Quantum statistics, superfluidity andsuperconductivity

Daniele Dominici

1

Contents

1 Bose Einstein and Fermi Dirac statistics 31.1 A gas of free fermions . . . . . . . . . . . . . . . . . . . . . . . 71.2 A gas of free bosons . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Introduction to superfluids 11

3 Bose Einstein condensation for an ideal gas 15

4 The Schrodinger field 18

5 Ginzburg Landau Model 20

6 Bogoliubov approach 23

7 Superconductivity 257.1 Study of the gap equation . . . . . . . . . . . . . . . . . . . . 317.2 Finite temperature . . . . . . . . . . . . . . . . . . . . . . . . 327.3 The BCS ground state . . . . . . . . . . . . . . . . . . . . . . 35

A Review of Statistical Mechanics 36A.1 Microcanonical Ensemble . . . . . . . . . . . . . . . . . . . . . 36A.2 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . 38A.3 Gran Canonical Ensemble . . . . . . . . . . . . . . . . . . . . 38

B Quantum statistical mechanics 39B.1 Microcanonical Ensemble . . . . . . . . . . . . . . . . . . . . 40B.2 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . 41B.3 Gran Canonical Ensemble . . . . . . . . . . . . . . . . . . . . 42B.4 Fundamental state of the theory . . . . . . . . . . . . . . . . . 42

C Bogoliubov transformation 44

2

1 Bose Einstein and Fermi Dirac statistics

Let us now first review the quantum statistics. We know from QuantumMechanics that there are two types of particles, bosons and fermions. Singlestates can be occupied by any number of bosons while for fermions a singlestate can be occupied at most by one fermion.

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 forfree bosons and fermions.Thermodynamic quantities are derived by the granpartition function Z, since Z is connected with the thermodynamic potentialvia

Ω = −kT logZ (1.1)

and the average number of particles and the gas pressure are given by:

N = −∂Ω∂µ

, p = −∂Ω∂V

(1.2)

Let H be the Hamiltonian

H =N∑i=1

~p2i2m

(1.3)

Let us suppose that for every momentum ~p there are n~p particles withsuch momentum. Since we are working in a box, ~p = 2π/L~m with ~m =(mx,my,mz) integers. Therefore we have

E =∑~p

n~pε~p ≡ E(n~p) N =∑~p

n~p (1.4)

The gran partition function is given by

Z(µ, V, T ) =∑N

∑n~p

g(n~p) exp(−βE(n~p) + βµN) (1.5)

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

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 (1.6)

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)] (1.7)

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)](1.8)

Note that in the boson case the series converges only if

exp β(µ− ε~p) < 1 (1.9)

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))] (1.10)

4

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(1.11)

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))] (1.12)

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))] (1.13)

N =4πV

h3

∫ ∞

0

p2dp1

1 + exp(β(−µ+ p2

2m))

(1.14)

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(1.15)

where z = exp (βµ):

p

kT=

1

λ3f5/2(z)

N

V=

1

λ3f3/2(z) (1.16)

5

where

λ =

√2π~2mkBT

(1.17)

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))] (1.18)

N = V4π

h3

∫ ∞

0

p2dp1

−1 + exp(β(−µ+ p2

2m))

+N0 (1.19)

where

N0 =exp(βµ)

1− exp(βµ)≡ z

1− z(1.20)

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(1.21)

We have

p

kT=

1

λ3g5/2(z)−

1

Vln(1− z)

N

V=

1

λ3g3/2(z) +

N0

V(1.22)

6

1.1 A gas of free fermions

Let us now rewrite the equations (1.16) for the pressure and concentrationfor a gas of fermions:

p

kT=

1

λ3f5/2(z)

N

V=

1

λ3f3/2(z) (1.23)

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

λ3

v= f3/2(z) (1.24)

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+ · · · (1.25)

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) (1.26)

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

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

In this case the thermal length 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+ . . . (1.27)

one gets

z =λ3

v+

1

23/2(λ3

v)2 + . . .) (1.28)

and the equation of state becomes

pV

kTN=

v

λ3(z − z2

25/2+ . . . = 1 +

1

25/2λ3

v+ . . . (1.29)

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

+ . . .) (1.30)

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 (1.31)

Therefore we getz = exp (βεF ) (1.32)

where the chemical potential εF is called Fermi energy

εF =~2

2m

[6π2

v

]2/3(1.33)

Let us now consider < n~p >, defined in eqs.(1.11)

< n~p >=1

eβ(ε~p−εF ) + 1(1.34)

When T → 0, β → +∞ we have

< n~p >T=0= 1 (1.35)

for ε~p < εF and< n~p >T=0= 0 (1.36)

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 a

8

sphere of radius pF , the Fermi surface. One defines also a Fermi temperatureor degeneracy temperature TF such that

kTF = εF (1.37)

Finally we can compute the internal energy

U =∑~p

ε~pn~p =V

h34π

2m

∫ ∞

0

dpp4 < n~p > (1.38)

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

(1.39)

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]

(1.40)

From the internal energy one can derive the specific heat

CV = NkBπ2

2

kT

εF(1.41)

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]

(1.42)

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 .

