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Файл:Introduction to superfluidity and superconductivity. Учебное пособие
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11
1.4. Operators in second-quantised form
As one can see, the procedure of second quantisation established in the
previous two sections consists in the replacement of the state vector with an
operator. It is in this sense that the term “second quantisation” is used (when we
perform the first quantisation, we assign a vector to each state and replace the
classical variables with operators that act on these vectors). We now need to
rewrite the familiar operators in the language of second quantisation.
The simplest operator is the number of particles. Indeed, we have already
seen that regardless of the statistics of the particles the eigenvalue of the operator
is the number of particles in the state α. Therefore the operator
(30)
is the total number of particles in the system†. From the above discussion
we can easily see that a ket representing a many-particle state is an eigenket of the
number-of-particles operator corresponding to the eigenvalue that is the total
number of particles in the system. Indeed,
,, ,,
(31)
Since the total number of particles in the system cannot depend on
our choice of the basis kets that are used to represent the state ket, the operator
(30) must be invariant under the transformation of the basis. To check if this is
indeed the case, let's go over to a different set of basis vectors and the
corresponding set of operators:
(32)
Since the basis kets form a complete set, we have
,
(33)
which provides us with the law of transformation between the two sets of
operators:
,
(34)
So the structure of the number of particles is indeed invariant:
,
(35)
†
In this sectrion we shall be putting hats on operators in order to avoid confusing them with c-numbers.
12
where we have made use of the completeness and orthonormality of the
bases. This shows that the definition of the number of particles adopted at the
beginning is reasonable.
Next we define operators that create of destroy a particle at a
particular position. They are called field operators:
x
x
(36)
It is left as an exercise to the reader to prove the transformation formulas
x
x
x,
x
x
(37)
In particular, transformation from the momentum basis to the coordinate
basis will be
k
x
x k
xx
x
kx
,
x
k
k x
k
k
k
kx
,
(38)
where the momentum eigenfunctions are normalised so that the particle can
be found with certainty in the box of volume Ω.
Using the transformation formulas (37) and the commutation
relations for fermion and boson operators one can prove the following the
commutation relations for the field operators:
x
x
x
x,
xxxx,
x
x
xxx x,
(39)
where the upper sign is for fermions, and the lower for bosons.
We can also express the number operator in terms of the field
operators as follows:
x
x
xxxx
xxx x
xxx
xxxx,
(40)
where the number-density operator has been introduced:
x
xx
(41)
This operator is Hermitian, as it should be. If we express the field operators
in terms of operators in the momentum representation, we shall have
x
k, k'
k'kx
k
k'
k, q
qx
k + q
kq
qx
q
,
(42)
where we have introduced the number-density operator in the momentum
representation:
13
q
k
k + q
k;
q
q
(43)
The next simplest operator is the kinetic energy. It is easiest to write
it down if we take the momentum eigenkets as our basis. Then
k
k
k
k
(44)
Indeed, allowing it to act on a ket of a many-particle system we have
k
k
k
k
,,
k
k
k
,,
(45)
It is left as an exercise to the reader to show that in terms of the field
operators the kinetic energy takes the form
x
x
x,
(46)
which, except for the hats, looks just like the average of the kinetic-energy
operator in the Schroedinger representation.
This resemblance suggests that we should define the potential energy
of the system in an external force field in the form
xxxxx
xx,
(47)
where V(x) is the potential energy of a single particle in the external force
field. If we go over to the momentum basis
k
k x,the potential
energy becomes
xxk
k
kx
kk
kx
kk
xx
kkx
k
kkqq
k + q
k
,
(48)
where we have introduced the Fourier transform of the external potential
q
xx
qx
(49)
and changed the summation variable. Eq. (48) describes the process in
which a particle of momentum k is destroyed and a particle of momentum k + q
is created. Physically, this means that the particle's momentum has changed from
k to k + q as a result of interaction with the external potential. This can be
represented pictorially by a Feynman diagram shown in Fig. 1.
14
We still need, however, to make sure that the definition (47) makes
sense. The way to do this is to compute matrix elements in a basis. The form (49)
suggests that we should take the eigenkets of the momentum operator as such a
basis. Then, on the one hand,
k
k
k
kqq
kq
k
k
kq
q
k
kq
k
k
kq
q
k
kq
kq
k
kk
k
k
k
k
,
(50)
where the upper sign is for bosons and the lower for fermions, and we have
used the fact that annihilation operators acting on the vacuum state produce zero.
On the other hand, we can compute the matrix element directly
k
k
xx
k
xxxxxk
xx
kxkx
xx x
k
k
(51)
So, the two representations of the external potential energy lead to the same
matrix elements, which justifies the definition (47).
