Lecture Note on Solid State Physics
Ginzburg-Landau Theory for Superconductivity
Masatsugu Suzuki and Itsuko S. Suzuki
Department of Physics, State University of New York at Binghamton, Binghamton, New
York 13902-6000
(Date January 16, 2007)
Abstract
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.
This article was originally prepared for a lecture (by M.S.) of the Solid State Physics
course (1989). Since then, the note has been revised many times. The use of Mathematica
was very useful in preparing this note. The students studying the solid state physics will
avail themselves of the Mathematica for solving problems and visualizing the
propositions. In particular, the variational packages (the VariationalD and FirstIntegrals
for the variational calculation) are very useful for the derivation of the GL equation and
the current density from the Helmholtz free energy.
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
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
1
3.4
Flux quantization
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 ξ
5.3 Magnetic field penetration depth λ
5.4 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: κ = 1 / 2
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«λ and d«ξ.
8.5.2 The case of d»λ and d»ξ.
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
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
4.
2
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 κ = 1 / 2 .
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 ψ = ψ e iθ ). The amplitude ψ 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 ψ and its spatial
derivative of ψ . In general this is valid in the vicinity of Tc below Tc, where the
amplitude ψ is small and the length scale for spatial variation is long.
The order parameter ψ 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 ψ (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 ψ (r ) is
introduced as a complex order parameter. The parameter ψ (r ) represents the local
2
density of superconducting electrons, ns (r ) . The macroscopic behavior of
superconductors (in particular the type II superconductors) can be explained well by this
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 κ is the ratio of λ to ξ, where λ is the
magnetic-field penetration depth and ξ is the coherence length of the superconducting
phase. The limiting value κ = 1 / 2 separating superconductors with positive surface
3
energy (κ < 1 / 2 ) (type-I) from those with negative surface energy (κ > 1 / 2 ) (typeII), is properly identified. For the type-II superconductor, the superconducting and normal
regions coexist. The normal regions appear in the cores (of size ξ) of vortices binding
individual magnetic flux quanta Φ 0 = 2π c / q* on the scale λ, with the charge
q* = 2 e appearing in Φ0 a consequence of the pairing mechanism. Since λ>ξ, 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
∇ ⋅ E = 4πρ
1 ∂B
∇×E = −
c ∂t
,
(2.1)
∇
⋅
B
=
0
4π
1 ∂E
∇ × B =
j+
c
c ∂t
B: magnetic induction (the microscopic magnetic field)
E: electric field
ρ: charge density
J: current density
c: the velocity of light
H: the applied external magnetic field
The Lorentz force is given by
1
F = q[E + ( v × B )] .
(2.2)
c
The Lorentz force is expressed in terms of fields E and B (gauge independent, see the
gauge transformation below)).
The equation of continuity
∇⋅ j +
∂ρ
= 0,
∂t
(2.3)
B = ∇× A ,
1 ∂A
− ∇φ ,
E=−
c ∂t
where A is a vector potential and φ: scalar potential.
2.2
(2.4)
(2.5)
Lagrangian of particles with mass m* and charge q* in the presence of
electric and magnetic field
4
The Lagrangian L is given by
1
1
L = m*v 2 − q* (φ − v ⋅ A ) ,
(2.6)
2
c
where m* and q* are the mass and charge of the particle. Canonical momentum
∂L
q*
= m* v + A .
(2.7)
p=
∂v
c
Mechanical momentum (the measurable quantity)
q*
*
= m v =p− A.
(2.8)
c
The Hamiltonian is given by
q*
1
1
q* 2
H = p ⋅ v − L = (m* v + A) ⋅ v − L = m* v 2 + q*φ =
p
−
A ) + q*φ .
(
*
c
2
2m
c
(2.9)
The Hamiltonian formalism uses A and φ, and not E and B, directly. The result is that the
description of the particle depends on the gauge chosen.
Gauge transformation: Analogy from classical mechanics15,16
When E and B are given, φ and A are not uniquely determined.
If we have a set of possible values for the vector potential A and the scalar potential φ, we
obtain other potentials A’ and φ’ which describes the same electromagnetic field by the
gauge transformation,
A' = A + ∇χ ,
(2.10)
1 ∂χ
φ'= φ −
,
(2.11)
c ∂t
where χ 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,
r ' = r and
v' = v ,
or
(2.12)
'= ,
*
q
Since = m* v = p − A , we have
c
*
q
q*
p'− A ' = p − A ,
(2.13)
c
c
or
q*
q*
p' = p + (A '− A ) = p + ∇χ .
(2.14)
c
c
In the Hamilton formalism, the value at each instant of the dynamical variables
describing a given motion depends on the gauge chosen.
2.3
2.4
Gauge invariance in quantum mechanics
In quantum mechanics, we describe the states in the old gauge and the new gauge as
5
We denote ψ and ψ ' 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.
ψ ' rˆ ψ ' = ψ rˆ ψ (gauge invariant),
(2.15)
ψ ' ˆ ψ ' = ψ ˆ ψ (gauge invariant),
(2.16)
*
q
∇χ ψ .
(2.17)
c
We now seek a unitary operator Û which enables one to go from ψ to ψ ' :
ψ ' pˆ ψ ' = ψ pˆ +
ψ ' = Û ψ .
(2.18)
From the condition, ψ ' ψ ' = ψ ψ , we have
Uˆ +Uˆ = UˆUˆ + = 1̂ .
From the condition, ψ ' rˆ ψ ' = ψ rˆ ψ ,
(2.19)
Uˆ + rˆUˆ = rˆ ,
(2.20)
ˆ
∂U
.
[rˆ ,Uˆ ] = 0 = i
∂pˆ
(2.21)
or
Û is independent of p̂ .
q
From the condition, ψ ' pˆ ψ ' = ψ pˆ + ∇χ ψ ,
c
q
Uˆ +pˆ Uˆ = pˆ + ∇χ ,
(2.22)
c
or
q ˆ
∂ ˆ
ˆ
ˆ
[p, U ] = + U∇χ =
U.
(2.23)
c
i ∂rˆ
Now we consider the form of Û . The unitary operator Û commutes with r̂ and χ:
iq* χ
Uˆ = exp( ) .
(2.24)
c
The wave function is given by
iq * χ
(2.25)
r ψ ' = r Uˆ ψ = exp( ) r ψ .
c
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
q*
q*
(2.26)
ψ ' pˆ − A' ψ ' = ψ pˆ − A ψ ,
c
c
q*
q*
q*
q*
ˆ
ˆ
ˆ
ψ ' p − A ' ψ ' = ψ p + ∇χ − ( A + ∇χ ) ψ = ψ p − A ψ ,
c
c
c
c
(2.27)
since
6
ψ ' pˆ ψ ' = ψ pˆ +
q*
∇χ ψ ,
c
(2.28)
and
ψ'
2.5
q*
q*
q*
A' ψ ' = ψ Uˆ + A 'Uˆ ψ = ψ
( A + ∇χ ) ψ .
c
c
c
Hamiltonian under the gauge transformation
∂
i
ψ = Hˆ ψ ,
∂t
∂
i
ψ ' = Hˆ ' ψ ' ,
∂t
∂ ˆ
i
U ψ = Hˆ 'Uˆ ψ ,
∂t
(2.29)
(2.30)
(2.31)
(2.32)
or
∂Uˆ
∂
ψ + Uˆ ψ ] = Hˆ 'Uˆ ψ .
∂t
∂t
*
∂Uˆ iq ∂χ ˆ
= Since
U,
∂t
c ∂t
q* ∂χ ˆ
∂
U ψ + Uˆi ψ = Hˆ 'Uˆ ψ ,
−
c ∂t
∂t
or
q* ∂χ ˆ
U ψ + UˆHˆ ψ = Hˆ 'Uˆ ψ ,
−
c ∂t
or
q* ∂χ ˆ ˆ ˆ
Hˆ 'Uˆ = −
U + UH .
c ∂t
Thus we have
q* ∂χ ˆ ˆ ˆ +
Hˆ ' = −
+ UHU .
c ∂t
Note that
q
Uˆ +pˆ Uˆ = pˆ + ∇χ ,
c
or
q
pˆ Uˆ = Uˆ (pˆ + ∇χ ) ,
c
or
q
pˆ Uˆ = Uˆpˆ + Uˆ∇χ ,
c
or
q
Uˆpˆ Uˆ + = (pˆ − ∇χ ) .
c
i [
7
(2.33)
(2.35)
(2.36)
(2.37)
(2.38)
(2.39)
(2.40)
(2.41)
(2.42)
From this relation, we also have
q*
q*
q*
q*
+
ˆ
ˆ
U (pˆ − A)U = (pˆ − A − ∇χ ) = (pˆ − A' ) ,
(2.43)
c
c
c
c
q*
q*
Uˆ (pˆ − A) 2Uˆ + = (pˆ − A ' ) 2 .
(2.44)
c
c
Then
q* ∂χ ˆ 1
q* 2
ˆ
ˆ
H '= −
(2.45)
+ U [ * (p − A) + q*φ ]Uˆ + ,
c ∂t
2m
c
or
q* ∂χ
1
q*
1
q*
2
*
ˆ
ˆ
Hˆ ' = −
+
(
p
−
A
'
)
+
=
p
−
A ' )2 + q*φ ' . (2.46)
q
φ
(
*
*
c ∂t 2m
c
2m
c
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.
1
q*
J = * Re[ ψ pˆ − A ψ ] ,
(2.47)
m
c
q*
q*
+
ˆ
ˆ
ˆ
(2.48)
ψ ' p − A ' ψ ' = ψ U (p − A ' )Uˆ ψ ,
c
c
q*
q*
q*
q*
(2.49)
Uˆ + (pˆ − A' )Uˆ = pˆ + ∇χ − A ' = pˆ − A ,
c
c
c
c
q*
q*
1
1
J ' = * Re[ ψ ' pˆ − A ' ψ ' ] = * Re[ ψ pˆ − A ψ ] .
(2.50)
m
c
m
c
The density is also gauge independent.
ρ'= r ψ '
2
= ρ = rψ
2
.
(2.51)
3.
3.1
Basic concepts
London’s equation
We consider an equation given by
q*
*
p = m v + A.
c
We assume that p = 0 or
(3.1)
q*
A =0.
c
The current density is given by
m* v +
(3.2)
q* ψ
J = q ψ v = − * A . (London equation)
(3.3)
mc
This equation corresponds
to a London
equation.
From this equation, we have
2
2
2
2
q* ψ
q* ψ
∇×J = − * ∇×A = − * B .
(3.4)
mc
mc
2
*
2
2
8
Using the Maxwell’s equation
4π
∇×B =
J , and ∇ ⋅ B = 0 ,
c
we get
4π
4πn*q*
∇× J = − * 2 B,
c
mc
2
∇ × (∇ × B ) =
(3.5)
where
n* = ψ = constant (independent of r)
2
λL =
m*c 2
2
4πn*q*
2
: λL penetration depth.
Then
∇ × (∇ × B ) = ∇(∇ ⋅ B ) − ∇ 2B = −
1
λL 2
B,
(3.6)
or
∇ 2B =
1
λL 2
B.
(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
q ψ
q*
J=
A.
[ψ *∇ψ − ψ∇ψ * ] −
*
2m i
m*c
Now we assume that
ψ = ψ e iθ .
Since
*2
2
ψ *∇ψ − ψ∇ψ * = 2iψ ∇θ ,
2
we have
(3.8)
(3.9)
(3.10)
q*
q*
2
2
J = * ψ (∇θ − A) = q * ψ v s
m
c
(3.11)
or
q*
(3.12)
A + m* v s .
c
This equation is generally valid. Note that J is gauge-invariant. Under the gauge
transformation, the wave function is transformed as
iq * χ
ψ ' (r ) = exp(
)ψ (r ) .
(3.13)
c
This implies that
q* χ
θ → θ '= θ + ,
(3.14)
c
Since A' = A + ∇χ , we have
∇θ =
9
q*
q* χ
q*
q*
A' ) = [∇(θ +
) − (A + ∇χ )] = (∇θ − A) . (3.15)
c
c
c
c
So the current density is invariant under the gauge transformation.
Here we note that
ψ (r ) = ψ (r ) exp[iθ (r )] .
(3.16)
and
J ' = (∇θ '−
pψ (r ) =
∇{exp[iθ (r )]ψ (r ) } = [iψ (r )∇θ (r ) + exp[iθ (r )]∇ψ (r ) ] .
i
(3.17)
If ψ (r ) is independent of r, we have
i
pψ (r ) = [ ∇θ (r )]ψ (r )
(3.18)
or
p = ∇θ (r ) .
Then we have the following relation
q*
p = ∇θ = A + m* v s
c
when ψ (r ) is independent of r.
(3.19)
(3.20)
3.3. London gauge
2
*
Here we assume that ψ = ns = constant. Then we have
q*
p = ∇θ = A + m* v s ,
c
2
*
*
J s = q ψ v s = q*ns v s .
(3.21)
(3.22)
Then
p = ∇θ =
q*
m*
A + * * Js .
c
q ns
(3.23)
We define
Λ=
m*
.
* 2
ns q *
Now the above equation is rewritten as
q*
p = ∇θ = A + q*ΛJ s ,
c
*
q
∇ × p = (∇ × A ) + q*Λ (∇ × J s ) = 0 ,
c
or
1
∇ × ( A + Js ) = 0 ,
cΛ
or
1
B + ∇ × Js = 0 .
cΛ
10
(3.24)
(3.25)
(3.26)
(3.27)
(3.28)
From the expression
q*
p=
A + q * ΛJ s
c
and p = 0, we have a London equation
1
Js = −
A.
cΛ
(3.29)
(3.30)
1
∇ ⋅ A = 0 or
cΛ
(3.31)
∇⋅A = 0.
From the condition that no current can pass through the boundary of a superconductor, it
is required that J s ⋅ n = 0 , or
(3.32)
A ⋅n = 0.
The conditions ( ∇ ⋅ A = 0 and A ⋅ n = 0 ) are called London gauge.
We now consider the gauge transformation: A' = A + ∇χ
1
(3.33)
Js '= −
A' ,
cΛ
1
1
(3.34)
∇ ⋅ Js '= −
∇ ⋅ A' = −
(∇ ⋅ A + ∇ 2 χ ) = 0 ,
cΛ
cΛ
1
1
J s '⋅n = −
A '⋅n = −
( A ⋅ n + ∇χ ⋅ n ) .
(3.35)
cΛ
cΛ
The London equation is gauge-invariant if we throw away any part of A which does not
satisfy the London gauge. A' = A + ∇χ , where ∇ 2 χ = 0 and ∇χ ⋅ n = 0 are satisfied,
For the supercurrent to be conserved, it is required that ∇ ⋅ J s = −
3.4
Flux quantization
We start with the current density
q*
q*
2
2
J s = * ψ (∇θ − A) = q * ψ v s .
m
c
2
*
Suppose that ns = ψ =constant, then we have
∇θ =
m*
q*
+
J
A ,
* s
c
q* ns
(3.36)
(3.37)
or
m* q* (3.38)
d
J
⋅
l
+
A ⋅ dl .
s
*
c
q* ns
The path of integration can be taken inside the penetration depth where J s =0.
∇θ ⋅ dl =
q*
q*
q*
q*
A ⋅ dl =
(∇ × A ) ⋅ da =
B ⋅ da =
Φ ,
c
c
c
c
where Φ is the magnetic flux. Then we find that
q*
∆θ = θ 2 − θ1 = 2πn = Φ
c
∇θ ⋅ dl =
11
(3.40)
(3.41)
where n is an integer. The phase θ of the wave function must be unique, or differ by a
multiple of 2π at each point,
2πc
Φ = * n.
(3.42)
q
The flux is quantized. When |q*| = 2|e|, we have a magnetic quantum fluxoid;
2πc
ch
= 2.06783372 × 10-7 Gauss cm2
(3.43)
Φ0 =
=
2e
2e
4.
4.1
Ginzburg-Landau theory-phenomenological approach
The postulated GL equation
We introduce the order parameter ψ (r ) with the property that
ψ * (r )ψ (r ) = ns (r ) ,
(4.1)
which is the local concentration of superconducting electrons. We first set up a form of
the free energy density Fs(r),
2
q*
B2
1
1
4
ψ
Fs (r ) = FN + α ψ + β ψ +
∇
−
A
+
(
)
,
c
2
2m* i
8π
2
(4.2)
where β is positive and the sign of α 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 ψ(r) and the
vector potential
A(r). We set
ℑ = Fs (r )dr
(4.3)
where the integral is extending over the volume of the system. If we vary
ψ (r ) → ψ (r ) + δψ (r ) and A (r ) → A(r ) + δA (r ) ,
(4.4)
we obtain the variation in the free energy such that
ℑ + δℑ .
By setting δℑ = 0 , we obtain the GL equation
2
1 q* αψ + β ψ ψ + * ∇ − A ψ = 0 ,
2m i
c
and the current density
2
q ψ
q*
J s = * [ψ *∇ψ − ψ∇ψ * ] −
A
2m i
m*c
*2
(4.5)
2
(4.6)
or
q*
q*
q*
*
∇ −
∇ −
[
ψ
(
A
)
ψ
+
ψ
(
−
A)ψ * ] .
(4.7)
*
2m
i
c
i
c
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.
n ⋅ Js = 0 .
(4.8)
Js =
4.2.1 Derivation of GL equation by variational method
12
((Mathematica program-1))
Variational method using “VarialtionalD” of the Mathematica 5.2.
(*Derivation of Ginzburg Landau equation*)
<<Calculus`VariationalMethods`
Needs["Calculus`VectorAnalysis`"]
SetCoordinates[Cartesian[x,y,z]]
Cartesian[x,y,z]
A={A1[x,y,z],A2[x,y,z],A3[x,y,z]}
{A1[x,y,z],A2[x,y,z],A3[x,y,z]}
eq1
1
2m
x, y, z
c x, y, z
1
2
x, y, z
2
c x, y, z
2
q A x, y, z .
c
q
Grad c x, y, z
Expand;
A c x, y, z
c
eq2 VariationalD eq1, c x, y, z , x, y, z
Expand
Grad x, y, z
q2 A1 x, y, z 2 x, y, z 2 c2 m
q2 A2 x, y, z 2 x, y, z q2 A3 x, y, z 2 x, y, z 2 c2 m
2 c2 m
q x, y, z A3 0,0,1 x, y, z x, y, z 2 c x, y, z 2cm
x, y, z q A3 x, y, z 0,0,1 x, y, z cm
q
2
2
x, y, z A2 0,1,0 x, y, z 2cm
0,2,0
0,0,2
x, y, z 2m
q A2 x, y, z 0,1,0 x, y, z cm
A1 1,0,0
x, y, z q
x, y, z x, y, z 2m
2cm
2 2,0,0 x, y, z q A1 x, y, z 1,0,0 x, y, z cm
2m
eq3=VariationalD[eq1, [x,y,z],{x,y,z}]//Expand
q2 A1 ! x, y, z " 2 c ! x, y, z "
#
c ! x, y, z " #
2 c2 m
q2 A2 ! x, y, z " 2 c ! x, y, z "
q2 A3 ! x, y, z " 2 c ! x, y, z "
#
#
2 c2 m
2 c2 m
)
q ' c ! x, y, z " A3 (0,0,1 ! x, y, z "
$
c ! x, y, z " 2 % &
! x, y, z "
2cm
)
&
q ' A3 ! x, y, z " c (0,0,1 ! x, y, z "
cm
%
'
%
&
)
&
'
&
q'
2
c ! x, y, z " A2 (0,1,0 ! x, y, z "
2cm
2
%
)
c (0,0,2 ! x, y, z "
2m
%
)
q ' A2 ! x, y, z " c (0,1,0 ! x, y, z "
cm
)
)
c (0,2,0 ! x, y, z " % & q ' c ! x, y, z " A1 (1,0,0 ! x, y, z "
2m
2cm
)
)
q ' A1 ! x, y, z " c (1,0,0 ! x, y, z " % ' 2 c (2,0,0 ! x, y, z "
cm
2m
(*We need to calculate the following*)
13
%
%
OP1 :
OP2 :
D #, y
OP3 :
D #, x
D #, z
q
A1 x, y, z # &
c
q
A2 x, y, z # &
c
q
A3 x, y, z # &
c
eq4 x, y, z x, y, z 2 c x, y, z 1
OP1 OP1 x, y, z OP2 OP2 x, y, z 2m OP3 OP3 x, y, z Expand
q2 A1 x, y, z 2 x, y, z 2 c2 m
q2 A2 x, y, z 2 x, y, z q2 A3 x, y, z 2 x, y, z 2 c2 m
2 c2 m
q x, y, z A3 0,0,1 x, y, z 2 x, y, z c x, y, z 2cm
x, y, z q A3 x, y, z 0,0,1 x, y, z cm
q
2
x, y, z A2 0,1,0 x, y, z 2cm
0,2,0
2
0,0,2
x, y, z 2m
q A2 x, y, z 0,1,0 x, y, z cm
A1 1,0,0
x, y, z q
x, y, z x, y, z 2m
2cm
2 2,0,0 x, y, z q A1 x, y, z 1,0,0 x, y, z cm
2m
eq2-eq4//Simplify
0
4.2.2. Derivation of current density by variational method
((Mathematica Program-2))
(*Derivation of Current density variational method*)
<<Calculus`VariationalMethods`;Needs["Calculus`VectorAnalysi
s`"]
SetCoordinates[Cartesian[x,y,z]];A={A1[x,y,z],A2[x,y,z],A3[x
,y,z]};H={H1,H2,H3}
{H1,H2,H3}
eq1 1
2m
q
A x, y, z .
c
q
A c x, y, z Grad c x, y, z c
1
1
Curl A . Curl A H.Curl A Expand;
8 4
eq2 VariationalD eq1, A1 x, y, z , x, y, z! Expand
Grad x, y, z 14
q2 A1 x, y, z x, y, z c x, y, z A1 0,0,2 x, y, z c2 m
4
0,2,0
1,0,0
A1 x, y, z q c x, y, z x, y, z 4
2cm
q x, y, z c 1,0,0 x, y, z A3 1,0,1 x, y, z A2 1,1,0 x, y, z 2cm
4
4
(*Current density in Quantum Mechanics*)
eq3 1 c
q2 Curl Curl A c x, y, z A x, y, z c 4
mc
q c x, y, z Grad x, y, z x, y, z Grad c x, y, z ;
2m
eq4 eq3 1 Expand
q2 A1 x, y, z x, y, z c x, y, z A1 0,0,2 x, y, z c2 m
4
0,2,0
1,0,0
A1 x, y, z q c x, y, z x, y, z 4
2cm
q x, y, z c 1,0,0 x, y, z A3 1,0,1 x, y, z A2 1,1,0 x, y, z 2cm
4
4
eq2-eq4//Simplify
0
5
5.1
Basic properties
GL free energy and Thermodynamic critical field Hc
A = 0 and ψ = ψ ∞ (real) has no space dependence. Why ψ is real?
