Quantum statistics and superfluids

Daniele Dominici

1

Contents

1 Review of Statistical Mechanics 31.1 Microcanonical Ensemble . . . . . . . . . . . . . . . . . . . . . 31.2 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . 51.3 Gran Canonical Ensemble . . . . . . . . . . . . . . . . . . . . 5

2 Quantum statistical mechanics 62.1 Microcanonical Ensemble . . . . . . . . . . . . . . . . . . . . 72.2 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . 82.3 Gran Canonical Ensemble . . . . . . . . . . . . . . . . . . . . 9

3 Bose Einstein and Fermi Dirac statistics 9

4 A gas of free fermions 13

5 A gas of free bosons 16

6 Introduction to superfluids 18

7 Bose Einstein condensation for an ideal gas 21

8 The Schrodinger field 24

9 Ginzburg Landau Model 269.1 Fundamental state of the theory . . . . . . . . . . . . . . . . . 29

A Bogoliubov transformation 31

2

1 Review of Statistical Mechanics

Let us start by briefly reviewing the classical statistical ensembles, see forinstance [1, 2]. The statistical mechanics is the branch of physics that stud-ies the properties of matter in equilibrium with the aim of obtaining themacroscopic properties of a system starting from the microscopic laws.

Let us consider a system given by N particles in a volume V . Let ussuppose the system as isolated so that the energy is constant. The state ofthe system is characterized by the 3N coordinates q1, · · · , qN and the corre-sponding momenta p1, · · · , pN satisfying the equations of motion

qi =∂H∂qi

, pi = −∂H∂pi

(1.1)

For large N it is impossible to solve the problem; SM searches to obtain themacroscopic properties and laws of the system.

Since we are considering a system which is isolated (or weakly interactingwith the environment) we will ask for the energy

E < H < E +∆ (1.2)

Every state corresponds to a point of the phase space and in general therewill be an infinite set of states satisfying (1.2). Therefore it is convenientto imagine to have an infinite set of system, each in a different microscopicstate but corresponding to the same macroscopic condition (1.2). In otherwords we represent all these systems as a cloud of points distributed in thephase space with a given density:

ρ(q, p)d3Nqd3Np (1.3)

1.1 Microcanonical Ensemble

The energy is constant since the system is isolated.

The postulate of equal probability:

ρ(q, p) = 1 if E < H < E +∆ (1.4)

3

ρ(q, p) = 0 otherwise (1.5)

If f is an observable one can compute the average as

< f >=

∫d3Nqd3Npf(q, p)ρ(q, p)∫

d3Nqd3Npρ(q, p)(1.6)

Let us then define the volume of the phase space as

Γ(E) =

∫E<H<E+∆

d3Nqd3N (1.7)

and the entropy asS = k log Γ(E) = S(V,E) (1.8)

where k is the Boltzmann constant. From (1.8) we get

1

T=∂S

∂Ep = T

∂S

∂V(1.9)

or

dS =1

T(dE + pdV ) (1.10)

By inverting (1.8) we can recover the internal energy

E(S, V ) ≡ U(S, V ) (1.11)

from whichdE = TdS − pdV (1.12)

Let us recall also the Helmholtz free energy, defined as

F = U − TS (1.13)

By differentiating, we get

dF = dE − dTS − TdS = −pdV − SdT (1.14)

Let us finally recall the Gibbs potential

G = F + pV (1.15)

dG = dF + pdV + dpV = −SdT + V dp (1.16)

4

When the number of particles is not constant one defines the thermodynamicpotential as

Ω− F − µN (1.17)

where µ is the chemical potential. Furthermore

dE = TdS − pdV − µdN (1.18)

Therefore

dΩ = dE − dTS − TdS − dµN − µdN = −SdT − pdV −Ndµ (1.19)

and so

S = −∂Ω∂T

, p = −∂Ω∂V

, S = −∂Ω∂µ

, (1.20)

1.2 Canonical Ensemble

To describe a non isolated system one make use of the canonical ensemble.The partition function is defined as

Z(V, T ) =1

N !h3N

∫d3Nqd3Np exp [−βH(q, p)] ≡ exp [−βF ] (1.21)

where F is the Helmholtz free energy and β = 1kT

k being the Boltzmannconstant.

p = −∂F∂V

S = −∂F∂T

(1.22)

If the number of particles is not constant then one define

1.3 Gran Canonical Ensemble

The gran canonical ensemble with partition function

Z(µ, V, T ) =∑N

1

N !h3N

∫d3Nqd3Np exp [−βH(q, p) + βµN ] (1.23)

with µ the chemical potential.

5

When passing from classical physics to Quantum Mechanics it is impos-sible to measure simultaneously q and p and therefore define ρ(q, p). It wasshown by Von Neumann (1927-30) that it is possible to substitute the classi-cal density with the matrix density. When one considers a macroscopic stateas in the classical case several microscopic states correspond to the samemacroscopic state. Therefore one considers a collection of systems identi-cal to the given one, by supposing that they exhaust all the wave functionscompatible with the macroscopic system. In the case of Statistical QuantumMechanics the macroscopic state is the result of two averages: the quantumaverage coming from the probabilistic interpretation of the wave functionand the statistical average coming from the impossibility of exactly knowingthe system.