Finally, consider two-particle interaction U(x, x'). We argue that it the
language of field operators it can be represented as follows:
xxx,xxx
xxx,x
xx
xx,
(52)
which may look plausible: but for the hats the expression would look just
like the potential energy of two charge distributions. The factor ½ is needed to
allow for counting the same pair of particles twice. However, if we use the
commutation relations (39) for field operators, we shall have
Fig. 1. A Feynman diagram giving a pictorial representation of the scattering of the particle
from a state with momentum k into a state with momentum k + q via interaction with the
potential V(x).
k k + q
V
q
15
xxx,x
x
xxxx xx
xxx,x
x
xxx
xx,xx,
(53)
with the upper sign for bosons and the lower for fermions, and where in the
second line we used the fact that two annihilation operators commute or
anticommute depending on whether they are boson operators or fermion
operators, respectively. The last term in (53) is the self-interaction energy and of
no interest to us. We thus keep only the first term and write
xxx,x
x
xxx,
(54)
or, in the momentum basis
k k' q
q
kq
kq
k
k
,
(55)
as the reader is invited to verify. Physically, (55) means that two particle
with initial momenta k and k' interact via the potential Uq and emerge from the
interaction with momenta k – q and k' + q. The total momentum is thus conserved,
as should be expected. This process can also be represented by a Feynman
diagram, as shown in Fig. 2.
The matrix elements of the mutual-interaction potential with respect
to the two particle basis are the same regardless of the way they are computed:
either directly as in (51) or in the representation of second quantisation (50). This
makes the definition (54) reasonable .
Fig. 2. A Feynman diagram giving a pictorial representation of interaction of two particles: the
particles scatter from states with momentuma k and k' into states with momenta k + q and
k' + q respectively via the the potential U(x, x'). Note that the total momentum is conserved, as
it should be.
k k – q
U
q
k' + qk'
16
Exercises
1. Prove that the matrix elements of the mutual-interaction potential with respect
to the two particle basis are the same regardless of the way they are computed:
either directly as in (51) or in the representation of second quantisation (50). This
makes reasonable the definition (54).
2. Prove the transformation formulas
x
x
x,
x
x
3. Using the result of Exercise 2 that the field operators satisfy the commutation
relations:
x
x
x
x,
xxxx,
x
x
xxx x,
where the upper sign is for fermions, and the lower for bosons.
4. Show that in terms of the field operators the kinetic energy takes the form
x
x
x
5. Define the momentum operator
P
kkk
Show that this definition is reasonable by acting with this operator on a state of a
definite momentum. Next show that in terms of field operators the momentum
operators takes the form
P
x
xx
6. Prove that in the momentum representation the two-particle interaction energy
is given by
k k' q
q
kq
kq
kk
17
2. THE UNIFORM WEAKLY INTERACTING BOSE GAS
In 1937 P. Kapitza and J. Allen discovered the phenomenon of superfluidity
in 4He. When 4He is cooled, a few remarkable effects can be observed. If cooled
down below a temperature of 2.17 K, called the critical temperature and set to
flow along a pipe 4He does not experience friction (hence the term superfluidity)
until its velocity reaches a certain critical value. The heat capacity as a function
of temperature shows a discontinuity at 2.17 K. Owing to the heat capacity profile
resembling the Greek letter λ the critical temperature is also called the λ-point. If
we put He4 in a cylindrical container above the λ-point and start to rotate it at an
angular velocity of the order of less than 10
-4
rad/sec, we shall at first observe that
it rotates as if it were an ordinary viscous liquid displaying a meniscus. The
moment the temperature is lowered below the λ-point, the rotation of the helium
ceases. This is called the Hess-Fairbank effect. If, however, we start above the λ-
point but with a sufficiently high angular velocity, then after the temperature is
lowered the helium in the container will maintain its meniscus, as would be the
case for an ordinary liquid. Additional experiments show that in this case
superfluidity does not disappear altogether, and the helium is penetrated by
vortices arranged in a regular pattern (called vortex lattice). The same phenomena
were observed quite recently in experiments with ultra-cold gases of alkaline
metals. The experiments described above suggest the existence of an energy gap
in the energy spectrum of elementary excitations. A theory developed for a
superfluid should be able to explain the observed phenomena described above.
An adequate theory for liquid helium is difficult to develop owing to very strong
interaction between its atoms. However, such theories are available for the weakly
interacting (dilute) Bose-Einstein condensate (BEC). One was formulated by
N.Bogoliubov in 1947 for a uniform BEC, and the other by E. Gross and L.
Pitaevskii in 1961 for a non-uniform one. The following two chapters are
dedicated to the theory of superfluidity in a weakly interacting Bose gas.
2.1. The ideal Bose gas
We start with the ideal Bose gas in a constant external potential. Consider
a collection of free non-interacting spin-zero Bose particles in a box of volume Ω.
