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Файл:Introduction to superfluidity and superconductivity. Учебное пособие
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21
Then the energy of the liquid in the K'-frame is
p
(75)
where E
gnd
is the energy of the liquid in the absence of any excitations. In
the K-frame we have
ppV
,
(76)
where M is the total mass of the liquid. In this frame in the absence of
excitation we should have
,
(77)
The state of the liquid with a single excitation can only be energetically
stable if
ppV ,
(78)
which means that the minimum macroscopic velocity of the liquid that
favours an excitation is (after dropping the primes that are now unimportant)
crit
min
p
(79)
This leads to the condition
(80)
For a non-interacting Bose gas the dispersion law is quadratic, and the
condition (80) is only satisfied at p = 0, which means that the critical velocity is
zero. This means that we must allow for some interaction among the particles in
the hope of obtaining the correct dispersion law.
2.2. Elements of scattering theory
Before proceeding to the weakly interacting Bose gas, it is instructive to
recall a few facts from the quantum-mechanical scattering theory. In the simplest
Fig. 3. A schematic of the scattering process: a particle in a state with momentum k directed
parallel to the z-axis is scattered into a state with momentum kS. The potential V(x') only acts
over a finite rage.
k
k
S
dΩ
dS
O
V(x')
z
scattering region
22
scattering experiments a stream of particles is propagated to interact with a steady
target (scattering centre). The particles emerging from the scattering region are
registered, and their angular distribution gives information about the properties of
the target (scatterer). A schematic of the scattering process is shown in Fig. 3.
Scattering experiments can be analysed within the time-independent
Schroedinger equation
xxxx,
(81)
where E is the energy of the incident particle far from the scattering region,
and m is the reduced mass. Realistic potentials go to zero with distance, so E > 0.
Rewrite the above equation as
x
xx;
(82)
We can now solve this equation using the method of Green functions. By
definition, a Green function of equation (82) is the solution of the inhomogeneous
equation with the delta-function on the right-hand side:
xxxxx x
(83)
Suppose we have found the Green function of our equatio. Then we can
write
xx xx xxx
xxxxx
(84)
Thus, the solution of (81) can be constructed as follows:
x
xxxxxx,
(85)
where φ (x) is a plane-wave solution of the homogeneous equation.
It is left as an exercise to the reader to prove that the Green function
of equation (82) corresponding to outgoing waves is
xx
xx
x x
,
(86)
so the solution is
xx
x
xx
x x
xx
(87)
Consider distances large compared to the size of the scattering region:
x xx
xx
x
xx
x
,x
xx
x x
kSx
,
(88)
wherek
S
xis the wave-vector of the scattered particle (the scattering is
assumed to be elastic). If we take the z-axis in the direction of the propagation of
23
the incident particle, then
x
,
(89)
where we have defined the scattering amplitude
x
kSx
xx
(90)
Substitution ofxk xforxinto (90) produces the first Born
approximation for the scattering amplitude:
x
kkSx
x
(91)
Thus, the scattering amplitude in the first Born approximation is
proportional to the Fourier-transform of the potential energy.
If the energy of the scattered particle is sufficiently low so that we
may put k k' ,equation (91) reduces to a constant called the scattering length:
xx
(92)
Thus if the potential is attractive, the scattering length is positive, and vice
versa.
In the simple case of contact interaction the scattering amplitude A
(1)
is also a constant:
xx
(93)
The differential scattering cross-section is the ratio of the flux of
particles scattered into an elementary solid angle dΩ to the flux density of incident
particles:
scat
S
inc
,
(94)
where the flux density of scattered particles far from the scattering region
is:
J
scat
S
S
*
S
*
S
S
*
S
x
x
,
(95)
and the flux density of incident particles is computed in a similar manner:
J
inc
z
(96)
Combining (94), (95) and (96) produces
(97)
The total cross-section in the case of contact interaction is
(98)
24
This is precisely the total cross-section for low-energy scattering off a rigid
sphere of radius a, which is why the scattering amplitude A
(1)
in the first Born
approximation is called the scattering length.
