Quantum Field Theory and Particle Physics

Home , Effective action, Quartic interaction, Scalar field theory

Badis Ydri

Département de Physique, Faculté des Sciences Université d’Annaba Annaba, Algerie 2 YDRI QFT Contents

1 Introduction and References 1

I Path Integrals, Gauge Fields and Renormalization Group 3

2 Path Integral Quantization of Scalar Fields 5 2.1 FeynmanPathIntegral...... 5 2.2 ScalarFieldTheory...... 9 2.2.1 PathIntegral ...... 9 2.2.2 The Free 2 Point Function ...... 12 − 2.2.3 Lattice Regularization ...... 13 2.3 TheEffectiveAction ...... 16 2.3.1 Formalism...... 16 2.3.2 Perturbation Theory ...... 20 2.3.3 Analogy with Statistical Mechanics ...... 22 2.4 The O(N) Model...... 23 2.4.1 The 2 Point and 4 Point Proper Vertices ...... 24 − − 2.4.2 Momentum Space Feynman Graphs ...... 25 2.4.3 Cut-off Regularization ...... 27 2.4.4 Renormalization at 1 Loop...... 29 − 2.5 Two-Loop Calculations ...... 31 2.5.1 The Effective Action at 2 Loop ...... 31 − 2.5.2 The Linear Sigma Model at 2 Loop ...... 32 − 2.5.3 The 2 Loop Renormalization of the 2 Point Proper Vertex . . . . 35 − − 2.5.4 The 2 Loop Renormalization of the 4 Point Proper Vertex . . . . 40 − − 2.6 Renormalized Perturbation Theory ...... 42 2.7 Effective Potential and Dimensional Regularization ...... 45 2.8 Spontaneous Symmetry Breaking ...... 49 2.8.1 Example: The O(N) Model ...... 49 2.8.2 Glodstone’s Theorem ...... 55

3 Path Integral Quantization of Dirac and Vector Fields 57 3.1 FreeDiracField ...... 57 3.1.1 Canonical Quantization ...... 57 3.1.2 Fermionic Path Integral and Grassmann Numbers ...... 58 4 YDRI QFT

3.1.3 The Electron Propagator ...... 64 3.2 Free Abelian Vector Field ...... 65 3.2.1 Maxwell’s Action ...... 65 3.2.2 Gauge Invariance and Canonical Quantization ...... 67 3.2.3 Path Integral Quantization and the Faddeev-Popov Method . . . . 69 3.2.4 The Photon Propagator ...... 71 3.3 GaugeInteractions...... 72 3.3.1 Spinor and Scalar Electrodynamics: Minimal Coupling ...... 72 3.3.2 The Geometry of U(1) Gauge Invariance ...... 74 3.3.3 Generalization: SU(N) Yang-Mills Theory ...... 79 3.4 Quantization and Renormalization at 1 Loop...... 85 − 3.4.1 The Fadeev-Popov Gauge Fixing and Ghost Fields ...... 85 3.4.2 Perturbative Renormalization and Feynman Rules ...... 89 3.4.3 The Gluon Field Self-Energy at 1 Loop...... 92 − 3.4.4 The Quark Field Self-Energy at 1 Loop...... 103 − 3.4.5 The Vertex at 1 Loop...... 105 − 4 The Renormalization Group 113 4.1 Critical Phenomena and The φ4 Theory ...... 113 4.1.1 Critical Line and Continuum Limit ...... 113 4.1.2 Mean Field Theory ...... 118 4.1.3 Critical Exponents in Mean Field ...... 123 4.2 The Callan-Symanzik Renormalization Group Equation ...... 128 4.2.1 Power Counting Theorems ...... 128 4.2.2 Renormalization Constants and Renormalization Conditions . . . . 133 4.2.3 Renormalization Group Functions and Minimal Subtraction . . . . 136 4.2.4 CS Renormalization Group Equation in φ4 Theory ...... 140 4.2.5 Summary ...... 146 4.3 Renormalization Constants and Renormalization Functions at Two-Loop . 149 4.3.1 The Divergent Part of the Eﬀective Action ...... 149 4.3.2 Renormalization Constants ...... 154 4.3.3 Renormalization Functions ...... 157 4.4 CriticalExponents ...... 158 4.4.1 Critical Theory and Fixed Points ...... 158 4.4.2 Scaling Domain (T > Tc) ...... 164 4.4.3 Scaling Below Tc ...... 168 4.4.4 Critical Exponents from 2 Loop and Comparison with Experiment 171 − 4.5 The Wilson Approximate Recursion Formulas ...... 175 4.5.1 Kadanoﬀ-Wilson Phase Space Analysis ...... 175 4.5.2 Recursion Formulas ...... 177 4.5.3 The Wilson-Fisher Fixed Point ...... 187 4.5.4 The Critical Exponents ν ...... 192 4.5.5 The Critical Exponent η ...... 196 YDRI QFT 5