As is known from statistical mechanics, the average number of particles in a state
of energyis given by the Bose-Einstein distribution:
,
(56)
18
where μ is the chemical potential and T is the temperature. In general, for a
Bose gas in an external potential, the chemical potential and temperature may be
position-dependent. Since the average number of particles cannot be negative or
infinite, the requirement is that the chemical potential should be greater than the
minimum energy. It should also be noted that the Bose-Einstein distribution is
only valid for non-interacting particles. Indeed, if there is interaction, the energy
of an individual particle will depend on the distribution of the other particles, and
the simple constraints under which the Bose-Einstein distribution is obtained no
longer hold.
Since the particles are free and non-interacting, the dispersion law
has a simple quadratic form:
k
k
(57)
Without loss of generality, let the box be a parallelepiped with dimensions
a, b, c along the axes x, y and z, respectively. We can either require that the wave
function be zero at the faces of the box, or we can impose periodic boundary
conditions. In what follows the size and shape of the box are immaterial, so the
choose the periodic boundary conditions:
;;
(58)
The wave functions of the particles are plane waves normalised so that a
particle is found with certainty somewhere in the box:
x
kx
,
(59)
where the components of the wave vector are quantised owing to the
conditions (58):
;
;
,
(60)
where nx, ny, nz are integers. It follows then that the volume per state in k-
space is equal to 8π3/Ω, and the density of states is
k
,
(61)
which allows us to convert sums over k into integrals over k-space:
kkkkk
k
k
,
(62)
where Fk is a function of the wave-vector k.
1
With the energy of the particle as a parameter, equation (57) defines
1
Equations (61) and (62) remain valid even for interacting particles in an external potential because the
wave function can be expanded in plain waves with the same boundary conditions, which will impose
the same constraints on the components of the wave-vector.
19
a sphere in k-space. The total number of states in a sphere of a constant energy is
equal to:
(63)
Then the density of states per unit volume as a function of energy is
(64)
Had we required that the solutions to the Schroedinger equation for
a particle in the box should be zero at the faces of the box, we should have arrived
at the same expression for the density of states.
Then the total number of particles in the box is given by the integral
(65)
but we must be careful not to lose particles in the lowest-energy state with
because D(0) = 0. Indeed, since we are dealing with bosons, at low
temperatures we should expect them all to occupy the lowest-energy state (the
particles are said to condense in the lowest-energy state with k = 0). Therefore, to
avoid losing particles in this state when replacing the sum with the integral, we
write out explicitly the term N0 that gives the number of particles in the lowestenergy state (condensate):
ex
(66)
where Nex is the number of particles in excited states (i.e. states with k ≠ 0).
As the temperature is lowered, more and more particles accumulate in the lowestenergy state, so at sufficiently low temperatures we expect
,
(67)
which means that below a certain critical temperature TC the chemical
potential becomes vanishingly small (of order 1/N). On the other hand, we may
argue that above the critical temperature the number of particles in excited states
becomes of order N:
ex
(68)
Inverting this equation, we obtain the critical temperature
20
(69)
Below T
C
, where μ = 0, the number of excited particles is given by
ex
(70)
and the number of particles in the condensate is
ex
(71)
We should like to explain the phenomenon of superfluidity, and what
we have so far achieved does not seem to through any light upon it. Let's try to
understand on the conceptual level what superfluidity, or the absence of viscosity,
is. If we set an ordinary (non-superfluid) liquid in motion along a pipe, it will soon
stop because of friction, as will stop a gently pushed billiard ball before it reaches
the cushion of the table. Recalling Newton's second law, we shall say that the
liquid (or the ball) gradually stops because it loses momentum to the walls of the
pipe (to the table). This loss of momentum manifests itself as heat, and heat on
the microscopic level means vibrations of the atoms of the pipe walls or the table
surface. So, when the viscous liquid moves along the pipe, it creates excitations
of certain momentum p and energyp
Consider two inertial frames K and K', and let the K-frame be tied to
our laboratory on the platform, and the K'-frame to our other laboratory in the
railway carriage moving in a straight line with a constant velocity V with respect
to the platform. In what follows the unprimed quantities refer to the K-frame and
the primed to the K'-frame. The Lagrangian of a particle in the moving frame is
v
x
vVx
vV
x
(72)
The second and the third terms in the last equality can be written as a time
derivative of a function of coordinates and time and so we can neglect them
obtaining the Lagrangian
x,
(73)
where we have taken into account that the potential energy is invariant
under Galilean transformations. The result (73) is no surprise if one recalls that
all inertial frames are equivalent. For the energy of the particle in the moving
frame we get:
v
v
x
xp V
p V
(74)
Returning to our liquid moving along the pipe, let the K-frame refer
to the pipe and the K'-frame to the liquid. Suppose an excitation has been created.
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