9

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 (1.43)

or

pF = (N

V)1/3(6π2)1/3~ (1.44)

εF =1

2m(N

V)2/332/3h2 =

~2

2m

[6π2

v

]2/3(1.45)

1.2 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 + · · · (1.46)

with

g3/2(1) =∞∑l=1

1

l3/2= ζ(

3

2) = 2.612... (1.47)

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.(1.22) for the averagenumber

λ3N0

V=λ3

v− g3/2(z) (1.48)

with

v =V

N(1.49)

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

λ3

v> g3/2(1) (1.50)

In fact

0 <λ3

v− g3/2(1) <

λ3

v− g3/2(z) (1.51)

10

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 (1.52)

or

Tc =1

kB

2π~2/m[vg3/2(1)]2/3

=1

kB

2π~2/m[ζ(3/2)]2/3

(N

V)2/3 (1.53)

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 (1.15)in eq.(1.16) or (1.46) in eq.(1.22) neglecting N0 we obtain

N

V=

1

λ3z (1.54)

from which we get

µ = kT lnλ3N

V(1.55)

Therefore the classical limit corresponds to

λ3N

V=N

V

(2π~2

mkT

)3/2

<< 1 (1.56)

Notice that in the Bose case the expression for µ given in eq.(1.55) cannotextrapolate to T → 0. In fact decreasing T µ becomes positive and thendiverges.

2 Introduction to superfluids

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

11

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

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 (2.1)

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 phase,HeII, Fig. 1 The phase transition, of the second order, is signaled by a peakin the specific heat (Fig. 2 ). The first who discovered these properties of theliquid Helium below 2.2 K was Kamerlingh Onnes (1908) in the experimentsleading tio discovering superconductivity. After this observation, it took 30years to understand that the new phase was a superfluid phase: the fluid canflow without any friction and viscosity.

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

12

Figure 2: Specific heat (from Huang)

(1938).

It is therefore natural to associate the λ transition to the Bose Einsteincondensation (London 1938) modified by the molecular interactions. In factas we have seen the critical temperature for the condensation in a non inter-acting gas of bosons is 3.14 K, very close to Tλ. In a boson gas below Tc afraction N0/N of bosons condense in the state p = 0.

The He3 becomes also superfluid but only at temperatures of the order10−3K (1972, Osheroff, et al): in this case fermions condensate. The theo-retical suggestion of fermion condensate was done by Pitayevski. H2 solidifiesat higher temperature because of stronger interactions among his molecules(Huang p.317).

At low temperature (T << Tc) the specific heat varies as T3, as shown in

Fig. 3 ..... 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 (2.2)

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 as

13

the 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. 4. Approximately one has

εk = ~ck k << k0 (2.3)

with c = (239± 5) m/s

εk = ∆+~2(k − k0)

2

2σk ∼ k0 (2.4)

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 cold neutron scattering on Helium su-perfluid (see for example, Palewsky et al, Physical Review 112, (1958), 11).The reason is that cold neutrons have momentum close to the momentum ofthe excitations (the energy of the neutron is ∼ 50K.

Making use of the conservation laws

1

2m(~p2i − ~p2f ) = ε(k) (2.5)

~pi − ~pf = ~k (2.6)

where pi(f) are the momenta of the incoming (outgoing) neutron one canobtain the spectrum of the excitations.

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 (2.7)

14

Therefore

(~V · ~k)~ = εk +~2k2

2M≥ c~k (2.8)

orV k ≥ |~V · ~k| ≥ ck (2.9)

implyingV ≥ c (2.10)

Therefore if V ≤ c the process is prohibited. The reasoning depends in anessential way from the linear spectrum of the phonons. If the spectrum isquadratic the minimum threshold for loosing energy is zero.

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

(~V · ~k0)~ = εk0 +~2k2

2M≥ εk0 (2.11)

orV k0 ≥ |V k0~ cos θ| ≥ εk0 = ∆ (2.12)

orV ≥ vc (2.13)

with vc = ∆/(~k0) = 8.65 1.38 10−23J/(1.055 10−34Js1.92 × 10−8 cm) ∼58 m/s. The rotons are the elementary excitations associated to vortex inthe fluid. At low temperature the roton effects are negligible due to theBoltzmann factor exp (−∆/KT ). At the thermal equilibrium elementaryexcitations have energies close to the minimum of ε that is zero. In presenceof a purely phonon spectrum the critical velocity is c = 239 m/sec, whenrotons are included the critical velocity drops to vc = 58 m/sec (observed inHe4 under pressure).

3 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.(1.19):

N = N0 +V

h3

∫d3p

1

exp [β(ε~p − µ)]− 1(3.1)

15

Figure 3:

16

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 (3.2)

or

N = 0 +V

h3

∫d3p

1

exp[ε~pkTc

]− 1

=V

λ3cg3/2(1) (3.3)

For T < Tc µ remains zero and using eq. (3.1), 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

(3.4)

orN −N0

N=

(T

Tc

)3/2

(3.5)

orN0

N= 1−

(T

Tc

)3/2

(3.6)

andN0

V=N

V

[1−

(T

Tc

)3/2]T < Tc (3.7)

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.

