Aft Notes

Advanced Quantum Field TheoryProf Arttu Rajantie

2014-15 Second Term

Contents

I Path Integral Quantization 3

1 Derivation of the Path Integral 31.1 Heuristic Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Many Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Scalar Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Correlation Functions 82.1 Path Integral Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 Generating Functional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Scalar Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4 Complex Scalar Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.5 Dirac Propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Gauge Fields 173.1 Abelian Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Non-Abelian Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 Gauge Fixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.4 Gauge and Ghost Propagators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.5 Summary of Propagators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Interaction Vertices 254.1 Scalar Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Connected and 1PI Correlators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3 Complex Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.4 QED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.5 QCD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

II Renormalisation 38

5 Perturbative Renormalisation 385.1 Ultraviolet Divergences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.2 Renormalised Couplings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415.3 Field Renormalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.4 Power Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6 Renormalised Perturbation Theory 476.1 Counterterms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2 Dimensional Regularisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.3 Minimal Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

1

7 Renormalisation Group 557.1 Callan-Symanzik Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.2 Wilsonian Renormalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587.3 Renormalisation Group Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 627.4 Renormalisation Group Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

8 Renormalisation of QCD 678.1 Counterterms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678.2 Correlation Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

8.2.1 Gluon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 718.2.2 Quark . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758.2.3 Ghost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.2.4 Quark-gluon coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 778.2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

8.3 Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 788.4 Renormalisation Group Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Index 81

2

Part I

Path Integral Quantization

1 Derivation of the Path Integral

1.1 Heuristic Argument

Double slit experiment:

S

O

A1

A2

- Light emitted at source S, observed at detector O- Probability given by the modulus of the amplitude squared:

p(S → O) = |A(S → O)|2 (1)

- Superposition principle: Total amplitude is sum over two holes

A(S → O) = A(S → A1 → O) +A(S → A2 → O) (2)

- Drill n holes:

S

O

A1

An

A(S → O) =

n∑i=1

A(S → Ai → O) (3)

- Add another screen:

3

S

O

A1,1

A1,n

A2,1

A2,n

A(S → O) =

n∑i,j=1

A(S → A1,i → A2,j → O) (4)

- Add infinite number of screens, drill infinite number of holes in each:

S

O

⇒ Sum over all paths:A(S → O) =

∑all paths

A(S → (path)→ O) (5)

1.2 Quantum Mechanics

- Non-relativistic particle in 1D potential:

H =p2

2m+ V (q) (6)

- Amplitude from point qa to point qb in time T :

U(qa, qb;T ) = 〈qb|U(T )|qa〉, (7)

where the time evolution operator is

U(t) = exp(−itH/~) (8)

- Split interval T into N shorter ones of duration ε = T/N , so that U(T ) = U(ε)N

U(qa, qb;T ) = 〈qb|U(ε)U(ε) · · · U(ε)U(ε)|qa〉, (9)

4

U(qa, qb;T ) =

∫ N−1∏j=1

dqj

N−1∏k=0

dpk2π~

exp

[i

~

N−1∑k=0

(pk(qk+1 − qk)− εH(pk, qk+1))

](17)

- Integral over all values of p and q at all times- One more p integral than q integral- Denote integral by shorthand Dq(t)Dp(t) and note that qk+1 − qk → εq

U(qa, qb;T ) =

∫ q(T )=qb

q(0)=qa

Dq(t)Dp (t) exp

[i

~

∫ T

0

dt (p q −H(p, q))

], (18)

- Boundary conditions: q(0) = qa and q(T ) = qb

- Hamiltonian path integral: Very general, works for any H- Usually p only appears as a square: In our case p2/2m

⇒ one can do the p integrals in Eq. (17) explicitly∫dp

2π~exp

[i

~(p∆q − εH(p, q))

]= exp

(− iε

~V (q)

)∫dp

2π~exp

[− iε

2m~

(p2 − 2m∆q

εp

)]= e

iε~ ( m

2ε2∆q2−V (q))

∫dp

2π~e−

iε2m~ (p−m∆q

ε )2

= eiε~ ( m

2ε2∆q2−V (q))

∫dp

2π~exp

(− iε

2m~p2

)

- We need to evaluate the Gaussian integral over p:- Strictly speaking not defined, because does not converge- Use analytic continuation∫

dp

2πexp

(−1

2cp2

)=

1√2πc

, c =iε

m~(19)

⇒ ∫dp

2π~exp

[i

~(p∆q − εH(p, q))

]=

√m

2π~iεexp

[iε

~

( m2ε2

∆q2 − V (q))]

(20)

- Thus, the amplitude is

U(qa, qb;T ) =( m

2π~iε

)N/2 ∫ N−1∏j=1

dqj

exp

[i

~

N−1∑k=0

ε( m

2ε2(qk+1−qk)2 − V (qk+1)

)]

→∫Dq(t) exp

[i

~

∫ T

0

dt

(1

2mq2 − V (q)

)]

=

∫ q(T )=qb

q(0)=qa

Dq(t) exp

[i

~

∫ T

0

dtL(q, q)

]=

∫ q(T )=qb

q(0)=qa

Dq(t)eiS/~ (21)

- This is the path integral expression of the amplitude- Note: When action is large, S =

∫dtL ~, integrand highly oscillatory

- Stationary phase approximation:Only paths where S is roughly stationary contribute significantly: δS/δq = 0

= Classical Lagrangian mechanics

6

1.3 Many Particles

- Generalise to M distinguishable particles:- Operators qα and pα, with α = 1, . . . ,M

- For notational simplicity, define M -component vectors q = (q1, . . . , qM ) and p = (p1, . . . , pM )

- Consider Hamiltonian H(p, q)

- Everything is more or less identical to the one-particle case:

U(qa,qb;T ) = 〈qb|U(ε)U(ε) · · · U(ε)U(ε)|qa〉 (22)

- Different components of the position operator q commute, so a complete set of states is

1 =

∫ ( M∏α=1

dqαk

)|qk〉〈qk| (23)

- The analogue of Eq. (11) is

U(qa,qb;T ) =

∫ N−1∏j=1

M∏α=1

dqαj

N−1∏k=0

〈qk+1|U(ε)|qk〉 (24)

- The matrix elements of the Hamiltonian are [cf. Eq. (13)]

〈qk+1|H(p, q)|qk〉 =

∫ ( M∏α=1

dpαk

)H(pk,qk+1)〈qk+1|pk〉〈pk|qk〉 (25)

- The scalar product between eigenstates of p and q is

〈p |q〉 = (2π~)−M/2 exp

(− i~

M∑α=1

p αqα

)(26)

- Consequently, for an infinitesimal time evolution we find

〈qk+1|U(ε)|qk〉 =

∫ ( M∏α=1

dpαk2π~

)exp

[i

~

(∑α

pαk (qαk+1 − qαk )− εH(pk,qk+1

)], (27)

and for finite T , in analogy with Eq. (17)

U(qa,qb;T ) =

∫ ∏α

∏j

dqαj∏k

dpαk2π~

exp

[i

~∑k

(∑α

pαk (qαk+1 − qαk )− εH(pk,qk+1)

)](28)

- Assume that the Hamiltonian is

H(p, q) =∑α

(pα)2

2m+ V (q) (29)

and do the p integrals using Eq. (19)

U(qa,qb;T ) =

∫ ∏α

√ m

2π~iε∏j

√m

2π~iεdqαj

exp

[i

~∑k

ε

(m

2

∑α

(qαk+1−qαk )2

ε2− V (qk+1)

)]

→∫Dq(t) exp

[i

~

∫ T

0

dt

(1

2m∑α

qαqα − V (q)

)](30)

7

1.4 Scalar Field Theory

- Consider now the Hamiltonian of a 1+1D relativistic scalar field

H =

∫dx

1

2π2 +

1

2

(dφ

dx

)2

+ V (φ)

(31)

- Discretize: Lattice spacing δx⇒ x = αδx

φ(x)→ φα

dφ/dx→ (φα+1 − φα)/δx

- Because the canonical commutation relation for fields is

[φ(t, x), π(t, y)] = i~δ(x− y) = i~δαβδx

, (32)

we define pα = π(x)δx

- Then we have [φα, pβ ] = i~δαβ as for particles- This gives the Hamiltonian

H(p, φ) =1

2δx

∑α

pαpα + V(φ), V(φ) =∑α

δx

[1

2δx2

(φα+1 − φα

)2

+ V (φα)

](33)

which is of the same form as Eq. (29) but with m = δx

- The amplitude from field configuration φa(x) to φb(x) in time T is therefore

U(φa, φb;T ) =

∫Dφα(t) exp

[i

~

∫ T

0

dt

(∑α

δx1

2φαφα − V(φ)

)]

=

∫Dφα(t) exp

[i

~

∫ T

0

dt∑α

δx

(1

2φαφα − 1

2

(φα+1 − φα)2

δx2− V (φα)

)]

→∫Dφ (t, x) exp

[i

~

∫ T

0

dt

∫dx

(1

2φ2 − 1

2

(∂φ

∂x

)2

− V (φ(t, x))

)]

=

∫Dφ exp

[i

~

∫d2x

(1

2∂µφ∂

µφ− V (φ)

)]=

∫ φ(T,x)=φb(x)

φ(0,x)=φa(x)

Dφ eiS/~ (34)

- The integral is over all functions φ(t, x) of two-dimensional Minkowski spacetime:µ = 0, 1, ηµν = diag(1,−1)

- Action S =∫d2xL, where L is the Lagrangian density

- Boundary conditions φ(0, x) = φa(x), φ(T, x) = φb(x)

- Spacetime integral from 0 to T- Lorentz invariant apart from boundary conditions

2 Correlation Functions

2.1 Path Integral Representation

- Practical applications:Correlation functions such as DF (t1 − t2, x1 − x2) ≡ 〈0|T φ(t1, x1)φ(t2, x2)|0〉 rather than amplitudes

8

- Heisenberg picture: Field operator φ(t, x) = U(−t)φ(0, x)U(t)

- Start with a slightly different problem: Calculate 〈φb;T |φ(t1, x1)φ(t2, x2)|φa;−T 〉:- |φa; t〉 denotes the eigenstate of φ(t, x) with eigenvalue φa(x)

φ(t, x)|φa; t〉 = φa(x)|φa; t〉 (35)

- Note: The t-dependence of |φa; t〉 is NOT Schrodinger picture time evolution- Instead: |φa; t〉 = U(−t)|φa; 0〉

- At initial time t = −T , the system is in field eigenstate with eigenvalue φa(x),and at final time t = T in another field eigenstate with eigenvalue φb(x)

- To simplify notation, write |φa〉 ≡ |φa; 0〉- Consider the matrix element

〈φb;T |φ(t1, x1)φ(t2, x2)|φa;−T 〉 = 〈φb|U(T − t1)φ(x1) U(t1 − t2)φ(x2)U(t2 + T )|φa〉 (36)

- Assume that t1 > t2 so that we can repeat the steps in Section 1⇒ Path integral representation- Assume now that ti = kε:

In Eq. (11) we obtain a factor

〈φk|φ(x)U(ε)|φk−1〉 = φk(x)〈φk|U(ε)|φk−1〉 (37)

- Thus, all it gives is an extra factor of φ(t, x) in the integrand,

〈φb;T |φ(t1, x1)φ(t2, x2)|φa;−T 〉 =

∫ φ(T,x)=φb(x)

φ(−T,x)=φa(x)

Dφ eiSφ(t1, x1)φ(t2, x2), (38)

- Note: RHS independent of the order of φs, LHS notWe assumed t1 > t2, so the path integral gives the time-ordered correlation function∫ φ(T,x)=φb(x)

φ(−T,x)=φa(x)

Dφ eiSφ(xµ)φ(yµ) = 〈φb;T |T φ(xµ)φ(yµ)|φa;−T 〉 (39)

- Easy to generalise to arbitrary operators O[φ]:∫ φ(T,x)=φb(x)

φ(−T,x)=φa(x)

Dφ eiSO[φ] = 〈φb;T |T O[φ]|φa;−T 〉 (40)

- We need correlation functions in vacuum |0〉:Take T → (1− iε)T , and T →∞

- Insert a complete set of energy eigenstates |n〉 with eigenvalues En, i.e., H|n〉 = En|n〉:

|φa;−T 〉 = U(T )|φa〉 = e−iHT |φa〉 =∑n

e−iEnT |n〉〈n|φa〉

→∑n

e−(ε+i)EnT |n〉〈n|φa〉 = e−(ε+i)E0T 〈0|φa〉|0〉+O(e−εE1T

)= e−(ε+i)E0T 〈0|φa〉|0〉 ×

(1 +O

(e−ε(E1−E0)T

))- Inverting this, we obtain

|0〉 = limT→∞

e(ε+i)E0T

〈0|φa〉|φa;−T 〉 (41)

and correspondingly

9

〈0| = limT→∞

e(ε−i)E0T

〈φb|0〉〈φb;T | (42)

- Thus, we find that for arbitrary O[φ] expressed in terms of operators φ(x),

〈0|T O[φ]|0〉 = limT→∞

e2εE0T

〈φb|0〉〈0|φa〉

∫ φb

φa

Dφ eiSO[φ] (43)

- In particular, this is true for the identity operator 1, so we can write

〈0|T O[φ]|0〉 =〈0|T O[φ]|0〉〈0|0〉

=

∫Dφ eiSO[φ]∫Dφ eiS

≡ 〈O[φ]〉 (44)

- Constants hidden in the integration measure Dφ cancel- In principle, the action is integrated over complex time path

S =

∫ (1−iε)∞

−(1−iε)∞dt

∫dxL (45)

- Dependence on the boundary conditions of φ has dropped out(unless there are degenerate vacua, e.g. spontaneous symmetry breaking!)

2.2 Generating Functional

- In order to calculate correlators (44), it is useful to define a generating functional

Z[J ] =

∫Dφ exp

[iS + i

∫ddxJ(x)φ(x)

](46)

where J(x) is an arbitrary function of spacetime- Correlation functions are given by functional derivatives with respect to J ,

〈φ(x1) · · ·φ(xN )〉 =(−i)N

Z[0]

δ

δJ(x1)· · · δ

δJ(xn)Z[J ]

∣∣∣∣J=0

(47)

- Consider free scalar field (in d dimensions) with V (φ) = 12m

2φ2

- The action is

S =

∫ddx

[1

2∂µφ∂

µφ− 1

2m2φ2

](48)

- By integrating by parts, the exponent in the path integral becomes

iS = −1

2

∫ddxddyφ(x)M(x, y)φ(y), (49)

- where M(x, y) is the operator

M(x, y) = iδ(x− y)[∂µ∂

µ +m2]

(50)

- More generally the action of any free field can be written in the form (49)- Eq. (46) is the continuum version of a multivariate Gaussian integral

Z(J) =

∫dNq exp

(−1

2qTMq + JTq

)(51)

where q is an N -component vector and M is a symmetric N ×N matrix- The moments of the Gaussian integral,

10

〈qi1 · · · qin〉 ≡∫dNq qi1 · · · qin exp

(− 1

2qTMq)∫

dNq exp(− 1

2qTMq) , (52)

are given by derivatives

〈qi1 · · · qin〉 =1

Z(0)

∂

∂Ji1· · · ∂

∂JinZ(J)

∣∣∣∣J=0

(53)

- We can evaluate Z(J) by diagonalising the matrix M:

M = ΛMΛT , Mij = miδij =

m1 0

m2

. . .0 mN

(54)

and writing q = Λq, J = ΛJ.Because Λ is orthogonal, the integral becomes

Z(J) =

∫dNq exp

(−1

2qTMq + JT q

)=∏i

[∫dqi exp

(−1

2miq

2i + Jiqi

)]

=∏i

[√2π

miexp

(J2i

2mi

)]=

(2π)N/2√det M

exp

(1

2JTM−1J

)=

(2π)N/2√det M

exp

(1

2JTM−1J

)(55)

- The second moment is

〈qjqj〉 =1

Z(0)

∂

∂Ji

∂

∂JjZ(J)

∣∣∣∣J=0

=1

Z(0)

∂

∂Ji

[(M−1)jkJkZ(J)

]∣∣∣∣J=0

= (M−1)ij (56)

- In an nth order moment 〈qi1 · · · qin〉- Half of the derivatives must act on the exponent, bringing down factors of Ji- The other half has to act on these factors, removing them- Each pair i, j gives a factor (M−1)ij = 〈qiqj〉- Thus,

〈qi1 · · · qin〉 = 〈qi1qi2〉 · · · 〈qin−1qin〉+ (all other pairings) (57)

- For example,

〈qiqjqkql〉 = 〈qiqj〉〈qkql〉+ 〈qiqk〉〈qjql〉+ 〈qiql〉〈qjqk〉 (58)

- This means that to evaluate the moments of a Gaussian integral all we need to know isthe “two-point function” 〈qiqj〉

- In the continuum, one finds in analogy with Eq. (56),

〈φ(x)φ(y)〉 = (M−1)(x, y) (59)

where the inverse M−1(x, y) is defined as∫d4yM(x, y)M−1(y, z) = δ(x− z) (60)

- Similarly, Eq. (57) becomes Wick’s theorem:The N -point correlator is given by the sum of all possible contractions in terms of products of

11

two-point functions

2.3 Scalar Propagator

- Consider a free real scalar field in 3+1 dimensions

L =1

2∂µφ∂

µφ− 1

2m2φ2 (61)

- By integrating by parts, we can write the action as

iS = − i2

∫d4xd4y φ(x)δ(x− y)

(∂µ∂

µ +m2)φ(y) (62)

- This is of the quadratic form (49), with

M(x, y) = iδ(x− y)(∂µ∂

µ +m2)

(63)

- The two-point function (i.e. the Feynman propagator), is given by Eq. (59)

