Basics of Thermal Field Theory1

Mikko Laine

February 6, 2008

1Notes based on lectures delivered at the University of Bielefeld, Germany, during the summer semester

2004 and during the winter semester 2007-2008.

Contents

1 Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Basic structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Canonical partition function . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Path integral for the partition function . . . . . . . . . . . . . . . . . . . . . 3

1.4 Evaluation of the path integral for harmonic oscillator . . . . . . . . . . . . 5

1.5 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Free scalar fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1 Path integral for the partition function . . . . . . . . . . . . . . . . . . . . . 12

2.2 Fourier representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Evaluation of thermal sums . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.4 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.5 Low-temperature expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.6 High-temperature expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.7 Properties of the Euler gamma and Riemann zeta functions . . . . . . . . . 25

2.8 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Interacting scalar fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.1 Weak-coupling expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Wick theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.3 Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.4 Naive free energy density to O(): ultraviolet divergences . . . . . . . . . . 35

3.5 Naive free energy density to O(2): infrared divergences . . . . . . . . . . . 35

3.6 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.7 Proper free energy density to O(): ultraviolet renormalization . . . . . . . 39

3.8 Proper free energy density to O(3

2 ): infrared resummation . . . . . . . . . 42

1

3.9 Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.1 Path integral for the partition function of a fermionic harmonic oscillator . 46

4.2 The Dirac field at finite temperature . . . . . . . . . . . . . . . . . . . . . . 49

4.3 Fermionic thermal sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.4 Exercise 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 Gauge fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.1 Path integral for the partition function . . . . . . . . . . . . . . . . . . . . . 56

5.2 Gauge fixing and ghosts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.3 Feynman rules for Euclidean continuum QCD . . . . . . . . . . . . . . . . . 61

5.4 Exercise 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.5 Thermal gluon mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

5.6 Free energy density to O(g3) . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.7 Exercise 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6 Low-energy effective field theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6.1 The infrared problem of thermal field theory . . . . . . . . . . . . . . . . . 75

6.2 A simple example of an effective field theory . . . . . . . . . . . . . . . . . 76

6.3 Dimensionally reduced effective field theory for hot QCD . . . . . . . . . . 80

6.4 Exercise 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

7 Finite density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

7.1 Complex scalar field with a finite chemical potential . . . . . . . . . . . . . 86

7.2 Effective potential and Bose-Einstein condensation . . . . . . . . . . . . . . 88

7.3 Dirac fermion with a finite chemical potential . . . . . . . . . . . . . . . . . 90

7.4 How about chemical potentials for gauge symmetries? . . . . . . . . . . . . 91

7.5 Exercise 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

8 Real-time observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

8.1 Different Greens functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

8.2 From Euclidean correlator to spectral function . . . . . . . . . . . . . . . . 104

8.3 Exercise 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

8.4 Hard Thermal Loop effective theory . . . . . . . . . . . . . . . . . . . . . . 110

8.5 Relation to classical kinetic theory . . . . . . . . . . . . . . . . . . . . . . . 118

8.6 Exercise 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

2

9 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

9.1 Thermal phase transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

9.2 Bubble nucleation rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

9.3 Exercise 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

9.4 Particle production rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

9.5 Dark matter abundance in cosmology . . . . . . . . . . . . . . . . . . . . . 145

9.6 Appendix: relativistic Boltzmann equation . . . . . . . . . . . . . . . . . . 151

9.7 Transport coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

10 Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

3

1. Quantum Mechanics

1.1. Basic structure

The properties of the system can be described by a Hamiltonian, which for non-relativistic spin-0particles in one dimension takes the form

H =p2

2m+ V (x) , (1.1)

where m is the particle mass. The dynamics is governed by the Schrodinger equation,

i~

t| = H| . (1.2)

Formally, the time evolution can be solved in terms of the time-evolution operator:

|(t) = U(t; t0)|(t0) , (1.3)

where, for a time-independent Hamiltonian,

U(t; t0) = e i

~H(tt0) . (1.4)

For the state vectors |, various bases can be chosen. In the |x-basis,

x|x|x = xx|x = x (x x) , x|p|x = i~xx|x = i~x (x x

) . (1.5)

In the energy basis,H |n = En|n . (1.6)

In the classical limit, the system of Eq. (1.1) can be described by the Lagrangian

LM =1

2mx2 V (x) . (1.7)

A Legendre transform leads to the classical Hamiltonian:

p =LMx

, H = xp LM =p2

2m+ V (x) . (1.8)

A most important example of a quantum mechanical system is provided by a harmonic oscillator:

V (x) 1

2m2x2 . (1.9)

In this case the energy eigenstates can be found explicitly:

En = ~(n+

1

2

), n = 1, 2, 3, . . . . (1.10)

All states are non-degenerate.

It will turn out to be useful to view (quantum) mechanics formally as (0+1)-dimensional (quan-

tum) field theory: the operator x can be viewed as the field operator at a certain point,

x (0) . (1.11)

In quantum field theory operators are usually represented in the Heisenberg picture rather than inthe Schrodinger picture; then

xH(t) H(t,0) . (1.12)

In the following we use an implicit notation whereby showing the time coordinate t as an argumentimplies automatically the Heisenberg picture, and the corresponding subscript is left out.

1

1.2. Canonical partition function

Taking now our quantum mechanical system to a finite temperature T , the basic quantity tocompute is the partition function Z. We employ the canonical ensemble, whereby Z is a functionof T . Introducing units where kB = 1 (i.e., There kBTSI-units), the partition function is definedby

Z(T ) Tr [eH ] , 1

T. (1.13)

From the partition function, other observables are obtained, for instance the free energy F , theentropy S, and the average energy E:

F = T lnZ , (1.14)

S = F

T= lnZ +

1

TZTr [HeH ] =

F

T+E

T, (1.15)

E =1

ZTr [HeH ] . (1.16)

Let us now compute these quantities for the harmonic oscillator. This can be trivially done inthe energy basis:

Z =

n=0

n|eH |n =

n=0

e~(12 +n) =

e~/2

1 e~=

1

2 sinh(

~2T

) . (1.17)

Consequently,

F = T ln

(e

~2T e

~2T

)=

~

2+ T ln

(1 e~

)(1.18)

T~

~

2(1.19)

T~ T ln

( T~

), (1.20)

S = ln

(1 e~

)+

~

T

1

e~ 1(1.21)

T~

~

Te

~T (1.22)

T~ 1 + ln

T

~, (1.23)

E = F + TS = ~

[1

2+

1

e~ 1

](1.24)

T~

~

2(1.25)

T~ T . (1.26)

Here we have also shown the behaviours of the various functions at low temperatures T ~ andat high temperatures T ~. Note how in most cases one can identify the contribution of theground state, and of the thermal states, with their Bose-Einstein distribution function.

Note also that the average energy rises linearly with T at high temperatures, with a coefficientcounting the number of degrees of freedom (i.e. the degeneracy).

2

1.3. Path integral for the partition function

In the case of the harmonic oscillator, energy eigenvalues are known, and Z can easily be evaluated.In many other cases, however, En are difficult to compute. A more useful representation of Z isobtained by writing it as a path integral.

In order to get started, let us recall some basic relations. First of all,

x|p|p = px|p = i~xx|p x|p = Aeipx

~ , (1.27)

where A is some constant. Second, we will need completeness relations, which we write asdx |xx| = 1 ,

dp

B|pp| = 1 , (1.28)

where B is another constant. The choices of A and B are not independent. Indeed,

1 =

dx

dp

B

dp

B|pp|xx|pp| =

dx

dp

B

dp

B|p|A|2e

i(pp)x~ p| (1.29)

=

dp

B

dp

B|p|A|22~(p p)p| =

2~|A|2

B

dp

B|pp| =

2~|A|2

B1 . (1.30)

Thereby B = 2~|A|2; we choose A 1, so that B = 2~.

We then move to the evaluation of the partition function. We do this in the x-basis, whereby

Z = Tr [eH ] =

dx x|eH |x =

dx x|e

H~ e

H~ |x . (1.31)

Here we have split eH into a product of N 1 different pieces, and defined ~/N .

The trick now is to insert

1 =

dpi2~

|pipi| , i = 1, . . . , N (1.32)

on the left side of each exponential, with i increasing from right to left; and

1 =

dxi |xixi| , i = 1, . . . , N (1.33)

on the right side of each exponential, with again i increasing from right to left.

Thereby we are left to consider matrix elements of the type

xi+1|pipi|e

~H(p,x)|xi = e

ipixi+1~ pi|e

~H(pi,xi)+O(

2)|xi

= exp

{

~

[p2i2m

ipixi+1 xi

+ V (xi) +O()

]}. (1.34)

Moreover we need to note that on the very right, we have

x1|x = (x1 x) , (1.35)

which allows to carry out the integral over x; and that on the very left, the role of xi+1| is playedby the state x| = x1|. Finally, we remark that the O()-correction in Eq. (1.34) can be eliminatedby sending N .

In total, then, we can write the partition function as

Z = limN

[ Ni=1

dxidpi2~

]exp

{1

~

Nj=1

[p2j2m

ipjxj+1 xj

+ V (xj)

]}xN+1x1,~/N

.

(1.36)

3

Oftentimes this is symbolically written as a continuum path integral, as

Z =

x(~)=x(0)

DxDp

2~exp

{1

~

~0

d

[[p()]2

2m ip()x() + V (x())

]}. (1.37)

Note that in this form, the integration measure is well-defined (as a limit of that in Eq. (1.36)).

The integral over the momenta pi is gaussian, and can be carried out explicitly:

dpi2~

exp

{

~

[p2i2m

ipixi+1 xi

]}=

m

2~exp

[m(xi+1 xi)

2

2~

]. (1.38)

Thereby Eq. (1.36) becomes

Z = limN

[ Ni=1

dxi2~/m

]exp

{1

~

Nj=1

[m

2

(xj+1 xj

)2+ V (xj)

]}xN+1x1,~/N

.

(1.39)We may also try to write this in a continuum form, like in Eq. (1.37). Of course, the measurecontains then a factor which appears quite divergent at large N ,

C

(m

2~

)N/2= exp

[N

2ln

(mN

2~2

)]. (1.40)

This factor is, however, completely independent of the properties of the potential V (xj). Therebyit contains no dynamical information, and we actually do not need to worry too much about theapparent divergence in any case, we will return to C from another angle in the next section.For the moment, then, we can again write a continuum form for the functional integral,

Z = C

x(~)=x(0)

Dx exp

{1

~

~0

d

[m

2

(dx()

d

)2+ V (x())

]}. (1.41)

Let us end by giving an interpretation to the result in Eq. (1.41). We recall that the usualpath integral at zero temperature has the exponential

exp

(i

~

dtLM

), LM =

m

2

(dx

dt

)2 V (x) . (1.42)

We note that Eq. (1.41) could be obtained with the following recipe:

(i) Carry out a Wick rotation, denoting it.