17

4 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 (4.1)

con H = ~p2/2m. Quantum states can be built by working in the Fock space:by postulating the existence of creation and destruction 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′ (4.2)

In general, the one particle state is given by

|ψ >=∑~k

ψ~ka†~k|0 > (4.3)

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 (4.4)

We can now consider the following field

Φ(~x, 0) =1√V

∑~k

ei~k·~xa~k (4.5)

such that

|ψ >=∫d3xψ(x)Φ†(~x, 0)|0 > (4.6)

In general

Φ(~x, t) =1√V

∑~k

ei(~k·~x−ω~k

t)a~k (4.7)

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

i~∂

∂tΦ = −~2

∇2

2mΦ (4.8)

18

Let us now write the Lagrangian associated to the Schrodinger equation

L = i~Φ†Φ− ~21

2m∇Φ† · ∇Φ (4.9)

and the corresponding Hamiltonian density

H = ΠΦ− L = ~21

2m∇Φ† · ∇Φ (4.10)

with Π = ∂L/∂Φ = i~Φ†. The commutation relations (4.2) imply

[Φ(~x, t),Π(~y, t)] = i~δ(~x− ~y) (4.11)

The Hamiltonian of the field is given by

H =

∫d3xH =

∑~k

~ω~ka†~ka~k (4.12)

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′1a†~k′2a~k1a~k2 (4.13)

where ~~k1, ~~k2 (~~k′1, ~k′2) represent the momenta of incoming (outgoing) bosons

and ~~k′1 − ~~k1 the transfer momentum.

The function W (k) is the Fourier transform of the four boson interaction

W (k) =

∫d~rW (r)ei

~k·~r (4.14)

Before discussing the Bogoliubov approach let us first review the LandauGinzburg one.

19

5 Ginzburg Landau Model

In the Ginzburg-Landau model the states of the system are described by ascalar field which can be interpreted as the wave function of the superfluid.

The Hamiltonian of the model (the free energy), which was proposed asan effective description of field theory for phase transitions is given by

Heff =~2

2m(∇φ)†(∇φ)− µφ†φ+

1

2g(φ†φ)2 (5.1)

with µ and g positive constants. In particular µ < 0 for T > Tc and passesthrough zero at Tc.

Therefore the potential can be identified as

V (φ) = −µφ†φ+1

2g(φ†φ)2 (5.2)

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. 5. Amicroscopic interpretation of the parameters µ and g can be found in [4].They can be related to the strength 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 and a minimum for

φ†φ =µ

g(5.3)

or

φ(x) = φ0 exp (iψ) =

õ

gexp (iψ), (5.4)

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 (5.5)

20

0.5 1.0 1.5 2.0ÈΦÈ

-0.2

-0.1

0.1

0.2

0.3

V

Figure 4: Potential V corresponding to eq.(5.2) for µ = 0.4 and g = 0.3 as a

function of |φ| =√φ†φ.

Let us notice that the Hamiltonian (5.1) is invariant under the transformation

φ(x) → φ(x) exp (iα), α ∈ [0, 2π) (5.6)

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.

By substituting eq.(5.5) in the potential (5.2) one gets, by expanding tosecond order 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) (5.7)

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

21

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)

](5.8)

The Hamiltonian (5.8) is not diagonal in the basis of occupation numbersbecause of the bi-linear terms a e in a†. However it is possible to find atransformation (Bogoliubov transformation, see Appendix C) from a~k(a

†~k) to

the operators A~k(A†~k), defined as

A~k = cosh(θk2)a~k + sinh(

θk2)a†

−~k(5.9)

withtanh θk =

µ

µ+ ~2~k22m

(5.10)

One has[A~k, A

†~k′] = δ~k,~k′ (5.11)

FurthermoreHeff = E0 +

∑~k 6=0

ε(k)A†~kA~k (5.12)

with

E0 = −µ2

2gV −

∑~k 6=0

ε(k) sinh2(θk2) (5.13)

and

ε(k) =

√√√√(µ+~2~k22m

)2

− µ2 =

√√√√ µ

m~2~k2 +

(~2~k22m

)2

(5.14)

The fundamental state is defined as

A~k|φ0 >= 0 (5.15)

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 > (5.16)

22

with energy

ε(k) =

√√√√ µ

m~2~k2 +

(~2~k22m

)2

(5.17)

Therefore the spectrum is linear for small k while for large k behaves ask2.

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.

6 Bogoliubov approach

Let us first rewrite the Bogoliubov interaction Hamiltonian

HI =1

2

∑~k1+~k2=~k′1+

~k′2

W (|~k1 − ~k′1|)a†~k′1a†~k′2a~k1a~k2 (6.1)

For T < Tc all bosons tend to belong to the state k = 0, therefore N0 ∼ Nand

N −N0

N0

<< 1 (6.2)

Since N0 >> 1 one can assume the operator a0 to be c-number a0 ∼ a†0 ∼√N0. So we consider the operators a~k and a†~k small with respect to a0 and

a†0, expanding the interaction Hamiltonian keeping only the terms which arelinear or bi-linear in N0. In other words Bogoliubov separates the condensatein the expansion of the field:

ψ ∼√N0 +

∑~k 6=0

a~k exp [i(~k · ~x)] (6.3)