DF (x− y) = 〈φ(x)φ(y)〉 = (M−1)(x, y) (64)

- We can invert the matrix M by taking its Fourier transform

M(k, q) =

∫d4yd4x eikµx

µ

M(xµ, yµ)eiqµyµ

= i

∫d4x eikµx

µ (∂µ∂

µ +m2)eiqµx

µ

= i

∫d4x ei(kµ+qµ)xµ

(−qµqµ +m2

)= −i(2π)4δ(k + q)

(k2 −m2

)(65)

- Inserting this into the Fourier transform of Eq. (60),∫d4q

(2π)4M(k, q)M−1(−q,−p) = (2π)4δ(k − p), (66)

we find

(M−1)(k, q) = (2π)4δ(k + q)i

k2 −m2(67)

- Transforming back to coordinate space, we have

DF (x− y) = (M−1)(x, y) =

∫d4k

(2π)4

d4q

(2π)4e−ikµx

µ−iqµyµM−1(k, q)

=

∫d4k

(2π)4

ie−ikµ(xµ−yµ)

k2 −m2=

∫d4k

(2π)4


k20 − ~k2 −m2

(68)

- Two problems:(i) M is imaginary→ Integral (51) does not converge(ii) The integral over k0 in Eq. (68) crosses the poles at k0 = ±

√~k2 +m2

The complex time used in Eq. (45) solves these problems:- When we rotate t→ (1− iε)t, we also have to change

12

dt→ (1− iε)dt, ∂0 =∂

∂t→ 1

1− iε∂

∂t(69)

- Therefore the action becomes

S = (1− iε)∫d4x

[1

2(1− iε)2(∂0φ)

2 − 1

2(∂iφ)

2 − 1

2m2φ2

], (70)

and the matrix M becomes

M(x, y) = (ε+ i)δ(x− y)

[∂2

0

(1− iε)2− ∂2

i +m2

](71)

- Real part ε⇒ Gaussian integral becomes well defined- Frequency k0 turns into k0/(1− iε) and the Feynman propagator becomes

DF (x− y) = (M−1)(x, y) =1

1− iε

∫d4k

(2π)4


(1− iε)−2k20 − ~k2 −m2

(72)

- This shifts the poles away from the real axis to k0 = ±(1− iε)√~k2 +m2

- Note: We are taking ε→ 0

- The residues are finite, so we can ignore the prefactor 1/(1− iε)- For the poles, it only matters which way we go around them, so we can modify the expression to

DF (x− y) =

∫d4k

(2π)4


k2 −m2 + iε, (73)

- This moves the poles to k0 = ±√~k2 +m2 − iε

- Lorentz invariant- Finally, it is illuminating to note that if we had made time fully imaginary in Eq. (70), we would have found that

iS → −∫d4x

[1

2(∂0φ)

2+

1

2(∂iφ)

2+

1

2m2φ2

]≡ −SE (74)

- The path integral becomes real ∫Dφe−SE (75)

- Same form as classical canonical partition function∑

exp(−βH)

- 3+1D QFT↔ 4D classical statistical mechanics

2.4 Complex Scalar Field

- Consider now a complex scalar field φ, with Lagrangian

L = ∂µφ∗∂µφ−m2φ∗φ (76)

(Note the different normalisation! No 1/2s!)- Write φ in terms of real and imaginary parts

φ =1√2

(φR + iφI) (77)

- The Lagrangian becomes

13

L =1

2∂µφR∂

µφR +1

2∂µφI∂

µφI −1

2m2φ2

R −1

2m2φ2

I (78)

- Two uncoupled real scalar fields!- The Feynman propagator is

〈φ∗(x)φ(y)〉 =1

2[〈φR(x)φR(y)〉+ i〈φR(x)φI(y)〉 − i〈φI(x)φR(y)〉+ 〈φI(x)φI(y)〉]

=1

2[〈φR(x)φR(y)〉+ 〈φI(x)φI(y)〉] (79)

where we have used 〈φR(x)φI(y)〉 = 0

- Fields φR and φI are identical and their two-point functions are given by Eq. (72), so we find thatthe propagator for the complex field is the same as for the real field

〈φ∗(x)φ(y)〉 = DF (x− y) =

∫d4k

(2π)4


k2 −m2 + iε(80)

2.5 Dirac Propagator

- Fermions: Equal-time anticommutation relations

ψα(~x), ψ†β(~y) = δ(~x− ~y)δαβ , ψα(~x), ψβ(~y) = ψ†α(~x), ψ†β(~y) = 0 (81)

- Cannot be represented by normal path integrals:Real/complex numbers commute

〈ψα(~x), ψβ(~y)〉 = 2〈ψα(~x)ψβ(~y)〉 6= 0 (82)

- Instead: Use path integrals over Grassmann numbers- Anticommuting: For any pair θ and η,

θη = −ηθ (83)

⇒ θ2 = −θ2 = 0

- Generators of Grassmann algebra:- Vector space: Addition and scalar multiplication- Multiplication rule fixed by the rule for single Grassmann numbers- Elements of algebra not anticommuting in general: (θ1η1)(θ2η2) = (θ2η2)(θ1η1).

- General function of a Grassman number θ can be written as f(θ) = A+ θB

(Taylor expansion: θ2 = 0)- A and B can be any elements of the algebra

- Derivativedθ

dθ= 1,

dconstdθ

= 0 ⇒ d

dθf(θ) = B (84)

- Acts on the first factord

dθηθ = − d

dθθη = −

(d

dθθ

)η = −η (85)

14

- Define integral over θ as an analog of∫∞−∞ dx

- Should be linear: For any elements C and D of the Grassmann algebra∫dθ [f(θ)C + g(θ)D] =

(∫dθf(θ)

)C +

(∫dθg(θ)

)D (86)

- This implies that for f(θ) = A+ θB∫dθf(θ) =

(∫dθ

)A+

(∫dθθ

)B (87)

- Invariant under arbitrary shift θ → θ + η:∫dθ θ =

∫dθ(θ + η) =

∫dθ θ +

∫dθη (88)

- This implies (∫dθ

)η = 0 (89)

and because η is arbitrary,∫dθ = 0

- Finally, we choose the overall normalisation to be∫dθ θ = 1 (90)

so that for a general function f(θ) = A+ θB we have∫dθ f(θ) = B (91)

- Note: Integral = Derivative!- Complex Grassman numbers

θ =θR + iθI√

2, θ∗ =

θR − iθI√2

(92)

- Integration ∫dθ∗dθ = −i

∫dθRdθI (93)

- One-dimensional Gaussian integral (with complex b)∫dθ∗dθ exp(−θ∗bθ) =

∫dθ∗dθ(1− θ∗bθ) =

∫dθ∗dθ(1 + θθ∗b) = b (94)

- n-dimensional complex Gaussian integral (i = 1, . . . , n, Hermitean Mij , sum over i,j )

∫ ( n∏i=1

dθ∗i dθi

)exp

−∑ij

θ∗iMijθj

=

∫ (∏i

dθ∗i dθi

)exp (θjθ

∗iMij)

=

∞∑k=0

1

k!

∫ (∏i

dθ∗i dθi

)(θjθ

∗iMij)

k (95)

- Only the k = n term survives:For k > n some θi has to appear twice – Vanishes because θ2

i = 0

For k < n some θi will not appear at – Vanishes because∫dθi = 0

15

∫ (∏i

dθ∗i dθi

)exp (−θ∗iMijθj) =

1

n!

∫ (∏i

dθ∗i dθi

)(θjθ

∗iMij)

n (96)

- Each θi can only appear once: n! such termsFactors θjθ∗i commute⇒ Put θs in decreasing numerical order

1

n!(θjθ

∗iMij)

n=(θnθ∗inMin,n

)· · ·(θ1θ∗i1Mi1,1

)(97)

- Put the θ∗i factors in the same order: Gives a factor εi1,...,in(Levi-Civita tensor, ε12...n = 1, fully antisymmetric)

1

n!(θjθ

∗iMij)

n= εi1,...,inMi1,1 · · ·Min,nθnθ

∗n · · · θ1θ

∗1 (98)

and therefore∫ (∏i

dθ∗i dθi

)exp (−θ∗iMijθj) = εi1,...,inMi1,1 · · ·Min,n

∫ (∏i

dθ∗i dθi

)θnθ∗n · · · θ1θ

∗1 (99)

- The integral is equal to 1, and the prefactor is just the definition of a determinant

det M = εi1,...,inMi1,1 · · ·Min,n (100)

and therefore we have ∫ (∏i

dθ∗i dθi

)exp (−θ∗iMijθj) = det M (101)

(cf. normal complex numbers (2π)N/ det M)- External field η (Grassmannian)

Z(η∗, η) ≡∫ (∏

i

dθ∗i dθi

)exp (−θ∗iMijθj + η∗i θi + θ∗i ηi) (102)

- Shift θi → θi + (M−1)ijηj

θ∗i → θ∗i + η∗j (M−1)ji

Z(η∗, η) =

∫ (∏i

dθ∗i dθi

)exp

(−θ∗iMijθj + η∗i (M−1)ijηj

)= eη

∗i (M−1)ijηj det M (103)

- The second moments can be calculated as derivatives

〈θiθ∗j 〉 =1

Z(0, 0)

∂

∂ηj

∂

∂η∗iZ(η∗, η)

∣∣∣∣η∗=η=0

=∂

∂ηj

∂

∂η∗ieη∗i (M−1)ijηj

∣∣∣∣η∗=η=0

= (M−1)ij (104)

- The same result as for normal numbers (but keep track of the order)- Dirac Lagrangian

L = ψα(i /∂ −m)αβψβ (105)

16

where ψ = ψ†γ0, /∂ = γµ∂µ,and α, β = 1 . . . 4 are spinor indices- The action is

iS =

∫d4xd4yψα(x)iδ(x− y)(i /∂ −m)αβψβ(y) (106)

⇒Matrix M

Mαβ(x, y) = −iδ(x− y)(i /∂ −m)αβ (107)

- Feynman propagator

SF,αβ(x− y) ≡ 〈ψα(x)ψβ(y)〉 = (M−1)αβ(x, y) =

∫d4k

(2π)4

(ie−ikµ(xµ−yµ)

/k −m+ iε

)αβ

(108)

3 Gauge Fields

3.1 Abelian Symmetry

- The QED Lagrangian is

L = −1

4FµνF

µν + ψ(i /D −m)ψ, (109)

where ψ is the electron field (and ψ = ψ†γ0)Fµν = ∂µAν − ∂νAµ is the field strength tensorDµ = ∂µ + ieAµ is the covariant derivative

- Path integral:

〈O〉 =

∫DAµDψDψeiSO[Aµ, ψ, ψ]∫

DAµDψDψeiS. (110)

- To derive the photon propagator, let us focus on the Maxwell term

S =

∫d4x

(−1

4FµνF

µν

); Fµν = ∂µAν − ∂νAµ

=1

2

∫d4xAµ(x)

(∂2gµν − ∂µ∂ν

)Aν(x); gµν = diag(1,−1,−1,−1)

=1

2

∫d4k

(2π)4Aµ(k)

(−k2gµν + kµkν

)Aν(−k) (111)

- The matrix we need to invert is therefore

Mµν(k, q) = i(2π)4δ(k + q)(k2gµν − kµkν

)(112)

- Unfortunately, this matrix cannot be inverted:- Zero eigenvalue:

Mµν(k, q)kν = 0 (113)

⇒ singular- The reason is gauge invariance: Gauge transformations

17

ψ(x) → eiθ(x)ψ(x)

Aµ(x) → Aµ(x)− 1

e∂µθ(x) (114)

which in momentum space is

Aµ(k)→ Aµ(k)− i

ekµθ(k) (115)

does not change the action (or indeed any physical observable)- Momentum space: Longitudinal component ALµ ∝ kµ unphysical- The path integral will be infinite (even with complex time)- Will be cured by gauge fixing

3.2 Non-Abelian Symmetry

- This is a brief reminder of the basic features of non-Abelian gauge theories.For more details, see the Unification course, Chapter 15 in Peskin&Schroederor Chapter 9 in Bailin&Love

- SU(N ) gauge symmetry + fermions in fundamental representation (for QCD, N = 3)- Other gauge groups and scalar fields follow the same lines

- The Lagrangian is

L = −1

2TrFµνF

µν + ψ(i /D −m)ψ, (116)

where - ψ is an N -component vector consisting of spinors ψi(i = 1, . . . , N labels different “colours”)

- Dµ = ∂µ + igAµ is the covariant derivative(be careful with the sign of g: Peskin&Schroder use “-”)

- The gauge field Aµ is a traceless, self-adjoint(=Hermitean) N ×N matrix(element of the SU(N ) Lie algebra)

- The field strength tensor is

Fµν = − ig

[Dµ, Dν ] = ∂µAν − ∂νAµ + ig[Aµ, Aν ] (117)

- The Lagrangian is invariant under transformation of the form

ψ(x)→ U(x)ψ(x), (118)

where U(x) is an element of the Lie group SU(N )[= unitary (U†U = 1), special (detU = 1), N ×N matrices]provided that the covariant derivative transforms as

Dµ(x)→ U(x)Dµ(x)U†(x) (119)

[because then Fµν(x)→ U(x)Fµν(x)U†(x)]- From this, we can determine how the gauge field has to transform

- Let us write

18

Dµ → Dµ ≡ ∂µ + igAµ (120)

- According to Eq. (119), this is equal to

Dµ = UDµU† = U(∂µ + igAµ)U† = ∂µ + ig

(UAµU

† − i

gU∂µU

†)

(121)

- Therefore Aµ must transform as

Aµ → UAµU† − i

gU∂µU

† (122)

(Check that Aµ remains traceless and self-adjoint!)

- Let us then count the number of independent degrees of freedom in the field Aµ:- A general complex N ×N matrix has 2N2 real degrees of freedom- Self-adjoint: Removes half of them⇒ N2

- Traceless: Removes one⇒ Altogether 2N2 −N2 − 1 = N2 − 1 real degrees of freedom

- We can write Aµ as a linear combination of matrices ta

Aµ = Aaµta, a = 1, . . . , N2 − 1 (123)

where Aaµ are real numbers- The matrices ta are known as the generators of the group- In principle any set of linearly independent, traceless, self-adjoint N ×N matrices would do

In practice, it is conventional to choose them in such a way that

Trtatb =1

2δab (124)

- The matrices do not commute, but the commutator is traceless and anti-self-adjoint, so we can write

[ta, tb] = ifabctc, (125)

where the coefficients fabc are known as structure constants- Note that

Tr [ta, tb]tc = ifabdTr tdtc =i

2fabc, (126)

from which it follows that fabc is cyclic and fully antisymmetric- For SU(2) one can choose ta = σa/2, where σa are the Pauli matrices⇒ Then the structure constants are given by the Levi-Civita tensor fabc = εabc

- Furthermore, the structure constants satisfy the Jacobi identity

fadef bcd + f bdef cad + f cdefabd = 0 (127)

- Other representations: Any set of p× p matrices T a which satisfy Eq. (125)- ψ is then a p-component vector, and Dµψ = ∂µψ + igAaµT

aψ

- E.g. adjoint representation T abc = ifabc

- For more details on groups, representations etc. see Peskin&Schroeder 15.4(Note the standard set of SU(3) generators on p.502!)

19

- Yang-Mills Lagrangian

L = −1

2TrFµνF

µν (128)

- In terms of the components Aaµ, the field strength tensor is

Fµν = ∂µAν − ∂νAµ + ig[Aµ, Aν ]

= (∂µAaν − ∂νAaµ)ta + igAbµA

cν [tb, tc]

= (∂µAaν − ∂νAaµ − gfabcAbµAcν)ta (129)

- Thus, if we write Fµν = F aµνta, we have

F aµν = ∂µAaν − ∂νAaµ − gfabcAbµAcν (130)

- The Lagrangian Eq. (128) is

L = −1

4F aµνF

aµν

= −1

4(∂µA

aν − ∂νAaµ − gfabcAbµAcν)(∂µAa ν − ∂νAaµ − gfadeAd µAe ν)

= −1

4(∂µA

aν − ∂νAaµ)(∂µAaν − ∂νAaµ)

+g

2fabc(∂µA

aν − ∂νAaµ)AbµAcν − g2

4fabcfadeAbµA

cνA

d µAe ν (131)

- Not purely quadratic: Interacting theory- Limit g → 0: Free theory

- Simply (N2 − 1) copies of the Maxwell term (111)⇒ Singular, cannot be inverted: We need to fix the gauge

3.3 Gauge Fixing

- Consider the integral

IO =

∫DAµO[Aµ]eiS[Aµ] (132)

- Observables given by ratios 〈O〉 = IO/I1.- Denote gauge transformed field by

AUµ = UAµU† − i

gU∂µU

† (133)

- Gauge invariance: S[AUµ ] = S[Aµ]

⇒ (Gaussian) integral diverges- Fix the gauge:

- Impose a gauge condition G[Aµ] = 0 that removes the ambiguity- G[Aµ] is some functional of Aµ, which maps a field configuration Aµ(x) to

a Lie-algebra-valued function of x, with N2 − 1 real-valued components Ga[Aµ](x)

- For instance G[Aµ] = ∂µAµ (Lorenz gauge)

- For a given fixed field configuration Aµ(x), G[AUµ ] defines a map from gauge transformations U(x)

20

to Lie-algebra-valued functions of x- Write the unity as an integral over a delta function

1 =

∫DGδ(G) =

∫DUδ

(G[AUµ ]

)det

(δG[AUµ ]

δU

), (134)

where DU is the Haar measure (the unique invariant measure for the Lie group)and the Jacobian determinant when changing the integration variable from G to U- Invariant under multiplication by a constant group element V∫

DU f(U) =

∫DU f(UV ) (135)

- This determines the measure uniquely up to a multiplicative constant- The integral gets a contribution only from point U0, where G[AU0 ] = 0

⇒ Consider infinitesimal region around U0

- We write U = (1 + iα)U0 where α is infinitesimalself-adjoint (so that U is unitary)and traceless (so that detU = 1)

- We can therefore write α = αata

- For |α| 1, the integration measure becomes

DU →∏a

Dαa (136)

- We can write the Jacobian explicitly as

det

(δG[AUµ ]

δU

)→ det

(δGa[A

(1+iα)U0µ ]

δαb

)(137)

- Let us now consider a particular choice of the gauge fixing functional

G[Aµ] = ∂µAµ(x)− ω(x) (138)

where ω(x) is some fixed Lie-algebra-valued function- This means imposing the gauge condition ∂µAµ(x) = ω(x)

- To compute the Jacobian, we Taylor expand in α

δAU0µ = A(1+iα)U0

µ −AU0µ

= (1 + iα)AU0µ (1− iα)− i

g(1 + iα)∂µ(1− iα)−AU0

µ

= i[α,AU0µ ]− 1

g∂µα = −1

g

(∂µα

a + gfabcαb(AU0µ )c

)ta (139)

- Thus,

(δAUµ )a = −1

g

(∂µα

a + gfabcαb(AUµ )c)

= −1

g

(δab∂µ + gf cab(AUµ )c

)αb ≡ −1

g(DU

µ )abαb, (140)

where we can now think of δAUµ and α as (N2−1)-component vectors rather than N ×N matrices

21

- The operator (DUµ )ab is the covariant derivative in the adjoint representation

- Therefore we findδGa[AUµ ]

δαb=∂µ(δAUµ )a

δαb= −1

g∂µ(DU

µ )ab (141)

- We will omit the indices ab from now on. This operator can be thought of as a multidimensional“matrix” with indices labelled by a = 1, . . . , (N2−1) and spacetime points.