(ii) Introduce

LE LM ( = it) =m

2

(dx

d

)2+ V (x) . (1.43)

(iii) Restrict to the interval 0...~.

(iv) Require periodicity over .

With these steps, the exponential becomes

exp

(1

~SE

) exp

(1

~

~0

d LE

). (1.44)

where the subscript E stands for Euclidean, in contrast to Minkowskian. It will turn out thatthis recipe works, almost without modifications, also in field theory, and even for spin-1/2 andspin-1 particles, although the derivation of the path integral itself looks quite different in thosecases.

4

1.4. Evaluation of the path integral for harmonic oscillator

As a crosscheck, we would now like to evaluate the path integral in Eq. (1.41) for the caseof a harmonic oscillator, and check that we get the correct result in Eq. (1.17). To make theexercise more interesting, we carry out the evaluation in Fourier space (with respect to the timecoordinate ) rather than in configuration space. Moreover, we would like to see if we can deducethe information contained in the divergent constant C without making use of its actual value inEq. (1.40).

Let us start by representing an arbitrary function x(), 0 < < ~, with the property x(~) =x(0), as a Fourier sum:

x() T

n=

xnein , (1.45)

where the factor T is a convention. Periodicity requires

ein~ = 1 , i.e. n~ = 2pin , n Z . (1.46)The values n = 2piTn/~ are called Matsubara frequencies.

Apart from periodicity, we also impose reality on x():

x() R x() = x() xn = xn . (1.47)If we write xn = an + ibn, it follows that

xn = an ibn = xn = an + ibn {

an = anbn = bn (1.48)

In particular, b0 = 0, and xnxn = a2n + b

2n. Thereby, we now have the representation

x() = T

{a0 +

n=1

[(an + ibn)e

in + (an ibn)ein]}

. (1.49)

Here, a0 is called (the amplitude of) the Matsubara zero-mode.

With the representation of Eq. (1.49), general quadratic structures in configuration space can bewritten as

1

~

~0

d x()y() = T 2m,n

xnym1

~

~0

d ei(n+m)

= T 2m,n

xnym1

Tn,m = T

n

xnyn . (1.50)

In particular, the argument of the exponential in Eq. (1.41) becomes

1~

~0

dm

2

[dx()

d

dx()

d+ 2x()x()

]

Eq. (1.50)= mT

2

n=

xn

[in in +

2]xn

n=n= mT

2

n=

(2n + 2)(a2n + b

2n)

= mT2

2a20 mTn=1

(2n + 2)(a2n + b

2n) . (1.51)

5

Next, we need to consider the integration measure. Let us make a change of variables from x(), (0, ~), to the Fourier components an, bn. As we have seen, the independent variables are thena0 and {an, bn}, n 1.This change of variables introduces a determinant,

Dx() =det

[x()

xn

] da0 [n1

dan dbn

]. (1.52)

The change of bases is purely kinematical, however, and independent of the potential V (x). Thuswe can define

C Cdet

[x()

xn

] , (1.53)and consider now C as an unknown coefficient.

Making use of the gaussian integral

dx exp(cx2) =pi/c, the expression in Eq. (1.41) now

becomes

Z = C

da0

[n1

dan dbn

]exp

[12mT2a20 mT

n1

(2n + 2)(a2n + b

2n)

]

= C

2pi

mT2

n=1

pi

mT (2n + 2)

, n =2piTn

~. (1.54)

It remains to determine C. How to do this?

Since C is independent of , we can determine it in the limit = 0, whereby the systemsimplifies.

The integral over the zero-mode a0 in Eq. (1.54) is, however, divergent for 0. We callsuch a divergence an infrared divergence: the zero-mode is the lowest-energy mode.

But we can still take 0 if we momentarily regulate the integration over the zero-modein some other way. We note from Eq. (1.49) that

1

~

~0

d x() = Ta0 , (1.55)

so that Ta0 represents the average value of x(). In terms of Eq. (1.31), we can identify theaverage value with the boundary condition x, over which we integrate.

Let us then simply regulate the system by putting it in a box, i.e. by restricting the valuesof x to some (asymptotially wide by finite) interval x, and those of a0 to the interval x/T .

With this setup, we can proceed to match for C.

Side A: effective theory computation. In the presence of the regulator, Eq. (1.54) becomes

lim0

Zregulated = Cx/T

da0

[n1

dan dbn

]exp

[mT

n1

2n(a2n + b

2n)

]

= Cx

T

n=1

pi

mT2n, n =

2piTn

~. (1.56)

6

Side B: full theory computation. In the presence of the regulator, and in the absence ofV (x), Eq. (1.31) can be computed in a very simple way:

lim0

Zregulated =x

dx x|e p2

2mT |x

=

x

dx

dp

2pi~x|e p

2

2mT |pp|x

=

x

dx

dp

2pi~e

p2

2mT x|pp|x

= x1

2pi~

2pimT . (1.57)

Matching the two sides. Equating Eqs. (1.56) and (1.57), the regulator x drops out, and wefind

C =T

2pi~

2pimT

n=1

mT2npi

. (1.58)

Since the infrared regulator has dropped out, we may called C an ultraviolet coefficient.

Now we can continue with the full Eq. (1.54). Inserting C from Eq. (1.58), we get

Z = T~

n=1

1

1 + 2

2n

(1.59)

=T

~

1n=1

[1 + (~/2piT )

2

n2

] . (1.60)Making use of

sinhpix

pix=

n=1

(1 +

x2

n2

), (1.61)

then yields directly Eq. (1.17): the result is correct!

Thus, we have indeed managed to reproduce the correct result from the path integral, withoutever making recourse to Eq. (1.40) or, for that matter, to the discretization that was present inEqs. (1.36), (1.39).

Let us end with a couple of final remarks. First of all, in Quantum Mechanics, the partition func-tion Z, and all other observables, are certainly finite and well-defined functions of the parametersT,m, and , if computed properly. We saw that with path integrals this is not always obvious atevery intermediate step, but at the end does work out. In Quantum Field Theory, on the contrary,divergences may remain, even if we compute everything correctly. These are then taken care ofby renormalization. It is important to realise, however, that the ambiguity of the functionalintegration measure (through C) is not in itself the source of these divergences, as our quantummechanical example has demonstrated!

As the second remark, it is appropriate to stress that in many physically relevant observables,the coefficient C drops out completely, and the procedure is thereby simpler. An example of sucha quantity is discussed in Exercise 1.

As a final remark, it should be noted that many of the concepts and techniques that wereintroduced with this simple example zero-modes, infrared divergences, their regulation, matchingcomputations, etc will also play a role in much less trivial quantum field theoretic exampleslater on, so it is important to master them as early as possible.

7

1.5. Exercise 1

Defining

x() e H~ xe H~ , 0 < < ~ , (1.62)we define a 2-point Greens function as

G() 1ZTr[eH x()x(0)

]. (1.63)

The corresponding path integral reads

G() =

x(~)=x(0)Dxx()x(0) exp[SE/~]

x(~)=x(0)Dx exp[SE/~], (1.64)

whereby the coefficient C has dropped out. The task is to compute explicitly G() for the harmonicoscillator, by making use of

(a) the canonical formalism [expressing H , x in terms of a, a].

(b) the path integral formalism, in Fourier space.

8

Solution to Exercise 1

(a) In terms of a, a, we can write

H = ~(aa+

1

2

), x =

~

2m(a+ a) , [a, a] = 1 . (1.65)

In order to construct x(), we make use of the expansion

eABeA = B + [A, B] +1

2![A, [A, B]] +

1

3![A, [A, [A, B]]] + ... . (1.66)

In particular,

[H, a] = ~[aa, a] = ~a ,[H, [H, a]] = (~)2a ,

[H, a] = ~[aa, a] = ~a ,

[H, [H, a]] = (~)2a , (1.67)

and so forth, so that we can write

eH

~ xeH

~ =

~

2m

{a

[1 + 1

2!()2 + ...

]+ a

[1 + +

1

2!()2 + ...

]}

=

~

2m

(ae + ae

). (1.68)

Inserting Z from Eq. (1.17), we then get

G() = 2 sinh(~

2

) n=0

n|e~(n+12 ) ~2m

(ae + ae

)(a+ a

)|n . (1.69)We now use a|n = n+ 1|n+1 and a|n = n|n 1 to identify the non-zero matrix elements,

n|aa|n = n+ 1 , n|aa|n = n . (1.70)Thereby

G() =~

msinh

(~2

)exp

(~

2

) n=0

e~n[e + n

(e + e

)]. (1.71)

The sums are simple,

n=0

e~n =1

1 e~ ,n=0

ne~n = 1~

d

d

1

1 e~ =e~

(1 e~)2 . (1.72)

In total, then,

G() =~

2m

(1 e~

)[e

1

1 e~ +(e + e

) e~(1 e~)2

]

=~

2m

1

1 e~[e + e(~)

]

=~

2m

cosh[(

~2

)]

sinh[~2

] . (1.73)

9

(b) Integration measure:

C

da0

[n1

dan dbn

]. (1.74)

Exponential:

exp

[12mT2a20 mT

n1

(2n + 2)(a2n + b

2n)

]= exp

[12T

n=

m(2n + 2)|xn|2

]. (1.75)

Fourier representation:

x() = T

{a0 +

k=1

[(ak + ibk)e

ik + (ak ibk)eik]}

, (1.76)

x(0) = T

{a0 +

l=1

2al

}. (1.77)

Observable:

G() = x()x(0) da0

n1 dan dbn x()x(0) exp[ ]

da0

n1 dan dbn exp[ ]. (1.78)

Since the exponential is quadratic in a0, an, bn R, we havea0ak = a0bk = akbl = 0 , akal = bkbl kl . (1.79)

Thereby

G() = T 2

a20 +

k=1

2a2k(eik + eik

). (1.80)

Here

a20 =da0 a

20 exp

( 12 mT2a20)da0 exp

( 12 mT2a20)= 2

m2d

dT

[ln

da0 exp

(12mT2a20

)]= 2

m2d

dT

[ln

2pi

m2T

]

=1

m2T, (1.81)

a2k =dak a

2k exp

[mT (2k + 2)a2k]dak exp [mT (2k + 2)a2k]

=1

2m(2k + 2)T

. (1.82)

Inserting into Eq. (1.80), we get

G() =T

m

(1

2+

k=1

eik + eik

2k + 2

)=

T

m

k=

eik

2k + 2, (1.83)

where k = 2pikT/~.