2 Quantum statistical mechanics

In QM the wave function of an isolated system is given by

ψ =∑n

cnφn (2.1)

where φn are the eigenfunctions of the stationary Schrodinger equation

Hφn = Enφn (2.2)

and cn = cn(t). If the system is not isolated the ψ will depend also on thecoordinates of the external world. Therefor the cn will depend also on thecoordinates of the external world. The average value of an observable O isgiven by

< ψ|O|ψ >< ψ|ψ >

=

∑n,m c

∗ncm < φn|O|φm >∑

n |cn|2(2.3)

In general the product c∗ncm depends on the time and when performing alaboratory measurement one averages over a time interval:

< ψ|O|ψ >< ψ|ψ >

=

∑n,m c

∗ncm < φn|O|φm >∑

n |cn|2=

∑n,m c

∗ncm < φn|O|φm >∑

n |cn|2(2.4)

where the last identity comes from the fact that the total Hamiltonian ishermitian.

The postulates of Statistical Quantum Mechanics are postulates over thecoefficients < c∗ncm >.

6

2.1 Microcanonical Ensemble

Postulate: Equal probability

c∗ncm = 1, if E < En < e+∆

= 0, otherwise (2.5)

Postulate of the casual phases

c∗ncm = 0, if n 6= m (2.6)

Therefore all this is equivalent to postulate the wave function

ψ =∑n

bnφn (2.7)

with

|bn|2 = 1, if E < En < E +∆

= 0 otherwise (2.8)

In this way one takes into account of the external world by neglecting theinterference. Let us introduce the matrix ρ with matrix elements

ρnm =< φn|ρ|φm >= δnm|bn|2 (2.9)

The average over the statistical ensemble of a operator O is given by

< O >=

∑n < φn|Oρ|φm >∑n < φn|ρ|φm >

=trρO

trρ(2.10)

We can write also ρ as

ρ =∑

E<En<En+∆

|φn > |bn|2 < φn| (2.11)

Therefore (assuming |bn|2 = 1)

trρ =∑

(number ofstates with E<En<E+∆)

≡ Γ(E) (2.12)

From Γ(E, V ) one obtains the entropy as

S(E, V ) = k log Γ(E, V ) (2.13)

7

2.2 Canonical Ensemble

The ρ matrix is defined as

ρnm = δnm exp(−βEn) (2.14)

orρ =

∑n

|φn > exp(−βEn) < φn| (2.15)

and the partition function Z(V, T ) as

Z(V, T ) = trρ =∑n

exp(−βEn) (2.16)

where the sum is over all the states. Let us then define the free energy

F = − 1

βlnZ (2.17)

The average over the ensemble of an operator O is obtained as

< O >=1

Ztr(O exp(−βH)) =

tr(Oρ)

trρ=

∑nOn exp(−βEn)∑

n exp(−βEn)(2.18)

One can show that F is the free energy by taking the derivative withrespect to T :

∂F

∂T= −k logZ − kT

∂β

∂T

d

∂βlogZ

=F

T+

1

TZ

∑n

(−En) exp(−βEn)

=F

T+

1

T(−E) = S (2.19)

where E is now the average value over the ensemble.

8

2.3 Gran Canonical Ensemble

In this case the number of particles N is not constant and one defines thegran partition function

Z(µ, V, T ) =∑N

∑n

exp(−βEn + βNµ) =∑N

tr exp(−βH + βµN) (2.20)

The internal thermodynamic potential is obtained as

Ω = −kT logZ(µ, V, T ) (2.21)

Let us check that eq.(2.21) defines the thermodynamic potential. In fact

∂Ω

∂T= −k logZ − kT

∂β

∂T

d

∂βlogZ

=Ω

T+

1

TZ∑N,n

(−En +Nµ) exp(−βEn + βNµ)

=Ω

T+

1

T(−E +Nµ) (2.22)

where E and N are now the average values over the ensemble. Therefore theidentification of Ω with the thermodynamic potential is correct:

∂Ω

∂T= −S =

Ω

T+

1

T(−E +Nµ) (2.23)

orΩ = E − TS − µN (2.24)

3 Bose Einstein and Fermi Dirac statistics

Let us now consider the aspect of indistiguishability aspect of quantum me-chanical particles. We know from Quantum Mechanics that there are twotypes of particles, bosons and fermions. Single states can be occupied by anynumber of bosons while for fermions a single state can be occupied at mostby one fermion.

9

Since atoms are composed of spin 1/2 particles (neutrons, protons andelectrons) there are atoms which are bosons (H1, He4) and atoms which arefermions (H2, He3). Let us now compute the gran partition function for freebosons and fermions.

