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Файл:Introduction to superfluidity and superconductivity. Учебное пособие
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51
Differentiating (230) with respect to temperature produces
k
k
k
k
,
(232)
where the factor of 2 appears in front of the integral to take into account
two branches of the energy of excitations as a function of the wave vector.
At low temperatures, such that
,the energy gap is almost
temperature independent and is nearly equal to its zero-temperature value. We
then have
;
(233)
At T > TC the energy gap is zero and
,
(234)
and we get the normal-metal electronic specific heat.
4.6. The Meissner effect
mag
p
A
p A A p
k, k'
k' p A A pk
k'
k
k, k'
k' A k k'k
k'
k
k k'
k, k'
k' A k
k'
k
(235)
We now turn to the case of a superconductor in an external magnetic field.
Let's take this field to be weak enough so that we may neglect the term in the
Hamiltonian that is quadratic in the vector potential. Thus
The matrix element is calculated as follows (we use 'the box'
normalisation):
k'Ak xxk' xxxxxk
xx
kxk'x
Axx xA
k'k
(236)
So after relabelling we finally get
mag
A
q
k, q
k q
k + q
k
(237)
52
To get the Meissner effect we need to show that the current that is induced
in response to the external magnetic field is proportional to the vector potential.
Recall the charge-conservation equation written in operator form
J
,
(238)
where we have explicitly written the electric charge in order to work with
the number-density operator. Taking the Fourier transform, and using the
Heisenberg equations of motion we write
q J
q
q
q
(239)
Writing the Hamiltonian as
kin
int
mag
,we obtain
kin
q
q k q
k
k
k + q
,
int
q,
mag
q
A
q
q k
k
k
(240)
The above commutation relations are left to the reader as an exercise. The
second relation is true provided the Fourier transform of the interaction potential
is an even function, i.e.
q
q This condition is satisfied if the interaction
potential depends only on the distance between the interacting particles, which is
a reasonable assumption. Then (239) and (240) lead to
J
q
k
k q
k
k + q
A
q
k
k
(241)
To obtain the current that is induced as a result of the magnetic field,
we take the expectation value of the current operator. We should however note
that the initial state of the superconductor has been perturbed by the magnetic
field:
mag
mag
(242)
Thus, up to terms of second order in the perturbation, the current is
J
q
J
q
J
q
mag
mag
(243)
Note that the subscript '0' does not refer to the ground state of the
superconductor. It is an indicator of the unperturbed state that is an eigenstate of
the operator
In order to proceed further one needs to express the current operator
and the perturbation Hamiltonian in terms of the γ-operators taking care of the
spin as well as momentum. Since we are interested in the response of the
superconductor to a nearly uniform external magnetic field, the most contribution
53
to the current will come from the terms with a small value of q. The intermediate
result is
J
q
A
q
Aqkkk
k + q
k
k + qk
,
(244)
where
k
k
k
,and we have made a convention that the sign of the spin
projection coincides with the sign on the momentum of the particle. If we
consider the case of a thermal equilibrium at a low temperature, we note that the
expectation value in the formula above is equal to
k
kAt low temperatures the
energy of the excitations is close to the gap, so
Converting the sum
into an integral over k-space and choosing the direction of the vector potential as
the polar axis, we obtain upon integrating over the angles and converting to
integration over energy
J
q
Aq
F
F
A
q
(245)
At T > TC the gap is zero and integral in the brackets is
,
(246)
and the current is zero. Thus, above the superconducting transition there is
no Meissner effect, as one should expect.
When the temperature is much lower than the critical temperature,
the gap is equal to its zero-temperature value and for the same integral in the
brackets we obtain
,
(247)
and the Meissner effect is restored.
