Ginzburg-Landau Theory for Superconductivity Masatsugu Sei...

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Ginzburg-Landau Theory for Superconductivity

Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton, Binghamton

(Date April 29, 2012)

__________________________________________________________________

Vitaly Lazarevich Ginzburg ForMemRS (Russian: Виталий Лазаревич Гинзбург; October 4, 1916 – November 8, 2009) was a Soviet theoretical physicist, astrophysicist, Nobel laureate, a member of the Russian Academy of Sciences and one of the fathers of Soviet hydrogen bomb. He was the successor to Igor Tamm as head of the Department of Theoretical Physics of the Academy's physics institute (FIAN), and an outspoken atheist.

http://en.wikipedia.org/wiki/Vitaly_Ginzburg __________________________________________________________________ Alexei Alexeyevich Abrikosov (Russian: Алексей Алексеевич Абрикосов; born June 25, 1928) is a Soviet and Russian theoretical physicist whose main contributions are in the field of condensed matter physics. He was awarded the Nobel Prize in Physics in 2003.

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http://en.wikipedia.org/wiki/Alexei_Alexeyevich_Abrikosov _______________________________________________________________________ Abstract

The original article (2006) were put in my web site http://www2.binghamton.edu/physics/docs/ginzburg-landau.pdf

After six years, I realize that there are some mistakes in this article. In Spring 2012, I have an opportunity to teach Solid State Physics (Phys.472/572). The present article is the revised version of the original article. The Mathematica programs are also revised, since they become old.

Here we discuss the phenomenological approach (Ginzburg-Landau (GL) theory) of the superconductivity, first proposed by Ginzburg and Landau long before the development of the microscopic theory [the Bardeen-Cooper-Schrieffer (BCS) theory]. The complicated mathematical approach of the BCS theory is replaced by a relatively simple second-order differential equation with the boundary conditions. In principle, the GL equation allows the order parameter, the field and the currents to be calculated. However, these equations are still non-linear and the calculations are rather complicated and in general purely numerical. The time dependent GL theory is not included in this article.

So far there have been many excellent textbooks on the superconductivity.1-7 Among them, the books by de Gennes,1 Tinkham,6 and Nakajima5 (the most popular textbook in Japan, unfortunately in Japanese) were very useful for our understanding the phenomena of superconductivity. There have been also very nice reviews,8,9 on the superconducting phenomena. We also note that one will find many useful mathematical techniques in the Mathematica book written by Trott.10 Content

1. Introduction 2. Background

2.1 Maxwell’s equation 2.2 Lagrangian of particles (mass m* and charge q*) in the presence of electric and

magnetic fields

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2.3 Gauge transformation: Analogy from classical mechanics 2.4 Gauge invariance in quantum mechanics 2.5 Hamiltonian under the gauge transformation 2.6 Invariance of physical predictions under the gauge transformation

3. Basic concepts 3.1 London’s equation 3.2 The quantum mechanical current density J 3.3 London gauge 3.4 Flux quantization

4. Ginzburg-Landau theory-phenomenological approach 4.1 The postulated GL free energy 4.2 Derivation of GL equation and current density from variational method

4.2.1 Derivation of GL equation by variational method (Mathematica) 4.2.2 Derivation of current density by variational method (Mathematica)

5. Basic properties 5.1 GL free energy and thermodynamic critical field Hc 5.2 Coherence length 5.3 Magnetic field penetration depth Parameters related to the superconductivity.

6. General theory of GL equation 6.1 General formulation 6.2 Special case

7 Free energy 7.1 Helmholtz and Gibbs free energies 7.2 Derivation of surface energy 7.3 Surface energy calculation for the two cases 7.4 Criterion between type-I and type-II superconductor: 2/1

8. Application of the GL equation 8.1 Critical current of a thin wire or film 8.2. Parallel critical field of thin film 8.3. Parallel critical fields of thick films 8.4 Superconducting cylindrical film 8.5 Little Parks experiments

8.5.1 The case of d« and d«. 8.5.2 The case of d» and d».

9. Isolated filament: the field for first penetration 9.1 Critical field Hc1 9.2 The center of the vortex 9.3 Far from the center of the vortex

10. Properties of one isolated vortex line (H = Hc1) 10.1 The method of the Green’s function 10.2 Vortex line energy 10.3. Derivation of Hc1 10.4. Interaction between vortex lines

11. Magnetization curves 11.1 Theory of M vs H

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11.2 Calculation of M vs H using Mathematica 12. The linearized GL equation (H = Hc2).

12.1. Theory 12.2 Numerical calculation of wave functions

13. Nucleation at surfaces: Hc3

13.1 Theory 13.2 Numerical solution by Mathematica 13.3 Variational method by Kittel

14. Abrikosov vortex states 14.1 Theory 14.2 Structure of the vortex phase near H = Hc2 14.3 Magnetic properties near H = Hc2

14.4 The wall energy at 2/1 . 15. Conclusion References Appendix

A1. Parameters related to the superconductivity A2 Modified Bessel functions

1. Introduction

It is surprising that the rich phenomenology of the superconducting state could be quantitatively described by the GL theory,11 without knowledge of the underlying microscopic mechanism based on the BCS theory. It is based on the idea that the superconducting transition is one of the second order phase transition.12 In fact, the universality class of the critical behavior belongs to the three-dimensional XY system such as liquid 4He. The general theory of the critical behavior can be applied to the superconducting phenomena. The order parameter is described by two components (complex number ie ). The amplitude is zero in the normal phase above a

superconducting transition temperature Tc and is finite in the superconducting phase below Tc. In the presence of an external magnetic field, the order parameter has a spatial variation. When the spatial variation of the order parameter is taken into account, the free energy of the system can be expressed in terms of the order parameter and its spatial derivative of

. In general this is valid in the vicinity of Tc below Tc, where the amplitude is small

and the length scale for spatial variation is long. The order parameter is considered as a kind of a wave function for a particle of

charge q* and mass m*. The two approaches, the BCS theory and the GL theory, remained completely separate until Gorkov13 showed that, in some limiting cases, the order parameter )(r of the GL theory is proportional to the pair potential )(r . At the same time this also shows that q* = 2e (<0) and m* = 2m. Consequently, the Ginzburg-Landau theory acquired their definitive status.

The GL theory is a triumph of physical intuition, in which a wave function )(r is

introduced as a complex order parameter. The parameter 2

)(r represents the local

density of superconducting electrons, )(rsn . The macroscopic behavior of

superconductors (in particular the type II superconductors) can be explained well by this

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GL theory. This theory also provides the qualitative framework for understanding the dramatic supercurrent behavior as a consequence of quantum properties on a macroscopic scale.

The superconductors are classed into two types of superconductor: type-I and type-II superconductors. The Ginzburg-Landau parameter is the ratio of to , where is the magnetic-field penetration depth and is the coherence length of the superconducting phase. The limiting value 2/1 separating superconductors with positive surface

energy )2/1( (type-I) from those with negative surface energy )2/1( (type-II), is properly identified. For the type-II superconductor, the superconducting and normal regions coexist. The normal regions appear in the cores (of size ) of vortices binding

individual magnetic flux quanta *0 /2 qcℏ on the scale , with the charge eq 2*

appearing in 0 a consequence of the pairing mechanism. Since >, the vortices repel and arrange in a so-called Abrikosov lattice. In his 1957 paper, Abrikosov14 derived the periodic vortex structure near the upper critical field Hc2, where the superconductivity is totally suppressed, determined the magnetization M(H), calculated the field Hc1 of first penetration, analyzed the structure of individual vortex lines, found the structure of the vortex lattice at low fields. 2. Background

In this chapter we briefly discuss the Maxwell’s equation, Lagrangian, gauge transformation, and so on, which is necessary for the formulation of the GL equation. 2.1 Maxwell's equation

The Maxwell’s equations (cgs units) are expressed in the form

tcc

tc

EjB

B

BE

E

140

14

, (2.1)

B: magnetic induction (the microscopic magnetic field) E: electric field : charge density J: current density c: the velocity of light H: the applied external magnetic field

The Lorentz force is given by

)](1

[ BvEF c

q . (2.2)

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The Lorentz force is expressed in terms of fields E and B (gauge independent, see the gauge transformation below)).

j t

0 , (equation of continuity) (2.3)

AB , (2.4)

tc

AE

1, (2.5)

where A is a vector potential and is scalar potential. 2.2 Lagrangian of particles with mass m* and charge q* in the presence of electric

and magnetic field

The Lagrangian L is given by

)1

(2

1 *2* Avv c

qmL , (2.6)

where m* and q* are the mass and charge of the particle. Canonical momentum

Avv

pc

qm

L*

*

. (2.7)

Mechanical momentum (the measurable quantity)

Apvπc

qm

** . (2.8)

The Hamiltonian is given by

*2*

**2*

** )(

2

1

2

1)( q

c

q

mqmL

c

qmLH ApvvAvvp .

(2.9) The Hamiltonian formalism uses A and , and not E and B, directly. The result is that the description of the particle depends on the gauge chosen. 2.3 Gauge transformation: Analogy from classical mechanics15,16

When E and B are given, and A are not uniquely determined. If we have a set of possible values for the vector potential A and the scalar potential , we obtain other potentials A’ and ’ which describes the same electromagnetic field by the gauge transformation,

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AA' , (2.10)

tc

1

' , (2.11)

where is an arbitrary function of r.

The Newton’s second law indicates that the position and the velocity take on, at every point, values independent of the gauge. Consequently,

rr ' and vv ' , or

ππ ' , (2.12)

Since Apvπc

qm

** , we have

ApApc

q

c

q**

'' , (2.13)

or

c

q

c

q**

)'(' pAApp . (2.14)

In the Hamilton formalism, the value at each instant of the dynamical variables describing a given motion depends on the gauge chosen. 2.4 Gauge invariance in quantum mechanics

In quantum mechanics, we describe the states in the old gauge and the new gauge as We denote and ' the state vectors relative to these gauges. The analogue of the

relation in the classical mechanics is thus given by the relations between average values.

rr ˆ'ˆ' (gauge invariant), (2.15)

ππ ˆ'ˆ' (gauge invariant), (2.16)

c

q*

ˆ'ˆ' pp . (2.17)

We now seek a unitary operator U which enables one to go from to ' :

8

U' . (2.18)

From the condition, '' , we have

1ˆˆˆˆ UUUU . (2.19)

From the condition, rr ˆ'ˆ' ,

rr ˆˆˆˆ UU , (2.20)

or

pr

ˆ

ˆ0]ˆ,ˆ[

U

iU ℏ . (2.21)

U is independent of p .

From the condition, c

qpp ˆ'ˆ' ,

c

qUU pp ˆˆˆˆ , (2.22)

or

Ui

Uc

qU ˆ

ˆˆ]ˆ,ˆ[

rp

ℏ

. (2.23)

Now we consider the form of U . The unitary operator U commutes with r and :

)exp(ˆ*

ℏc

iqU

. (2.24)

The wave function is given by

rrr )exp(ˆ'*

ℏc

iqU . (2.25)

For the wave function, the gauge transformation corresponds to a phase change which varies from one point to another, and is not, therefore, a global phase factor. Here we show that

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ApApc

q

c

q**

ˆ''ˆ' , (2.26)

ApApApc

q

c

q

c

q

c

q****

ˆ)(ˆ''ˆ' ,

(2.27) since

c

q*

ˆ'ˆ' pp , (2.28)

and

)(ˆ'ˆ'''***

AAA

c

qU

c

qU

c

q. (2.29)

2.5 Hamiltonian under the gauge transformation

Ht

i ˆℏ , (2.30)

''ˆ' Ht

i ℏ , (2.31)

UHUt

i ˆ'ˆˆ ℏ , (2.32)

or

UHt

Ut

Ui ˆ'ˆ]ˆ

ˆ[

ℏ . (2.33)

Since Utc

iq

t

U ˆˆ *

ℏ,

UHt

iUUtc

q ˆ'ˆˆˆ*

ℏ , (2.35)

or

UHHUUtc

q ˆ'ˆˆˆˆ*

, (2.36)

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or

HUUtc

qUH ˆˆˆˆ'ˆ

*

. (2.37)

Thus we have

UHUtc

qH ˆˆˆ'ˆ

* . (2.38)

Note that

c

qUU pp ˆˆˆˆ , (2.39)

or

)ˆ(ˆˆˆ c

qUU pp , (2.40)

or

Uc

qUU ˆˆˆˆˆ pp , (2.41)

or

)ˆ(ˆˆˆ

c

qUU pp . (2.42)

From this relation, we also have

)'ˆ()ˆ(ˆ)ˆ(ˆ****

ApApApc

q

c

q

c

qU

c

qU , (2.43)

2*

2*

)'ˆ(ˆ)ˆ(ˆ ApApc

qU

c

qU . (2.44)

Then

Uqc

q

mU

tc

qH ˆ])ˆ(

2

1[ˆ'ˆ *2

*

*

*

Ap , (2.45)

or

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')'ˆ(2

1)'ˆ(

2

1'ˆ *2

*

**2

*

*

*

qc

q

mq

c

q

mtc

qH

ApAp . (2.46)

Therefore the new Hamiltonian can be written in the same way in any gauge chosen. 2.6 Invariance of physical predictions under a gauge transformation

The current density is invariant under the gauge transformation.

]ˆRe[1 *

* ApJc

q

m , (2.47)

Uc

qU

c

q ˆ)'ˆ(ˆ''ˆ'**

ApAp , (2.48)

ApApApc

q

c

q

c

qU

c

qU

****

ˆ'ˆˆ)'ˆ(ˆ , (2.49)

]ˆRe[1

]''ˆ'Re[1

'*

*

*

* ApApJc

q

mc

q

m . (2.50)

The density is also gauge independent.

22'' rr . (2.51)

3. Basic concepts

3.1 London’s equation

We consider an equation given by

Avpc

qm

** . (3.1)

We assume that 0p or

0*

* Avc

qm . (3.2)

The current density is given by

AvJcm

qq *

22*2*

. (London equation) (3.3)

This equation corresponds to a London equation. From this equation, we have

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BAJcm

q

cm

q*

22*

*

22* . (3.4)

Using the Maxwell’s equation

JBc

4 , and 0 B ,

we ge

BJB 2*

2**44)(

cm

qn

c

, (3.5)

where

*n 2

= constant (independent of r)

2**

2*2

4 qn

cmL

: (L; penetration depth)

Then

BBBB 22 1

)()(L

, (3.6)

or

BB 22 1

L . (3.7)

In side the system, B become s zero, corresponding to the Meissner effect. 3.2 The quantum mechanical current density J

The current density is given by

AJcm

q

im

q*

22***

*

*

][2

ℏ. (3.8)

Now we assume that

ie . (3.9)

Since

13

2** 2i , (3.10)

we have

sqc

q

m

qvAJ

2**

2

*

*

)( ℏ

ℏ (3.11)

or

smc

qvA

**

ℏ . (3.12)

This equation is generally valid. Note that J is gauge-invariant. Under the gauge transformation, the wave function is transformed as

)()exp()('*

rr

c

iq

ℏ . (3.13)

This implies that

c

q

ℏ

*

' , (3.14)

Since AA' , we have

)()]()([)''('****

AAAJℏ

ℏℏℏ

ℏℏ

ℏc

q

c

q

c

q

c

q

. (3.15)

So the current density is invariant under the gauge transformation.

Here we note that

)](exp[)()( rrr i . (3.16)

and

])()](exp[)()([})()]({exp[)( rrrrrrrp iii

ii

ℏℏ.

(3.17) If )(r is independent of r, we have

)()]([)( rrrp ℏ (3.18)

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or

)(rp ℏ . (3.19) Then we have the following relation

smc

qvAp

**

ℏ (3.20)

when )(r is independent of r.

3.3. London gauge

Here we assume that *2

sn = constant. Then we have

smc

qvAp

**

ℏ , (3.21)

ssss nqq vvJ**2* . (3.22)

Then

s

s

Jnq

m

c

q**

**

Ap ℏ . (3.23)

We define

2**

*

qn

m

s

. (3.24)

Now the above equation is rewritten as

sqc

qJAp *

*

ℏ , (3.25)

0)()( **

sqc

qJAp , (3.26)

or

0)1

(

sc

JA , (3.27)

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or

01

s

cJB . (3.28)

From the expression

sqc

qJAp *

*

(3.29)

and p = 0, we have a London equation

AJ

c

s

1. (3.30)

For the supercurrent to be conserved, it is required that

01

AJc

s

or

0 A . (3.31)

From the condition that no current can pass through the boundary of a superconductor, it is required that

0nJ s ,

or

0nA . (3.32) The conditions ( 0 A and 0nA ) are called London gauge.

We now consider the gauge transformation:

AA'

'1

' AJ

c

s , (3.33)

0)(1

'1

' 2

AAJcc

s , (3.34)

16

)(1

'1

' nnAnAnJ

cc

s . (3.35)

The London equation is gauge-invariant if we throw away any part of A which does not satisfy the London gauge.