Let H be the Hamiltonian

H =N∑i=1

~p2i2m

(3.1)

Let us suppose that for every momentum ~p there are n~p particles withsuch momentum. Since we are working in a box, ~p = h/L~m with ~m =(mx,my,mz) integers. Therefore we have

E =∑n~p

n~pε~p ≡ E(n~p) N =∑~p

n~p (3.2)

The partition function is given by

Z(µ, V, T ) =∑N

∑n~p

g(n~p) exp(−βE(n~p) + βµN) (3.3)

with g(n~p) = 1 since all particles are identical. Fer fermions n~p = 0, 1 whilefor bosons n~p = 0, 1, 2, · · ·. So we have

Z(µ, V, T ) =∞∑

N=0

∑n~p

g(n~p) exp(−βE(n~p) + βµN)

=∞∑

N=0

∑n~p

exp(−βE(n~p) + βµN)

=∞∑

N=0

∑n~p

exp[−β∑~p

(n~pε~p − µn~p)]

=∞∑

N=0

∑n~p

Π~p [exp(β(µ− ε~p))]n~p

=∑n0

∑n1

[exp(β(µ− ε0))]n0 [exp(β(µ− ε1))]

n1 · · ·

= Π~p

∑n~p

[exp(β(µ− ε~p))]n~p

= Π~pZ~p (3.4)

10

Let us now consider a gas of fermions, then

ZF~p =

∑n~p=0,1

[exp(β(µ− ε~p))]n~p = 1 + exp[β(µ− ε~p)] (3.5)

For a boson gas we have

ZB~p =

∑n~p=0,1,2,...

[exp(β(µ− ε~p))]n~p =

1

1− exp[β(µ− ε~p)](3.6)

Note that in the boson case the series converges only if

exp β(µ− ε~p) < 1 (3.7)

Therefore if the ground level is for ε0 = 0 the chemical potential must benegative. Finally we can calculate the thermodynamic potential:

ΩF = −kT∑~p

ln[1 + exp(β(µ− ε~p))]

ΩB = kT∑~p

ln[1− exp(β(µ− ε~p))] (3.8)

Given the energy ε~p we can calculate all the thermodynamic quantities. Letus first compute the average number:

NF = −∂ΩF

∂µ≡∑~p

< n~p >=∑~p

1

exp(β(ε~p − µ)) + 1

NB = −∂ΩB

∂µ≡∑~p

< n~p >=∑~p

1

exp(β(ε~p − µ))− 1(3.9)

At low temperature bosons tend to accumulate in the ground state (~p = 0),only thermal fluctuations can invert the process. In fact NB increases whenε~p − µ→ 0.

The classical limit, the Boltzmann distribution, is obtained for exp(β(ε~p−µ)) >> 1 or exp(β(µ − ε~p)) << 1. This corresponds to exp(µ/kT ) <<exp(ε~p/kT ) or large T and µ/kT → −∞.

Let us now compute the fermion partition function ΩF for a free particlegas by going in the continuum (V → ∞):

ΩF = −kT 4πV

h3

∫ ∞

0

p2dp ln[1 + exp((β(µ− p2

2m))] (3.10)

11

We can now derive average pressure and number of fermions as

p = −∂ΩF

∂V= kT

4π

h3

∫ ∞

0

p2dp ln[1 + exp(β(µ− p2

2m))] (3.11)

N =4πV

h3

∫ ∞

0

p2dp1

1 + exp(β(−µ+ p2

2m))

(3.12)

All the results can be expressed in terms of the functions

f3/2(z) = zd

dzf5/2(z) =

∞∑l=1

(−)l+1 zl

l3/2

f5/2(z) =4√π

∫ ∞

0

dxx2 ln(1 + z exp(−x2)) =∞∑l=1

(−)l+1 zl

l5/2(3.13)

where z = exp (βµ):

p

kT=

1

λ3f5/2(z)

N

V=

1

λ3f3/2(z) (3.14)

where

λ =

√2π~2mkT

(3.15)

For a Bose gas the partition function is singular when ~p = 0 and µ → 0or z → 1. Therefore it is convenient to separate in the sum the term with~p = 0 before passing to the continuum.

We get

p = −kT 4π

h3

∫ ∞

0

p2dp ln[1− exp(β(µ− p2

2m))]− kT

Vln[1− exp(βµ))] (3.16)

N = V4π

h3

∫ ∞

0

p2dp1

−1 + exp(β(−µ+ p2

2m))

+N0 (3.17)

where

N0 =exp(βµ)

1− exp(βµ)≡ z

1− z(3.18)

12

N0 denotes the number of particles in the ~p = 0 state.

The results can now be written in terms of the functions

g3/2(z) = zd

dzg5/2(z) =

∞∑l=1

zl

l3/2

g5/2(z) = − 4√π

∫ ∞

0

dxx2 ln(1− z exp(−x2)) =∞∑l=1

zl

l5/2(3.19)

We have

p

kT=

1

λ3g5/2(z)−

1

Vln(1− z)

N

V=

1

λ3g3/2(z) +

N0

V(3.20)

4 A gas of free fermions

Let us now rewrite the equations (3.14) for the pressure and concentrationfor a gas of fermions:

p

kT=

1

λ3f5/2(z)

N

V=

1

λ3f3/2(z) (4.1)

The equation of state is obtained by eliminating z from the two equations.Let us start with the equation

λ3

v= f3/2(z) (4.2)

Therefore it is convenient to study the function f3/2 as a function of z. f3/2is a function monotone in z. For small z

f3/2(z) = z − z2

23/2+

z3

33/2− z4

43/2+ · · · (4.3)

For large z (Huang p.246)

f3/2(z) =4

3√π[(ln z)3/2 +

π2

8

1

(ln z)1/2] +O(1/z) (4.4)

13

Therefore for every positive value of λ a solution for z exists.