There are various ways to evaluate the sum in Eq. (1.83). We will encounter one generic methodin the later sections, so let us here present a different approach. We start by noting that(

d2

d2+ 2

)G() =

T

m

k=

eik =~

m( mod~) , (1.84)

10

where we made use of a standard summation formula.1

We now first solve Eq. (1.84) for 0 < < ~:( d

2

d2+ 2

)G() = 0 G() = Ae +Be , (1.85)

where A,B are unknown constants. The definition, Eq. (1.83), indicates that G(~ ) = G(),which allows to related A and B:

G() = A[e + e(~)

]. (1.86)

The remaining unknown A can be obtained by approaching the limit 0+. Then, from (1.83),

G(0) = A(1 + e~

)=

T

m

k=

1

2k + 2=

T

m

~2

(2piT )2

k=

1

k2 +(

~2piT

)2=

~

2m

cosh(

~2T

)sinh

(~2T

) , (1.87)where we made use of

k= 1/(k

2 + x2) = pi/x tanh(pix). Solving for A, and inserting intoEq. (1.86), yields the important final result:

G() =~

2m

cosh[(

~2

)]

sinh[~2

] . (1.88)

1Proof: ClearlyP

k= eikx = C(xmod 2pi), where C is some constant. In order to determine C, let us

integrate both sides of this equation from 0 to 2pi + 0. On the left-hand side we get 2pik,0, on the right-handside C. Thus, C = 2pi. Replacing now x by 2piT/~, and using (ax) = (x)/|a| on the right-hand side, yieldsP

k= eik = ~( mod~).

11

2. Free scalar fields

2.1. Path integral for the partition function

We start by deriving a path integral representation for scalar field theory, by making use of theresult obtained for the quantum mechanical harmonic oscillator (HO) in the previous section.

In quantum field theory, the form of the theory is most economically defined in terms of thecorresponding classical Lagrangian LM , rather than the Hamiltonian H (for instance, Lorentzsymmetry is explicit only in LM ). Let us therefore start from Eq. (1.7), and re-interpret x as aninternal degree of freedom , situated at the origin 0 of d-dimensional space, like in Eq. (1.11):

SHOM =

dtLHOM , (2.1)

LHOM =m

2

((t,0)

t

)2 V ((t,0)) . (2.2)

Let us compare this with the usual action of scalar field theory (SFT) in d-dimensional space:

SSFTM =

dt

ddxLSFTM , (2.3)

LSFTM =1

2 V () =

1

2(t)

2 1

2(i)(i) V () , (2.4)

where we assume that repeated indices are summed over (irrespective of whether they are up anddown, or both at the same altitude), and the metric is (+).

Comparing Eq. (2.2) with Eq. (2.4) we see that formally, scalar field theory is nothing but acollection (sum) of almost independent harmonic oscillators with m = 1, one at every x. Theseoscillators only interact via the derivative term (i)(i) which, in the language of statisticalphysics, couples nearest neighbours:

i (t,x + i) (t,x)

, (2.5)

where i is a unit vector in the direction i.

We then realise, however, that such a coupling does not change the derivation of the pathintegral (for the partition function) in Sec. 1.3 in any essential way: it was only important thatthe Hamiltonian was quadratic in the canonical momenta, p = mx t. In other words, thederivation of the path integral is only concerned with objects having to do with the time coordinate(or time derivatives), and these appear identically in Eqs. (2.2) and (2.4). Therefore, we can directlytake over the result from Eqs. (1.41)(1.44):

ZSFT(T ) =

(~,x)=(0,x)

x

[C D(,x) ] exp

[1

~

~0

d

ddxLSFTE

], (2.6)

LSFTE = LSFTM (t i) =

1

2

(

)2+

di=1

1

2

(

xi

)2+ V () . (2.7)

We will drop out the superscript SFT in the following and also, for brevity, mostly write LE inthe form LE =

12 + V ().

12

2.2. Fourier representation

We now parallel the strategy in Sec. 1.4, and rewrite the path integral in Fourier representation. Inorder to simplify the notation somewhat, we measure time in units where ~ = 1.05 1034 Js = 1.Then the dependence of can be expressed as

(,x) = T

n=

(n,x)ein , n = 2Tn , n Z . (2.8)

For the space coordinates, it is useful to momentarily take each direction finite, of extent Li, andimpose periodic boundary conditions, just like for the time direction. Then the dependence on xican be represented as

f(xi) =1

Li

ni=

f(ni)eikixi , ki =

2niLi

, ni Z , (2.9)

where 1/Li plays the same role as T in the time direction. In the infinite volume limit, the sum inEq. (2.9) goes over to the usual Fourier integral,

1

Li

ni

=1

2

ni

kiLi

dki2

, (2.10)

so that the finite volume is really just an intermediate regulator. The whole function in Eq. (2.8)now becomes

(,x) = Tn

1

V

k

(n,k)ein+ikx , V L1L2...Ld . (2.11)

Like in Sec. 1.4, the reality of (,x) implies that the Fourier modes satisfy

[(n,k)

]= (n,k) . (2.12)

Thereby only half of the Fourier-modes are independent; we can choose, for instance,

(n,k) , n 1 ; (0,k) , k1 > 0 ; (0, 0, k2, ...) , k2 > 0 ; . . . ; and (0,0) (2.13)

as the integration variables. Note again the presence of a zero-mode.

Quadratic forms can be written as

0

d

ddx1(,x)2(,x) = T

n

1

V

k

1(n,k)2(n,k) . (2.14)

In particular, in the free case, i.e. for V () 12 m22, the exponent in Eq. (2.6) becomes

exp(SE) = exp(

0

d

ddxLE

)

= exp

[1

2Tn

1

V

k

(2n + k2 +m2)|(n,k)|

2

]

=k

{exp

[

T

2V

n

(2n + k2 +m2)|(n,k)|

2

]}. (2.15)

13

The exponential here is precisely the same as the one in Eq. (1.75), with the replacements

m(HO) 1

V, 2(HO) k2 +m2 , |x2n|

(HO) |(n,k)|

2 . (2.16)

The result thus factorises into a product of harmonic oscillator partition functions, for which weknow the answer already. In fact, rewriting Eqs. (1.54), (1.59), (1.18) for the case ~ = 1, theharmonic oscillator partition function can be represented as

ZHO = C

n0

dxn

exp

[mT

2

n=

(2n + 2)|xn|

2

](2.17)

=T

n=1

2n2 + 2n

(2.18)

= T

n=

(2n + 2)

1

2

n=

(2n)1

2 (2.19)

= exp

{1

T

[

2+ T ln

(1 e

)]}, (2.20)

where n means that the zero-mode n = 0 is omitted. Note in particular that all dependence onm(HO) has dropped out.

Combining Eq. (2.15) with Eqs. (2.17)(2.20), we obtain two useful representations for ZSFT.First of all, denoting

Ek

k2 +m2 , (2.21)

Eq. (2.19) yields

ZSFT = exp

(F SFT

T

)=

k

{Tn

(2n + E2k)

1

2

n

(2n)1

2

}(2.22)

= exp

{k

[lnT +

1

2

n

ln2n 1

2

n

ln(2n + E2k)

]}. (2.23)

Taking then the infinite-volume limit, the free-energy density, F/V , can be written as

limV

F SFT

V=

ddk

(2)d

[Tn

1

2ln(2n + E

2k) T

n

1

2ln(2n) T

1

2ln(T 2)

]. (2.24)

Second, making directly use of Eq. (2.20), we get

ZSFT = exp

(F SFT

T

)=

k

{exp

[1

T

(Ek2

+ T ln(1 eEk

))]}, (2.25)

limV

F SFT

V=

ddk

(2)d

[Ek2

+ T ln(1 eEk

)]. (2.26)

We will return to the evaluation of the momentum integration in Secs. 2.5 and 2.6.

14

2.3. Evaluation of thermal sums

Due to the previously established equality between Eqs. (2.19), (2.20), we have arrived at twodifferent representations for the free energy density of free scalar field theory, Eqs. (2.24), (2.26).The purpose of this section is to take the step from Eq. (2.24) to (2.26) directly, and learn on theway how to carry out thermal sums such as those in Eq. (2.24) also in more general cases.

In fact, the very sum in Eq. (2.24) is a somewhat special case: it includes a physical term,the first one, which depends on the energy (and thus on the mass of the scalar particle); as wellas unphysical subtractions, the second and third terms, which are independent of the energy,but are needed in order to make the sum convergent. Only the result from energy-dependent termsurvives in Eq. (2.26). In order not to lose focus on the peculiarities of this special case, we willmostly concentrate on another, convergent sum:

i(E) 1E

dj(E)

dE= T

n

1

2n + E2, (2.27)

from which the first term appearing in Eq. (2.24),

j(E) Tn

1

2ln(2n + E

2) , (2.28)

can be obtained by integration with respect to E, apart from an integration constant, which shouldbe chosen according to the 2nd and 3rd terms in Eq. (2.24).

To begin with, we want to be completely general. Let f(p) be a function which is analytic in thecomplex plane, and regular on the real axis. We then consider the sum

S Tn

f(n) . (2.29)

Consider now the auxiliary function

i nB(ip) iexp(ip) 1 . (2.30)

This function has poles at ip = 2n, n Z, i.e. p = n. Expanding in Laurent series around anypole we get

i nB(i[n + z]) =i

exp(i[n + z]) 1 =i

exp(iz) 1 T

z+O(1) . (2.31)

Therefore, the residue at any pole is T . This means that we can replace the sum in Eq. (2.29) bya complex integral:

S =

dp

2if(p) inB(ip)

+i0+i0+

dp

2f(p)nB(ip) +

+i0+++i0+

dp

2f(p)nB(ip) , (2.32)

where, as indicated, the integration contour runs clockwise around the real axis of the complexp-plane.

In the latter term in Eq. (2.32), we can furthermore substitute integration variables as p p,and note that

nB(ip) = 1exp(ip) 1 =

exp(ip) 1 + 11 exp(ip) = 1 nB(ip) . (2.33)

15

We thereby get the final formula

S =

+i0+i0+

dp

2

{f(p) + [f(p) + f(p)]nB(ip)

}

=

+

dp

2f(p) +

+i0+i0+

dp

2[f(p) + f(p)]nB(ip) , (2.34)

where in the first term we were able to return to the real axis (because there are no singularitiesthere), and substitute once again p p. Thus we have managed to convert the sum in Eq. (2.29)to a complex integral.

The first term in Eq. (2.34) is temperature-independent: it gives the zero-temperature vacuumcontribution. The latter term determines how thermal effects change the result.

As a technical point let us note, furthermore, that in the lower half plane,

|nB(ip)| p=xiy= 1eixey 1

yT ey yx e|p| . (2.35)Therefore, if the function f(p) grows slower than e|p| at large |p| (in particular, polynomially), theintegration contour can be closed in the lower half plane, and the result is determined by the polelocations and residues of the function f(p) + f(p). Physically, we therefore say that the thermalcontribution to S is related to on-shell particles.