AA' , where 02 and 0 n are satisfied, 3.4 Flux quantization

We start with the current density

ss qc

q

m

qvAJ

2**

2

*

*

)( ℏ

ℏ. (3.36)

Suppose that 2* sn =constant, then we have

AJℏℏ c

q

nq

ms

s

*

**

*

, (3.37)

or

lAlJl dc

qd

nq

md s

sℏℏ

*

**

*

. (3.38)

The path of integration can be taken inside the penetration depth where sJ =0.

ℏℏℏℏ c

qd

c

qd

c

qd

c

qd

****

)( aBaAlAl , (3.40)

where is the magnetic flux. Then we find that

ℏc

qn

*

12 2 (3.41)

where n is an integer. The phase of the wave function must be unique, or differ by a multiple of 2 at each point,

nq

c*

2 ℏ . (3.42)

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The flux is quantized. When |q*| = 2|e|, we have a magnetic quantum fluxoid;

e

ch

e

c

22

20

ℏ = 2.06783372 × 10-7 Gauss cm2 (3.43)

4. Ginzburg-Landau theory-phenomenological approach

4.1 The postulated GL equation

We introduce the order parameter )(r with the property that

)()()(* rrr sn , (4.1)

which is the local concentration of superconducting electrons. We first set up a form of the free energy density Fs(r),

8)(

2

1

2

1)(

22*

*

42 BAr

c

q

imFF Ns

ℏ

where is positive and the sign of is dependent on temperature. 4.2 The derivation of GL equation and the current density by variational method

We must minimize the free energy with respect to the order parameter (r) and the vector potential A(r). We set

drFs )(r

where the integral is extending over the volume of the system. If we vary

)()()( rrr and )()()( rArArA , (4.4) we obtain the variation in the free energy such that

By setting 0 , we obtain the GL equation

02

12*

*

2

A

c

q

im

ℏ, (4.5)

and the current density

AJcm

q

im

qs *

22***

*

*

][2

ℏ (4.6)

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or

])()([2

***

**

*

AAJc

q

ic

q

im

qs

ℏℏ. (4.7)

At a free surface of the system we must choose the gauge to satisfy the boundary condition that no current flows out of the superconductor into the vacuum.

0 sJn . (4.8)

4.2.1 Derivation of GL equation by variational method

((Mathematica program-1)) Variational method using “VarialtionalD” of the Mathematica. Note that c is the complex conjugate of .

______________________________________________________________________

______________________________________________________________________

Derivation of Ginzburg Landau equation

Clear@"Global`∗"D; << "VariationalMethods`"; Needs@"VectorAnalysis`"D;SetCoordinates@Cartesian@x, y, zDD;A = 8A1@x, y, zD, A2@x, y, zD, A3@x, y, zD<;

eq1 = α H ψ@x, y, zD ψc@x, y, zD L +1

2β Iψ@x, y, zD2 ψc@x, y, zD2 M +

1

2 m

—

�Grad@ψ@x, y, zDD −

q

cA ψ@x, y, zD .

−—

�Grad@ψc@x, y, zDD −

q

cA ψc@x, y, zD êê Expand;

eq2 = VariationalD@eq1, ψc@x, y, zD, 8x, y, z<D êê Expand;

We need to calculate the following

OP1 :=—

�D@ , xD −

q

cA1@x, y, zD & ; OP2 :=

—

�D@ , yD −

q

cA2@x, y, zD & ;

OP3 :=—

�D@ , zD −

q

cA3@x, y, zD & ;

eq3 =

α ψ@x, y, zD + β ψ@x, y, zD2 ψc@x, y, zD +

1

2 mHOP1@OP1@ψ@x, y, zDDD + OP2@OP2@ψ@x, y, zDDD + OP3@OP3@ψ@x, y, zDDDL êê

Expand;

eq4 = eq2 − eq3

0

19

________________________________________________________________________ 4.2.2. Derivation of current density by variational method

((Mathematica Program-2)) ______________________________________________________________________

______________________________________________________________________

5 Basic properties

5.1 GL free energy and Thermodynamic critical field Hc

A = 0 and (real) has no space dependence. Why is real? We have a gauge transformation;

AA' and )()exp()('*

rr

ℏc

iq . (5.1)

We choose

AAA ' with = 0 = constant. Then we have

)()exp()(' 0*

rr

ℏc

iq ,

Derivation of Current density variational method

Clear@"Global`∗"D;<< "VariationalMethods`"; Needs@"VectorAnalysis`"D;SetCoordinates@Cartesian@x, y, zDD;A = 8A1@x, y, zD, A2@x, y, zD, A3@x, y, zD<; H = 8H1, H2, H3<;

eq1 =1

2 m

—


q

cA ψ@x, y, zD . −

—


q

cA ψc@x, y, zD +

1

8 π[email protected]@ADL −

1

4 πH.Curl@AD êê Expand;

eq2 = VariationalD@eq1, A1@x, y, zD, 8x, y, z<D êê Expand;

Current density in Quantum Mechanics

eq3 =

1

c

c

4 πCurl@Curl@ADD +

q2

m cψ@x, y, zD ψc@x, y, zD A −

q —

� 2 mHψc@x, y, zD Grad@ψ@x, y, zDD − ψ@x, y, zD Grad@ψc@x, y, zDDL ;

eq4 = eq3@@1DD êê Expand;

eq2 − eq4 êê Simplify

0

20

or

)(')exp()( 0*

rr

ℏc

iq .

Even if ' is complex number, can be real number.

42

2

1 Ns FF

When 0

F

, Fs has a local minimum at

2/12/1 )/()/( . (5.3)

Then we have

82

22c

Ns

HFF , (5.4)

from the definition of the thermodynamic critical field:

)1)(0(4

2

22/12

c

ccT

THH

(parabolic law). Suppose that is independent of T, then

)1()1)(0(4

2)1)(0(4 02

2

cc

c

c

cT

T

T

TH

T

TH

where

)0(4

20 cH

.

The parameter is positive above Tc and is negative below Tc. Note that >0. For T<Tc, the sign of is negative:

420 2

1)1( tFF Ns

21

where 0>0 and t = T/Tc is a reduced temperature.

Fig.1 The GL free energy functions expressed by Eq.(5.7), as a function of .

0 = 3. = 1. t is changed as a parameter. t = T/Tc. (t = 0 – 2) around t = 1.

Fig.2 The order parameter as a function of a reduced temperature t = T/Tc. We

use 0 = 3 and = 1 for this calculation.

5.2 Coherence length

We assume that A = 0. We choose the gauge in which is real.

y¶

f

O

t=0

0.2

0.6

0.8

0.4

1

2

0.0 0.2 0.4 0.6 0.8 1.0t0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

y¶

22

02 2

2

*

23

dx

d

m

ℏ. (5.8)

We put f .

02

32

2

*

2

ffdx

fd

m ℏ

. (5.9)

We introduce the coherence length

2*

2*

*

2

*

22 2

22c

s

Hm

n

mm

ℏℏℏ

, (5.10)

where

82

22cH

, sn

2 , (5.11)

or

2/1

0*

2

|1|2

cT

T

m

ℏ

Then we have

032

22 ff

dx

fd , (5.13)

with the boundary condition f = 1, df/dx = 0 at x = ∞ and f = 0 at x = 0.

0)( 32

22

dx

dfff

dx

fd

dx

df , (5.14)

or

)24

(2

2422ff

dx

d

dx

df

dx

d

, (5.15)

or

23

2222

)1(41

2f

dx

df

, (5.16)

or

)1(2

1 2fdx

df

. (5.17)

The solution of this equation is given by

2tanh

xf . (5.18)

______________________________________________________________________

((Mathematica))

______________________________________________________________________

Ginzburg-Landau equation; coherence length

Clear@"Global`∗"D; eq1 = f'@xD ==1

2 ξ

I1 − f@xD2M;

eq2 = DSolve@8eq1, f@0D 0<, f@xD, xD;f@x_D = f@xD ê. eq2@@1DD êê ExpToTrig êê Simplify

TanhA x

2 ξE

0 1 2 3 4 5xêx0.0

0.2

0.4

0.6

0.8

1.0

1.2f HxL

24

Fig.3 Normalized order parameter f(x) expressed by Eq.(5.18) as a function of x/.

Fig.4 Derivative )(' xf as a function of x/. 5.3 Magnetic field penetration depth

AJcm

q

im

qs *

22***

*

*

][2

ℏ. (5.19)

We assume that (real).

AJcm

qs *

22*

(London’s equation), (5.20)

BAJcm

q

cm

qs *

22*

*

22*

, (5.21)

where

sc

JB4

, AB , 0 B . (5.22)

Then we have

BBcm

qc*

22*

)4

(

, (5.23)

0 1 2 3 4 5xêx0.0

0.5

1.0

1.5

2.0f 'HxL

25

or

BBBB 2*

22*2 4

)()(cm

q

. (5.24)

Then we have London’s equation

BB 22 1

, (5.25)

where

2*

22*2 4

cm

q

. (5.26)

is the penetration depth

2*

2*

22*

2*

44 q

cm

q

cm

. (5.27)

The solution of the above differential equation is given by

)/exp()0()( xxBxB ZZ , (5.28) where the magnetic field is directed along the z axis.

)0),(,0(4

)(0044

xBx

c

xB

zyx

ccz

z

zyx

eee

BJ . (5.29)

The current J flows along the y direction.

)/exp(4

)0(

x

xcBJ z

y

. (5.30)

26

Fig.5 The distribution of the magnetic induction B(x) (along the z axis) and the

current density (along the y axis) near the boundary between the normal phase and the superconducting phase. The plane with x = 0 is the boundary.

5.4 Parameters related to the superconductivity

Here the superconducting parameters are listed for convenience.

2*

sn ,

2

2*2 444 sc nH , (5.31)

42 4 cH (5.32)

2

2

4

cH, 4

2

4

cH,

*0

2

q

cℏ , (5.33)

2*

2*

4 q

cm

*2m

ℏ

*

*

2 q

cm

ℏ

Then we have

2**

2*

4 qn

cm

s

*

*2

mH

n

c

s ℏ

**

*

22 qn

Hcm

s

c

ℏ , (5.35)

cH

22

1

0

, ℏc

Hq c

*

2

2

, cH

2202 , (5.36)

27

*2 q

cH c

ℏ . (5.37)

We also have

0

222

cH ,

22 0

cH . (5.38)

Note:

eq 2* (>0) mm 2* , 2/*ss nn .

((Mathematica Program-5)) ______________________________________________________________________

28

______________________________________________________________________

n1 = ns*, q1 = q*, a1 = Abs[a], m1 = m* *

rule1 = :α1 →Hc2

4 π n1, β →

Hc2

4 π n12, Φ0 →

2 π — c

q1>; λ =

m1 c2 β

4 π q12 α1;

ξ =—

2 m1 α1

;

κ =λ

ξêê PowerExpand

c m1 β

2 π q1 —

λ1 = λ ê. rule1 êê PowerExpand

c m1

2 n1 π q1

ξ1 = ξ ê. rule1 êê PowerExpand

n1 2 π —

Hc m1

λ ξ

Φ0ê. rule1 êê PowerExpand

1

2 2 Hc π

κ1 = κ ê. rule1 êê PowerExpand

c Hc m1

2 2 n1 π q1 —

κ

λ2ê. rule1 êê PowerExpand

2 Hc q1

c —

κ

λ2Φ0 ê. rule1 êê PowerExpand

2 2 Hc π

29

6. Formulation of GL equation

6.1 General theory2

In summary we have the GL equation;

02

12*

*

2

A

c

q

im

ℏ, (6.1)

and

AAAAJcm

q

im

qccs *

22***

*

*2 ][

2])([

4)(

4

ℏ,

(6.2)

AB . (6.3) The GL equations are very often written in a form which introduces only the following dimensionless quantities.

f , ρr , cH2

Bh , (6.4)

where

2*

22*2 4

cm

q

,

*

2

*

22

22 mm

ℏℏ ,

2

2 4cH ,

2 ,

(6.5)

**0

22

q

c

q

c ℏℏ (since q* = 2e<0), (6.6)

02

2

2

1

cH,

22 02

cH ,

. (6.7)

We have

ffffi

22

~1

A, (6.8)

where

30

AA0

2~

. (6.9)

Here we use the relation

AAAB

h~~

22

1

2

1

20

ccc HHH, (6.10)

or

Ah~

, (6.11)

A

AAAJ

cm

q

im

q

ccs

*

22***

*

*

2

][2

])([4

)(4

ℏ

, (6.12)

AA~

2][

12

)~

(24

20*

22***2

*

*0

2 fcm

qffff

im

qc

ℏ,

(6.13)

AA~

][2

)~

(2** fffff

i

, (6.14)

since

2*

22*2 4

cm

q

(6.15)

and

11

2

2

1

2

2

4

1

2

241

2

24

*0

*0

22*

2*

*

*

0

22

*

*

0

2

q

c

q

c

q

cm

m

q

cm

q

c

ℏ

ℏℏℏ

, (6.16)

2

2

*

2*

4

c

cm

q . (6.17)

31

In summary we have the following equations:

ffffi

22

~1

A, (6.18)

AA~

][2

)~

(2** fffff

i

, (6.19)

Ah~

. (6.20)

Gauge transformation:

0AA , (6.21)

)()exp()( 0

*

rc

iqr

ℏ , (6.22)

AA0

2~

, f , (6.23)

)2

(1~12~2~~

00

00

00

AAAA , (6.24)

fi

fc

iqf )

2exp()exp(

00

*

ℏ. (6.25)

Here we assume that 00

2

. Then we have

00 )exp( fif , 00

1~~

AA . (6.26)

We can choose the order parameter f as a real number f0 such that f0 is real, f0 is a constant.

0

~~AA .

3000

2

0

~1fff

i

A, (6.27)

02

00

~)

~( AhA f , (6.28)

0

~Ah , (6.29)

32

)]~

()~

([1~1

~1

00000

2

002

2

3000

2

0

ffi

ff

fffi

AAA

A

. (6.30)

Using the formula of the vector analysis

02

000002

002

0

02

00

~~2

~~

)~

(0)]~

([

AAAA

AA

fffff

f (6.31)

or

0000

~~2 AA ff , (6.32)

We also have

000000

~~)

~( AAA fff . (6.33)

We have

0

2

002

2

00000

2

002

2

300

~1

]~

2~

[1~1

ff

ffi

ffff

A

AAA

, (6.34)

or

3000

2

002

2

~1ffff A

. (6.35)

We also have

02

0

~Ah f , (6.36)

hA 20

0

1~

f, (6.37)

)(1

)(1

]1

[~

20

20

20

0

hh

hAh

ff

f, (6.38)

33

or

)(1

)(2

20

030

hhh f

ff

, (6.39)

or

)()(2

00

20 hhh f

ff . (6.40)

In summary,

300

2

30

02

2

11ff

ff h

, (6.41)

)()(2

00

20 hhh f

ff , (6.42)

hA 20

0

1~

f. (6.43)

6.2 Special case

We now consider the one dimensional case. For convenience we remove the subscript

and r =(x, y, z) instead of . We also use A for 0

~A and f for f0. f is real.

32

32

2

11ff

ff h

, (6.44)

)()(22 hhh ff

f , (6.45)

hA 2

1

f, (6.46)

Ah . (6.47)

34

Fig.6 The direction of the magnetic induction h(x). x, y and z are dimensionless.

)0,,0(

)(00dx

dh

xh

zyx

zyx

eee

h , (6.48)

zdx

hdehhhh 2

222)()( , (6.49)

32

32

2

2

11ff

dx

dh

fdx

fd

, (6.50)

2

22 2

dx

hd

dx

dh

dx

df

fhf for the z component, (6.51)

or

)1

(2 x

h

fxh

, (6.52)

ydx

dh

feA

2

1 . (6.53)

(a) <<1 (type I superconductor)

01

2

2

23

dx

fdff

. (6.54)

The solution of this equation is already given above.

35

(b) >>1 (type-II superconductor)

01

2

33

x

h

fff . (6.55)

7 Free energy

7.1 Helmholtz and Gibbs free energies

The analysis leading to the GL equation can be done either in terms of the Helmholtz free energy (GL) or in terms of the Gibbs free energy. (a) The Helmholtz free energy

It is appropriate for situations in which B =<H>; macroscopic average is held constant rather than H, because if B is constant, there is no induced emf and no energy input from the current generator.

(b) The Gibbs free energy It is appropriate for the case of constant H.

The analysis leading to the GL equations can be done either in terms of the Helmholtz

free energy or in terms of the Gibbs energy. The Gibbs free energy is appropriate for the case of constant H.

HB 41

fg (Legendre transformation). (7.1)

B is a magnetic induction (microscopic magnetic field) at a given point of the superconductors.