Low density and high temperature, λ3/v << 1

In this case the thermic lenght λ ∼ ~/p ∼ ~/√mKT is much smaller than

the average distance among the particles v1/3, therefore quantum effects arenegligible. From

λ3

v= z − z2

23/2+ . . . (4.5)

one gets

z =λ3

v+

1

23/2(λ3

v)2 + . . .) (4.6)

and the equation of state becomes

pV

kTN=

v

λ3(z − z2

25/2+ . . . = 1 +

1

25/2λ3

v+ . . . (4.7)

Therefore one obtains quantum corrections to the classic case. Recalling thevirial expansion

p

kT=N

V(1 +B(T )

N

V+ C(T )

(N

V

)2

+ . . .) (4.8)

one can identify the virial coefficients.

High density and low temperature, λ3/v >> 1

In this case the thermal distance is much larger than the average distanceso the quantum effects become relevant. The leading term is now

λ3

v=

4

3√π(ln z)3/2 (4.9)

Therefore we getz = exp (βεF ) (4.10)

where the chemical potential εF is called Fermi energy

εF =~2

2m

[6π2

v

]2/3(4.11)

14

Let us now consider < n~p >, defined in eqs.(3.9)

< n~p >=1

eβ(ε~p−εF ) + 1(4.12)

When T → 0, β → +∞ we have

< n~p >T=0= 1 (4.13)

for ε~p < εF and< n~p >T=0= 0 (4.14)

for ε~p > εF .

Therefore at zero temperature the fermions occupy all the lowest levels upto εF . Because of the Pauli principle they cannot occupy all the ground stateand therefore they fill all the states up to the highest energy εF . Such a stateis called a degenerate Fermi gas. In the momentum space the particles fill asphere of radius pF , the Fermi surface. One defines also a Fermi temperatureor degeneracy temperature TF such that

kTF = εF (4.15)

Finally we can compute the internal energy

U =∑~p

ε~pn~p =V

h34π

2m

∫ ∞

0

dpp4 < n~p > (4.16)

By part integration we get

U =V

4π2m~3

∫ ∞

0

dpp5

5

(− ∂

∂pn~p

)=

βV

20π2m2~3

∫ ∞

0

dpp6eβε~p−βµ

(eβε~p−βµ + 1)2

(4.17)

The integrand has a peak at p = pF . The asymptotic behavior of theintegral is (see Huang)

U =3

5NεF

[1 +

5

12π2(

kT

εF)2]

(4.18)

From the internal energy one can derive the specific heat

CV = Nkπ2

2

kT

εF(4.19)

15

which goes to zero when T → 0 (Third law of thermodynamics) and thepressure

p =2

3

U

V=

2

5

εFv

[1 +

5

12π2(

kT

εF)2]

(4.20)

Notice that even at T = 0 as a consequence of Pauli principle the gas hasa non vanishing pressure. This pressure is responsible for the gravitationalstability of white dwarfs and neutron stars. A white dwarf can be thought as agas of ionized helium and electrons. The gravitational stability is guaranteedby the degenerate electron pressure. In the case of neutron stars the stabilityis guaranteed by the pressure of the degenerate gas of neutrons.

Note Alternative way to compute εF .

An alternative way of computing εF is to fill all the states in the momen-tum space up to pF :

N =4πV

h3

∫ pF

0

p2dp =4πV

3h3p3F =

V

6π2~3p3F (4.21)

or

pF = (N

V)1/3(6π2)1/3~ (4.22)

εF =1

2m(N

V)2/332/3h2 =

~2

2m

[6π2

v

]2/3(4.23)

5 A gas of free bosons

Let us now study with some detail the Bose case. The function g3/2 for zsmall can be studied as a series

g3/2(z) = z +1

23/2z2 +

1

33/2z3 + · · · (5.1)

with

g3/2(1) =∞∑l=1

1

l3/2= ζ(

3

2) = 2.612... (5.2)

16

where ζ is the Riemann function. As we have already noticed for a boson gasµ < 0 then 0 ≤ z ≤ 1 and g3/2 ≤ 1. Rewriting the eq.(3.20) for the averagenumber

λ3N0

V=λ3

v− g3/2(z) (5.3)

with

v =V

N(5.4)

we see that N0/V > 0 if temperature and specific volume v are such that

λ3

v> g3/2(1) (5.5)

In fact

0 <λ3

v− g3/2(1) <

λ3

v− g3/2(z) (5.6)

This means that the ground state is occupied by a macroscopic fraction ofbosons. The critical temperature for the Bose condensation is defined by

λ3cv

=1

v

(2π~2

mkTc

)3/2

= g3/2(1), or µ = 0, N0 = 0 (5.7)

or

Tc =1

k

2π~2/m[vg3/2(1)]2/3

=1

k

2π~2/m[ζ(3/2)]2/3

(N

V)2/3 (5.8)

At the critical temperature the bosons start to occupy the ~p = 0 state and ifthe temperature decreases more and more bosons occupy such a state. ForT < Tc the chemical potential µ remain zero. Inserting the values ρHe4 =0.145g/cm3 ∼ mHe4N/V with mHe4 = 4mp, mp = 4 × 1.67 × 10−27Kg,~ = 1.05510−34 J s, the Boltzmann constant k = 1.38 10−23J/K) we getTc ∼ 3.14 K. This temperature is very close to the critical temperature ofliquid Helium, Tλ = 2.17 K, below which the helium becomes superfluid.