We have a gauge transformation;
iq* χ
(5.1)
A' = A + ∇χ and ψ ' (r ) = exp(
)ψ (r ) .
c
We choose A ' = A + ∇χ = A with χ = χ0 = constant. Then we
have
iq* χ 0
iq* χ 0
)ψ (r ) , or ψ (r ) = exp( −
)ψ ' (r ) . Even if ψ ' is complex number, ψ
c
c
can be real number.
1
2
4
Fs = FN + αψ ∞ + βψ ∞ .
(5.2)
2
∂F
When
= 0 , Fs has a local minimum at
∂ψ ∞
ψ ' (r ) = exp(
ψ ∞ = (−α / β )1 / 2 = ( α / β )1 / 2 .
Then we have
2
Hc
α2
Fs − FN = −
=−
,
2β
8π
from the definition of the thermodynamic critical field:
$
Hc =
"
"
#
4πα 2
β
!
1/ 2
T2
= H c (0)(1 − 2 ) .
Tc
(5.3)
(5.4)
(5.5)
15
Suppose that β is independent of T, then
β
β
T
T
T2
α =−
H c (0)( − 1) = α 0 ( − 1) ,
H c (0)(1 − 2 ) ≈ 2
4π
4π
Tc
Tc
Tc
(5.6)
β
H c (0) . The parameter α is positive above Tc and is negative below Tc.
4π
Note that β>0. For T<Tc, the sign of α is negative:
where α 0 = 2
1
βψ ∞ 4 ,
2
where α0>0 and t = T/Tc is a reduced temperature.
Fs − FN = α 0 (t − 1)ψ ∞ +
(5.7)
2
((Mathematica Program-3))
(* GL free energy*)
f 0 t 1 2 4; f1 f . 0 3, 1 ;
Plot Evaluate Table f1, t, 0, 2, 0.2 , , 2, 2 ,
PlotRange 2, 2 , 3, 3 ,
PlotStyle Table Hue 0.1 i , i, 0, 10 ,
Prolog AbsoluteThickness 2 , Background GrayLevel 0.7 ,
AxesLabel " ", "f" f
3
2
1
-2
-1.5
-1
-0.5
0.5
1
1.5
2
-1
-2
-3
Graphics
Fig.1 The GL free energy functions expressed by Eq.(5.7), as a function of ψ ∞ . α0 = 3.
β = 1. t is changed as a parameter. t = T/Tc. (t = 0 – 2) around t = 1.
(* Order parameter T dependence*)
eq1=Solve[D[f, ] 0, ]//Simplify
0 t 0
0 ,
2
0 t 0
,
2
= /.eq1[[3]]
!! !! !! !! !! !! !! !!
"
"
#
0 t 0
! ! !!
2
! $! !!
Plot[ /.{ 0 3, 1},{t,0,1},PlotRange {{0,1},{0,1.5}},PlotS
tyle Hue[0],Prolog AbsoluteThickness[2],Background GrayLeve
l[0.7],AxesLabel {"t"," "}]
%
&
' &
&
&
&
&
&
16
1.4
1.2
1
0.8
0.6
0.4
0.2
t
0.2
0.4
0.6
0.8
1
Graphics
Fig.2 The order parameter ψ ∞ as a function of a reduced temperature t = T/Tc. We use
α0 = 3 and β = 1 for this calculation.
5.2
Coherence length ξ
We assume that A = 0. We choose the gauge in which ψ is real.
2
d2
αψ + βψ − * 2 ψ = 0 .
2m dx
We put ψ = ψ ∞ f .
3
(5.8)
d2 f
+ f − f 3 =0.
2m *α dx 2
We introduce the coherence length
*
2
2
2πns 2
2
ξ =− * = * = * 2 ,
2 m α 2m α
m Hc
where
α
α 2 Hc2
=
,
ψ ∞ 2 = = ns ,
2β
8π
β
or
−
2
ξ=
2
|1−
T −1 / 2
| .
Tc
2m α 0
Then we have
d2 f
ξ2 2 + f − f 3 = 0,
dx
with the boundary condition f = 1, df/dx = 0 at x = and f = 0 at x = 0.
df d 2 f
df
ξ2
+ ( f − f 3)
= 0,
2
dx dx
dx
or
*
d ξ 2 df d f4 f2
= ( − ),
dx 2 dx
dx 4
2
(5.9)
(5.10)
(5.11)
(5.12)
(5.13)
(5.14)
2
(5.15)
or
17
ξ 2 df
2 dx
2
1
= (1 − f 2 ) 2 ,
4
(5.16)
or
df
1
=
(1 − f 2 ) .
dx
2ξ
The solution of this equation is given by
x
.
f = tanh
2ξ
(5.17)
(5.18)
((Mathematica Program-4))
(* Ginzburg-Landau equation; coherence length*)
eq1
1
f' x
1 fx
2
. eq2 1
2
; eq2 DSolve eq1, f 0
0 ,f x ,x ;
f x_
Tanh
fx
ExpToTrig
Simplify
x
2
(*tangential line at x = 0*)
eq3=D[f[x],x]//Simplify
Sech
2
x
2
!
"
# $ $ $$
2
%
eq4=eq3/.x 0
&
1
'
( ( ((
2
)
Plot Evaluate f x
*
PlotStyle
Background
PlotRange
1
1
/
,
-
,
.
.
/ 0
-
1
1,
1
x , x, 0, 5 ,
2
Hue 0.7 , Hue 0 , Prolog AbsoluteThickness 2 ,
GrayLevel 0.7 , AxesLabel
"x ", "f x " ,
0, 5 , 0, 1.2
* +
2
,
3
5 6
4 4 44
- 2
/
2
1
,
-
8
1
,
/ /
2
-
/
1
/
. 0
7
2
2 2 6
f ;x <
1
0.8
0.6
0.4
0.2
x9
1
2
3
4
:
5
Graphics
Fig.3 Normalized order parameter f(x) expressed by Eq.(5.18) as a function of x/ξ.
=
=
Plot[eq3/. 1,{x,0,5},PlotStyle Hue[0.7],Prolog AbsoluteThi
ckness[2],Background GrayLevel[0.7],AxesLabel {"x/ ","f'(x)
"}, PlotRange {{0,5},{0,0.8}}]
> &
&
&
&
&
&
18
>
f' ; x <
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
x9
1
2
3
4
:
5
Graphics
Fig.4 Derivative f ' ( x ) as a function of x/ξ.
=
5.3
=
Magnetic field penetration depth λ
q ψ
q*
[ψ *∇ψ − ψ∇ψ * ] −
Js =
A.
*
2m i
m*c
We assume that ψ = ψ ∞ (real).
*2
2
(5.19)
q* ψ ∞
Js = −
A (London’s equation),
m*c
(5.20)
q* ψ ∞
q* ψ ∞
∇ × Js = −
∇×A = −
B,
m*c
m*c
(5.21)
2
2
2
where
∇×B =
2
2
4π
Js ,
c
B = ∇× A ,
2
∇⋅B = 0 .
(5.22)
Then we have
c
q* ψ ∞
∇ × ( ∇ × B) = −
B,
4π
m*c
2
2
(5.23)
or
4πq * ψ ∞
∇ × (∇ × B ) = ∇(∇ ⋅ B ) − ∇ B = −
B.
m*c 2
Then we have London’s equation
1
∇ 2B = 2 B ,
λ
where
2
2
2
4πq* ψ ∞
λ =
.
m*c 2
λ is the penetration depth
2
−2
λ=
2
(5.25)
2
m*c 2
4πq* ψ ∞
(5.24)
2
=
(5.26)
m*c 2 β
4πq * α
2
.
(5.27)
The solution of the above differential equation is given by
BZ ( x) = BZ ( x = 0) exp( − x / λ ) ,
19
(5.28)
where the magnetic field is directed along the z axis.
ex e y
ez
c
c ∂
∂
∂
c
∂
J=
∇× B =
=
(0,− Bz ( x ),0) .
4π
4π ∂x ∂y
∂z
4π
∂x
0
0 Bz ( x)
The current J flows along the y direction.
cB ( x = 0)
Jy = z
exp(− x / λ ) .
4πλ
(5.29)
(5.30)
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.
α
n =ψ ∞ = ,
β
2
4
H c = 4πψ ∞ β
α =
λ=
Hc
H c = 4πn α = 4π α ψ ∞
2
4πψ ∞
*
s
β=
,
Hc
2
4πψ ∞
4
Φ0 =
,
m*c 2 β
4πq α
*2
2
(5.31)
(5.32)
2
2
α
= 4π
,
β
2
2
*
s
,
ξ=
2m* α
2π c
,
q*
(5.33)
,
κ=
λ cm* β
=
.
ξ
2π q *
(5.34)
,
κ=
cm* H c
,
*
2 2π ns q *
(5.35)
Then we have
λ=
λξ
m*c 2
4πns q*
*
2
,
1
=
,
Φ 0 2 2πH c
ξ=
2πns
*
H c m*
2 q* H c
κ
=
,
λ2
c
20
κ
Φ = 2 2πH c ,
λ2 0
(5.36)
c
.
2 q * λξ
Hc =
(5.37)
We also have
Φ0
2 2πλ2 H c
2Hc =
,
.
κ=
2πλξ
Φ0
Note:
*
q* = 2e (>0) m* = 2m ,
ns = ns / 2 .
((Mathematica Program-5))
(*n1=ns*, q1=Abs[q*],
rule1 1
Hc2
4 n1
,
Hc2
(5.38)
1=Abs[ ], m1=m*
*)
2 c ;
q1
, 0
4 n12
m1 c2 ;
4 q12 1
;
2 m1 1
c m1 2
PowerExpand
q1 1= /.rule1//PowerExpand
c m1
2 n1
q1
1= /.rule1//PowerExpand
n1
2
Hc
m1
#
!
. rule1
$
0"
PowerExpand
$ $
1
2 % &2
& && Hc '
1=( /.rule1//PowerExpand
(
c Hc m1
2 ) *2
* ** n1 + q1 ,
.
-
0
4
3
2
/
. rule1 / / PowerExpand
2 Hc q1
1 1 11 c 2
2
5
0 6 . rule1 6 6 PowerExpand
27 82
8 88 Hc 9
6.
6.1
Formulation of GL equation
General theory2
In summary we have the GL equation;
?
<
2
q* :
1 =
αψ + β ψ ψ + * => ∇ − A ; : ψ = 0 ,
c
2m @i
2
and
21
(6.1)
q ψ
c
c
q*
[ψ *∇ψ − ψ∇ψ * ] −
Js =
∇ × (∇ × A) =
[∇(∇ ⋅ A) − ∇ 2 A] =
A,
*
4π
4π
2m i
m*c
(6.2)
(6.3)
B = ∇×A .
The GL equations are very often written in a form which introduces only the following
dimensionless quantities.
B
ψ =ψ ∞ f ,
r = λ ,
h=
,
(6.4)
2Hc
where
*2
2
2
2
α
4πq* ψ ∞
4πα 2
α
2
2
2
,
=
−
=
,
=− ,
H
=
, ψ∞ =
λ =
ξ
c
* 2
*
*
β
β
β
2m α 2m α
mc
(6.5)
2π c
2π c
Φ 0 = * = − * (since q* = 2e<0),
(6.6)
q
q
2
2
−2
1
2π
,
=
2
Φ 0κ
2H cλ
2 H c λ2 =
Φ0
κ,
2π
κ=
λ
.
ξ
(6.7)
We have
2
1
~ 2
∇ρ + A f = f − f f ,
iκ
(6.8)
where
~ 2πξ
A=
A.
Φ0
Here we use the relation
1
B
h=
=
∇×A =
2Hc
2Hc
or
~
h = ∇ρ × A ,
Js =
(6.9)
Φ0
1
~
~
∇ρ × A = ∇ρ × A ,
2 H c 2πλξ
(6.10)
(6.11)
c
c
[∇(∇ ⋅ A ) − ∇ 2 A ]
∇ × (∇ × A) =
4π
4π
q ψ
q*
[ψ *∇ψ − ψ∇ψ * ] −
=
A
*
2m i
m*c
*2
2
,
(6.12)
~
q* ψ ∞ Φ 0 2 ~
q*
Φ0
2 1
*
*
f A,
(
)
f
f
f
f
[
]
A
ψ
∇
×
∇
×
=
∇
−
∇
−
∞
ρ
ρ
ρ
m*c 2πξ
4πλ2 2πξ
2m*i
λ
(6.13)
~
i
2~
∇ ρ × (∇ ρ × A) =
[ f *∇ ρ f − f∇f * ] − f A ,
(6.14)
2κ
c
2
2
since
4πq* ψ ∞
λ =
m*c 2
2
−2
2
(6.15)
22
and
4πλ2 2πξ q *
4πλ2 2πξ q * 1 m * c 2
2πξ c 1
2 1
=
=
ψ∞
*
*
2 2
*
Φ 0 2q * λ
c Φ 0 2m
c Φ 0 2m λ 4πq λ
λ
, (6.16)
2π c 1
1
=−
=−
Φ0 2 q* κ
κ
2
q*
c
ψ ∞2 =
.
*
mc
4πλ2
In summary we have the following equations:
2
1
~ 2
∇ρ + A f = f − f f ,
iκ
~
i
2~
∇ ρ × (∇ ρ × A) =
[ f *∇ ρ f − f∇f * ] − f A ,
2κ
~
h = ∇ρ × A .
Gauge transformation:
A = A 0 + ∇χ ,
ψ (r ) = exp(
(6.17)
iq* χ
)ψ 0 (r ) ,
c
(6.18)
(6.19)
(6.20)
(6.21)
(6.22)
~ 2πξ
A=
A,
ψ =ψ ∞ f ,
Φ0
~ ~
2πξ
~
2πξ 1
~
1
2π
χ) ,
A = A0 +
∇χ = A 0 +
∇ ρ χ = A0 + ∇ ρ (
Φ0
Φ0 λ
κ
Φ0
(6.23)
(6.24)
iq* χ
i 2π
(6.25)
) f 0 = exp( −
χ) f .
c
Φ0
2π
Here we assume that −
χ = ϕ 0 . Then we have
Φ0
~ ~
1
A = A 0 − ∇ ρϕ0 .
(6.26)
f = exp(iϕ 0 ) f 0 ,
κ
We can choose the order parameter f as a real number f0 such that f0 is real, f0 is a
~ ~
constant. A = A 0 .
f = exp(
2
1
~ 3
∇ ρ + A0 f0 = f0 − f0 ,
iκ
~
2~
∇ ρ × (∇ ρ × A 0 ) = ∇ ρ × h = − f 0 A 0 ,
~
h = ∇ ρ × A0 ,
(6.27)
(6.28)
(6.29)
2
1
~ 3
∇ ρ + A0 f0 = f0 − f0
iκ
.
1
1
~ 2
~
~
2
= − 2 ∇ ρ f 0 + A 0 f 0 + [∇ ρ ⋅ ( A 0 f 0 ) + ( A 0 ⋅ ∇ ρ f 0 )]
iκ
κ
( )
23
(6.30)
Using the formula of the vector analysis
~
2~
∇ ρ ⋅ [∇ ρ × (∇ ρ × A 0 )] = 0 = −∇ ρ ⋅ ( f 0 A 0 )
~
~
~
2 ~
2
2
= −∇ ρ f 0 ⋅ A 0 − f 0 ∇ ρ ⋅ A 0 = −2 f 0 ∇ ρ f 0 ⋅ A 0 − f 0 ∇ ρ ⋅ A 0
or
~
~
− 2 A 0 ⋅ ∇ ρ f 0 ⋅ = f 0∇ ρ ⋅ A 0 ,
We also have
~
~
~
∇ ρ ⋅ ( A 0 f 0 ) = A 0 ⋅ ∇ ρ f 0 + f 0∇ ρ ⋅ A 0 .
We have
1
1
~ 2
~
~
3
2
f 0 − f 0 = − 2 ∇ ρ f 0 + A 0 f 0 + [ f 0∇ ρ ⋅ A 0 + 2A 0 ⋅ ∇ ρ f 0 ]
iκ
κ
,
1
~ 2
2
= − 2 ∇ ρ f0 + A0 f0
(6.31)
(6.32)
(6.33)
( )
κ
or
−
1
( )
( )
~ 2
2
3
∇ρ f0 + A0 f0 = f0 − f0 .
(6.35)
κ
We also have
2~
∇ ρ × h = − f0 A0 ,
~
1
A0 = − 2 ∇ρ × h ,
f0
~
1
h = ∇ ρ × A 0 = −∇ ρ × [ 2 ∇ ρ × h]
f0
2
= −∇ ρ
(6.34)
(6.36)
(6.37)
1
1
× (∇ ρ × h) − 2 ∇ ρ × (∇ ρ × h)
2
f0
f0
,
(6.38)
or
h=
2
1
∇ ρ f 0 × (∇ ρ × h) − 2 ∇ ρ × (∇ ρ × h) ,
3
f0
f0
(6.39)
2
∇ ρ f 0 × (∇ ρ × h) − ∇ ρ × (∇ ρ × h) .
f0
(6.40)
or
f0 h =
2
In summary,
1
1
2
2
3
− 2 ∇ ρ f 0 + 3 (∇ ρ × h ) = f 0 − f 0 ,
κ
f0
2
2
f 0 h = ∇ ρ f 0 × (∇ ρ × h) − ∇ ρ × (∇ ρ × h) ,
f0
~
1
A0 = − 2 ∇ρ × h .
f0
6.2
Special case
24
(6.41)
(6.42)
(6.43)
We now consider the one dimensional case. For convenience we remove
~
and r =(x, y, z) instead of ρ. We also use A for A 0 and f for f0. f is real.
1
1
2
− 2 ∇ 2 f + 3 (∇ × h ) = f − f 3 ,
f
κ
2
f 2h = ∇f × (∇ × h) − ∇ × (∇ × h) ,
f
1
A = − 2 ∇×h ,
f
h = ∇×A.
the subscript ρ
(6.44)
(6.45)
(6.46)
(6.47)
Fig.6 The direction of the magnetic induction h(x).
x, y and z are dimensionless.
ex
∂
∇×h =
∂x
0
ey
∂
∂y
0
ez
dh
∂
= (0,− ,0) ,
dx
∂z
h( x)
(6.48)
d 2h
∇ × (∇ × h) = ∇(∇ ⋅ h) − ∇ h = −∇ h = − 2 e z ,
dx
2
1 d2 f
1 dh
− 2
+ 3
2
f dx
κ dx
f 2h = −
or
2
(6.49)
2
= f − f3,
2 df dh d 2 h
+
f dx dx dx 2
(6.50)
for the z component,
∂ 1 ∂h
(
),
∂x f 2 ∂x
1 dh
A= 2
ey .
f dx
h=
(6.51)
(6.52)
(6.53)
25
(a) κ«1 (type I superconductor)
1 d2 f
f − f3+ 2
=0.
κ dx 2
The solution of this equation is already given above.
(6.54)
(b) κ»1 (type-II superconductor)
1 ∂h
f −f − 3
f ∂x
3
2
= 0.
(6.55)
7
7.1
Free energy
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.
1
B ⋅ H (Legendre transformation).
g= f −
(7.1)
4π
B is a magnetic induction (microscopic magnetic field) at a given point of the
superconductors.
The Helmholtz free energy is given by
2
1
1
q*
B2
4
(
)
ψ
fs = fn + α ψ + β ψ +
∇
−
A
+
,
2
2m* i
8π
c
The Gibbs free energy is expressed by
2
(7.2)
2
1
1
q*
1
B2
4
g s = fn + α ψ + β ψ +
( ∇ − A)ψ +
−
B ⋅ H . (7.3)
*
2
2m i
c
8π 4π
2
(i)
At x =- (normal phase)
2
H
f (−∞) = f n + c ,
8π
2
2
2
Hc
Hc
Hc
g (−∞) = g n = f n +
−
= fn −
.
8π
4π
8π
(ii)
At x = (superconducting phase)
2
1
H
α2
2
4
g s = g s (∞) = f s = f n + αψ ∞ + βψ ∞ = f n −
= fn − c .
2
2β
8π
The interior Gibbs free energy
(7.4)
(7.5)
26
(7.6)
E g = dr[ g s (r ) − g n ]
= dr[ f n − g n + α ψ
or
1
+ βψ
2
2
4
q*
1
+
(
∇
−
A )ψ
c
2m * i
2
+
B2
1
BH c ]
−
8π 4π
(7.7)
2
q*
1
1
1
4
E g = dr[α ψ + β ψ +
(
∇
−
A)ψ +
(B − Hc )2 ]
*
c
2
2m i
8π
2
7.2
(7.8)
Derivation of surface energy
We have a GL equation;
2
1
q*
αψ + β ψ ψ + * ∇ − A ψ = 0 .
2m i
c
2
(7.9)
If one multiplies the GL equation by ψ * and integrates over all r by parts, one obtain the
identity
2
1 *
q*
dr[α ψ + β ψ +
ψ
∇
−
A ψ]=0,
2m*
i
c
which is equal to
2
4
(7.10)
2
1
q*
dr[α ψ + β ψ +
(
A)ψ ] = 0 ,
∇
−
2 m* i
c
Here we show that the quantity I defined
2
4
I = dr[ψ
*
2
2
q*
q*
∇ − A ψ − ( ∇ − A)ψ ] .
i
c
i
c
(7.11)
(7.12)
is equal to 0, which is independent of the choice of ψ and ψ*. The variation of I is
calculated as
δI = dr[Λδψ + Γδψ * ]
(7.13)
using the Mathematica 5.2 (VariationalD). We find that Λ = Γ = 0 for any ψ and ψ *. I is
independent of the choice of ψ and ψ*. When we choose ψ = ψ* = 0. Then we have I = 0.