The formula (245) and two limiting cases can be neatly written as
follows:
J
q
A
q
,
;
(248)
54
4.7. Tunnelling
4.7.1. A simple problem
Consider a simple problem of two identical potential wells separated by a
barrier that allows tunnelling of particles from the left well to the right one and
backwards. Since the problem is symmetric, the eigenfunctions of the
Hamiltonian are even and odd with respective energies
e
e
e
;
o
o
o
(249)
As will be known from the solution of this rather simple quantummechanical problem, Eo > Ee. Suppose that the states
o
and
e
are the only
possible states of our system (i.e. we have a two-level system). Then the
Hamiltonian can be written as
e
e
e
o
o
o
e
e
eo
o
o
,
(250)
where at the last step we introduced creation and annihilation operators for
the particle (excitation) corresponding to the two levels of the system. Define now
new state vectors and corresponding operators
e
o
,
e
o
;
e
o
,
e
o
(251)
The new state vectors describe the particle occupying either left ('g') or right
('d') well. With this transformation we rewrite the Hamiltonian in a form that
allows the discussion of tunnelling
T
;
eo,
eo
(252)
The first term of the Hamiltonian is the energy of the system when
the particle is in one of the wells. Since the problem is symmetric, the energy is
invariant with respect to interchange of left and right. Now we ask ourselves,
'What is the probability for the particle to go from left to right, or vice versa?' The
answer is given by Fermi's golden rule that says that this probability, per unit time,
is equal
T
dg
dg,
(253)
where the delta-function ensures energy conservation.
55
4.7.2. NN-contact
Let's start with a case of two identical normal metals (N) separated
by a thin layer of a dielectric and maintained at a constant voltage difference V.
The total current will be a sum of the current caused by electrons tunnelling from
left to right (LR) and the reverse current (RL). For simplicity we assume that the
matrix element that appears in equation (253) does not depend on the energy of
the electron. The probabilities of tunnelling for a single electron are
LR
LRRL
kk'k'k;
RL
RLLR
k'kkk';
(254)
here we have chosen the left metal to be at a potential V with respect to the
right one and took into account the occupation of the states that the electron is
going from and to. The current is then equal
kk'
kk'k'kk'kkk'
kk'
kk'
k'k
F,
(255)
where we have used the fact that the density of states is a slow-varying
function of the energy and the applied voltage is not too large; we have also taken
account of the electron spin by multiplying the current by a factor of two. Thus
we observe Ohm's law. Note that the dimension of the coefficient in front of the
voltage is cm/sec, which is the dimension of the conductance.
4.7.3. NS-contact
Let now the metal on the left remain normal (N) and the one on the
right turn superconducting (S). The part of the Hamiltonian responsible for
tunnelling is
T
,
(256)
where we again use the subscript α to refer to the wave-vector and the spin
of the electron with the convention
k,kSince the metal on the
right is superconducting we need to express the d-operators in terms of the γoperators according to (202). This gives
56
T
(257)
Then the probabilities of tunnelling from left to right and backwards are
LR
;
RL
(258)
Before proceeding further with the calculations, recall that u and v
are functions of the electron energy measured with respect to the Fermi energy
(see (207) with the upper signs). If it is necessary to express them as functions of
the energy of the elementary excitation, then they should be double valued:
,
(259)
and similarly for v. This means that when converting the sums into integral
we should sum over the two branches. But
,,
(260)
and so they drop out of the integral and we have
F
(261)
Recalling that the energy of an elementary excitation is greater than or equal
to the gap energy, we see that the square bracket is non-zero in the intervals
(– eV, – Δ) and (Δ, eV). Therefore
F
F
(262)
Thus while the applied voltage is below the gap voltage Δ/e the current
cannot flow. At higher voltages the current-voltage characteristic is given by
(262). When the applied voltage becomes much greater than the gap voltage we
return to Ohm's law (255). A schematic representation of the dependence (262) is
shown in Fig. 10.
57
Exercises
1. Show that the normalised ground state of a BCS superconductor is
2. Prove the commutation relations (240).
Fig. 10. A chematic current-voltage characteristic of the SN tunnel junction at zero temperature:
no current is flowing untul the voltage reaches the gap voltage Δ/e. At non-zero temperature
this sharp transition is smeared owing to non-zero probability of appearing unpaired electrons.
The dashed assymptotes correspod to Ohm's law.