Note Let us briefly discuss the classical limit. Using the expansion (3.13)in eq.(3.14) or (5.1) in eq.(3.20) neglecting N0 we obtain

N

V=

1

λ3z (5.9)

from which we get

µ = kT lnλ3N

V(5.10)

17

Therefore the classical limit corresponds to

λ3N

V=N

V

(2π~2

mkT

)3/2

<< 1 (5.11)

Notice that in the Bose case the expression for µ given in eq.(5.10) cannotextrapolate to T → 0. In fact decreasing T µ becomes positive and thendiverges.

6 Introduction to superfluids

There are two stable isotopes of the helium. The first He4 was discoveredi n 1871 in solar spectrum, while the He3 was discovered in 1933. The He4

gas at low temperatures, close to the absolute zero, and ordinary pressures(atmosphere) remain liquid and solidifies at 25 atmosferes, as shown in Fig. 1.

The isotope He4, which has a nucleus composed by two protons and twoneutrons, is a boson (spin 0) while He3 composed by two protons and oneneutron, is a fermion (spin 1/2). Both liquids have, at low temperatures, lowdensities

ρ3 = 0.081g/cm3, ρ4 = 0.145g/cm3 (6.1)

The two liquids behave in different way because the Pauli principle keepsHe3 fermions far each other. The He4, below Tλ =2.17 K enters in a newphase, HeII, Fig.1. The phase transition, of the second order, is signalled bya peak in the specific heat (Fig. 2).

The transition line λ is the separation between HeI and HeII liquid, thefirst normal, the second superfluid. In this phase the liquid is capable offlowing through narrow capillaries without friction. Experiments to provesuperfluidity where performed first by Kapitza (1938) and Allen and Misener(1938).

The He3 becomes also superfluid but only at temperatures of the order10−3K (1972, Osheroff, et al): in this case fermions condensate. It is there-fore natural to associate the λ transition to the Bose Einstein condensation(London 1938) modified by the molecular interactions. In fact as we haveseen the critical temperature for the condensation in a non interacting gas of

18

Figure 1: Phase diagram of superfluid He-4 (from J.C.Davis Group, Cornell)

Figure 2: Specific heat (from Huang)

19

bosons is 3.14 K, very close to Tλ. In a boson gas below Tc a fraction N0/Nof bosons condense in the state p = 0.

He2 solidifies at higher temperature because of stronger interactions amonghis molecules (Huang p.317).

At low temperature (T << Tc) the specific heat varies as T3, as shownin Fig. 6. To explain such a behavior Landau (1941) suggested to interpretthe quantum states of the liquid as a phonon gas with the linear dispersionrelation

εk = ~ck (6.2)

where c is the velocity of propagation of the sound. The main idea is that thebody moving in the liquid excites sound waves which are collective motionsin the liquid. Furthermore the liquid at low temperature must be treated as aquantum liquid. Therefore the excitations are treated as a gas of phonons asthe vibrations of a crystal. A body moving in the helium at low temperaturedo not loose energy transferring to single atoms but excites collective quantaas phonons.

The helium dispersion relation curve is measured by neutron scattering,see Fig. 6. Approximately one has

εk = ~ck k << k0 (6.3)

with c = (239± 5) m/s

εk = ∆+~2(k − k0)

2

2σk ∼ k0 (6.4)

and with ∆/kB = (8.65 ± 0.04) K, with kB the Boltzmann constant, k0 =(1.92±0.01)108 cm−1, σ/m = 0.16±0.01 withm the mass of the helium atom.Therefore the dispersion relation is linear for small k and has a minimum atk = k0.

The nature of the excitations in the helium is studied by measuring thechange in energy and momentum in the scattering and using the conservationlaws

1

2m(~p2i − ~p2f ) = ε(k) (6.5)

~pi − ~pf = ~k (6.6)

20

where pi(f) are the momenta of the incoming (outcoming) neutron.

Let us now see how it is possible that an object can move in a superfluidwithout loosing energy. Let us consider an object of mass M moving in asuperfluid. The only way in which it can loose energy is by emission of aphonon

~P 2

2M− (~P − ~~k)2

2M= −~2~k2

2M+~P · ~k~M

= εk (6.7)

Therefore

(~V · ~k)~ = εk +~2k2

2M≥ c~k (6.8)

or~V · ~k ≥ ck, V cos θ ≥ c (6.9)

Therefore if V ≤ c the process is prohibited. The reasoning depends in anessential way from the linear spectrum of the phonons.