Let us now apply the general formula in Eq. (2.34) to the particular example of Eq. (2.27). Infact, without any additional cost, we can consider a slight generalization,

i(E; c) Tn

1

(n + c)2 + E2, c C . (2.36)

In terms of Eq. (2.29), we then have

f(p) =1

(p+ c)2 + E2=

i

2E

[1

p+ c+ iE 1

p+ c iE]

(2.37)

f(p) + f(p) = i2E

[1

p+ c+ iE+

1

p c+ iE 1

p+ c iE 1

p c iE]. (2.38)

For Eq. (2.34), we need the poles in the lower half plane; for | Im c| < E, these are at p = c iE.According to Eq. (2.38), the residue at each pole is i/2E. The vacuum term in Eq. (2.34) thenproduces

1

2(2i) i

2E=

1

2E, (2.39)

while the matter part yields

1

2(2i) i

2E

[1

e(Eic) 1 +1

e(E+ic) 1]. (2.40)

In total, then,

i(E; c) =1

2E

[1 + nB(E ic) + nB(E + ic)

]. (2.41)

We note, first of all, that the result is periodic in c c+2Tn, n Z, as it must be according toEq. (2.36); and that the appearance of ic resembles that of a chemical potential. Indeed, as we willsee in Exercise 2, setting ic corresponds to a situation where we have averaged over a particle(chemical potential ) and an antiparticle (chemical potential .)

16

To conclude this section, let us integrate the result of Eq. (2.41) with respect to E, in order toobtain a generalization of the function in Eq. (2.28),

j(E; c) Tn

1

2ln[(n + c)

2 + E2] . (2.42)

The relation given in Eq. (2.27) continues to hold in the presence of c. Noting that

1

ex 1 =ex

1 ex =d

dx

(1 ex

), (2.43)

we immediately get

j(E; c) = const. +E

2+

T

2

{ln[1 e(Eic)

]+[1 e(E+ic)

]}, (2.44)

where the constant can depend on T and c.

Setting now c = 0, and comparing with Eqs. (2.24), (2.26), we notice that we have been lucky:the extra terms in Eq. (2.24) are such that they precisely cancel against the integration constantin Eq. (2.44). So, the full physical result containing j(E; 0) can be deduced directly from i(E; 0).That the same is true even for ic 6= 0, if we interpret j(E; c) as a free energy density averagedover a particle and an antiparticle (in a harmonic oscillator potential, i.e. a free field), is shown inExercise 2.

17

2.4. Exercise 2

Compute the partition function of the harmonic oscillator in the presence of a chemical potential,

eF (T,) Z(T, ) Tr[e(HN)

], (2.45)

and show that the expression1

2

[F (T, ic) + F (T,ic)

](2.46)

agrees with the E-dependent part of Eq. (2.44).

Solution to Exercise 2

Trivially,

n|(H N)|n = ~(n+

1

2

) n = (~ )n+ ~

2. (2.47)

Evaluating the partition function in the energy basis yields

ZHO =n=0

exp

(~2T

~ T

n

)=

exp(~2T

)

1 exp(~T

) . (2.48)

Putting ~ 1, E, ic, we can rewrite the result as

ZHO = exp{ 1T

[E

2+ T ln

(1 eEicT

)]}. (2.49)

Reading from here F (T, ) according to Eq. (2.45), and computing 12

[F (T, ic)+F (T,ic)

], yields

directly the E-dependent part of Eq. (2.44).

18

2.5. Low-temperature expansion

Our next goal is to carry out the momentum integration in Eqs. (2.24), (2.26). We denote

J(m,T )

ddk

(2)d

[Ek2

+ T ln(1 eEk

)](2.50)

= Tn

ddk

(2)d

[1

2ln(2n + E

2k) const.

], (2.51)

I(m,T ) 1m

d

dmJ(m,T ) (2.52)

=

ddk

(2)d1

2Ek

[1 + 2nB(Ek)

](2.53)

= Tn

ddk

(2)d1

2n + E2k

, (2.54)

where d is the space dimension, Ek =

k2 +m2, and we made use of the fact that m1d/dm =E1

kd/dEk. In order to simplify the notation in the following, we will denote

K

Tn

ddk

(2)d,

K

T

n

ddk

(2)d,

k

ddk

(2)d, (2.55)

where K (n,k), and a prime denotes that the zero-mode is omitted. The tilde in K is areminder for Euclidean metric.

At low temperatures, T m, we expect that the results resemble those of the zero-temperturetheory. Therefore we can write

J(m,T ) = J0(m) + JT (m) , I(m,T ) = I0(m) + IT (m) , (2.56)

where J0 is the temperature-independent vacuum energy density,

J0(m) k

Ek2

, (2.57)

and JT is the thermal part of the free energy density,

JT (m) k

T ln(1 eEk

). (2.58)

The sum-integral I(m,T ) can be divided in a similar way. It is clear that J0(m) is ultravioletdivergent, and can only be evaluated in the presence of a regularization; as indicated by Eq. (2.55),we will mostly be employing dimensional regularization here. In contrast, the integrand in JT isexponentially small for |k| T , and therefore the integral is well convergent.Let us start by evaluating J0(m). Writing the mass dependence explicitly, the task is to evaluate

J0(m) =

ddk

(2)d1

2(k2 +m2)

12 , (2.59)

and subsequently insert d = 32. For generality and future reference, we will in fact first compute

F (m, d,A)

ddk

(2)d1

(k2 +m2)A, (2.60)

and obtain then J0 as J0(m) =12 F (m, d, 12 ).

19

Since the integrand only depends on |k|, angular integrations can be carried out, and the inte-gration measure obtains the known form2

ddk =d/2

(d/2)(k2)

d2

2 d(k2) , (2.61)

where (s) is the Euler gamma function (we reiterate some of its basic properties in Sec. 2.7).Substituting k2 z m2t in Eq. (2.60), we get

F (m, d,A) =d/2

(d/2)

1

(2)d

0

dz zd2

2 (z +m2)A

=md2A

(4)d/2(d/2)

0

dt td/21(1 + t)A . (2.62)

A further substitution t 1/s 1, dt ds/s2 yields

F (m, d,A) =md2A

(4)d/2(d/2)

10

ds sAd/21(1 s)d/21 . (2.63)

We now recognize a standard integral that can be expressed in terms of the gamma function:

F (m, d,A) =

ddk

(2)d1

(k2 +m2)A=

1

(4)d/2(A d/2)

(A)

1

(m2)Ad/2. (2.64)

We now return to J0(m) in Eq. (2.59), i.e. set A = 12 , d = 32, A d2 = 2+ in Eq. (2.64),and multiply by 12 . The basic property (s) = s

1(s+1) allows to transport the argument of thegamma function to the vicinity of 1/2 or 1, where a Taylor expansion is easily carried out (basicformulae can be found in Sec. 2.7):

(2 + ) = 1(2 + )(1 + )(1 + ) (2.65)

=1

2

(1 +

2

)(1 +

)(1 E) +O() , (2.66)

(12) = 2(1

2) = 2 . (2.67)

The other parts of Eq. (2.64) are written as

(4)32

+ =2

(4)2

[1 + ln(4)

]+O(2) , (2.68)

(m2)2 = m42(2

m2

)= m42

(1 + ln

2

m2

)+O(2) , (2.69)

where is an arbitrary scale parameter, introduced as 1 = 22.

Collecting now everything together, we get

J0(m) = m4

6422

[1

+ ln

2

m2+ ln(4) E + 3

2+O()

]. (2.70)

It is convenient at this point to introduce the MS scheme scale parameter , through

ln 2 ln2 + ln(4) E . (2.71)2On one hand,

Rddk exp(tk2) = [

R

dk1 exp(tk21)]

d = (/t)d/2. On the other,Rddk exp(tk2) =

c(d)R

0dk kd1 exp(tk2) = c(d)td/2

R

0dxxd1ex

2

= c(d)(d/2)/2td/2 . Thereby c(d) = 2d/2/(d/2).

20

Thereby

J0(m) = m42

642

[1

+ ln

2

m2+

3

2+O()

]. (2.72)

For I0(m), we obtain from Eqs. (2.52), (2.72) the expression

I0(m) =

k

1

2Ek= m

22

162

[1

+ ln

2

m2+ 1 +O()

]. (2.73)

Incidentally, note thatdk0/(2) 1/(k20 + E2k) = 1/(2Ek), so that I0(m) can also be written

as

I0(m) =

dd+1k

(2)d+11

k2 +m2. (2.74)

We then move to the finite-temperature parts, JT (m) and IT (m). As already mentioned, the cor-responding integrals are finite. Therefore, we can normally set d = 3 to begin with.3 Substituting|k| Tx in Eq. (2.58), and taking the derivative in Eq. (2.52), we find

JT (m) =T 4

22

0

dxx2 ln(1 e

x2+y2

)ym/T

, (2.75)

IT (m) =T 2

22

0

dxx2x2 + y2

1

ex2+y2 1

ym/T

. (2.76)

Unfortunately, these integrals cannot be expressed in terms of elementary functions. In retrospec-tive, though, this may even be understandable: as we will see, they contain so much importantphysics that it would be unrealistic to find it in any simple well-behaved analytic function! At thesame time, Eqs. (2.75), (2.76) can of course be numerically evaluated without problems.

Even though Eq. (2.75) cannot be evaluated exactly, we can still find approximate expressionvalid in various limit. In this section we are interested in low temperatures, i.e. y = m/T 1. Letus thus evaluate the leading term of Eq. (2.75) in an expansion in exp(y) and 1/y. We can write

0

dxx2 ln(1 e

x2+y2

)=

0

dxx2 ex2+y2 +O(e2y)

wx2+y2

= y

dwww2 y2ew +O(e2y)

vwy= ey

0

dv(v + y)2vy + v2ev +O(e2y)

= 2y

32 ey

0

dv v12

(1 +

v

y

)(1 +

v

2y

) 12

ev +O(e2y)

= 2(

3

2)y

32 ey

[1 +O

(1y

)+O

(ey

)], (2.77)

where (3/2) =/2.

Inserting into Eq. (2.58), we thus obtain

JT (m) = T 4(

m

2T

) 32

em

T

[1 +O

(T

m

)+O

(em/T

)]. (2.78)

The derivative in Eq. (2.52) subsequently yields

IT (m) =T 3

m

(m

2T

) 32

em

T

[1 +O

(T

m

)+O

(em/T

)]. (2.79)

3In multiloop computations, JT (m) or IT (m) could get multiplied by a divergent term, 1/, in which casecontributions of O() would be needed as well. They could then be obtained by noting from Eq. (2.61) that2ddk/(2)d = {1 + [ln(2/4k2) + 2] +O(2)}d3k/(2)3, for d = 3 2 and an integrand only depending on k2.