A Exercises 199

B References 205 6 YDRI QFT 1 Introduction and References

• M.E.Peskin, D.V.Schroeder, An Introduction to Quantum Field Theory.

• J.Strathdee,Course on Quantum Electrodynamics, ICTP lecture notes.

• C.Itzykson, J-B.Zuber, Quantum Field Theory.

• J.Zinn-Justin,Quantum Field Theory and Critical Phenomena.

• A.M.Polyakov, Gauge Fields and Strings.

• C.Itzykson, J-M.Drouﬀe, From Brownian Motion to Renormalization and Lattice Gauge Theory.

• D.Griﬃths, Introduction to Elementary Particles.

• W.Greiner, Relativistic Quantum Mechanics.Wave Equations.

• W.Greiner, J.Reinhardt, Field Quantization.

• W.Greiner, J.Reinhardt, Quantum Electrodynamics.

• S.Randjbar-Daemi, Course on Quantum Field Theory, ICTP lecture notes.

• V.Radovanovic, Problem Book QFT. 2 YDRI QFT Part I

Path Integrals, Gauge Fields and Renormalization Group

2 Path Integral Quantization of Scalar Fields

2.1 Feynman Path Integral

We consider a dynamical system consisting of a single free particle moving in one dimension. The coordinate is x and the canonical momentum is p = mx˙. The Hamiltonian is H = p2/(2m). Quantization means that we replace x and p with operators X and P satisfying the canonical commutation relation [X, P ] = i~. The Hamiltonian becomes H = P 2/(2m). These operators act in a Hilbert space . The quantum states which de- H scribe the dynamical system are vectors on this Hilbert space whereas observables which describe physical quantities are hermitian operators acting in this Hilbert space. This is the canonical or operator quantization. We recall that in the Schrödinger picture states depend on time while operators are independent of time. The states satisfy the Schrödinger equation, viz ∂ H ψ (t) >= i~ ψ (t) > . (2.1) | s ∂t| s Equivalently

i ψ (t) >= e− ~ H(t−t0) ψ (t ) > . (2.2) | s | s 0 Let x > be the eigenstates of X, i.e X x >= x x >. The completness relation is | | | dx x> in this basis are . Thus | | | s | s R ψ (t) > = dx x > . (2.3) | s | s | Z i = | s | | s 0 = dx G(x,t; x ,t ) . (2.4) 0 0 0 0| s 0 Z 6 YDRI QFT

In above we have used the completness relation in the form dx x >< x = 1. The 0| 0 0| Green function G(x,t; x ,t ) is deﬁned by 0 0 R

i G(x,t; x ,t ) = . (2.5) 0 0 | | 0 In the Heisenberg picture states are independent of time while operators are dependent of time. The Heisenberg states are related to the Schrödinger states by the relation

i i x,t >= e ~ Ht x > , x ,t >= e ~ Ht0 x > . (2.7) | | | 0 0 | 0 From the facts X(t) x,t >= x x,t > and X x>= x x> we conclude that the Heisenberg | | | | operators are related to the Schrödinger operators by the relation

i i X(t)= e ~ HtXe− ~ Ht. (2.8)