- Thus, we have found that

1 =

∫DUδ

(G[AUµ ]

)det

(−1

g∂µDU

µ

)∝∫DUδ

(G[AUµ ]

)det(i∂µDU

µ

), (142)

where we dropped the factor i/g because it gives an uninteresting constant factor

- Inserting Eq. (142) into Eq. (132) (and changing the order of integrations) we obtain

IO =

∫DUDAµδ

(G[AUµ ]

)det(i∂µDU

µ

)O[Aµ]eiS[Aµ] (143)

- We can change the integration variable from Aµ to Aµ ≡ AUµ , and the measure remains unchanged

DAµ = DAµ (144)

because this only involves a unitary rotation and a shift- Note that Aµ = AU

†

µ

- Because of gauge invariance, S[Aµ] = S[AU†

µ ] = S[Aµ]

- Assuming that the correlator we are calculating is also gauge-invariant,i.e., O[Aµ] = O[AU

†

µ ] = O[Aµ], we have

IO =

∫DUDAµδ

(G[Aµ]

)det(i∂µDµ

)O[Aµ]eiS[Aµ] (145)

- We can now relabel Aµ → Aµ and find

IO =

∫DUDAµδ (G[Aµ]) det (i∂µDµ)O[Aµ]eiS[Aµ] (146)

- The integrand is now independent of U , and the integral factorizes

IO =

∫DU ×

∫DAµδ (G[Aµ]) det (i∂µDµ)O[Aµ]eiS[Aµ] (147)

- Integral over U gives a constant: Volume of the gauge group!- Finite for compact groups such as SU(N)

(Gaussian integral diverges, but it is only an approximation)- Will cancel from any correlator, because they are ratios of these integrals

- Writing the gauge constraint explicitly, we have

IO ∝ IO[ω] ≡∫DAµδ

(∂µAaµ − ωa

)det (i∂µDµ)O[Aµ]eiS[Aµ] (148)

- The integral IO[ω] is independent of ωa

- We average over ωa(x) with a Gaussian weight

22

IO ∝∫Dω IO[ω] exp

(−i∫d4x

ωaωa

2ξ

)∝

∫DωDAµ exp

(−i∫d4x

ωaωa

2ξ

)δ(∂µAaµ − ωa

)det (i∂µDµ)O[Aµ]eiS[Aµ]

=

∫DAµ exp

(−i∫d4x

(∂µAaµ)2

2ξ

)det (i∂µDµ)O[Aµ]eiS[Aµ] (149)

where we changed the order of the integrations- We have exchanged the delta function to an extra Gaussian factor- Can be absorbed into the action as a gauge fixing term

IO ∝∫DAµ det (i∂µDµ)O[Aµ]e

i

(S[Aµ]−

∫d4x

(∂µAaµ)2

2ξ

)(150)

- Physical quantities are independent of the gauge fixing parameter ξ: Useful check!

- According to Eq. (101), a determinant can be written as a Gaussian Grassmann integral- Thus, we introduce a (N2 − 1)-component Grassmann-valued field ca, and write

det (i∂µDµ) =

∫Dc∗Dc exp

(−i∫d4xc∗∂µDµc

)=

∫Dc∗Dc exp

(i

∫d4x (∂µca∗)

(∂µc

a − gfabcAbµcc))

(151)

- This field is:- Anticommuting⇒ fermion field- Scalar (and thus violates the spin-statistics theorem)- In the adjoint representation- Known as the Faddeev-Popov ghost field

- Thus, we have been able to write the integral IO as

IO =

∫DAµDc∗DcO[Aµ]ei

∫d4xLξ , (152)

with the gauge-fixed Lagrangian

Lξ = −1

4(∂µA

aν − ∂νAaµ)(∂µAaν − ∂νAaµ)

+g

2fabc(∂µA


4fabcfadeAbµA

cνA

d µAe ν

− 1

2ξ(∂µAaµ)2 + ∂µca∗∂µc

a − gfabc∂µca∗Abµcc (153)

- By construction, this Lagrangian gives the same expectation values for gauge-invariant operatorsas the original Lagrangian, so it describes exactly the same physics, but it is not gauge invariant

- The result can be directly applied to the Abelian case, too- The U(1) group has only one generator t1 = 1

23

- The colour index a has only one possible value a = 1

- The only structure constant vanishes f111 = 0

- All the interaction terms vanish, including the photon-ghost interaction- Therefore the ghost decouples and can be ignored

3.4 Gauge and Ghost Propagators

- For photons (i.e. Abelian gauge symmetry), the gauge-fixed action is

Sξ =

∫d4x

(−1

4FµνF

µν − 1

2ξ(∂µAµ)2

)=

∫d4x

1

2Aµ

(∂2gµν −

(1− 1

ξ

)∂µ∂ν

)Aν (154)

- Our matrix M becomes

Mµν(k, q) = i(2π)4δ(k + q)

[k2gµν −

(1− 1

ξ

)kµkν

](155)

- This is invertible and gives the photon propagator

〈Aµ(x)Aν(y)〉 =(M−1

)µν

(x, y) =

∫d4k

(2π)4

−ie−ikµ(xµ−yµ)

k2 + iε

[gµν − (1− ξ)kµkν

k2

]≡ DF

µν(x− y) (156)

- Convenient choices for ξ:ξ = 0 Landau gauge (Dµν

F transverse)ξ = 1 Feynman gauge (Dµν

F looks like scalar propagator)

- For gluons (i.e., non-Abelian gauge symmetry), each component has the same propagator as the photon

〈Aaµ(x)Abν(y)〉 = δabDFµν(x− y) =

∫d4k

(2π)4e−ikµ(xµ−yµ)−iδab

k2

[gµν − (1− ξ)kµkν

k2

](157)

- For ghosts, the free action is

S =

∫d4xca∗

(−δab∂2

)cb, (158)

so the matrix to invert is

Mab(k, q) = −i(2π)4δ(k + q)δabk2 (159)

- Therefore, the propagator is

〈ca(x)cb∗(y)〉 =

∫d4k

(2π)4e−ikµ(xµ−yµ) iδab

k2 + iε(160)

24

3.5 Summary of Propagators

- Propagators in momentum space:- Scalar field:

DF (k) =i

k2 −m2 + iε(161)

- Dirac fermion:

SF (k) =i

/k −m+ iε(162)

- Photon:

DµνF (k) =

−ik2 + iε

[gµν − (1− ξ)k

µkν

k2

](163)

- Gluon:

δabDµνF (k) = δab

−ik2 + iε

[gµν − (1− ξ)k

µkν

k2

](164)

- Ghost:

δabDF (k) =i

k2 + iε(165)

4 Interaction Vertices

4.1 Scalar Theory

- Consider now a (weakly) interacting scalar field theory

L =1

2∂µφ∂

µφ− 1

2m2φ2 − 1

4!λφ4 = L0 + LI (166)

where L0 is the free Lagrangian (61) and LI is the interaction partbut this is now non-Gaussian

- We can write full expectation values 〈O〉I in terms of free ones,

〈O〉I ≡∫DφO[φ] exp

[i∫d4x(L0 + LI)

]∫Dφ exp

[i∫d4x(L0 + LI)

]=〈O[φ] exp

[i∫d4xLI

]〉0

〈exp[i∫d4xLI

]〉0

(167)

where 〈· · ·〉0 is the expectation value in the free theory- Therefore, we need to calculate free expectation values of the form

〈O[φ] exp

[i

∫d4xLI

]〉0 =

∑k

ik

k!〈O[φ]

(∫d4xLI

)k〉0, (168)

which are given by Gaussian path integrals

25

- If O is polynomial, all we need are terms like1

k!

(−iλ4!

)k ∫d4y1 · · · d4yk〈φ(x1) · · ·φ(xn)φ(y1)4 · · ·φ(yk)4〉0 (169)

- Gaussian expectation value:Sum of all possible pairings into products of 〈φ(x)φ(y)〉0 = DF (x− y)

- In particular, we want to calculate n-point correlation functions in the interacting theory

Gn(x1, . . . , xn) ≡ 〈φ(x1) · · ·φ(xn)〉I (170)

- As an example, consider the two-point function G2(x1, x2) ≡ 〈φ(x1)φ(x2)〉I- To linear order in λ, we have

G2(x1, x2) =〈φ(x1)φ(x2)

(1− iλ

4!

∫d4yφ(y)4

)〉0

〈1− iλ4!

∫d4yφ(y)4〉0

+O(λ2)

= 〈φ(x1)φ(x2)〉0 −iλ

4!〈φ(x1)φ(x2)

∫d4yφ(y)4〉0

+iλ

4!〈φ(x1)φ(x2)〉0〈

∫d4yφ(y)4〉0 +O(λ2) (171)

- In the second term, we have to consider pairings of 〈〈φ(x1)φ(x2)φ(y)φ(y)φ(y)φ(y)〉0:- If x1 is connected to x2,

〈φ(x1)φ(x2)φ(y)φ(y)φ(y)φ(y)〉0 (172)

we obtain

− iλ4!× 3×DF (x1 − x2)

∫d4yDF (y − y)2 (173)

where the factor 3 arises because the first φ(y) can choose between 3 other φ(y)s- If x1 is connected to y,

〈φ(x1)φ(x2)φ(y)φ(y)φ(y)φ(y)〉0 (174)

we have

− iλ4!× 4× 3×

∫d4yDF (x1 − y)DF (x2 − y)DF (y − y) (175)

where the factor 4× 3 = 12 arises because x1 can choose between 4 and x2 between 3 φ(y)s- The third term only has one possible contraction,

〈φ(x1)φ(x2)〉0〈φ(y)φ(y)φ(y)φ(y)〉0 (176)

and givesiλ

4!× 3×DF (x1 − x2)

∫d4yDF (y − y)2 (177)

which cancels Eq. (173)- Thus the whole two-point function is

26

G2(x1, x2) = DF (x1 − x2)− iλ

2DF (0)

∫d4yDF (x1 − y)DF (x2 − y) (178)

- We can give this result a simple pictorial representation in terms of Feynman diagrams:- Spacetime points xi are represented by points in the diagram- Propagator DF (x− y) is represented by a dashed line connecting points x and y:x y

- The two-point function looks like

G2(x1, x2) =− iλ

4!

3× + 4× 3×

+iλ

4!

× 3×

= − iλ

2 (179)

- Consider then the four-point function G4(x1, x2, x3, x4)

- Two-to-two scattering process- To linear order in λ, we have

G4(x1, x2, x3, x4) = 〈φ(x1)φ(x2)φ(x3)φ(x4)〉0

− iλ4!〈φ(x1)φ(x2)φ(x3)φ(x4)

∫d4yφ(y)4〉0

+iλ

4!〈φ(x1)φ(x2)φ(x3)φ(x4)〉0〈

∫d4yφ(y)4〉0 +O(λ2) (180)

- The leading term gives

〈φ(x1)φ(x2)φ(x3)φ(x4)〉0 = DF (x1 − x2)DF (x3 − x4) +DF (x1 − x3)DF (x2 − x4)

+DF (x1 − x4)DF (x2 − x3)

= + + (181)

- Particles propagating between points without interacting

27

- The second term consists of diagrams with five points: four lines end at y- The ones in which all xis disconnected from y,

〈φ(x1)φ(x2)φ(x3)φ(x4)φ(y)φ(y)φ(y)φ(y)〉0, (182)

giving + permutations (183)

are again cancelled by the third term: Generally the case for disconnected bubbles!- There are six diagrams in which one pair of xis is connected to each other and the rest to y,

〈φ(x1)φ(x2)φ(x3)φ(x4)φ(y)φ(y)φ(y)φ(y)〉0 (184)

corresponding to + permutations (185)

Each gives a contribution

− iλ4!× 4× 3×DF (x1 − x2)

∫d4yDF (x3 − y)DF (x4 − y)DF (y − y) (186)

where the factor 4× 3 arises because x3 can choose between 4 legs of y andx4 can choose between 3 remaining legs

- Finally there is one diagram in which all xis are connected to y,

〈φ(x1)φ(x2)φ(x3)φ(x4)φ(y)φ(y)φ(y)φ(y)〉0 (187)

corresponding to (188)

and giving

− iλ4!× 4× 3× 2× 1×

∫d4yDF (x1 − y)DF (x2 − y)DF (x3 − y)DF (x4 − y) (189)

The factor 4× 3× 2× 1 = 4! arises because x1 can choose between 4 φ(y)s etc.- In summary, the diagrammatic expression for G4(x1, x2, x3, x4) is

28

G4(x1, x2, x3, x4) = + +− iλ

2

+ ++ + + −iλ (190)

- It is easy to read the full expression from the diagrams using these rules:

• The n-point function is given by the sum of all diagrams with n external legs, arbitrary number offour-point vertices and no disconnected “bubbles” (as we will see later)

• Each four-point vertex gives ↔ − iλ4!

∫d4y (191)

• Each line gives x y ↔ DF (x− y) (192)

• Multiply by the number of contractions leading to the same diagram

• Multiply by 1/(number of vertices)! from the Taylor expansion of the exponential

- For example, the so-called sunset diagram

x1 x2

(193)

corresponds to

1

2!

(− iλ

4!

)2

× 8× 4× 3× 2

∫d4y1d

4y2DF (x1 − y1)DF (y2 − y1)3DF (y2 − x2)

=(−iλ)2

6

∫d4y1d

4y2DF (x1 − y1)DF (y2 − y1)3DF (y2 − x2) (194)

- The numerical factors are usually combined to one symmetry factor:- If the diagram consist of k vertices, the integral is divided by

29

k!(4!)k

number of different contractions(195)

- Here the numerator counts the total number of possible permutations of the legs of the internal vertices:Any such permutation will give the same topology (so represented by the same diagram)

- When this is divided by the number of different contractions, it gives the number of permutationsthat give the same contraction:This is the order of symmetry of the diagram

- For example, the symmetry factor for the sunset diagram (193) is 1/6 because it has a six-foldsymmetry under permutations of internal lines

- It is generally more convenient to work in momentum space, because the propagator is simpler- Writing the interaction term in terms of the Fourier transformed field

φ(x) =

∫d4k

(2π)4e−ikxφ(k), (196)

we find

iSI = i

∫d4xLI = − iλ

4!

∫d4xφ(x)4

= − iλ4!

