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Файл:Introduction to superfluidity and superconductivity. Учебное пособие
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61
we also need to change the phase of the order parameter:
This
will ensure that the current does not change. The reader is invited to show that the
first GL equation is also gauge invariant in the above sense: a change of gauge
should be accompanied by the proper change of the phase of the order parameter.
5.3. Characteristic lengths of the Ginzburg-Landau theory
As an application of the GL equations consider a superconductor occupying
the region x > 0 in zero magnetic field. This problem is analogous to the problem
that was solved in section 3.2 for a BEC. The order parameter only depends on x,
the current is zero (on physical grounds), so the order parameter can be chosen
real (up to a constant phase factor). Therefore, the first GL equation becomes
(279)
As in section 3.2, the structure of the equation suggests that we define
the coherence length
C
,
C
,
(280)
which, as we shall soon see, is a characteristic length-scale of spatial
variations of the order parameter. Far from the boundary we expect the order
parameter not to vary at all. This gives us a natural boundary condition at infinity
(281)
We also require that at the boundary the order parameter be zero. In
practice, this can be obtained by covering the surface of the superconductor with
a ferromagnetic material. Then the solution of (279) is
(282)
The second GL equation can be recast in a different form if we recall
Maxwell's equation connecting the current with the magnetic field and apply the
gauge A This yields
A
A
(283)
Consider the one-dimensional case of nearly constant order parameter.
Then the last equation becomes
A
A,
(284)
suggesting another characteristic length-scale, this time of variation of the
magnetic field:
62
C
(285)
This is called the penetration depth of a weak magnetic field.
As the temperature increases towards TC, the coherence length and
the penetration depth diverge but their ratio
C
,
(286)
called the Ginzburg-Landau parameter, remains constant.
5.4. Flux quantisation
Consider a superconducting ring with walls much thicker than the
penetration depth in a weak magnetic field (Fig. 11). Rewrite the second GL
equation:
J
A
(287)
Now integrate this equation along a contour deep inside the superconductor,
where the current is nearly zero. This will produce:
Cl
A
(288)
The integral of the vector potential along the contour is equal to the flux
through the ring. Reasoning as we did in section 3.5 we come to the conclusion
that the magnetic flux is quantised:
,,
(289)
where we have defined the flux quantum
Fig. 11. A superconducting ring in a weak magnetic field. The walls of the ring are much thicker
than the penetration depth.
B
0
r
B
0
λ
C
63
(290)
Recalling the connection (286) of the GL parameter and measurable
characteristics of the superconductor we have
C
(291)
5.5. Energy of the SN interface
With respect to their response to a magnetic field, all superconductors can
be divided into two types: type-I and type-II. As the magnetic field is increased, a
type-I material goes into the so-called intermediate state, when it is separated into
normal and superconducting phases; in a type-II material one observes vortices
with a normal core forming a triangular lattice and each carrying a quantum of
magnetic flux. In both cases the magnetic field in the normal phase is equal to the
critical field.
We would now like to compute the energy of the interface between the
superconducting and normal phases of a metal. Taking the SN interface as the zyplane and the superconductor occupying the half-space x < 0, we write the Gibbs
free energy per unit area of the boundary
C
,
(292)
where f(x) is the Helmholtz free energy. Fig. 12 shows the geometry of the
problem and also the coordiate dependence of the order parameter and the
magnetic filed for type-I and type-II superconductors. Let the magnetic field be
directed along the z axis and the vector potential along the y axis. Since the
material is simply connected, we can take the order parameter to be real.
If the whole material were superconducting, the Gibbs energy per unit
volume would be
S
N
C
Therefore the energy of the interface will be the
Fig. 12. The geometry of the problem of the surface energy of an SN interface (a), and the
coordinate dependence of the order parameter and the magnetic field in in type-I (b) and type
II (c) superconductors.
B
C
x
z
y
0
N
S
B(x
)
ψ(x)
λ
ξ
x
0
B(x
)
ψ(x)
λ
ξ
x
0
(a)
(b) (c)
64
difference between (292) and the Gibbs energy of the uniform superconductor.