The Helmholtz free energy is given by

8)(

2

1

2

1 22*

*

42 BA

c

q

imff ns

ℏ, (7.2)

The Gibbs free energy is expressed by

HBB

A

4

1

8)(

2

1

2

1 22*

*

42

c

q

imfg ns

ℏ. (7.3)

(i) At x =-∞ (normal phase)

8)(

2c

n

Hff , (7.4)

848)(

222c

ncc

nn

Hf

HHfgg . (7.5)

36

(ii) At x = ∞ (superconducting phase)

822

1)(

2242 c

nnnsss

Hffffgg . (7.6)

The interior Gibbs free energy

]4

1

8)(

2

1

2

1[

])([

22*

*

42

cnn

nsg

BHB

c

q

imgfd

ggdE

Ar

rr

ℏ

(7.7) or

])(8

1)(

2

1

2

1[ 2

2*

*

42

cg HBc

q

imdE

Ar

ℏ (7.8)

7.2 Derivation of surface energy

We have a GL equation;

02

12*

*

2

A

c

q

im

ℏ. (7.9)

If one multiplies the GL equation by * and integrates over all r by parts, one obtain the identity

0]2

1[

2**

*

42

Ar

c

q

imd

ℏ, (7.10)

which is equal to

0])(2

1[

2*

*

42 Ar

c

q

imd

ℏ, (7.11)

Here we show that the quantity I defined by

])([2*2*

* AArc

q

ic

q

idI

ℏℏ. (7.12)

37

is equal to 0, which is independent of the choice of and *. The variation of I is calculated as

][ * rdI (7.13)

using the Mathematica (VariationalD). We find that 0 for any and *. I is independent of the choice of and *. When we choose = * = 0. Then we have I = 0. ________________________________________________________________________

((Mathematica Program-6))

________________________________________________________________________

Subtracting Eq.(7.11) from Eq.(7.8), we obtain the concise form

Clear@"Global`∗"D; << "VariationalMethods`"; Needs@"VectorAnalysis`"D;SetCoordinates@Cartesian@x, y, zDD;A = 8A1@x, y, zD, A2@x, y, zD, A3@x, y, zD<;eq1 =

1

2 m

—


q

cA ψ@x, y, zD .

−—


q

cA ψc@x, y, zD êê Expand;

OP1 :=—

�D@, xD −

q

cA1@x, y, zD & ;

OP2 :=—

�D@, yD −

q

cA2@x, y, zD & ;

OP3 :=—

�D@, zD −

q

cA3@x, y, zD & ;

eq2 =

1

2 mψc@x, y, zD

HOP1@OP1@ψ@x, y, zDDD + OP2@OP2@ψ@x, y, zDDD + OP3@OP3@ψ@x, y, zDDDL êêExpand;

eq3 = eq1 − eq2 êê Expand;

VariationalD@eq3, ψc@x, y, zD, 8x, y, z<D0

VariationalD@eq3, ψ@x, y, zD, 8x, y, z<D0

38

])(8

1

2

1[ 24

cg HBdE r . (7.14)

Suppose that the integrand depends only on the x axis. Then we can define the surface energy per unit area as :

])(8

1

2

1[ 24

cHBdx

, (7.15)

which is to be equal to

2

8

1cH

. (7.16)

Then we obtain a simple expression

])1([ 24

4

cH

Bdx

. (7.17)

The second term is a positive diamagnetic energy and the first term is a negative condensation energy due to the superconductivity. When f and B/Hc are defined by

f , cH

Bh

2 . (7.18)

Here we use xx ~ ( x~ is dimensionless). For simplicity, furthermore we use x instead of x~ . Then we have

]2)1(2

1[2])21([ 2424 hhfdxhfdx . (7.19)

7.3 Surface energy calculation for the two cases

(a) <<1 (type-I superconductor)

When h = 0 for x>0 and 44

4

f

with

2tanh

xf , we have

8856.13

24

3

24)1( 4

fdx >0, (7.20)

a result first obtained by Ginzburg and Landau. So the surface energy is positive. _______________________________________________________________________

39


_____________________________________________________________________ (b) >>1 (type-II superconductor) Here we assume that is much larger than .

224 )1(

x

hff .

As h must decrease with increasing x, we have

2/122

)1(1

fx

h

f

, (7.21)

and

2/122

)1()1

( fxx

h

fxh

. (7.22)

Here f obeys the following differential equation. From Eq.(7.22) we have

2/122

2

)1( fxx

h

. (7.23)

Using Eq.(7.21),

22/122/122

2

)1()1( fffxx

h

, (7.24)

or

2/322/1222/122/122

2

)1()1()11()1()1( fffffx

. (7.25)

When 2/12 )1( fu , we have

Clear@"Global`∗"D;I1 = ‡

0

∞1 − TanhB x

2 κ

F4 �x êê Simplify@, κ > 0D &

4 2 κ

3

40

32

2

uuz

u

, (7.26)

constuuz

u

42

2

2

1. (7.27)

When u =0,

0dx

du (at f = 1 (x = ∞) at 0

dx

du).

So we have

)2

11(

2

1 22422

uuuudx

du

, (7.28)

or

2/12)2

11( uu

dx

du , (7.29)

since du/dx must be negative. We solve the problem with the boundary condition; u = 1 at x = 0. The solution for u is

x

x

e

eu

2223

)22(2

.

which is obtained from the Mathematica Program-9 (see below).

]2)1(2

1[ 24 hhfdx , (7.30)

or

])2

1(2)2

1(2[ 2/122

2 uu

uudx , (7.31)

or

dudu

dxuu

uudu

])2

1(2)2

1(2[ 2/122

2 , (7.32)

41

du

uu

uu

uudu

2/12

0

1

2/122

2

)2

11(

1])

21(2)

21(2[

(7.33)

or

0)12(3

4]2)

21(2[

1

0

2/12

duu

udu , (7.34)

Thus, for >>1 the surface energy is negative. _____________________________________________________________________ ((Mathematica Program-8))

_____________________________________________________________________ ((Mathematica Program-9)) Negative surface energy

Clear@"Global`∗"D;

I1 = ‡0

12 u 1 −

u2

2− 2 �u

−4

3I−1 + 2 M

42

Negative surface energy

Clear@"Global`∗"D; eq1 = u'@xD −u@xD 1 −u@xD22

1ê2;

eq2 = DSolve@8eq1, u@0D 1<, u@xD, xD êê Simplify;

u@x_D = u@xD ê. eq2@@1DD êê Simplify

2 I−2 + 2 M �x

−3 + 2 2 − �2 x

h1 = Plot@u@xD, 8x, 0, 5<, PlotStyle → 8Thick, Hue@0D<, PlotPoints → 100,

Background → LightGray, AxesLabel → 8"x", "u"<D;h2 = PlotB 1 − u@xD2 , 8x, 0, 5<, PlotStyle → 8Thick, Hue@0D<, PlotPoints → 100,

PlotRange → 880, 5<, 80, 1<<, Background → LightGray, AxesLabel → 8"x", "f"<F;

F = 2 u@xD2 1 −u@xD22

− 2 u@xD 1 −u@xD22

1ê2êê Simplify;

h3 = Plot@F, 8x, 0, 10<, PlotStyle → 8Thick, Hue@0D<, PlotPoints → 100,

Background → LightGray, AxesLabel → 8"x", "F"<D;Integrate@F, 8x, 0, ∞<D−4

3I−1 + 2 M

% êê N

−0.552285

43

_____________________________________________________________________

Fig.7 Plot of 2/12 )1( fu vs x. x (dimensionless parameter) is the ratio of the

distance to .

F1 = D@F, xD;h4 = Plot@F1, 8x, 0, 10<, PlotStyle → 8Thick, Hue@0D<, PlotPoints → 100,

Background → LightGray, AxesLabel → 8"x", "dFdx"<, PlotRange → AllD;FindRoot@F1 0, 8x, 1, 2<D8x → 1.14622<pt1 = 8h1, h2, h3, h4<; Show@GraphicsGrid@Partition@pt1, 2DDD

0 1 2 3 4 5x

0.2

0.4

0.6

0.8

1.0u

0 1 2 3 4 5x0.0

0.2

0.4

0.6

0.8

1.0f

2 4 6 8 10x

-0.25

-0.20

-0.15

-0.10

-0.05

F

2 4 6 8 10x

-0.3

-0.2

-0.1

0.1

dFdx

1 2 3 4 5x

0.2

0.4

0.6

0.8

1.0

u

44

Fig.8 Plot of 21 uf vs x.

Fig.9 Plot of the surface energy 2/122

2 )2

1(2)2

1(2u

uu

us as a function of

x. ).(xFs

0 1 2 3 4 5x0.0

0.2

0.4

0.6

0.8

1.0f

2 4 6 8 10x

-0.25

-0.20

-0.15

-0.10

-0.05

F

45

Fig.10 Plot of the derivative ds/dx = dF/dx with respect to x. 7.4 Criterion between type-I and type-II superconductor: 2/1

The boundary between the type-II and type-I superconductivity can be defined by finding the value of which corresponds to a surface energy equal to zero.

]2)1(2

1[

224 hhfdx

. (7.35)

It is clear that this value is zero if

hhf 2)1(2

1 24 , (7.36)

or

0)]1(2

1)][1(

2

1[ 22 fhfh (7.37)

or

)1(2

1 2fh (7.38)

Substituting this into the two equations

011

2

32

2

23

x

h

fdx

fdff

(7.39)

2 4 6 8 10x

-0.3

-0.2

-0.1

0.1

dFdx

46

dx

dh

dx

df

fdx

hdhf

22

22 (7.40)

From the calculation by Mathematica, we conclude that 2/1 . _____________________________________________________________________ ((Mathematica Program-10))

_____________________________________________________________________ 8. Application of the GL equation

8.1 Critical current of a thin wire or film1,4

Determination of κ = 1í 2

Clear@"Global`∗"D;eq1 = f@xD − f@xD3 + 1

κ2f''@xD −

1

f@xD3 h'@xD2 0;

eq2 = f@xD2 h@xD h''@xD −2

f@xD f'@xD h'@xD;

rule1 = :h →1

2

I1 − f@ D2M & >;

eq3 = eq1 ê. rule1 êê Simplify;

eq31 = Solve@eq3, f''@xDD êê Expand;

eq32 = f′′@xD ê. eq31@@1DD−κ2 f@xD + κ2 f@xD3 +

2 κ2 f′@xD2f@xD

eq4 = eq2 ê. rule1 êê Simplify;

eq41 = Solve@eq4, f''@xDD êê Expand;

eq42 = f′′@xD ê. eq41@@1DD−f@xD2

+f@xD32

+f′@xD2f@xD

Solve@eq32 eq42, κD99κ → −

1

2=, 9κ →

1

2==

47

Fig.11 The direction of the current density. The direction of current is the x axis. The direction of the thickness is the z axis. The following three equations are valid in general.

ss qc

q

m

qvAJ

2**

2

*

*

)( ℏ

ℏ, (8.1)

8)(

2

1

2

1 22*

*

42 BA

c

q

imff ns

ℏ, (8.2)

)()()( rrr ie . (8.3)

We consider the case when

*2)( snr = constant (8.4)

Then we have

)()()( ***

smc

q

c

q

ivAA ℏ

ℏ, (8.5)

and

822

1 222

*42 B

sns vm

ff . (8.6)

We now consider a thin film. We assume that d<<(T) in order to have *2

sn =

constant. has the same value everywhere. The minimum of fs with respect to is

obtained for

48

02

2*

2 sv

m , (8.7)

for a given vs, where )8/(2 B is neglected. Here and f with

//2 . Then we have

)1(2

2222*

ffvm

s . (8.8)

The corresponding current is a function of f,

22*

2*2* 12

ffm

qvqJ ss

. (8.9)

We assume that 221 1 ffJ . This has a maximum value when 0/1 fJ ;

33

2max1 J at

32

f ,

*

2**

2*

33

22

33

2

mq

mqJ s

ℏ

(8.10)

_____________________________________________________________________ ((Mathematica Program-11))

_____________________________________________________________________

Critical current of a thin wire

Clear@"Global`∗"D; J1 = f2 I1 − f2M1ê2;k1 = Plot@J1, 8f, 0, 1<, PlotStyle → 8Red, Thick<,

Background → LightGray, AxesLabel → 8"f", "J1"<D;H1 = D@J1, fD êê Simplify; H2 = Solve@H1 0, fD

98f → 0<, 9f → −2

3=, 9f →

2

3==

max1 = J1 ê. H2@@3DD2

3 3

49

Fig.12 The normalized current density J1 vs f. 221 1 ffJ .

8.2. Parallel critical field of thin film

Fig.13 Configuration of the external magnetic field H and the current density J. Helmholtz free energy:

822

1 222

*42 B

sns vm

ff . (8.11)

Gibbs free energy:

HBB

4

1

822

1 222

*42

sns vm

fg , (8.12)

or

0.2 0.4 0.6 0.8 1.0f

0.1

0.2

0.3

J1

50

88

)(

22

1 2222

*42 H

vm

fg sns

HB

. (8.13)

So the Gibbs free energy per unit area of film is

2/

2/

2222

*42

2/

2/

]88

)(

22

1[)(

d

d

sn

d

d

ss dzH

vm

fdzzgG

HB

,

(8.14)

xxx vqAc

q

xm

qJ

2**

2

*

*

)(

ℏ

ℏ. (8.15)

When is constant,

)(*

*

zAcm

qv xx . (8.16)

)0,)(

,0()0),(,0(z

zAzB x

y

AB . (8.17)

or

z

zAzB x

y

)(

)( , (8.18)

or

Hz

zAzB x

y

)()( , (8.19)

or

HzzAx )( , Hzcm

qvx *

*

, (8.20)

2/

2/

22

2

*

*2

*242

2/

2/

]8

)(

282

1[

)(

d

d

n

d

d

ss

dzHB

zcm

HqmHf

dzzgG

,

(8.21) When the last term is negligibly small, we obtain

51

32

*

*2

*242

23

2

2)

82

1(

d

cm

Hqmd

HfG ns

, (8.22)

or

]24

)82

1[(

2

2*

222*242

cm

dHqHdFG ns . (8.24)

Similarly,

2/

2/

222

*42

]822

1[

d

d

sns dzB

vm

FF

, (8.25)

2

2*

322*242

24)

82

1(

cm

dHqBdFF ns . (8.26)

Minimizing the expression of sF with respect to 2

, we find

024 2*

222*2

cm

dHq , (8.27)

241

241

241

2

2

2

2

2

2

2

2*

222*

2

2

cc H

Hd

H

H

cm

dHq

, (8.28)

or

22

2

124

fH

H

c

, (8.29)

where

d . (8.30)

This is rewritten as

2//

22 1

cH

Hf , (8.31)

52

where //cH is defined as

c

c

HH 62// . (8.32)

This parallel critical field //cH can exceed the thermodynamic critical field Hc when is

lower than 899.462 . We now consider the difference between Gs and Gn

dH

FG nn 8

2

(8.33)

where

nn dfF , and nn dgG

The difference is given by

]24

)82

1[()

8(

2

2*

222*242

2

cm

dHqHHGG sn . (8.34)

or

]24

)2

1[(

2

2*

222*42

cm

dHqGG sn . (8.35)

Noting that

024 2*

222*2

cm

dHq , (8.36)

we obtain

02

1

2

1 444 sn GG . (8.37)

The superconducting state is energetically favorable. 8.3. Parallel critical fields of thick films1

We now consider the case when an external magnetic field is applied to a thick film with a thickness d. The field is applied along the plane. We solve the GL equation with the appropriate boundary condition. The magnetic induction and the current density are given by

53

)2

cosh(

)cosh()(

f

fz

HzBy

, (8.38) and

)2

cosh(

)sinh(

4

)(

4)(

f

fz

cfH

dz

zdBczJ

y

sx

. (8.39) Note that

f,

22*

2*2

4

q

cm

, (8.40)

)2

cosh(

)sinh()()()(

*

*

22*2* f

fz

cfm

Hq

fq

zJ

q

zJzv sxsx

sx

, (8.41)

)2

(cosh

)1)sinh(

(

2

1)]([

1

2222*

22*22/

2/

22

f

f

f

fcm

qHdzzv

dv

d

d

sxsx

. (8.42)

Helmholtz free energy:

822

1 222

*42 B

sns vm

ff ’ (8.43)

On minimizing this with respect to 2

,

02

sf , (8.44)

02

2*

2 sv

m , (8.45)

or

2*

2

2

*

2

2*

22

2

21

212

ss

s

vm

vm

vm

f

, (8.46)

54

)2

(cosh

)1)sinh(

(1

4

11

)2

(cosh

)1)sinh(

(1

41

22

2

22

2

2*

222*2

f

f

f

fH

H

f

f

f

fH

H

cm

Hqf

cc

c

,

(8.47)

]1)sinh(

[

)2

(cosh)1(4

2

22

2

f

f

f

ffH

H

c

. (8.48)