∫d4k1

(2π)4

d4k2

(2π)4

d4k3

(2π)4

d4k4

(2π)4(2π)4δ(k1+k2+k3+k4)φ(k1)φ(k2)φ(k3)φ(k4) (197)

- The free two-point function is

〈φ(p)φ(q)〉0 =i

p2 −m2(2π)4δ(p+ q) (198)

- The delta functions simply enforce momentum conservation at all vertices- Fixes all but loop momenta- Sign choice: Positive momentum = inward

- Thus, repeating the same arguments as in the coordinate space, we find the Feynman rules

1. The n-point function is given by the sum of all diagrams with n external legs, arbitrary number offour-point vertices and no disconnected “bubbles”

2. Each four-point vertex gives

k3

k1

k4

k2

↔ −iλ (momentum conservation k1+k2+k3+k4 = 0) (199)

More generally, write iS in momentum space, and the vertex is the coefficient of the correspondingterm multiplied by the number of permutations of legs

3. Each line givesp q ↔ i

p2 −m2(momentum conservation p+q = 0) (200)

4. Multiply by the symmetry factor (i.e., divide by the order of symmetry)

30

5. Integrate over loop momenta

- Example: Sunset diagram (193)

(−iλ)2

6

i

p2 −m2

i

q2 −m2(2π)4δ(p+ q)

∫d4k1

(2π)4

d4k2

(2π)4

i

k21 −m2

i

k22 −m2

i

(p−k1−k2)2 −m2(201)

4.2 Connected and 1PI Correlators

- Consider the generating functional Z[J ] defined in Eq. (46)- Diagrammatic expansion:

- External field gives one-point vertex with coupling J(x)

φ ↔ iJ(x) (202)

- Z[J ] is the sum of all diagrams with- Arbitrarily many separate pieces- Arbitrarily many one-point vertices iJ(x)

- No external legs- Any such diagram factorises:

- full diagram = (bubbles) × (J-dependent piece)where bubbles are diagrams with no one-point vertices

- Therefore the whole sum factorises, too

Z[J ] =∑

(all diags) =∑

(bubbles)×∑

(J-dependent) (203)

- The sum of bubbles is simply Z[0], so Z[J ]/Z[0] is given by all diagrams with no bubbles- Therefore the same applies to correlators GN given by Eq. (47), as well

- Now, consider diagram with k separate pieces- Whole integral = Product of integrals corresponding to each individual piece- Symmetry: (permutations of pieces) × (symmetry of each connected piece)

- Organise the sum as a sum over k:

Z[J ] = sum over all diagrams

=

∞∑k=1

1

k!× (sum over all connected diagrams)k

= exp(sum over all connected diagrams) (204)

- Define generating functional of connected correlators

E[J ] = i lnZ[J ] (205)

31

- Given by sum over all connected diagrams- Analogous to free energy in statistical physics (F = −kBT lnZ)

- Define connected correlator GN (x1, . . . , xN ) as functional derivative

GN (x1, . . . , xN ) = 〈φ(x1) · · ·φ(xN )〉conn = (−i)N+1 δ

δJ(x1)· · · δ

δJ(xN )E[J ]

∣∣∣∣J=0

(206)

- Given by diagrams in which all vertices are connected to each other- Examples:

- Connected two-point correlator

G2(x, y) = iδ

δJ(x)

δ

δJ(y)E[J ]

∣∣∣∣J=0

= − δ

δJ(x)

δ

δJ(y)ln Z[J ]

∣∣∣∣J=0

= − 1

Z[J ]

δ

δJ(x)

δZ[J ]

δJ(y)

∣∣∣∣J=0

+1

Z[J ]

δZ[J ]

δJ(x)

1

Z[J ]

δZ[J ]

δJ(y)

∣∣∣∣J=0

= 〈φ(x)φ(y)〉 − 〈φ(x)〉〈φ(y)〉 (207)

- If the one-point function 〈φ(x)〉 vanishes, as it does in our theory, then G2(x, y) = G2(x, y)

- Connected four-point correlator

G4(k1, k2, k3, k4) = G4(k1, k2, k3, k4)−G2(k1, k2)G2(k3, k4)

−G2(k1, k3)G2(k2, k4)−G2(k1, k4)G2(k2, k3) (208)

- More generally, subtract all possible ways of expressing in terms of products of lower correlators- To quadratic order in λ

G4 = + 4× + 3×! (209)

- The second diagram factorises- In fact, any diagram with separate parts that are connected by only one propagator factorises

into a product of two separate integrals

"k1

k2

k3

k4

k5

k6

= #k1

k2

k3

−k′ × i

k′2 +m2× $k′

k4

k5

k6

(210)

with k′ = k1 + k2 + k3 etc- Define one-particle irreducible (1PI) diagrams:

Cutting any single line will not split the diagram into two disconnected pieces- 1PI correlator Γn(k1, . . . , kn):

- Sum of all 1PI diagrams with n external legs- Generating functional: Effective action Γ[φc(x)]

- Defined as the Legendre transform

32

Γ[φc] = −E[J ]−∫d4xJ(x)φc(x),

δE[J ]

δJ(x)= −φc(x) (211)

- Correlators as derivatives

Γn(x1, . . . , xn) = iδ

δφc(x1)· · · δ

δφc(xn)Γ[φc] (212)

- The external leg propagators are not included, e.g. in Eq. (193)i

p2 −m2

i

q2 −m2(213)

- Also conventionally leave out the momentum conservation delta when in momentum space, e.g.

(2π)4δ(p+ q) (214)

⇒% =(−iλ)2

6

∫d4k1

(2π)4

d4k2

(2π)4

i

k21 −m2

i

k22 −m2

i

(p−k1−k2)2 −m2(215)

- Connected and full correlators can be expressed in terms of 1PI correlators- When calculating them, remember to include

- A propagator for each external leg- Overall momentum conservation delta function

- Connected n-point correlator Gn (black circle):- Given by sum of all tree level (i.e. no loops) diagrams with 1PI correlators Γi with i ≤ n as vertices (shaded circles)

- First example: two-point function- All diagrams consist of an arbitrary number of Γ2’s in a row

& ='+(+)+ · · · (216)

- Thus,

G2(p, q) = G2(p, q) = (2π)4δ(p+ q)i

p2 −m2

∞∑k=0

(i

p2 −m2Γ2(p)

)k= (2π)4δ(p+ q)

i

p2 −m2

(1− i

p2 −m2Γ2(p)

)−1

= (2π)4δ(p+ q)i

p2 −m2 − iΓ2(p)(217)

- Second example: four-point function- Symmetry: Odd n-point correlators vanish⇒ The only contributing 1PI vertices are Γ2 and Γ4

- The only way to make a 4-point diagram: One Γ4 and any number of Γ2’s in legs⇒ Legs are just connected two-point functions

33

G4 =* =+ (218)

- If the 1PI three-point function Γ3 was non-zero, it would also contribute and we would have

G4 =, =-+. (219)

- The 1PI contribution is

Γ4 =/ + 3×0 (220)

- The second diagram gives the integral1 =(−iλ)2

2

∫d4p

(2π)4

i

p2 −m2

i

(p+ k1 + k2)2 −m2(221)

4.3 Complex Scalar

- Interacting Lagrangian

L = ∂µφ∗∂µφ−m2φ∗φ− λ

4(φ∗φ)

2 (222)

- Note that the normalisation of the interaction term is different from the real scalar:There are only 4 rather than 4! equivalent permutations of the factors φ and φ∗

- Interaction term

i

∫d4xLI = − iλ

4

∫d4xφ∗(x)φ∗(x)φ(x)φ(x)

= − iλ4

∫d4x

d4k1

(2π)4· · · d

4k4

(2π)4ei(k1+k2−k3−k4)·xφ∗(k1)φ∗(k2)φ(k3)φ(k4)

= − iλ4

∫d4k1

(2π)4· · · d

4k4

(2π)4δ(k1 + k2 − k3 − k4)φ∗(k1)φ∗(k2)φ(k3)φ(k4) (223)

- Feynman rules:- Propagator 2φ∗(q) φ(k)

i

k2 −m2(mom cons k − q = 0) (224)

- Vertex

34

3φ(k3)

φ∗(k1)

φ(k4)

φ∗(k2)

− iλ (mom cons k1 + k2 − k3 − k4) (225)

- Note:- Lines have a direction:

- On vertices, outgoing lines↔ creation operators φ∗

incoming lines↔ annihilation operators φ- A propagator line goes from a creation operator to an annihilation operator φ- Thus: The φ∗ end of a line is attached to a φ∗ leg of a vertex etc.

- Momentum is defined in the direction of the arrow- Example: Sunset in complex theory

4k1

k2

−p+k1+k2p q ↔ 〈φ∗(p)φ(q) φ∗φ∗φφ φ∗φ∗φφ〉

=1

2!

(− iλ

4

)2

× 4× 2× 2

∫d4k1

(2π)4

d4k2

(2π)4

i

k21 −m2

i

k22 −m2

i

(p− k1 − k2)2 −m2

=(−iλ)2

2

∫d4k1

(2π)4

d4k2

(2π)4

i

k21 −m2

i

k22 −m2

i

(p− k1 − k2)2 −m2(226)

where we have again omitted the external propagators- The numerical factor 1/2: Symmetry between the top and bottom arcs

4.4 QED

- Theory of fermions coupled to photon field

L = ψ(i /D −m)ψ − 1

4FµνF

µν (227)

where Dµ = ∂µ + ieAµ and Fµν = ∂µAν − ∂νAµ- Free part

L0 = ψ(i /∂ −m)ψ − 1

4FµνF

µν (228)

- Interaction part (with spinor indices α, β written explicitly)

LI = −eψγµψAµ = −eψαγµαβψβAµ (229)

- Momentum space

i

∫d4xLI = −ieγµαβ

∫d4k1

(2π)4

d4k2

(2π)4

d4k3

(2π)4(2π)4δ(k1 − k2 − k3)ψα(k1)ψβ(k2)Aµ(k3) (230)

- Feynman rules:- Interaction vertex

35

5k2

k3

k1

ψβ

ψα

Aµ − ieγµαβ (momentum cons k1 − k2 − k3 = 0) (231)

- Propagators6kAµ(k) Aν(−k) ↔ Dµν

F (k) =−ik2

[gµν − (1− ξ)k

µkν

k2

](232)

7kψβ(k) ψα(k) ↔ SαβF (k) =

(i

/k −m

)αβ

(233)

- Trace over each fermion loop- Factor (−1) for each fermion loop

(Need to anticommute one pair)- Example:

8p+ k

k

Aµ(p) Aν(−p)αβ

δγ

↔ 〈Aµ(p)Aν(−p) ψψA ψψA〉

=1

2!× (−2)

∫d4k

(2π)4

(−ieγµαβ

)SβγF (k)

(−ieγνγδ

)SδαF (p+ k)

= −(−ie)2

∫d4k

(2π)4tr

(γµ

i

/k −mγν

i

/p+ /k −m

)(234)

- Third line: Trace over spinor indicesOrder of matrix factors opposite to the direction of the arrow in loop

4.5 QCD

- SU(N) gauge field + fundamental fermions- After gauge fixing, the Lagrangian (116) becomes L = L0 + LI , with

L0 =1

2Aaµ

(∂2gµν −

(1− 1

ξ

)∂µ∂ν

)A2ν − ca∗∂2ca + ψi (i /∂ −m)ψi (235)

and

LI =g

2fabc(∂µA


4fabcfadeAbµA

cνA

d µAe ν

−gfabc∂µca∗Abµcc − gγµαβt

aijA

aµψiαψjβ (236)

36

- There are three different fields:- Gluons (gauge field): Each component has the same propagator as the photon9k

Aaµ(k) Abν(−k) ↔ δabDµνF (k) =

−iδab

k2

[gµν − (1− ξ)k

µkν

k2

](237)

- Quarks (fermions): Each component has the same propagator as the electron:kψjβ(k) ψiα(k) ↔ δijS

αβF (k) = δij

(i

/k −m

)αβ

(238)

- Ghosts: Just like charged scalars;kcb∗(k) ca(k) ↔ iδab

k2(239)

- There are also four types of vertices:- The three-gluon vertex arises from the term

iS = . . .+

∫d4x

ig

2fabc(∂νAaρ − ∂ρAaν)AbνA

cρ = . . .+

∫d4x

ig

2fabc(gρµ∂νAaµ − gµν∂ρAaµ)AbνA

cρ

(240)

- Going to momentum space ∂νAaµ → −ikν1Aaµ(k1), and we obtain∫d4k1

(2π)4

d4k2

(2π)4

d4k3

(2π)4(2π)4δ(k1 + k2 + k3)

g

2fabc(gρµkν1 − gµνk

ρ1)Aaµ(k1)Abν(k2)Acρ(k3) (241)

- In the symmeric form, the integrand isg

6fabc [gµν(kρ2 − k

ρ1) + gνρ(kµ3 − k

µ2 ) + gµρ(kν1 − kν3 )]Aaµ(k1)Abν(k2)Acρ(k3) (242)

- Including a factor 3! counting permutations of legs, the Feynman rule is

<Aaµ(k1)

Abν(k2) Acρ(k3)

↔ gfabc [gµν(kρ2 − kρ1) + gνρ(kµ3 − k

µ2 ) + gµρ(kν1 − kν3 )] (243)

- The four-gluon vertex is given by the term

−g2

4fabef cdeAaµA

bνA

c µAd ν = −g2

4fabef cdegµρgνλAaµA

bνA

cρA

dλ

= −g2

24

[fabef cde

(gµρgνλ − gµλgνρ

)+facef bde

(gµνgρλ − gµλgνρ

)+fadef bce

(gµνgρλ − gµρgνλ

)]AaµA

bνA

cρA

dλ (244)

and including the factors i and 4! we have the rule

37

=Aaµ Abν

Acρ Adλ

↔ −ig2[fabef cde


)

+facef bde(gµνgρλ − gµλgνρ

)+fadef bce


)](245)

- The ghost-gluon interaction term is

−gfabc∂µca∗Abµcc → −igfabckµ1 ca∗(k1)Abµ(k2)cc(k3) (246)

and therefore we have

>Abµ(k2)

ca∗(k1) cc(k3)

↔ gfabckµ1 (247)

- Finally, the quark-gluon coupling is given by the term

−gγµαβtaijA

aµψiαψjβ (248)

whereby

?Aaµ

ψiα ψjβ

− igγµαβtaij (249)

- In addition, we have the usual rules that we integrate over all loop momentaand have a factor (−1) for each fermion (quark or ghost) loop

Part II

Renormalisation

5 Perturbative Renormalisation

5.1 Ultraviolet Divergences

- 1PI two-point correlator Γ2 at one loop

38

Γ2(p) =@ =iλ

2

∫d4k

(2π)4

i

k2 −m2(250)

- Need to evaluate the integral

I1(m) =

∫d4k

(2π)4

1

k2 −m2= (2π)−4

∫d3k

∫dk0

k20 − ~k2 −m2 + iε

(251)

- The k0 integrand has two poles at k0 = ±(√~k2 +m2 − iε)

- Rotate the integration contour

Re

Im

∫ ∞−∞

dk0

k20 − ω2 + iε

= −iR∫ π/2

0

dθeiθ

R2e2iθ − ω2

[k0 = −Reiθ

]+i

∫ ∞−∞

dkE−k2

E − ω2[k0 = ikE ]

−iR∫ π/2

0

dθeiθ

R2e2iθ − ω2

[k0 = Reiθ

]→ −i

∫ ∞−∞

dkEk2E + ω2

(252)

- This implies

I1(m) = −i(2π)−4

∫d3k

∫dkE

k2E + ~k2 +m2

= −i∫

Eucl

d4k

(2π)4

1

k2 +m2(253)

where the subscript ”Eucl” indicates a four-dimensional Euclidean integral- In spherical polar coordinates

I1(m) = − i

(2π)4

∫dΩ4

∫ ∞0

k3dk

k2 +m2(254)

where∫dΩ4 is the integral over the angles, i.e., the volume of S3

- We can calculate it by considering

39

πd/2 =

(∫ ∞−∞

dxe−x2

)d=

∫ddxe−~x

2

=

∫dΩd

∫ ∞0

dr rd−1e−r2

=1

2

∫dΩd

∫ ∞0

dz zd/2−1e−z =Γ(d/2)

2

∫dΩd

⇒∫dΩd =

2πd/2

Γ(d/2)(255)

where Γ(x) is the gamma function and satisfies xΓ(x) = Γ(x+ 1) and Γ(n) = (n− 1)! for n ∈ Z- Since Γ(2) = 1, we find

I1(m) = − i

8π2

∫ ∞0

k3dk

k2 +m2(256)

- The k integral is obviously divergent- Introduce a ultraviolet cutoff Λ

I1(m) = − i

8π2

∫ Λ

0

k3dk

k2 +m2= − i

16π2

∫ Λ2

0

udu

u+m2= − i

16π2

∫ Λ2

0

du

(1− m2

u+m2

)= − i

16π2

(Λ2 −m2 ln

Λ2 +m2

m2

)= − i

16π2

(Λ2 − 2m2 ln

Λ

m+O(m4/Λ2)

)(257)

- Diverges at Λ→∞- Has both quadratic (Λ2) and logarithmic (ln Λ) divergences

- Therefore, the two-point function is

Γ2(p) = −i λ

32π2

(Λ2 − 2m2 ln

Λ

m

)(258)

- Let us then look at the four-point function Γ4

Γ4(k1, k2, k3, k4) =A + 3×B (259)

where the factor 3 in the second term stands for the three inequivalent permutations of external legs- The second diagram gives the integral

C =(−iλ)2

2

∫d4p

(2π)4

i

p2 −m2

i

(p+ k1 + k2)2 −m2≡ − (−iλ)2

2I2(k1 + k2,m) (260)

- At high p, we have∫d4p/p4 = log Λ: Logarithmic divergece

- Momentum cutoff not well suited for this calculation because it breaks shift invariance p→ p+ const

- We’ll use a better approach later- To find the divergence, let us calculate this for the special case k1 + k2 = 0

40

I2(0,m) =

∫d4p

(2π)4

1

(p2 −m2)2= i

∫Eucl

d4p

(2π)4

1

(p2 +m2)2=

i

8π2

∫ Λ

0

p3dp

(p2 +m2)2

=∂

∂m2I1(m) =

i

8π2

[log

Λ

m− 1

2

]+O

(m2/Λ2

)(261)

- Since the divergence comes from high p, it is independent of k1 and k2, and we have

I2(k,m) =i

8π2

[log

Λ

m+ (finite piece)

](262)

- The 1PI four-point function is therefore

Γ4 = −iλ+ 3× λ2

2

i

8π2log

Λ

m= −iλ

(1− 3λ

16π2log

Λ

m+ finite

)(263)

5.2 Renormalised Couplings

- Our theory seems to give divergent results, but is this real?- Look at physical, observable quantities such as scattering cross section

dσ

dΩ=

1

64π2E2cm

|M|2 (264)

where the scattering amplitude is

iM = (sum of amputated 4-point diagrams) = Γ4 (265)

- This was indeed divergent,

iM = −iλ[1− 3λ

16π2

(log

Λ

m+ (finite piece)

)]+O(λ3) (266)

- Assume there is a finite physical cutoff Λ

- Theory not applicable above Λ

- Makes everything finite- The finite piece in Eq. (266) is a function of k1, k2, k3, k4 and m

- Lorentz invariant, only dependent on Lorentz-invariant combinations (Mandelstam variables)s = (k1 + k2)2, t = (k1 + k3)2, u = (k1 + k4)2

- Dimensionless: Can only depend on dimensionless ratios s/m2, t/m2 and u/m2

iM = −iλ[1− 3λ

16π2

(log

Λ

m+ f

(s

m2,t

m2,u

m2

))]+O(λ3), (267)

where f is a function that we can calculate by evaluating the full integral I2(k,m)

-M is now finite, but dependent on Λ

- If an experimentalist want to test the theory, (s)he needs to know the coupling λ- This can be determined by measuringM0 ≡M(s0, t0, u0) for a given set of Mandelstam variables- According to our theory, this is equal to

iM0 = −iλ[1− 3λ

16π2

(log

Λ

m+ f0

)]+O(λ3) (268)

41

where we have defined

f0 = f

(s0

m2,t0m2

,u0

m2

)(269)

- We can find λ by inverting this,

λ = −M0

[1− 3M0

16π2

(log

Λ

m+ f0

)]+O(M3

0) (270)

- Once we have this, we can write Eq. (267) in terms ofM0 as

iM(s, t, u) = iM0

[1 +

3M0

16π2

(f

(s

m2,t

m2,u

m2

)− f0

)]+O(M3

0), (271)

- This expression is independent of Λ, andM0 is finite: The divergence has disappeared!