Let us first rewrite the interaction Hamiltonian by taking into accountthe momentum conservation

HI =1

2

∑~k1,~k2,~k′1

W (|~k1 − ~k′1|)a†~k′1a†~k1+~k2−~k′1

a~k1a~k2 (6.4)

23

Let us first list all the cases where the indices are zero. One has four zeroindices for ~k1 = ~k2 = ~k′1 = 0. One has two zero indices when (we list also thetransfer momentum)

~k1 = ~k′1 = 0 ~k1 − ~k′1 = 0

~k1 = ~k2 = 0 ~k1 − ~k′1 = −~k′1~k2 = ~k′1 = 0 ~k1 − ~k′1 =

~k1~k2 = ~k1 − ~k′1 = 0 ~k1 − ~k′1 = 0

~k1 = ~k2 − ~k′1 = 0 ~k1 − ~k′1 = −~k′1~k′1 =

~k1 + ~k2 = 0 ~k1 − ~k′1 =~k1 (6.5)

Therefore neglecting all terms of order O(a4~k) we obtain

HI =1

2

[W (0)(a†0a0)

2

+ W (0)a†0a0∑~k2

a†~k2a~k2) + a0a0

∑~k′1

W (k′1)a†~k′1a†−~k′1

+ a†0a0∑~k1

W (k1)a†~k1a~k1 +W (0)a†0a0

∑~k1

a†~k1a~k1

+ a†0a0∑~k′1

W (k′1)a†~k′1a~k′1

+ a†0a†0

∑~k1

W (k1)a−~k1a~k1

]=

1

2

[W (0)N2

0 + 2N0

∑~k 6=0

(W (0) +W (k))a†~ka~k

+ N0

∑~k 6=0

W (k)(a†~ka†−~k

+ a~ka−~k)]

(6.6)

Furthermore the number operator

N = a†0a0 +∑~k 6=0

a†~ka~k (6.7)

so that neglecting order O(a4~k)

N20 ∼ N2 − 2N0

∑~k 6=0

a†~ka~k (6.8)

24

By substituting eq.(6.8) in eq.(6.6) we get

HI =1

2[W (0)N2 + 2N

∑~k 6=0

W (k)a†~ka~k

+ N∑~k 6=0

W (k)(a†~ka†~k+ a~ka~k)

](6.9)

The total Hamiltonian, obtained adding to HI the kinetic term, can be di-agonalized as shown in Appendix C by the Bogoliubov transformation givenin eq.(C.3)

tanh θk =NW (k)

NW (k) + ~2k22m

(6.10)

The total Hamiltonian can be rewritten as∑~k

ε(k)A†~kA~k (6.11)

with

ε(k) =

√(NW (k) +

~2k22m

)2 − (NW (k))2

=

√(~2k22m

)2 +NW (k)~2k2m

(6.12)

The choiceW (k) = c1 − c2k (6.13)

is able to reproduce not only the phonon part of the spectrum but also theroton part.

Finally, by comparison with the Landau Ginzburg Hamiltonian, we obtainthe identification

µ ∼ NW (0) (6.14)

7 Superconductivity

Let us now consider, as a second application of non relativistic quantum fieldtheory, the phenomenon of superconductivity. Superconductivity is charac-terized by two main properties:

25

• In various metals, for example lead, tin, below a critical temperatureTc ∼ few K, resistivity drops to zero (the discovery was made workingat low temperature with mercury by Kamerlingh Onnes, 1911)

• Meissner effect: exclusion of magnetic fields from superconducting re-

gions in distances of order 500A (Meissner, Ochsenfeld, 1933)

The theoretical explanation is based on the formation of Cooper pairs:below the critical temperature the interaction between electrons close to theFermi surface and the phonons of the ion lattice can compensate for theCoulomb repulsion and provides the mechanism for the formation of Cooperpairs. Cooper showed (1956) that the Fermi sea of electrons is unstableagainst the formation of Cooper pairs.

One can show that the excitations of such a system have a spectrumwhich has a minimum corresponding to a finite energy gap and therefore anelectron moving in the metal cannot loose energy if its energy is below thegap. Therefore the current flows without resistivity.

In the following we will follow the Bardeen, Cooper, Schrieffer approach(1957). We consider a non relativistic spinor field

ψσ(~x, t) =1

V

∑~k,σ

c~k,σuσ exp [−i(ωkt− ~k · ~x)] (7.1)

where uσ, σ = 1, 2 are the two orthogonal two dimensional spinors and theoperators c~k,σ, c

†~k′,σ′ satisfy the anticommutation relations

[c~k,σ, c†~k′,σ′ ]+ = δ~k,~k′δσ,σ′ (7.2)

The Bardeen, Cooper and Schrieffer (1957) Hamiltonian is given by thegrand canonical Hamiltonian which includes a term −µN where µ ∼ µF =p2F/2m. In other words the chemical potential is approximated by its valueat the Fermi surface. We have

H =∑~k,σ

ξ~kc†~k,σc~k,σ −

1

V

∑~k,~l

W~k,~lc†~k,↑c†−~k,↓

c−~l,↓c~l,↑ (7.3)

26

with

ξ~k = ε~k − εF =~2k2

2m− ~2k2F

2m(7.4)

and W~k~l = W only for electrons close to the Fermi surface

|ξ~k|, |ξ~l| ≤ ~ωD (7.5)

where ωD is the Debye frequency. Otherwise W~k~l = 0. This can be under-stood from the fact that only electrons close to the Fermi states can scatterfrom a phonon and find a different not occupied final state.