At energies around k0 the object looses energy by emitting the so-calledrotons:

(~V · ~k0)~ = εk0 +~2k2

2M≥ εk0 (6.10)

orV k0~ cos θ ≥ εk0 = ∆ (6.11)

orV ≥ vc/ cos θ (6.12)

with vc = ∆/(~k0) = 8.65 1.38 10−23J/(1.055 10−34Js1.92× 10−8 cm) ∼ 58m/s. The rotons are the elementary excitations associated to vortex in thefluid. At low temperature the roton effects are negligible. At the thermalequilibrium elementary excitations have energies close to the minimum of εthat is zero. In presence of a purely phono spectrum the critical velocity isc = 239 m/sec, when rotons are included the critical velocity drops to vc = 58m/sec (observed in He4 under pressure).

7 Bose Einstein condensation for an ideal gas

Let us now recall the average number of an ideal gas of N non interactingbosons, derived by using the grand partition function, and given by eq.(3.17):

N = N0 +V

h3

∫d3p

1

exp [β(ε~p − µ)]− 1(7.1)

21

Figure 3:

22

At T = 0 all bosons occupy the state with p = 0 (Bose condensation). Atfinite temperature, T 6= 0, only a fraction N0/N of bosons remain in thestate p = 0. For T >> Tc, where Tc is the critical temperature, there is nocondensate, N0 = 0, n = N/V requiring µ < 0 because of the singularity forµ > 0. When T decreases, for fixed N/V , the absolute value of the chemicalpotential decreases until for temperatures sufficiently low reaches the valueof 0. The condensation starts when µ = 0

N0

N= 0 µ = 0 (7.2)

or

N = 0 +V

h3

∫d3p

1

exp[ε~pkTc

]− 1

=V

λ3cg3/2(1) (7.3)

For T < Tc µ remains zero and using eq. (7.1), with µ = 0 we get

N −N0

V=

∫d3p

h31

exp [ε~p/kT ]− 1

=1

λ3g3/2(1) =

(mkT

2π~2

)3/2

g3/2(1) =N

V

(T

Tc

)3/2

(7.4)

orN −N0

N=

(T

Tc

)3/2

(7.5)

orN0

N= 1−

(T

Tc

)3/2

(7.6)

andN0

V=N

V

(1− T

Tc

)3/2

T < Tc (7.7)

In conclusion below Tc a fraction of particles occupy the state with p = 0.Therefore for T < Tc we have a condensate with a macroscopic number ofparticles in the same quantum state with p = 0. Bose Einstein condensationprovides only a qualitative description of superfluidity. For example thespecific heat of a non interacting boson gas vanishes as T 3/2 while the specificheat of He4 behaves as T 3. Furthermore we have superfluidity only for zerovelocity of the atoms.

23

8 The Schrodinger field

Let us now see how to reformulate the non relativistic Schrodinger theory inthe second quantization formalism, or in the quantum field theory language.Let us consider a non relativistic particle with zero spin which satisfies theSchrodinger equation:

i~∂

∂tψ = Hψ (8.1)

con H = ~p2/2m. Quantum states can be built by working in the Fock space:by postulating the existence of creation and distruction operators a†~k e a~kwhich satisfy the commutation relations

[a~k, a~k′ ] = [a†~k, a†~k′] = 0 [a~k, a

†~k′] = δ~k,~k′ (8.2)

In general the one particle state is given by

|ψ >=∑~k

ψ~ka†~k|0 > (8.3)

In this basis the wave function is given by

ψ(x) =< x|ψ >=∑~k

ψ~k < x|a†~k|0 >==∑~k

ψ~k

1√Vei~k·~x (8.4)

We can now consider the following field

Φ(~x, 0) =1√V

∑~k

ei~k·~xa~k (8.5)

such that

|ψ >=∫d3xψ(x)Φ†(~x, 0)|0 > (8.6)

In general

Φ(~x, t) =1√V

∑~k

ei(~k·~x−ω~k

t)a~k (8.7)

with ω~k = ~~k2/2m. The field Φ(~x, t) satisfies the Schrodinger equation

i~∂

∂tΦ = −~2

∇2

2mΦ (8.8)

24

Let us now write the Lagrangian associated to the the Schrodinger equation

L = i~Φ†Φ− ~21

2m∇Φ† · ∇Φ (8.9)

and the corresponding Hamiltonian density

H = ΠΦ− L = ~21

2m∇Φ† · ∇Φ (8.10)

with Π = ∂L/∂Φ = i~Φ†. The commutation relations (8.2) imply

[Φ(~x, t),Π(~y, t)] = i~δ(~x− ~y) (8.11)

The Hamiltonian of the field is given by

H =

∫d3xH =

∑~k

~ω~ka†~ka~k (8.12)

As an application of the non relativistic field theory we will consider thesuperfluidity theory.

Bogoliubov (1947) studied the fundamental state of a dilute gas of weaklyinteracting bosons and their excitations using the second quantization of amany body system and assuming the following interaction Hamiltonian, see[3, 4]:

HI =1

2

∑~k1+~k2=~k′1+

~k′2

W (|~k1 − ~k′1|)a†(~k′1)a†(~k′2)a(~k1)a(~k2) (8.13)

and W (k) is the Fourier transform of the interaction

W (k) =

∫d~rW (r)ei

~k·~r (8.14)

In the following we follow instead the use of the effective Hamiltonian ap-proach to phase transition of Ginzburg Landau.