- SinceM0 rather than λ measures the real strength of the interaction, it makes sense to definethe “physical” or renormalised coupling λR ≡ −M0

iM(s, t, u) = −iλR[1− 3λR

16π2

(f

(s

m2,t

m2,u

m2

)− f0

)]+O(λ3

R), (272)

- The original coupling λ is called the “bare” coupling and denoted by λB- We can rewrite Eq. (268) as

λR = λB

[1− 3λB

16π2

(log

Λ

m+ f

(s0

m2,t0m2

,u0

m2

))]+O(λ3

B) (273)

- Inverting this to order λ2R, we have

λB = λR

[1 +

3λR16π2

(log

Λ

m+ f

(s0

m2,t0m2

,u0

m2

))]+O(λ3

R) (274)

- Note that for fixed physical observablesλB - depends on Λ and diverges in the limit Λ→∞

- is the actual coeffient of φ4 in the Lagrangian- is given by the fundamental theory, if Λ is a real cutoff

λR - is finite and independent of Λ

- depends on which values (s0, t0, u0) were used define it= scale and scheme dependence

- Measures (roughly) the strength of interaction at low energies

- Consider then the two-point function- The 1PI two-point function diverges [see Eq. (258)]

Γ2(k) = − iλ

32π2

(Λ2 −m2 log

Λ2 +m2

m2

)+O(λ2) (275)

- According to Eq. (217), the full two-point function is

G2(k, q) = 〈φ(k)φ(q)〉 = (2π)4δ(k + q)i

k2 −m2 − iΓ2(k)

= (2π)4δ(k + q)i

k2 −[m2 + λ

32π2

(Λ2 −m2 log Λ2+m2

m2

)] +O(λ2) (276)

- The physical mass of a particle corresponds to the pole of its propagator,

42

so an experimentalist would measure

m2meas = m2 +

λ

32π2

(Λ2 −m2 log

Λ2 +m2

m2

)+O(λ2) ≡ m2

R (277)

- We call this the renormalised mass m2R and the original mass the bare mass m2

B = m2,

m2B = m2

R −λ

32π2

(Λ2 −m2

R logΛ2 +m2

R

m2R

)+O(λ2) (278)

- Again, the bare mass is divergent but the two-point function G2(k, q) is finite

5.3 Field Renormalisation

- In scalar theory, the leading-order expression for Γ2(k) in Eq. (275) is independent of k- Not true generally: O(λ2) term is k-dependent

O(e2) term in QED- Lorentz invariance: Can only depend on k2, i.e., Γ2 = Γ2(k2)

- The k2 term is generally logarithmically divergent- The general two-point function is

G2(k, q) = 〈φ(k)φ(q)〉 = (2π)4δ(k + q)i

k2 −m2 − iΓ2(k2)(279)

- Again, define m2R as the location of the pole

m2R = m2 + iΓ2(k2 = m2

R) (280)

- Expand Γ2(k2) around k2 = m2R in powers of (k2 −m2

R)

iΓ2(k2) =

∞∑i=0

bi(k2 −m2

R)i (281)

where dimensional analysis tells us that generally b0 ∼ Λ2 and b1 ∼ log Λ

- Combining Eqs. (281) and (280), we find

b0 = iΓ2(k2 = m2R) = m2

R −m2 (282)

- In general bi for i > 0 are non-zero- The full two-point function then has the expansion

1

k2 −m2 − iΓ2(k2)=

1

−(m2 −m2R + b0) + (1− b1)(k2 −m2

R)−∑∞i=2 bi(k

2 −m2R)i

=1

(1− b1)(k2 −m2R)−

∑∞i=2 bi(k

2 −m2R)i

=Z

k2 −m2R

+

∞∑i=0

ai(k2 −m2

R)i, (283)

where ai are calculable coefficients, and Z = 1/(1− b1)

- Thus, we have

G2(k, q) = (2π)4δ(k + q)

[iZ

k2 −m2R

+ (regular terms)]

(284)

- In general the residue of the pole Z is not one

43

⇒ Cannot identify the pole with a canonically normalised particle- Rescale, or renormalise, the field

φR = Z−1/2φ (285)

- This implies

GR2 (k, q) ≡ 〈φR(k)φR(q)〉 = Z−1G2(k, q) = (2π)4δ(k + q)

[i

k2 −m2R

+ (regular terms)]

(286)

which has the correct normalisation- Similarly, any n-point function Gn ≡ 〈φn〉 is generally divergent,

but GRn ≡ 〈φnR〉 = Z−n/2Gn is finite- In terms of the renormalised field, the Lagrangian is

L =1

2Z∂µφR∂

µφR −1

2Zm2

Bφ2R −

1

4!Z2λBφ

4R (287)

- To higher order in λ:- We must solve the equations for λR, m2

R and Z simultaneously:Can be messy because the results can be complicated functions of λ and m2

- The power series makes little sense because λ is large (divergent)- There could also be more divergences to take care of, so we need to identify them systematically first

5.4 Power Counting

- Consider a 1PI diagram with E external legs, L loops, V vertices and I internal linesin d spacetime dimensions- Each internal line corresponds to a propagator D(k)

(use˜ for quantities in momentum space)- Each loop corresponds to integration over ddk- Each vertex gives one power of coupling λ

D ∼ λV∫ddLk D(k)I (288)

- In the UV (i.e. k →∞), D(k) will go like some power of k, so the integrand will behave like

∼ λV∫ Λ

0

dk kD−1 ∝ λV ΛD, (289)

where we have defined the superficial degree of divergence D to parameterise the behaviour- If D ≥ 0, the diagram is superficially divergent

(though not always actually divergent)

- We can determine D with dimensional analysis- Denote the dimensionality of X by [X]: If X has dimensions of massn then [X] = n

44

- First, because the diagram contributes to the momentum-space 1PI correlator ΓE(k), Eq. (289) implies[ΓE(k)

]=[λV ΛD

]= V [λ] +D, (290)

from which we obtain

D =[ΓE(k)

]− V [λ] (291)

- Now, we need to find the dimensionality of ΓE(k)

- The full momentum space correlator GE has “trivial” factors, which are absent from the 1PI correlator ΓE :- Momentum conservation delta function δd(k1 + · · ·+ kE)

- External propagators Di(k)[GE(k)

]=[ΓE(k)

]+[δd(k1 + · · ·+ kE)

]+

E∑i=1

[Di

]=[ΓE(k)

]− d+

E∑i=1

[Di

](292)

- Therefore we find [ΓE(k)

]=[GE(k)

]+ d−

E∑i=1

[Di

](293)

- Take the Fourier transform to go to coordinate space[GE(k)

]=

[∫ddx1 · · · ddxEeikixiGE(x)

]= dE[x] + [GE(x)] = [GE(x)]− dE (294)

- This gives [ΓE(k)

]= [GE(x)]− dE + d−

E∑i=1

[Di

](295)

- The dimensionality of the coordinate space correlator is

[GE(x)] = [〈φ1(x1) · · ·φE(xE)〉] =

E∑i=1

[φi(x)], (296)

where we each φi could be a different field- To calculate [Di], we note that it is just the free two-point function,

G2,i = 〈φi(k)φi(q)〉 = (2π)dδd(k + q)Di(k) (297)

- Using Eq. (294), this becomes[Di

]= d+

[G2,i

]= d+ [G2,i]− 2d = 2 [φi]− d (298)

- Substituting this into Eq. (295), we find[ΓE(k)

]= d− [GE(x)] , (299)

and finally, comparing with Eq. (290),

D = d− [GE(x)]− V [λ] (300)

- Furthermore, if one expands the correlator in powers of external momenta k,each power reduces the superficial degree of divergence of the corresponding term by one:

45

ΓE(k) ∼∑n

knΛD−n (301)

- Three possibilities:[λ] > 0: “superrenormalisable”

Degree of divergence decreases with increasing VOnly finite number of divergent diagrams (assuming [φ] > 0)

[λ] = 0: “renormalisable”Degree of divergence independent of VInfinite number of divergent diagramsFinite number of divergent correlators (assuming [φ] > 0)Examples: QED, QCD, EW theory in four dimensions

[λ] < 0: “non-renormalisable”Degree of divergence grows with VInfinite number of divergent diagramsInfinite number of divergent correlators (in fact all of them)Examples: gravity [GN ] = −2, Fermi theory of weak interactions [GF ] = −2

- Caveat: The actual degree of divergence may not be the same as D- The leading divergence may cancel, especially if the theory has symmetries (e.g. SUSY)- The actual divergence may be higher if there is a divergent subdiagram:

D

E = −2 (302)

but the diagram has a divergent subdiagram (tadpole loop)- The divergence comes from the two-point function, and disappears as it is renormalised

- Therefore it is enough to make sure all superficially divergent correlators are finite

- Example: Scalar theory- The action is dimensionless 0 = [S] = [

∫ddxL] = d[x] + [L], so [L] = d

- The Lagrangian contains the derivative term [∂µφ∂µφ] = 2 + 2[φ] = [L] = d, so

[φ(x)] = d/2− 1 (303)

- The Lagrangian also contains the interaction term [λφ(x)4] = [λ] + 4[φ(x)] = [L] = d,which implies

[λ] = d− 4[φ(x)] = 4− d (304)

- The theory is non-renormalisable in d > 4 dimensions- In particular, the scalar coupling is dimensionless in d = 4, so the theory is renormalisable

- The divergent correlators are E ≤ 4, which means:Γ0 ∼ Λ4 – vacuum energy (cosmological constant)Γ2 ∼ Λ2 + k2 log Λ – two-point function

46

Γ4 ∼ log Λ – four-point function

- In general, determine every parameter in the Lagrangian (including kinetic terms) from measurements- If the number of divergent correlators is less than or equal to the number of coefficients,

the theory can be renormalised- Symmetries: Fewer divergences, fewer parameters

- Renormalisable theories:- Eq. (300)⇒ Superficial degree of divergence of expectation value 〈O〉:

D (〈O〉) = d− [O] (305)

- Compare with the corresponding term in the Lagrangian:

L = · · ·+ gO (306)

- [L] = d ⇒ [O] = d− [g] ⇒ D (〈O〉) = [g]

⇒ The superficial degree of divergence is the same as the dimensionality of the couplingfor the corresponding term in L

- Divergences are in one-to-one correspondence with renormalisable terms- Just enough parameters to account for all divergences

- Non-renormalisable theories:- Infinite number of divergences⇒ infinite number of parameters and measurements required⇒ No predictive power!

- We conclude that we should write down the most general renormalisable Lagrangian subject to symmetries- Number of couplings = Number of divergences- Determine each coupling from experiment⇒ Absorbs all divergences- When the theory has symmetries, there are fewer divergences and fewer couplings

6 Renormalised Perturbation Theory

6.1 Counterterms

- In principle we could follow Section 5 and do our calculations in terms of λB and m2B

- Two problems:- To renormalise, we would calculate λR and m2

R, and invert the expressionsbut the expressions can be complicated if one works at high order in perturbation theory

- Our expansion parameter would be λB which is divergent- Therefore we want to organise the perturbative expansion in powers of λR from the beginning

- The general philosophy in perturbative calculations:- Take a simple model (L0) that you can solve and is a good approximation to the full theory- Do a Taylor expansion in powers of LI = L − L0

- This should be a good approximation if the subleading terms in the expansion are small- In bare perturbation theory, the two-point function in the free theory is

47

G2(k, q) = (2π)4δ(k + q)i

k2 −m2B

(307)

where m2B is divergent, while the interacting theory gives Eq. (284), where m2

R is finite but Z is divergent⇒L0 not a very good approximation!

- Clearly, it would be better to perturb around

L0 =1

2∂µφR∂

µφR −1

2m2Rφ

2R (308)

because then the free two-point function is

G2(k, q) = ZGR2 (k, q) = (2π)4δ(k + q)iZ

k2 −m2R

(309)

- This is a good approximation to Eq. (284), and the expansion parameter is λR- To do this, we write

Zm2B = m2

R + δm2

Z2λB = λR + δλ

Z = 1 + δZ (310)

where δm2, δλ and δZ are known as counterterms- Our Lagrangian then becomes

L =1

2∂µφR∂

µφR −1

2m2Rφ

2R −

1

4!λRφ

4R

+1

2δZ∂µφR∂

µφR −1

2δm2φ2

R −1

4!δλφ4

R (311)

- Note that we haven’t changed the Lagrangian, we have only written it in a different way- We can now use the first two terms as our free theory L0 and the rest as interaction LI- The last three terms are known as counterterms

- The interaction vertices are F ↔ −iλR

G ↔ i(k2δZ − δm2)

H ↔ −iδλ (312)

- The counterterms are treated as next order in λR:

δm2 = O(λR)

δλ = O(λ2R)

δZ = O(λR) (313)

- We also need to specify the renormalisation scheme, i.e., how we choose m2R, λR and Z

48

- Our previous renormalisation scheme consisted of(i) The scattering amplitude at (s0, t0, u0) is equal to −λR

Γ4(s0, t0, u0) = −iλR (314)

(ii) The two-point function has a pole at k2 = m2R

Γ2(m2R) = 0 (315)

(iii) The residue of the pole is one∂

∂k2Γ2(k2)

∣∣∣∣k2=m2

R

= 0 (316)

- Use these conditions order by order to determine the values of the counterterms δm2, δλ and δZ- As we will see later, there are other alternatives that can be more convenient

- Let us see how this works- The 1PI two-point function is given by

Γ2(k2) =I +J +O(λ2R)

= − iλR32π2

(Λ2 −m2

R logΛ2 −m2

R

m2R

)+ i(k2δZ − δm2

)(317)

- This has to satisfy the renormalisation conditions (316)∂

∂k2Γ2(k2)

∣∣∣∣k2=m2

R

= iδZ = 0 (318)

and (315)

Γ2(m2R) = − iλR

32π2

(Λ2 −m2

R logΛ2 −m2

R

m2R

)− iδm2 = 0

⇒ δm2 = − λR32π2

(Λ2 −m2

R logΛ2 −m2

R

m2R

)(319)

- The four-point function is given by

Γ4(s0, t0, u0) =K + 3×L +M +O(λ3R)

= −iλR[1− 3λR

16π2

(log

Λ

mR+ f(s0, t0, u0)

)]− iδλ = −iλR

⇒ δλ =3λ2

R

16π2log

Λ

mR+ f(s0, t0, u0) (320)

- We could now, in principle, move to the next order- Calculate Γ2(k2) and Γ4 to next order in λR- Determine the values of the counterterms to next order from Eqs. (314), (315) and (316)- This would cancel all divergences to that order in λR

49

- BHPZ (Bogoliubov-Parasiuk-Hepp-Zimmermann) theorem:All correlators finite to all orders in λR

6.2 Dimensional Regularisation

- The cutoff regularisation Λ is concrete and easy to understand but usually impractical- Multi-loop integrals become very cumbersome- Incompatible with gauge invariance

- An elegant (but abstract) alternative: Dimensional regularisation- Consider a loop integral In(m) in d dimensions (cf Eq. (251)

In(m) =

∫ddk

(2π)d1

(k2 −m2)n= (−1)niIEn (m) (321)

where we used the deformed complex integration path to turn it into a Euclidean integral

IEn (m) =

∫E

ddk

(2π)d1

(k2 +m2)n= (2π)d

∫dΩd

∫ ∞0

kd−1dk

(k2 +m2)n(322)

- Using Eq. (255), we can write this as

IEn (m) =2

(4π)d/2Γ(d/2)

∫ ∞0

kd−1dk

(k2 +m2)n=

1

(4π)d/2Γ(d/2)

∫ ∞0

(k2)d/2−1dk2

(k2 +m2)n(323)

- Now, let us change the integration variable to

x =m2

k2 +m2, (324)

so that the integral becomes

IEn (m) =md−2n

(4π)d/2Γ(d/2)

∫ 1

0

xn−d/2−1(1− x)d/2−1dx (325)

- Now, recall the definition of the beta function∫ 1

0

xα−1(1− x)β−1dx = B(α, β) =Γ(α)Γ(β)

Γ(α+ β)(326)

- This implies that

IEn (m) =md−2n

(4π)d/2Γ(d/2)

Γ(n− d/2)Γ(d/2)

Γ(n)=

md−2n

(4π)d/2Γ(n− d/2)

Γ(n)(327)

- Consequently, the Minkowskian integral is equal to

In(m) = (−1)nmd−2n

(4π)d/2Γ(n− d/2)

Γ(n)(328)

- The Gamma function Γ(z) has poles at z = 0,−1,−2, . . .