I
V
Δ/e
−Δ/e
58
5. THE NON-UNIFORM SUPERCONDUCTING STATE
NEAR T
C
5.1. Thermodynamics of superconductors
Before discussing the Ginzburg-Landau theory in any detail it is worth
while to review thermodynamics of magnetised media. Additional energy
acquired by a system in a magnetic field per unit volume is
H B
(263)
This is also the work produced by an external circuit on a magnetised
medium at constant temperature. Therefore, we should associate (263) with an
increase of the Helmholtz free energy per unit volume of our magnetic material:
H B
(264)
The Helmholtz free energy is then seen to be a function of the B-field. The
B-field, however, is not a very good independent variable because it results from
the free currents supplied by the external current source and the magnetisation and
polarisation currents. The H-field, on the contrary, is produced by external
currents and is thus a good independent variable because it is controlled (through
the current) in a real experiment. We are therefore led to define a new
thermodynamic potential (called the Gibbs free energy):
H B
(265)
Differentiating we see that the Gibbs free energy is a function of the Hfield:
B H
(266)
Consider now a superconductor that displays the Meissner effect. Let
this material be non-magnetic in the normal state, meaning that B = H, which
simplifies the discussion. Then the contribution of the magnetic energy to the
Gibbs free energy is
B
H
, C,
C
;
C
, C,C
(267)
How do we interpret this result? The first and second lines in (267)
correspond to the superconducting and normal state respectively. If we put H = 0
in the second line, we shall get the Gibbs free energy of the normal state below
the critical temperature at zero magnetic field (provided that the normal state
under such circumstances is stable). This means that below the critical
59
temperature the energy of the superconducting state is lower by
C
per unit
volume than the energy of the normal state:
SN
C
,
(268)
where the subscripts “S” and “N” refer to the superconducting and normal
state respectively.
5.2. The Ginzburg-Landau equations
The BCS model developed in the previous section did not allow for spatial
dependence. A generalisation to the inhomogeneous case can be made within the
framework of Bogoliubov equations, but a simpler solution is offered by the
Ginzburg-Landau (GL) theory. At the heart of this approach is the complex order
parameterxnormalised to give the total number of “superconducting” electrons
(i.e. the total number of paired electrons):
x
x
S
,
(269)
where the subscript “S” refers to the “superconducting” state. As the
temperature approaches the critical value, the number of Cooper pairs decreases,
so the order parameter is expected to be small near TC, and the Helmholtz free
energy can be expanded in a Taylor series in
x
and its spatial derivative.
Consider first a uniform superconductor at a temperature slightly
below the critical temperature. Then the Gibbs free energy per unit volume is
equal
SNx
x
(270)
where gN is the Helmholtz free energy per unit volume of the normal
material. We choose the coefficients as follows:
CT
(271)
This means that above the superconducting transition the the minimum of
the free energy is only achieved at
xbut below the transitions there is a
local minimum at xIndeed, differentiating (270) with respect to
x
and
equating the result to zero we obtain
x
min
S
min
N
(272)
Comparing (268) and (272) we find the connection between the
phenomenological coefficients and the measurable critical magnetic field:
C
C
(273)
60
Let's now add a magnetic field and take account of spatial variations
of the order parameter. Since in experiments it is the external magnetic field that
is fixed, we should now construct the Gibbs free energy, rather than the Helmhotz
free energy. Thus,
S Nxx
x
Axx
x
H
Bx
(274)
where GN is the Gibbs free energy of the normal state, H0 is the external
magnetic field and B(x) is the magnetic field inside the material related to the
vector potential in the usual way:BxAxNote thatBxis the true
microscopic field, which allows us to write the magnetic energy per unit volume
as
x
Bearing in mind that in a superconductor the current is carried by
Cooper pairs, we put 2e and 4m as coefficients in the “kinetic-energy” term.
The reader is invited to show that varying G
S
with respect to*xand
A(x) produces two GL equations connecting the order parameter and the vector
potential:
Axxxx
x,
J
x
Axx
(275)
Note that the second GL equation is the usual quantum-mechanical
expression for the current in the presence of a magnetic field. These two equations
should be augmented by a suitable boundary condition. The reasonable
requirement is that the current should not flow normal to the boundary between
the superconductor and vacuum or the superconductor and a dielectric. It is left as
an exercise to the reader to show that this leads to the condition
n
Axxx,
(276)
where u is an arbitrary real number. In particular, if u = 0, we have
n
Axx
(277)
Writing the order parameter in exponential form allows recasting the
second GL equation in such a way that its gauge invariance becomes nearly
obvious. Indeed, if
,then
J
A
(278)
A particular choice of the gauge should not affect the current – it is an
observable. Supposing that the vector potential has been changed according to
A A ,where f(x, t) is an arbitrary function. Then, as one can see from (278),
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