We now discuss the Helmholtz free energy

822

12

22*

42 B sns v

mff , (8.49)

with

02

2*

2 sv

m . (8.50)

Then we have

82

1

8)(

2

1

24

22242

Bf

Bff

n

ns

, (8.51)

or

88

2

42 B

fH

ff cns , (8.52)

22/

2/

22

)]cosh(1[

)sinh()]([

1H

ff

ffdzzB

dB

d

d

y

, (8.53)

and

)2

tanh(2

)(1 2/

2/

f

f

HdzzB

dB

d

d

y

. (8.54)

55

Gibbs free energy:

4884

2

42 HBB

fH

fHB

fg cnss , (8.55)

)]2

tanh(4

)]cosh(1[)sinh(

[88

24

2f

fff

ffHf

Hfg c

ns

. (8.56)

Since

848

22H

fBHH

fg nnn , (8.57)

where B = H. From the condition that ns gg at the critical field H = Hl,

22242 )

2tanh(

14

)]cosh(1[

)sinh(H

f

fH

ff

ffHfHc

, (8.58)

or

4

2

)}2

tanh(4

)]cosh(1[

)sinh(1{ f

f

fff

ff

H

H

c

, (8.59)

or

)2

tanh(4

)]cosh(1[

)sinh(1

42

f

fff

ff

f

H

H

c

l

. (8.60)

In the limit of f →0,

642

222

2

][)350

81

5

24()

1

5

1(24

24fOff

H

H

c

l

. (8.61)

From the above two equations, we get

)2

tanh(4

)]cosh(1[

)sinh(1]1

)sinh([

)2

(cosh)1(4

42

22

f

fff

ff

f

f

f

f

ff

, (8.62)

56

or

ff

ff

f

f

)sinh(

]1)[cosh(31

)1(61 2

2

. (8.63)

In the limit of f →0,

644

22

][)126006

1()

51(

6

1fOff

Using Mathematica, 1. we determine the value of f as a function of . 2. we determine the value of 2/ cl HH as a function of f or f2 for each . There occurs the first order transition because of discontinuous change of order parameter. ((Mathematica Program-12))

57

Here we discuss the problem dscribed in the book of de Gennes, page 189

Clear@"Global`∗"D; hy@z_D = H

CoshBz f

λF

CoshBε f

2F; rule1 = :ε →

d

λ>;

Jx@z_D = −c

4 πD@hy@zD, zD êê Simplify

−c f H SechA f ε

2E SinhA f z

λE

4 π λ

ψ = f ψ0, ψ02 =m c2 λ−2

4 π q2

rule2 = :ψ0 →m c2 λ−2

4 π q2>; vsx@z_D =

Jx@zDq f2 ψ02

ê. rule2

−H q λ SechA f ε

2E SinhA f z

λE

c f m

Vavsq =1

d‡−dê2dê2

vsx@zD2 �z êê Simplify; eq1 = f2 1 +m

2 αVavsq;

rule3 = 9H2 → x=; eq2 = eq1 ê. rule3

f2 1 +

q2 x λ2 I−d f + λ SinhA d f

λEM

2 c2 d f3 m α H1 + Cosh@f εDL

eq3 = Solve@eq2, xD ê. d → ε λ êê Simplify

::x → −2 c2 f3 I−1 + f2M m α ε H1 + Cosh@f εDL

q2 λ2 Hf ε − Sinh@f εDL >>

58

Hc2 =4 π α2

β, ψ∞2 = α êβ,

c2 m β

q2 2 π H−αL λ2= 2

Calculation of H2 ë Hc2, β = −4 π q2 α λ2

c2 m

rule4 = :β →4 π H−αL q2 λ2

c2 m>; eq4 =

x ê. eq3@@1DD4 π α2

β

ê. rule4

2 f3 I−1 + f2M ε H1 + Cosh@f εDLf ε − Sinh@f εD

Hsqave =1

d‡−dê2dê2

hy@zD2 �z ê. d → ε λ êê Simplify

H2 SechA f ε

2E2 Hf ε + Sinh@f εDL2 f ε

B1 =1

d‡−dê2dê2

hy@zD �z ê. d → ε λ êê Simplify

2 H TanhA f ε

2E

f ε

Fs = Fn −Hc2

8 πf4 +

Hsqave

8 π;

59

Gs = Fs −B1 H

4 π;

Gn = Fn −H2

8 π;

eq6 = Gs − Gn ê. H2 → y Hc2

−f4 Hc2

8 π+Hc2 y

8 π+Hc

2y SechA f ε

2E2 Hf ε + Sinh@f εDL16 f π ε

−Hc2 y TanhA f ε

2E

2 f π ε

eq7 = Solve@eq6 0, yD::y →

2 f5 ε

2 f ε + f ε SechA f ε

2E2 + SechA f ε

2E2 Sinh@f εD − 8 TanhA f ε

2E >>

eq8 = y ê. eq7@@1DD êê FullSimplify

f5 ε H1 + Cosh@f εDLf ε H2 + Cosh@f εDL − 3 Sinh@f εD

We now consider the solution of de Gennes model

K@f_, ε_D := 1 +f2

6 I1 − f2M −f ε HCosh@ε fD − 1L3 H−f ε + Sinh@f εDL ;

Evaluation of critical magnetic field as a function of the thickness e

Series@K@f, εD, 8f, 0, 5<D1

6−

ε2

30f2 +

I2100 + ε4M f412600

+ O@fD6

L1@ε_D := Module@8f1, g1, ε1, f2<, ε1 = ε;

g1 = FindRoot@K@f1, ε1D 0, 8f1, 0.2, 0.99<D; f2 = f1 ê. g1@@1DDDs1 = Table@8ε, L1@εD<, 8ε, 0.22, 10, 0.01<D;h1 = ListPlot@s1, PlotStyle → 8Hue@0D, Thick<, Background → LightGray,

AxesLabel → 8"ε", "f"<D;

60

h2 = PlotBL1@εD, :ε, 5 , 3>, PlotStyle → 8Hue@0D, [email protected]<,Background → [email protected], PlotRange → 882.1, 3<, 80, 0.7<<,AxesLabel → 8"ε", "f"<F;

Evaluation of HHê HcL2 vs ε

Comparison with the approximation for HHê HcL2 vs ε : 24ë ε2

Hcrsq@ε_, f_D :=2 f3 I−1 + f2M ε H1 + Cosh@f εDL

f ε − Sinh@f εDSeries@Hcrsq@ε, fD, 8f, 0, 5<D24

ε2+ K 24

5−24

ε2O f

2 + −24

5+81 ε2

350f4 + O@fD6

h3 = PlotB: 24ε2

, Hcrsq@ε, L1@εDD>, :ε, 5 , 8>,PlotStyle → 88Hue@0D, [email protected]<, [email protected], [email protected]<<,Background → [email protected], PlotRange → 882, 8<, 80, 6<<F;

Evaluation of f2vs HHêHcL2 with ε as a parameter.

Approximation for HHêHlL2 vs ε : 24ëε2

<< "ErrorBarPlots`"; << "PlotLegends`"

fsq1@ε_, hsq1_D := Module@8f1, g1, ε1, f2, hsq2<, ε1 = ε; hsq2 = hsq1;

g1 = FindRoot@hsq2 == Hcrsq@ε1, f1D, 8f1, 0.6, 1<D; f2 = f1 ê. g1@@1DDD

61

The value of f for H/Hc= 1

h4 = Plot@fsq1@ε, 1D, 8ε, 2, 10<, AxesLabel → 8"ε", "f"<,PlotStyle → 8Hue@0D, [email protected]<, Background → [email protected];

f2vs HHêHcL2 with ε = 5 , 2.5, 3, 4, 4, 5, 5, 7, 9, 11, 20, and 30

X1@ε_D := TableA9Hcrsq@ε, fsq1@ε, hsq1DD, fsq1@ε, hsq1D2=,8hsq1, 0, 1, 0.01<E;

h5 = ListPlotB:X1B 5 F, [email protected]`D, X1@3D, X1@4D, [email protected]`D, X1@5D,X1@7D, X1@9D, X1@11D, X1@20D, X1@30D>,AxesLabel → 8"HHêHc\!\H\∗SuperscriptBox@\HL\L, \H2\LD\L","\!\H\∗SuperscriptBox@\Hf\L, \H2\LD\L"<F;

hny@t_, f_, ε_D =

CoshBz f

λF

CoshBε f

2F

ê. :λ →d

ε, z → t d>;

Jnx@t_, f_, ε_D = f SechBf ε

2F SinhBf z

λF ê. :λ →

d

ε, z → t d>;

Magnetic field distribution

h6 = Plot@Evaluate@Table@hny@t, L1@εD, εD, 8ε, 2.5, 10.0, 0.2<DD,8t, −1ê2, 1ê2<, PlotStyle → Table@[email protected] iD, 8i, 0, 20<D,Prolog → [email protected], Background → [email protected],AxesLabel → 8"xêd", "h"<D;

Current density distribution

h7 = Plot@Evaluate@Table@Jnx@t, L1@εD, εD, 8ε, 2.5, 10.0, 0.2<DD,8t, −1ê2, 1ê2<, PlotStyle → Table@[email protected] iD, 8i, 0, 20<D,Prolog → [email protected], Background → [email protected],AxesLabel → 8"xêd", "J"<D;

p1 = 8h1, h2, h3, h4, h5, h6, h7<;Show@GraphicsGrid@Partition@p1, 3DDD

0 2 4 6 8 10e0.0

0.20.40.60.8

f

2.22.4 2.62.8 3.0e0.00.10.20.30.4

0.50.60.7f

2 3 4 5 6 7 80123456

2 4 6 8 10e0.9400.9450.9500.9550.9600.9650.970

f

0.00.20.40.60.81.0HHêHcL20.880.900.920.940.960.981.00

f 2

-0.4-0.20.00.20.4xêd

0.20.40.60.81.0

h

62

_______________________________________________________________________

Fig.14(a) The order parameter f vs (= d/). f tends to 1 in the large limit of →∞.

Fig.14(b) Detail of Fig.14(a). The order parameter f vs (= d/). f reduces to zero at

5 .

2 4 6 8 10e

0.2

0.4

0.6

0.8

f

2.2 2.4 2.6 2.8 3.0e0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7f

63

Fig.15 2)/( cHH vs (red line) and the approximation 24/2 vs (blue line).

These agree well only in the vicinity of 5 .

Fig.16 Plot of f at H/Hc = 1 as a function of .

2 3 4 5 6 7 80

1

2

3

4

5

6

2 4 6 8 10e

0.940

0.945

0.950

0.955

0.960

0.965

0.970

f

64

Fig.17 Plot of f2 vs (H/Hc)2 with = 5 , 2.5, 3, 4, 4.5, 5, 7, 9, 11, 20, and 30.

Fig.18 Plot of (a) the magnetic field distribution and (b) the current density as a function of x/, where e is changed as a parameter. = 2.5 – 10.0. = 0.2.

8.4 Superconducting cylindrical film1

We consider a superconducting cylindrical film (radius R) of thickness (2d), where 2d«R.

A small axial magnetic field is applied to the cylinder.

0.2 0.4 0.6 0.8 1.0HHêHcL2

0.90

0.92

0.94

0.96

0.98

1.00

f 2

-0.4 -0.2 0.2 0.4xêd

0.2

0.4

0.6

0.8

1.0

h

65

Fig.19 Geometrical configuration of a cylinder (a radius R) with small thickness.

R>>d.

ss qc

q

m

qvAJ

2**

2

*

*

)( ℏ

ℏ. (8.65)

Suppose that 22* sn =constant, then we have

AJ s

snq

cm

q

c*2*

*

* ℏ

. (8.67)

Integrating the current density around a circle running along the inner surface of the cylinder,

lAlJl ddc

d s

20 4

2

, (8.68)

and

iddd aBaAlA )( , (8.69)

where i is the flux contained within the core of the cylinder. The magnetic flux f is

defined as

ldf 2

0 . (8.70)

Then we have

66

isf dc

lJ24

, (8.71)

or

)(4 2 ifs

cd

lJ . (8.72)

The phase of the wave function must be unique, or differ by a multiple of 2 at each point,

nq

cf *

2 ℏ . (8.73)

In other words, the flux is quantized.

When the radius is much larger than the thickness 2d, the magnetic field distribution is still described by

)(1

)( 22

2

xBxBdx

dzz

, (8.74)

with the boundary condition, iz HdB )( and 0)( HdBz . The origin x = 0 is shown in

the above Figure.

)cosh(

)cosh(

2)sinh(

)sinh(

2)( 00

d

x

HH

d

x

HHxB ii

z

. (8.75)

The current density Jy(x) is given by

)2

sinh(

)cosh()cosh(

4)(

4)(

0

d

xdH

xdH

c

dx

xdBcxJ

iz

y

. (8.76)

The value of current flowing near the inner surface is

)2

sinh(

)2

cosh(

4)(

0

d

Hd

Hc

dJi

y

. (8.77)

67

Then we have

)(4

)(2 2 ifys

cdRJd

lJ , (8.78)

or

)(4)

2sinh(

)2

cosh(

42 2

0

if

i c

d

Hd

Hc

R

, (8.79)

or

)

2coth(

21

)2

sinh(

2

2

0

d

RR

d

RH

H

f

i . (8.80)

The Gibbs free energy is given by

])(8

1)(

8[ 2

02

2

HBBr

dG , (8.81)

or

d

d

zz dxHxBx

xBG }])([

)(

8{ 2

0

22

, (8.82)

where the constant terms are neglected and Hi is given by Eq.(8.80). For a fixed H0, we need to minimize the Gibbs free energy with respect to the fluxsoid f .

We find that the solution of 0 fG f leads to f = i , where R/ = 10 – 50 and

d/<1. To prove this, we use the Mathematica for the calculation. ________________________________________________________________________


68

________________________________________________________________________

de Gennes book p. 195 Cylinder

Clear@"Global`∗"D; Bz =H0 − Hi

2

SinhB x

λF

SinhB d

λF

+H0 + Hi

2

CoshB x

λF

CoshB d

λF;

rule1 = :Hi →

Φf +2 H0 π R λ

Sinh B2 dλ

F

π R2 1 +2 λ

R

Cosh B2 dλ

FSinh B2 d

λF

>; Bz1 = Bz ê. rule1 êê Simplify;

C1 = D@Bz1, xD êê Simplify; G1 =λ2

8 πC12 +

1

8 πHBz1 − H0L2 êê Simplify;

G2 = ‡−dê2dê2

G1 �x êê Simplify;

G3 = D@G2, ΦfD êê Simplify;

G4 = Solve@G3 0, ΦfD êê Simplify; F1 = Φf ê. G4@@1DD êê Simplify;

rule1 = 8d → ε λ, R → µ λ<;F2 = F1 ë IH0 π R2M ê. rule1 êê Simplify

Sech@2 εD Iµ − µ CoshA 3 ε

2E + µ CoshA 5 ε

2E − 2 SinhA 3 ε

2E + 2 SinhA 5 ε

2EM

µ

Φf is the local m inimum value of the Gibbs free energy. Plot

of Φf ëIH0 π R2M vs ε = dê λ = 0 − 1, where µ = Rê λ = 10 − 50

g1 = Plot@Evaluate@Table@F2, 8µ, 10, 50, 10<DD, 8ε, 0, 1<,PlotRange → 880, 1<, 80.8, 1.5<<, PlotPoints → 100,

PlotStyle → Table@8Thick, [email protected] iD<, 8i, 0, 5<D,Background → LightGray , AxesLabel → 8"ε", "F"<D;

g2 = Graphics@8Text@Style@"µ=10", Black, 12D, 80.6, 1.2<D,Text@Style@"µ=50", Black, 12D, 80.6, 1.07<D<D;

g3 = Show@g1, g2D

69

Fig.20 Plot of the normalized magnetic flux F = /(R2H0) as a function of ( =

0 – 1), where = 10 – 50. = 10. 8.5 Little Parks experiments1,4

8.5.1 The case of d<< and d<<,The thickness d of the superconducting cylinder is much shorter than the radius R. It

is also shorter than and . Since d«, the change of B inside the superconducting layer is slight. The current density (proportional to derivative of B) is assumed to be constant. The absolute value of the order parameter is constant in the superconducting layer. The following three equations are valid in general.

ss qc

q

m

qvAJ

2**

2

*

*

)( ℏ

ℏ, (8.83)

8)(

2

1

2

1 22*

*

42 B

c

q

imff ns A

ℏ, (8.84)

)()()( r

rr ie . (8.85)

We consider the case when

*2)( snr = constant or )()( r

r ie . (8.86)

Then we have

)()()( ***

smc

q

c

q

ivAA ℏ

ℏ, (8.87)

m=10

m=50

0.0 0.2 0.4 0.6 0.8 1.0e0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5F

70

and

822

1 222

*42 B

vm

ff sns , (8.88)

2*q

ss

Jv , sm

c

qvA

**

ℏ , (8.89)

Then we have

lvlAl dmdc

qd s

**

][ ℏℏ , (8.90)

where ][ represents the change in phase after a complete revolution about the cylinder.

n 2][ (n; arbitrary integer).