((Mathematica Program-6))
(*Surface energy calculation -integral*)
<<Calculus`VariationalMethods`
Needs["Calculus`VectorAnalysis`"]
SetCoordinates[Cartesian[x,y,z]]
Cartesian[x,y,z]
27
A A1 x, y, z , A2 x, y, z , A3 x, y, z ;
eq1
1 q
A x, y, z .
Grad x, y, z
2m
c
q
A c x, y, z
Grad c x, y, z
c
q
A1 x, y, z # & ;
OP1 : D #, x
c
q
OP2 : D #, y
A2 x, y, z # & ;
c
q
OP3 : D #, z
A3 x, y, z # & ;
c
eq2
1
c x, y, z
2m
OP1 OP1 x, y, z
OP2 OP2 x, y, z
Expand;
General ::spell1 : Possible spelling error :
new symbol name " c" is similar to existing
Expand;
OP3 OP3 x, y, z
symbol " ". More…
eq3=eq1-eq2//Expand
q
x, y, z c x, y, z A3 0,0,1 x, y, z 2cm
q A3 x, y, z c x, y, z 2cm
q A3 x, y, z 2
0,0,1
q
0,0,1
c
x, y, z 2m
q
0,1,0
x, y, z 2
1,0,0
0,1,0
0,1,0
x, y, z x, y, z 1,0,0
2
x, y, z x, y, z x, y, z c x, y, z 0,2,0
2m
x, y, z x, y, z c 1,0,0 x, y, z 2cm
x, y, z c 1,0,0 x, y, z 2m
0,0,2
2m
x, y, z c x, y, z A1 1,0,0 x, y, z 2cm
c x, y, z x, y, z c 0,1,0 x, y, z 2cm
x, y, z c
2m
q A1 x, y, z x, y, z q A1 x, y, z c x, y, z 2cm
2
x, y, z c x, y, z A2 0,1,0 x, y, z 2cm
q A2 x, y, z 0,0,1
x, y, z c 0,0,1 x, y, z 2cm
q A2 x, y, z c x, y, z 2cm
2
2
c x, y, z 2,0,0
2m
VariationalD[eq3, c[x,y,z],{x,y,z}]
0
VariationalD[eq3, [x,y,z],{x,y,z}]
0
________________________________________________________________________
28
Subtracting Eq.(7.11) from Eq.(7.8), we obtain the concise form
1
1
4
(B − Hc )2 ] .
(7.14)
E g = dr[ − β ψ +
2
8π
Suppose that the integrand depends only on the x axis. Then we can define the surface
energy per unit area as γ:
∞
1
1
4
(7.15)
γ = dx[ − β ψ + ( B − H c ) 2 ] ,
2
8π
−∞
which is to be equal to
1
2
γ =δ
Hc .
(7.16)
8π
Then we obtain a simple expression
ψ
B
(7.17)
δ = dx[− 4 + (1 − ) 2 ] .
H
ψ
c
−∞
∞
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
ψ
B
f =
,
h=
.
(7.18)
ψ∞
2H c
Here we use x = λ~
x is dimensionless). For simplicity, furthermore we use x instead
x (~
~
of x .
Then we have
∞
4
∞
∞
−∞
−∞
1
2
δ = λ dx[− f 4 + (1 − 2h) 2 ] = 2λ dx[ (1 − f 4 ) + h 2 − 2h] .
(7.19)
7.3
Surface energy calculation for the two cases
(a) κ«1 (type-I superconductor)
ψ
= f 4 with f = tanh
When h = 0 for x>0 and
4
ψ∞
4
x
, we have
2κ
∞
4 2
4 2
(7.20)
κλ =
ξ = 1.8856ξ >0,
3
3
−∞
a result first obtained by Ginzburg and Landau. So the surface energy is positive.
δ ≈ λ dx(1 − f 4 ) =
((Mathematica Program-7))
Simplify
0
4
2
3
1 Tanh
x
4 2
x,
0
_______________________________________________________
(b) κ»1 (type-II superconductor)
Here we assume that λ is much larger than ξ.
29
∂h
f (1 − f ) =
.
∂x
As h must decrease with increasing x, we have
1 ∂h
= −(1 − f 2 )1 / 2 ,
2
f ∂x
and
∂ 1 ∂h
∂
h = ( 2 ) = − (1 − f 2 )1 / 2 .
∂x
∂x f ∂x
Here f obeys the following differential equation,
From Eq.(7.22) we have
∂h
∂2
= − 2 (1 − f 2 )1 / 2 .
∂x
∂x
Using Eq.(7.21),
∂h
∂2
= − 2 (1 − f 2 )1 / 2 = −(1 − f 2 )1 / 2 f 2 ,
∂x
∂x
or
∂2
(1 − f 2 )1 / 2 = (1 − f 2 )1 / 2 ( f 2 − 1 + 1) = (1 − f 2 )1 / 2 − (1 − f 2 )3 / 2 .
2
∂x
When u = (1 − f 2 )1 / 2 , we have
4
2
2
∂ 2u
= u − u3 ,
∂z 2
∂u
∂z
(7.22)
(7.23)
(7.24)
(7.25)
(7.26)
2
1
= u 2 − u 4 + const .
2
du
du
When u =0,
= 0 (at f = 1 (x = ) at
= 0 ), so we have
dx
dx
(7.21)
(7.27)
du
dx
2
1
1
= u 2 − u 4 = u 2 (1 − u 2 ) ,
2
2
(7.28)
or
du
1
= −u (1 − u 2 )1 / 2 ,
dx
2
since du/dx must be negative.
We solve the problem with the boundary condition; u = 1 when x=0.
∞
1
δ = λ dx[ (1 − f 4 ) + h 2 − 2h] ,
2
−∞
or
∞
u2
u2
δ = λ dx[2u 2 (1 − ) − 2u (1 − )1 / 2 ] ,
2
2
−∞
or
(7.29)
(7.30)
30
(7.31)
∞
u2
u2
dx
) − 2u (1 − )1 / 2 ] du ,
2
2
du
δ = λ du[2u 2 (1 −
−∞
0
(7.32)
u2
u2 1/ 2
1
δ = λ du[2u (1 − ) − 2u (1 − ) ]
du
1
2
2
1
− u (1 − u 2 )1 / 2
2
2
(7.33)
or
1
δ = λ du[2u (1 −
0
u 2 1/ 2
4
) − 2 ]du = −λ ( 2 − 1) < 0 ,
3
2
(7.34)
Thus, for κ»1 the surface energy is negative.
((Mathematica Program-8))
1 0
2u
4
3
u2
2
1
1
2
u
2
((Mathematica Program-9))
(*Negative surface energy*)
Clear[u]
!
eq1 u' x
" #
u x
ux
#
$
%
&
u x
$
'
u
1
1
#
ux
2
2 12
x$ 2
( ( ( ( (&( ( ( ( ( ( ( ( ( ( ( ( ( (
2
eq2=DSolve[{eq1,u[0] 1},u[x],x]//Simplify
)
2
u x
+
,
2
. /
0
1
x
4
2
2 2 22 3
2 2
.
, u x
-
+
3 2
/
* *
0
1
2
2 2 22
/
4
2x
5
,
3 2
0
*
0
1
2
2 2 22 3
4
x
-
2
1 2 2 22
0
4
2x
5 5
u[x_]=u[x]/.eq2[[1]]//Simplify
2
7
2
8
9 : : :: ;
3 2
9
: : ::
6 7
8
2
2
7
x
<
<
2x
Plot[u[x],{x,0,5},PlotStyle Hue[0],Prolog
AbsoluteThickness[2],
PlotPoints 100,Background GrayLevel[0.7],
AxesLabel {"x","u"}]
=
=
=
=
=
u
1
0.8
0.6
0.4
0.2
x
1
>
Graphics
2
3
>
31
4
5
Fig.7 Plot of u = (1 − f 2 )1 / 2 vs x. x (dimensionless parameter) is the ratio of the distance
to λ.
2
u
x
, x, 0, 5 , PlotStyle Hue 0 , PlotPoints 100,
Plot 1
Prolog AbsoluteThickness 2 , PlotRange 0, 5 , 0, 1 ,
Background GrayLevel 0.7 , AxesLabel "x", "f" f
1
0.8
0.6
0.4
0.2
x
1
>
Graphics
2
3
4
5
>
Fig.8 Plot of f = 1 − u 2 vs x.
u x 2 u x 2 1 2 u x
F 2ux 1 Simplify;
2
1 2
2
Plot F, x, 0, 10 , PlotStyle Hue 0 ,
Prolog AbsoluteThickness 2 , PlotPoints 100,
Background GrayLevel 0.7 , AxesLabel "x", "F" 2
F
x
2
4
6
8
10
-0.05
-0.1
-0.15
-0.2
-0.25
>
Graphics
>
Fig.9 Plot of the surface energy σ s = 2u 2 (1 −
u2
u2
) − 2u (1 − )1 / 2 as a function of x.
2
2
Integrate[F,{x,0, }]
4
3
1
2!
%//N
-0.552285
F1=D[F,x]; Plot[F1,{x,0,10},PlotStyle Hue[0],Prolog
AbsoluteThickness[2],
PlotPoints 100,Background GrayLevel[0.7],
AxesLabel {"x","dFdx"}]
=
=
=
=
=
32
dFdx
0.1
x
2
4
6
8
10
-0.1
-0.2
-0.3
Graphics
Fig.10 Plot of the derivative dσs/dx with respect to x.
>
>
FindRoot[F1 0,{x,1,2}]
{x 1.14622}
)
Criterion between type-I and type-II superconductor: κ = 1 / 2
The boundary between the type-II and type-I superconductivity can be defined by
finding the value of κ which corresponds to a surface energy equal to zero.
δ ∞ 1
(7.35)
= dx[ (1 − f 4 ) + h 2 − 2h] .
2 −∞ 2
It is clear that this value is zero if
1
(1 − f 4 ) + h 2 − 2h ,
(7.36)
2
or
1
1
[h −
(1 − f 2 )][h −
(1 + f 2 )] = 0
(7.37)
2
2
or
1
h=
(1 − f 2 )
(7.38)
2
Substituting this into the two equations
7.4
2
1 d2 f
1 ∂h f − f3+ 2
−
=0
κ dx 2 f 3 ∂x
(7.39)
d 2 h 2 df dh
f h= 2 −
dx
f dx dx
(7.40)
2
From the calculation by Mathematica, we conclude that κ = 1 / 2 .
((Mathematica Program-10))
Determination of
1
2
Clear[f,h]
eq1 f x
f x 3 f x
f x 3 1
2
h x 2
f x 3
f'' x
f x
2
1
h' x 2 0
f x 3
0
33
eq2 f x 2 h x
h'' x
f x 2 h x
2 f x h x
f x 1
rule1 h 2
f' x h' x
f x
h x
1 f #
& 2
2
1 ! f " #1 # 2
h
$ % % %%
2
(
&&'((( )
eq3=eq1/.rule1//Simplify
--
f *x+
-
f *x +
,
.
2
/
f *x+ 3 ,
2 f *x+ 2
f * x+
Solve[eq3,f''[x]]
0 0
2
11 2
f x3
4
67
5
2
2
1 2
f x 3 2 8 f x 3 4 8 2 f x 3 29
2
f x3
: :
eq4=eq2/.rule1//Simplify
>
f ; x < 4 = 2 f ; x< 2 ? f ;x < @ f ; x <
>>
=
2 f ;x< A
Solve[eq4,f''[x]]//Expand
B B
D
CC D
f xE
8.
8.1
F
D
C D
f x E 2 H f xE 4 H 2 f x E 2
G
D
I
2 f xE
I
Application of the GL equation
Critical current of a thin wire or film1,4
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.
q*
q*
2
2
J s = * ψ (∇θ − A) = q * ψ v s ,
m
c
J
(8.1)
J
2
1
1
q*
B2
4
,
( ∇ − A )ψ +
f s − fn = α ψ + β ψ +
2
2m* i
8π
c
2
K
ψ (r ) = ψ (r ) eiθ (r ) .
We consider the case when
2
ψ (r ) = ns * = constant
Then we have
(8.2)
(8.3)
(8.4)
34
q*
q*
( ∇ − A)ψ = ( ∇θ − A )ψ = (m* v s )ψ ,
i
c
c
(8.5)
and
1
m* 2 2 B 2
2
4
fs − fn = α ψ + β ψ +
.
(8.6)
ψ vs +
2
2
8π
2
*
We now consider a thin film. We assume that d«ξ(T) in order to have ψ = ns =
constant. ψ has the same value everywhere. The minimum of fs with respect to ψ is
obtained for
m*
2
α + β ψ + vs 2 = 0 ,
(8.7)
2
for a given vs, where B 2 /(8π ) is neglected.
Here α = − α and ψ = ψ ∞ f with ψ ∞ = −α / β = α / β . Then we have
2
m* 2
2
vs = −α − βψ ∞ f 2 = α (1 − f 2 ) .
2
The corresponding current is a function of f,
J s = q * ψ vs = q *ψ ∞
2
We assume that J 1 = f
2
2α
(8.8)
f 2 1− f 2 .
*
m
(8.9)
1− f 2 .
2
This has a maximum value when ∂J 1 / ∂f = 0 ; J 1 max =
Js =
2
3 3
q *ψ ∞
2α
2
m
*
=
2
3 3
2
3 3
at f =
2
3
q *ψ ∞
2
(8.10)
m *ξ
((Mathematica Program-11))
(* Critical current of a thin wire*)
J1 f2 1 f2 1 2
2
f
f2 1
Plot[J1,{f,0,1},
PlotStyle Hue[0.7],Background GrayLevel[0.7],
Prolog AbsoluteThickness[2], AxesLabel {"f","J1"}]
J1
0.3
0.2
0.1
f
0.2
0.4
0.6
Graphics 35
0.8
1
Fig.12 The normalized current density J1vs f. J1 = f 2 1 − f 2
H1=D[J1,f]//Simplify;H2=Solve[H1 0,f]
2
2
f 0 , f , f 3 3 max1=J1/.H2[[3]]
2
3 3
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:
1
m* 2 2 B 2
2
4
fs − fn = α ψ + β ψ +
.
(8.11)
ψ vs +
2
2
8π
Gibbs free energy:
1
m* 2 2 B 2 1
2
4
ψ vs +
gs = fn + α ψ + β ψ +
−
B⋅H ,
(8.12)
2
2
8π 4π
or
m * 2 2 (B − H ) 2 H 2
1
2
4
ψ vs +
gs = fn + α ψ + β ψ +
−
.
(8.13)
2
2
8π
8π
So the Gibbs free energy per unit area of film is
d / 2
d / 2
2
1
4
m * 2 2 (B − H ) 2 H 2
−
]dz ,
ψ vs +
Gs = g s ( z )dz = [ f n + α ψ + β ψ +
2
2
8π
8π
−d / 2
−d / 2
(8.14)
*
*
q
q
2 ∂θ
2
(8.15)
J x = * ψ ( − Ax ) = q * ψ vx .
∂x c
m
When θ is constant,
q*
vx = − * Ax ( z ) .
(8.16)
mc
∂A ( z )
B = (0, B y ( z ),0) = ∇ × A = (0, x ,0) .
(8.17)
∂z
36
or
By ( z) =
∂Ax ( z )
,
∂z
(8.18)
By ( z ) =
∂Ax ( z )
≈H,
∂z
(8.19)
or
or
Ax ( z ) = Hz ,
vx = −
q*
Hz ,
m*c
(8.20)
d /2
Gs =
g s ( z )dz
−d / 2
d /2
=
[ fn + α ψ
2
−d / 2
1
+ βψ
2
4
H 2 m*
−
+
ψ
8π
2
2
q* H
m* c
2
,
(B − H ) 2
z +
]dz
8π
(8.21)
2
When the last term is negligibly small, we obtain
H2
m*
1
4
Gs ≈ ( f n + α ψ + β ψ −
ψ
)d +
2
8π
2
2
2
q* H
m*c
2
2 d
3 2
3
,
(8.22)
or
2
1
H2
q* H 2 d 2 2
4
)+
ψ ].
Gs ≈ Fn + d [(α ψ + β ψ −
2
8π
24m*c 2
Similarly,
d/2
2
4
1
m* 2 2 B 2
ψ vs + ]dz ,
Fs = Fn +
[α ψ + β ψ +
2
2
8π
−d / 2
2
(8.24)
(8.25)
2
1
B2
q* H 2 d 3 2
4
βψ + ) +
ψ .
2
8π
24m*c 2
2
Minimizing the expression of Fs with respect to ψ , we find
Fs = Fn + d (α ψ +
2
(8.26)
2
q* H 2 d 2
α + βψ +
= 0,
24m*c 2
(8.27)
ψ
q* H 2d 2
H2 d2
H2 ε2
= 1−
,
= 1− 2
= 1− 2
2
24m*c 2 α
H c 24λ
H c 24
ψ ∞2
(8.28)
2
2
2
or
H
Hc
2
=
24
ε
2
(1 − f ),
2
(8.29)
where
ε=
d
(8.30)
λ
This is rewritten as
37
H2
,
(8.31)
2
H c //
where H c // is defined as
H
H c // = 2 6 c .
(8.32)
ε
This parallel critical field H c // can exceed the thermodynamic critical field Hc when ε is
f 2 = 1−
lower than 2 6 = 4.899 .
We now consider the difference between Gs and Gn.
H2
Gn = Fn −
d
8π
where
Fn = df n ,
and
Gn = dg n
The difference is given by
(8.33)
2
H2
1
H2
q * H 2d 2 2
2
4
)+
ψ ].
Gn − Gs = (−
) − [(α ψ + β ψ −
8π
2
8π
24m*c 2
(8.34)
or
2
Gn − Gs = −[(α ψ +
2
1
q* H 2 d 2 2
4
βψ ) +
ψ ].
2
24m*c 2
(8.35)
Noting that
2
q* H 2 d 2
α + βψ +
= 0,
24m*c 2
we obtain
1
1
4
4
4
Gn − Gs = − β ψ + β ψ = β ψ > 0 .
2
2
The superconducting state is energetically favorable.
2
(8.36)
(8.37)
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
f
cosh( z )
λ ,
(8.38)
By ( z ) = H
f
cosh(ε )
2
and
fz
sinh( )
dB
(
z
)
c
cfH
y
λ .
(8.39)
J sx ( z ) = −
=−
4π dz
4πλ cosh( fε )
2
Note that
38
ψ =ψ∞ f ,
ψ ∞2 =
m *c 2
4πq* λ2
2
,
(8.40)
fz
)
J sx ( z )
q λH
λ
,
vsx ( z ) = * 2 = * 2 2 = − *
q f ψ∞
m cf cosh( εf )
q ψ
2
sinh( εf )
(
− 1)
d/2
1
1 H 2 q*2λ2
εf
2
2
.
vsx =
[vsx ( z )] dz =
d −d / 2
2 m*2c 2 f 2 cosh 2 ( εf )
2
Helmholtz free energy:
1
m* 2 2 B 2
2
4
fs = fn + α ψ + β ψ +
’
ψ vs +
2
2
8π
2
On minimizing this with respect to ψ
J sx ( z )
∂f s
∂ψ
2
sinh(
*
(8.41)
(8.42)
(8.43)
= 0,
(8.44)
m* 2
α + β ψ + vs = 0 ,
2
2
(8.45)
or
ψ
2
= f
ψ ∞2
f 2 =1−
2
=
α −
m*
2
vs
2
=1−
2
βψ ∞
q λ Hc H
4 α m*c 2 H c
2
*2 2
2
m*
vs
2
=1−
2 βψ ∞
sinh( εf )
(
− 1)
1
1 H
εf
= 1−
2
εf
f
4 Hc
cosh 2 ( )
2
cosh 2 (
2
H
Hc
2
= 4 f 2 (1 − f 2 )
εf
1
m*
4
ψ
f s = fn + α ψ + β ψ +
2
2
2
)
2 .
sinh( εf )
[
− 1]
εf
We now discuss the Helmholtz free energy
m*
2
vs ,
2α
2
1
f2
(
(8.46)
sinh( εf )
− 1)
εf
,
εf
cosh 2 ( )
2
(8.47)
(8.48)
2
2
vs
2
B
+
,
8π
(8.49)
with
m*
α + βψ +
vs
2
Then we have
2
2
=0.
(8.50)
39
B2
1
4
2
2
f s = f n + α ψ + β ψ + ψ (−α − β ψ ) +
2
8π
,
2
B
1
4
= fn − β ψ +
2
8π
2
(8.51)
or
2
2
Hc 4 B
fs = fn −
f +
,
8π
8π
d /2
1
εf + sinh( εf )
B2 =
[ B y ( z )]2 dz =
H2,
d −d / 2
εf [1 + cosh(εf )]
(8.52)
(8.53)
and
εf
1
2H
B y ( z )dz =
tanh( ) .
εf
d −d / 2
2
Gibbs free energy:
2
B2
B H
B H
Hc 4
g s = fs −
= fn −
f +
−
,
4π
8π
8π
4π
2
εf
H
H 2 εf + sinh( εf )
4
g s = fn − c f 4 +
[
− tanh( )] .
8π
8π εf [1 + cosh(εf )] εf
2
Since
H 2 BH
H2
gn = fn +
−
= fn −
,
8π
4π
8π
where B = H.
From the condition that g s = g n at the critical field H = Hl,
1
εf + sinh( εf )
εf
2
− 4 H 2 tanh( ) = − H 2 ,
− Hc f 4 + H 2
2
εf [1 + cosh(εf )]
εf
or
d/2
B =
H
Hc
2
{1 +
εf + sinh( εf )
εf
4
− tanh( )} = f 4 ,
2
εf [1 + cosh(εf )] εf
(8.54)
(8.55)
(8.56)
(8.57)
(8.58)
(8.59)
or
2
f4
=
.