2.3. Costructing the Hamiltonian
The Hamiltonian of an interacting Bose gas in the language of second
quantisation can be constructed as follows:
kk
k
k
k k' q
q
kq
kq
kk
(99)
Here
k
is the kinetic energy of a particle, and Ω is the box
volume introduced for future convenience. In an ideal Bose gas, at zero
temperature all the particles occupy the same state with a zero momentum. This
minimised the total energy of the gas. When particles interact, they exchange
momenta. This means that there must always be particles in states with a non-zero
momentum. If, however, the interaction is weak (the exact criterion will be
established later), we expect that these states will not be macroscopically
occupied, unlike the zero-momentum state, so that
ex
(100)
Further, if
is an arbitrary state of our system, then
,
(101)
So, to within an error of order 1/N the operators
and
commute and may
be treated as c-numbers equal to
Besides, in view of (100) we need only retain
terms of order N0 and greater in the Hamiltonian (99):
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
(102)
The two terms on the second line can be rewritten as follows
k
k
k
k
k
k
k
k
k
ex
,
(103)
where we have used (100). Since the interaction is weak, we do not expect
significant change of the momenta of any two interacting particles, and so we may
25
safely put Uk = U0 everywhere in the formulae below. Finally, we have the
Hamiltonian of a weakly interacting Bose gas:
k
k
k
k
k
k
k
k
,
(104)
where n0 is the number density of condensate particles.
But for the second term in the brackets, we could immediately write
down the energy levels of the system. In other words, the Hamiltonian of a weakly
interacting Bose gas is not diagonal in the basis of eigenkets of the single-particle
momentum operator. This is similar to the problem of vibrations in a solid. There,
interactions between the nearest neighbours do not allow the classical
Hamiltonian to be written down as a sum of separate Hamiltonians. If, however,
we perform a coordinate transformation and go over to normal coordinates, the
Hamiltonian breaks down into a sum of Hamiltonians, each representing a simple
harmonic oscillator quite independent of the others. Quantisation of the system
then leads to
vib
k s
k s
k s
k s
k s
k s
k s
,
(105)
where
k s
and
k s
are creation and annihilation operators for phonons with
momentum k and polarisation s. One says that the Hamiltonian (105) is diagonal,
which means that it has non-zero matrix elements between the states with the same
number of phonons of each kind. Now recall that a phonon does not represent a
vibration of a particular atom, but rather a particular mode of vibration. By going
over to normal coordinates we have gone over from a system of interacting atoms
to a collection of independent particles that we call phonons.
2.4. Bogoliubov transformations
So the programme is now to find the basis in which the Hamiltonian
(104) is diagonal. An elegant way is to perform the Bogoliubov transformation to
new creation and annihilation operators. Since our Hamiltonian is bilinear in the
original operators, the Bogoliubov transformation must be linear:
kkkk
k
,
k
k
k
kk
(106)
We require that the functions uk and vk be real, and depend only on the
absolute value of the particle's momentum. Since we are dealing with bosons, the
other requirement is that the new operators kand
k
should satisfy the
commutation relations for boson operators, analogous to (29). This places a
constraint on these functions:
26
,
(107)
which allows us to invert (106) and obtain
kk
k
,
k
k
k
(108)
Putting (108) into (104) produces the Hamiltonian in terms of the new
operators:
k
k
k
k
k
k
k
k
k
k
kk
k
k
(109)
Since (107) is the only constraint placed on the functions uk and vk , we can
require that the second sum in the Hamiltonian (109) should be zero to diagonalise
our Hamiltonian. It is left as an exercise to the reader to prove that the result is
,
,
(110)
where
,
(111)
and the Hamiltonian becomes
k
k
k
k
(112)
Thus, the quantity Ek defined above is the energy of an elementary
excitation in the weakly interacting Bose gas, with two limiting cases:
,
,
,
,
(113)
where
Physically, this means that for low momenta, when the
kinetic energy of a particle is much smaller when the “potential energy” n0U0, the
excitations are sound waves, or phonons. In the opposite case the excitations
behave like free particles. Combining these results with the condition (79), we see
that the critical velocity is finite for a weakly interacting Bose gas.
Expressed in the new γ-operators, the number-of-particles takes the
form
k
k
k
k
k
k
k
kk
(114)
27
If we compute the average value of this operator with respect to an arbitrary
state of our gas, the first term will produce the number of particles in the
condensate; the third the number of thermal excitations; the fourth will produce
zero because the creation and annihilation operators do not balance each other in
each term. How are we to interpret the second term? The only thing that has not
yet been taken into account are non-thermal excitations caused by interactions and
present even at zero temperature. These will be produced by the second term. The
calculation that is left as an exercise to the reader produces
ex
,
(115)
where we have used the definition of the scattering length (93). The gas is
weakly interacting if the number of zero-temperature excitations is much smaller
than the total number of particles, so the criterion of weak interaction is
(116)
Physically, this means that within the range of the interaction potential there
are barely any particle.