25

9 Ginzburg Landau Model

The Hamiltonian del model of Ginzburg Landau, which was proposed as aneffective description of field theory for phase transitions is given by

Heff =~2

2m(∇φ)†(∇φ)− µφ†φ+

1

2g(φ†φ)2 (9.1)

with µ and g positive constants. Therefore the potential can be identified as

V (φ) = −µφ†φ+1

2g(φ†φ)2 (9.2)

and is dominated for low density by the chemical potential and at largedensity by the g term. The form of the potential is shown in Fig. 9.2. Amicroscopic interpretation of the parameters µ and g can be found in [4].They can be related to the strenght of the four boson interaction and to thedensity of the condensate.

Let us now assume that the chemical potential depend on the tempera-ture, or µ < 0 for T > Tc, Tc being the critical temperature, and µ > 0 forT < Tc. The potential has in |φ| = 0 a maximum an a minimum for

φ†φ =µ

g(9.3)

or

φ(x) = φ0 exp (iψ) =

õ

gexp (iψ), (9.4)

The minimum is degenerate varying ψ ∈ [0, 2π). For simplicity let us choosethe minimum at ψ = 0. The series of the field in normal modes can beperformed with respect to the new minimum in φ0

φ(x) = φ0 + φ(x) = φ0 +∑~k 6=0

1

Va~ke

i~k·~x (9.5)

Let us notice that the Hamiltonian (9.1) is invariant under the transformation

φ(x) → φ(x) exp (iα), α ∈ [0, 2π) (9.6)

while the minimum state is not (φ0 → φ0 exp (iα)). When the Hamiltonianis symmetric under a transformation while the state of minimum energy isnot one speaks of Spontaneous symmetry breaking.

26

-2 -1 1 2ÈΦÈ

-0.2

-0.1

0.1

0.2

0.3

0.4

V

Figure 4: Potential V corresponding to eq.(9.2) for µ = 0.4 and g = 0.3 as a

function of |φ| =√φ†φ.

By substituting eq.(9.5) in the potential (9.2) one gets, by expanding tosecond order sviluppando in φ

V = −µ[φ20 + φ0(φ+ φ†) + φ†φ] +

1

2g[φ40 + φ2

0(φ+ φ†)2 + 2φ30(φ+ φ†) + 2φ2

0φ†φ]

+O(φ3, φ4)

= −gφ20

[φ20 + φ0(φ+ φ†) + φ†φ

]+

1

2gφ4

0 +1

2gφ2

0(φ+ φ†)2 + gφ30(φ+ φ†)

+gφ20φ

†φ+O(φ3, φ4)

= −1

2gφ4

0 +1

2gφ2

0(φ+ φ†)2 +O(φ3, φ4)

= −µ2

2g+

1

2µ(φ+ φ†)2 +O(φ3, φ4) (9.7)

Let us now quantize the scalar field φ, by requiring standard commutationrelations a~k e a†~k. By substituting the normal mode series and integrating in

d3x Hamiltonian, one obtains

Heff = −µ2

2gV +

∑~k 6=0

[(µ+

~2~k2

2m

)a†~ka~k +

µ

2(a~ka−~k + a†~ka

†−~k)

](9.8)

The Hamiltonian (9.8) is not diagonal in the basis of occupation numbersbecause of the bilinear terms a e in a†. However it is possible to find a

27

transformation (Bogoliubov transformation, see Appendix A) from a~k(a†~k) to

the operators A~k(A†~k), defined as

A~k = cosh(θk2)a~k + sinh(

θk2)a†

−~k(9.9)

withtanh θk =

µ

µ+ ~2~k22m

(9.10)

One has[A~k, A

†~k′] = δ~k,~k′ (9.11)

FurthermoreHeff = E0 +

∑~k 6=0

ε(k)A†~kA~k (9.12)

with

E0 = −µ2

2gV −

∑~k 6=0

ε(k) sinh2(θk2) (9.13)

and

ε(k) =

√√√√(µ+~2~k22m

)2

− µ2 =

√√√√ µ

m~2~k2 +

(~2~k22m

)2

(9.14)

The fundamental state is defined as

A~k|φ0 >= 0 (9.15)

Starting from this new vacuum state one build the new Fock space with theoperators A†

~k. For example, the first excited state (or quasi-particle) is given

byA†

~k|φ0 > (9.16)

with energy

ε(k) =

√√√√ µ

m~2~k2 +

(~2~k22m

)2

(9.17)

Therefore the spectrum is linear for small k while for large k behaves as k2.

In conclusion the Ginzburg Landau is capable to explain the spectrum ofthe liquid helium at low k but does not reproduce the local minimum due tothe rotons.

