Добавил:
ivanov666
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Introduction to superfluidity and superconductivity. Учебное пособие
.pdf
41
;
x
*
*
x
xx
x
Write the wave function in the formx
x
x
and put it into the action.
By and by varying S with respect to the amplitude and phase obtain Eqns (157).
3. Show that the energy of interaction of two vortices separated by a distance a in
a region with a characteristic size R, per unit length, is equal to
int
42
4. THE UNIFORM SUPERCONDUCTING STATE
Superconductivity was discovered in 1911 by H. Kamerlingh Onnes. As the
temperature is lowered below a certain value, called the critical temperature, the
resistivity of some metals drops to zero. This is equivalent to the frictionless flow
of 4He. If a sufficiently weak magnetic field is applied above the critical
temperature, it will be expelled from the bulk of the material as the the
temperature is lowered through the critical value. The experiment can be reversed:
one can first cool the material below the critical temperature and then apply a
weak magnetic field. The result will be the same: the magnetic field is expelled.
This is called the Meissner effect. Some superconducting meterials, however,
respond differently to a magnetic field: as the field increases, it enters the
superconductor in the form of vortices. The electronic heat capacity of a
superconductor has a dicontinuity at the critical temperature, suggestive of a gap
in the energy spectrum of elementary excitations (electrons). The existence of an
energy gap is supported by the dependence of the absorption coefficient on the
radiation frequency and also by non-linear current-voltage characteristics of
tunnel junctions. In the vicinity of the critical temperature superconductors can be
treated within the phenomenological theory put forth by V. Ginzburg and L.
Landau in 1950; uniform superconductors are described by a microscopic theory
formulated by J. Bardeen, L. Cooper and R. Schrieffer in 1957. These are the
simplest theories and they will be discussed in the sections that follow.
4.1. The ideal Fermi gas
We begin by recalling a few basic facts about the ideal Fermi gas. Just as in
the case of the ideal Bose gas, imagine a box of volume Ω with impenetrable walls
that contains N non-interacting electrons. The density of states is obtained in
exactly the same way as in section 2.1 and is given by the expression (64)
multiplied by a factor of 2 to take into account the spin: each energy level can
carry two electrons with opposite projections of the spin. At zero temperature the
levels are filled starting from the lowest energy. The top-most occupied level is
called the Fermi level, corresponding to the Fermi energy equal to
F
F
,
F
(188)
Typically, the Fermi energy in metals is of the order of 10 eV, which means
that at any reasonable temperature excitations above the Fermi level are small.
The energy of an electron with the momentum relative to the Fermi energy (about
equal to the chemical potential) is then
43
F
F
FFF
(189)
This means that if the momentum of the electron is greater than the Fermi
momentum, its energy relative to the Fermi energy is positive; if the momentum
is less than the Fermi momentum, the energy is negative. However, if an excitation
is created leaving a hole below the Fermi surface, then the energy required for
such a process is equal:
>
<
>
F
F
<
FF
(190)
That is, the energy of an excitation is positive, as it should be. The
subscripts “>” and “<” on indicate that the wave vectors are greater or less than
the Fermi momentum, respectively. Fig. 7 shows a schematic dependence of the
energy of an excitation on the wave vector.
4.2. Constructing the Hamiltonian
The discovery of the isotope effect in superconductors, which consists in
the connection of the critical temperature and the ion mass, suggests that electronphonon interaction must play a role in the phenomenon of superconductivity.
Further, this interaction must be such that when combined with the Coulomb
repulsion it will produce an overall attraction between the electrons. Indeed, if the
electron-phonon interaction were of the repulsive type, it would only enhance the
Coulomb repulsion and we could not expect to observe superconductivity.