We immediately obtain the Heisenberg equation of motion

dX(t) i ∂X i i = e ~ Ht e− ~ Ht + [H, X(t)]. (2.9) dt ∂t ~

The Green function (2.5) can be put into the form

G(x,t; x ,t ) = < x,t x ,t > . (2.10) 0 0 | 0 0

This is the transition amplitude from the point x0 at time t0 to the point x at time t which is the most basic object in the quantum theory. We discretize the time interval [t ,t] such that t = t + jǫ, ǫ = (t t )/N, j = 0 j 0 − 0 0, 1, ..., N, tN = t0 + Nǫ = t. The corresponding coordinates are x0,x1, ..., xN with xN = x. The corresponding momenta are p0,p1, ..., pN−1. The momentum pj corresponds to the interval [xj,xj+1]. We can show

G(x,t; x ,t ) = < x,t x ,t > 0 0 | 0 0 = dx < x,t x ,t >< x ,t x ,t > 1 | 1 1 1 1| 0 0 Z N−1 = dx1dx2...dxN−1 . (2.11) j=0 | Z Y YDRI QFT 7

We compute (with

= exp( ipx/~)/√2π~) | − i = j+1 j+1| j j j+1| − ~ | j i = dp j j+1| j j| − ~ | j Z i = dp (1 H(p ,x )ǫ) j − ~ j j j+1| j j| j Z dpj i i i = (1 H(p ,x )ǫ) e ~ pj xj+1 e− ~ pj xj 2π~ − ~ j j Z dpj i = e ~ (pj x˙ j −H(xj ,pj))ǫ. (2.12) 2π~ Z In above x˙ = (x x )/ǫ. Therefore by taking the limit N , ǫ 0 keeping j j+1 − j −→ ∞ −→ t t = ﬁxed we obtain − 0 − dp0 dp1dx1 dpN−1dxN−1 i N 1(p x˙ −H(p ,x ))ǫ G(x,t; x ,t ) = ... e ~ Pj=0 j j j j 0 0 2π~ 2π~ 2π~ Z i t ~ R ds(px˙−H(p,x)) = p x e t0 . (2.13) D D Z Now x˙ = dx/ds. In our case the Hamiltonian is given by H = p2/(2m). Thus by performing the Gaussian integral over p we obtain 1

i t ~ R dsL(˙x,x) G(x,t; x ,t ) = x e t0 0 0 N D Z i = x e ~ S[x]. (2.14) N D Z In the above equation S[x] = dt L(x, x˙) = m dt x˙ 2/2 is the action of the particle. As it turns out this fundamental result holds for all Hamiltonians of the form H = R R 2 p2/(2m)+ V (x) in which case S[x]= dt L(x, x˙)= dt (mx˙ 2/2 V (x)) . − This result is essentially the principle of linear superposition of quantum theory. R R The total probability amplitude for traveling from the point x0 to the point x is equal

1Exercise: • Show that

2 2 −ap +bp π b dpe = e 4a . Z r a

• Use the above result to show that

2 2 i t ds(px˙ − p ) i t ds m x˙ (s) Dp e ~ Rt0 2m = N e ~ Rt0 2 . Z Determine the constant of normalization N .

2Exercise: Repeat the analysis for a non-zero potential. See for example Peskin and Schroeder 8 YDRI QFT

to the sum of probability amplitudes for traveling from x0 to x through all possible paths connecting these two points. Clearly a given path between x0 and x is deﬁned by a conﬁguration x(s) with x(t0) = x0 and x(t) = x. The corresponding probability i amplitude (wave function) is e ~ S[x(s)]. In the classical limit ~ 0 only one path (the −→ classical path) exists by the method of the stationary phase. The classical path is clearly the path of least action as it should be. We note also that the generalization of the result (2.14) to matrix elements of operators is given by 3

i < x,t T (X(t )...X(t )) x ,t > = x x(t )...x(t ) e ~ S[x]. (2.15) | 1 n | 0 0 N D 1 n Z The T is the time-ordering operator deﬁned by

T (X(t1)X(t2)) = X(t1)X(t2) if t1 >t2. (2.16)

T (X(t1)X(t2)) = X(t2)X(t1) if t1

i e−itEn+it0Em < n T (X(t )...X(t )) m> = x x(t )...x(t ) e ~ S[x]. | | 0 | 1 n | N D 1 n n,m X Z (2.18)