⇒ In 4D, In(m) diverges for n = 1, 2, as we have seen before- But for non-integer d it is always finite!- We can use Eq. (328) to define the integral for d ∈ R- In practice, we consider d = 4− ε with ε 1

- For integer l ≥ 0 one has

50

Γ(−l + ε/2) =(−1)l

l!

[2

ε+ ψ1(l + 1) +O(ε)

], (329)

where

ψ1(l + 1) =

l∑k=1

1

l− γ = 1 +

1

2+ . . .+

1

l− γ, (330)

and γ ≈ 0.577216 is the Euler-Mascheroni constant- For n ≥ 3, the result is finite (and therefore agrees with momentum cutoff)- For n = 2, we have

I2(m) =im−ε

(4π)2−ε/2 Γ(ε/2) =i(1− ε logm)

(4π)2(1− ε

2 log 4π) (2

ε− γ)

+O(ε)

=i

16π2

(2

ε+ log

4π

m2− γ)

+O(ε) (331)

- Comparison this with Eq. (261) shows that the two results agree if we identify

log Λ↔ 1

ε+ finite terms (332)

- For n = 1, we find

I1(m) = − im2−ε

(4π)2−ε/2 Γ(−1 + ε/2) = − im2

(4π)2

1− ε logm

1− ε2 log 4π

(−1)

(2

ε+ 1− γ

)+O(ε)

=im2

16π2

(2

ε+ log

4π

m2+ 1− γ

)+O(ε) (333)

- Comparing this with Eq. (257),

I1(m) = − i

16π2

(Λ2 −m2 ln

Λ2

m2

)+O

(m2

Λ2

), (334)

we see that the logarithmic terms agree again, but there is no analogue of the quadratic term Λ2

- Generally true: Only log divergences are present- Simplifies calculations- Can be sometimes misleading!

- In our results (333) and (331), we seem to have logs of dimensionful quantities- Solution: In d = 4− ε, the coupling has dimensions [λ] = ε

- Let us therefore write the 4− ε dimensional coupling as

λε = λµε, (335)

where [λ] = 0 and [µ] = 1

- Consider a diagram with L loops and V vertices- The integral is (schematically) proportional to

λV µεV(∫

ddk

(2π)d

)L(336)

- Using the results in Problem Sheet 4, we find that the number of loops L in a diagram is

51

L = V − E/2 + 1 (337)

- Using this, our integral becomes

λV µε(L+E/2−1)

(∫ddk

(2π)d

)L= µε(E/2−1)λV

(µε∫

ddk

(2π)d

)L(338)

- Each loop comes therefore with a factor µε

- Also an overall factor µ(E/2−1)ε, if the whole correlator is finite, this goes to 1 as ε→ 0

- Thus, when going to d = 4− ε dimensions, we should replace the integration measure by∫d4k

(2π)4→ µε

∫ddk

(2π)d(339)

and we have

In(m) = µε∫

ddk

(2π)d1

(k2 −m2)n= (−1)n

(4πµ2

m2

)ε/2im4−2n

(4π)2

Γ(n− 2 + ε/2)

Γ(n)(340)

- Then, Eq. (331) becomes

I2(m) =

(4πµ2

m2

)ε/2i

(4π)2Γ(ε/2) =

i(

1 + ε2 log 4πµ2

m2

)(4π)2

(2

ε− γ)

+O(ε)

=i

16π2

(2

ε+ log

4πµ2

m2− γ)

+O(ε) (341)

and Eq. (331) becomes

I1(m) = −(

4πµ2

im2

)ε/2m2

(4π)2Γ(−1 + ε/2) = − im2

(4π)2

(1 +

ε

2log

4πµ2

m2

)(−1)

(2

ε+ 1− γ

)+O(ε)

=im2

16π2

(2

ε+ log

4πµ2

m2+ 1− γ

)+O(ε) (342)

- Finally, consider a massless (m = 0) integral IEn (0)

- If d > 2n, Eq. (328) implies that IEn (0) = 0

- If d < 2n, it diverges- However, we define it by analytically continuing from high enough dIEn (0) = 0 for any n and d

- More generally, any scale-free integral is set to vanish

- This procedure is known as dimensional regularisation- Go to d dimensions by replacing the integrals according to Eq. (339)- Calculate the integral using Eq. (340)- If the integral is scale-free, it vanishes- This way power-law divergences disappear, and log divergences appear as

logΛ2

m2∼ 2

ε+ log

4πµ2

m2(343)

- To use dimensional regularisation in practice, we need to show that this can be applied to an arbitrary

52

Minkowskian integral of the form

µε∫

ddk

(2π)dkµ1 · · · kµm

(k2 −m21)((k + p2)2 −m2

2) · · · ((k + pn)2 −m2n)

(344)

- First, use the identity

1

(k2 −m21)((k + p)2 −m2

2)=

∫ 1

0

dx

(k2 + x(2k · p+ p2 −m22 +m2

1)−m21)2

=

∫ 1

0

dx

((k + xp)2 + x(1− x)p2 − x(m22 −m2

1)−m21)2

=

∫ 1

0

dx

(k′2 −∆)2(345)

where x is known as a Feynman parameter, k′ = k + xp and

∆ = −x(1− x)p2 + x(m22 −m2

1) +m21 (346)

- Furthermore, we can write1

(k′2 −∆)2=

∂

∂∆

(1

k′2 −∆

)(347)

- Using these steps iteratively, we can write the general loop integral (344) in terms of

Iµ1···µmn (∆) ≡

∫µε

ddk

(2π)dkµ1 · · · kµm(k2 −∆)n

(348)

(Of course, the expression will involve complicated derivatives, and an n− 1-dimensional integralover Feynman parameters x1, . . . , xn!)

- To deal with the numerator, note that Iµ1···µm1 (∆) is a symmetric tensor of rank m

- Depends only on a scalar quantity ∆, and therefore has to be Lorentz invariant- The only symmetric tensor available is gµν , and therefore the tensor structure must

be a combination of gµνs only!- Note, odd number of Lorentz indices⇒ vanish- For example,

Iµν1 (∆) = Cgµν , (349)

with some Lorentz scalar C- Contracting this with gµν , we find

gµνIµν1 (∆) = µε

∫ddk

(2π)dk2

k2 −∆= Cgµνg

µν = dC, (350)

and therefore

Iµν1 (∆) =gµν

dµε∫

ddk

(2π)dk2

k2 −∆= gµν

∆

dµε∫

ddk

(2π)d1

k2 −∆= gµν

∆

dI1(∆1/2), (351)

where we used the result that any scale-free integral vanishes- This way, we can write an arbitrary loop integral as a multi-dimensional integral over Feynman

parameters (which can still be extremely difficult to evaluate!)- We must do it consistently in d = 4− ε dimensions and only take ε→ 0 at the very end

53

- As an example, let us calculate

I2(p,m) = µε∫

ddk

(2π)d1

k2 −m2

1

(k + p)2 −m2(352)

- Using Feynman parameters (345) we can write this as

I2(p,m) = µε∫

ddk

(2π)d

∫ 1

0

dx

((k + xp)2 + x(1− x)p2 −m2)2

=

∫ 1

0

dxµε∫

ddk′

(2π)d1

(k′2 −∆(x))2, (353)

where

∆(x) = −x(1− x)p2 +m2 (354)

- Going to Euclidean space, this becomes

I2(p,m) = i

∫ 1

0

dxµε∫

ddkE(2π)d

1

(k2E + ∆(x))2

= i

∫ 1

0

dx IE2 (∆1/2)

=i

16π2

∫ 1

0

dx

(2

ε+ log

4πµ2

∆(x)− γ)

=i

16π2

(2

ε+

∫ 1

0

dx log4πµ2

−x(1− x)p2 +m2− γ)

(355)

- The integral over x can be done analytically, but leads to a long expression which we omit here

6.3 Minimal Subtraction

- In dimensional regularisation, the one-loop 1PI correlators are

Γ2(k2) =N +O=

iλRm2R

32π2

(2

ε+ log

4πµ2

m2R

+ 1− γ)

+ i(k2δZ − δm2

)(356)

and

Γ4 =P + 3×Q +R= −iλR

[1− 3λR

32π2

(2

ε+ log

4πµ2

m2R

− γ + f(s, t, u)

)]− iδλ (357)

- To renormalise the theory, we need to specify how we split the bare parameters torenormalised parameters and counterterms [see Eq. (310)]- This split is completely artificial and does not affect the final result

(i.e. relationships between physical observables)

54

- Earlier we used the “physical” scheme (314)–(316), which has a clear physical interpretationbut is rather clumsy

- Simpler alternative: Minimal subtraction- Choose counterterms in such a way that they cancel the divergences, but only contain the 1/ε pole

- Even more convenient: MS (modified minimal subtraction)- For each pole, include the combination

1

ε+

1

2

(log

4πµ2

M2− γ)

(358)

in the counterterm (because these terms always appear together)- The mass scale M (renormalisation scale) is arbitrary

- The value does not affect physical results- Analogous to the energy scales s0, t0 and u0 in Eq. (314)- One often identifies the scales M = µ, but we keep them separate:

- µ⇔ regularisation- M ⇔ renormalisation

- Using MS and Eqs. (356) and (357), we find

δZ = 0

δm2 =λRm

2R

32π2

(2

ε+ log

4πµ2

M2− γ)

δλ =3λ2

R

32π2

(2

ε+ log

4πµ2

M2− γ)

(359)

- Substituting these back in Eqs. (356) and (357), we find

Γ2(k2) =iλRm

2R

32π2

(log

M2

m2R

+ 1

)+O(λ2

R)

Γ4 = −iλR[1− 3λR

32π2

(log

M2

m2R

+ f(s, t, u)

)]+O(λ3

R) (360)

- We could even take a shortcut and simply “subtract the divergences” from Eqs. (356) and (357)and replace µ→M

- Now m2R and λR do not have any direct physical interpretation

- They are dependent on the renormalisation scale M- Their scale dependence cancels the explicit M -dependence in Eq. (360)

7 Renormalisation Group

7.1 Callan-Symanzik Equation

- Renormalised parameters m2R, λR depend on the renormalisation scale M

but physical observables don’t⇒ Explicit dependence cancelled by the implicit dependence of the renormalised couplings

- For fixed bare coupling, the bare n-point function GBn (p1, . . . , pn) is independent of M- On the other hand, the renormalised n-point function GRn (p1, . . . , pn) is a finite function of the

55

renormalised couplings and the scale M

GRn (p1, . . . , pn;m2R(M), λR(M);M) (361)

- Field renormalisation relates the two

GBn (p1, . . . , pn) = Zn/2(M)GRn (p1, . . . , pn;m2R(M), λR(M);M) (362)

- The M dependence on the RHS must cancel- Let us differentiate GBn with respect to logM

0 =∂

∂ logMGBn

∣∣∣∣B

= Zn/2(M)

(n

Z1/2

∂Z1/2

∂ logM

∣∣∣∣B

+∂m2

R

∂ logM

∣∣∣∣B

∂

∂m2R

∣∣∣∣R

+λR

∂ logM

∣∣∣∣B

∂

∂λR

∣∣∣∣R

+∂

∂ logM

)GRn (m2

R, λR;M), (363)

where |B and |R show whether we keep the bare parameters m2B or λB or the renormalised

parameters m2R, λR fixed when taking the partial derivative

- We then define

γφ =1

Z1/2

∂Z1/2

∂ logM

∣∣∣∣B

=1

2Z

∂Z

∂ logM

∣∣∣∣B

β =∂λR

∂ logM

∣∣∣∣B

γ2 =1

m2R

∂m2R

∂ logM

∣∣∣∣B

(364)

where γφ is known as the anomalous dimension and β as the beta function- This leads to the Callan-Symanzik equation(

∂

∂ logM+ β

∂

∂λR+ γ2m

2R

∂

∂m2R

+ nγφ

)GRn (m2

R, λR;M) = 0, (365)

- It shows how the renormalised correlation functions depend on the renormalisation scale Mwhen the fundamental (“bare”) theory is kept fixed

- This is useful, because we can optimise perturbative calculations by a good choice of M :- We want our renormalised perturbation theory to converge as fast as possible- In general, this happens when M is comparable with the characteristic energy scale of the

process we are looking at- For instance, the scattering amplitude for φφ→ φφ scattering is given by Γ4 in Eq. (360):

- If we choose M appropriately, the quantum correction becomes small, and λR isa good approximation of the actual scattering amplitude- This typically means that M is close to the most relevant energy scales for the process

⇒ The beta function β tells how the interaction strength depends on the characteristic energy scale

- According to Eq. (310), we can write the renormalised parameters as

56

Z(M) = 1 + δZ(M)

λR(M) = Z(M)2λB − δλ(M)

m2R(M) = Z(M)m2

B − δm2(M) (366)

- This means that to leading order we have

γφ =1

2Z

∂δZ

∂ logM

∣∣∣∣B

≈ 1

2

∂δZ

∂ logM

∣∣∣∣R

β = 2ZλB∂Z

∂ logM

∣∣∣∣B

− ∂δλ

∂ logM

∣∣∣∣B

≈ 4λRγφ −∂δλ

∂ logM

∣∣∣∣R

γ2 =m2B

m2R

∂Z

∂ logM

∣∣∣∣B

− 1

m2R

∂δm2

∂ logM

∣∣∣∣B

≈ 2γφ −∂δm2

∂ logM

∣∣∣∣R

(367)

- In the MS scheme, we obtain the counterterms in terms of the renormalised couplings- The counterterms are are proportional to

2

ε+ log

4π2µ2

M2− γ, (368)

with no other (explicit) M dependence- The coefficients γφ, β and γ2 are therefore M -independent- In fact, we can write the counterterms as

δZ = −γφ(

2

ε+ log

4π2µ2

M2− γ)

δm2 =

(1

2γ2 − γφ

)m2R

(2

ε+ log

4π2µ2

M2− γ)

δλ =

(1

2β − 2λRγφ

)(2

ε+ log

4π2µ2

M2− γ)

(369)

- Comparing with Eq. (359), we find

γφ = 0

β =3λ2

R

16π2

γ2 =λR

16π2(370)

- Positive β ⇒ λR grows with energy- Explicit solution

λR(M) =16π2

3 log(Λ/M)(371)

where Λ is an integration constant- Note that Λ is really what parameterises the strength of the interaction- λR blows up at M = Λ: “Landau pole”

57

7.2 Wilsonian Renormalisation

- So far we have taken the “experimentalist’s view”, expressing everything in terms of renormalisedparameters and essentially ignoring the bare ones

- But bare parameters are what actually appear in the Lagrangian, and they can give us important insight- Therefore, let us look at the dependence of the bare theory on the cutoff,

keeping physical observables fixed- Consider a “fundamental” theory with action S[φ] and a physical cutoff Λ (e.g. Planck scale)

- Observables are given by derivatives of the generating functional

Z[J ] =

∫DφeiS+i

∫ddxJ(x)φ(x) (372)

- Cutoff: Integrate only over those φ(x) whose Fourier transform φ(k) vanishes for |k2| > Λ2

- Does not work well in Minkowski space:Allows arbitrarily high energies when momentum is nearly lightlike

- Wick rotation t = −iτ

iS =

∫dtd3x

[1

2(∂tφ)2 − 1

2(∂iφ)2 − V (φ)

]= −

∫dτd3x

[1

2(∂τφ)2 +

1

2(∂iφ)2 + V (φ)

]= −

∫Eucl

d4x

[1

2(∂µφ)2 + V (φ)

]≡ −SE (373)

- Write τ = x4: Four-dimensional Euclidean integral- Exactly the same rotation k0 → ikE we did in Eq. (252)- The cutoff becomes meaningful and we have

Z[J ] =

∫0<k<Λ

Dφe−SE−∫ddxJ(x)φ(x) =

∫ ∏|k|<Λ

dφ(k)

e−SE (374)

- Compare with classical statistical mechanics

Z =

∫all configs

e−βH ⇔ βH ↔ SE (375)

- 3+1D QFT = 4D classical statistical mechanics- In statmech, the cutoff is physical: Atomic spacing Λ = 1/δ in ferromagnets- Usually relevant length scales ξ comparable to δ- At critical (phase transition) point ξ →∞

Critical phenomena ⇔ QFTξ δ ⇔ m Λ

Universality ⇔ Renormalisability(Independence of microscopic details) (Independence of UV physics)

- Now, we want to move the cutoff from Λ to Λ′ < Λ, but still describe the same physics- This is known as coarse-graining, or integrating out momenta Λ′ < k < Λ

- We can do this by splitting the integration into two pieces: k < Λ′ and Λ′ < k < Λ

- Write φ→ φ+ φ where

58

φ(k) =

nonzero for k < Λ′

0 for Λ′ < k < Λ0 for k > Λ

, φ(k) =

0 for k < Λ′

nonzero for Λ′ < k < Λ0 for k > Λ

(376)

- Let us assume that we are only interested in observables with momenta less than the new cutoff Λ′:J(k) = 0 for k > Λ′

- Then we have

Z[J ] =

∫0<k<Λ′

Dφ∫

Λ′<k<Λ

Dφe−S[φ+φ]−∫ddxJ(x)φ(x) (377)

and defining the effective action SΛ′

eff [φ] as

SΛ′

eff [φ] = − log

∫Λ′<k<Λ

Dφ e−S[φ+φ] (378)

we can write this as

Z[J ] =

∫0<k<Λ′

Dφ e−SΛ′eff [φ]−

∫ddxJ(x)φ(x) (379)

- Any correlation function can be obtained from Z[J ]