BRdddAA

2)( aBaAlA . (8.91)

represents the flux contained in the interior of the cylinder.

)(0

*

nRm

vs

ℏ, (8.92)

where

*0

2

q

cℏ . (8.93)

Here we see from the free energy density that the free energy will be minimized by

choosing n such that |vs| is minimum since f is proportional to 22*

2 svm

.

|]|min[0

*

nRm

vs

ℏ. (8.94)

71

Fig.21 Plot of the velocity vs as a function of 0/ .

The current density is equal to zero when 0 n .

Since vs is known, we can minimize the free energy with respect to and find

0/2 f , (8.95)

or

02

1 2*2 svm , (8.96)

or

0/)2

1( 2*2

svm , (8.97)

or

2*

2

1svm or 2*

2

1svm . (8.98)

-3 -2 -1 0 1 2 3FêF0

0.1

0.2

0.3

0.4

0.5vsêH2m

*RL

72

Fig.22 Plot of vs2 vs 0/ . The horizontal lines denote y for several .

The transition occurs when

]/)(1[2

1

00

2*cs TTvm

, (8.99)

where )0(4

20 cH

.

In the above figure, the intersection of the horizontal line of at fixed T and the curve

2sv gives a critical temperature. In the range of 0/ where the horizontal line is located

above the curve 2sv , the system is in the superconducting phase. In the range of 0/

where the horizontal line is located below the curve 2sv , on the other hand, the system is

in the normal phase. 8.5.2 The case of d>> and d>>.

The thickness d of the superconducting cylinder is longer than and . The current density (proportional to derivative of B) is zero. The absolute value of the order parameter is constant in the superconducting layer.

0)(2*

*2

*

*

ss qc

q

m

qvAJ

ℏ

ℏ, (8.100)

-3 -2 -1 0 1 2 3FêF0

0.05

0.10

0.15

0.20

0.25v2

2êH2m*RL2

73

or

Aℏc

q*

, (8.101)

c

qd

c

qd

c

qd

***

][ aBlAl ℏℏ , (8.102)

or

0*

2 nn

q

cℏ. (8.103)

Then the magnetic flux is quantized. We consider a superconducting ring. Suppose that A’=0 inside the ring. The vector potential A is related to A’ by

0' AA , (8.104) or

A . (8.105) The scalar potential is described by

iextdadadL BAAl )()0()( , (8.106)

where is the total magnetic flux.

We now consider the gauge transformation. A’ and A are the new and old vector potentials, respectively. ’ and are the new and old wave functions, respectively.

0)(' AA , (8.107)

)()exp()('*

rr

c

iq

ℏ . (8.108)

Since A’ = 0, ’ is the field-free wave function and satisfies the GL equation

0'2

''' 2*

22

m

ℏ. (8.109)

In summary, we have

74

]2exp[)(']exp[)(']exp[)(')(0

**

ic

iq

c

iq rrrr

ℏℏ, (8.110)

since 0

*2

c

q

ℏ and q*<0.

9. Isolated filament: the field for first penetration2

9.1 Critical field Hc1

We have the following equations;

ffffi

22

~1

A, (9.1)

AA~

][2

)~

(2** fffff

i

, (9.2)

Ah~

. (9.3)

The vector potential in cylindrical co-ordinates takes the form

)0,,0( AA , cH2

Bh , AA

0

2~

, ρr (9.4)

Stokes theorem:

rAddd 2)( rAaAaB , (9.5)

or

A

rA

~

220

(9.6)

Far from the center of the vortex,

111~

000

r

A , (9.7)

for 0 .

Near the center of the vortex,

75

2)0(2 rHrA , or )0(~

0

HA

. (9.8)

We assume that

ief )( , (9.9) where )( is a real positive function of .

)

~~

(1

)~

(1~

A

AA zz eeAh , (9.10)

)]~

(1

[)]

~~

(11

[ 2

2

2

AAA

A

eeh . (9.11)

GL equation can be described by

322

)~1

()(11

Ad

d

d

d. (9.12)

The current equation: e component.

0)~1

()]~

(1

[ 2

AA

d

d

d

d. (9.13)

The above equations can be treated approximately when »1. We assume that )( =1 in the region /1 .

0)~1

)](1

1(1

[ 22

2

A

d

d

d

d (9.14)

The solution of this differential equation is

)(~1

11 KcA (9.15)

or

)(1~

11 KcA (9.16)

76

]))(1

()(1

[1

)~

~(

1 11211

d

dKcKc

AAhz

, (9.17)

or

)()]([]))(

)([ 01011

11

KcKc

d

dKK

chz . (9.18)

or

)(01 Kchz , (9.19)

from the recursion formula

)()()(

011

KxKd

dK , (9.20)

What is the value of c1? When 0 , A~

should be equal to zero.

01

)(1~ 1

11

cKcA , (9.21)

because /1)(1 K . Then we have

1

1 c . (9.22)

In fact,

)(1

)(22

)(2

22

0020

0

20

KKH

KHH

Bh

ccc

, (9.23)

or

)(1

0

Kh , (9.24)

since

2

2 02 cH . (9.25)

77

In summary, we have

)(1

)(11~

0

1

Kh

KA

, (9.26)

((Note)) The total magnetic flux through a circle of radius of »1 is given in the following way

1

)(11~

1 KA . (9.27)

Then we have

000 1

22

~

22)(2

AA . (9.28)

9.2 The center of vortex4

First we consider the behavior at the center of vortex, as →0.

322

)~1

()(11

Ad

d

d

d, (9.29)

with

)0(~

0

2

HA

, (9.30)

or

32

0

2

2 )]0(1

[)(11

Hd

d

d

d. (9.31)

We assume that can be described by series

...)( 33

2210 aaaak , (9.32)

where a0≠0. Using the Mathematica 5.2, we determine the parameters;

78

0

])0(

1[8

])0(2

1[8

0

1

3

2

2

00

22

02

1

a

H

Ha

Haa

a

k

c

, (9.33)

where

20

2 2

cH .

Then we have

)])0(

1(8

1[)(2

22

0cH

Ha

. (9.34)

The rise of )( starts to saturate at /2 . ((Summary)) If the vortex contains 1 flux quantum, we have

)()( ief , (9.35)

...}])0(

1[8

1{)(2

22

0 cH

Ha

. (9.37)

9.3 Far from the center of the vortex

32

)(11

d

d

d

d, (9.38)

with

1~

A . (9.39)

This differential equation is simplified as

0)()()]('1

)("[1 3

2

. (9.40)

79

In the region where 1)( , for simplicity we assume that

)(1)( , (9.41)

2

3 )(")(')(1)](1[

. (9.42)

This equation can be approximated by

2

)(")(')(1)(31

, (9.43)

or

0)(2)('1

)(" 2

. (9.44)

The solution is given by

)2()2()( 0201 KCIC . (9.45)

Since )2(0 I increases with increasing , C1 should be equal to zero.

Then we have

)2()( 02 KC . (9.46)

For large ,

x

exK

x

2

)(0

,

x

exK

x

2

)(1

. (9.47)

10. Properties of one isolated vortex line (H = Hc1)

10.1 The method of the Green’s function

ss q vJ2* , (10.1)

2

2*

*

42

8

1

2

1

2

1B

c

q

imff ns

A

ℏ, (10.2)

When ie and *2

sn = independent of r,

80

822

1 22*

*2** B

ssssns vnm

nnff . (10.3)

**s

ss

nq

Jv and s

cJB

4 . (10.4)

Then the net energy can be written by

])([8

1 2220 BB

drEE

When

BBB , , EEE , (10.6)

BBBr

])([4

1 2dE . (10.7)

The integrand can be calculated by using VariationalD of the Mathematica 5.2. ________________________________________________________________ ((Mathematica Program-14)) Variation method

81

________________________________________________________________ Demanding that E be a minimum leads to the London equation; E = 0,

0)(2 BB . (10.8) This is called the second London’s equation. ((Note))

BBBB 222 , (10.9)

)()(2)()( 22 BBBB . (10.10) Then we have

)]()([4

1 2BBBB

drE

Since

)]([)]([)()( BBBBBB , (10.12)

Derivation of London' s equation

Clear@"Global`∗"D;<< "VariationalMethods`"

Needs@"VectorAnalysis`"D;h = 8h1@x, y, zD, h2@x, y, zD, h3@x, y, zD<;eq1 = h.h + λ2 [email protected]@hDL êê Expand

h1@x, y, zD2 + h2@x, y, zD2 + h3@x, y, zD2

eq3 = VariationalD@eq1, h2@x, y, zD, 8x, y, z<D ê2 êê Expand;



London's equation

eq5 = h + λ2 Curl@Curl@hDD;A1 = 8eq2 − eq5@@1DD, eq3 − eq5@@2DD, eq3 − eq5@@2DD< êê Simplify

80, 0, 0<

82

)]([4

1)]([

4

1 2 BBaBBBr

ddE . (10.13)

The second term is equal to zero. Here we assume that (i) >> (ii) The hard core of radius is very small, and we shall neglect completely its

contribution to the energy

Fig.23 Top view of the geometrical configuration of a single vortex. The line energy is given by

rd ])([

8

1 2222 BBr . (10.14)

Demanding that be a minimum leads to the London’s equation.

0)]([2 BB , (10.15) for r . In the interior of hard core, we can try to replace the corresponding singularity

by a 2D delta function.

)()]([ 202 reBB z , (10.16)

0202 )()]([ areaBaB ddd z , (10.17)

Using Stokes theorem,

83

0202 )()( arelBaB ddd z . (10.18)

Fig.24 The path of line integral If the circle of the path integral has a radius r ( <<r<<), the first term I1 is on the order

of 220 /r and is negligibly small. In the second term I2,

rdd el , (10.19)

)(rBdr

deB . (10.20)

Then we have

02

2

0

222

)(2

)]([)(

rBdr

dr

rdrBdr

ddI

eelB

, (10.21)

or

rdr

rdB2

0

2

)(

, (10.22)

or

])[ln(2

)( 20 const

rrB

, (10.23)

for r . We now consider the exact solution

)()]([ 202 reBB z , and 0 B . (10.24)

Note that

84

BBBB 22)()( . (10.25) Thus we have

)(1

220

22

reBB z

, (10.26)

where 2(r) is a 2D delta function. ((Note)) 2D case Green function (i) Modified Helmholtz

)(),()( 2122122

1 rrrr Gk , (10.27)

)(2

1),( 21021 rrrr kKG

. (10.28)

(ii) Modified Bessel function

)()()( 22 rr fk , (10.29)

')'()',()( rrrrr dfG , (10.30)

)(')'()'(

')'()',()()()( 2222

rrrrr

rrrrr

fdf

dfGkk

, (10.31)

The solution of the equation

)(1

220

22

reBB z

, (10.32)

is that

)(2

)'()',()( 020

220

r

KG zz

errrerB , (10.33)

or

)(2

)( 020

r

KrB

(in general). (10.34)

((Asymptotic form)) For <r<<,

85

]115932.0)[ln(2

)( 20

r

rB . (10.35)

For r>>,

)exp(22

)( 20

r

rrB

. (10.36)

The magnetic flux )(r inside the circle with a radius r is

)](1[)()'

('')'(''2)( 10

/

0

00

0

020

0

r

Kr

xxdxKr

KdrrrBdrrr

rrr

.

(10.37) The current density

)(44

rBdr

dcc

eBJ . (10.38)

Fig.25 Schematic diagram for the plots of the magnetic induction B (blue line) and the magnitude of the order parameter (red line) as a function of x (the distance from the center of the vortex) for an isolated vortex line in a mixed phase of type-II superconductor.

_______________________________________________________________________


-4 -2 2 4x

0.2

0.4

0.6

0.8

1.0

y,B

86

_______________________________________________________________________

Fig.26 Plot of the normalized magnetic induction )/(2

/)( 020

rKrB

as a

function of /r .

the radial distribution of h(r)

Clear@"Global`∗"D;h1 = PlotABesselK@0, xD, 8x, 0, 2<, PlotPoints → 100,

PlotStyle → 8Thick, Red<, Background → LightGray,

AxesLabel → 9"rêλ", "hêHΦ0ê2πλ2"=E;

Φ = ‡0

xBesselK@0, yD y �y êê Simplify@ , x > 9D &

1 − x BesselK@1, xD

h2 = Plot@Φ, 8x, 0, 10<, PlotPoints → 100, PlotStyle → 8Red, Thick<,PlotRange → 880, 10<, 80, 1<<, Background → LightGray,

AxesLabel → 8"rêλ", "ΦêΦ0"<D;

0.5 1.0 1.5 2.0rêl

0.5

1.0

1.5

2.0

2.5

hêHF0ê2pl2

87

Fig.27 Magnetic flux )](1[/)( 10 r

Kr

r as a function of x = r/. 0/)( r

reduces to 1 as x . 10.2. Vortex line energy

0 2 4 6 8 10rêl0.0

0.2

0.4

0.6

0.8

1.0FêF0

88

Cylinder with a radius .

Fig.28 Geometrical configuration of a single vortex. The vortex is approximated

by a cylinder.

Now let us find the line tension or free energy per unit length. neglecting the core, we have only the contributions

])([8

1 22231 BBr

d . (10.39)

Vector identity

)]([)]([)( 2 BBBBB . (10.40)

)]([8

1)]([

8

1 23231 BBrBBBr

dd , (10.41)

or

)]([8

)]([8

1 2

203

1 BBaBrer dd z

. (10.42)

We use the Stokes theorem for the second term. We now neglect the core contribution (the

first term). The integral ad is to be taken over the surface of the hard core (cylinder of

radius ).

89

)]([8

2

1 BBad

. (10.43)

Note that

eBB

)()()( z

z

BB and dad ea (10.44)

]2

)()([

8)](

)()([

8

22

1z

zz

z

BBda

BB ee .

(10.45) Since

1

2

)(2

0

d

dBz (10.46)

we have

]115932.0)[ln(4

)(8

2

001

zB . (10.47)

More exactly

)()(4 10

2

01

KK

. (10.48)

(see the derivation of this equation in the Mathematica Program-16). _____________________________________________________________________ ((Mathematica Program-16))

90

_____________________________________________________________________

Fig.29 Plot of )/1()/1(1

4// 102

01 xKxKx

[blue line] as a function of

/x . For comparison, the approximation given by (lnx+0.115932) is also shown by a red line.

10.3. Derivation of Hc1

vortex line energy

Clear@"Global`∗"D; << "VectorAnalysis`"; SetCoordinates@Cylindrical@ρ, φ, zDD;h = 80, 0, hz@ρD<; Cross@h, Curl@hDD êê Simplify;

rule1 = :hz →Φ0

2 π λ2BesselKB0,

λF & >;

ε1 = −λ2

8 π2 π ξ hz@ξD D@hz@ξD, ξD ê. rule1 êê Simplify

ξ Φ02 BesselKA0, ξ

λE BesselKA1, ξ

λE

16 π2 λ3

f1 =1

xBesselKB0, 1

xF BesselKB1, 1

xF; Limit@f1 − Log@xD, x → ∞D êê N

0.115932

graph, Comparison with ln (l/x) + 0.115932

h1 = PlotA8Log@xD + 0.115932, f1<, 8x, 1, 10<, PlotPoints → 100,

PlotRange → 880, 10<, 80, 2.5<<, PlotStyle → 88Red, Thick<, 8Blue, Thick<<,Background → LightGray, AxesLabel → 9"λêξ", "ε1êHΦ02ê16π2λ2"=E;

0 2 4 6 8 10lêx0.0

0.5

1.0

1.5

2.0

2.5e1êHF0

2ê16p2l2

91

r3

4hd

HFG

. (10.49)

By definition, when H = Hc1, the Gibbs free energy must have the same value whether the first vortex is in or out of the system;

Gs|no flux = Gs|first vortex,` (10.50) where

Gs|no flux = sF . (10.51)

Gs|first vortex = 01

1 4 L

HLF c

s , (10.52)

where L is the length of the vortex line in the system and

03 LBdaLBd r . (10.53)

Then we have

01

1 4 L

HLFF c

ss , (10.54)

or

0

11

4

cH , (10.55)

or

)ln(2

)ln(4 2

01

c

c

HH

. (10.56)

10.4. Interaction between vortex lines

Suppose that there are two vortices in the 2 D plane.