21

Thereby we have arrived at the main conclusion of this section: at low temperatures, T m,finite-temperature effects in a theory with a mass gap are exponentially suppressed by the Boltz-mann factor, exp(m/T ). In fact the results agree with those in non-relativistic classical statisticalmechanics. Consequently, the results for the functions J(m,T ), I(m,T ) can be well approximatedby their zero-temperature limits, J0(m), I0(m), given in Eqs. (2.72), (2.73), respectively.

2.6. High-temperature expansion

We now move to the opposite limit than in the previous section, i.e. T m or, in terms ofEq. (2.75), y = m/T 1. It appears obvious that the procedure then should be a Taylorexpansion in y2, around y2 = 0. The zeroth order term, for instance, yields

JT (0) =T 4

22

0

dxx2 ln(1 e

x2)=

2T 4

90, (2.80)

which is nothing but the free-energy density (minus the pressure) of black-body radiation with onemassless degree of freedom. A term of order O(y2) can also be computed exactly.However, that is as far as it works: trying to go to the second order, O(y4), one finds that

the integral determining the coefficient of y4 is power-divergent for small x! In other words, thefunction JT (m) is not analytic around the origin in the variable m

2.

Nevertheless, Eq. (2.75) can still be expanded in a generalized sense, as we will see. The resultwill in fact read

JT (m) = 2T 4

90+m2T 2

24m

3T

12 m

4

2(4)2

[ln

(meE

4T

) 34

]+

m6(3)

3(4)4T 2+O

(m8

T 4

)+O() , (2.81)

where m (m2)1/2. It is the cubic term in Eq. (2.81) which first indicates that JT (m) is notanalytic in m2 around the origin (because there is a branch cut); this term plays a very importantrole in certain physical contexts, as we will see later on.

Our goal now is to derive the expansion in Eq. (2.81). A classic derivation, starting directlyfrom the definition in Eq. (2.75), can be found in a paper by Dolan and Jackiw, Phys. Rev. D 9(1974) 3320. It will be easier, and ultimately more useful, to carry out another type of derivation,however: we start from Eq. (2.51) rather than Eq. (2.50), and now carry out first the integrationk, and only then the sum

n.

Of course, Eq. (2.51) contains inconvenient additional constant terms, which appears problem-atic. Fortunately we already know that mass-independent correct value of J(0, T ): it is given inEq. (2.80). Therefore it is enough to study I(m,T ), in which case the starting point, Eq. (2.54),is simple enough, and subsequently integrate for J(m,T ) as

J(m,T ) =

m0

dmm I(m, T ) + J(0, T ) . (2.82)

Proceeding now with I(m,T ) in Eq. (2.54), the essential insight is to split the sum into thecontribution of the zero-mode, n = 0, and that of the non-zero modes, n 6= 0. Using thenotation of Eq. (2.55), we thus write

K

= K

+ T

k

. (2.83)

Let us first compute the contribution of the last term, which will be denoted by I(n=0).

22

To start with, let us return to the infrared divergences alluded to above. Trying naively a Taylorexpansion in m2, we would get

I(n=0) = T

k

1

k2 +m2= T

ddk

(2)d

[1

k2 m

2

k4+

m4

k6+ . . .

]. (2.84)

For d = 3 2, the first term is ultraviolet divergent, i.e. grows at large |k|; while the secondand subsequent terms are infrared divergent, i.e. explode at small |k| too fast to be integrable.(Of course, in dimensional regularization, every term in Eq. (2.84) appears strictly speaking to bezero. The total result is non-zero, however, as we will see: thus the Taylor expansion in Eq. (2.84)is really not justified, whichever way one looks at it.)

We now compute the integral in Eq. (2.84) properly. The result can be directly read fromEq. (2.64), by just setting d = 3 2, A = 1:

I(n=0) = TF (m, 32, 1) = T 1(4)3/2

( 12 + )(1)

1

(m2)1/2+( 1

2)=2= Tm

4+O() . (2.85)

This result is quite remarkable: a linearly divergent integral over a manifestly positive function isfinite and negative in dimensional regularization! According to Eq. (2.52), the corresponding termin J (n=0) reads

J (n=0) = Tm3

12+O() . (2.86)

Given the importance of the result, and its somewhat counter-intuitive appearance, it is worth-while to demonstrate that Eq. (2.85) is not an artifact of dimensional regularization. Indeed, letus compute it with cutoff regularization, by restricting |k| to be smaller than an upper bound, :

I(n=0) = T4

(2)3

0

dkk2

k2 +m2

=T

22

[m2

0

dk

k2 +m2

]=

T

22

[m arctan

( m

)]

= T

[

22 m

4+O

(m2

)]. (2.87)

We now observe that, due to the first term, Eq. (2.87) is indeed positive. This term is unphysical,however: it must be cancelled by similar terms emerging from the non-zero modes, since thetemperature-dependent part of Eq. (2.53) is manifestly finite. Representing a power divergence, itdoes not appear in dimensional regularization at all. The second term in Eq. (2.87) is the physicalone; it indeed agrees with Eq. (2.85). The remaining terms in Eq. (2.87) vanish when we take thecutoff to infinity, and are the analogy of the O()-terms of Eq. (2.85).We next turn to the contribution of the non-zero Matsubara modes, which will be denoted by

I (m,T ). It is important to realise that in this case, a Taylor expansion in m2 can be carried out:the integrals will be of the type

k

(m2)n

(2n + k2)n+1

, n 6= 0 , (2.88)

and thus the integrand remains finite for small |k|, i.e., there are no infrared divergences. (Therecould be ultraviolet divergences at small n, but these are taken care of by the regularization.)

More explicitly,

I (m,T ) = Tn

ddk

(2)d1

2n + k2 +m2

23

Taylor= 2T

n=1

ddk

(2)d

l=0

(1)l m2l

[(2nT )2 + k2]l+1

Eq. (2.64)= 2T

n=1

l=0

(1)lm2l 1(4)d/2

(l + 1 d2 )(l + 1)

1

(2nT )2l+2d

= 2T1

(4)d/2(2T )2d

l=0

[ m2(2T )2

]l(l + 1 d2 )(l + 1)

(2l + 2 d) , (2.89)

where in the last step we have interchanged the orders of the two summations, and identified the n-sum as a Riemann zeta function, (s) n=1 ns. Some of the properties of (s) are summarisedin Sec. 2.7.

Let us work out the orders l = 0, 1, 2 explicitly. For d = 32, the order l = 0 requires evaluating( 12 + ) and (1 + 2); l = 1 requires evaluating (12 + ) and (1 + 2); and l = 2 requiresevaluating (32 + ) and (3+2). We give some more details in Sec. 2.7 and in Exercise 3. In anycase, a straightforward computation yields

I (m,T ) =T 2

12 2m

2

(4)22

[1

2+ ln

(eE

4T

)]+

2m4(3)

(4)4T 2+O

(m6

T 4

)+O() . (2.90)

Adding the zero-mode contribution, Eq. (2.85), we get

I(m,T ) =T 2

12 mT

4 2m

2

(4)22

[1

2+ ln

(eE

4T

)]+

2m4(3)

(4)4T 2+O

(m6

T 4

)+O() . (2.91)

Subtracting Eq. (2.73), finally yields

IT (m) =T 2

12 mT

4 2m

2

(4)2

[ln

(meE

4T

) 1

2

]+

2m4(3)

(4)4T 2+O

(m6

T 4

)+O() . (2.92)

Note how the divergences and have cancelled from IT (m), as must be the case.

We can now transport these results to various versions of the function J , by making use ofEqs. (2.80) and (2.82). From Eq. (2.90), we get

J (m,T ) = 2T 4

90+

m2T 2

24 m

4

2(4)2

[1

2+ ln

(eE

4T

)]+

m6(3)

3(4)4T 2+O

(m8

T 4

)+O() . (2.93)

Adding the zero-mode contribution from Eq. (2.86) leads to

J(m,T ) = 2T 4

90+

m2T 2

24 m

3T

12 m

4

2(4)2

[1

2+ ln

(eE

4T

)]+

m6(3)

3(4)4T 2+O

(m8

T 4

)+O() .

(2.94)Subtracting the zero-temperature part, J0(m) in Eq. (2.72), leads finally to the expansion forJT (m), given in Eq. (2.81). Note again the cancellation of 1/ and from JT (m). The numericalconvergence of the high-temperature expansion is inspected in Exercise 3.

24

2.7. Properties of the Euler gamma and Riemann zeta functions

(s)

The function (s) is to be viewed as a complex function, with a complex argument s. ForRe(s) > 0, it can be defined as

(s) =

0

dxxs1ex , (2.95)

while for Re(s) 0, the values can be obtained through iterative use of the relation

(s) =(s+ 1)

s. (2.96)

On the real axis, Im(s) = 0, (s) is regular at s = 1; as a consequence of Eq. (2.96), it then hasfirst order poles at s = 0,1,2, ... .In practical applications, the argument s is typically either close to an integer, or close to a

half integer. In the former case, we can use Eq. (2.96) to relate the desired value to the values of(s) and its derivatives around s = 1; these can then be worked out from the convergent integralrepresentation in Eq. (2.95). In particular,

(1) = 1 , (1) = E , (2.97)

where E is the Euler constant, E = 0.577215664901... . In the latter case, we can use Eq. (2.96)to relate the desired value to the value of (s) and its derivatives around s = 12 ; these can thenbe worked out from the convergent integral representation in Eq. (2.95). In particular,

(12

)= ,

(12

)=(E 2 ln 2) . (2.98)

The values required for Eq. (2.90) thus become

(12+

)= 2 +O() , (2.99)

(12+

)=

[1 (E + 2 ln 2) +O(2)

], (2.100)

(32+

)=

2+O() . (2.101)

We have gone one order deeper in the middle one, because the result turns out to be multipliedby 1/ (cf. Eq. (2.111)).

(s)

The function (s) is to be viewed as a complex function, with a complex argument s. ForRe(s) > 1, it can be defined as

(s) =n=1

ns =1

(s)

0

dxxs1

ex 1 . (2.102)

The equivalence of the two forms in Eq. (2.102) can be seen by writing 1/(ex 1) = ex/(1 ex) =

n=1 e

nx, and using the definition of the gamma function in Eq. (2.95). The remarkableproperties of (s) follow from the fact that by writing

1

ex 1 =1

(ex/2 1)(ex/2 + 1) =1

2

[1

ex/2 1 1

ex/2 + 1

], (2.103)

25

and then substituting integration variables, x 2x, we find an alternative integral representationfor (s),

(s) =1

(1 21s)(s) 0

dxxs1

ex + 1. (2.104)

The integral here is defined for Re(s) > 0. Moreover, even though it diverges at s 0, the function(s) also diverges at the same point, and consequently (0) will be finite and regular around s = 0:

(0) = 12

[=

n=1

!