Since at low temperature the phonon electron interaction generates acondensate with pairs of electron of opposite spin and momentum the newvacuum (fundamental state) of the theory must be such that

< c−~k,↓c~k,↑ > 6= 0 (7.6)

Therefore the new vacuum |BCS > 6= |0 > since the standard vacuum |0 >satisfies

c~k,σ|0 >= 0 (7.7)

Let us now see whether it is possible to find a new vacuum |BCS > suchthat

< BCS| : c−~k,↓c~k,↑ : |BCS >= 0 (7.8)

and< BCS|c−~k,↓c~k,↑|BCS >= Qk 6= 0 (7.9)

orc~k,↑c−~k,↓ = Q~k+ : c~k,↑c−~k,↓ : (7.10)

As we have done in superfluidity we perform the transformation from c~k,↑c−~k,↓to a new pair of operators A~k, B~k

A~k = u~kc~k,↑ − v~kc†−~k,↓

B~k = u~kc−~k,↓ + v~kc†~k,↑

(7.11)

We assume u~k, v~k real.

By requiring the anticommutation relation for A~k, B~k

[A~k, A†~k′]+ = [B~k, B

†~k′]+ = δ~k,~k′ (7.12)

27

we getu2~k + v2~k = 1 (7.13)

which can be satisfied assuming

u~k = cos θ~k , v~k = sin θ~k (7.14)

We require alsoA~k|BCS >= B~k|BCS >= 0 (7.15)

We have also the inverse relations

c~k,↑ = u~kA~k + v~kB†~k

c−~k,↓ = −v~kA†~k+ u~kB~k (7.16)

Furthermore

c−~k,↓c~k,↑ = (−v~kA†~k+ u~kB~k)(u~kA~k + v~kB

†~k) = u~kv~k+ : c−~k↓c~k↑ : (7.17)

with

: c−~k↓c~k↑ : = −u~kv~k(A†~kA~k +B†

~kB~k)

+ u2~kB~kA~k − v2~kA†~kB†

~k(7.18)

We can now perform the transformation in the Hamiltonian: first the kineticterm∑

~k

ξ~k(c†~k↑c~k↑ + c†~k↓c~k↓) = 2

∑~k

ξ~kv2~k+∑~k

ξ~k(u2~k− v2~k)(A

†~kA~k +B†

~kB~k)

− 2∑~k

ξ~ku~kv~k(B†~kA†

~k+ A~kB~k) (7.19)

and then the interaction, neglecting terms of order O(c4~k),

− 1

V

∑~k,~l

W~k~lc†~k,↑c†−~k,↓

c−~l,↓c~l,↑ = − 1

V(∑~k,~k′

W~k~k′u~kv~ku~k′v~k′)

− 1

V

∑~k

u~kv~k

∑~k′

W~k~k′(: c†~k′,↑c†−~k′,↓

: + : c−~k′,↓c~k′,↑ :)

(7.20)

28

Summing eq.(7.19) and eq.(7.20) we get

2∑~k

ξ~kv2~k+∑~k

ξ~k(u2~k− v2~k)(A

†~kA~k +B†

~kB~k)

− 2∑~k

ξ~ku~kv~k(A~kB~k +B†~kA†

~k)

− 1

V(∑~k~k′

W~k~k′u~kv~ku~k′v~k′)

−∑~k,~k′

W~k~k′(−2)u~kv~ku~k′v~k′(A†~k′A~k′ +B†

~k′B~k′)

−∑~k,~k′

W~k~k′u~kv~k(u2~k′− v2~k′)(B~k′A~k′ + A†

~k′B†

~k′) (7.21)

Requiring the vanishing of the term A~kB~k +B†~kA†

~kone gets

2ξ~ku~kv~k =1

V

∑~k′

W~k~k′(u2~k− v2~k)u~k′v~k′ (7.22)

so that the total Hamiltonian is

H =∑~k

E~k(A†~kA~k +B†

~kB~k) + E0 (7.23)

with

E~k = ξ~k(u2~k− v2~k) +

2

V

∑~k′

W~k~k′u~k′v~k′u~kv~k (7.24)

and

E0 = 2∑~k

ξ~kv2~k− 1

V

∑~k~k′

W~k~k′u~kv~ku~k′v~k′ (7.25)

The eq.(7.22) can be rewritten as

ξ~k sin 2θ~k =1

2V

∑~k′

sin 2θ~k′W~k~k′ cos 2θ~k (7.26)

29

The eq.(7.24), using eq.(7.26), can be rewritten as

E~k = ξ~k cos 2θ~k + ξk(sin 2θ~k)

2

cos 2θ~k

=ξ~k

cos 2θ~k(7.27)