28

9.1 Fundamental state of the theory

Let us now discuss the properties of the new vacuum state |φ0 >. The usualform of quantum field theory cannot be used since the fundamental state fora system of N bosons is given by

|φ0(N) >= |N, 0, · · · 0 > (9.18)

that means that all the particles are in the lowest energy state (k = 0).Therefore the annihilation operator does not annihilate the minimum energystate but

a0|φ0(N) >= N1/2|φ0(N − 1) > (9.19)

ea†0|φ0(N) >= (N + 1)1/2|φ0(N + 1) > (9.20)

The minimum energy state is the coherent state

|φ0 >= A1/2 exp[√V φ0a

†0]|0 > (9.21)

which satisfiesa0|φ0 >=

√V φ0|φ0 > (9.22)

anda~k|φ0 >= 0 ~k 6= 0 (9.23)

n0 =< N(k = 0) >

V=

1

V< φ0|a†0a0|φ0 >=

1

VV φ2

0 (9.24)

In other words the expectation value of N is V φ20. The normalization is given

by

A1/2 = exp[−1

4V φ2

0] (9.25)

Therefore n0 is the boson density in the state k = 0. The vacuum expectationvalue |φ0 > of the field φ(x)

< φ0|φ(x)|φ0 >= φ0 =√n0 =

√< N(k = 0) >

V(9.26)

is related to the density of the condensate. The true vacuum state is howeverdefined as

A~k|φ0 >=

[cosh(

θk2)a~k + sinh(

θk2)a†

−~k

]|φ0 >= 0 (9.27)

29

with

|φ0 >= N exp [−1

2

∑k 6=0

tanh(θk/2)a†~ka†−~k]|φ0 > (9.28)

The solution is given by |φ0 > defined by (9.21). This means that the truevacuum state contains pair of bosons with opposite momenta.

Exercise. Verify the at eq. (9.27) is satisfied by the new vacuum (9.28).

30

A Bogoliubov transformation

Let us now derive the Bogoliubov transformation. Let us start considering∑k 6=0

[αa†kak +

µ

2(aka−k + a†ka

†−k)]

(A.1)

where

α = µ+~2k2

2m(A.2)

Let us considerAk = βak + γa†−k, (A.3)

with β, γ ∈ R. Then we get

[Ak, A†k′ ] = [βak + γa†−k, βa

†k′ + γa−k′ ] = (β2 − γ2)δkk′ (A.4)

In order to get standard commutation relations, let us require

β2 − γ2 = 1 (A.5)

It is convenient to define

β = cosh

(θk2

), γ = sinh

(θk2

)(A.6)

The inverse transformations are

ak = βAk − γA†−k, a†k = βA†

k − γA−k (A.7)

In factβAk − γA†

−k = β(βak + γa†−k)− γ(βa†−k + γak) = ak (A.8)

Substituting in eq.(A.1) one obtains∑k 6=0

[αa†kak +

µ

2(aka−k + a†ka

†−k)]

=∑k 6=0

[α(βA†

k − γA−k)(βAk − γA†−k)

+µ

2

((βAk − γA†

−k)(βA−k − γA†k)

+(βA†k − γA−k)(βA

†−k − γAk)

)]=

∑k 6=0

[(β2 + γ2)α− 2βγµ)A†kAk +

(−βγα+µ

2(β2 + γ2))(AkA−k + A†

kA†−k)

+αγ2 − βγµ] (A.9)

31

By requiring the vanishing of the coefficient of AkA−k + A†kA

†−k we get

tanh θk =2βγ

β2 + γ2=µ

α(A.10)

Then the coefficient of A†kAk becomes, using (A.10) and (A.6)

(β2 + γ2)α− 2βγµ = (β2 + γ2)α− 4β2γ2α

β2 + γ2=

(β2 − γ2)2α

β2 + γ2=

α

β2 + γ2

=α

cosh θk= α

√1− tanh2 θk =

√α2 − µ2

≡ ε(k) (A.11)

with ε(k) given by eq.(9.17). Finally

αγ2 − βγµ = αγ2γ2 − β2

β2 + γ2= − αγ2

β2 + γ2= −ε(k) sinh2

(θk2

)(A.12)

where use has been made of eq.(A11).

References

[1] K. Huang, Statistical Mechanics, Wiley, 1987

[2] D. J. Amit and Y. Verbin, Statistical Physics, An introductory course,World Scientific 1999

[3] E.M. Lifshitz andL.P.Pitaevskii, Landau and Lifshitz, Course of Theo-retical Physics, Statistical Physics, part 2, Pergamon Press

[4] A.L. Fetter and J.D. Walecka, Quantum Theory of Many-Particle Sys-tems, McGraw-Hill 1971

[5] J. D. Jackson, Classical Electrodynamics, Wiley 1998

[6] S. J. Chang, Introduction to Quantum Field Theory, World Scientific,Singapore 1990.

[7] F. Mandl and G. Shaw, Quantum Field Theory, John Wiley and sons,1984

32

[8] R. Casalbuoni, Introduction to Quantum Field Theory, World ScientificPublishing, Singapore 2011

[9] H. R. Glyde, Excitations in Liquid and Solid Helium, Clarendon Press,Oxford 1994

[10] F. Gross, Relativistic QuantumMechanics and Field Theory, JohnWileyand sons, 1993

[11] J.J. Sakurai, Advanced Quantum Mechanics, Addison Wesley pub.Company, 1967

[12] C. Cohen-Tannoudji, J. Dupont-Roc, G. Grynberg, Photons & Atoms,Introduction to Quantum Electrodynamics, John Wiley and Sons NewYork 1989

33