], (2.105)

(0) = 12ln(2) . (2.106)

Finally, for Re(s) 0, an analytic continuation is obtained through

(s) = 2ss1 sin(s2

)(1 s)(1 s) . (2.107)

On the real axis, Im(s) = 0, (s) has a pole only at s = 1. Its values at even arguments areeasy; in fact, at negative even integers, Eq. (2.107) implies that

(2n) = 0 , n = 1, 2, 3, ... , (2.108)

while at positive even integers the values can be related to the Bernoulli numbers,

(2) =2

6, (4) =

4

90, ... . (2.109)

Negative odd integers can be related to the positive even ones through Eq. (2.107), which equationalso allows to determine the behaviour around the pole at s = 1. In contrast, odd positive integerslarger that unity, i.e. s = 3, 5, ..., yield new transcendental numbers.

The values required for Eq. (2.90) become

(1 + 2) = 122

(2)(2) +O() = 112

+O() , (2.110)

(1 + 2) = 21+22[sin(2

)+ cos

(2

)]( 12

)(1 2)(2)

= 2(1 + 2 ln 2)(1 + 2 ln)(1)

( 12

)(1 + 2E)

(12

)(1 2 ln 2) +O()

=1

2+ E +O() , (2.111)

(3 + 2) = (3) +O() 1.2020569031...+O() , (2.112)

where in the first two cases we made use of Eq. (2.107), and in the second also of Eqs. (2.105),(2.106).

26

2.8. Exercise 3

(a) Complete the derivation leading to Eq. (2.90).

(b) Inspecting JT (m), Eq. (2.81), sketch the regimes where the low-temperature and high-temperature expansions are numerically accurate.

Solution

(a) For the term l = 0 in Eq. (2.89), we make use of the values in Eqs. (2.99), (2.110):

I (m,T )|l=0 = 2T1

(4)3/2(2T )

21

( 112

)+O() = T

2

12+O() . (2.113)

For the term l = 1 in Eq. (2.89), we make use of the values in Eqs. (2.100), (2.111):

I (m,T )|l=1 = 2T(4)

(4)3/2(2T )12

[ m2(2T )2

][1 (E + 2 ln 2)

] 12(1 + 2E) +O()

1=22

= m2

(4)22

{1

+ ln

2

T 2+ ln(4) E + 2[E ln(4)]

}+O()

= m2

(4)22

{1

+ ln

2

T 2+ 2 ln

(eE4

)}+O() , (2.114)

where in the last step we introduced the MS scheme scale parameter through Eq. (2.71). For theterm l = 2 in Eq. (2.89), we make use of the values in Eqs. (2.101), (2.112):

I (m,T )|l=2 = 2T1

(4)3/2(2T )

m4

(2T )4

12

2(3) +O() = 2m

4(3)

(4)4T 2+O() . (2.115)

(b) We again denote y m/T , and inspect then the function

J (y) JT (m)T 4

=1

22

0

dxx2 ln(1 e

x2+y2

). (2.116)

Apart from evaluating this expression numerically, we also consider the low-temperature resultfrom Eq. (2.78),

J (y) y1 (

y

2

) 32

ey +O() , (2.117)

as well as the high-temperature result from Eq. (2.81),

J (y) y1 = 2

90+

y2

24 y

3

12 y

4

2(4)2

[ln

(yeE

4

) 3

4

]+

y6(3)

3(4)4+O() . (2.118)

The results of the comparison are shown in Fig. 1. We observe that if we keep terms up to y6 inthe high-temperature expansion, its numerical convergence is reasonable for y 6. In between, anumerical evaluation is needed.

27

Figure 1: The exact result from Eq. (2.116); the low-temperature approximation from Eq. (2.117);and five orders of the high-temperature approximation from Eq. (2.118).

3. Interacting scalar fields

In order to move from a free to an interacting theory, we now choose

V () 12m22 +

1

44 (3.1)

in Eq. (2.4), where > 0 is a dimensionless coupling constant. Thereby the Minkowskian andEuclidean Lagrangians become

LM = 12 1

2m22 1

44 , (3.2)

LE = 12+

1

2m22 +

1

44 , (3.3)

where repeated indices are again summed over, irrespective of whether they are up and down, orboth at the same altitude; and the case with all indices down implies the use of Euclidean metric(i.e. no minus signs), like in Eq. (2.7).

Now, in the presence of > 0, it is no longer possible to exactly determine the partition functionof the system. We therefore need to develop approximation methods, which could in principle beeither analytic or numerical. In the following we restrict our attention to the simplest analyticprocedure which, as we will see, already teaches us quite a lot about the nature of the system.

28

3.1. Weak-coupling expansion

In the weak-coupling expansion the theory is solved by assuming that 1, and by expressingthe result for the physical observable in question as a (generalized) Taylor series in .

The physical observable we are interested in, is the partition function in Eq. (2.6). Defining thefree and interacting parts of the Euclidean action as

S0

0

d

V

d3x

[1

2+

1

2m22

], (3.4)

SI

0

d

V

d3x

[1

44], (3.5)

the partition function can be written as

ZSFT(T ) = CD eS0SI

= C

D eS0

[1 SI + 1

2S2I

1

6S3I + . . .

]

= ZSFT(0)[1 SI0 + 1

2S2I 0

1

6S3I 0 + . . .

], (3.6)

where

ZSFT(0) = CD eS0 (3.7)

is the free partition function that we determined in Sec. 2, and the expectation value 0 isdefined as

0 CD [ ] exp(S0)CD exp(S0) . (3.8)

The free energy density then reads

F SFT(T, V )

V= T

VlnZSFT

=F SFT(0)

V TV

ln

[1 SI0 + 1

2S2I 0

1

6S3I 0 + . . .

](3.9)

=F SFT(0)

V TV

{SI0 + 12

[S2I0 SI20

]

16

[S3I0 3SI0

S2I0+ 2

SI30

]+ . . .

}, (3.10)

where we have Taylor-expanded the logarithm, ln(1 x) = x x2/2 x3/3+ ... . The first term,F SFT(0) /V , is given in Eq. (2.26), while the subsequent terms correspond to corrections of orders

O(), O(2), and O(3). As we will see, the combinations that appear within the square bracketsin Eq. (3.10) have a very specific significance: Eq. (3.10) is actually simpler than Eq. (3.9)!

Inserting Eq. (3.5) into the various terms in Eq. (3.10), we are lead to evaluate expectation valuesof the type

(x1)(x2) . . . (xn)0 . (3.11)These can be reduced to products of free two-point correlators, (xk)(xl)0, through the Wicktheorem, as we now recall.

29

3.2. Wick theorem

The Wick theorem states that free (Gaussian) expectation values of any number of integrationvariables can be reduced to products of two-point correlators, according to

(x1)(x2) . . . (xn1)(xn)0 =

all combinations

(x1)(x2)0 (xn1)(xn)0 . (3.12)

Before applying this to the terms in Eq. (3.10), we briefly recall, for completeness, how the theoremcan be derived with (path) integration techniques.

Let us assume that we discretise the space-time such that the coordinates x only take a finitenumber of values (provided that the volume is finite as well). Then we can collect the values(x), x, into a single vector v. The free action can be written as S0 = 12 vTAv, where A is amatrix; we assume that A1 exists and that A is symmetric, AT = A.

The trick allowing to evaluate various integrals weighted by exp(S0) is to introduce a sourcevector b, and to take derivatives with respect to its components. Specifically, we introduce

exp[W (b)

]

dv exp

[12viAijvj + bivi

]

vivi+A1

ij bj= exp

[12biA

1ij bj

] dv exp

[12viAijvj

], (3.13)

where we made use of a substitution of integration variables. We then obtain

vkvl...vn0 =dv (vkvl...vn) exp

[ 12 viAijvj

]dv exp

[ 12 viAijvj

]

=

{d

dbkd

dbl... ddbn exp

[W (b)

]}b=0

exp[W (0)

]=

{ ddbk

d

dbl...

d

dbnexp

[12biA

1ij bj

]}b=0

=

{d

dbk

d

dbl...

d

dbn

[1 +

1

2biA

1ij bj +

1

2

(12

)2biA

1ij bjbrA

1rs bs + . . .

]}b=0

. (3.14)

Taking the derivatives in Eq. (3.14), it is clear that:

10 = 1. If there is an odd number of components of v in the expectation value, the result is zero. vkvl0 = A1kl . vkvlvmvn0 = A1kl A1mn +A1kmA1ln +A1knA1lm

= vkvl0vmvn0 + vkvm0vlvn0 + vkvn0vlvm0 . And that, in general, we are lead to a discretized version of Eq. (3.12).

Since all of the operations carried out are purely combinatorial in nature, it is clear that removingthe discretization does not modify the result, and that thereby Eq. (3.12) indeed also holds in thecontinuum limit.

30

Let us now apply Eq. (3.12) to Eq. (3.10). We will denote

f(T ) limV

F (T, V )

V, (3.15)

where we have dropped out the superscript SFT for simplicity. From Eqs. (2.26), (2.51), (3.10),the leading order result is the familiar one,

f(0)(T ) = J(m,T ) . (3.16)

At the first order, we get

f(1)(T ) = limV

T

VSI0 = lim

V

T

V

0

d

V

d3x

4(x)(x)(x)(x)0 . (3.17)

Here we can use the Wick theorem, Eq. (3.12). Noting furthermore that (x)(y)0 can onlydepend on x y, due to translational invariance, the space-time integral is trivial, and we get

f(1)(T ) =3

4(0)(0)0(0)(0)0 . (3.18)

Finally, at the second order, we get

f(2)(T ) = limV

{ T2V

[S2I 0 SI20

]}

= limV

{ T2V

[x,y

(4

)2(x)(x)(x)(x)(y)(y)(y)(y)0

x

4(x)(x)(x)(x)0

y

4(y)(y)(y)(y)0

]}, (3.19)

where we have denoted x

0

d

V

d3x . (3.20)

Carrying out contractions according to Eq. (3.12), the role of the subtraction term, the secondone in Eq. (3.19), now becomes clear: it cancels all disconnected contractions, i.e. contractionsof the type where all fields at point x are contracted with other fields at the same point. Inother words, the combination in Eq. (3.19) amounts to taking into account only the connectedcontractions. This miraculous combinatorial fact is brought about by the logarithm in Eq. (3.10),i.e., by going from the partition function to the free energy!