In the limit t and t we observe that only the ground state with energy 0 −→ −∞ −→ ∞ E0 contributes, i.e. the rapid oscillation of the ﬁrst exponential in this limit forces n = m = 0 4. Thus we obtain in this limit

i eiE0(t0−t) < 0 x >< 0 T (X(t )...X(t )) 0 > = x x(t )...x(t ) e ~ S[x]. | | 0 | 1 n | N D 1 n Z (2.20)

We write this as

′ i < 0 T (X(t )...X(t )) 0 > = x x(t )...x(t ) e ~ S[x]. (2.21) | 1 n | N D 1 n Z 3Exercise:Verify this explicitly. See for example Randjbar-Daemi lecture notes. 4We consider the integral ∞ I = dx F (x)eiφ(x). (2.19) Z−∞

The function φ(x) is a rapidly-varying function over the range of integration while F (x) is slowly-varying by comparison. Evaluate this integral using the method of stationary phase. YDRI QFT 9

In particular

′ i < 0 0 > = x e ~ S[x]. (2.22) | N D Z Hence

i S[x] x x(t1)...x(tn) e ~ < 0 T (X(t1)...X(tn)) 0 > = D i . (2.23) | | x e ~ S[x] R D We introduce the path integral Z[J] in the presence ofR a source J(t) by

i i Z[J] = x e ~ S[x]+ ~ R dtJ(t)x(t). (2.24) D Z This path integral is the generating functional of all the matrix elements < 0 T (X(t )...X(t )) 0 >. | 1 n | Indeed 1 ~ n δnZ[J] < 0 T (X(t )...X(t )) 0 > = . (2.25) | 1 n | Z[0] i δJ(t )...δJ(t )|J=0 1 n From the above discussion Z[0] is the vacuum-to-vacuum amplitude. Therefore Z[J] is the vacuum-to-vacuum amplitude in the presence of the source J(t).

2.2 Scalar Field Theory

2.2.1 Path Integral A field theory is a dynamical system with N degrees of freedom where N . The −→ ∞ classical description is given in terms of a Lagrangian and an action principle while the quantum description is given in terms of a path integral and correlation functions. In a scalar field theory the basic field has spin j = 0 with respect to Lorentz transformations. It is well established that scalar field theories are relevant to critical phenomena and to the Higgs sector in the standard model of particle physics. µ 2 2 µ We start with the relativistic energy-momentum relation p pµ = M c where p = (p0, ~p) = (E/c, ~p). We adopt the metric (1, 1, 1, 1), i.e. p = (p , ~p) = (E/c, ~p). − − − µ 0 − − Next we employ the correspondence principle p i~∂ where ∂ = (∂ , ∂ ) and apply µ −→ µ µ 0 i the resulting operator on a function φ. We obtain the Klein-Gordon equation M 2c2 ∂ ∂µφ = m2φ, m2 = . (2.26) µ − ~2 As a wave equation the Klein-Gordon equation is incompatible with the statistical interpretation of quantum mechanics. However the Klein-Gordon equation makes sense as an equation of motion of a classical scalar field theory with action and Lagrangian S = dtL, L = d3x where the lagrangian density is given by L L R R 1 1 = ∂ φ∂µφ m2φ2. (2.27) L 2 µ − 2 10 YDRI QFT

So in summary φ is not really a wave function but it is a dynamical variable which plays the same role as the coordinate x of the free particle discussed in the previous section. The principle of least action applied to an action S = dtL yields (with the assump- tion δφ = 0) the result 5 |xµ=±∞ R δS δ δ = L ∂µ L = 0. (2.28) δφ δφ − δ(∂µφ)

It is not difficult to verify that this is the same equation as (2.26) if L = d3x and 6 L L is given by (2.27) . R The free scalar field theory is a collection of infinite number of decoupled harmonic oscillators. To see this fact we introduce the fourier transform φ˜ = φ˜(t,~k) of φ = φ(t, ~x) as follows

3 d ~k ~ ~ φ = φ(t, ~x)= φ˜(t,~k) eik~x , φ˜ = φ˜(t,~k)= d3~xφ(t, ~x) e−ik~x. (2.29) (2π)3 Z Z Then the Lagrangian and the equation of motion can be rewritten as

d3~k 1 1 L = ∂ φ∂˜ φ˜∗ ω2φ˜φ˜∗ . (2.30) (2π)3 2 0 0 − 2 k Z

2 ˜ 2 ˜ 2 ~ 2 2 ∂0 φ + ωkφ = 0 , ωk = k + m . (2.31)