εf + sinh( εf )
εf
4
1+
− tanh( )
εf [1 + cosh(εf )] εf
2
In the limit of f 0,
Hl
Hc
(8.60)
2
Hl
24
1 1
24 81ε 2 4
) f + O[ f ]6 .
= 2 + 24( − 2 ) f 2 + (− +
ε
Hc
5 ε
5
350
From the above two equations, we get
40
(8.61)
4 f 2 (1 − f 2 )
cosh 2 (
εf
)
f4
2
=
,
εf + sinh( εf )
εf
sinh( εf )
4
[
− 1] 1 +
− tanh( )
εf
εf [1 + cosh(εf )] εf
2
(8.62)
or
1 εf [cosh(εf ) − 1]
f2
.
(8.63)
=
2
6(1 − f ) 3 sinh( εf ) − εf
In the limit of f 0,
1
ε2
1
ε4
(1 − ) f 2 + ( +
(8.64)
) f 4 + O[ f ]6 .
6
5
6 12600
Using Mathematica,
1.
we determine the value of f as a function of ε.
2
2.
we determine the value of (H l / H c ) as a function of f or f2 for each ε.
There occurs the first order transition because of discontinuous change of order
parameter.
1+
((Mathematica Program-12))
Here we discuss the problem dscribed in the book of de Gennes, page 189
<<Graphics`ImplicitPlot`
Cosh z
hy z_
H
c f H Sech f Sinh 2
4
0
H q Sech f Sinh 2
Vavsq
f2 3 1 4
rule3 B
2
f
1
d
(
1H
Hc2 e
H
fz
=
5
4
df
Sinh 8
9
1 4 Cosh > f ?
[
g
: ;
@ A
x ; eq2 B eq1 F . rule3
2 c2 f3
h
Jx % z&
. rule2
q f2 02 '
vsx , z- 2 . z / / Simplify; eq1 ( f2 0 1 1
+
) *
JK
Sinh L df
M
N O
1 H Cosh R f S
T U
df H
2 c2 d f3 m P
4f
d2
2 c2 d fE 3 m <
2D
Y Y
;
Simplify
Q
I
eq3=Solve[eq2,x]/.d
xZ
c f+ m
q2 x I 2
G
fz
d2
H2 q2 5 2 6 7 d f
C
m c2 2
4 q2
2
mc ! "2
; vsx % z_&
4 # q2 $
f 0, 02
rule2 ; rule1 c
D hy z , z
4
d
f
2
Cosh Jx z_
f
X
V
W
//Simplify
1 ] f2 ^ m _ ` \ 1 ] Cosh a f `
2
q c 2 \ f ` [ Sinh a f ` b ^
\
[
2
,ij
2
e
k
g
k
Calculationof H2 l Hc2,
l h
h
e
,
n
b
^
d d
h
c2 m
q2 2 f m n g o
4 f q2 g p 2
c2 m
e
p
2
2
41
m
Vavsq
22
rule4
1
d
d2
2
2
B1 -
. rule4
hy z
'
&
f(
2
z .d !
"
Simplify
#
)
Sinh * f (
+ ,
2f (
0
d2
2
3
hy 1 z
0
. /
4 2
d2
H2 Sech $ f %
1
d
x . eq4 1
q2 2 x . eq4 1
c2 m Hsqave q2 2 ; eq5
c2 m
4
z 4 . d5
6
7
Simplify
4 4
d2
f9
2 :
2 H Tanh 8
f;
Hc2 4 ? Hsqave
f
8>
8>
Fs < Fn =
f4 Hc2
8A
Fn @
fD
2
2
F
E
fG
16 f A
B
Sinh H f G
I J
Sinh V f U
W X
B
G
B1 H
4M
Gs K Fs L
f4 Hc2
8O
Fn N
H2 Sech C
H2 Sech Q
fR
2
2
T
fU
S
P
16 f O
P
H2 Tanh Q
N
2fO
U
fR
2
S
U
H2
8[
Gn Y Fn Z
H2
8]
Fn \
eq6 ^ Gs _ Gn ` . H2 a y Hc2
b
f4 Hc2
8c
Hc2 y
8c
d
2
Hc2 y Sech e f f
2
h
fi
g
16 f c
d
Sinh jf i
Hc2 y Tanh e
k l
d
i
2fc
b
ff
2
g
i
eq7=Solve[eq6 0,y]
m
2 f5 p
n n
yo
2f
p
q
p
f Sech r
fs
2
2q
t
Sech r
fs
2
x x
2
t
Sinh uf
p v
w
8 Tanh r
fs
2
t
eq8=y/.eq7[[1]]//FullSimplify
f
y
z
f5 y z1 { Cosh | f y } ~
2 { Cosh | f y } ~  3 Sinh | f y
}
We now consider the solution of de Gennes model
‚
K € f_,  _ :ƒ 1 „
f2
‡
6 … 1 † f2
‚
†
‡
f  … Cosh €  f † 1
‚ ‡
3 … † f  „ Sinh € f 
Evaluation of critical magnetic field as a function of the thickness
Series[K[f, ],{f,0,5}]
ˆ
‰
2
1
Š 6
Œ
Œ
‹

Ž
30
‘
‘

2’
f

1
Š 6
Œ
Œ
‹
4
’
Ž
12600
‘
‘

f4 ’ O “ f ” 6

L1[ _]:=Module[{f1,g1, 1,f2}, 1= ;g1=FindRoot[K[f1, 1] 0,{f
1,0.2,0.99}];f2=f1/.g1[[1]]]
‰
‰
‰
‰
‰
m
s1=Table[{ ,L1[ ]},{ ,0.22,10,0.01}];ListPlot[s1,PlotStyle
{Hue[0],PointSize[0.015]},Background GrayLevel[0.7],
AxesLabel {" ","f"}]
‰
‰
‰
•
•
‰
42
•
f
0.8
0.6
0.4
0.2
2
4
6
8
10
Graphics
Fig.14(a)
The order parameter f vs ε (= d/λ). f tends to 1 in the large limit of ε
Plot L1 , , 5 , 3 , PlotStyle
Hue 0 , Thickness 0.015 ,
Background GrayLevel 0.7 , PlotRange
2.1, 3 , 0, 0.7 ,
" ", "f"
AxesLabel
f
0.6
0.5
0.4
0.3
0.2
0.1
2.2
2.4
2.6
2.8
Graphics
Fig.14(b) Detail of Fig.14(a). The order parameter f vs ε (= d/λ). f reduces to zero at
ε = 5.
Evaluationof H Hc 2 vs
Comparison with the approximation for H Hc 2 vs : 24
2 f3 1 f2
1 Cosh f
Hcrsq _, f_ :
f
Sinh f
2
Series[Hcrsq[ ,f],{f,0,5}]
‰
24
2
Plot
24
5
24
) *
+
2
!
24
2
f
2
#
#
"
+
24
5
, Hcrsq , L1
,
PlotStyle
Background
1
+ - - .
,
-
*
+
,
/
.
5, 8 ,
0 0 00
- 3
-
- 3 3
Hue 0 , Thickness 0.01 , Hue 0.7 , Thickness 0.01
GrayLevel 0.7 , PlotRange
2, 8 , 0, 6
2 2
1
,
$
81 2 && 4
O 'f( 6
% f
350
,
,
,
-
2
1
,
2 2
43
3
,
2
3 3 4
,
.
6
5
4
3
2
1
3
4
5
6
7
8
Graphics
Fig.15 ( H / H c )2 vs ε (red line) and the approximation 24/ε2 vs ε (blue line). These agree
well only in the vicinity of ε = 5 .
Evaluationoff2 vs H Hc 2 with asa parameter.
Approximation for H Hl 2 vs : 24 2
<<Graphics`MultipleListPlot`
fsq1[ _,hsq1_]:=Module[{f1,g1, 1,f2,
hsq2}, 1= ;hsq2=hsq1;g1=FindRoot[hsq2==Hcrsq[ 1,f1],{f1,0.6,
1}];f2=f1/.g1[[1]]]
The value of f for H/Hc= 1
Plot[fsq1[ ,1],{ ,2,10},AxesLabel {" ","f"},
PlotStyle {Hue[0],Thickness[0.01]},
Background GrayLevel[0.7]]
‰
‰
‰
‰
‰
‰
‰
•
‰
•
•
f
0.97
0.965
0.96
0.955
0.95
0.945
4
6
8
10
Graphics
Fig.16 Plot of f at H/Hc = 1 as a function of ε.
f2 vs H Hc 2 with 5 , 2.5, 3, 4, 4, 5, 5, 7, 9, 11, 20, and 30
X1 _ : Table Hcrsq , fsq1 , hsq1 , fsq1 , hsq1 2 ,
hsq1, 0, 1, 0.01 MultipleListPlot X1 5
, X1 2.5 , X1 3 , X1 4 , X1 4.5 , X1 5 ,
!
X1 7 , X1 9 , X1 11 , X1 20 , X1 30 , AxesLabel " H Hc 2", "f2"
44
f2
H Hc 2
0.2
0.4
0.6
0.8
1
0.98
0.96
0.94
0.92
0.9
0.88
Graphics
Fig.17 Plot of f2 vs (H/Hc)2 with ε =
hny t_, f_, _
Cosh z
Cosh
Cosh f t
f
Sech
2
f
2
.
d
,z
td
Jnx t_, f_, _
f Sech
f
5 , 2.5, 3, 4, 4.5, 5, 7, 9, 11, 20, and 30.
f
2
fSech
Sinh f t
"
f
2
Sinh
fz
.
d
,z
td
#
!
Magnetic field distribution
Plot[Evaluate[Table[hny[t,L1[ ], ],{ ,2.5,10.0,0.2}]],{t,1/2,1/2},PlotStyle Table[Hue[0.05
i],{i,0,20}],Prolog AbsoluteThickness[1.5],
Background GrayLevel[0.7],AxesLabel {"x/d","h"}]
$
$
$
%
%
%
%
h
1
0.8
0.6
0.4
0.2
x& d
-0.4
-0.2
0.2
0.4
Graphics
Current density distribution
Plot[Evaluate[Table[Jnx[t,L1[ ], ],{ ,2.5,10.0,0.2}]],{t,1/2,1/2},PlotStyle Table[Hue[0.05
i],{i,0,20}],Prolog AbsoluteThickness[1.5],
Background GrayLevel[0.7],AxesLabel {"x/d","J"}]
$
$
$
%
%
%
%
45
J
1
0.5
x d
-0.4
-0.2
0.2
0.4
-0.5
-1
Graphics
Fig.18 Plot of (a) the magnetic field distribution and (b) the current density as a function
of x/λ, where e is changed as a parameter. ε = 2.5 – 10.0. ∆ε = 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.
Fig.19 Geometrical configuration of a cylinder (a radius R) with small thickness. R»d.
q*
q*
2
2
J s = * ψ (∇θ − A) = q * ψ v s .
m
c
2
*
2
Suppose that ns = ψ = ψ ∞ =constant, then we have
(8.65)
c
m*c
∇
θ
=
J +A.
(8.67)
2 * s
q*
q * ns
Integrating the current density around a circle running along the inner surface of the
cylinder,
46
−
4πλ2
Φ0
∇θ ⋅ dl =
J s ⋅ dl + A ⋅ dl ,
2π
c
(8.68)
A ⋅ dl = (∇ × A ) ⋅ da = B ⋅ da = Φ i ,
(8.69)
and
where Φ i is the flux contained within the core of the cylinder. The magnetic flux Φf is
defined as
Φ
Φ f = − 0 ∇ θ ⋅ dl .
(8.70)
2π
Then we have
4πλ2
Φf =
J s ⋅ dl + Φ i ,
(8.71)
c
or
c
J s ⋅ dl =
(Φ f − Φ i ) .
(8.72)
4πλ2
The phase θ of the wave function must be unique, or differ by a multiple of 2π at each
point,
2πc
Φf = * n .
(8.73)
q
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
d2
1
(8.74)
Bz ( x) = 2 Bz ( x ) ,
2
dx
λ
with the boundary condition, Bz (− d ) = H i and Bz (d ) = H 0 . The origin x = 0 is shown in
the above Figure.
x
x
sinh( )
cosh( )
+
H 0 − Hi
H
H
λ + 0
λ .
i
(8.75)
Bz ( x ) =
d
d
2
2
sinh( )
cosh( )
λ
λ
The current density Jy(x) is given by
J y ( x) = −
c dBz ( x )
c
=
4π dx
4πλ
H i cosh(
d−x
λ
) − H 0 cosh(
sinh(
2d
The value of current flowing near the inner surface is
2d
H i cosh( ) − H 0
c
λ
.
J y (−d ) =
2d
4πλ
sinh( )
λ
d+x
λ
)
.
(8.76)
)
(8.77)
λ
Then we have
J s ⋅ dl = 2πRJ y (− d ) =
c
4πλ2
(Φ f − Φ i ) ,
47
(8.78)
or
2πR
H i cosh(
c
4πλ
) − H0
λ
sinh(
or
2d
=
)
λ
c
(Φ f − Φ i ) ,
4πλ2
(8.79)
2 H 0πRλ
2d
sinh( )
Φf +
Hi =
2d
λ
2λ
2d
coth( )
πR 2 1 +
R
λ
.
(8.80)
or
The Gibbs free energy is given by
1
λ2
∆G = dr[ (∇ × B) 2 +
(B − H 0 ) 2 ] ,
8π
8π
or
(8.81)
λ2 ∂Bz ( x)
∆G = {
8π
∂x
−d
d
2
+ [ Bz ( x ) − H 0 ]2}dx ,
(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 Φ f .
We find that the solution of ∂∆G ∂Φ f = 0 Φ f leads to Φ f = Φ i , where R/λ = 10 – 50
and d/λ<1.
To prove this, we use the Mathematica for the calculation.
((Mathematica Program-13))
de Gennes book p. 195 Cylinder
Bz
Hi Sinh
2
Sinh
H0
x
H0
1
x
d
H0 Hi Cosh
Sech
2
rule1
Hi
$
!
'
'
%
'
'
&
R2 1
!
f 2 H0 R Csch 2d
/
,
Hi
-
0
.
1
2
3
+
#
Cosh " 2d
!
#
Sinh " 2d
2!
R
d
2 H0 R !
Sinh " 2d
x
1
d
x
H0 Hi Csch
Sinh
2
f
d
Hi Cosh
2
Cosh
(
*
*
*
*
)
#
=
4
3
0
7
7
5
7
7
6
R2 1
3
9
2 Coth 8 2d
/
R
<
<
<
<
;
:
Bz1=Bz/.rule1//Simplify
>
Csch
>
2d
?
@
A
H0 R2 Sinh
B
d x
C
@
2 H0 R Cosh
B
?
d x
>
C
?
@
@
A F F G
C
A
D
>
B
f Sinh
d x
E
?
@
C
C1=D[Bz1,x]//Simplify
48
C
2d
@
R R 2 Coth
A
?
@
A F F
2d
Csch f Cosh R
G1 1
8
C12 d x
2 Sinh 2d
8
R 2 Coth 2
d x
H0 R R Cosh d x
Bz1 H0 2 Simplify
1
8 3
2
2
2d
Csch d x
2 Sinh d x
f Cosh d x
2 H0 R Cosh R2 Sinh 2d
H0 Csch !
f Sinh H0 R R Cosh d x
H0
2
2d
R R 2 Coth d x
2 !
d x
R2 R 2 Coth 2! !
2d
! !
%
d2
G2 "
# $
G1 & x ' ' Simplify; G3 " D ( G2, ) f*
%
d2
+,
+
Simplify
3d
2, d . 2
,
. 0 1 f Cosh 2
+
,
,
5
d
3d
5d
H0 / R R 0 R Cosh - , . 2 2 Sinh - , . 0 2 Sinh - , .
2
2
2
5
8
+
,
2
2d
2d
7
:
3 2
7
R 2 Cosh - , . 0 R Sinh - , . 3 9:
64 /
Sinh -
d
,
' '
.
H0 / R2 Cosh -
3 3 3 4
G4=Solve[G3 0, f]//Simplify
;
<
= = >
f ? H0 @ R Sech A
5d
B
2
R Cosh A
2d
B
C
D
B
E
C
3d
B
2
R E R Cosh A
2 Sinh A
3d
B
2
C
F
B
C F
2 Sinh A
5d
B
2
C G H H
F1= f/.G4[[1]]//Simplify
<
H0 I R Sech J
2d
K
M
3d
5d
K L O R Cosh J
K L
2 S
2S
R N R Cosh J
rule1={d
{d
,R
Q
U
V
W
L
U
R
W
X
]
F2 Y F1 Z [ H0 \ R2
1
^
_
Sech ` 2 a
, R
}
Z
Q
T
K
N
2 Sinh J
3d
K L
2
K
O
2 Sinh J
5d
K
2
L P
}
. rule1 Z Z Simplify
b
3a
5a c
3a
5a
^
Cosh d
2 Sinh d
2 Sinh d
2 e f
2 e
2 e f
2 egg
f is the local m inimum value of the Gibbs free energy. Plot
of f i j H0 k R2 l vs m n d o p n 0 q 1, where r n R o p n 10 q 50
h
^
c
_
^
Cosh d
Plot[Evaluate[Table[F2,
{ ,10,50,10}]],{ ,0,1},PlotRange {{0,1},{0,1.5}},
PlotPoints 100, PlotStyle Table[Hue[0.2 i], {i,0,5}],
Prolog AbsoluteThickness[1.5], Background GrayLevel[0.7] ]
h
T
R
Q
Q
Q
Q
Q
49
1.4
1.2
1
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1
Graphics
Fig.20 Plot of the normalized magnetic flux Φ/(πR2H0) as a function of ε (ε = 0 – 1),
where µ = 10 – 50. ∆µ = 10.
8.5
Little Parks experiments1,4
8.5.1 The case of d«λ and d«ξ,
The thickness d of the superconducting cylinder is much shorter than the radius R. It
is also shorter than λ and ξ. Since d«λ, 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.
q*
q*
2
2
J s = * ψ (∇θ − A) = q * ψ v s ,
(8.83)
m
c
2
1
1
q*
B2
4
,
(
)
ψ
f s − fn = α ψ + β ψ +
∇
−
A
+
2
2m* i
8π
c
2
ψ (r ) = ψ (r ) eiθ (r ) .
We consider the case when
2
ψ (r ) = ns * = constant
or
ψ (r ) = ψ e iθ (r ) .
Then we have
q*
q*
( ∇ − A)ψ = ( ∇θ − A )ψ = (m* v s )ψ ,
i
c
c
and
1
m* 2 2 B 2
2
4
ψ vs +
f s = fn + α ψ + β ψ +
,
2
2
8π
Js
q*
vs = * 2 ,
∇θ = A + m* v s ,
c
qψ
(8.84)
(8.85)
(8.86)
(8.87)
(8.88)
(8.89)
Then we have
∇θ ⋅ dl = [θ ] =
q*
A ⋅ dl + m * v s ⋅ dl ,
c
(8.90)
where [θ ] represents the change in phase after a complete revolution about the cylinder.
[θ ] = 2πn (n; arbitrary integer).
50
(∇ × A) ⋅ da = B ⋅ da =Φ = πR 2 B .
A ⋅ dl =
A
(8.91)
A
Φ represents the flux contained in the interior of the cylinder.
Φ
vs = * (n +
),
Φ0
mR
where
2πc
Φ0 = * .
q
(8.92)
(8.93)
Here we see from the free energy density that the free energy will be minimized by
m* 2 2
choosing n such that |vs| is minimum since f is proportional to
ψ vs .
2
Φ
vs = min[ * | n +
|] .
(8.94)
mR
Φ0
vs 2m R 0.5
0.4
0.3
0.2
0.1
0
-3
-2
-1
1
Fig.21 Plot of the velocity vs as a function of Φ / Φ 0 .
2
3
The current density is equal to zero when Φ = nΦ 0 .
Since vs is known, we can minimize the free energy with respect to ψ and find
∂f / ∂ψ = 0 ,
(8.95)
1
2
(8.96)
1
2
(8.97)
2
or
α + β ψ + m*vs 2 = 0 ,
2
or
ψ = −(α + m*vs 2 ) / β ≥ 0 ,
2
or
−α ≥
1 * 2
m vs
2
or
1
2
α ≥ m*vs 2 .
51
(8.98)
vs 2
2m R 2
0.25
0.2
0.15
0.1
0.05
-3
Fig.22 Plot of
vs2
-2
-1
1
2
vs Φ / Φ 0 . The horizontal lines denote y = α for several α.
The transition occurs when
Φ
1
α = m*vs 2 = α 0[1 − T ( ) / Tc ] ,
2
Φ0
where α 0 = 2
0
3
(8.99)
β
H c (0) .
4π
In the above figure, the intersection of the horizontal line of α at fixed T and the curve
vs gives a critical temperature. In the range of Φ / Φ 0 where the horizontal line is located
2
2
above the curve vs , the system is in the superconducting phase. In the range of
Φ / Φ 0 where the horizontal line is located below the curve vs , on the other hand, the
system is in the normal phase.
2
8.5.2 The case of d»λ and d»ξ.
The thickness d of the superconducting cylinder is longer than λ and ξ. The current
density (proportional to derivative of B) is zero. The absolute value of the order
parameter is constant in the superconducting layer.
q*
q*
2
2
J s = * ψ (∇θ − A) = q* ψ v s = 0 ,
(8.100)
m
c
or
q*
∇θ =
A,
(8.101)
c
q*
q*
q*
A ⋅ dl =
B ⋅ da = Φ ,
∇θ ⋅ dl = [θ ] =
(8.102)
c
c
c
or
Φ=
2πc
n = nΦ 0 .
q*
(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
52
A ' = A − ∇χ = 0 ,
(8.104)
or
A = ∇χ .
The scalar potential χ is described by
χ ( L) − χ (0) = dl ⋅ A = da ⋅ (∇ × A) = da ⋅ B = Φ = Φ ext + Φ i ,
(8.105)
(8.106)
where Φ is the total magnetic flux.