We are thus lead to write the Hamiltonian of the electron system as
eff
k
k
k
k, k', q
q
k' + q
k - q
kk'
(191)
The energy in the first term is measured with respect to the chemical
potential. In writing down this Hamiltonian we have implicitly taken the model in
which electrons interact via phonons, so that the interaction does not involve the
spins of the electrons. The maximum energy change of two electrons interacting
via phonons is the Debye energy
D
Let's simplify the model by choosing the
matrix element of the interaction to be a non-zero negative constant for electrons
with energies within
D
of the Fermi level, and zero otherwise (Bardeen-Cooper-
Schrieffer model):
q
,
k + qk
D
;
, otherwise
(192)
44
We make one further simplification by assuming that the electrons are
paired with equal and opposite momenta and opposite spins. The former
assumption means that pairs have zero net momentum, and hence their kinetic
energy is zero, while the latter ensures that the coordinate part of the wave
function is symmetric and hence the electrons can be found with a finite
probability at the same place. Fig. 8 shows a Feynman diagram of this process.
With these simplifications the Hamiltonian becomes
eff
k
k
k
k, k'
k'
k'
kk
(193)
To make the notation tidier we shall use only one letter to keep track of the
momentum and spin of the electron:k
,k Indeed, we can perform
the summation over the spin indices in (193) and use the fact that the momenta
take on positive and negative values:
eff
K.E. P.E
(194)
Since we are now dealing with fermions, we cannot proceed exactly
along the same lines as we did with the Bose gas. However, we hope to find a
transformation analogous to the Bogoliubov transformation employed for the
weakly interacting Bose gas that would diagonalise the Hamiltonian. We
introduce new operators:
Fig. 8. A Feynman diagram showing scattering of two electrons with equal and opposite
momenta and oppsite spins in the BCS model of superconductivity.
g
|
k ,↑
⟩
|
k',↑
⟩
|
−k,
↓
⟩
|
−k',
↓
⟩
Fig. 7. The dependence of the energy of an excitation on the wave vector neat the Fermi surface
in a normal metal.
ϵ
k
kk
F
45
,
,
(195)
Where we take the functions uα and vα real. We require that the new
operators should satisfy the same commutation relations as the old ones. For
example:
(196)
This leads to
,
(197)
which means that one of the functions uα and vα is an odd with respect to
the change of sign of its index. We choose uα to be even and vα to be odd:
,
(198)
The other commutation relation gives
,
(199)
whence we obtain the other condition on uα and vα:
(200)
Thus we have the other set of transformation formulas:
,
(201)
The definitions (195) and (201) can be inverted to obtain
,
;
,
;
(202)
and express the model Hamiltonian in terms of the new operators:
K.E.
;
P.E.
(203)
We have neglected in the interaction part of the Hamiltonian terms that are
proportional to the product of four γ-operators and are either identically zero in
the ground state (to be defined below) or small. As can be seen from the structure
of the transformed Hamiltonian, the new operators describe excitations that are
neither electrons nor holes, and the ground state of the electron system with
effective attraction is the vacuum state with respect to these new operators:
(204)
Let's start with a relatively simple case of the ground state. In this
46
state terms such as
are zero since there are no excitations. Therefore the
ground-state Hamiltonian has a much simpler form:
gnd
,
(205)
Since there is only one constraint, equation (200), on the functions uα and
v
α
, we may chose them such as to eliminate the term in
from the
ground-state Hamiltonian. This leads to the equation
(206)
The solution is
,
(207)
To decide on the sign, let's write down the number-of-particles
operator:
(208)
and compute the total number of particles in the ground state. We obtain
(209)
If there were no interaction, the parameter Δ would be zero and the number
of particles in the ground state would be
(210)
The ground state of the fermion system in the absence of excitations is
realised at zero temperature, with all the states with energies below the Fermi
energy occupied and the rest empty. Therefore
and we must choose the upper
sign in (207), (209) and (210).
It is left as an exercise to the reader to show that exactly the same
result (207) can be obtained if one computes the expectation value of the
Hamiltonian (203) in the ground state (204) and then minimises it with respect to
the functions uα and vα bearing in mind the relation (200).
47
4.3. Excitations and the energy gap
D
D
D
,
(211)
Now that the functions uα and vα have been determined we can compute the
ground-state energy of the system:
where we have invoked the convention (190) converted summation into
integration with the use of the density of states. At the last step we took into
account that the density of states is a slowly varying function of energy and
D
,so it can be taken outside the integral and evaluated at the chemical
potential.