)]()([1

221220

22

rrrr

BB . (10.57)

Using the Green function

)'

(2

1)',( 0

rrrr

KG . (10.58)

Then we have

92

)]()([)',(' 221220 rrrrrrr

GdB , (10.59)

or

)]()()['

('2 221202

0 rrrrrr

r

KdB , (10.60)

or

)()()]()([2

)( 212

01

020 rr

rrrrr BBKKB

. (10.61)

where

)(2

)( 102

01

rrr

KB , (10.62)

)(2

)( 202

02

rrr

KB . (10.63)

)()( 1221 rr BB . (10.64)

The total increase in free energy per unit length can be rewritten as

)(8

2)(8

2

)}]()({)}()([{8

210

110

222112110

rr

rrrr

BB

BBBB

The first tem is the energy of individual vortex line. We have the interaction energy

)(8

)(4

12022

20

210

12 r

KBU

r , (10.66)

where 2112 rr r

The interaction is repulsive, in which the flux has the same sense in both vortices. 11. Magnetization curves

11.1 Theory of M vs H

We consider the Gibbs function given by

93

410

BHUnGG

ji

ijLs

. (11.1)

The first term is an individual energy of the lines. nL is the number of lines per unit area (cm2).

0 LnB . (11.2)

The second term is a repulsive interaction between vortices.

)(8 022

20

ij

ij

rKU

, (11.3)

where || jiijr rr

One can distinguish three regimes between Hc1 and Hc2. (a) 1cHH : 2Ln <<1, r

(b) 2Ln >>1 r (c) 2cHH : 12 Ln r

410

0

BHU

BGG

ji

ijs

(11.4)

(a) At very low densities (low B) the interaction is small and we neglect it completely.

BH

GG s )4

(0

10

. (11.5)

When

40

1 H

or H<Hc1, the lowest G is obtained for B = 0. When

40

1 H

, we can

lower G by choosing B≠0. There is some flux penetration.

)ln(4

42

0

0

11

cH . (11.6)

(b) When 1cHH

d: distance between neighboring lines (d>). (i) Square lattice

94

20

1

4

s

L

s

d

Bn

z

(11.7)

or

Bds

0 (11.8)

(ii) Triangular lattice

20

2

3

1

6

t

L

t

d

Bn

z

(11.9)

or

BBdt

00

4/1

07457.134

(11.10)

We neglect all contributions except those of the nearest neighbors.

)(16

)(82 022

0022

20

0 d

KBzd

KzB

Uji

ij

(11.11)

Then we have

)](22

1)[(

4 020

10 d

KzHHB

GG cs

(11.12)

or

)]()1(2

[4

)](

221

)([

422

1 01

10

20

1

20

0

dK

H

H

zH

HBdK

z

HHB

z

GG

cc

ccs

,

(11.13) or

)]()1(2

[4

2

01

10

dK

H

H

zH

HB

zH

GG

cc

c

c

s

, (11.14)

95

or

)]()1(2

[4

2

2/1

000

1

10

BK

H

H

zH

HB

zH

GG

cc

c

c

s

, (11.15)

where

cH22

20

. (11.16)

The B vs H curve can be obtained from

.0

B

G (11.17)

or

01

)('21

)()1(2

001

1 B

dK

ddK

H

H

zH

H

cc

c

(11.18)

or

0)(1

21

)()1(2

2/1

001

2/1

00

2/1

000

1

1

B

KBBB

KH

H

zH

H

cc

c

(11.19) Here we note that

c

c

zH

H 10

2 , (11.20)

and

2/1

00

B

d

(11.21)

where 0 = 1 for the square lattice and 1.07457 for the triangular lattice.

)()(' 10 xKxK (11.22)

11.2 Calculation of M vs H using Mathematica

96

We make a plot of the free energy vs B and B vs H using the Mathematica 5.2. _______________________________________________________________________


_______________________________________________________________________

(a) a = 0.4

(b) a = 0.4

Gibbs free energy; h=K/Hc1

F@a_, b_D :=B

4 πBesselKB0, a

B

F − b Hh − 1L ;

K@a_, b_D := Plot@Evaluate@Table@F@a, bD, 8h, 1, 3, 0.1<DD, 8B, 0, 5<,PlotStyle → Table@[email protected] iD, Thick<, 8i, 0, 10<D, Background → LightGrayD;

G@a_, b_D := DB B

4 πBesselKB0, a

B

F − b Hh − 1L , BF êê Simplify;

p11 = ContourPlot@Evaluate@G@1, 1D 0, 8h, 1, 1.15<, 8B, 0.01, 0.15<D,PlotPoints → 50, ContourStyle → 8Hue@0D, Thick<, Background → LightGray,

AxesLabel → 8"h", "B"<D;p12 = ContourPlot@Evaluate@[email protected], 1D 0, 8h, 1, 1.03<, 8B, 0.01, 0.08<D,PlotPoints → 50, ContourStyle → 8Hue@0D, Thick<, Background → LightGray,

AxesLabel → 8"h", "B"<D;

1 2 3 4 5

0.2

0.4

0.6

97

(b) a = 1.2

Fig.30 Gibbs free energy given by )]()1([4 0

B

aKhb

B

as a function of B,

where h (=H/Hc1) is changed as a parameter (h = 1 – 3, h = 0.1). b (= 1) is fixed.. The parameter a is chosen appropriately. (a) a = 0.4, (b) a = 0.8, and (c) a = 1.2.

1 2 3 4 5

-0.2

0.2

0.4

1 2 3 4 5

-0.4

-0.2

0.2

98

Fig.31 The magnetic induction B vs h (=H/Hc1) above h = 1. h = 1 (H = Hc1).

Fig.32 The detail of the magnetic induction B vs h (=H/Hc1) near h = 1 (H = Hc1).

1.00 1.02 1.04 1.06 1.08 1.10 1.12 1.14

0.02

0.04

0.06

0.08

0.10

0.12

0.14

1.000 1.005 1.010 1.015 1.020 1.025 1.030

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

99

12. The linearized GL equation (H = Hc2).

12.1. Theory

When the term 2

can be dropped in the GL equation, we obtain linearized GL

equation

02

12*

*

A

c

q

im

ℏ. (12.1)

This omission will be justified only if /22 .

Em

m

mc

q

im

2*

2

2

*

*

22*

* 2

2

22

1 ℏ

ℏ

ℏℏA , (12.2)

where

*

22

2m

ℏ . and 2*

2

2 mE

ℏ .

When B = H (//z), we choose the gauge

)0,,0( HxA . (12.3) Note that we use 2/)( rHA (-Hy/2, Hx/2, 0). = Hxy/2,

)0,,0( Hx AA (gauge transformation). This problem is reduced to the eigenvalue problem in the quantum mechanics.

EH ˆ , (12.4)

]ˆ)ˆˆ(ˆ[2

1ˆ 22*

2zyx pxH

c

qpp

mH . (12.5)

This Hamiltonian H commutes with yp and zp .

0]ˆ,ˆ[ ypH and 0]ˆ,ˆ[ zpH (12.6)

zynzy kknEkknH ,,,,ˆ , (12.7)

100

and

zyyzyy kknkkknp ,,,,ˆ ℏ , zyzzyz kknkkknp ,,,,ˆ ℏ , (12.8)

zyyzyy kknzyxkkknpzyx ,,,,,,ˆ,, ℏ ,

zyzzyz kknzyxkkknpzyx ,,,,,,ˆ,, ℏ , (12.9)

or

zyyzy kknykkknyyi

,,,, ℏℏ

, yik

zyyekkny ,, . (12.10)

zyzzy kknzyxkkknzyxzi

,,.,,,,, ℏℏ

zik

zyzekknzyx ,,,, .

(12.11)

Schrödinger equation for zy kknzyxzyx ,,,,),,( ,

),,(),,(])()()[(2

1 22

*

2* zyxEzyx

yiHx

c

q

yixim

ℏℏℏ

, (12.12)

)(),,( xezyxzikyik zy , (12.13)

x , with H

c

c

Hqm

ℓℏℏ

1***

and cm

Hq

c *

*

* , (12.14)

yHy

ykk

Hq

c

Hq

kcℓ

ℏℏ

**0 . (12.15)

We assume the periodic boundary condition along the y axis.

),,(),,( zyxzLyx y , (12.16)

or

1yy Like , (12.17)

or

101

y

y

y nL

k2

(ny: integers). (12.18)

Then we have

)()]2()[()(" 22*

*

20 zkEm

Hq

cℏ

ℏ . (12.19)

We put

*

22*

2)

2

1(

m

knE z

c

ℏℏ . (12.20)

where n = 0, 1, 2, 3, … (Landau level), or

)2

1(

2)

2

1(22

*22**22* n

c

HqknmkEm zcz

ℏℏℏℏ , (12.21)

or

)2

1(

2 *

*2

*

2

ncm

Hqk

mE z

ℏℏ. (12.22)

Finally we get a differential equation for )(

0)(])(12[)(" 20 n . (12.23)


)()!2()( 02

)(2/1

20

n

n

n Hen , (12.24)

with

yHy kkHq

cℓ

ℏ

*0 , (12.25)

Hq

cH *

ℏℓ , (12.26)

102

yHH kx2

00

0 ℓℓ

(12.27)

The coordinate x0 is the center of orbits. Suppose that the size of the system along the x axis is Lx. The coordinate x0 should satisfy the condition, 0<x0<Lx. Since the energy of the system is independent of x0, this state is degenerate.

xyHH Lkx 20

000 ℓℓ

, (12.28)

or

xyH

y

yH LnL

k 22 2ℓℓ

, (12.29)

or

22 ℓyx

y

LLn , (12.30)

)21

(2

)21

(22 *

*2

*

2

*

*2

*

2

2*

2

ncm

Hqk

mn

cm

Hqk

mmE zz

ℏℏℏℏℏ

. (12.31)

)2

1(

4)

2

1(

21

0

*2

2

n

Hn

c

Hqkz

ℏ

. (12.32)

or

)2

1(

212

2

2 nk

H

zℓ

, (12.33)

where

*0

2

q

cℏ ,

HHq

cH 2

0*

2

ℏℓ . (12.34)

Then we have

103

2

20 1

)2/1(4 zkn

H

. (12.35)

((Ground state))

If kz=0 and n = 0, this has its highest value at H = H 20

2 2

cH , (12.36)

or

cc HH 22 . (12.37)

This is the highest field at which superconductivity can nucleate in the interior of a large sample in a decreasing external field.

022

211

H

H

ℓ

. (12.38)

12.2 Numerical calculation of wave finctions

_______________________________________________________________________ ((Mathematica Program-18))

(a) n = 0 (even parity)

Simple Harmonics wave function: plot of jn[z]

In[31]:= Clear@"Global`∗"D; conjugateRule = 8Complex@re_, im_D � Complex@re, −imD<;Unprotect@SuperStarD; SuperStar ê: exp_ ∗ := exp ê. conjugateRule;

Protect@SuperStarD;ψ@n_, ζ_, ζ0_D := 2−nê2 π−1ê4 Hn !L−1ê2 ExpB− Hζ − ζ0L2

2F HermiteH@n, ζ − ζ0D

In[34]:= pt1 = Plot@Evaluate@Table@ψ@0, ζ, ζ0D, 8ζ0, −2, 2, 0.4<DD, 8ζ, −4, 4<,PlotLabel → 80<, PlotPoints → 100, PlotRange → All,

PlotStyle → Table@8Thick, [email protected] iD<, 8i, 0, 10<D, Background → LightGrayD;pt2 = Plot@Evaluate@Table@ψ@1, ζ, ζ0D, 8ζ0, −2, 2, 0.4<DD, 8ζ, −4, 4<,PlotLabel → 81<, PlotPoints → 100, PlotRange → All,

PlotStyle → Table@8Thick, [email protected] iD<, 8i, 0, 10<D, Background → LightGrayD;pt3 = Plot@Evaluate@Table@ψ@2, ζ, ζ0D, 8ζ0, −2, 2, 0.4<DD, 8ζ, −6, 6<,PlotLabel → 82<, PlotPoints → 100, PlotRange → All,

PlotStyle → Table@8Thick, [email protected] iD<, 8i, 0, 10<D, Background → LightGrayD;

104

(b) n = 1 (odd parity)

(c) n = 2 (even parity)

-4 -2 2 4

0.2

0.4

0.6

80<

-4 -2 2 4

-0.6

-0.4

-0.2

0.2

0.4

0.6

81<

-6 -4 -2 2 4 6

-0.4

-0.2

0.2

0.4

0.682<

105

Fig.33. Plot of the wave function with the quantum number (a) n = 0 (an even parity), (b) n = 1 (an odd parity), and (c) n = 2 (even parity) centered around = 0, as a function of . The position 0 is changed as aparameter between -2 and 2.

13. Nucleation at surfaces: Hc3

13.1 Theory

)()]1

()[()(" 2

2

220

zH k ℓ , (13.1)

and the boundary condition: Js|n = 0 at the boundary ( = 0).

0/)( dd at = 0. (13.2) For 00 Hx ℓ , the harmonic oscillator function is nearly zero at = 0 and the

boundary condition is satisfied. The problem is reduced to a model with a 1D parabolic potential well, where x0 is a

constant corresponding to the center of orbit. Let us now consider a double well consisting of two equal parabolic well lying symmetrically to the plane = 0: the wave function is either an even or an odd function). The corresponding wave function should be an even function since 0/)( dd at = 0. The ground level of a particle in the double well is below that in the single wall. The eigenvalue associated with the double well potential is smaller than the eigenvalues associated with the potential 2

0 )( .

)()])[()(" 0

20 g . (13.3)

0)(' at = 0. (13.3)

where

)1

( 2

2

20 zH kg

ℓ . (13.4)

When kz = 0,

)12(2

2

0 ng H

ℓ

, (13.5)


106

Fig.34 The harmonic potential (denoted by red line) and the symmetrical potential

(denoted by dark blue line). From the book of Eugene Merzbacher,16 we have

0)(])(12[)(" 20 . (13.6)

We put

)(2 0 z , (13.7)

where 00 . Then the differential equation can be rewritten as

0)()42

1()("

2

zz

z , (13.8)

where is generally not an integer.

Clear@"Global`∗"D; f1 = Hx − 1L2;f2@x_D := Hx − 1L2 ê; x > 0

f2@x_D := Hx + 1L2 ê; x < 0

Plot@8f1, f2@xD<, 8x, −2, 2<,PlotStyle → 88Thick, Hue@0D<, [email protected], Thick<<,PlotRange → 88−2, 2<, 80, 2<<, AxesLabel → 8"ζêζ0", "VHζL"<,Background → LightGrayD

-2 -1 0 1 2zêz0

0.5

1.0

1.5

2.0VHzL

107

The boundary condition:

(1) 0)( z .

(2) 0)2(' 0 z . (13.9)


)]2

;23

;2

1(

)2/()2/1(

2)

2;

21

;2

(]2/)1[(

)2/1([2)(

2

11

2

114/2/ 2 z

Fzz

FezDz

,

(13.10) where 11 F is the confluent hypergeometric (or Kummer) function. For even )(z the boundary condition is given by

0)2(' 0 zD . (13.11)

Imagine the potential well about = 0 to be extended by a mirror image outside the surface. Lowest eigenfunction of a symmetric potential is itself symmetric; 0)(' at the surface. (i) 0

Wave function )(z will be localized around = 0, and nearly zero at the surface. Therefore the boundary condition will automatically be satisfied.

(ii) 00 .

The function still satisfies the boundary condition and Eq.(13.8). 13.2 Numerical solution by Mathematica

_______________________________________________________________________ ((Mathematica Program-20))

108

_______________________________________________________________________

Fig.35 Plot of as a function of 0, satisfying the condition. 0)2(' 0 zD .

= 0 at 0 = 0, corresponding to the ground-state wave function for the simple harmonic potential.

Nucleationof surface superconductivity

ψ@z_D =

2νê2 ExpB− z2

4F

Gamma@1ê2DGammaB 1−ν

2F

Hypergeometric1F1B− ν

2,1

2,z2

2F +

z

2

Gamma@−1ê2DGammaB −ν

2F

Hypergeometric1F1B1 − ν

2,

3

2,z2

2F ;

eq1 = ψ'@zD êê Simplify;

p11 = ContourPlotBEvaluateBeq1 0 ê. z → − 2 ζ0, 8ζ0, 0, 3<, 8ν, −0.3, 0.1<F,PlotPoints → 50, ContourStyle → 8Red, Thick<, Background → LightGray,

AspectRatio → 1, AxesLabel → 8"ζ0", "ν"<F;

0.0 0.5 1.0 1.5 2.0 2.5 3.0

-0.3

-0.2

-0.1

0.0

0.1

109

Fig.36 Detail of Fig.38. The plot of as a function of 0 around 0 = 0.768 ( =

= -0.20494), satisfying the condition 0)2(' 0 zD .

Fig.37 Plot of as a function of 0, satisfying the condition. 0)2(' 0 zD .

= 2 and = 4 at 0 = 0, corresponding to for the wave function (n = 2

and n = 4). The solid line denotes the curve given by . 2/)1( 20 .