We now consider the gauge transformation. A’ and A are the new and old vector
potentials, respectively. ψ’ and ψ are the new and old wave functions, respectively.
A ' = A + ∇( − χ ) = 0 ,
(8.107)
iq* χ
)ψ (r ) .
(8.108)
c
Since A’ = 0, ψ’ is the field-free wave function and satisfies the GL equation
ψ ' (r ) = exp(
αψ '+ β ψ ' ψ '−
2
2
*
2m
∇ 2ψ ' = 0 .
(8.109)
In summary, we have
iq*
iq*
Φ
ψ (r ) = ψ ' (r ) exp[ − χ ] = ψ ' (r ) exp[− Φ ] = ψ ' (r ) exp[ 2πi ] , (8.110)
c
c
Φ0
since
q*
c
=
2π
and q*<0.
Φ0
9.
Isolated filament: the field for first penetration2
9.1
Critical field Hc1
We have the following equations;
1
~
∇ρ + A
iκ
2
f = f − f f,
2
(9.1)
~
i
2~
∇ ρ × (∇ ρ × A) =
[ f *∇ ρ f − f∇f * ] − f A ,
2κ
~
h = ∇ρ × A .
The vector potential in cylindrical co-ordinates takes the form
B
~ 2πξ
, A=
A = (0, Aφ ,0) , h =
A,
r=λ
Φ0
2Hc
Stokes theorem
B ⋅ da = (∇ × A ) ⋅ da = A ⋅ dr = 2πrAφ = Φ ,
or
Aφ =
Φ
Φ ~
= 0 Aφ
2πr 2πξ
(9.2)
(9.3)
(9.4)
(9.5)
(9.6)
or
Far from the center of the vortex,
53
~ ξ Φ
ξ Φ 1 Φ 1 1
=
Aφ =
=
=
,
r Φ 0 λρ Φ 0 κ Φ 0 ρ κρ
for Φ = Φ 0 .
Near the center of the vortex,
~ πκρ
Aφ =
H (0) .
or
Aφ 2πr = H (0)πr 2 ,
Φ0
We assume that
f = ψ ( ρ )e − iφ ,
where ψ ( ρ ) is a real positive function of ρ.
~
∂Aφ
~
1 ∂
~
1 ~
h = ∇ρ × A = ez
),
( ρAφ ) = e z ( Aφ + ρ
ρ ∂ρ
ρ
∂ρ
~
~
∂ 2 Aφ
∂ 1 ∂
1
1 ∂Aφ
~
∇ ρ × h = eφ [ 2 Aφ − (
+ ρ 2 )] = −eφ
[
( ρAφ )] .
∂ ρ
ρ
ρ ∂ρ
∂ρ ρ ∂ρ
GL equation can be described by
1 1 d
1 ~ 2
dψ
(ρ
) + (−
− 2
+ Aφ ) ψ = ψ − ψ 3 .
κ ρ dρ
κρ
dρ
The current equation: eφ component.
d 1 d
~
1 ~ 2
−
[
( ρAφ )] + (−
+ Aφ )ψ = 0 .
κρ
dρ ρ dρ
The above equations can be treated approximately when κ»1.
We assume that ψ ( ρ ) =1 in the region ρ > 1 / κ .
1 d
1
1 ~
d2
+
− (1 + 2 )](−
+ Aφ ) = 0
2
ρ dρ
ρ
κρ
dρ
The solution of this differential equation is
1 ~
−
+ Aφ = c1K1 ( ρ )
[
κρ
(9.7)
(9.8)
(9.9)
(9.10)
(9.11)
(9.12)
(9.13)
(9.14)
(9.15)
or
~
1
Aφ =
+ c1K1 ( ρ )
(9.16)
~
∂Aφ
1 1
1
1 ~
dK ( ρ ))
)= [
],
hz = ( Aφ + ρ
+ c1K1 ( ρ ) + ρ ( − 2 + c1 1
ρ
ρ κρ
κρ
dρ
∂ρ
(9.17)
κρ
or
hz =
c1
ρ
[ K1 ( ρ ) + ρ
dK1 ( ρ )) c1
] = [− ρK 0 ( ρ )] = −c1 K 0 ( ρ ) .
dρ
ρ
(9.18)
or
hz = −c1K 0 ( ρ ) ,
from the recursion formula
dK ( ρ )
+ K1 ( x) = − ρK 0 ( ρ ) ,
ρ 1
dρ
(9.19)
(9.20)
54
What is the value of c1?
~
When ρ ≈ 0 , Aφ should be equal to zero.
~
1
1 c1
+ c1K1 ( ρ ) ≈
− =0,
Aφ =
(9.21)
κρ
κρ ρ
because K1 ( ρ ) ≈ 1 / ρ . Then we have
c1 =
1
κ
.
(9.22)
In fact,
Φ0
2
B
Φ0
1
h=
= 2πλ K 0 ( ρ ) =
K0 ( ρ ) = K 0 ( ρ ) ,
2
κ
2H c
2H c
2π 2 H c λ
(9.23)
or
h=
1
κ
K0 (ρ ) ,
(9.24)
since
Φ0
κ.
2π
In summary, we have
~
1 1
Aφ =
− K1 ( ρ )
2 H c λ2 =
κρ
h=
1
κ
κ
(9.25)
,
(9.26)
K0 (ρ )
((Note))
The total magnetic flux through a circle of radius of ρ »1 is given in the following way.
~
1 1
1
Aφ =
− K1 ( ρ ) ≈
.
(9.27)
κρ
κ
κρ
Then we have
Φ = 2π (λρ ) Aφ = 2πλρ
Φ0 ~
Φ 1
= Φ0 .
Aφ = 2πλρ 0
2πξ
2πξ κρ
9.2
The center of vortex4
First we consider the behavior at the center of vortex, as ρ 0.
dψ
1 1 d
1 ~ 2
− 2
+ Aφ ) ψ = ψ − ψ 3 ,
(ρ
) + (−
κ ρ dρ
κρ
dρ
with
~ πκρξ 2
Aφ =
H (0) ,
Φ0
or
1 1 d
1 πκξ 2 ρ
dψ
(ρ
) + [−
− 2
+
H (0)]2ψ = ψ − ψ 3 .
κρ
κ ρ dρ
dρ
Φ0
55
(9.28)
(9.29)
(9.30)
(9.31)
We assume that ψ can be described by series
ψ = ρ k (a0 + a1ρ + a2 ρ 2 + a3 ρ 3 + ...) ,
where a0 0. Using the Mathematica 5.2, we determine the parameters;
k =1
a1 = 0
a2 = − a0
κ2
8
[1 +
2πξ 2 H (0)
κ2
H (0) ,
] = −a0 [1 +
]
Φ0
8
Hc2
(9.32)
(9.33)
a3 = 0
Φ0
.
2πξ 2
Then we have
where H c 2 =
κ 2ρ 2
H (0)
)] .
Hc2
8
The rise of ψ ( ρ ) starts to saturate at ρ = 2 / κ .
ψ ( ρ ) = a0 ρ[1 −
(1 +
((Summary))
If the vortex contains 1 flux quantum, we have
f ( ρ ) = e − iφψ ( ρ ) ,
ψ ( ρ ) = a0 ρ{1 −
9.3
κ ρ
2
8
2
[1 +
H (0)
] + ...} .
Hc2
(9.34)
(9.35)
(9.37)
Far from the center of the vortex
dψ
1 1 d
− 2
(ρ
) = ψ −ψ 3 ,
κ ρ dρ
dρ
(9.38)
~
1
Aφ =
.
(9.39)
with
κρ
This differential equation is simplified as.
1
1
[ψ " ( ρ ) + ψ ' ( ρ )] + ψ ( ρ ) − ψ ( ρ )3 = 0 .
2
κ
ρ
In the region where ψ ( ρ ) → 1 , for simplicity we assume that
ψ (ρ ) = 1 − ϕ(ρ ) ,
ϕ ' ( ρ ) + ρϕ "( ρ )
[1 − ϕ ( ρ )]3 = 1 − ϕ ( ρ ) −
.
κ 2ρ
This equation can be approximated by
ϕ ' ( ρ ) + ρϕ " ( ρ )
1 − 3ϕ ( ρ ) = 1 − ϕ ( ρ ) −
,
2
(9.40)
(9.41)
(9.42)
κ ρ
(9.43)
ϕ ' ( ρ ) − 2κ 2ϕ ( ρ ) = 0 .
(9.44)
or
ϕ"( ρ ) +
1
ρ
56
The solution is given by
ϕ ( ρ ) = C1 I 0 ( 2 ρκ ) + C 2 K 0 ( 2 ρκ ) .
(9.45)
Since I 0 ( 2 ρκ ) increases with increasing ρ, C1 should be equal to zero.
Then we have
ϕ ( ρ ) = C 2 K 0 ( 2 ρκ ) .
For large ρ,
π e− x
π e− x
,
.
K 0 ( x) ≈
K1 ( x) ≈
2 x
2 x
10.
10.1
(9.46)
(9.47)
Properties of one isolated vortex line (H = Hc1)
The method of the Green’s function
2
J s = q* ψ v s ,
1
1
4
f s = fn + α ψ + β ψ +
2
2m*
2
(10.1)
2
q*
1 2
∇− A ψ +
B ,
i
c
8π
(10.2)
When ψ = ψ eiθ and ψ = ns = independent of r,
2
*
( )
1
m* * 2 B 2
* 2
f s − f n = αns + β ns +
ns vs +
.
(10.3)
2
2
8π
J
4π
Js .
vs = * s *
and
∇×B =
(10.4)
c
q ns
Then the net energy can be written by
1
E = E0 +
dr[B 2 + λ2 (∇ × B)2 ] .
(10.5)
8π
When
(10.6)
B → B + δB , , E → E + δE ,
1
δE =
dr[B + λ2∇ × (∇ × B)] ⋅ δB .
(10.7)
4π
The integrand can be calculated by using VariationalD of the Mathematica 5.2.
*
((Mathematica Program-14)) Variation method
(*Derivation of London's equation*)
<<Calculus`VariationalMethods`
Needs["Calculus`VectorAnalysis`"]
SetCoordinates[Cartesian[x,y,z]]
Cartesian[x,y,z]
h h1 x, y, z , h2 x, y, z , h3 x, y, z ;
eq1 h.h 2 Curl h . Curl h
Expand
2
h1 x, y, z 2 h2 x, y, z 2 h3 x, y, z 2 2 h1 0,0,1 x, y, z
2
2
2
2
2 2 h2 0,0,1 x, y, z h3 0,1,0 x, y, z
h2 0,0,1 x, y, z
h1 0,1,0 x, y, z
2
2
0,1,0
2
0,1,0
h3
x, y, z
2 h1
x, y, z h2 1,0,0 x, y, z
2
2
2
2
h2 1,0,0 x, y, z
h3 1,0,0 x, y, z
2 2 h1 0,0,1 x, y, z h3 1,0,0 x, y, z
eq3=VariationalD[eq1,h2[x,y,z],{x,y,z}]/2//Expand;eq4=Variat
57
ionalD[eq1,h3[x,y,z],{x,y,z}]/2//Expand;eq2=VariationalD[eq1
,h1[x,y,z],{x,y,z}]/2//Expand;
(*London's equation*)
eq5 h 2
h1 x, y, z h2 x, y, z h3 x, y, z Curl Curl h 2
h1 0,0,2 x, y, z h1 0,2,0 x, y, z h3 1,0,1 x, y, z h2 1,1,0 x, y, z ,
h2 0,0,2 x, y, z h3 0,1,1 x, y, z h1 1,1,0 x, y, z h2 2,0,0 x, y, z ,
h2 0,1,1 x, y, z h3 0,2,0 x, y, z h1 1,0,1 x, y, z h3 2,0,0 x, y, z 2
2
eq2-eq5[[1]]//Simplify
0
eq3-eq5[[2]]//Simplify
0
eq4-eq5[[3]]//Simplify
0
Demanding that E be a minimum leads to the London equation; δE = 0,
B + λ2∇ × (∇ × B ) = 0 .
This is called the second London’s equation.
((Note))
B 2 → B 2 + 2B ⋅ δB ,
(∇ × B )2 → (∇ × B) 2 + 2(∇ × B) ⋅ (∇ × δB ) .
Then we have
1
δE =
dr[B ⋅ δB + λ2 (∇ × B) ⋅ (∇ × δB)] .
4π
Since
(∇ × B ) ⋅ (∇ × δB) = δB ⋅ [∇ × (∇ × B)] + ∇ ⋅ [δB × (∇ × B)] ,
1
1
δE =
dr[B + λ2∇ × (∇ × B)] ⋅ δB +
da ⋅ [δB × (∇ × B)] .
4π
4π
The second term is equal to zero.
Here we assume that
(i) λ»ξ.
(ii) The hard core of radius ξ is very small, and we shall neglect
contribution to the energy
58
(10.8)
(10.9)
(10.10)
(10.11)
(10.12)
(10.13)
completely its
Fig.23 Top view of the geometrical configuration of a single vortex.
The line energy is given by
1
ε=
d 2r [B 2 + λ2 (∇ × B) 2 ] .
(10.14)
r >ξ
8π
Demanding that ε be a minimum leads to the London’s equation.
B + λ2 [∇ × (∇ × B)] = 0 ,
(10.15)
for r > ξ .
In the interior of hard core, we can try to replace the corresponding singularity by a 2D
delta function.
B + λ2 [∇ × (∇ × B)] = Φ 0e zδ 2 (r ) ,
(10.16)
B ⋅ da + λ2 [∇ × (∇ × B )] ⋅ da = Φ 0e zδ 2 (r ) ⋅ da = Φ 0 ,
Using Stokes theorem,
B ⋅ da + λ2 (∇ × B) ⋅ dl = Φ 0e zδ 2 (r ) ⋅ da = Φ 0 .
(10.17)
(10.18)
Fig.24 The path of line integral
where ξ < r << λ .
If the circle of the path integral has a radius r ( ξ <<r<<λ), the first term I1 is on the order
of Φ 0 r 2 / λ2 and is negligibly small. In the second term I2,
dl = eφ rdφ ,
(10.19)
∇ × B = −eφ
d
B(r ) .
dr
(10.20)
59
Then we have
I2 = λ
(∇ × B) ⋅ dl = λ
2
2π
2
[ −e φ
0
d
B(r )] ⋅ e φ rdφ
dr
,
(10.21)
d
= −λ2 2πr B(r ) = Φ 0
dr
or
dB(r )
Φ0
=−
,
dr
2πλ2 r
or
B(r ) =
(10.22)
Φ0
λ
[ln( ) + const ] ,
2
2πλ
r
(10.23)
for ξ < r << λ
We now consider the exact solution
B + λ2 [∇ × (∇ × B)] = Φ 0e zδ 2(r ) , and
Note that
∇ × (∇ × B) = ∇(∇ ⋅ B) − ∇ 2B = −∇ 2B .
Thus we have
Φ
1
∇ 2B − 2 B = − 20 e zδ 2(r ) ,
λ
∇⋅B = 0 .
(10.24)
(10.25)
(10.26)
λ
where δ2(r) is a 2D delta function.
((Note)) 2D case Green function
Modified Helmholtz
(i)
2
(∇1 − k 2 )G (r1 , r2 ) = −δ 2(r1 − r2 ) ,
1
G (r1, r2 ) =
K 0 (k r1−r2 ) .
2π
Modified Bessel function
(ii)
(∇ 2 − k 2 )ψ (r ) = − f (r ) ,
(10.27)
(10.28)
(10.29)
ψ (r ) = G (r, r ' ) f (r ' )dr ' ,
(10.30)
(∇ 2 − k 2 )ψ (r ) = (∇ 2 − k 2 )G (r , r ' ) f (r ' )dr '
= − δ (r − r ' ) f (r ' )dr ' = − f (r )
The solution of the equation
1
Φ
∇ 2B − 2 B = − 20 e zδ 2(r ) ,
λ
B(r ) = e z G (r , r ' )
Φ0
λ
2
δ 2(r ' ) = e z
(10.31)
(10.32)
λ
is that
,
Φ0
r
K0 ( ) ,
2
2πλ
λ
or
60
(10.33)
Φ0
r
K 0 ( ) (in general).
2
2πλ
λ
((Asymptotic form))
For ξ<r«λ,
Φ
r
B(r ) = 0 2 [ln( ) + 0.115932] .
2πλ
λ
For r»λ,
Φ 0 πλ
r
exp( − ) .
B(r ) =
2
2πλ 2r
λ
The magnetic flux Φ (r ) inside the circle with a radius r is
B(r ) =
r
Φ (r ) = 2πr ' dr ' B(r ' ) =
0
Φ0
λ
r
r'
r ' dr ' K 0 ( ) = Φ 0
2
λ
0
(10.34)
(10.35)
(10.36)
r/λ
xdxK 0 ( x ) = Φ 0[1 −
0
r
λ
r
K1 ( )] .
λ
(10.37)
The current density
c
c d
J=
∇ × B = −e φ
B(r ) .
4π
4π dr
(10.38)
,B
1
0.8
0.6
0.4
0.2
x
-4
Fig.25
-2
2
4
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))
(* the radial distribution of h(r)*)
Plot BesselK 0, x , x, 0, 2 , PlotPoints 100, PlotStyle Hue 0.5 ,
Prolog AbsoluteThickness 2 , Background GrayLevel 0.7 ,
AxesLabel
"r ", "h 0 2 2"
61
h
0 2
2
5
4
3
2
1
r
0.5
Graphics
1
1.5
2
Fig.26 Plot of the normalized magnetic induction B(r ) /
x = r/λ.
Φ0
= K 0 (r / λ ) as a function of
2πλ2
x
BesselK 0, y y y
0
1-x BesselK[1,x]
Plot[ ,{x,0,10},PlotPoints 100, PlotStyle Hue[0.5],
PlotRange {{0,10},{0,1}},Prolog AbsoluteThickness[2],Backgr
ound GrayLevel[0.7],AxesLabel {"r/ "," / 0"}]
0
1
0.8
0.6
0.4
0.2
r
2
Graphics
4
6
8
10
Fig.27 Magnetic flux Φ(r ) / Φ 0 = [1 −
reduces to 1 as x → ∞ .
r
λ
r
K1 ( )] as a function of x = r/λ. Φ (r ) / Φ 0
λ
10.2. Vortex line energy
62
Cylinder with a radius ξ.
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
1
ε 1 = d 3r [B 2 + λ2 (∇ × B)2 ] .
(10.39)
8π
Vector identity
(∇ × B )2 = B ⋅ [∇ × (∇ × B)] + ∇ ⋅ [B × (∇ × B )] .
(10.40)
1
1
ε1 =
d 3r[B + λ2∇ × (∇ × B )] ⋅ B +
d 3rλ2∇ ⋅ [B × (∇ × B )] , (10.41)
8π
8π
63
or
λ2
1
3
ε1 =
(10.42)
d r[Φ 0e zδ 2 (r )] ⋅ B +
da ⋅ [B × (∇ × B)] .
8π
8π
We use the Stokes theorem for the second term.
We now neglect the core contribution (the first term). The integral da is to be taken
over the surface of the hard core (cylinder of radius ξ).
λ2
da ⋅ [B × (∇ × B)] .
ε1 =
8π
Note that
B × (∇ × B) = Bz ( ρ )
(10.43)
∂Bz ( ρ )
eρ
∂ρ
da = −e ρ da
and
(10.44)
∂Bz ( ρ )
∂B ( ρ )
λ2
λ2
e ρ ⋅ (−e ρ )]da ⋅ = − [ Bz ( ρ ) z
ε1 =
[ Bz ( ρ )
2πρ ]ρ =ξ .
∂ρ
∂ρ
8π
8π
(10.45)
Since
Φ 1
dBz ( ρ )
= − 02
2πλ ρ
dρ
we have
(10.46)
λ
Φ
Φ0
[ln( ) + 0.115932] .
ε1 = 0 Bz (ξ ) =
8π
4πλ
ξ
More exactly
2
ξ
ξ
Φ0 ξ
K 0 ( ) K1 ( ) .
ε1 =
4πλ λ
λ
λ
(see the derivation of this equation in the Mathematica Program-16).
2
((Mathematica Program-16))
(*vortex line energy*)
<<Calculus`VectorAnalysis`
SetCoordinates[Cylindrical[ , ,z]]
Cylindrical[ , ,z]
h={0,0,hz[ ]}
{0,0,hz[ ]}
Cross[h,Curl[h]]//Simplify
hz
rule1
hz
hz
)
)
)
)
' *
)
(
&
%
, 0, 0
0
2
BesselK 0,
2
+
,
-
&
2
.
9
hz
/
# $
&
"
#1
0 BesselK 0,
2
#
!
2
2
2
02
23
1
2
1
4
hz
=
>
5
6
2
7
8
(10.47)
8
: 9 ;
D hz
:
: 9 ;
,
9 ;
<
. rule1
< <
8
02 BesselK 0,
?
A@
B
BesselK 1,
?
A@
B
16 2 3
C
D
64
Simplify
(10.48)
f1
1
1
1
BesselK 0,
BesselK 1,
x
x
x
BesselK 0, 1 BesselK 1, 1
x
x
x
Limit[f1-Log[x],x ]//N
0.115932
(*graph, Comparison with ln( / )+0.115932*)
Plot Log x 0.115932, f1 , x, 1, 10 , PlotPoints 100,
PlotRange 0, 10 , 0, 2.5 , PlotStyle Hue 0 , Hue 0.5 ,
Prolog AbsoluteThickness 2 , Background GrayLevel 0.7 ,
AxesLabel " ", " 1 02 16 2 2" 1
02 16
2.5 2 2
2
1.5
1
0.5
2
!
Graphics
4
6
8
10 !
Fig.29 Plot of ε 1 / (Φ 0 / 4πλ ) =
1
K 0 (1 / x) K1 (1 / x ) [blue line] as a function of x = λ / ξ .
x
For comparison, the approximation given by (lnx+0.115932) is also shown by a
red line.
2
10.3. Derivation of Hc1
H
G=F−
hd 3r .
(10.49)
4π
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 = Fs .