By substituting the previous equation in (7.26) we obtain

E~k cos 2θ~k sin 2θ~k =1

2V

∑~k′

sin 2θ~k′W~k~k′ cos 2θ~k (7.28)

or

E~k sin 2θ~k =1

2V

∑~k′

sin 2θ~k′W~k~k′ (7.29)

By defining∆~k = E~k sin 2θ~k (7.30)

∆~k =1

2V

∑~k′

W~k~k′∆~k′

E~k′(7.31)

withE~k =

√ξ2~k +∆2

~k(7.32)

Using the form of W~k~k′ , we see that ∆ does not depend on ~k

∆ =W

2V

∑~k′

∆

E~k′(7.33)

We can now study the gap equation (7.33) which has the gapless trivialsolution ∆ = 0. Looking for a solution with ∆~k 6= 0 we obtain

1 =1

2V

∑~k′

W1√

ξ2~k′ +∆2(7.34)

This equation can be studied by going into the continuum

1 =W

2VV

1

(2π)3

∫d3k

1√ξ(k)2 +∆2

(7.35)

where the integral is performed around the Fermi surface |ξ(k)| ≤ ~ωD.Notice that there is no solution for W < 0, that is in case of repulsion.

30

7.1 Study of the gap equation

Let us now study the gap equation

1 =W

2

1

(2π)3

∫|ξ(k)|≤~ωD

dΩk2dk1√

(ε(k)− εF )2 +∆2

=W

2

1

(2π)3

∫|ξ(k)|≤~ωD

dΩk2dk

dε

dε√(ε− εF )2 +∆2

∼ W

4ρF

∫|ξ(k)|≤~ωD

dε√(ε− εF )2 +∆2

=W

4ρF

∫ +~ωD

−~ωD

dξ√(ξ2 +∆2

=W

2ρFarcsinh

~ωD

∆(7.36)

where we have introduced the density of states at the Fermi surface

ρF = 24π

(2π)3k2dk

dε|kF

=1

π2

k2

dεdk

|kF

=1

π2

kFm

~2|kF

(7.37)

where we have introduced the Fermi momentum ~kF

Inverting eq.(7.36), we get the gap energy in the form, if WρF/2 << 1,as it is

∆ = 2~ωD exp (− 2

WρF) = 2~ωD exp (− 2π2~2

WmkF) (7.38)

For typical methals WρF ∼ 0.3 − 0.6, see p.448 [4]. Then, considering ~ωD

100 K and WρF ∼ 0.6, we get ∆ ∼ 4 K.

Therefore the energy of the first excitation is given by

E~k =

√(~k2

2m− εF )2 +∆2 (7.39)

31

The spectrum has a gap, meaning that one cannot create excitation witharbitrary small energy. The magnitude of this gap is ∆. The quasiparticlesare mixure of electrons and holes (see eq.(7.11)). Furthermore since thequasiparticles have spin 1/2 the quasiparticles must appear in pairs, so theminimim energy is 2∆.

7.2 Finite temperature

Let us now compute how the gap ∆ depends on the temperature. Startingagain from the gap equation, recall that

∆~k =1

V

∑~k′

W~k~k′u~k′v~k′

=1

V

∑~k′

W~k~k′ < BCS|c−~k′,↓c~k′,↑|BCS >

=1

V

∑~k′

W~k~k′u~k′v~k′ < BCS|(1− (A†~k′A~k′ +B†

~k′B~k′)|BCS >

(7.40)

At T = 0, since there is no quasi particles, we recover eq.(7.31) and theformula for ∆ ≡ ∆(T = 0). However this method can be extended at finitetemperature T . Taking the average over a statistical ensemble at temperatureT we have

∆~k =1

V

∑~k′

W~k~k′u~k′v~k′ < (1−(A†~kA~k+B

†~kB~k) >=

1

V

∑~k′

W~k~k′u~k′v~k′(1−2f(E~k′))

(7.41)where f(E~k) is the probability to have an excitations with energyE~k at tem-perature T :

f(E~k) =1

1 + eβE~k(7.42)

Therefore the gap equation at finite temperature becomes

∆~k =1

2V

∑~k′

W~k~k′∆~k′

E~k′(1− 2f(E~k′)) (7.43)

32

using the explicit expression for W one gets

1 =W

2V

∑~k′

1

E~k′(1− 2f(E~k′)) (7.44)

or

−1 +W

2V

∑~k′

1

E~k′=

1

V

∑~k′

W

E~k′f(E~k′) (7.45)

Passing to the continuum

−1 +1

2V

V

(2π)3W

∫d3k

1√ξ2(k) + ∆(T )2

=1

V

V

(2π)3W

∫d3kf(E(k))

1

E(k)

(7.46)Let us first compute the l.h. side. Proceeding as before, we get

l.h.side = −1 +1

2WρFasinh

~ωD

∆(T )

= −1 +1

2WρF ln

2~ωD

∆(T )

= −1− ln2~ωD

∆(T )

1

ln ∆(0)2~ωD

=1

2WρF ln

∆(0)

∆(T )(7.47)

Summing up we have

1

2WρF ln

∆(0)