0.75 0.76 0.77 0.78 0.79 0.80

-0.20495

-0.20490

-0.20485

-0.20480

z0

n

0 1 2 3 4 5

-1

0

1

2

3

4

5

6

110

Fig.38 The plot of the ground state wave function with 768.00 and = -

0.20494.

Fig.39 Overview of Fig.41 for the same wave function with 768.00 , = -

0.20494

-0.010 -0.005 0.000 0.005 0.010z

0.10

0.15

0.20

0.25

0.30Dy HarbitraryL

-4 -2 0 2 4z

0.2

0.4

0.6

0.8

1.0

1.2

y

111

It is clear by inspection that this new surface eigenfunction must have a lower eigenfunction that the interior ones, because it arises from a potential curve that is lower and broader than the simple parabora about 0 .

The exact solution shows that this eigenvalue is lower by a factor of 0.59.

).2(695.1695.159.0 2

23 cc

cc HH

HH (13.12)

)21

()21

(2 *

*

*2*

2

cm

Hq

mE

ℏℏ

ℏ. (13.13)

When = n = 0,

cm

Hq

m

c

*

2*

2*

2

22

ℏℏ

(13.14)

or

20

2*2 2

q

cH c

ℏ (13.15)

where

*0

2

q

cℏ .

((Surface superconductivity)) When = -0.20494, 0 = 0.768,

29506.0)5.020494.0(2 *

3*

*

*

2*

2

cm

Hq

cm

Hq

m

cℏℏℏ

(13.16)

or

332*5901.0)229506.0( cc HH

q

c

ℏ

(13.17)

or

322*5901.0 cc HH

q

c

ℏ

, (13.18)

112

or

23 695.1 cc HH . (13.19)

13.3 Variational method by Kittel

We assume a trial function given by

)exp(),( 2axeyxyik y , (13.20)

The parameter a and ky are determined variationally so as to minimize the Gibbs free energy per unit area

2*

*

42)(

2

1

2

1 A

c

q

imfg ns

ℏ, (13.21)

or

])21

(1

[2

2

0

2

2*

2

A

im

fg ns

ℏ, (13.22)

with )0,,0( HxA .

H

kx

y

20

0

. (13.23)

We now calculate

0

2

0

2

2*

2

])21

(1

[2

dxim

G

Aℏ

. (13.24)

After the calculation of G, we put

22

1

a . (13.25)

})]2(4[4{4

20

4220

230

2/322

0

xxHG

We take a derivative of G with respect to x0.

113

20

022

0

)(4

xH

x

G

We have

0x , (13.26)

220

3 6589.122 cc HH

. (13.27)


114

________________________________________________________________________

14. Abrikosov vortex states14

14.1 Theory

(1) All k are orthogonal because of the different ikye factor.

(2) Each of these solutions is equally valid exactly at H = Hc2 and all give the same Hc2.

(3) 4

term is not longer negligible. We expect a crystalline array of vortices to have

lower energy than a random one.

Kittel' s method for the surface superconductivity: Variational method

Clear@"Global`∗"D;conjugateRule = 8Complex@re_, im_D � Complex@re, −imD<;Unprotect@SuperStarD; SuperStar ê: exp_ ∗ := exp ê. conjugateRule;

Protect@SuperStarD;

πx :=1

�D@, xD & ; πy :=

1

�D@, yD +

2 π

Φ0H x & ;

ψ@x_, y_D := ExpA−a x2E Exp@� ky yD; K1 = πx@ψ@x, yDD êê Simplify;

K2 = πy@ψ@x, yDD êê Simplify;

K3 = −1

ξ2ψ@x, yD∗ ψ@x, yD + IK1 K1∗ + K2 K2∗M ê. :ky →

−2 π H x0

Φ0> êê

Simplify;

G1 = ‡0

∞K3 �x; G2 = Simplify@G1, a > 0D; G3 = G2 ê. a →

1

2 ξ2êê Simplify;

G4 = Simplify@G3, ξ > 0D; G5 = D@G4, x0D êê Simplify;

G6 = G4 ê. x0 →ξ

π

êê Simplify

π I4 H2 H−2 + πL π ξ4 − Φ02M4 ξ Φ02

Solve@G6 0, HD êê Simplify

99H → −Φ0

2 H−2 + πL π ξ2=, 9H →

Φ0

2 H−2 + πL π ξ2==

π

π − 2êê N

1.6589

115

y

yy

y

y

yy

y

k H

kyik

k

k

kyik

kLl

xxeCeCyx ]

2

)(exp[]

2

)(exp[),( 2

22 ,

(14.1)

with

Hlx , 20*

2

2

HHq

clH

ℏ, (14.2)

yy kHk lx yHk kly ,

b

nlknlklx HHyHk y

2222 , (14.3)

knnb

ky 2

, (14.4)

where

bk

2 , (14.5)

y

y

k H

Hikny

kLl

knlxeCyx ]

2

)(exp[),( 2

22

, (14.6)

n H

HnL inky

l

nklxCyx ]

2

)(exp[),( 2

22

, (14.7)

Here we assume that nC is independent of n. We define the period in the x direction.

22 2

HH lb

kla

, (14.9)

n H

L inkyl

naxCyx ]

2

)(exp[),( 2

2

, (14.10)

n H

L inkyl

naaxCyax ]

2

)(exp[),( 2

2

, (14.11)

),(])1())1((2

1exp[),( 2

2 yxekynianxl

Ceyax L

iky

n H

iky

L ,

116

(14.12)

n H

Lb

iny

l

naxCyx ]

2

2

)(exp[),( 2

2 . (14.13)

________________________________________________________________________ ((Mathematica Program 22)) Square vortex lattice

________________________________________________________________________

Fig.40 The spatial configuration of 2

near H = Hc2 for a square vortex lattice. a

= 1. b = 1. lH = 0.4 This is a general solution to the linearized GL equation at H = Hc2, periodic by construction. The form of condition chosen by Abrikosov is given by (see also the paper by Kleiner et al.)17

Clear@"Global`∗"D; rule1 = 8a → 1, b → 1, LH → 0.40<;f@m_D := SumBExpB− Hx − n aL2

2 LH2+

2 π � n y

bF, 8n, −m, m<F ê.

rule1

ContourPlotAEvaluateATableAAbs@f@200DD2 α, 8α, 0, 1, 0.05<EE,8x, −1.5, 1.5<, 8y, −1.5, 1.5<, AspectRatio → 1,

ContourStyle → Table@[email protected] iD, Thick<, 8i, 0, 20<D,PlotPoints → 100E

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

117

n H

nnL inky

l

xxCyx ]

2

)(exp[),( 2

2

(14.14)

nn CC ( = integer), (14.15)

n H

nnL inky

l

xxxCyxx ]

2

)(exp[),( 2

2

, (14.16)

or

),(]2

)(exp[),( 2

2)(

yxel

xxeCeyxx L

kyi

n H

nkyni

n

kyi

L

,

(14.17)

Fig.41 Rectangular vortex lattice Each unit cell of the periodic array carriers quanta of flux.

00 2

.2

k

kbHx (4.18)

HH 2

02 ℓ (4.19)

H

H

20

2

2

2

ℓ

(4.20)

22

11

Hℓ

(4.21)

Triangular lattice:

nn CC 2 , iCiCC 01 (4.22)

118

oddn

H

H

evenn

H

H

inkynklxl

i

inkynklxl

Cyx

])(2

1exp[

])(2

1exp[[),(

22

2

22

2

(4.23)

where bk /2 . We define the period in the x direction

22 42 HH l

bkla

(14.25)

We note that ab 3 for a triangular lattice.

Fig.42 Triangular vortex lattice, where red circles denote the centers of vortices.

evenn

H

H

evenn

H

H

kyniknlxl

i

inkynklxl

Cyx

])1())1((2

1exp[

])(2

1exp[[),(

22

2

22

2

(14.26)

or

119

]})1())1((2

1exp[

])(2

1{exp[),(

22

2

22

2

kyniknlxl

i

inkynklxl

Cyx

H

H

evenn

H

H

]})1()2

)1((

2

1exp[

])2

(2

1{exp[),(

2

2

22

kynian

xl

i

inkyna

xl

Cyx

H

evenn H

]})12()2

)12((

2

1exp[

]2)2

2(

2

1{exp[),(

2

2

22

kynian

xl

i

nkyina

xl

Cyx

H

n H

]}))2/1((2

1exp[)exp(

])(2

1){exp[2exp(),(

22

22

anxl

ikyi

naxl

nkyiCyx

H

n H

(14.27)

or

]}))2/1((2

1exp[)

2exp(

])(2

1){exp[

4exp(),(

2

2

22

anxlb

iyi

naxlb

inyCyx

H

n H

(14.28)

________________________________________________________________________ ((Mathematica Program 23)) Triangular vortex lattice

________________________________________________________________________

Clear@"Global`∗"D; rule1 = :a → 1, b → 3 , LH → 0.31>;f@m_D :=

SumBExpB 4 π � n y

bF ExpB− Hx − n aL2

2 LH2F + � ExpB 2 π � y

bF ExpB−

Jx − Jn +1

2N aN2

2 LH2F ,

8n, −m, m<F ê. rule1

ContourPlotAEvaluateATableAAbs@f@200DD2 α, 8α, 0, 2, 0.1<EE, 8x, −1.5, 1.5<,8y, −1.5, 1.5<, AspectRatio → 1,

ContourStyle → Table@[email protected] iD, Thick<, 8i, 0, 20<D, PlotPoints → 100E

120

Fig.43 The spatial configuration of 2

near H = Hc2 for a triangular vortex . a =

1, 3b , and lH = 0.31. 14.2 Structure of the vortex phase near H = Hc2

We start with the following equation. This equation was derived previously.

0])(2

1[

2*

*

42 LLL

c

q

imd Ar

ℏ, (14.29)

where L is a linear combination of the solutions of the linearized GL equation

10 AAA .

A0 is the vector potential which would exist in the presence of the field Hc2. Ginzburg-Landau equation for L is given by

02

12

0

*

*

LL

c

q

im A

ℏ, (14.30)

and

0*

22***

*

*

][2

AJcm

q

im

q L

LLLLL

ℏ, (14.31)

or

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

121

)]()([2

*0

**

0

**

*

*

LLLLLLLc

iq

c

iq

im

q AAJ

ℏℏ

ℏ , (14.32)

or

])()([2

*0

*

0

**

*

*

LLLLLc

q

ic

q

im

q AAJ

ℏℏ. (14.33)

JL is the current associated with the unperturbed solution. Denoting quantities of the form

...1

V by a bar (V is a volume),

0)(2

12*

*

42 LLL

c

q

im A

ℏ. (14.34)

Expanding Eq.(14.34) to the first order in A1, we obtain

01

)(2

11

2

0

*

*

42 LLLL

cc

q

imJAA

ℏ. (14.35)

((Note))

LLLLLm

q

cc

q

imc

q

im *2

1*

2*

1

2

0

*

*

2

10

*

* 2

1)(

2

1)((

2

1AJAAAA

ℏℏ

. (14.35’)

If the right term is negligibly small, we get

LLLcc

q

imc

q

imJAAAA 1

2

0

*

*

2

10

*

*

1)(

2

1)((

2

1

ℏℏ.

((Mathematica Program 24))

122

((Note))

If one multiplies the linearized GL equation by *L and integrates over all the volume by

parts, one obtain the identity

Proof

Clear@"Global`∗"D; << "VariationalMethods`";

Needs@"VectorAnalysis`"D;A0 = 8A01@x, y, zD, A02@x, y, zD, A03@x, y, zD<;A1 = 8A11@x, y, zD, A12@x, y, zD, A13@x, y, zD<;eq1 =

1

2 m

—


q

cHA0 + A1L ψ@x, y, zD .

−—


q

cHA0 + A1L ψc@x, y, zD êê Expand;

eq2 =

1

2 m

—


q

cA0 ψ@x, y, zD .

−—


q

cA0 ψc@x, y, zD êê Expand;

eq12 = eq1 − eq2 êê Simplify;

Current density

Jeq1 =

−q2

m cψ@x, y, zD ψc@x, y, zD A0 +

q —

� 2 mHψc@x, y, zD Grad@ψ@x, y, zDD −

ψ@x, y, zD Grad@ψc@x, y, zDDL êê Simplify;

seq2 = −1

cJeq1.A1 êê Simplify;

eq12 − seq2 êê Simplify

q2 IA11@x, y, zD2 + A12@x, y, zD2 + A13@x, y, zD2M ψ@x, y, zD ψc@x, y, zD2 c2 m

123

0]2

1[

2

0

**

*

2

rA d

c

q

imLLL

ℏ (14.36)

or

0]2

1[

2

0

*

*

2

rA d

c

q

imLL

ℏ (14.37)

or

0)(2

12

0

*

*

2 LL

c

q

im A

ℏ. (14.38)

Then we obtain

01

1

4 LL

cJA , (14.39)

)]()([2

10

**

0

**

*

*

1 AAJAc

iq

c

iq

cim

q

cLLLLLℏℏ

ℏ . (14.40)

Note that

L

s

cJB

4)( , 1)1( AB , (14.41)

)]()([4

1

4

11 )(11

)()(11

sss

Lc

BAABBAJA

,

(14.42) or

)](4

1

4

11 )(1

)1()(1

ss

Lc

BABBJA

. (14.43)

We use the formula of vector analysis

)(11

)()(1 )()( sss BAABBA (14.44)

We note that

124

aBArBA ddss )()( )(

1)(

1 (Stokes theorem).

This value becomes zero on the boundary surface of the system. Then we have

04

1 )1()(4 BB s

L . (14.45)

What are the expressions for )1(

B and )(sB ?

)(

2)1( s

c BHHB , (14.46)

where )(s

B gives the effect of the supercurrent. We introduce the canonical operator,

0

*

ˆ ApΠc

q , (14.47)

which has the commutation relation

Hc

iqyx

ℏ*

]ˆ,ˆ[ . (14.48)

We introduce the raising (creation) and lowering (destruction) operators.

yx iˆˆˆ , yx i

ˆˆˆ , (14.49)

zyx Bc

qi

ℏ*

2]ˆ,ˆ[2]ˆ,ˆ[ . (14.50)

The linearized GL equation can be described by

0ˆˆ L . (14.51) ((Note))

02

12

0

*

*

LL

c

q

im A

ℏ, (14.52)

0)ˆˆ(2

1 22

* LyxLm

. (14.53)

Since

125

zyx Bc

q*22 ˆˆˆˆ ℏ

, (14.54)

zyx Bc

q*

22 ˆˆˆˆ ℏ , (14.55)

we have

0)2(2

1ˆˆ2

1 **

** LzL mBc

q

mm

ℏ, (14.56)

with

)2

(22 *

*

*

*

*

*

q

cmB

cm

qB

cm

qzz

ℏ

ℏℏ . (14.57)

We also note that

220

2**

*

2

2cH

q

c

q

cm

ℏ

ℏ, (14.58)

0)(2

ˆˆ2

12*

*

* LczL HBcm

q

m

ℏ. (14.59)

When ,02 cz HB

0ˆˆ L . (14.60)

The ground state has the property

0 L . (14.61) or

0)A()A( 0

*

0

*

L

yLL

xLL

c

q

yii

c

q

xi

ℏℏ, (14.62)

or

0AA 0

*

0

*

L

yLL

xL

c

qi

yc

q

xi

ℏℏ . (14.63)

126

The substitution of i

LL e into the above equation leads to

00

*

0

*

x

ixyy

iAc

iqA

c

q L

L

L

LL

y

L

x

ℏℏℏℏ .

(14.64) The real part of Eq.(14.64) is

00

*

xyA

c

qL

L

L

x

ℏℏ , (14.65)

or

yA

c

q

x

Lx

L

)( 0

*

ℏ. (14.66)

The imaginary part of Eq.(14.64) is

00

*

yxA

c

qL

L

L

y

ℏℏ , (14.67)

or

xA

c

q

y

Ly

L

)( 0

*

ℏ. (14.68)

Then the current density is given by

ym

q

ym

qA

c

q

xm

qJ

LL

L

x

LLx

2

*

*

*

*

0

*2

*

*

2)(

ℏℏ

ℏ

ℏ, (14.69)

xm

q

xm

qA

c

q

ym

qJ

LL

L

y

LLy

2

*

*

*

*

0

*2

*

*

2)(

ℏℏ

ℏ

ℏ. (14.70)

We see that the current lines are the lines of constant (2

L =const). We note that 2

L

is perpendicular to the lines of constant (2

L =const).

)(2 *

222

yLxxLyLyLxL LJq

m

yxeeee

ℏℏ

(14.71)

and

127

0)(2

)()(2 **

2 LyLxLyLxyLxxLyyLyxLxLL LLJJ

q

mLJLJ

q

m

ℏℏℏℏeeeeJ .