As far as the connected contractions are concerned, the Wick theorem tells that

(x)(x)(x)(x)(y)(y)(y)(y)0,c= 4 (x)(y)0 (x)(x)(x)(y)(y)(y)0,c ++ 3 (x)(x)0 (x)(x)(y)(y)(y)(y)0,c= 4 3 (x)(y)0 (x)(y)0 (x)(x)(y)(y)0,c ++ 4 2 (x)(y)0 (x)(x)0 (x)(y)(y)(y)0,c ++ 3 4 (x)(x)0 (x)(y)0 (x)(y)(y)(y)0,c= 4 3 2 (x)(y)0 (x)(y)0 (x)(y)0 (x)(y)0 ++ (4 3 + 4 2 3 + 3 4 3)(x)(x)0 (x)(y)0 (x)(y)0 (y)(y)0 ,

(3.21)

where the subscript (...)c refers to connected.

31

Inspecting the two-point correlators in Eq. (3.21), we note that they either depend on x y, oron neither x nor y (in the cases where fields at the same point are contracted). Thereby one ofthe spacetime integrals is again trivial (just substitute x x+ y, and note that (x+ y)(y)0 =(x)(0)0), and cancels against the factor T/V = 1/(V ) in Eq. (3.19). In total, then

f(2)(T ) = (4

)2[12

x

(x)(0)40 + 36(0)(0)20x

(x)(0)20]. (3.22)

One could go on with the third-order terms in Eq. (3.10): again, it could be verified that thesubtraction terms cancel all disconnected contractions, so that only the connected ones contributeto f(T ); and that one spacetime integral cancels against the explicit factor T/V . These facts are,in fact, of general nature, and hold at any order in the weak-coupling expansion.

To summarise, the Wick theorem has allowed us to reduce the terms in Eq. (3.10) to variousstructures made of the two-point correlator (x)(0)0 . We now turn to its properties.

3.3. Propagator

The two-point correlator (x)(y)0 is usually called the free propagator. Denoting

(P + Q) x

ei(P+Q)x = pn+qn,0(2)d(d)(p+ q) , (3.23)

where P (pn,p), and pn are bosonic Matsubara frequencies; and employing the Fourier-spacepresentation

(x) P

(P )eiP x , (3.24)

we recall from basic quantum field theory that the propagator can be written as

(P )(Q)0 = (P + Q) 1P 2 +m2

, (3.25)

(x)(y)0 =P

eiP (xy)1

P 2 +m2. (3.26)

Before inserting these expressions into Eqs. (3.18), (3.22), we briefly review their derivation, aswell as some basic properties of (x)(y)0 .In order to carry out the derivation, we return for a moment to a finite volume V , and proceed

as in Sec. 2.2. Let us start by inserting Eq. (3.24) into the definition of the propagator,

(x)(y)0 =P ,Q

eiP x+iQy(P )(Q)0 . (3.27)

In order to compute this expectation value, we insert Eq. (3.24) also into the free action, S0, finding

S0 =1

2

P

(P )(P 2 +m2)(P ) = 12

P

(P 2 +m2)|(P )|2 . (3.28)

We write (P ) = a(P ) + i b(P ), with a(P ) = a(P ), b(P ) = b(P ); only half of the Fouriercomponents are independent, and we could choose the ones specified in Eq. (2.13).

Restricting the sum to independent components, and making use of the symmetry properties ofa(P ) and b(P ), Eq. (3.28) becomes

S0 =T

V

Pindep.

(P 2 +m2)[a2(P ) + b2(P )] . (3.29)

32

The Gaussian integral, dxx2 exp(cx2)dx exp(cx2) =

1

2c, (3.30)

and the symmetries of a(P ) and b(P ), thus imply that

a(P ) b(Q)0 = 0 , (3.31)

a(P ) a(Q)0 = (P ,Q + P ,Q)V

2T

1

P 2 +m2, (3.32)

b(P ) b(Q)0 = (P ,Q P ,Q)V

2T

1

P 2 +m2, (3.33)

where the delta-functions are of Kronecker-type in finite volume. Thereby the propagator becomes

(P )(Q)0 = a(P ) a(Q) + i a(P ) b(Q) + i b(P ) a(Q) b(P ) b(Q)0= P ,Q

V

T

1

P 2 +m2= pn+qn,0V p+q,0

1

P 2 +m2. (3.34)

In the infinite-volume limit,

1

V

p

ddp

(2)d, V p,0 (2)d(d)(p) , (3.35)

and we recover Eq. (3.25) which, in combination with Eq. (3.27), in turn leads to Eq. (3.26).

We would now like to learn something more about the behaviour of the propagator (x)(y)0 ,at short and large separations x y. We note, first of all, that as shown by Eqs. (1.84), (1.73),

Tpn

eipn

p2n + E2=

1

2E

cosh[(

2

)E]

sinh[E2

] . (3.36)

This equation is valid for 0 ; as is obvious from the left-hand side, we can extend thevalidity to by replacing by | |. Thereby the propagator from Eq. (3.26) nowbecomes

G0(x y) (x)(y)0 =

ddp

(2)deip(xy)

1

2Ep

cosh[(

2 |x0 y0|

)Ep

]

sinh[Ep

2

]Ep

p2+m2

. (3.37)

Since G0 only depends on the separation x y, we set y = 0 in the following.Consider first short distances, |x|, |x0| 1/T, 1/m. We may expect the dominant contribution in

the Fourier transform in Eq. (3.37) to come from the regime |p||x| 1, so let us assume |p| T,m.Then Ep |p|, and Ep |p|/T 1. Consequently,

cosh[(

2 |x0|

)Ep

]

sinh[Ep

2

] exp[(

2 |x0|

)Ep

]

exp[Ep

2

] e|x0||p| . (3.38)

We note that1

2|p|e|x0||p| =

dp02

eip0x0

p20 + p2, (3.39)

whereby

G0(x)

dd+1P

(2)d+1eiP x

P 2, (3.40)

33

with P (p0,p).At this point we can make use of rotational symmetry, in order to choose x in the direction of

the component p0. Then

dd+1P

(2)d+1eiP x

P 2=

ddp

(2)d

dp02

eip0|x|

p20 + p2

=

ddp

(2)de|x||p|

2|p|

=1

(2)dd/2

(d/2)

0

d|p||p|d2e|x||p|

=(d 1)

(4)d/2(d/2)|x|d1 , (3.41)

where we made use of Eq. (2.61). Inserting d = 3 and (3/2) =/2, we find

G0(x) 142|x|2 , |x|

1

T,1

m. (3.42)

We note that this behaviour is independent of T and m: at short enough distances (in the ultra-violet regime), temperature and masses do not play a role, and the propagator diverges.

Consider, on the other hand, large distances, |x| 1/T (the temporal coordinate x0 remains byconstruction always small, i.e. is at most 1/T ). We expect the Fourier-transform in Eq. (3.37)to now be dominated by small momenta, |p| T . If we simplify the situation further by assumingthat we are also at very high temperatures, T m, then Ep 1 in the relevant regime, and wecan expand the hyperbolic functions in Taylor series, cosh() 1, sinh() . Then

G0(x) T

ddp

(2)deipx

p2 +m2. (3.43)

Restricting for simplicity to d = 3, we can write4

G0(x) T(2)2

+11

dz

0

d|p| |p|2 ei|p||x|z

|p|2 +m2

=T

(2)2

0

d|p| |p|2|p|2 +m2

ei|p||x| ei|p||x|i|p||x|

=T

(2)2i|x|

dp p eip|x|

p2 +m2

= Tem|x|

4|x| , |x| 1

T, (3.44)

where the last integral was carried out by closing the contour in the upper half plane.

We note from Eq. (3.44) that at large distances (in the infrared regime), thermal effects modifythe behaviour of the propagator in an essential way. In particular, if we were to set the mass tozero, then Eq. (3.42) would be the exact behaviour at zero temperature, both at small and atlarge distances, while Eq. (3.42) shows that a temperature would slow down the long-distancedecay to T/(4|x|). In other words, we can say that at a non-zero temperature the theory is moresensitive to infrared physics than at zero temperature.

4For a general d,R ddp

(2pi)deipx

p2+m2= 1

(2pi)d/2( m|x|

)d/21Kd/21(m|x|), where K is a modified Bessel function.

34

3.4. Naive free energy density to O(): ultraviolet divergences

We now return to the free energy density of scalar field theory, given by Eqs. (3.16), (3.18), (3.22).Noting from Eqs. (2.54) and (3.26) that G0(0) = I(m,T ), we obtain to O() that

f(T ) = J(m,T ) +3

4[I(m,T )]2 . (3.45)

According to Eqs. (2.72), (2.73),

J(m,T ) = m4

6422

[1

+ ln

2

m2+

3

2+O()

]+ JT (m) , (3.46)

I(m,T ) = m2

1622

[1

+ ln

2

m2+ 1 +O()

]+ IT (m) , (3.47)

where the functions JT (m) and IT (m) are finite, and were evaluated in various limits in Eqs. (2.78),(2.79), (2.81), (2.92).

Inserting Eqs. (3.46), (3.47) into Eq. (3.45), we note that the result is, in general, ultravioletdivergent. For instance, restricting for simplicity to very high temperatures, T m, and makinguse of Eq. (2.92),

IT (m) T 2

12mT

4+O(m2) , (3.48)

the dominant term at 0 reads

f(T )2

642

{m4 +

[1

2T 2m2

3

2Tm3 +O(m4)

]+O(2)

}+O(1) . (3.49)

This result is obviously non-sensical: the divergences appear even to depend on the temperature.A proper procedure requires renormalization; we return to this in Sec. 3.7, but identify first alsoanother problem with the naive result.

3.5. Naive free energy density to O(2): infrared divergences

Let us consider the second order contribution to Eq. (3.45), given in Eq. (3.22). With the notationof Eqs. (3.20), (3.37), we can write it as

f(2)(T ) = 3

42x

[G0(x)]4

9

42[I(m,T )]2

x

[G0(x)]2 . (3.50)

The question we would like to answer is, what happens if we take the limit that the particle massm is very small, m T . As Eqs. (3.45), (2.81), (3.48) show, at O() this limit is perfectly well-defined. Consider then the first term in Eq. (3.50). We know from Eq. (3.42) that the behaviourof G0(x) is independent of m at small x; thus nothing particular happens for |x| T

1. On theother hand, for large |x|, G0(x) is given by Eq. (3.42). We can estimate the contribution as

|x|>T

1

[G0(x)]4

0

d

|x|>T

1

d3x

(Tem|x|

4|x|

)4. (3.51)

This integral is convergent even for m 0; so in the first term of Eq. (3.50), taking m 0 doesnot cause any divergences.