Thus
SN
x
y
C
(293)
Now turn to the GL equations. For the given geometry the equations
become:
,
(294)
It is convenient to define dimensionless quantities
,
,
C
,
C
(295)
and rewrite equations (294) in terms of them (dropping the tildes):
,
(296)
Far from the boundary in the superconducting half-space the order
parameter is equal to its equilibrium value given by (281), the magnetic field and
the vector potential are zero. Far from the boundary in the normal region the order
parameter is zero and the magnetic field has the critical value. We therefore have
the following boundary conditions for equations (296):
;
(297)
It is left as an exercise to the reader to demonstrate that if we multiply
the first and second of the equations (296) by
and respectively and
combine them, and also use the boundary conditions (297), we shall get
(298)
Rewrite the surface energy (293) in terms of scaled variables:
SN
C
(299)
If we take into account (298), we finally arrive at
SN
C
C
,
(300)
65
where, in view of the boundary conditions (297), we have put the upper
limit of the integral equal to zero. If the penetration depth in much smaller
that the coherence length, the main contribution to the integral comes from the
first term in the integrand. On physical grounds it is clear that the order parameter
decreases towards the boundary but does not turn zero abruptly at x = 0. However,
we may still use the solution (282) to obtain a good estimate of the surface energy
in this case. The reader is invited to show that the result is
SN
C
,
(301)
In the opposite case, when the penetration depth is large compared to
the coherence length, we may neglect the first term in the integrand and use
The result, which is left to the reader to prove, is
SN
C
,
(302)
5.6. Critical field of a thin film
Consider a superconducting film of a thickness d much smaller than the
coherence length and the penetration depth in an external magnetic field H0
applied parallel to the film. Let the faces of the film be parallel to the yz-plane and
given by the equations x = ± d/2, and the magnetic field be directed along the z-
axis (Fig. 13). To satisfy the conditionA the vector potential must be directed
along the y-axis. Thus we need to solve the GL equations (296) subject to the
boundary condition
(303)
Because of the condition d << ξ we can solve the second GL equation
taking the order parameter constant. The solution satisfying the boundary
conditions is
,
(304)
where we have used the fact x << 1 (meaning, in regular units, that x << λ,
ξ). Inserting this solution into the first GL equation, where we must put zero the
second derivative of the order parameter, we have
Fig. 13. The geometry of the problem of the critical field of a thin film.
x
y
z
d
≪λ,ξ
H
0
66
(305)
This solution may seem to be at odds with our initial approximation of the
constant order parameter, but it is not. It simply gives the first-order
approximation to the solution of the problem (the zeroth-order approximation is a
constant). Let's define the parallel critical field of a thin film by requiring that at
this field the average order parameter
(306)
should turn zero. The calculation yields
C, film
C
(307)
5.7. Critical current of a thin film
Consider the same film as in the previous section carrying a small current.
Let the current be directed along the y-axis (Fig. 14). Now the vector potential is
an even function of x and the magnetic field an odd function. Thus, we need to
solve the GL equations (296) subject to the condition
,
(308)
where H0 is yet to be determined.
As in the previous section we take the order parameter to be a
constant, which, together with the condition (308) gives the solution for the vector
potential:
,
(309)
where we have again used the fact x << 1. The vector potential is thus a
constant in this approximation. Inserting this solution into the first equation (296)
and putting to zero the second derivative of the order parameter again we arrive
at the following equation:
(310)
67
The right-hand side of this equation turns to zero atand ,and has
stationary points atand
Below the transition temperature at zero
magnetic field the equilibrium value of the order parameter is unity (in the units
of (295)). Therefore the maximum sustainable magnetic field corresponds to
Using this value in equation (310) yields the connection between the
maximum field and the parameters of the film:
(311)
The current density corresponding to this field is found from
Maxwell's equation:
,
(312)
which in combination with (311) gives the critical current of the thin film:
C
C
(313)
5.8. Vortex solutions
As has been remarked above, the second GL equation is the usual quantummechanical expression for the current in the presence of a magnetic field.