∆(T )=

1

2WρF

∫ ~ωD

−~ωD

dξ1√

ξ2 + (∆(T ))21

eβ√

ξ2+(∆(T ))2 + 1

∼ 1

2WρF

∫ ∞

−∞dx

1√x2 + u2

1

e√x2+u2 + 1

(7.48)

with u = β∆(T ). The integral has been extended to (−∞,∞) because of itsrapid convergence. So:

ln∆(0)

∆(T )= 2

∫ ∞

0

dx1√

x2 + u21

e√x2+u2 + 1

(7.49)

This integral is discussed in [3]. For small ∆ ,

ln∆(0)

∆(T )∼ ln

πkBT

γ∆(T )+

7ζ(3)

8π2

(∆(T ))2

(kBT )2(7.50)

33

Tc(K) ~ωD/kB Wρ/2 ∆(T = 0)/kBTc

BCS 1.76Cd 0.56 164 0.18 1.6Al 1.2 375 0.18 1.3-2.1Sn 3.75 195 0.25 1.6Pb 7.22 96 0.39 2.2

Table 1: Some superconductor properties (From [4])

where γ/π ∼ 0.57 and ζ(3) ∼ 1.2. For ∆ = 0 we get the critical temperature:

kBTc =γ

π∆(0) ∼ 0.57∆(0) (7.51)

Using eq. (7.50), expanding for small T − Tc, one gets that by increasingthe temperature the gap become smaller and vanishes at TC as

∆(T ) =

√8π

7ζ(3)kBTc(1−

T

Tc)12 ∼ 3.07kBTc(1−

T

Tc)12 (7.52)

As shown in table 7.2, the prediction of BCS theory ∆(T = 0) ∼ 1.76kBTcis quite well satisfied.

Let us finally show that E0 < 0, so that |BCS > is the real ground state.Using (7.27) we can write

ξ~k = E~k cos 2θ~k (7.53)

and using (7.29) we can rewrite E0:

E0 = 2∑~k

ξ~kv2~k− 1

V

∑~k~k′

W~k~k′u~kv~ku~k′v~k′

= 2∑~k

E~k cos 2θ~k sin2 θ~k −

1

2

∑~k

E~k sin2 2θ~k

= 2∑~k

E~k(cos 2θ~k sin2 θ~k − sin2 θ~k cos

2 θ~k)

= −2∑~k

E~k sin4 θ~k (7.54)

34

7.3 The BCS ground state

Let us now study the BCS ground state. It is based on the idea that electronsform Cooper pairs. The BCS vacuum is given by

|BCS >= Π~k(u~k + v~kc†~k↑c†−~k↓

)|0 > (7.55)

Thus it is a superposition of states of Cooper pairs.

This state is normalized:

1 =< BCS|BCS > (7.56)

and< BCS|c†~k↑c

†−~k↓

|BCS >= v~ku~k (7.57)

Proof:

< BCS|BCS > = < 0|Π~k(u~k + v~kc−~k↓c~k↑)Π~k′(u~k′ + v~k′c†~k′↑c†−~k′↓

)|0 >

= < 0|Π~ku~kΠ~k′u~k′ +Π~kv~kc−~k↓c~k↑Π~k′v~k′c†~k′↑c†−~k′↓

|0 >

= Π~ku2~k+ < 0|Π~k v

2~kc−~k↓c~k↑Π~k′v~k′c

†~k′↑c†−~k′↓

δ~k,~k′ |0 >

= Π~k(u2~k+ v2~k) = 1 (7.58)

where use has been made of anticommutation relations:

< 0|c−~k↓c~k↑c†~k↑c†−~k↓

|0 >=< 0|c−~k↓(1−c†~k↑c~k↑)c

†−~k↓

|0 >=< 0|c−~k↓c†−~k↓

−c−~k↓c†~k↑c~k↑c

†−~k↑

|0 >= 1

(7.59)Furthermore

< BCS|c†−~k↓

c†~k↑|BCS > = < 0|Π~k′(u~k′ + v~k′c−~k′↓c~k′↑)c†~k↑c†−~k↓

Π~k′′(u~k′′ + v~k′′c†~k′′↑c†−~k′′↓

)|0 >= v~ku~k (7.60)

35

A 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

(A.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 +∆ (A.2)

Every state corresponds to a point of the phase space and in general therewill be an infinite set of states satisfying (A.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 (A.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 (A.3)

A.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 +∆ (A.4)

36

ρ(q, p) = 0 otherwise (A.5)

If f is an observable one can compute the average as

< f >=

∫d3Nqd3Npf(q, p)ρ(q, p)∫

d3Nqd3Npρ(q, p)(A.6)

Let us then define the volume of the phase space as

Γ(E) =

∫E<H<E+∆

d3Nqd3N (A.7)

and the entropy asS = k log Γ(E) = S(V,E) (A.8)

where k is the Boltzmann constant. From (A.8) we get

1

T=∂S

∂Ep = T

∂S

∂V(A.9)

or

dS =1

T(dE + pdV ) (A.10)

By inverting (A.8) we can recover the internal energy

E(S, V ) ≡ U(S, V ) (A.11)

from whichdE = TdS − pdV (A.12)

Let us recall also the Helmholtz free energy, defined as

F = U − TS (A.13)

