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1
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.
2
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
3
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
4
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
5
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)
6
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,
7
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
9
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)
10
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
11
')'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
12
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)
14
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)
15
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)
17
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)
18
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
—
�Grad@ψ@x, y, zDD −
q
cA ψ@x, y, zD . −
—
�Grad@ψc@x, y, zDD −
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
—
�Grad@ψ@x, y, zDD −
q
cA ψ@x, y, zD .
−—
�Grad@ψc@x, y, zDD −
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
((Mathematica Program-7))
_____________________________________________________________________ (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. ________________________________________________________________________
((Mathematica Program-13))
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;
eq4 = VariationalD@eq1, h3@x, y, zD, 8x, y, z<D ê2 êê Expand;
eq2 = VariationalD@eq1, h1@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.
_______________________________________________________________________
((Mathematica Program-15))
-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. _______________________________________________________________________
((Mathematica Program-17))
_______________________________________________________________________
(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)
The solution of this differential equation is
)()!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)
((Mathematica Program-19))
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)
The solution of this differential equation is
)]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)
((Mathematica Program-21))
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
—
�Grad@ψ@x, y, zDD −
q
cHA0 + A1L ψ@x, y, zD .
−—
�Grad@ψc@x, y, zDD −
q
cHA0 + A1L ψc@x, y, zD êê Expand;
eq2 =
1
2 m
—
�Grad@ψ@x, y, zDD −
q
cA0 ψ@x, y, zD .
−—
�Grad@ψc@x, y, zDD −
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