(14.72) Since

0,,

00

4

x

B

y

B

B

zyxc

ss

s

zyx

Ls

eee

JB

, (14.73)

we have

x

B

cxm

q

y

B

cym

q

sL

sL

4

2

4

22

*

*

2

*

*

ℏ

ℏ

(14.74)

or 2

*

*2Ls

cm

qB

ℏ (14.75)

or

2

222

*

*

2

2L

cLs f

H

cm

qB

ℏ (14.76)

where

L

Lf (14.77)

2*

*2cH

q

cm

ℏ

. and

cm

qHc

*

*

22

2

2

ℏ , (14.78)

sc BHHB 21 , (14.79)

0][4

12

4 scsL BHHB

, (14.80)

0)(4

1 22

44 sscL BBHHf

. (14.81)

128

Since

2

22

2 Lc

s fH

B

, (14.82)

we have

0]42

)([4

1 4

4

222

22

2

44 Lc

Lc

cL fH

fH

HHf

, (14.83)

or

0]4

141

21

)1(41 4

4

2

22

44

22

LL

c

L

c

ffH

Hf

H

, (14.84)

or

0)1()2

18(

2

2

4

2

4

22

2 L

c

L

c

fH

Hf

H

. (14.85)

Noting that

222

2

2

2

242 22488

cH , and 2

1

2

c

c

H

H,

(14.86) we get the final form

0)1()2

11(

2

2

4

2 L

c

L fH

Hf

. (14.87)

The quantities 2

Lf and 4

Lf may be calculated, depending on the vortex lattice symmetry

and the lattice spacing. It turns out that near Hc2, the ratio

22

4

L

L

A

f

f (14.88)

is independent of H. Then the above equation can be rewritten by

129

)2

11(

1

2

22

A

cL

H

H

f (14.89)

and

22

2

22

24

)2

11(

1

A

c

LAL

H

H

ff (14.90)

((Note)) Here we follow the discussion given by Tinkham. We assume that

Lfc1 (14.91)

where c1 ia an adjustable parameter. Then the free energy is expressed by

441

221

42

2

1

2

1LLNs fcfcff

Minimizing this with respect to c1

2, we find that

4

22

1

L

L

f

fc

(14.93)

Then the minimum free energy is given by

A

c

AL

L

Ns

H

f

f

ff

18

122

22

4

22

2

If fL is constant, A = 1. If fL is not constant, A is larger than 1. The more A increases, the less favorable is the energy. It is found that A = 1.18 for the square lattice and A = 1.16 for the triangular lattice. The latter case is slightly more stable. 14.3 Magnetic properties near H = Hc2

1

We consider the magnetic properties near H = Hc2 based on the discussion given in the previous section. (a) The magnetic induction B . It is given by

130

22

22

22 Lc

Lc

s fH

HfH

HBHB

, (14.,95)

or

)12()2

11(2

12

2

2

22

A

c

A

c HHH

HHHB . (14.96)

Note that HB at H = Hc2, (b) The magnetization M . It is described by

)12(4

22

A

c HHBHM . (14.97)

The magnetization M is negative as expected (since superconductors are diamagnetic) for H<Hc2. It vanishes at H = Hc2. The transition at H = Hc2 is of second order.

(c) The deviation 22)( BBB

2B ,

2B and

22BB are expressed as follows.

4

4

222

22222

222

4)

2( L

cL

cL

c fH

fHH

HfH

HB

, (14.98)

2

2

4

222

22222

222

4)

2(

L

cL

cL

c fH

fHH

HfH

HB

. (14.99)

Then the deviation B is calculated as

2

2222 )1

2)( LA

c fH

BBB

, (14.100)

or

)12(1)(

22

A

cA

HHB . (14.101)

(d) The Helmholtz free energy F is given by

]8

1)(

2

1

2

1[ 2

2*

*

42

0 BAr

c

q

imdFF

ℏ, (14.102)

131

or

240

8

1

2

1~B

V

FFF L

(14.103)

242244

8

1

8

1

8

1

2

1~BfHBF LcL

, (14.104)

Here we also note that

]2

1)1([

22

22

2

2

22

22 L

c

cLc

cc fH

HHf

HHHHB

. (14.105)

Since

2

2

21)

2

11(

c

LAH

Hf

(14.106)

we have

2222

2 )]12(1[2 LA

cc f

HHB

(14.107)

or

222

2

2

242

2

)]12(1[

4

Ac

cL

H

HBf . (14.108)

Using this relation, we obtain the expression

])1(4

[8

1

28

1~ 22

4

222

22

2

22

L

cL

c fH

BfH

F

)]21(4

[8

1 22

2

4

222

L

c fH

B , (14.109

or

]

)12(1 [

8

1~2

2

22

A

cHBBF . (14.110)

Using this relation, we have

132

H

HBHBBF

B A

cA

A

c

)12(1

)12(

)12(1

~4

22

2

22

(14.111)

since

HHB AcA )]12(1[)12( 22

2 (14.112)

So we can verify the thermodynamic relation as expected. (e) The Gibbs free energy G

~

The Gibbs free energy G~

is related to the Helmholtz free energy F~

by

B

HHBBB

HFG

A

c

4]

)12(1 [

8

1

4~~

2

2

22

. (14.113)

Using the relation

)12( 22

A

c HHHB , (14.114)

we find that

])12(

)([

81~ 2

2

22 H

HHG

A

c

. (14.115)

It is clear that the lower the value of A, the lower the value of G

~. Thus the triangular

lattice is more stable than the square lattice.

14.4 The wall energy at 2/1 .1

We show that the wall energy in a type I superconductor vanishes for 2/1 . We consider the 2D GL equation, where the magnetic field is directed along the z axis.

0ˆˆ2

1 222* yx

m, (14.116)

where

xxxx Ac

qpA

c

q

xi

**

ˆˆ ℏ,

yyyy Ac

qpA

c

q

yi

**

ˆˆ ℏ,

(14.117)

133

yx iˆˆˆ , yx i

ˆˆˆ . (14.118)

We note that

]ˆ,ˆ[2]ˆ,ˆ[ yxi , (14.119)

or

],ˆ[],ˆ[]ˆ,ˆ[]ˆ,ˆ[****

yyyxyyxxyx Apc

qAp

c

qA

c

qpA

c

qp . (14.120)

For any arbitrary function f, we have,

fxi

AfAxi

fAxi

fAp yyyyx

ℏℏℏ

,],ˆ[

fx

A

ix

f

iAf

x

A

ix

fA

i

y

y

y

y

ℏℏℏℏ

. (14.121)

Similarly, we have

fy

A

ifAp x

xy

ℏ

],ˆ[ . (14.122)

Then we get the relation

zxy

yx Bc

qi

y

A

ix

A

ic

q**

)(]ˆ,ˆ[ℏℏℏ

, (14.123)

where

y

A

x

AB xy

z

.

Using this commutation relation, we get

zyx Bc

qi

ℏ*

2]ˆ,ˆ[2]ˆ,ˆ[ . (14.124)

Since

]ˆ,ˆ[ˆˆ)ˆˆ)(ˆˆ(ˆˆ 22yxyxyxyx iii , (14.125)

134

we get

zyx Bc

q*

22 ˆˆˆˆ ℏ . (14.126)

Then the GL equation can be rewritten as

02

ˆˆ2

1 2

*

*

* zBcm

q

m

ℏ. (14.127)

We now look for the solution such that 0ˆ .

0ˆ**

yx Ac

q

yiiA

c

q

xi

ℏℏ, (14.128)

or

)(*

yx iAAc

q

xi

y

ℏ. (14.129)

When 0ˆ , the GL eq. reduces to

02

2

*

*

zBcm

qℏ. (14.130)

Note that the current density is given by

])()([2

***

**

*

AAJc

q

ic

q

im

qs

ℏℏ (14.131)

Then we have

]))1

()1

([2

0,, ***

**

*

AABℏℏ

ℏ

c

q

ic

q

icm

q

x

B

y

B zz

,

(14.132) or

])1

()1

([2 *

***

*

*

xxz A

c

q

xiA

c

q

xicm

q

y

B

ℏℏ

ℏ

, (14.133)

and

135

***

**

*

)1

()1

([2

yyz A

c

q

yiA

c

q

yii

cm

q

x

Bi

ℏℏ

ℏ

. (14.134)

Then we have

])(2

)()([2 *

****

*

*

yxzz iAA

c

q

yxi

xi

ycm

q

x

Bi

y

B

ℏ

ℏ,

(14.135) or

)]()([2 **

**

*

xi

yxi

ycm

q

x

Bi

y

B zz

ℏ

, (14.136)

This relation must agree with the relation given by Using the relation (14.130), we have

)](2

[)](2

[ **

**

*

*

q

cm

xi

q

cm

yx

Bi

y

B zz

ℏℏ, (14.137)

or

0)]()(2

****

*

x

iyx

iyx

Bi

y

B

cm

q zzℏ, (14.138)

From the comparison between Eqs.(14.137) and (14.138), we find that

2*

2*

2

2

m

q

c

ℏ . (14.139)

Since is defined by

*

*

2 q

cm

ℏ

we obtain

2

1

2 *

*

*

*

m

q

cq

cm ℏ

ℏ

. (14.141)

((Note))

136

)2

(2

)(2

)(2 2

2*

2

*

*2

*

*2

*

*

mq

cm

q

cm

q

cmBz

ℏ

ℏℏℏ, (14.142)

or

22*

*2

2*

2

*

* 42)

2(

2

cm

qH

mq

cmB cz

ℏℏ

ℏ , (14.143)

where

cc HHc

c

c

qmq

cm

2

22

1

2

222

02

02*2*

2

*

*

ℏ

ℏ

ℏℏ

ℏ, (14.145)

and

22*

*22

2*2

2*2

*

*2

*

* 4222

cm

q

mc

q

q

cm

q

cm ℏℏ

ℏℏ . (14.146)

One can estimate the surface energy given by

])(8

1

2

1[ 24

cz HBdx

. (14.147)

In our case,

22*

*42

cm

qHB cz

ℏ , (14.148)

with

22*

2*2

22cm

qℏ

Then the integrand of is described by

4

22*

2*2

2

2

22*

*24

2 4)12(

828

cm

q

cHm

qHHB

c

cczℏℏ

. (14.150)

This is equal to 0 when 2/1 , leading to = 0.

137

15. Conclusion

We show that the phenomena of the superconductivity can be well explained in terms of the Ginzburg-Landau theory. The superconductors are classed into two types of

superconductors, type I and type-II. The GL parameter has a limiting value 2/1

separating superconductors with positive surface energy ( 2/1 ) (type-I

superconductors) from those with negative surface energy at large 2/1 (type-II superconductors. The instability noticed by Ginzburg and Landau allowed the magnetic field to enter the superconductor without destroying it. This leads to the appearance of the mixed phase (the Abrikosov phase), where superconducting and normal regions coexist. The normal regions appear in the cores (of size ) of vortices binding individual magnetic flux quanta 0 = hc/2e on the scale , with the charge '2e' appearing in 0 a consequence of the pairing mechanism; since >, the vortices repel and arrange in a stable lattice. In his 1957 paper, Abrikosov derived the periodic vortex structure near the upper critical field Hc2. In 2003, Abrikosov and Ginzburg got the Nobel prize in physics for their "pioneering contributions to the theory of superconductors. References

1. P. G. Gennes, Superconductivity of Metals and Alloys (W.A. Benjamin, Inc. 1966). 2. D. Saint-Janes, E.J. Thomas, and G. Sarma, Type II Superconductivity (Pergamon

Press, Oxford, 1969). 3. S. Nakajima, Introduction to Superconductivity (in Japanese) (Baifukan, Tokyo,

1971). 4. M. Tinkham, Introduction to Superconductivity (Robert E. Grieger Publishing Co.

Malabar, Florida 1980). 5. J.B. Ketterson and S.N. Song, Superconductivity (Cambridge University Press

1999). 6. L.M. Lifshitz and Pitaefski, Statistical Physics part 2 Theory of the Condensed State,

Vol.9 of Course of Theoretical Physics (Pergamon Press, Oxford, 1991). 7. T. Tsuneto, Superconductivity and Superfluidity (Cambridge University Press 1998). 8. M. Cyrot, Reports on Prog. Phys. 36, 103 (1973). 9. G. Blatter, M.V. Feigel’man, V.B. Geshkenbein, A.I. Larkin, and V.M. Vinokur,

Rev. Mod. Physics 66, 1125 (1994). 10. M. Trott, The Mathematica Guide Book Vol.1-4 (Springer Verlag, 2006). 11. V.L. Ginzburg and L.D. Landau, Zh. Exp. Theor. Fiz 20, 1064 (1950). 12. L.D. Landau, Phys. Z. der Sowjet Union , 11, 26 (1937); ibid. 11, 129 (1937). 13. L.P. Gorkov, Sov. Phys. JETP 9, 1364 (1959). 14. A.A.A. Abrikosov, Soviet Physics, JETP 5, 1174 (1957). 15. J.J. Sakurai, Modern Quantum Mechanics Revised Edition (Addison-Wesley,

Reading, Massachusetts, 1994). 16. E. Merzbacher, Quantum Mechanics 3rd edition (John Wiley & Sons, New York

1998). 17. W.H. Kleiner, L.M. Roth, and S.H. Autler, Phys. Rev. 113, A1226 (1964). Appendix

A. Superconducting parameters

138

The definition of the parameters are given in the text.

Thermodynamic field 24

cH .

Order parameter /*2 sn .

The parameters *

2

4 s

c

n

H

, 2*

2

4 s

c

n

H

.

Quantum fluxoid *0

2

q

cℏ .

Magnetic field penetration depth

2*

2*

4 q

cm 2**

2*

4 qn

cm

s

Coherence length

*2m

ℏ

*

*2

mH

n

c

s ℏ

Ginzburg-Landau parameter *

*

2 q

cm

ℏ

**

*

22 qn

Hcm

s

c

ℏ .

Some relations between Hc, Hc1, and Hc2

cH

22

1

0

, ℏc

Hq c

*

2

2

.

cH

2202 , 0

222

cH

.

22 0

cH , 2

1

2

c

c

H

H.

Upper critical field (type II) 20

2 2

cH .

Lower critical field (type II) )ln(2

)ln(4 2

01

c

c

HH

.

139

Surface-sheath field 23 695.1 cc HH .

A.2. Modified Bessel functions

The differential equation for the modified Bessel function is given by

0)(''' 222 ynxxyyx ,

0)1('1

'' 2

2

yx

ny

xy ,

The solution of this equation is

)()( 21 xKcxIcy nn ,

where

)()( ixJixI n

n

n

,

)(2

)]()([2

)( )1(11 ixHixiNixJixK n

n

nn

n

n

,

Recursion formula:

)()]([

)()]([

)()()('

)()()('

)(2

)()(

1

1

1

1

11

xKxxKxdx

d

xKxxKxdx

d

xKx

nxKxK

xKx

nxKxK

xKx

nxKxK

n

n

n

n

n

n

n

n

nnn

nnn

nnn

,

The parameter is the Euler-Mascheroni constant: = 0.57721.

xxK

xxx

xK

1)(

11593718.0ln2lnln)2

ln()(

1

0

,

for 0x .

140

_______________________________________________________________________ ((Mathematica program-25))

________________________________________________________________________

Fig. Modified Bessel function In(x) with n = 0, 1, 2, 3, 4, and 5.

Clear@"Global`∗"D;p1 = Plot@Evaluate@Table@BesselI@n, xD, 8n, 0, 5<DD, 8x, 0.1, 4<,

PlotPoints → 50, AxesLabel → 8"x", "BesselI@n,xD"<,PlotStyle → Table@8Thick, [email protected] iD<, 8i, 0, 5<D, Background → LightGray,

Epilog → 8Text@Style@"n=0", Black, 12D, 81.8, 2<D,Text@Style@"n=1", Black, 12D, 82.3, 2<D,Text@Style@"n=2", Black, 12D, 83, 2<D,Text@Style@"n=3", Black, 12D, 83.5, 2<D,Text@Style@"n=4", Black, 12D, 83.5, 0.8<D,Text@Style@"n=5", Black, 12D, 83.5, 0.2<D<D;

p2 = Plot@Evaluate@Table@BesselK@n, xD, 8n, 0, 1<DD, 8x, 0, 1.5<,PlotPoints → 50, PlotStyle → Table@8Thick, [email protected] iD<, 8i, 0, 5<D,Background → LightGray, AxesLabel → 8"x", "BesselK@n,xD"<,Epilog → 8Text@Style@"n=0", Black, 12D, 80.4, 0.8<D,

Text@Style@"n=1", Black, 12D, 80.4, 3<D<D;

1 2 3 4x

1

2

3

4

5

6

BesselI@n,xD

n=0 n=1 n=2 n=3

n=4

n=5

141

Fig. Modified Bessel function Kn(x) with n = 0 and 1.

0.2 0.4 0.6 0.8 1.0 1.2 1.4x

1

2

3

4

5BesselK@n,xD

n=0

n=1

Ginzburg-Landau Theory for Superconductivity Masatsugu Sei...

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