We argue that the ground state is only realised at zero temperature.
When the bath temperature is raised, excitations are created. Neglecting the terms
proportional to the product of four γ-operators, we can write down the
Hamiltonian as
(212)
To diagonalise the Hamiltonian we require that the term in the square
brackets should be zero:
,
(213)
This condition leads to the same equations as (207) except that now we
should drop the subscript “0”.
Since the excitations are fermions, the expectation values
,which, together with the condition (213) simplifies the Hamiltonian
considerably:
,
(214)
and we immediately recognise the energy of an elementary excitation:
48
,
;
,
(215)
In writing (215) we have reinstalled the wave vector index, since the energy
of the excitation does not depend on the spin. A plot of the energy of an excitation
is presented in Fig. 9. We have also taken into account that the lengths of the wave
vectors of the electrons are very close to the length of the Fermi wave vector. The
equation (215) shows that there is a gap in the energy spectrum of elementary
excitations.
Until now the ground state
has only been defined as the vacuum
state with respect to the γ-operators. We now need to connect it to the vacuum
state of the c-operators, the state of no fermions present. First we note that
(216)
for an arbitrary state. Therefore we make a guess that the ground state with
respect to excitations can be built as follows:
,
(217)
where C is a normalisation constant. The reader is invited to prove that, up
to the insignificant phase factor, the normalisation constant is equal
to
,leading to
(218)
Suppose our system is subjected to an external perturbation with a
Hamiltonian of the usual type:
Fig. 9. The dependence of the energy of an elementary excitation on the wave vector in a
superconductor. The dashed lines are assymptotes, corresponding to the energy of an
elementary excitation in a normal metal.
E
k
k
Δ
k
F
49
ext
(219)
Allowing it to act on the ground state produces
ext
,
(220)
which means that excitations are always created in pairs, so the minimum
energy required to create an excitation is equal
min
ex
(221)
4.4. The critical temperature
In thermal equilibrium the average occupation numbers of
excitations will be given by the Fermi-Dirac distribution:
,
(222)
regardless of the spin since the energy of an excitation is spin-independent.
Note the absence of the chemical potential because the number of excitations is
not conserved. Recalling the definition of the energy gap (213) at non-zero
temperature and the equations (207) for the functions uα and vα, we arrive at the
self-consistent equation for the temperature dependence of the energy gap:
k
D
D
(223)
In writing (223) we have employed the definition (193) and converted the
summation over wave vectors into the integration over energies measured relative
to the Fermi energy using the density of states.
At zero temperature equation (223) gives
D
D
,
(224)
whence
D
(225)
If the effective attractive interaction is weak enough that the
condition
is satisfied (the so-called weak-coupling limit), then
weak
D
(226)
As the temperature increases we expect the energy gap to decrease
50
until it reaches zero. We thus define the critical temperature by the condition
and write
D
D
D
(227)
When going from the second line to the third we have used the fact that
typically
This has allowed us to replace the hyperbolic tangent with unity
and the upper limit of the integral with infinity. The last integral can only be
evaluated numerically. Note that the condition of weak coupling that has lead us
to (226) is consistent with the assumption
Combining (226) and (227) we
arrive at the connection between the zero-temperature energy gap and the critical
temperature in the weak-coupling limit:
weak
(228)
4.5. The discontinuity of the electronic heat capacity
We would now like to show that the microscopic theory of
superconductivity can explain the discontinuity of the electronic heat capacity that
is observed in the experiment. By definition
,
(229)
where the angle brackets denote the expectation value in a given state.
Then, with the Hamiltonian (212), the definition of the energy gap (213) and the
Fermi-Dirac distribution for the occupation numbers (222) we can write for the
energy of the system at temperature T:
k
k
k
k
k
k
kkkk
k
kk
k
(230)
When performing calculations, it is often convenient to define the
density of states with respect to excitations:
, ;
,
(231)
Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]