Consider it for the case when the penetration depth is much larger than the
coherence length so that as far as variations of the magnetic field are concerned
the order parameter may be regarded constant and equal to its value (272). The
second GL equation is then
B
A
A
,
(314)
where we have used the definition of the penetration depth (285). Taking
the curl of both sides and recalling that the B-field has zero divergence we have
B B
,
(315)
where, drawing on our experience in superfluidity (see sec. 3.5), we do not
rush to put the curl of the gradient of the phase to zero. If the superconductor is
simply connected then the circulation of the gradient of phase is zero for any
contour. If, however, the superconductor is multiply connected (the simplest
Fig. 14. The geometry of the problem of the critical current of a thin film.
x
y
z
d
≪λ,ξ
J
68
example being a superconducting ring), the phase of the order parameter is
allowed to change by a multiple of 2π as we traverse a contour enclosing the 'hole',
or an area where the order parameter is zero. Physically, just as in the case of
superfluidity, this means that we can set up a current around the hole. This is called
a vortex solution of the GL equations. We have therefore
l S n,
(316)
where we have used Stokes's theorem.
Consider the axially symmetric case, with the axis of symmetry in
the z-direction. Then we may rewrite the previous equality in differential form:
zr,,
(317)
which brings us to the following equation for the magnetic field created by
a vortex directed along the z-axis:
B B zr
†
(318)
In the simplest case of a single quantum of flux the second GL
equation and the boundary condition for the magnetic field are
r,
(319)
If we introduce new variables
,
,equation (319) with
a zero right-hand side will turn into
,
(320)
where the primes denote differentiation with respect to x. The solution to
this equation that approaches zero at infinity is the modified Bessel function K0(x)
(also called the McDonald function) with the following asymptotic behaviour:
, ,
,
(321)
Therefore, the solution to (319) is
,
(322)
where 2π has been added to the denominator to ensure the correct total flux.
Recall that we are considering the case where the coherence length
is much smaller than the penetration depth. This means that at small distances
†
The two-dimensional delta-function is defined as follows:
r,
r
, ,;
, ,
r
69
from the axis of the vortex, at its core the magnetic field is equal
,
(323)
The structure of the vortex solution is sketched in Fig. 15 for the case when
the penetration depth is much larger than the coherence length.
We would now like to compute the energy associated with a single
vortex. It is a sum of the energy of the magnetic field and the kinetic energy of the
flow of the Cooper pairs of mass 2m and density
with velocity v:
magkin
x
(324)
The current density isJ
v
B,which, in combination with the
definition (285) of the penetration depth, leads to
x
B
(325)
The reader is invited to show that with the help of equation (318) the above
equation gives for the energy of the vortex per unit length
v
,
(326)
It is now easy to calculate the magnetic field at which a vortex
appears in a uniform superconductor. The Gibbs free energy per unit length of a
superconductor with a single vortex in an external magnetic field H0 is
v
xxv
(327)
The value of the field at which the Gibbs energy turns to zero is called the
lower critical field. When this field is applied it becomes energetically favourable
for a vortex to enter the superconductor. Thus
C1
,
(328)
Fig. 15. (a) The structure of the vortex solution for the case when the penetration depth is much
larger than the coherence length; (b) a vortex array at a meagnetic field in excess of the upper
critical field.
B
∼ξ
J
r
⊥
ψ
B
0
∼ξ
H≥H
C2
J
(a)
(b)
70
Now suppose we have two vortices. In this case equation (318)
should be generalised as follows:
B B z
rr
rr
,
(329)
where l1 and l2 are integers. The energy of the two vortices can be computed
in exactly the same way as (326):
2v
Br
Br
z
(330)
Here
Br
z
(331)
is the field at the core of vortex 1. It is a vector sum of the field of vortex 1
at its core and the field produced by vortex 2 at the core of vortex 1; a is the
distance between the cores of the vortices. Similarly,
Br
z
(332)
is the field at the core of vortex 2. Therefore the energy of the two vortices
is
2v
(333)
The first term is the energy of the two vortices when they are wide apart;
the second is then the interaction energy.
From the expression (333) one can see that the energy is lower for
two vortices wide apart rotating in the same sense and each carrying a single
quantum of flux than for a single vortex carrying two quanta of flux. Therefore as
the external field increases more vortices with a single quantum of flux enter the
superconductor. It is interesting to estimate the value of the magnetic field when
the number of vortices is so large that their cores touch (this field is called the
second, or the upper, critical field). Choose a contour in the xy-plane (the vortices
are directed along the z-axis). It subtends an area A that contains approximately
A/ξ2 vortices. The flux through this area is (A/ξ
2
)Φ
0
= AH0. Therefore the upper
critical field is of order
C2
C
(334)
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