continuous case,

s =d2

�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

f = 1 +2

s2cos

�2eHdx

~c

�+

1

2s4cos

�4eHdx

~c

�+ : : :

under the condition

s2 =1

1� (d2=2�2?)

This condition determines the upper critical �eld

Hc2 =�0

2�d2

��?�k

�1p

1� (d2=2�2?)(272)

This solution is valid for s� 1, or �?(T )! d=p2.

In a situation such that �?(0) � d, 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 e�ectively

a two-dimensional superconductor at the temperature when �?(T ) = d=p2.

We observe that the upper critical �eld diverges as �?(T ) approaches d=p2 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 e�ect for a non-

singly-connected superconductor (Fig. 10) where the critical temperature is a periodic func-

tion 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:

vs =~

2m

�r�� 2e

~cA

�6= 0

Its circulation around the cylinder isZvs dl = 2�Rvs =

~

2m

�2�n� 2e�

~c

�

since the circulation of the phase can only be 2�n 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

�n� �

�0

�

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

vsmax =~

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

e�ect 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

~cAz

�(273)

where the quantity

jc =4e ?~

dj�1�2j

is called the critical Josephson current. Equation (273) has the same form as the current

through the short weak link.

χ

χ

2

1

FIG. 30. The Josephson junction between two superconductors.

Equation (273) describes the d.c. Josephson e�ect: 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(�ApFd=~), where A is a constant of order

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

H

xy

z

FIG. 31. A long Josephson junction in a magnetic �eld.

Deep in the superconductors

@�

@y=

2e

~cAy

due to the Meissner screening. The �eld decays as

hx(z) =

8><>:He�(z�d=2)=�2 z > d=2

He(z+d=2)=�1 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 d=2

0

hx dz �Z 1

d=2

hx dz = �Hd

2�H�2

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

hx dz +

Z d=2

�1

hx dz = H

�d

2+ �1

�

As a result

H =~c

2e(�1 + d=2)

@�1@y

= � ~c

2e(�2 + d=2)

@�2@y

so that

@'

@y= �2e(�1 + �2 + d)H

~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

4�curlz h = jc sin' or � c

4�

@H

@y= jc sin'

we obtain

~c2

8�e(�2 + �1 + d)

@2'

@y2= jc sin'

or

�2J@2'

@y2= sin' (275)

where

�J =

s~c2

8�ejc(�2 + �1 + d)(276)

is called the Josephson length. We can write Eq. (274) in the form

H = �4�jc�2J

c

@'

@y(277)

Equation (275) is similar to equation of motion for a pendulum. Indeed, the latter has

the form

108

d�2

dt2= �g

lsin �

where the angle � is measured from the bottom. Equation (275) is obtained from it by

replacing � = � � ' and putting ��2J = g=l. This means that the pendulum angle � is

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'

@y2= '

whence (see Fig. 33, curve 1)

' = '0e�y=�J

The magnetic �eld decays as

H(y) = H(0)e�y=�J

where we �nd from Eq. (277)

H(0) =4�jc�J'0

c(278)

Magnetic �eld decays into the junction in a way similar to the Meissner e�ect. The penetra-

tion 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 4�jc�J=c.

109

φ

−π

0

π

2π

Y0 2Y0 y1 2 3Y0

FIG. 33. The phase di�erence ' 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 2�n 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

�2J2

�@'

@y

�2

+ cos' = C (279)

where C is a constant. Eq. (279) gives

�Jp2

Z '

'0

d'pC � cos'

= y (280)

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

H(0) = �4�jc�2j

c

@'(0)

@y=

8�jc�jc

sin'0

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 =8�jc�Jc

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 inde�nitely from �� at the edge

through the values 2�n producing the so called solitons (Fig. 33, curve 2). The phase

solitons are also called the Josephson vortices: the phase di�erence 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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113