This is the equation of motion of a harmonic oscillator with frequency ωk. Using box normalization the momenta become discrete and the measure d3~k/(2π)3 becomes /V . Reality of the scalar ﬁeld φ implies that φ˜(t,~k) = φ˜∗(t, ~k) and by writing ~k R− φ˜ = √V (X + iY ) we end up with the Lagrangian P k k 1 L = ∂ φ∂˜ φ˜∗ ω2φ˜φ˜∗ V 0 0 − k kX1>0 kX2>0 kX3>0 = (∂ X )2 ω2X2 + (∂ Y )2 ω2Y 2 . (2.32) 0 k − k k 0 k − k k kX1>0 kX2>0 kX3>0 The path integral of the two harmonic oscillators Xk and Yk is immediately given by

i i Z[J , K ] = X Y e ~ S[Xk,Yk]+ ~ R dt(Jk(t)Xk (t)+Kk(t)Yk(t)). (2.33) k k D kD k Z The action S[Xk,Yk] is obviously given by

t−→+∞ S[X ,Y ]= ds (∂ X )2 ω2X2 + (∂ Y )2 ω2Y 2 (2.34) k k 0 k − k k 0 k − k k Zt0−→−∞ 5Exercise: Verify this statement. 6Exercise: Verify this statement. YDRI QFT 11

The definition of the measures X and Y must now be clear from our previous D k D k considerations. We introduce the notation X (t ) = x(k), Y (t ) = y(k), i = 0, 1, ..., N k i i k i i − 1,N with the time step ǫ = t t = (t t )/N. Then as before we have (with i − i−1 − 0 N , ǫ 0 keeping t t fixed) the measures −→ ∞ −→ − 0 N−1 N−1 X = dx(k) , Y = dy(k). (2.35) D k i D k i Yi=1 Yi=1 The path integral of the scalar field φ is the product of the path integrals of the harmonic oscillators Xk and Yk with different k = (k1, k2, k3), viz

Z[J, K] = Z[Jk, Kk] kY1>0 kY2>0 kY3>0 i = X Y exp S[X ,Y ] D kD k ~ k k Z kY1>0 kY2>0 kY3>0 kX1>0 kX2>0 kX3>0 i + ~ dt (Jk(t)Xk(t)+ Kk(t)Yk(t)) . (2.36) Z kX1>0 kX2>0 kX3>0 The action of the scalar ﬁeld is precisely the ﬁrst term in the exponential, namely

S[φ] = S[Xk,Yk] kX1>0 kX2>0 kX3>0 t−→+∞ 2 2 2 2 2 2 = ds (∂0Xk) ωkXk + (∂0Yk) ωkYk t0−→−∞ − − Z kX1>0 kX2>0 kX3>0 1 1 = d4x ∂ φ∂µφ m2φ2 . (2.37) 2 µ − 2 Z 3 3 i~k~x We remark also that (with J(t, ~x) = d ~k/(2π) J˜(t,~k) e , J˜ = √V (Jk + iKk)) we have R 3 (Jk(t)Xk(t)+ Kk(t)Yk(t)) = d xJ(t, ~x)φ(t, ~x). (2.38) kX1>0 kX2>0 kX3>0 Z We write therefore the above path integral formally as

i i 4 Z[J] = φ e ~ S[φ]+ ~ R d xJ(x)φ(x). (2.39) D Z This path integral is the generating functional of all the matrix elements < 0 T (Φ(x )...Φ(x )) 0 > | 1 n | (also called n point functions). Indeed − 1 ~ n δnZ[J] < 0 T (Φ(x )...Φ(x )) 0 > = | 1 n | Z[0] i δJ(x )...δJ(x )|J=0 1 n i S[φ] φ φ(x1)...φ(xn) e ~ = D i . (2.40) φ e ~ S[φ] R D R 12 YDRI QFT

The intercations are added by modifying the action appropriately. The only renormalizable interacting scalar ﬁeld theory in d = 4 dimensions is the quartic φ4 theory. Thus we will only consider this model given by the action