Consider then the second term in Eq. (3.50). Repeating the argument of Eq. (3.51), we get

|x|>T

1

[G0(x)]2

0

d

|x|>T

1

d3x

(Tem|x|

4|x|

)2. (3.52)

35

If we now attempt to take m 0, the integral is linearly divergent! Because the problem emergesfrom large distances, we call this an infrared divergence.

In fact, it is easy to be more precise about the form of the divergence. We can write

x

[G0(x)]2 =

x

P

eiP x

P 2 +m2

Q

eiQx

Q2 +m2

=P ,Q

(P + Q)1

(P 2 +m2)(Q2 +m2)

=P

1

[P 2 +m2]2

= d

dm2I(m,T ) . (3.53)

Inserting Eq. (3.48), we get

x

[G0(x)]2 =

1

2m

d

dmI(m,T ) =

T

8m+O(1) . (3.54)

Therefore, for m T , Eq. (3.50) evaluates to

f(2)(T ) = 9

42

T 4

144

T

8m+O(m0) , (3.55)

and indeed diverges for m 0.

It is clear that, like the ultraviolet divergence in Eq. (3.49), the infrared divergence in Eq. (3.55)must also be an artifact of some sort: a gas of weakly interacting massless scalar particles shouldcertainly have a finite pressure and other thermodynamic properties, just like a gas of masslessphotons has. We return to the resolution of this problem in Sec. 3.8.

3.6. Exercise 4

(a) Let us consider a scalar field theory where the interaction term 14 4 is replaced with 13

3.Derive the expression for f(T ) up to O(3).

(b) What kind of infrared divergences does this expression have, if we take the limit m 0?

Solution to Exercise 4

(a) According to Eq. (3.10), the radiative corrections to the free energy density can be compactlyrepresented by the formula

f(1)(T ) = 1

V

exp(SI)

0,connected

=SI

1

2S2I + . . .

0,connected,drop overall

Rx

. (3.56)

The term of O() obviously vanishes, since it contains an odd number of fields. The term ofO(2) reads

f(2)(T ) = 2

18

x

(x)(x)(x)(0)(0)(0)0,c . (3.57)

36

Here, according to the Wick theorem,

(x)(x)(x)(0)(0)(0)0,c

= 3 (x)(0)0 (x)(x)(0)(0)0,c +

+ 2 (x)(x)0 (x)(0)(0)(0)0,c

= 3 2 (x)(0)0 (x)(0)0 (x)(0)0 +

+ 3 1 (x)(0)0 (x)(x)0 (0)(0)0 +

+ 2 3 (x)(x)0 (x)(0)0 (0)(0)0

= 6 (x)(0)0 (x)(0)0 (x)(0)0 +

+ 9 (x)(x)0 (x)(0)0 (0)(0)0 , (3.58)

and we get

f(2)(T ) = 1

32x

[G0(x)]3

1

22[G0(0)]

2

x

G0(x) . (3.59)

It turns out, however, that the latter term in Eq. (3.59) should actually be neglected. Thereason is that once we add the interaction 13

3, then the ground state of the theory is nolonger at = 0, as we have (implicitly) assumed; to see this, just compute to O()! Onthe other hand, the problem can be rectified by also adding another term to the potentialV (), of the type . Then there are additional contributions both to and to f(T ):

(0) (0)

x

[(x) 1

3(x)(x)(x)]0

= [+ G0(0)]

x

G0(x) , (3.60)

f(T ) 1

2

x

2(x) (0) +

1

3[(x)(x)(x) (0) + (x) (0)(0)(0)]

0

= 1

2[2 + 2G0(0)]

x

G0(x) . (3.61)

Summing together Eq. (3.59) and (3.61), we get

f(2)(T ) = 1

32x

[G0(x)]3

1

2[+ G0(0)]

2

x

G0(x) . (3.62)

Hence, we note that if we choose the coefficient such as to make the expression in Eq. (3.60)vanish, as it should, we simultaneously need to remove part of the terms in f(2)(T ).

To summarise, a physically meaningful expression is given by

f(T ) = J(m,T )1

32x

[G0(x)]3 +O(4) . (3.63)

(b) Inspecting the large-|x| behaviour like in Eqs. (3.51), (3.52), we obtain

|x|>T

1

[G0(x)]3

0

d

|x|>T

1

d3x

(Tem|x|

4|x|

)3. (3.64)

The radial part of the integration yields

x0

dx

xe3mx =

x0

dx( ddx

lnx)e3mx

=[lnx e3mx

]x0

+ 3m

x0

dx lnx e3mx

37

= lnx0e3mx0 + 3m

0

dx lnx e3mx +O(x0m)

= lnx0 +

0

dx lnx

3mex +O(x0m)

= lnx0 ln(3m)

0

dx ex +

0

dx lnx ex +O(x0m)

= ln1

3x0m E +O(x0m) , (3.65)

where we denoted x |x|; inserted a lower bound x0 to the integration, assuming x0 1/m;and set x0 0 whenever that does not lead to an ultraviolet divergence.

We now see that Eq. (3.65) diverges logarithmically if we set m 0.

38

3.7. Proper free energy density to O(): ultraviolet renormalization

In Sec. 3.4 we attempted to compute the free energy density f(T ) of scalar field theory up toO(), but found a result which appeared divergent. Let us now show that, as must be the casein a renormalizable theory, the divergences disappear order-by-order in perturbation theory, if were-express f(T ) in terms of finite renormalized parameters.

In order to proceed properly, we need to first change the notation somewhat. The zero-temperatureparameters we employed before, m2, , will now be re-interpreted as bare parameters, m2B, B .(The temperature T , in contrast, is a physical property of the system, and is not subject to anymodifications.) The expansion in Eq. (3.45) can then be written as

f(T ) = (0)(m2B, T ) + B (1)(m2B, T ) +O(

2B) . (3.66)

As a second step, we introduce some renormalized parameters, m2R, R. These could either bedirectly physical quantities (say, the mass of the scalar particle, and the scattering amplitude withsome particular kinematics), or quantities which are not yet directly physical, but are relatedto physical quantities by finite equations (say, so-called MS scheme parameters). In any case,it is natural to choose the renormalized parameters such that in the limit of an extremely weakinteraction, R 1, they formally agree with the bare parameters. In other words,

m2B = m2R + R f

(1)(m2R) +O(2R) , (3.67)

B = R + 2R g

(1)(m2R) +O(3R) . (3.68)

Note that renormalized parameters are defined at zero temperature, so no T can appear in theserelations. The functions f (i) and g(i) are, in general, divergent in the limit that regularization isremoved; for instance, in dimensional regularization, they contain poles like 1/.

The idea now is simply to convert the expansion in Eq. (3.66) into an expansion in R, byinserting the expressions from Eqs. (3.67), (3.68), and Taylor-expanding in R:

f(T ) = (0)(m2R, T ) + R

[(1)(m2R, T ) +

(0)(m2R, T )

m2Rf (1)(m2R)

]+O(2R) . (3.69)

We note that to O(2R), only the mass parameter needs to be renormalized.

To carry out the renormalization in practice, we need to choose a scheme. We will here choose theso-called pole mass scheme, where m2R is taken to be the physical mass squared of the -particle,denoted by m2phys. In Minkowskian spacetime, this appears as the exponential time-evolutionfactor,

eiE0t eimphyst , (3.70)

in the propagator of a particle at rest, p = 0. In Euclidean spacetime, this corresponds to theexponential fall-off, exp(mphys), of the propagator. Therefore, in order to determine m

2phys to

O(R), we need to compute the full propagator, G(x), to O(R) at zero temperature.

The full propagator can be defined as the generalization of Eq. (3.37) to the interacting case:

G(x) (x)(0) exp(SI)0

exp(SI)0

=(x)(0)0 (x)(0)SI 0 +O(

2B)

1 SI0 +O(2B)

= (x)(0)0 [(x)(0)SI 0 (x)(0)0SI0

]+O(2B) . (3.71)

We may recall from Quantum Field Theory that the second term inside the square brackets servesto cancel disconnected contractions, just like the subtractions in Eq. (3.10) did for the free energy

39

density. Therefore, we will drop the second term in the following, and replace the expectationvalue in the first term by ...0,c, like we already did in Eq. (3.21).

Now, let us start by inspecting the leading (zeroth order) term in Eq. (3.71), in order learn howmphys can be conveniently extracted from the propagator. Introducing the notation

P

limT0

P

=

dd+1P

(2)d+1, (3.72)

the free propagator reads (cf. Eq. (3.26))

G0(x) = (x)(0)0 =

P

eiP x

P 2 +m2. (3.73)

For Eq. (3.70), we need to project to zero spatial momentum, p = 0; evidently, this can be achievedby taking a spatial average of G0(x):

ddx (,x)(0)0 =

dp02

eip0

p20 +m2. (3.74)

We see that we get an integral which can be evaluated with the help of the Cauchy theorem and,in particular, that the exponential fall-off of the correlation function is determined by the poleposition of the momentum-space propagator:

ddx (,x)(0)0 =1

22i

em

2im, 0 . (3.75)

Hence,m2phys

=0

= m2 , (3.76)

and, more generally, the physical mass can be extracted by determining the pole position of the fullpropagator in momentum space.

We then proceed to the second term in Eq. (3.71):

(x)(0)SI 0,c = B4

z

(x)(0) (z)(z)(z)(z)0,c

= B4

z

4 3 (x)(z)0 (z)(0)0 (z)(z)0

= 3BG0(0)

z

G0(z)G0(x z)

= 3B

P

1

P 2 +m2B

z

Q,R

eiQzeiR(xz)1

Q2 +m2B

1

R2 +m2B

= 3BI0(mB)

R

eiRx

(R2 +m2B)2. (3.77)

Summing together with Eq. (3.73), the full propagator reads

G(x) =

P

eiP x[

1

P 2 +m2B 3BI0(mB)

1

(P 2 +m2B)2+O(2B)

]

=

P

eiP x

P 2 +m2B + 3BI0(mB)+O(2B) , (3.78)

where we have effectively resummed a series of higher order corrections.

The same steps that lead us from Eq. (3.74) to (3.76) now produce

m2phys = m2B + 3BI0(mB) +O(

2B) . (3.79)

40

Recalling from Eq. (3.68) that m2B = m2R+O(R), B = R+O(

2R), this relation can be inverted,

to givem2B = m

2phys 3RI0(mphys) +O(

2R) . (3.80)

This corresponds precisely to Eq. (3.67). The function I0, given in Eq. (2.73), diverges in the limit 0,

I0(mphys) = m2phys162

2[1

+ ln

2

m2phys+ 1 +O()

], (3.81)

and we may hope that the divergence cancels the unphysical ones that we found in f(T ).

Indeed, let us take the step from Eq. (3.66) to Eq. (3.69), employing the explicit expression fromEq. (3.45),

f(T ) = J(mB, T ) +3

4B [I(mB, T