2.5. The two-fluid picture
As we have seen, the total particle density below the critical temperature
can be written down as a sum of two contributions: one coming from the
condensed (“superfluid”) particles and the other coming from excited (“normal”)
particles:
exsn
(117)
We expect that the total particle current can be represented as a sum of the
current carried by superfluid particles (“supercurrent”) and the current carried by
normal particles (“normal” current):
J JsJn
svsnvn
,
(118)
where we have introduced the mass densities and defined velocities of the
normal and superfluid components. Experiment demonstrates the existence of the
critical velocity of the superfluid flow above which the flow becomes normal.
Therefore we should expect the normal particle density to depend on the velocity
of the flow. Actually, the normal particle density must be a function of the relative
velocity of the two components because the theory must be invariant under
Galilean transformations.
A Bose gas in its ground state at rest carries no current, since by
definition the ground state is a zero-momentum state. If now we add excitations
that are distributed according to a certain equilibrium distribution function fp,
28
there will be a non-zero current
J
ex
p
p
p
p
pp
(119)
If we go to a frame of reference moving with velocity vs, i.e. if we
ride with the condensate, the expression (118) will transform to
J Jnnvnvsnu,u vnv
s
,
(120)
which is the normal current, and where we have introduces the velocity of
the normal component with respect to the superfluid. This current is the same as
(119) provided that the distribution function depends on the energy of an
excitation referred to the frame of reference where excitations as a whole are at
rest. If Ep is the energy of an excitation referred to the stationary condensate (as
defined by (111)), then Ep – p·u is the energy of an excitation in the frame of
reference where the excitations as a whole are at rest. Thus, the distribution
function is
p
p u
,
(121)
and the normal current is equal to
J
n
p
p
p u
(122)
The momentum of an excitation can be represented as a sum of the
component along u and the component perpendicular to it:
p
uu p
u u p
(123)
The perpendicular component vanishes upon integration over the angles
and so we have
J
n
p
uu p
u p
nuu,
(124)
in agreement with (120). We have defined the velocity dependent normal
particle density
nu
p
u p
u p
(125)
Let the relative velocity u be sufficiently small so that we can expand
the distribution function in p·u/T and keep only the linear term:
p
u p
u p
(126)
Since the distribution function f 0 depends on the absolute value of the
excitation momentum, only the second term in (126) will contribute to the normal
particle density:
29
nu
p
u p
,
(127)
where we have performed integration over angles and have extended the
upper limit of the integration with respect to the excitation momentum to infinity
in view of the rapidly decreasing exponential function. As can be seen from (127),
at sufficiently small velocities the normal particle density does not depend on the
relative velocities of the two components.
It is left as an exercise to the reader to show that at low temperatures,
when excitations are phonon-like and Ep = sp, the normal particle density is
n
,
;
(128)
while at high temperatures, when excitations are particle-like with the
dispersion law Ep = p2/(2m), the normal particle density is
n
,
(129)
Exercises
1. Determine the low-temperature behaviour of an ideal one-dimensional
Bose gas.
2. Find the temperature dependence of the chemical potential of an ideal
two-dimensional Bose gas.
3. For an ideal three-dimensional Bose gas show that:
4. Find the Green function of Eqn (82). Proceed as follows:
a) make the Fourier-transform of Eqn (83) noting that the Green function
only depends on the difference x – x'; you should obtain
b) make the inverse Fourier-transform; since the G(q) only depends on
the absolute value of q, you can perform integration over angles to arrive at
xx
x x
x x
The further integration can be performed with the use of the residue
theorem. Note, however, that the final result will depend on the way you choose
to go round the poles.
5. Show that if one requires that the new operators kand
k
that appear in
the Bogoliubov transformation should satisfy the commutation relations for boson
30
operators, then one obtains a condition:
6. Show that the number of non-thermal excitations caused by interactions
and present even at zero temperature is
ex
7. Prove that at low temperatures, when the excitations are phonon-like and
Ep = sp, the normal particle density is
n
,
;
while at higher temperatures, when the excitations are particle-like with the
dispersion Ep = p2/(2m), law the normal particle density is
n
,
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