(10.51)
H
Gs|first vortex = Fs + ε 1L − c1 LΦ 0 ,
(10.52)
4π
where L is the length of the vortex line in the system and
Bd 3r = L Bda = LΦ 0 .
(10.53)
"
#
#
Then we have
Fs = Fs + ε1L −
or
H c1 =
H c1
LΦ 0 ,
4π
(10.54)
4πε1
,
Φ0
(10.55)
65
or
H c1 =
H
Φ0
λ
ln( ) = c ln(κ ) .
2
ξ
4πλ
2κ
(10.56)
10.4. Interaction between vortex lines
Suppose that there are two vortices in the 2 D plane.
Φ
1
∇ 2 B − 2 B = − 20 [δ 2(r − r1 ) +δ 2(r − r2 )] .
λ
λ
Using the Green function
r − r'
1
G (r , r ' ) =
K0 (
).
2π
λ
Then we have
Φ
B = dr ' G (r, r ' ) 20 [δ 2(r − r1 ) +δ 2(r − r2 )] ,
λ
or
(10.57)
(10.58)
(10.59)
B=
r − r'
Φ0
dr ' K 0 (
)[δ 2(r − r1 ) +δ 2(r − r2 )] ,
2
λ
2πλ
(10.60)
or
B (r ) =
r − r1
r − r2
Φ0
[ K0 (
) + K0 (
)] = B1 (r ) + B2 (r ) .
2
2πλ
λ
λ
(10.61)
where
r − r1
Φ0
K0 (
(10.62)
),
2
2πλ
λ
r − r2
Φ
).
B2 (r ) = 0 2 K 0 (
(10.63)
2πλ
λ
B1 (r2 ) = B2 (r1 ) .
(10.64)
The total increase in free energy per unit length can be rewritten as
Φ
ε = 0 [{B1 (r1 ) + B2 (r1 )} + {B1 (r2 ) + B2 (r2 )}]
8π
.
(10.65)
Φ0
Φ0
B1 (r1 ) + 2
B1 (r2 )
=2
8π
8π
The first tem is the energy of individual vortex line. We have the interaction energy
2
Φ0
Φ0
r
U12 =
B1 (r2 ) = 2 2 K 0 ( 12 ) ,
(10.66)
4π
8π λ
λ
where r12 = r1 − r2
The interaction is repulsive, in which the flux has the same sense in both vortices.
B1 (r ) =
11.
Magnetization curves
11.1 Theory of M vs H
We consider the Gibbs function given by
66
BH
.
(11.1)
4π
i> j
The first term is an individual energy of the lines. nL is the number of lines per unit area
(cm2).
B = nL Φ 0 .
(11.2)
The second term is a repulsive interaction between vortices.
2
r
Φ0
U ij = 2 2 K 0 ( ij ) ,
(11.3)
8π λ
λ
where rij =| ri − r j |
One can distinguish three regimes between Hc1 and Hc2.
(a)
H ≈ H c1 : nL λ2 «1,
r>λ
G − Gs 0 = nLε1 +
(b)
(c)
U ij −
ξ <r<λ
nL λ2 »1
2
H ≈ H c 2 : nLξ ≈ 1
r ≈ξ
B
BH
G − Gs 0 =
ε 1 + U ij −
Φ0
4π
i> j
(11.4)
(a)
At very low densities (low B) the interaction is small and we neglect it completely.
ε
H
G − Gs 0 = ( 1 − ) B .
(11.5)
Φ 0 4π
H
ε
or H<Hc1, the lowest G is obtained for B = 0.
When 1 >
Φ 0 4π
ε
H
When 1 <
, we can lower G by choosing B 0. There is some flux penetration.
Φ 0 4π
4πε 1
Φ0
λ
H c1 =
=
ln( ) .
(11.6)
2
Φ0
ξ
4πλ
(b) When H > H c1
d: distance between neighboring lines (d>λ).
(i)
Square lattice
zs = 4
nL =
(11.7)
1
B
= 2
Φ0 ds
or
Φ0
B
Triangular lattice
ds =
(ii)
(11.8)
67
zt = 6
B
=
Φ0
nL =
1
3 2
dt
2
(11.9)
or
4
Φ0
Φ0
(11.10)
dt =
= 1.07457
3
B
B
We neglect all contributions except those of the nearest neighbors.
2
B z Φ0
d
BzΦ 0
d
(11.11)
U ij =
K0 ( ) =
K0 ( )
2 2
2 2
Φ 0 2 8π λ
λ 16π λ
λ
i> j
Then we have
B
1 Φ
d
G − Gs 0 =
[( H c1 − H ) + z 0 2 K 0 ( )]
(11.12)
4π
2 2πλ
λ
or
G − Gs 0
B ( H − H c1 )
d
B
d
2πκH c1 H
=
+ K 0 ( )] =
− 1) + K 0 ( )] ,
[−
[−
(
1 Φ0
1 Φ0
4π
zH c
H c1
λ
λ
4π
z
z
2
2 2πλ
2 2πλ2
(11.13)
or
2πκH c1 H
d
G − Gs 0
B
(11.14)
[−
(
=
− 1) + K 0 ( )] ,
zH c
4π
λ
zH c
H c1
2πκ
or
1/ 4
2πκH c1 H
Φ
G − Gs 0
B
=
[−
(
− 1) + K 0 ( ρ 0 0
zH c
zH c
H c1
B
4π
2πκ
1/ 2
)] ,
(11.15)
where
Φ0
λ
2
=
2 2H c
(11.16)
κ
The B vs H curve can be obtained from
∂G
= 0.
∂B
or
2πκH c1 H
d
1d
d 1
(
K0 '( ) = 0
−
− 1) + K 0 ( ) −
zH c
H c1
λ 2λ
λ B
or
Φ
2πκH c1 H
−
− 1) + K 0 ( ρ 0 0
(
zH c
H c1
B
1/ 2
1 1 Φ0
) + ρ0
2 B B
(11.17)
(11.18)
1/ 2
Φ
K1 ( ρ 0 0
B
1/ 2
)=0
(11.19)
Here we note that
68
2πκH c1
,
zH c
α0 =
(11.20)
and
Φ
= ρ0 0
λ
B
where ρ0 = 1 for the square lattice and 1.07457 for the triangular lattice.
K 0 ' ( x ) = − K1 ( x )
d
1/ 2
(11.21)
(11.22)
11.2 Calculation of M vs H using Mathematica
We make a plot of the free energy vs B and B vs H using the Mathematica 5.2.
((Mathematica Program-17))
(*Gibbs free energy
h=H/Hc1
*)
F a_, b_ :
B
4
BesselK 0,
a
b h 1
B
K[a_,b_]:=Plot[Evaluate[Table[F[a,b],{h,1,3,0.1}]],{B,0,5},P
lotStyle Table[Hue[0.1
i],{i,0,10}],Prolog AbsoluteThickness[1.5],Background GrayL
evel[0.7]]
Table[K[a,1],{a,0.4,1.2,0.4}]
0.4
0.3
0.2
0.1
1
2
3
4
5
1
2
3
4
5
-0.1
0.2
0.1
-0.1
-0.2
-0.3
69
0.2
0.1
1
2
3
4
5
-0.1
-0.2
-0.3
-0.4
{ Graphics , Graphics , Graphics }
B
a
[−b(h − 1) + K 0 (
) as a function of B, where h
Fig.30 Gibbs free energy given by
4π
B
(=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.
G a_, b_ :
B
D
4
BesselK 0,
a
B
b h 1 , B
Simplify
G[a,b]
1
8
2b 2bh 2 BesselK 0,
a
a BesselK 1,
a
B
B
B
!
!!
!!
!!
!!
!
<<Graphics`ImplicitPlot`
p11=ImplicitPlot[Evaluate[G[1,1] " 0,{h,1,1.15},{B,0.01,0.15}
],PlotPoints
50,PlotStyle {Hue[0],Thickness[0.015]},Background GrayLevel
[0.7],AxesLabel {"h","B"}]
70
B
0.14
0.12
0.1
0.08
0.06
0.04
0.02
h
1
1.02 1.04 1.06 1.08 1.1 1.12 1.14
ContourGraphics
Fig.31 The magnetic induction B vs h (=H/Hc1) above h = 1. h = 1 (H = Hc1).
p11=ImplicitPlot[Evaluate[G[1.2,1] " 0,{h,1,1.03},{B,0.01,0.0
8}],PlotPoints
50,PlotStyle {Hue[0],Thickness[0.015]},Background GrayLevel
[0.7],AxesLabel {"h","B"}]
B
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
h
1 1.0051.011.0151.021.0251.03
ContourGraphics
Fig.32 The detail of the magnetic induction B vs h (=H/Hc1) near h = 1 (H = Hc1).
71
12.
The linearized GL equation (H = Hc2).
12.1. Theory
2
When the term β ψ ψ can be dropped in the GL equation, we obtain linearized GL
equation
2
1
q*
αψ + * ∇ − A ψ = 0 .
2m i
c
(12.1)
This omission will be justified only if ψ << ψ ∞ = α / β .
2
2
2
2
2m* α
1
q*
∇
−
A
ψ
=
ψ
=
ψ = Eψ ,
2
2m* i
c
2m*
2m*ξ 2
2
(12.2)
where ξ =
2
2
. and E =
.
2m*ξ 2
2m* α
When B = H (//z), we choose the gauge
A = (0, Hx,0) .
(12.3)
Note that we use A = (H × r ) / 2 = (-Hy/2, Hx/2, 0). χ = Hxy/2,
A → A + ∇χ = (0, Hx,0) (gauge transformation).
This problem is reduced to the eigenvalue problem in the quantum mechanics.
Hˆ ψ = Eψ ,
(12.4)
2
1
q*
2
2
Hˆ =
[ pˆ x + ( pˆ y − Hxˆ ) 2 + pˆ z ] .
2m
c
This Hamiltonian Ĥ commutes with p̂ y and p̂ z .
[ Hˆ , pˆ y ] = 0 and
[ Hˆ , pˆ ] = 0
(12.5)
(12.6)
z
Hˆ n, k y , k z = En n, k y , k z ,
(12.7)
and
pˆ y n, k y , k z = k y n, k y , k z , pˆ z n, k y , k z = k z n, k y , k z ,
(12.8)
x, y, z pˆ y n, k y , k z = k y x, y, z n, k y , k z ,
x, y, z pˆ z n, k y , k z = k z x, y, z n, k y , k z ,
(12.9)
or
∂
ik y
y n, k y , k z = k y y n, k y , k z ,
y n, k y , k z ≈ e y .
(12.10)
i ∂y
∂
x, y, z n, k y , k z = k z x, y.z n, k y , k z
x, y, z n, k y , k z ≈ eik z z .
i ∂z
(12.11)
Schrödinger equation for ψ ( x, y, z ) = x, y, z n, k y , k z ,
*
∂ 2
∂ q
1
∂ 2
+
[(
) +(
Hx ) 2 + (
) ]ψ ( x, y, z ) = Eψ ( x, y , z ) , (12.12)
*
2 m i ∂x
i ∂y c
i ∂y
ψ ( x, y, z ) = e
ik y y + ik z z
φ ( x) ,
(12.13)
72
*
ς
m*ωc
x = , with β =
=
β
q* H
=
c
1
and ω c =
*
q* H
H
m*c
,
(12.14)
c ky
c
ς 0 = −β * = − * k y = −
q H
q H
H
ky .
(12.15)
We assume the periodic boundary condition along the y axis.
ψ ( x, y + Ly , z ) = ψ ( x, y, z ) ,
or
ik L
e y y = 1,
or
2π
n y (ny: integers).
Ly
Then we have
c
φ " (ς ) = [(ς − ς 0 ) 2 + *
( − 2m * E +
q H
ky =
(12.16)
(12.17)
(12.18)
2
k z )]φ (ς ) .
2
(12.19)
We put
2
2
1
kz
.
E = ωc ( n + ) +
2
2m*
where n = 0, 1, 2, 3, … (Landau level).
or
*
2m E =
*
2
1
k z + 2m ωc (n + ) =
2
*
2
*
(12.20)
2
kz +
2
2q
*
c
H
1
(n + ) ,
2
(12.21)
or
*
2
q H
1
(n + ) .
E=
k +
*
* z
2m
mc
2
Finally we get a differential equation for φ (ς )
2
φ " (ς ) + [2n + 1 − (ς − ς 0 ) 2 ]φ (ς ) = 0 .
(12.22)
(12.23)
The solution of this differential equation is
φn (ς ) = ( π 2 n!)
n
−1 / 2
e
−
( ς −ς 0 ) 2
2
H n (ς − ς 0 ) ,
(12.24)
with
c
ς 0 = − * ky = −
q H
H
=
x0 =
c
H
ky ,
(12.25)
*
q H
ς0
=
β
,
ς =−
H 0
(12.26)
2
H
ky
(12.27)
73
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.
ς0
=
β
0 < x0 =
or
2
H
2π
Ly
ky =
H
2
H
ς0 = −
2
H
k y < Lx ,
(12.28)
n y < Lx ,
(12.29)
or
ny <
Lx Ly
2π
2
,
(12.30)
*
*
2
q H
q H
1
1
2
(
)
(n + ) . (12.31)
=
+
+
=
n
k +
E=
k
*
*
* z
* 2
* z
2m ξ
2m
2
2m
2
mc
mc
2
2
2
*
2q H
1
4πH
1
(n + ) =
(n + ) .
− kz =
2
ξ
2
2
c
Φ0
1
2
(12.32)
or
1
ξ
where
2
− kz =
2
2
2
H
1
(n + ) ,
2
(12.33)
2π c
Φ0 = * ,
q
2
H
=
c
Φ
= 0 .
q H 2πH
(12.34)
*
Then we have
Φ0
1
2
− kz .
H=
2
4π (n + 1 / 2) ξ
(12.35)
((Ground state))
If kz=0 and n = 0, this has its highest value at H = H H c 2 =
Φ0
, (12.36)
2πξ 2
or
H c2 = κ 2 H c .
(12.37)
This is the highest field at which superconductivity can nucleate in the interior of a large
sample in a decreasing external field.
1
1
2πH
.
(12.38)
= 2 =
2
ξ
Φ0
H
12.2
Numerical calculation of wave finctions
74
((Mathematica Program-18))
(*Simple Harmonics wave function*)
(*plot of n[ ]*)
conjugateRule Complex re_, im_ Complex re, im ;
Unprotect SuperStar ; SuperStar : exp_ : exp . conjugateRule;
Protect SuperStar
{SuperStar}
n_, _, 0_ : 2
n2
1 4 n
1 2
Exp 0 2
HermiteH n, 2
0
pt1=Plot[Evaluate[Table[ [0, , 0],{ 0,-2,2,0.4}]],{ ,4,4},PlotLabel {0},PlotPoints 100,PlotRange All,PlotStyle T
able[Hue[0.1
i],{i,0,10}],Background GrayLevel[0.7],Prolog AbsoluteThick
ness[1.5]]
0
0.7
0.6
0.5
0.4
0.3
0.2
0.1
-4
-2
2
4
Graphics pt2=Plot[Evaluate[Table[ [1, , 0],{ 0,-2,2,0.4}]],{ ,4,4},PlotLabel {0},PlotPoints 100,PlotRange All,PlotStyle T
able[Hue[0.1
i],{i,0,10}],Background GrayLevel[0.7],Prolog AbsoluteThick
ness[1.5]]
0
0.6
0.4
0.2
-4
-2
2
4
-0.2
-0.4
-0.6
Graphics pt3=Plot[Evaluate[Table[ [2, , 0],{ 0,0,4,0.4}]],{ ,2,6},PlotLabel {0},PlotPoints 100,PlotRange All,PlotStyle T
able[Hue[0.1
i],{i,0,10}],Background GrayLevel[0.7],Prolog AbsoluteThick
ness[1.5]]
75
0
0.6
0.4
0.2
-2
2
4
6
-0.2
-0.4
Graphics Fig.32
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 ζ = ζ0, as a
function of ζ.
13.
13.1
Nucleation at surfaces: Hc3
Theory
1
φ "(ς ) = [(ς − ς 0 ) 2 + H 2 (− 2 + k z 2 )]φ (ς ) ,
ξ
(13.1)
and the boundary condition: Js|n = 0 at the boundary (ζ = 0).
dφ (ς ) / dς = 0 at ζ = 0.
(13.2)
For x0 = H ς 0 >> ξ , the harmonic oscillator function is nearly zero at ζ = 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 ζ = 0: the wave
function is either an even or an odd function). The corresponding wave function should
be an even function since dφ (ς ) / dς = 0 at ζ = 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 (ς − ς 0 ) 2 .
φ " (ς ) = [(ς − ς 0 )2 + g 0 )]φ (ς ) .
φ ' (ς ) = 0 at ζ = 0.
(13.3)
(13.3)
where
g0 =
2
H
(−
1
+ kz ) .
(13.4)
= −(2n + 1) ,
(13.5)
ξ
2
2
When kz = 0,
g0 = −
2
H
2
ξ
((Mathematica Program-19))
f1 x 1
2
1 x 2
76
f2 x_ : x 1 2 ; x 0
f2 x_ : x 1 2 ; x 0
Plot[{f1,f2[x]},{x,-2,2},
PlotStyle {Hue[0],Hue[0.7]},AxesLabel {" / 0","V( )"},Prolo
g AbsoluteThickness[2], Background GrayLevel[0.7]]
V
2
1.5
1
0.5
-2
Graphics
-1
1
0
2
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
ϕ " (ς ) + [2ν + 1 − (ς − ς 0 )2 ]ϕ (ς ) = 0 .
We put
z = 2 (ς − ς 0 ) ,
where ς 0 > 0
Then the differential equation can be rewritten as
2
1 z
ϕ "( z ) + (ν + − )ϕ ( z ) = 0 ,
2 4
where ν is generally not an integer.
The boundary condition:
(1) ϕ ( z = ∞) = 0 .
(13.6)
(13.7)
(13.8)
(2) ϕ ' ( z = − 2ς 0 ) = 0 .
(13.9)
The solution of this differential equation is
2
ν 1 z2
Γ (1 / 2)
1 −ν 3 z 2
z Γ(−1 / 2)
−
+
; ; )] ,
(
;
)
(
;
Dν ( z ) = 2ν / 2 e − z / 4[
F
F
1 1
1 1
Γ[(1 − ν ) / 2]
2 2 2
2 2 2
2 Γ (−ν / 2)
(13.10)
where 1 F1 is the confluent hypergeometric (or Kummer) function. For even ϕ (z ) the
boundary condition is given by
Dν ' ( z = − 2ς 0 ) = 0 .
(13.11)
Imagine the potential well about ζ = ζ0 to be extended by a mirror image outside the
surface. Lowest eigenfunction of a symmetric potential is itself symmetric; ϕ ' (ς ) = 0 at
the surface.
77
(i) ς 0 → ∞
Wave function ϕ (z ) will be localized around ζ = ζ0, and nearly zero at the surface.
Therefore the boundary condition will automatically be satisfied.
(ii) ς 0 → 0 .
The function still satisfies the boundary condition and Eq.(13.8).
13.2
Numerical solution by Mathematica
((Mathematica Program-20))
Nucleationof surface superconductivity
z_
2
2
Exp 1 z2 Gamma 1 2
Hypergeometric1F1
,
,
2 2 2
Gamma 12
2
2
1
Gamma 1 2
3 z2 ,
Hypergeometric1F1 ,
2
2 2 2 Gamma 2
z
z2 4
z2
4
2 z Hypergeometric1F1
Gamma " Hypergeometric1F1
Gamma "
2
,
1
2
,
3 , z2
2
2
!
2
#
1 , z2
2
2
1 2
$
%
!
'
'
'
'
'
''
&
#
eq1= '[z]//Simplify
<<Graphics`ImplicitPlot`
(
0 00 1 0,
p11 ) ImplicitPlot * Evaluate * eq1 + 0 , . z - . / 0 2
2 4 , . 0.3, 0.13 5 , PlotPoints - 50,
PlotStyle - 2 Hue 6 07 , Thickness 6 0.017 3 ,
Background - GrayLevel 6 0.77 , AspectRatio - 1,
AxesLabel - 2 "1 0", "4 "3 5
78
21
0, 0, 3 3 ,
0.1
0
-0.1
-0.2
-0.3
0
0
ContourGraphics
0.5
1
1.5
2
2.5
3
Fig.35 Plot of ν as a function of ζ0, satisfying the condition. Dν ' ( z = − 2ς 0 ) = 0 . ν = 0 at
ζ0 = 0, corresponding to the ground-state wave function for the simple harmonic
potential.
p111 ImplicitPlot Evaluate eq1 0 . z
2 0,
0, 0.75, 0.80 , , 0.20496, 0.2048 , PlotPoints
PlotStyle
Hue 0 , Thickness 0.01 ,
Background GrayLevel 0.7 , AspectRatio 1,
AxesLabel
" 0", " "
50,
-0.2048
-0.204825
-0.20485
-0.204875
-0.2049
-0.204925
-0.20495
0
0.75
0.76
0.77
0.78
0.79
0.8
ContourGraphics
Fig.36 Detail of Fig.38. The plot of ν as a function of ζ0 around ζ0 = 0.768 (ν = ν = 0.20494), satisfying the condition Dν ' ( z = − 2ς 0 ) = 0 .
79
p12 ImplicitPlot Evaluate eq1 0 . z 2
0 ,
, 0, 6 , PlotPoints 50,
0, 0, 5 ,
PlotStyle Hue 0 , Thickness 0.01 ,
Background GrayLevel 0.7 , AspectRatio 1,
2
0 1
AxesLabel " 0", " " ; p13 Plot
, 0, 0, 5 ;
2
Show p11, p12, p13
5
4
3
2
1
0
1
2
3
4
5
Graphics
Fig.37 Plot of ν as a function of ζ0, satisfying the condition. Dν ' ( z = − 2ς 0 ) = 0 . ν = 2
and ν = 4 at ζ0 = 0, corresponding to ν for the wave function (n = 2 and n = 4).
2
The solid line denotes the curve given by .ν = (ς 0 − 1) / 2 .