By differentiating, we get

dF = dE − dTS − TdS = −pdV − SdT (A.14)

Let us finally recall the Gibbs potential

G = F + pV (A.15)

dG = dF + pdV + dpV = −SdT + V dp (A.16)

37

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

Ω− F − µN (A.17)

where µ is the chemical potential. Furthermore

dE = TdS − pdV − µdN (A.18)

Therefore

dΩ = dE − dTS − TdS − dµN − µdN = −SdT − pdV −Ndµ (A.19)

and so

S = −∂Ω∂T

, p = −∂Ω∂V

, N = −∂Ω∂µ

, (A.20)

A.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 ] (A.21)

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

k being the Boltzmannconstant.

p = −∂F∂V

S = −∂F∂T

(A.22)

If the number of particles is not constant then one define

A.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 ] (A.23)

with µ the chemical potential.

38

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.

B Quantum statistical mechanics

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

ψ =∑n

cnφn (B.1)

where φn are the eigenfunctions of the stationary Schrodinger equation

Hφn = Enφn (B.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(B.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(B.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 >.

39

B.1 Microcanonical Ensemble

Postulate: Equal probability

c∗ncm = 1, if E < En < e+∆

= 0, otherwise (B.5)

Postulate of the casual phases

c∗ncm = 0, if n 6= m (B.6)

Therefore all this is equivalent to postulate the wave function

ψ =∑n

bnφn (B.7)

with

|bn|2 = 1, if E < En < E +∆

= 0 otherwise (B.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 (B.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(B.10)

We can write also ρ as

ρ =∑

E<En<En+∆

|φn > |bn|2 < φn| (B.11)

Therefore (assuming |bn|2 = 1)

trρ =∑

(number ofstates with E<En<E+∆)

≡ Γ(E) (B.12)

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

S(E, V ) = k log Γ(E, V ) (B.13)

40

B.2 Canonical Ensemble

The ρ matrix is defined as

ρnm = δnm exp(−βEn) (B.14)

orρ =

∑n

|φn > exp(−βEn) < φn| (B.15)

and the partition function Z(V, T ) as

Z(V, T ) = trρ =∑n

exp(−βEn) (B.16)

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

F = − 1

βlnZ (B.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)(B.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 (B.19)

where E is now the average value over the ensemble.

41

B.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) (B.20)

The internal thermodynamic potential is obtained as

Ω = −kT logZ(µ, V, T ) (B.21)

Let us check that eq.(B.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) (B.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) (B.23)

orΩ = E − TS − µN (B.24)

B.4 Fundamental state of the theory

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

|φ0(N) >= |N, 0, · · · 0 > (B.25)

42

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) > (B.26)

ea†0|φ0(N) >= (N + 1)1/2|φ0(N + 1) > (B.27)

To find the minimum energy state, let us first build the coherent state

|φ0 >= A1/2 exp[√V φ0a

†0]|0 > (B.28)

which satisfiesa0|φ0 >=

√V φ0|φ0 > (B.29)

anda~k|φ0 >= 0 ~k 6= 0 (B.30)

n0 =< N(k = 0) >

V=

1

V< φ0|a†0a0|φ0 >=

1

VV φ2

0 (B.31)

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

by

A1/2 = exp[−1

4V φ2

0] (B.32)

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

< φ0|φ(x)|φ0 >= φ0 =√n0 =

√< N(k = 0) >

V(B.33)

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 (B.34)

with

|φ0 >= N exp [−1

2

∑k 6=0

tanh(θk/2)a†~ka†−~k]|φ0 > (B.35)

The solution is given by |φ0 > defined by (B.28). This means that the truevacuum state contains pair of bosons with opposite momenta.

Exercise. Verify the at eq. (B.34) is satisfied by the new vacuum (B.35).

Exercise. verify that the state |φ0 > corresponds to a lower value of theenergy with respect to |φ0 >.

43

C 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)]

(C.1)

where

α = µ+~2k2

2m(C.2)

Let us considerAk = βak + γa†−k (C.3)

with β, γ ∈ R. Then we get

[Ak, A†k′ ] = [βak + γa†−k, βa

†k′ + γa−k′ ] = (β2 − γ2)δkk′ (C.4)

In order to get standard commutation relations, let us require

β2 − γ2 = 1 (C.5)

It is convenient to define

β = cosh

(θk2

), γ = sinh

(θk2

)(C.6)

The inverse transformations are

ak = βAk − γA†−k, a†k = βA†

k − γA−k (C.7)

In factβAk − γA†

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

Substituting in eq.(C.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] (C.9)

44

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

†−k we get

tanh θk =2βγ

β2 + γ2=µ

α(C.10)

Then the coefficient of A†kAk becomes, using (C.10) and (C.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) (C.11)

with ε(k) given by eq.(5.17). Finally

αγ2 − βγµ = αγ2γ2 − β2

β2 + γ2= − αγ2

β2 + γ2= −ε(k) sinh2

(θk2

)(C.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

45

[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

[13] A.A. Abrikosov, Quantum Field Theoretical Methods in StatisticalPhysics

[14] M. Tinkham, Introduction to Superconductivity, Robert. E. KriegerPub. Company, Malabar Florida

46