⇒ SΛ′

eff gives an effective theory, with a lower cutoff Λ′, with exactly the same low-energy (k < Λ′) physics- We say that we “integrate” out the high-momentum (k > Λ′) modes to obtain an effective field theory- The effective theory can then be used just as if it was the fundamental one

- Now, let us consider the scalar theory with Euclidean action

S =

∫ddx

[1

2(∂µφ)2 +

1

2m2φ2 +

1

4!λφ4

](380)

- The effective action (378) is then

SΛ′

eff [φ] = S[φ]− log

∫Λ′<k<Λ

Dφe−∆S[φ] (381)

where

∆S[φ] = S[φ+ φ]− S[φ]

=

∫ddx

[1

2(∂µφ)2 +

1

2m2φ2 + λ

(1

6φ3φ+

1

4φ2φ2+

1

6φφ3+

1

24φ4

)](382)

and the quadratic cross terms φφ cancel because of momentum conservation- Eq. (381) is essentially the generating functional for connected correlators E[J ] (205)

in theory with dynamical field φ and action ∆S[φ]

- The low-energy field φ plays roughly the role of the external field J(x)

- The path integral can be calculated using same perturbative techniques as before:Sum of connected diagrams

- Euclidean⇒ No imaginary units- Momenta restricted to Λ′ < k < Λ

- Represent φ by double lineS : Internal lines- Propagator: T ↔ θ(k)

k2 +m2(383)

59

where θ(k) = 1 if Λ′ < k < Λ and = 0 otherwise- Vertices:

U ↔ −λ6φ3

V ↔ −λ2φ2

W ↔ −λφ

X ↔ −λ

(384)

where dashed linesY represent factors of φ- Note that what we denote by “φn” is, strictly speaking

φn = φ(x)n →∫ Λ′

0

ddk1

(2π)d· · · d

dkn(2π)d

φ(k1) · · ·φ(kn)(2π)dδ(k1 + k2 + k3 + k4) (385)

- To order O(λ), the integral in Eq. (381) is(assuming m2 Λ′2 so that we can approximate m2 ≈ 0)

log

∫Dφe−∆S =Z +[ +O(λ2)

= const× λ− λ

4φ2

∫ Λ

Λ′

ddk

(2π)d1

k2+O(λ2)

= −λ

(1

4φ2(2π)−d

∫dΩd

∫ Λ

Λ′dk kd−3 + const

)+O(λ2)

= −λ(

1

2

Λd−2 − Λ′d−2

(d− 2)(4π)d/2Γ(d/2)φ2 + const

)+O(λ2) (386)

where the constant arises from the figure-eight diagram- Taking the logarithm, we find the effective Lagrangian

Leff = L+ λ

(1

2

Λd−2 − Λ′d−2

(d− 2)(4π)d/2Γ(d/2)φ2 + const

)+O(λ2)

=1

2(1 + ∆Z)(∂µφ)2 +

1

2

(m2 + ∆m2

)φ2 +

1

4!λφ4 +O(λ2) (387)

where we have dropped the uninteresting constant and defined

∆m2 = λΛd−2 − Λ′d−2

(d− 2)(4π)d/2Γ(d/2)

d=4−→ λ

32π2

(Λ2 − Λ′2

)(388)

60

- Thus, we have found an effective theory:- Lower cutoff Λ′ < Λ

- Equivalent to the original one to linear order in λ- To this order, we can compensate for the change in the cutoff by changing the bare mass

- To order O(λ2), we have

log

∫Dφe−∆S = . . .+ +\ +]

+^+_+`+a+b+c (389)

- The first row gives an uninteresting constant term- The sixth and eighth terms vanish because of momentum conservation- The logarithm in Eq. (381) removes the disconnected diagrams

- Thus, the effective action to order O(λ2) is

Seff [φ] = S[φ]−d −e −f−g −h (390)

- The first three diagrams are proportional to φ2 and give contributions to the mass term- Because the two-loop diagrams are subleading, we ignore them

- The fourth diagram gives a correction to the φ4 term (ignoring the external momenta):

∆λ = −3

2λ2

∫ Λ

Λ′

ddk

(2π)d

(1

k2

)2

= −3

2λ2 2πd/2

(2π)dΓ(d/2)

∫ Λ

Λ′dk kd−5

=

− 3λ2

(4π)d/2(d−4)Γ(d/2)

(Λd−4 − Λ′d−4

), when d 6= 4

− 3λ2

16π2 log ΛΛ′ , when d = 4

(391)

- The fifth diagram gives rise to a new, six-point interaction vertex

61

ip3

p2

p1

∼ λ2φ6

(p1 + p2 + p3)2θ(p1 + p2 + p3) ≈ λ2

Λ2φ6θ(p1 + p2 + p3) (392)

- This may look worrying: We wanted to understand why QFTs in particle physics are renormalisable,but instead, we found that even if we start from a renormalisable fundamental theory,it creates a non-renormalisable effective interaction⇒ Does it make any difference whether the “fundamental” theory is renormalisable

because it is equivalent to a non-renormalisable theory anyway?- Note, however, that the effective φ6 coupling is suppressed by Λ−2

- In summary, the effective Lagrangian with cutoff Λ′ in 4D is

LΛ′

eff ≈ 1

2(∂µφ)2 +

1

2

[m2 +

λ

32π2

(Λ2 − Λ′2

)]φ2 +

1

4!

[λ− 3λ2

16π2log

Λ

Λ′

]φ4 + #

λ2

Λ2φ6(393)

- Because we ignored the external momenta, we are missing derivative terms such as φ2(∂µφ)2

- It is easy to see that at higher order in λ, all terms allowed by symmetries appear- Because Λ is the only dimensionful parameter in loops, the coefficient of φn is ∝ Λ4−n

7.3 Renormalisation Group Transformation

- In the previous section, we changed the cutoff from Λ to Λ′

- In general, this gives an effective Lagrangian of the form

Leff =1

2(1 + ∆Z)(∂µφ)2 +

1

2(m2 + ∆m2)φ2 +

1

4!(λ+ ∆λ)φ4 + ∆C(∂µφ)4 + ∆Dφ6 + . . . , (394)

where we have included the contribution ∆Z, which is closely related to field renormalisation andtwo examples of non-renormalisable terms ∆C and ∆D

- Now, zoom out by the same amount: Rescale all lengths by factor b = Λ′/Λ

x → x′ = xb

k → k′ = k/b (395)

- This takes the cutoff back to its original value

Λ′ → Λ′

b= Λ (396)

- Basically, we are expressing everything in units of the cutoff- Coarse-graining + rescaling = renormalisation group transformation

- Maps the theory to a different theory in the same space (i.e. same cutoff)- The action becomes

62

Seff =

∫ddxLeff

=

∫ddx′b−d

[1

2(1 + ∆Z)b2(∂′µφ)2 +

1

2(m2 + ∆m2)φ2 +

1

4!(λ+ ∆λ)φ4

+∆Cb4(∂′µφ)4 + ∆Dφ6 + . . .], (397)

- As before, it is convenient to rescale the field back to the canonical normalisation

φ′ =√b2−d(1 + ∆Z)φ (398)

so that the kinetic term (and therefore the free propagator) has the canonical normalisation

Seff =

∫ddx′

[1

2(∂′µφ

′)2 +1

2

m2 + ∆m2

1 + ∆Zb−2φ′2 +

1

4!

λ+ ∆λ

(1 + ∆Z)2b4−dφ′4

+∆C

(1 + ∆Z)2bd(∂′µφ

′)4 +∆D

(1 + ∆Z)3b2d−6φ′6 + . . .

](399)

- The coefficient G of a general term with n powers of φ and m derivatives is

G′ =G+ ∆G

(1 + ∆Z)n/2b(d/2−1)n+m−d =

G+ ∆G

(1 + ∆Z)n/2b−[G] (400)

7.4 Renormalisation Group Flow

- Often useful to consider infinitesimal transformations- Write b = Λ′/Λ = 1− ε, with ε 1

- The corrections ∆m2, ∆λ, ∆Z etc. are then O(ε)

- The mass parameter changes

m′2 = m2 + dm2 =m2 + ∆m2

1 + ∆Z(1 + 2ε) , ⇒ dm2 = m2

(∆m2

m2−∆Z + 2ε

)(401)

- To linear order in λ we had (in 4D)

∆m2 = (1− b2)λ

32π2Λ2 =

λ

16π2Λ2ε

∆Z = 0 (402)

- Thus, writing

ε = d log1

b(403)

we have

dm2 =

(2m2 +

λ

16π2Λ2

)d log

1

b, (404)

which we can write as a differential equation

63

dm2

d log(1/b)= 2m2 +

λ

16π2Λ2 (405)

- For λ, we have

λ′ =λ+ ∆λ

(1 + ∆Z)2= λ− 3λ2

16π2d log

1

b, (406)

which impliesdλ

d log(1/b)= − 3λ2

16π2(407)

- Compare with Eq. (370):dλ

d log(1/b)= − 3λ2

16π2= −β(λ) +O(λ3) (408)

- The leading term is generally the same:

β =∂λR

∂ logM

∣∣∣∣B

= − ∂λR∂λB

∣∣∣∣M

∂λB∂ logM

∣∣∣∣R

=∂λR∂λB

∣∣∣∣M

∂λB∂ log Λ

∣∣∣∣R

≈ ∂λB∂ log Λ

∣∣∣∣R

= − ∂λB∂ log 1/b

∣∣∣∣R

(409)

where we used the fact that between the cutoff and dimensional regularisation, the divergencescorrespond to each other as

log Λ ∼ logµ

M(410)

and that to leading order λR = λB

- Higher orders depend on the renormalisation scheme

- We have a similar equation for every couplingdG

d log(1/b)= [G]G+

∆G− (n/2)∆Z

ε(411)

- Formally, we can write an infinite-dimensional vector ~g = (m2, λ, C,D, . . .),containing all the couplings, and write

d~g

d log(1/b)= f(~g) (412)

- This tells how the effective theory changes under gradual coarse-graining- Flow in the space of all possible theories

= Renormalisation group flow- First-order equation⇒ Trajectories cannot cross- There can be fixed points ~g∗ with f(~g∗) = 0

- They define scale-invariant (conformal) field theories

- Example: Gaussian fixed point m2 = λ = C = D = . . . = 0, i.e.,

L =1

2(∂µφ)2 (413)

64

- Let us look at the flow near the fixed point:- λ 1, m2/Λ2 1 just as in our perturbative calculation

- Leading order: Ignore corrections ∆m2, ∆λ etc⇒ Scaling

m′2 = m2b−2 → m2/b2 in 4Dλ′ = λbd−4 → λ

G′ = Gb−[G] (414)

- Following the flow toward smaller b, we find three different possibilities- [G] > 0: Coupling grows (unstable direction)

= Relevant operator- Corresponds to a superrenormalisable term- Only m2 in our theory

- [G] = 0: Coupling does not change= Marginal operator- Corresponds to a renormalisable term- Only λ in our theory

- [G] < 0: Coupling decreases (stable direction)= Irrelevant operator- Corresponds to a non-renormalisable term- All other terms

- Thus, if we start with some “fundamental” theory with all couplings of order 1

- Relevant operators grow- These must be fine tuned to remain small even at low energies in order to stay near the fixed point- This is known as the hierarchy problem

- Irrelevant operators decrease- Theory becomes automatically renormalisable

- This means that only relevant operators have to be fine-tuned!- Sometimes no renormalisable terms are compatible with symmetries

- Then the least irrelevant term determines the strength of the interaction- The strength of the interaction is suppressed by a power of b- Examples: Gravity, Fermi theory of weak interactions

- There may also be other fixed points, which would also define valid low-energy theories- However, these will be non-perturbative- For example, second-order phase transitions define non-trivial fixed points

- Let us then include the leading quantum corrections

dm2

d log(1/b)= 2m2 +

λ

16π2Λ2

dλ

d log(1/b)= − 3λ2

16π2(415)

- Flow is somewhat more complicated

65

-1 0 1 2

λ

-0.1

-0.05

0

0.05

0.1

m2

- Fixed point still at m2 = λ = 0

- Unstable direction along λ = 0

- Stable direction along m2 = m2c(λ) with

m2c(λ) = − λ

32π2Λ2 (416)

- The mass parameter of the “fundamental theory” has to be fine tuned to m2 ≈ m2c(λ)

- This is precisely what Eq. (278) does- Analogous to critical phenomena T ≈ Tc

- Coupling constant runs: Interactions become stronger at high energies

λ(Λ) =16π2

3 log(Λ0/Λ)(417)

- Diverges at a finite cutoff value Λ = Λ0 (Landau pole)⇒ The “continuum limit” Λ→∞ cannot be taken if β > 0

- Strictly speaking the perturbative expansion in λ becomes unreliable when λ ≈ 1

- It is possible that there is an ultraviolet fixed point at (λ∗,m2∗ = m2

c(λ∗))

- Non-perturbative: Could not be studied using perturbation theory- This would make it possible to define a continuum theory:

RG flow trajectory that starts infinitesimally close to (λ∗,m2∗)

- No evidence for such a fixed point exists- If no fixed point, the scalar theory is trivial:

Finite coupling λ in fundamental theory⇒ λ→ 0 at low energies= Free theory

- The same conclusions apply to QED:

β(e2) =Nf6π2

e4 ⇒ de2

d log(1/b)= −Nf

6π2e4 (418)

- Coupling e (=electron charge!) decreases at low energies (=long distances)- Can be seen as a screening effect:

66

- Virtual electron-positron pairs get polarised by the electric field of an electron:The positron is attracted and electron is repelled

- This screens some of the electric field- If we calculate the electron change from the field strenght at long distances, we obtain a smaller value

- Again, we find a Landau pole

e2(Λ) =6π2

Nf log(Λ0/Λ)(419)

⇒ QED does not exist as a (perturbative) continuum theory- Little experimental significance:

If Λ = me (the electron mass), e2(me) = 0.09 and Nf = 1, then

Λ0 ≈ 10300 GeV (420)

- QED can only exist as a part of some more fundamental theory (GUT?)- Most 4D theories have a Landau pole and are trivial

- Main exception: Non-Abelian gauge field theories

8 Renormalisation of QCD

8.1 Counterterms

- The gauge-fixed Lagrangian (153) has six parameters:- Three field normalisations: Aµ, c and ψ- Parameters: g, ξ and m

- Do we need to include more terms to renormalise the theory?- Generally all terms compatible with symmetries- Gauge invariance is broken by gauge fixing, so it cannot protect the Lagrangian- The gauge fixed theory has a more complicated BRST (Becchi-Rouet-Stora-Tyutin) symmetry

- Introduce auxiliary field Ba with Lagrangian

LB =1

2ξBaBa +Ba∂µAaµ (421)

- Integrating over Ba gives the gauge fixing term

67

∫DBei

∫ddxLB = const× ei

∫ddx(− 1

2ξ (∂µAaµ)2) (422)

- We can therefore replace in the Lagrangian (153)

− 1

2ξ(∂µAaµ)2 → 1

2ξBaBa +Ba∂µAaµ (423)

- The resulting Lagrangian has a peculiar BRST symmetry

Aaµ → Aaµ + εDacµ c

c

ψi → ψi − igεtaijcaψj

ca → ca +1

2gεfabccbcc

ca∗ → ca∗ + εBa

Ba → Ba (424)

- This symmetry does not allow any new terms, and therefore it is enough to considerLagrangians of the form (153)

- All four interactions have coupling g (or g2)- Related by gauge invariance, even in quantum theory- We treat them as independent at first

- To set up renormalised perturbation theory, we first rescale the fields by renormalisation constants:

Aaµ → Z1/2A Aaµ, ca → Z1/2

c ca, ψi → Z1/2ψ ψi (425)

- The Lagrangian (153) becomes

L = −1

4ZA(∂µA

aν − ∂νAaµ)(∂µAaν − ∂νAaµ)− 1

2

ZAξB

(∂µAaµ)2

+Zc∂µca∗∂µc

a + Zψψi (i /∂ −mB)ψi

+1

2gBZ

3/2A fabc(∂µA

aν − ∂νAaµ)AbµAcν − 1

4g2BZ

2Af

abcfadeAbµAcνA

d µAe ν

−gBZcZ1/2A ∂µca∗fabcAbµc

c − gBZψZ1/2A γµαβt

aijA

aµψiαψjβ , (426)

where we have also added the subscripts “B” to the parameters to indicate they are bare ones- We define renormalised parameters by introducing multiplicative renormalisation constants

ZAξB

=Zξξ

ZψmB = Zmm

Z3/2A gB = Z3g

Z2Ag

2B = Z4g

2

ZcZ1/2A gB = Z1g

ZψZ1/2A gB = Z2g (427)

- Note that the renormalisation constants Z1, Z2, Z3 and Z4 are related to each other, e.g.