Plot of the wave function in the ground state
F 2
. 0.20494, 0 0.768 ! ;
Abs 0
Plot Evaluate F , , 0.1, 0.1! ,
PlotRange 0.03, 0.03! , 1.21702, 1.217032 ! ! ,
PlotStyle Hue 0 , Thickness 0.01 ! ,
Background GrayLevel 0.7 , AspectRatio 1,
AxesLabel " ", " "! 80
1.21703
1.21703
1.21703
1.21703
1.21702
1.21702
-0.03
-0.02
-0.01
0.01
0.02
0.03
Graphics
Fig.38 The plot of the ground state wave function with ς 0 = 0.768 and ν = -0.20494.
Plot[Evaluate[F],{ ,-5,5}, PlotRange {{-5,5},{0,
1.2171}},PlotStyle {Hue[0],Thickness[0.01]},
Background GrayLevel[0.7],AspectRatio 1,
AxesLabel {" "," "}]
1.2
1
0.8
0.6
0.4
0.2
-4
-2
2
4
Graphics
Fig.39 Overview of Fig.41 for the same wave function with ς 0 = 0.768 , ν = -0.20494
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 ς = ς 0 .
The exact solution shows that this eigenvalue is lower by a factor of 0.59.
81
Hc2
= 1.695H c 2 = 1.695( 2κH c ).
0.59
2
q* H
1
1
*
= ω (ν + ) =
(ν + ) .
E=
* 2
*
2m ξ
2
2
mc
When ν = n = 0,
2
q* H c 2
=
2m*ξ 2
2m*c
or
c
Φ0
H c2 = * 2 =
2πξ 2
q ξ
H c3 =
where Φ 0 =
(13.12)
(13.13)
(13.14)
(13.15)
2πc
q*
((Surface superconductivity))
When ν = -0.20494, ζ0 = 0.768,
2
q* H
q* H c 3
=
(−0.20494 + 0.5) =
0.29506
2m*ξ 2
m*c
m*c
or
c
= (0.29506 × 2) H c 3 = 0.5901H c 3
q* ξ 2
or
(13.16)
(13.17)
c
= H c 2 = 0.5901H c 3 ,
q* ξ 2
(13.18)
H c 3 = 1.695H c 2 .
(13.19)
or
13.3 Variational method by Kittel
We assume a trial function given by
ik y
ψ ( x, y ) = e y exp( − ax 2 ) ,
(13.20)
The parameter a and ky are determined variationally so as to minimize the Gibbs free
energy per unit area
2
q*
1
1
4
g s = fn + α ψ + + β ψ +
∇
−
A)ψ ,
(
c
2
2m* i
2
(13.21)
or
2
1
2π
g s ≈ fn +
[− 2 ψ + ( ∇ +
A )ψ ] ,
*
ξ
2m
i
Φ0
with A = (0, Hx,0) .
2
1
2
82
(13.22)
x0 = −
k yΦ 0
.
2πH
We now calculate
(13.23)
∞
2
1
2π
∆G =
[− 2 ψ + ( ∇ +
A)ψ ]dx .
*
2m 0 ξ
i
Φ0
2
1
2
(13.24)
After the calculation of ∆G, we put
1
a= 2.
2ξ
We take a derivative of ∆G with respect to x0. We have
(13.25)
x0 =
ξ
,
π
(13.26)
H c3 =
π
Φ0
= 1.6589 H c 2 .
2
2πξ π − 2
(13.27)
((Mathematica Program-21))
conjugateRule Complex re_, im_
Complex re, im ;
Unprotect SuperStar ; SuperStar : exp_ : exp . conjugateRule;
Protect SuperStar
1
x:
1
D #, x & ; y :
2
0
x
. ky
General ::spell1 : Possible spelling error : new
symbol name " y" is similar to existing symbol
!
Simplify;
K3 x; G2 Simplify G1, a 0
"
$
!
%
&
2
H x0
0
" x". More…
#
G1
x, y
Hx# & ;
2
D #, y
x_, y_ : Exp a x Exp ky y ; K1
K2 y x, y
Simplify;
1
K3
x, y
x, y
K1 K1 K2 K2
2
{SuperStar}
'
0
H2
2
(
) *
8
a x0
+ , , ,,
-
(
+ , , , , , ,,
.
2
1 4 a x02
-
3
4 a3 2
G3 G2 . a
4
5
6
1
2 2
5 5
4 H2
A
A
A
A
?
@
1
/ 0
1
2-
a
(
+ , , , , , ,,
2
E
. *
2
1 a
-
2/
2
02
I
GI
I D
I
H
J
02
1
2 2 02
Simplify
7
9
: : ::
;
A
A
A
A
?
@
1
2
<
= = = = = ==
>
8
B
3 2 x0 <
;
4
1
2
= = = = = ==
>
4D
;
C
2 x02
2D
C
8
J
4 02
8
G4=Simplify[G3, >0]
K
O
4 H2
P
P
L
L
M
4
O
Q
3 2 x0 R 3
L
4
N N NN
R
S O
U
2
P
2 x02
R
2
S
R
4T
T
S
02
G5=D[G4,x0]//Simplify
4 H2
V
2
WX YV
Y YY
]
x0
G6 G4 . x0
^
_
Z
[ \
b
a
cc cc
d
[
02
`
_ _
Simplify
83
U
02
4F
C
T
I
GI
I
I
H
Simplify;
2 4 02
4 H2
4
Solve[G6 0,H]
H
2
2
0
2
2
02 2
, H
2
0
2
2
2
N
1.6589
14.
Abrikosov vortex states14
14.1 Theory
(1) All ψ k are orthogonal because of the different e iky factor.
(2) Each of these solutions is equally valid exactly at H = Hc2 and all give the same Hc2.
4
(3) β ψ term is not longer negligible. We expect a crystalline array of vortices to have
lower energy than a random one.
2
(ς − ς k y )
( x − xk y )2
ik y y
ik y y
] = Ck y e exp[ −
],
ψ L ( x, y ) = Ck y e exp[ −
2
2
2lH
ky
ky
(14.1)
with
ς
c
Φ
2
x = = lH ς ,
lH = * = 0 = ξ 2 ,
(14.2)
β
q H 2πH
xk y = lH ς k y
ky =
where
k=
ς k = lH k y ,
y
xk y = lH k y = lH kn = lH
2
2π
n = kn ,
b
2
2πn
,
b
(14.3)
(14.4)
2π
,
b
ψ L ( x, y ) =
2
(14.5)
ky
( x − lH kn) 2
],
2
2lH
2
Ck y eikny exp[−
( x − lH nk )2
ψ L ( x, y ) = Cn exp[−
+ inky ] ,
2
2lH
n
Here we assume that Cn is independent of n.
We define the period in the x direction.
2π 2
2
a = lH k =
lH ,
b
( x − na) 2
ψ L ( x, y ) = C exp[−
+ inky ] ,
2
2lH
n
(14.6)
2
84
(14.7)
(14.9)
(14.10)
( x + a − na )2
(14.11)
+ inky ] ,
2
2lH
n
1
ψ L ( x + a, y ) = Ce iky exp[ − 2 ( x − (n − 1)a )2 + i (n − 1)ky ] = eikyψ L ( x, y ) ,
2lH
n
(14.12)
2
( x − na )
2πiny
(14.13)
ψ L ( x, y ) = C exp[ −
+
].
2
b
2lH
n
ψ L ( x + a, y ) = C
exp[ −
((Mathematica Program 22)) Square vortex lattice
rule1={a 1,b 1,LH 0.40}
{a 1,b 1,LH 0.4} x na 2
2 ny
f m_ : Sum Exp , n, m, m . rule1
2 LH2
b
ContourPlot Evaluate Abs f 200 2 , x, 1.5, 1.5 ,
y, 1.5, 1.5 , AspectRatio 1, PlotPoints 100
1.5
1
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
0.5
1
1.5
ContourGraphics Fig.40
The spatial configuration of ψ
= 1. b = 1. lH = 0.4
2
near H = Hc2 for a square vortex lattice. a
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
( x − xn ) 2
(14.14)
ψ L ( x, y ) = Cn exp[ −
+ inky ]
2
2lH
n
Cn +ν = Cn (ν = integer),
(14.15)
ψ L ( x + xν , y ) =
n
Cn exp[ −
( x + xν − xn ) 2
+ inky ] ,
2
2lH
85
(14.16)
or
ψ L ( x + xν , y ) = e
iνky
n −ν
Cn −ν e
i ( n −ν ) ky
( x − xn −ν ) 2
exp[ −
] = eiνkyψ L ( x, y ) ,
2
2lH
(14.17)
Fig.41 Rectangular vortex lattice
Each unit cell of the periodic array carriers ν quanta of flux.
kΦ 0 2π
Hxν b = ν
= νΦ 0
.
2π k
2
Φ0
H =
2πH
2
Φ0
H
=
2
λ
2πλ2 H
1
1
= 2
2
ξ
(4.18)
(4.19)
(4.20)
(4.21)
H
Triangular lattice:
Cn + 2 = Cn , C1 = iC0 = iC
ψ ∆ ( x, y ) = C[
n = even
exp[−
(4.22)
1
2
( x − l H nk ) 2 + inky ]
2
2l H
1
2
exp[ −
+i
( x − l H nk ) 2 + inky ]
2
2l H
n = odd
where k = 2π / b .
We define the period in the x direction
4π 2
2
a = 2l H k =
lH
b
We note that b = 3a for a triangular lattice.
86
(4.23)
(14.25)
Fig.42 Triangular vortex lattice, where red circles denote the centers of vortices.
ψ ∆ ( x, y ) = C[
exp[ −
n = even
1
2
( x − l H nk ) 2 + inky ] +
2
2l H
1
2
i
( x − l H (n + 1)k ) 2 + i (n + 1)ky ]
exp[ −
2
2l H
n = even
or
ψ ∆ ( x, y) = C {exp[ −
n = even
+ i exp[ −
1
2l H
n = even
1
2l H
ψ ∆ ( x, y) = C {exp[ −
n
+ i exp[ −
2
( x − l H ( n + 1) k ) 2 + i( n + 1) ky ]}
1
2l H
(x −
1
2l H
1
2l H
2l H
( x − l H nk ) 2 + inky ]
2
2
2
2
ψ ∆ ( x, y) = C {exp[ −
+ i exp[ −
1
2
2
(x −
2
(x −
na 2
) + inky ]
2
( n + 1) a 2
) + i( n + 1) ky ]}
2
(x −
2na 2
) + i 2nky ]
2
(2n + 1)a 2
) + i (2n + 1)ky ]}
2
87
(14.26)
ψ ∆ ( x, y) = C
exp( i 2nky ){exp[ −
n
+ i exp(iky ) exp[ −
or
ψ ∆ ( x, y) = C exp(
n
1
2l H
2
1
2l H
2
( x − na) 2 ]
(14.27)
( x − (n + 1 / 2)a ) 2 ]}
4πiny
1
){exp[ −
( x − na ) 2 ]
2
b
2l H
(14.28)
2πiy
1
+ i exp(
) exp[ −
( x − ( n + 1 / 2) a) 2 ]}
2
b
2l H
((Mathematica Program 23)) Triangular vortex lattice
rule1 a 1, b 3
, LH 0.31
, LH 0.31 a 1, b 3
f m_ :
4 ny
Sum Exp b
2
x n 1 a 2
2 y
Exp x n a
2 Exp Exp 2 LH2 b 2 LH2
,
n, m, m ! . rule1
"
ContourPlot # Evaluate # Abs $ f $ 200% % 2& , ' x, ( 1.5, 1.5) ,
' y, ( 1.5, 1.5) , AspectRatio * 1, PlotPoints * 100&
1.5
1
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
0.5
1
1.5
+ ContourGraphics +
Fig.43 The spatial configuration of ψ
2
near H = Hc2 for a triangular vortex . a = 1,
b = 3 , 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.
88
2
dr[α ψ L + β ψ L
2
4
1
q*
(
A)ψ L ] = 0 ,
+
∇
−
2m* i
c
(14.29)
where ψ L is a linear combination of the solutions of the linearized GL equation
A = A 0 + A1 .
A0 is the vector potential which would exist in the presence of the field Hc2.
Ginzburg-Landau equation for ψ L is given by
2
1
q*
αψ L + * ∇ − A 0 ψ L = 0 ,
2m i
c
(14.30)
and
q ψL
q*
*
*
JL =
A0 ,
[ψ L ∇ψ L − ψ L∇ψ L ] −
*
2m i
m*c
*2
2
(14.31)
or
q*
iq*
iq*
*
*
*
JL =
[
ψ
(
∇
ψ
−
A
ψ
)
−
ψ
(
∇
ψ
+
A 0ψ L )] ,
0 L
L
L
L
L
*
2m i
c
c
(14.32)
or
q*
q*
q*
*
*
[
ψ
(
∇
−
A
)
ψ
+
ψ
(
−
∇
−
A 0 )ψ L ] .
L
L
L
0
*
2m
i
c
i
c
JL is the current associated with the unperturbed solution.
1
Denoting quantities of the form
... by a bar (V is a volume),
V
JL =
(14.33)
2
1
q*
αψ L + βψ L +
(
∇
−
A)ψ L = 0 .
c
2m * i
Expanding Eq.(14.34) to the first order in A1, we obtain
2
4
(14.34)
2
αψ L + βψ L
2
4
q*
1
1
+
( ∇ − A 0 )ψ L − A1 ⋅ J L = 0 .
*
c
c
2m i
(14.35)
((Mathematica Program 24))
Proof of
1
q*
(
∇
−
A )ψ L
c
2m * i
2
2
1
q*
1
=
(
∇
−
A 0 )ψ L − A 1 ⋅ J L .
*
c
c
2m i
(14.35’)
Proof of -<A1 JL>=average of kinetic energy
<<Calculus`VariationalMethods`
Needs["Calculus`VectorAnalysis`"]
SetCoordinates[Cartesian[x,y,z]]
Cartesian[x,y,z]
A0={A10[x,y,z],A20[x,y,z],A30[x,y,z]};A={A1[x,y,z],A2[x,y,z],
89
A3[x,y,z]};A1={A11[x,y,z],A21[x,y,z],A31[x,y,z]};rule1={A1
(A10[#1,#2,#3]&)};rule2={A2 (A20[#1,#2,#3]&)};
rule3 A3 A30 #1, #2, #3
op1x : D #, x
op1y :
D #, y
op1z :
D #, z
eq1 1
2m
A30 #1, #2, #3
& ;
q
A10 x, y, z # & ;
c
q
A20 x, y, z # & ;
c
q
A30 x, y, z # & ;
c
q
A x, y, z .
c
q
Expand;
A c x, y, z
Grad c x, y, z c
eq2 VariationalD eq1, A1 x, y, z , x, y, z Expand
1
c2 m & ; rule3 A3 Grad x, y, z q2 x, y, z c x, y, z A11 x, y, z , A21 x, y, z , A31 x, y, z x, y, z q c x, y, z 1,0,0 x, y, z q x, y, z c 1,0,0 x, y, z 2cm
2cm
eq21=A11[x,y,z] eq2/.rule1//Simplify
1
q A11 x, y, z 2 q A10 x, y, z x, y, z c x, y, z !
2 c2 m %
%
"
c#
c x, y, z $1,0,0 x, y, z & x, y, z c $1,0,0 x, y, z '
' '
eq3=VariationalD[eq1,A2[x,y,z],{x,y,z}]//Expand
q2 A2 ( x, y, z )
,
* ( x,
y, z ) * c ( x, y, z )
c2 m
/
q - * c ( x, y, z ) * .0,1,0 ( x, y, z )
0
2cm
+
,
q-
* ( x,
/
y, z ) * c .0,1,0 ( x, y, z )
2cm
eq31=A21[x,y,z] eq3/.rule2//Simplify
1
q A21 x, y, z 2 q A20 x, y, z x, y, z c x, y, z !
2 c2 m %
%
"
c#
c x, y, z $0,1,0 x, y, z & x, y, z c $0,1,0 x, y, z '
' '
eq4=VariationalD[eq1,A3[x,y,z],{x,y,z}]//Expand
q2 A3 ( x, y, z )
,
* ( x,
y, z ) * c ( x, y, z )
c2 m
/
q - * c ( x, y, z ) * .0,0,1 ( x, y, z )
2cm
0
+
,
q-
* ( x,
/
y, z ) * c .0,0,1 ( x, y, z )
2cm
eq41=A31[x,y,z] eq4/.rule3//Simplify
1
q A31 x, y, z 2 q A30 x, y, z x, y, z c x, y, z !
2 c2 m %
%
"
c#
c x, y, z $0,0,1 x, y, z & x, y, z c $0,0,1 x, y, z '
seq1=eq21+eq31+eq41//Simplify
90
' '
1 q 2 q A10 x, y, z A11 x, y, z x, y, z c x, y, z 2 c2 m
2 q A20 x, y, z A21 x, y, z x, y, z c x, y, z 2 q A30 x, y, z A31 x, y, z x, y, z c x, y, z c A31 x, y, z c x, y, z 0,0,1 x, y, z c A31 x, y, z x, y, z c 0,0,1 x, y, z c A21 x, y, z c x, y, z 0,1,0 x, y, z c A21 x, y, z x, y, z c 0,1,0 x, y, z c A11 x, y, z c x, y, z 1,0,0 x, y, z c A11 x, y, z x, y, z c 1,0,0 x, y, z (*Current density*)
Jeq1 q2 x, y, z c x, y, z A0 mc
q c x, y, z Grad x, y, z 2m
Simplify
1
q 2 q A10 x, y, z ! x, y, z ! c x, y, z "
2cm &
&
1,0,0
1,0,0
x, y, z
x, y, z
! c x, y, z
! %
" ! x, y, z
! c%
c#
$
1
q 2 q A20 x, y, z ! x, y, z ! c x, y, z "
2cm &
&
0,1,0
0,1,0
c#
! c x, y, z
! %
x, y, z
" ! x, y, z
! c%
x, y, z
$
1
q 2 q A30 x, y, z ! x, y, z ! c x, y, z "
2cm &
c#
seq2 )
,
x, y, z Grad c x, y, z *
!
c x, y, z
0,0,1
! %
' ' '
,
' ' '
,
&
x, y, z
"
! x, y, z
!
c %0,0,1 x, y, z
' ' ' (
1 $
Jeq1.A1 + + Simplify
c
1 q 2 q A10 x, y, z A11 x, y, z x, y, z c x, y, z 2 c2 m
2 q A20 x, y, z A21 x, y, z x, y, z c x, y, z 2 q A30 x, y, z A31 x, y, z x, y, z c x, y, z ,
c A31 x, y, z c x, y, z 0,0,1 x, y, z c A31 x, y, z x, y, z c 0,0,1 x, y, z ,
c A21 x, y, z c x, y, z 0,1,0 x, y, z c A21 x, y, z x, y, z c 0,1,0 x, y, z ,
c A11 x, y, z c x, y, z 1,0,0 x, y, z c A11 x, y, z x, y, z c 1,0,0 x, y, z seq1-seq2
0
((Note))
*
If one multiplies the linearized GL equation by ψ L and integrates over all the volume by
parts, one obtain the identity
2
3
[α ψ L
2
/
2
1
q*
*
+
ψL
∇ − A 0 ψ L ]dr = 0
2m*
i
c
0
0
1
-
4
.
91
(14.36)
or
[α ψ L
2
1
+
2m*
2
q*
∇ − A 0 ψ L ]dr = 0
i
c
(14.37)
or
2
1
q*
(14.38)
α ψ L + * ( ∇ − A 0 )ψ L = 0 .
2m i
c
Then we obtain
1
4
β ψ L − A1 ⋅ J L = 0 ,
(14.39)
c
1
q*
iq*
iq*
*
*
− A1 ⋅ J L = − * [ψ L (∇ψ L −
A 0 ) − ψ L (∇ψ L +
A 0 )] . (14.40)
c
2m ci
c
c
Note that
4π
∇ × B( s) =
B (1) = ∇ × A1 ,
(14.41)
JL ,
c
1
1
1
− A1 ⋅ J L = −
A1 ⋅ ∇ × B ( s ) = − [B ( s ) ⋅ (∇ × A1 ) − ∇ ⋅ ( A1 × B ( s ) )] ,
c
4π
4π
(14.42)
or
1
1 ( s ) (1) 1
− A1 ⋅ J L = −
B ⋅B +
∇ ⋅ ( A1 × B ( s ) )] .
(14.43)
c
4π
4π
We use the formula of vector analysis
∇ ⋅ ( A1 × B ( s ) ) = B ( s ) ⋅ (∇ × A1 ) ⋅ − A1 ⋅ ∇ × B ( s )
(14.44)
We note that
∇ ⋅ ( A1 × B ( s ) ) ⋅ dr = ( A1 × B ( s ) ) ⋅ da (Stokes theorem). This value becomes zero on the
2
boundary surface of the system.
Then we have
1 ( s ) (1)
4
βψ L −
B ⋅B = 0.
4π
What are the expressions for B(1) and B( s ) ?
B (1) = H − H c 2 + B ( s ) ,
(14.45)
(14.46)
where B( s ) gives the effect of the supercurrent.
We introduce the canonical operator,
*
ˆ = p − q A0 ,
(14.47)
c
which has the commutation relation
*
ˆ x ,Π
ˆ y ] = iq H .
[Π
(14.48)
c
We introduce the raising (creation) and lowering (destruction) operators.
ˆ+ =Π
ˆ x + iΠ
ˆ y,
ˆ− =Π
ˆ x − iΠ
ˆ y,
Π
Π
(14.49)
92
q*
Bz .
c
The linearized GL equation can be described by
ˆ Π
ˆ ψ = 0.
Π
+
− L
ˆ −,Π
ˆ + ] = 2i[Π
ˆ x,Π
ˆ y ] = −2
[Π
(14.50)
(14.51)
((Note))
2
1 q*
αψ L + * ∇ − A 0 ψ L = 0 ,
2m i
c
αψ L +
(14.52)
1 ˆ 2 ˆ 2
( Π x + Π y )ψ L = 0 .