68

Z2 =ZψZc

Z1 (428)

- In terms of the renormalised fields and parameters, the Lagrangian is therefore

L = −1

4ZA(∂µA


2

Zξξ

(∂µAaµ)2

+Zc∂µca∗∂µc

a + ψi(iZψ /∂ − Zmm

)ψi

+1

2Z3gf

abc(∂µAaν − ∂νAaµ)AbµAcν − 1

4Z4g

2fabcfadeAbµAcνA

d µAe ν

−Z1g∂µca∗fabcAbµc

c − Z2gγµαβt

aijA

aµψiαψjβ , (429)

- Next we write the renormalisation constants ZX in terms of counterterms, ZX = 1 + δZX

- This gives

L = −1

4(∂µA


2ξ(∂µAaµ)2

+∂µca∗∂µca + ψi

(i /∂ −m

)ψi

+1

2gfabc(∂µA

aν − ∂νAaµ)AbµAcν − 1

4g2fabcfadeAbµA

cνA

d µAe ν

−g∂µca∗fabcAbµcc − gγµαβt

aijA

aµψiαψjβ

−1

4δZA(∂µA


2

δZξξ

(∂µAaµ)2

+δZc∂µca∗∂µc

a + ψi(iδZψ /∂ − δZmm

)ψi

+1

2δZ3gf

abc(∂µAaν − ∂νAaµ)AbµAcν − 1

4δZ4g

2fabcfadeAbµAcνA

d µAe ν

−δZ1g∂µca∗fabcAbµc

c − δZ2gγµαβt

aijA

aµψiαψjβ (430)

- The propagators were given in Eqs. (237)–(239)- The interaction vertices are Eqs. (243), (245), (247) and (249), together with the counterterm vertices

69

jAaµ(k) Abν(−k) ↔ −iδab[δZA

(k2gµν − kµkν

)− δZξ

ξkµkν

]

kψjβ(k) ψiα(k) ↔ iδij (δZψ /k − δZmm)αβlcb∗(k) ca(k) ↔ −iδabδZck2

mAaµ(k1)

Abν(k2) Acρ(k2)

↔ gδZ3fabc [gµν(kρ2 − k

ρ1) + gνρ(kµ3 − k

µ2 ) + gµρ(kν1 − kν3 )]

nAaµ Abν

Acρ Adλ

↔ −ig2δZ4

[fabef cde


)

+facef bde(gµνgρλ − gµλgνρ

)+fadef bce


)]

oAbµ(k2)

ca∗(k1) cb(k3)

↔ δZ1gfabckµ1

pAaµ

ψiα ψjβ

↔ −iδZ2gγµαβt

aij (431)

- To determine the counterterms, we have to find out which correlator gets a tree-level contributionfrom each counterterm

70

δZA, δZξ ↔qδZc ↔r

δZψ, δZm ↔sδZ1 ↔tδZ2 ↔uδZ3 ↔vδZ4 ↔w (432)

- We need to compute all the two-point functions, but because Z1–Z4 are related, it is enoughto compute one of the last four correlators- We choose the quark-gluon interaction δZ2, because it consists of fewer diagrams than

the gluon-gluon interactions, and it corresponds to a real physical process unlikethe ghost-gluon interaction

⇒We need four correlation functions to determine the six counterterms

8.2 Correlation Functions

8.2.1 Gluon

- We start by looking at the gluon two-point function

71

x =y +z ++| + (433)

- The quark loop is

~p+ k

p

Aaµ(k) Abν(−k)i α

j β

γ i

δ j=

1

2!× (−1)× 2×

(−igγµαβt

aij

) (−igγνδγtbji

)×

×µε∫

ddp

(2π)d

(i

/p+ /k −m

)γα

(i

/p−m

)βδ

= g2tr tatbµε∫

ddp

(2π)dTr

(i

/p+ /k −m

)γµ(

i

/p−m

)γν ,(434)

where tr is a trace over colour indices ij and Tr over spinor indices αβ- To proceed, we have to recall some identities involving gamma matrices

γµ, γν = 2gµν ⇒ 1

/p−m=

/p+m

p2 −m2

Tr γµγν =1

2Tr γµ, γν = gµνTr 1 = 4gµν

Tr γµγνγρ = 0

Tr γµγνγργλ = 4(gµνgρλ + gµλgνρ − gµρgνλ

)(435)

- Thus, we have = −g2tr tatbµε∫

ddp

(2π)dTr ( /p+ /k +m)γµ( /p−m)γν

((k + p)2 −m2)(p2 −m2)

= −g2tr tatb∫ 1

0

dxµε∫

ddp

(2π)dTr ( /p+ /k +m)γµ( /p−m)γν

((p+ xk)2 −∆(x))2, (436)

where on the second line we have used the Feynman parameter trick (345) and defined

∆(x) = −x(1− x)k2 +m2 (437)

- We shift the integration variable p+ xk → p,

72

= −g2tr tatb∫ 1

0

dxµε∫

ddp

(2π)dTr ( /p+ (1− x) /k +m)γµ( /p− x /k −m)γν

(p2 −∆(x))2(438)

= −g2tr tatb∫ 1

0

dxµε∫

ddp

(2π)d(p+ (1−x)k)ρ(p− xk)λTrγργµγλγν +m2Tr γµγν

(p2 −∆(x))2

= −g2tr tatb∫ 1

0

dxµε∫

ddp

(2π)d(pρpλ − x(1−x)kρkλ)Trγργµγλγν +m2Tr γµγν

(p2 −∆(x))2,

where in the last step we used the fact that terms with one power of p in the numerator vanishby Lorentz invariance

- As in Eq. (351), we have

µε∫

ddp

(2π)dpρpλ

(p2 −∆)2=gρλdµε∫

ddp

(2π)dp2

(p2 −∆)2=gρλd

(I1(∆) + ∆I2(∆)) , (439)

where

In(∆) = µε∫

ddp

(2π)d1

(p2 −∆)n(440)

- The values of the integrals are given by Eq. (328)

I1(∆) = −i∆d/2−1

(4π)d/2Γ(1− d/2)

I2(∆) = i∆d/2−2

(4π)d/2Γ(2− d/2) (441)

- The Gamma function satisfies zΓ(z) = Γ(z + 1), which implies

(2− d)I1(∆) = −2i∆d/2−1

(4π)d/2

(1− d

2

)Γ(1− d/2) = −2i

∆d/2−1

(4π)d/2Γ(2− d/2) = −2∆I2(∆) (442)

and further

µε∫

ddp

(2π)dpρpλ

(p2 −∆)2=gρλd

(2

d− 2∆I2(∆) + ∆I2(∆)

)=

gρλd− 2

∆I2(∆) (443)

- Using Eqs. (435), we find = −4g2tr tatb∫ 1

0

dx

[(gρλd− 2

∆I2(∆)− x(1− x)kρkλI2(∆)

)×

×(gµρgνλ + gµλgνρ − gµνgρλ

)+m2gµνI2(∆)

]= −4g2tr tatb

∫ 1

0

dx

[2− dd− 2

gµν∆I2(∆)

−x(1− x)(2kµkν − k2gµν

)I2(∆) +m2gµνI2(∆)

]= −4g2tr tatb

∫ 1

0

dx[(−∆ + x(1− x)k2 +m2

)gµν − 2x(1−x)kµkν

]I2(∆)

= −4g2tr tatb∫ 1

0

dx 2x(1−x)(k2gµν − kµkν

)I2(∆), (444)

73

where we used ∆ = m2 − x(1− x)k2

- Near four dimensions, when d = 4− ε, we have

I2(∆) ≈ i

16π2

(2

ε+ finite

), (445)

and therefore = − 4ig2

16π2tr tatb

(k2gµν − kµkν

)(2

ε+ finite

)∫ 1

0

dx 2x(1− x)

= − ig2

16π2tr tatb

4

3

(k2gµν − kµkν

)(2

ε+ finite

)(446)

- We could calculate the finite part, but for MS renormalisation it is not needed- The group factor tr tatb ≡ C(r)δab depends on the quark representation r

- For the fundamental representation (our case), Eq. (124) gives C(r) = 1/2

- For Nf fermion species, the total result is = − ig2

16π2

4

3NfC(r)δab

(k2gµν − kµkν

)(2

ε+ finite

)(447)

- The second diagram in Eq. (433) gives =ig2

16π2facdf bcd

[(25

12− ξ

2

)k2gµν −

(17

6− ξ)kµkν

](2

ε+ finite

)=

ig2

16π2C2(G)δab

[(25

12− ξ

2

)(k2gµν − kµkν

)+

(−3

4+ξ

2

)kµkν

]×

×(

2

ε+ finite

), (448)

where we have written facdf bcd = C2(G)δab in terms of the quadratic Casimir invariantof the adjoint representation (for SU(N ), C2(G) = N )

- The third diagram is = 0, (449)because the integral is scale-free

- The ghost loop (4th diagram) gives =ig2

16π2facdf bcd

(1

12k2gµν +

1

6kµkν

)(2

ε+ finite

)=

ig2

16π2C2(G)δab

[1

12

(k2gµν − kµkν

)+

1

4kµkν

](2

ε+ finite

), (450)

74

- Finally, the counterterm diagram gives = −iδZA(k2gµν − kµkν

)− i δZξ

ξkµkν (451)

- Putting all of these together, we find =ig2

16π2δab[(−4

3NfC(r) +

(13

6− ξ

2

)C2(G)

)(k2gµν − kµkν

)+ξ − 1

2C2(G)kµkν

](2

ε+ finite

)−iδZA

(k2gµν − kµkν

)− i δZξ

ξkµkν (452)

- Following the MS scheme, we must therefore choose

δZA =g2

16π2

[−4

3NfC(r) +

(13

6− ξ

2

)C2(G)

](2

ε+ log

4πµ2

M2− γ)

δZξξ

=g2

16π2

ξ − 1

2C2(G)

(2

ε+ log

4πµ2

M2− γ)

(453)

- There are some interesting points to note:- The counterterms depend on ξ, but that is not a problem because they are not physical observables- The gauge fixing parameter ξ gets renormalised:

It is therefore not consistent to simply fix its value to something convenient

8.2.2 Quark

- The quark two-point function consists of only two diagrams

= + (454)

- The loop diagram is

p+ k

−p

ψiα(k) ψjβ(k)a µ

i l

ν a

l j=

(−igγµγαtali

) (−igγνβδtajl

)×

×µε∫

ddp

(2π)dDµνF (p)SδγF (p+ k)

= −g2(tata)jiµε

∫ddp

(2π)dDµνF (p)

(γνSδγF (p+ k)γµ

)βα

(455)

75

- Note that the order of the matrices is opposite to the propagation of the quark- We have here the quadratic Casimir of the quark representation

(tata)ij = C2(r)δij (456)

- For the fundamental representation in SU(N ), C2(r) = (N2 − 1)/2N

- Doing the algebra and evaluating the integral, one finally obtains

=ig2

16π2δijC2(r) (ξ /k − (3 + ξ)m)βα

(2

ε+ finite

)(457)

- The counterterm diagram gives = i (δZψ /k − δZmm) δij (458)

- In the MS scheme, we therefore have

δZψ = − g2

16π2C2(r)ξ

(2

ε+ log

4πµ2

M2− γ)

δZm = − g2

16π2C2(r)(3 + ξ)

(2

ε+ log

4πµ2

M2− γ)

(459)

8.2.3 Ghost

- The ghost two-point function consists of

= + , (460)

where =ig2

16π2δabC2(G)

3− 5ξ

4k2

(2

ε+ finite

)(461)

and = −iδabk2δZc (462) -Therefore, we have

76

δZc =g2

16π2C2(G)

3− 5ξ

4

(2

ε+ log

4πµ2

M2− γ)

(463)

8.2.4 Quark-gluon coupling

- To compute δZ2, we need the quark-gluon interaction

Aaµ(q)

ψiα(k) ψjβ(k + q)

= + + , (464)

where

= − ig3

16π2

(C2(r)− 1

2C2(G)

)ξtajiγ

µβα

(2

ε+ finite

), (465)

= − ig3

16π2C2(G)

3(1 + ξ)

4tajiγ

µβα

(2

ε+ finite

), (466)

and

= −iδZ2gγµβαt

aji (467)

- This means that the counterterm is

δZ2 =g2

16π2

[−3(1 + ξ)

4C2(G)− ξC2(r) +

ξ

2C2(G)

](2

ε+ log

4πµ2

M2− γ)

= − g2

16π2

[3

4C2(G) +

(1

4C2(G) + C2(r)

)ξ

](2

ε+ log

4πµ2

M2− γ)

(468)

8.2.5 Summary

- In summary, the values of the counterterms are

77

δZA =g2

16π2

[(13

6− ξ

2

)C2(G)− 4

3NfC(r)

](2

ε+ log

4πµ2

M2− γ)

δZψ = − g2

16π2C2(r)ξ

(2

ε+ log

4πµ2

M2− γ)

δZc =g2

16π2C2(G)

3− 5ξ

4

(2

ε+ log

4πµ2

M2− γ)

δZξξ

=g2

16π2

ξ − 1

2C2(G)

(2

ε+ log

4πµ2

M2− γ)

δZm = − g2

16π2C2(r)(3 + ξ)

(2

ε+ log

4πµ2

M2− γ)

δZ2 = − g2

16π2

[3

4C2(G) +

(1

4C2(G) + C2(r)

)ξ

](2

ε+ log

4πµ2

M2− γ)

(469)

8.3 Beta Function

- We can now compute the QCD beta function

β(g) = M∂g

∂M

∣∣∣∣B

(470)

- In terms of the bare coupling and the counterterms, the renormalised coupling is

g =ZψZ

1/2A

Z2gB =

(1 + δZψ +

1

2δZA − δZ2

)gB +O(g5

B) (471)

- Therefore, we have

β(g) = 2g

(∂δZψ

∂ logM2+

1

2

∂δZA∂ logM2

− ∂δZ2

∂ logM2

)= 2g

[g2

16π2ξC2(r)

−1

2

g2

16π2

((13

6− ξ

2

)C2(G)− 4

3NfC(r)

)− g2

16π2

(3

4C2(G) +

(1

4C2(G) + C2(r)

)ξ

)]=

2g3

16π2

[(ξ − ξ)C2(r) +

(−13

12+ξ

4− 3

4− ξ

4

)C2(G) +

2

3NfC(r)

]=

g3

16π2

[−11

3C2(G) +

4

3NfC(r)

](472)

- The gauge (ξ) dependence has now disappeared- For SU(N ) and fundamental quarks we have C2(G) = N and C(r) = 1/2, and therefore

β(g) =g3

16π2

(−11N

3+

2

3Nf

)(473)

- If Nf < (11/2)N , the beta function is negative

78

⇒ The interaction gets weaker at higher energies (=asymptotic freedom)- This is precisely what deep inelastic scattering results showed already in 1970s- Interaction becomes strong at low energies: This is why we cannot see individual quarks and gluons- This calculation gave Gross, Politzer and Wilczek the 2004 Nobel prize

8.4 Renormalisation Group Flow

- Consider the flow in the Wilsonian approach (Section 7.4)- Use g and m/Λ as coordinates- How does the effective bare coupling g change as we lower the cutoff?- It is given by the equation

dg

d log Λ= β(g) =

g3

16π2

(−11N

3+

2

3Nf

)(474)

- If Nf < (11/2)N , β(g) < 0, and g is marginally unstable- This is easy to solve

1

g2=

1

8π2

(11N

3− 2

3Nf

)log

Λ

ΛQCD, (475)

where ΛQCD is an integration constant

0 0.5 1 g0

1

m/Λ

- Because the beta function is negative (not positive as in QED or scalar theory), the flow is directedaway from the Gaussian fixed point (circle)

- Continuum limit Λ→∞ corresponds to following the flow trajectory backwards- We can go arbitrarily far⇒ The theory has a continuum limit (unlike QED or scalar theory)

- Following the flow towards low energies, g diverges at Λ = ΛQCD

- Analogue of the Landau pole but at low energies- Physics does not really become singular:

It just becomes non-perturbative, so our assumptions fail- Below ΛQCD we need a different effective theory:

- The relevant degrees of freedom are protons and neutrons, not quarks and gluons- In a sense, the flow does not actually hit g =∞ but it leaves the two-dimensional plane- The appropriate theory is known as chiral perturbation theory

- This explains why we don’t see massless gluons or individual quarks- The hadron masses given by ΛQCD

79

- Account for most of the mass of everyday objects- Nothing to do with the Higgs field!

- In the Callan-Symanzik approach, the same equation tells how the renormalised coupling depends on M ,1

g2R

=1

8π2

(11N

3− 2

3Nf

)log

M

ΛQCD(476)

- Note that gR depends only on how far above ΛQCD we are:- There is no free dimensionless parameter- LEP:

αs(MZ) =g2(MZ)

4π= 0.1176± 0.002 ⇒ ΛQCD ≈ 250 MeV (477)

- Compare with classical theory- Classical Yang Mills:

- scale-invariant- interaction strength is given by dimensionless parameter g

- Quantised Yang-Mills:- scale-dependent- interaction strength given by dimensionful scale ΛQCD

- If we assume the quarks are massless, ΛQCD is the only scale in the theory⇒ There is nothing we could compare it with

- This means that Yang-Mills or QCD with massless quarks is a unique theory:There is no free parameter that we could vary

80

Index1PI correlator, 32

anomalous dimension, 56asymptotic freedom, 79

beta function, 56BHPZ theorem, 50BRST symmetry, 67

Casimir invariant, 74coarse-graining, 58conformal field theory, 64connected correlator, 32continuum limit, 66counterterm, 48critical phenomena, 66

dimensionality, 44

effective action, 59effective field theory, 59

Faddeev-Popov ghost, 23Feynman diagram, 27Feynman gauge, 24Feynman parameter, 53Feynman propagator, 12Feynman rules, 30fixed point, 64

gauge condition, 20gauge fixing parameter, 23gauge invariance, 17Gaussian fixed point, 64Grassmann algebra, 14Grassmann numbers, 14

Haar measure, 21

irrelevant operator, 65

Landau gauge, 24Landau pole, 66Lorenz gauge, 20

Mandelstam variables, 41

marginal operator, 65modified minimal subtraction (MS), 55

non-renormalisable, 46

one-particle irreducible, 32

relevant operator, 65renormalisable, 46renormalisation constant, 68renormalisation group transformation, 62renormalisation scale, 55renormalisation scheme, 49renormalised coupling, 42renormalised mass, 43running coupling, 66

scale-free integral, 52scattering cross section, 41structure constants, 19superficial degree of divergence, 44superrenormalisable, 46symmetry factor, 29

trivial theory, 66

ultraviolet cutoff, 40

Wick rotation, 58

81

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