2m*
(14.53)
Since
*
ˆ 2x + Π
ˆ 2y = Π
ˆ −Π
ˆ + + q Bz ,
Π
c
*
ˆ 2x + Π
ˆ 2y = Π
ˆ +Π
ˆ − − q Bz ,
Π
c
we have
1 ˆ ˆ
1
q*
Π + Π −ψ L +
Bz + 2m*α )ψ L = 0 ,
(−
*
*
2m
2m
c
with
q*
2m*c α
q*
α
−
+
=
(
−
).
B
B
z
z
2m*c
2m*c
q*
We also note that
2m*c α
c
Φ0
= * 2 =
= H c2 ,
*
q
2πξ 2
q ξ
(14.54)
(14.55)
(14.56)
(14.57)
(14.58)
q*
1 ˆ ˆ
( Bz − H c 2 )ψ L = 0 .
Π + Π −ψ L +
2m*
2m*c
When Bz − H c 2 = 0,
ˆ Π
ˆ ψ =0.
Π
+
− L
The ground state has the property
Π −ψ L = 0 .
or
∂ψ L q *
∂ψ L q *
x
y
− A0 ψ L ) = 0 ,
− A 0 ψ L ) − i(
Π −ψ L = (
i ∂x
c
i ∂y
c
or
q* x
q* y
∂ψ
∂ψ
L
L
−i
− A0 ψ L −
+ i A0 ψ L = 0 .
c
c
∂x
∂y
iθ
The substitution of ψ L = ψ L e into the above equation leads to
93
(14.59)
(14.60)
(14.61)
(14.62)
(14.63)
∂ψ L
∂ψ L
∂θ
∂θ
iq* y
q* x
= 0.
−i
+ ψL
−
A0 ψ L − i ψ L
A0 ψ L +
∂x
∂x
∂y
∂y
c
c
(14.64)
The real part of Eq.(14.64) is
∂ψ L
q*
∂θ
− A0x ψ L − + ψL
= 0,
(14.65)
c
∂y
∂x
or
∂ψ L
∂θ q *
.
(14.66)
ψ L ( − A0x ) =
∂y
∂x c
The imaginary part of Eq.(14.64) is
∂ψ L
q* y
∂θ
A0 ψ L − − ψL
(14.67)
= 0,
c
∂x
∂y
or
∂ψ
∂θ q*
(14.68)
ψ L ( − A0y ) = − L .
∂x
∂y c
Then the current density is given by
−
J Lx
∂ψ L
q*
q * ∂ψ L
2 ∂θ
q*
q* x
=
= * ψ L ( − A0 ) = * ψ L
,
∂y
∂x c
m
2 m* ∂y
m
(14.69)
J Ly
∂ψ L
q*
q* ∂ψ L
q*
q* y
2 ∂θ
.
=− *
− A0 ) = − * ψ L
= * ψL (
m
m
∂x
∂y c
2 m ∂x
(14.70)
2
2
We see that the current lines are the lines of constant ( ψ L =const). We note that ∇ψ L
2
is perpendicular to the lines of constant ( ψ L =const).
2
∇ψ L = e x
2
∂
2
∂
2
2m*
ψ L + e y ψ L = − ( J Ly e x − LLxe y )
∂x
∂y
q
(14.71)
and
J L ⋅ ∇ψ L = −
2
2 m*
2 m*
( J Lx e x + LLy e y ) ⋅ ( J Ly e x − LLxe y ) = −
( J Lx J Ly − LLx LLy ) = 0 .
q
q
(14.72)
Since
ex
∂
4π
∇ × Bs =
JL =
∂x
c
0
we have
ey
∂
∂y
0
ez ∂ ∂Bs
∂Bs
=
, −
, 0
,
∂z
∂y
∂x
Bs
94
(14.73)
2
q * ∂ψ L
2 m * ∂y
−
2
4π ∂Bs
=
∂y
c
q ∂ψ L
2 m * ∂x
*
2
(14.74)
∂B
4π
=− s
∂x
c
or
2πq*
2
Bs = * ψ L
mc
(14.75)
or
Bs = −
where
fL =
ψL = −
2
*
mc
Hc2
2
fL
2
2κ
(14.76)
ψL
ψ∞
2m*c α
2π q *
(14.77)
*
= Hc2 .
q*
2π q α
Hc2
=−
,
2
2κ
m * cβ
and
(14.78)
B1 = H − H c 2 + Bs ,
1
4
βψ L −
Bs ⋅ [ H − H c 2 + Bs ] = 0 ,
4π
1
4
2
βψ ∞ 4 f L −
( H − H c 2 ) Bs + Bs = 0 .
4π
(14.79)
(14.80)
(14.81)
Since
Bs = −
Hc2
2
fL ,
2
2κ
(14.82)
we have
βψ ∞ f L
4
or
β
Hc2
2
1
H
H
4
2
[ −( H − H c 2 ) c 22 f L + c 24 f L ] = 0 ,
−
4π
2κ
4κ
4
ψ ∞4 fL +
4
2
or
(8πκ 2
β
Hc2
2
(14.83)
1 H
1
2
1 1
4
− 1) 2 f L −
fL ] = 0 ,
(
4
4π H c 2
2κ
4π 4κ
ψ ∞4 −
1
2κ
) f L − (1 −
4
2
H
2
) fL = 0 .
Hc2
(14.84)
(14.85)
Noting that
α
α
2
= 8πκ β 2 = 4π
2κ 2 = H c 2κ 2 , and
β
β
2
8πκ βψ ∞
2
4
2
2
we get the final form
1
4
2
H
(1 − 2 ) f L − (1 −
) fL = 0 .
2κ
Hc2
Hc
1
,
=
Hc2
2κ
(14.86)
(14.87)
95
2
4
The quantities f L and f L may be calculated, depending on the vortex lattice
symmetry and the lattice spacing. It turns out that near Hc2, the ratio
βA =
4
fL
fL
2
(14.88)
2
is independent of H. Then the above equation can be rewritten by
H
1−
H c2
2
fL =
1
β A (1 − 2 )
2κ
and
fL = β A fL
4
2
2
2
H
1−
Hc2
=
1
β A (1 − 2 ) 2
2κ
(14.89)
(14.90)
((Note)) Here we follow the discussion given by Tinkham. We assume that
ψ = c1 f L
(14.91)
where c1 ia an adjustable parameter. Then the free energy is expressed by
( f s − f N ) = α ψ 2 + 1 β ψ 4 = αc12 f L 2 + 1 βc14 f L 4
(14.92)
2
2
Minimizing this with respect to c12, we find that
2
α fL
c =−
β f 4
2
1
(14.93)
L
Then the minimum free energy is given by
2
( fs − f N ) = − α
2β
fL
fL
2
2
4
2
α2 1
Hc 1
=−
=−
2β β A
8ι β A
(14.94)
If fL is constant, βA = 1. If fL is not constant, βA is larger than 1. The more βA increases,
the less favorable is the energy. It is found that βA = 1.18 for the square lattice and βA =
1.16 for the triangular lattice. The latter case is slightly more stable.
14.3 Magnetic properties near H = Hc21
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
H
H
2
2
B = H + Bs = H − c 22 f L = H − c f L ,
2κ
2κ
or
96
(14.,95)
H c2 − H
H c2 − H
1
.
=H−
2
2κ β (1 − 1 )
β A (2κ 2 − 1)
A
2κ 2
Note that B = H at H = Hc2,
(b)
B=H−
(14.96)
The magnetization M . It is described by
Hc2 − H
.
− 4π M = H − B =
β A (2κ 2 − 1)
(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 (∆B) = B 2 − B
2
2
2
B 2 , B and B 2 − B are expressed as follows.
2
H
HH c 2
Hc2
2
2
4
B 2 = ( H − c 22 f L )2 = H 2 −
f
+
fL ,
L
2
4
2κ
κ
4κ
2
2
H
HH c 2
Hc2
2
2
2
B = ( H − c 22 f L )2 = H 2 −
f
+
fL
L
2
4
2κ
κ
4κ
Then the deviation ∆B is calculated as
2
H
2
( ∆B ) = B 2 − B = c 22 β A − 1) f L ,
2κ
or
H c2 − H
.
(∆B) = β A − 1
β A (2κ 2 − 1)
(14.98)
2
.
(14.99)
(d)
(14.100)
(14.101)
The Helmholtz free energy F is given by
2
q*
1 2
1
1
4
F = F0 + dr[α ψ + β ψ +
∇
−
(
A)ψ +
B ],
*
c
8π
2
2m i
2
(14.102)
or
~ F − F0
1
1 2
4
F=
= − βψ L +
B
V
2
8π
~
1
1 2
1
1 2
4
4
4
2
F = − βψ ∞ ψ L +
B =−
Hc fL +
B ,
2
8π
8π
8π
Here we also note that
1
H
H
2
2
) − 2 fL ] .
B − H c 2 = H − H c 2 − c 22 f L = H c 2 [−(1 −
2κ
H c 2 2κ
Since
H
1
2
β A (1 − 2 ) f L = 1 −
Hc2
2κ
we have
97
(14.103)
(14.104)
(14.105)
(14.106)
B − Hc2 = −
H c2
2
[1 + β A ( 2κ 2 − 1)] f L
2
2κ
or
(
(14.107)
)
2
4κ 4 B − H c 2
fL
.
=
2
H c 2 [1 + β A (2κ 2 − 1)]2
Using this relation, we obtain the expression
2
2
2
2
1 Hc2
1 2 Hc2
2
~
2
F =−
β
f
+
[
B
+
(
β
−
1
)
f
]
L
L
8π 2κ 2
8π
4κ 4
2
2
1 2 Hc2
2
=
[B −
f
( − β + 1 + 2κ 2 β )] ,
L
4
8π
4κ
or
2
2
(
(14.108)
)
(14.109
2
B − H c2
~ 1 2
[B −
].
F=
8π
1 + β A ( 2κ 2 − 1)
Using this relation, we have
β A ( 2κ 2 − 1) B + H c 2
∂ ~
B − Hc2
4π
=
F = B−
=H
1 + β A ( 2κ 2 − 1)
1 + β A ( 2κ 2 − 1)
∂B
since
β A (2κ 2 − 1) B + H c 2 = [1 + β A (2κ 2 − 1)]H
So we can verify the thermodynamic relation as expected.
(
)
(14.110)
(14.111)
(14.112)
~
(e) The Gibbs free energy G
~
~
The Gibbs free energy G is related to the Helmholtz free energy F by
(
)
2
1 2
~ ~ H
B − H c2
H
[B −
]−
G=F−
B=
B.
2
4π
8π
1 + β A (2κ − 1) 4π
Using the relation
Hc2 − H
B=H−
,
β A (2κ 2 − 1)
we find that
~
1 (H − H c 2 )2
+ H 2].
G=− [
8π β A (2κ 2 − 1)
(14.113)
(14.114)
(14.115)
~
It is clear that the lower the value of βA, the lower the value of G . Thus the triangular
lattice is more stable than the square lattice.
14.4 The wall energy at κ = 1 / 2 .1
We show that the wall energy in a type I superconductor vanishes for κ = 1 / 2 . We
consider the 2D GL equation, where the magnetic field is directed along the z axis.
1 ˆ2 ˆ2
2
Π x + Π y ψ + αψ + β ψ ψ = 0 ,
(14.116)
*
2m
where
(
)
98
q*
q*
∂ q*
∂ q*
ˆ
ˆ
ˆ
ˆ
,
Πx =
− Ax = px − Ax Π y =
− Ay = p y − Ay ,
i ∂x c
i ∂y c
c
c
(14.117)
ˆ+ =Π
ˆ x + iΠ
ˆ y,
ˆ− =Π
ˆ x − iΠ
ˆ y.
Π
Π
(14.118)
We note that
ˆ −,Π
ˆ + ] == 2i[Π
ˆ x,Π
ˆ y ],
[Π
or
q*
q*
q*
q*
ˆ
ˆ
ˆ
ˆ
ˆ
[Π x , Π y ] = [ p x − Ax , p y − Ay ] = − [ px , Ay ] + [ pˆ y , Ay ] .
c
c
c
c
For any arbitrary function f, we have,
∂
∂
∂
[ pˆ x , Ay ] f =
, Ay f =
Ay f − Ay
f
i ∂x
i ∂x
i ∂x
∂Ay
∂Ay
∂f
∂f
f − Ay
f.
=
= Ay
+
i
i ∂x i ∂x
∂x i ∂x
Similarly, we have
∂Ax
[ pˆ y , Ax ] f =
f.
i ∂y
Then we get the relation
*
*
ˆ x,Π
ˆ y ] = − q ( ∂Ay − ∂Ax ) = i q Bz ,
[Π
c i ∂x i ∂y
c
∂A ∂A
where Bz = y − x .
∂x
∂y
Using this commutation relation, we get
*
ˆ −,Π
ˆ + ] = 2i[Π
ˆ x,Π
ˆ y ] = −2 q Bz .
[Π
c
Since
ˆ −Π
ˆ + = (Π
ˆ x − iΠ
ˆ y )(Π
ˆ x + iΠ
ˆ y ) == Π
ˆ 2x + Π
ˆ 2y + i[Π
ˆ x,Π
ˆ y ],
Π
we get
*
ˆ 2x + Π
ˆ 2y = Π
ˆ −Π
ˆ + + q Bz .
Π
c
Then the GL equation can be rewritten as
1 ˆ ˆ
q*
2
ψ
Bzψ + αψ + β ψ ψ = 0 .
Π
Π
+
−
+
*
*
2m
2m c
ˆ ψ =0.
We now look for the solution such that Π
+
(14.119)
(14.120)
(14.121)
(14.122)
(14.123)
&
#
'
&
(14.124)
(14.125)
(14.126)
(14.127)
#
'
∂ q*
∂ q*
ˆ ψ =
Π
+
−
ψ
− Ay ψ = 0 ,
A
i
+
x
i ∂x c
i ∂y c
(14.128)
∂ψ q*
∂ψ
= ( Ax + iAy )ψ .
−i
∂x c
∂y
(14.129)
$
$
!
!
%
"
$
$
!
!
%
"
or
(
99
ˆ ψ = 0 , the GL eq. reduces to
When Π
+
q*
2
Bz + α + β ψ = 0 .
(14.130)
*
2m c
Note that the current density is given by
q*
q*
q*
*
Js =
[ψ ( ∇ − A )ψ + ψ (− ∇ − A )ψ * ]
(14.131)
*
2m
i
c
i
c
Then we have
∂Bz
q*
q*
2π q* * 1
1
∂Bz
∇×B = , −
, 0 = * [ψ ( ∇ − A)ψ + ψ (− ∇ − A)ψ * ]) ,
∂x
mc
i
c
i
c
∂y
(14.132)
or
1 ∂ q*
∂Bz 2π q* * 1 ∂ q*
− Ax )ψ * ] ,
− Ax )ψ + ψ (−
= * [ψ (
(14.133)
∂y
mc
i ∂x c
i ∂x c
and
∂Bz 2π q*
q*
1 ∂ q*
* 1 ∂
−i
(14.134)
= * i[ψ (
− Ay )ψ + ψ (−
− Ay )ψ * .
∂x
mc
i ∂y c
i ∂y c
Then we have ∂Bz
∂B
∂ψ
∂ψ
∂ψ * ∂ψ *
2π q*
2q *
− i z = * [ψ * (
−i
−
) + ψ (i
)−
( Ax + iAy )ψ *ψ ] ,
∂y
∂x
∂y
∂x
∂x
∂y
mc
c
(14.135)
or
∂ψ *
∂ψ *
∂ψ
∂Bz 2π q*
∂Bz
* ∂ψ
i
i
[
(
)
(
)] ,
(14.136)
−
ψ
ψ
−
−
=
−
−i
m*c
∂x
∂y
∂x
∂y
∂x
∂y
This relation must agree with the relation given by
Using the relation (14.130), we have
∂Bz
∂B
∂ 2m*c
∂ 2m*c
− i z = − [ * (α + βψ *ψ )] + i [ * (α + βψ *ψ )] ,
(14.137)
∂y
∂x
∂y q
∂x q
or
∂Bz ∂
q* ∂Bz
∂
∂
* ∂
(14.138)
ψ − i ψ ) + βψ ( ψ * − i ψ * )] = 0 ,
−i
+ βψ (
* 2m c ∂y
∂x
∂x
∂y
∂x
∂y
From the comparison between Eqs.(14.137) and (14.138), we find that
β =π
2
q*
2
.
c 2 m *2
Since t κ is defined by
λ cm* β
κ= =
,
ξ
2π q*
(14.139)
(14.140)
we obtain
q*
cm*
1
=
κ=
π
.
*
*
cm
2
2π q
(14.141)
100
((Note))
2
2
2
2m*c
2m*c
2m*c
2
(
α
+
β
ψ
)
=
−
(
−
α
+
β
ψ
)
=
(
− βψ ) ,
Bz = −
*
* 2
*
*
q
q
q 2m ξ
(14.142)
or
2
4π q * 2 2
2m*c
2
κ ψ ,
Bz = * ( * 2 − β ψ ) = 2κH c −
m*c
q 2m ξ
(14.143)
where
2
Φ0 c
Φ0
2m*c
1 c
= * 2 =
=
= H c 2 = 2κH c ,
* 2
2
*
2πc ξ
2πξ 2
q 2m ξ
q ξ
(14.145)
and
2
2 *
4π q* 2 2
2m*c
2m*c 2π q
2
2
2
βψ = *
κ ψ =
κ ψ .
2
m*c
q*
q
c 2 m*
(14.146)
One can estimate the surface energy given by
∞
1
1
4
γ = dx[ − β ψ + ( Bz − H c ) 2 ] .
2
8π
−∞
In our case,
4π q* 2 2
Bz = 2κH c − * κ ψ ,
mc
with
β = 2κ
π
2
q*
(14.147)
(14.148)
2
.
2
m* c 2
Then the integrand of γ is described by
( Bz − H c )
2
8π
2
β
2
H
− ψ = c
2
8π
4
( 2κ − 1) −
(14.149)
4π q *
*
m cH c
2
κ 2ψ
2
−κ
π
2
2
q*
*2 2
m c
2
ψ . (14.150)
4
This is equal to 0 when κ = 1 / 2 , leading to γ = 0.
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 κ has a limiting value κ = 1 / 2
separating superconductors with positive surface energy ( κ < 1 / 2 ) (type-I
superconductors) from those with negative surface energy at large κ > 1 / 2 (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 ξ) of vortices binding individual magnetic
flux quanta Φ0 = hc/2e on the scale λ, with the charge '2e' appearing in Φ0 a consequence
101
of the pairing mechanism; since λ>ξ, 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.
Acknowledgement
We thank to Lyubov Anisimova for useful suggestions for preparing this article. All
the Mathematica programs presented in this article were made with the seminar with her.
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
The definition of the parameters are given in the text.
4πα 2
Thermodynamic field
Hc =
Order parameter
ψ ∞ 2 = ns* = α / β .
The parameters
α =
β
2
Hc
,
*
4πns
102
.
β=
Hc
2
4πns
*2
.
Quantum fluxoid
Φ0 =
Magnetic field penetration depth
λ=
2π c
.
q*
m*c 2 β
,
4πq * α
2
λ=
Coherence length
ξ=
Ginzburg-Landau parameter
κ=
2m* α
,
λ cm* β
=
,
ξ
2π q *
ξ=
κ=
m*c 2
4πns q*
*
2πns
2
.
*
H c m*
.
cm* H c
.
*
2 2π ns q *
Some relations between Hc, Hc1, and Hc2
λξ
1
=
,
Φ 0 2 2πH c
2 q* H c
κ
=
.
λ2
c
κ
Φ = 2 2πH c ,
λ2 0
κ=
Lower critical field (type II)
Surface-sheath field
2 2πλ2 H c
.
Φ0
Hc
1
.
=
Hc2
2κ
Φ0
,
2πλξ
Φ0
H c2 =
.
2πξ 2
Φ
λ
H
H c1 = 0 2 ln( ) = c ln(κ ) .
4πλ
ξ
2κ
H c3 = 1.695 H c 2 .
2Hc =
Upper critical field (type II)
A.2. Modified Bessel functions
The differential equation for the modified Bessel function is given by
x 2 y ' '+ xy '−( x 2 + n 2 ) y = 0 ,
1
n2
y ' '+ y '− (1 + 2 ) y = 0 ,
x
x
The solution of this equation is
y = c1I n ( x ) + c2 K n ( x) ,
where
I n ( x) = i − n J n (ix ) ,
K n ( x) =
π
i n +1[ J n (ix ) + iN n ( x)] =
2
Recursion formula:
π
2
i n +1H n(1) (ix) ,
103
2n
K n ( x)
x
n
K n ' ( x) = − K n −1 ( x) − K n ( x )
x
n
K n ' ( x) = − K n +1 ( x ) + K n ( x) ,
x
d n
[ x K n ( x)] = − x n K n −1 ( x)
dx
d −n
[ x K n ( x )] = − x − n K n +1 ( x)
dx
The parameter γ is the Euler-Mascheroni constant: γ = 0.57721.
x
K 0 ( x) = − ln( ) − γ = − ln x + ln 2 − γ = − ln x + 0.11593718
2
,
1
K1 ( x ) =
x
for x ≈ 0 .
K n +1 ( x ) = K n −1 ( x ) +
((Mathematica Program 25))
Plot[Evaluate[Table[BesselI[n,x],{n,0,5}]],{x,0.1,4},PlotPoi
nts 50,PlotStyle Table[Hue[0.1
i],{i,0,5}],Prolog AbsoluteThickness[2],Background GrayLeve
l[0.7]]
6
5
4
3
2
1
1
2
3
4
Graphics Plot[Evaluate[Table[BesselK[n,x],{n,0,1}]],{x,0.1,1.5},PlotP
oints 50,PlotStyle Table[Hue[0.2
i],{i,0,5}],Prolog AbsoluteThickness[2],Background GrayLeve
l[0.7]]
104
4
3
2
1
0.2
0.4
0.6
0.8
1
1.2
1.4
Graphics Fig.44 Plot of (a) In(x) [n = 0 (red), 1, 2, 3, 4 (blue)] and Kn(x) [n = 0 (red) and 1 (yellow)]
as a function of x.
105