1 1 λ S[φ]= d4x ∂ φ∂µφ m2φ2 φ4 . (2.41) 2 µ − 2 − 4! Z 2.2.2 The Free 2 Point Function − It is more rigorous to perform the diﬀerent computations of interest on an Euclidean spacetime. Euclidean spacetime is obtained from Minkowski spacetime via the so-called Wick rotation. This is also called the imaginary time formulation which is obtained by the substitutions t iτ, x0 = ct ix4 = icτ, ∂ i∂ . Hence ∂ φ∂µφ −→ − −→ − − 0 −→ 4 µ −→ (∂ φ)2 and iS S where − µ −→ − E 1 1 λ S [φ]= d4x (∂ φ)2 + m2φ2 + φ4 . (2.42) E 2 µ 2 4! Z The path integral becomes

1 1 4 Z [J]= φ e− ~ SE [φ]+ ~ R d xJ(x)φ(x). (2.43) E D Z The Euclidean n point functions are given by − n n 1 δ ZE[J] < 0 T (Φ(x1)...Φ(xn)) 0 >E = ~ J=0 | | Z[0] δJ(x1)...δJ(xn)| 1 − SE [φ] φ φ(x1)...φ(xn) e ~ = D 1 . (2.44) φ e− ~ SE [φ] R D The action of a free scalar ﬁeld is given by R

1 1 1 S [φ]= d4x (∂ φ)2 + m2φ2 = d4x φ ∂2 + m2 φ. (2.45) E 2 µ 2 2 − Z Z The corresponding path integral is (after completing the square)

1 1 4 Z [J] = φ e− ~ SE[φ]+ ~ R d xJ(x)φ(x) E D Z 1 R d4x(JKJ)(x) − 1 R d4x φ−JK −∂2+m2 φ−KJ = e 2~ φ e 2~ . (2.46) D Z In above K is the operator deﬁned by

K ∂2 + m2 = ∂2 + m2 K = 1. (2.47) − − YDRI QFT 13

After a formal change of variable given by φ φ KJ the path integral Z[J] is reduced −→ − to (see next section for a rigorous treatment)

1 R d4x(JT KJ)(x) 1 R d4xd4yJ(x)K(x,y)J(y) Z [J] = e 2~ = e 2~ . (2.48) E N N The is an unimportant normalization factor. The free 2 point function (the free N − propagator) is deﬁned by

1 − SE [φ] φ φ(x1)φ(x2) e ~ < 0 T (Φ(x1)Φ(x2)) 0 >E = D 1 | | φ e− ~ SE [φ] R D 2 1 2 δ ZE[J] = ~R J=0. (2.49) Z[0] δJ(x1)δJ(x2)| A direct calculation leads to

< 0 T (Φ(x )Φ(x )) 0 > = ~K(x ,x ) (2.50) | 1 2 | E 1 2 Clearly

∂2 + m2 K(x,y)= δ4(x y). (2.51) − − Using translational invariance we can write d4k K(x,y)= K(x y)= K˜ (k) eik(x−y). (2.52) − (2π)4 Z By construction K˜ (k) is the fourier transform of K(x,y). It is trivial to compute that 1 K˜ (k)= . (2.53) k2 + m2 The free euclidean 2 point function is therefore given by − d4k ~ < 0 T (Φ(x )Φ(x )) 0 > = eik(x−y) (2.54) | 1 2 | E (2π)4 k2 + m2 Z 2.2.3 Lattice Regularization The above calculation of the 2 point function of a scalar field can be made more explicit − and in fact more rigorous by working on an Euclidean lattice spacetime. The lattice provides a concrete non-perturbative definition of the theory. We replace the Euclidean spacetime with a lattice of points xµ = anµ where a is the lattice spacing. In the natural units ~ = c = 1 the action is dimensionless and hence the field is of dimension mass. We define a dimensionless field φˆn by the relation φˆn = aφn 2 2 2 where φn = φ(x). The dimensionless mass parameter is mˆ = m a . The integral over spacetime will be replaced with a sum over the points of the lattice, i.e

d4x = a4 , = ... . (2.55) n n n n Z X X X1 X4 14 YDRI QFT

The measure is therefore given by

φ = dφn , = ... . (2.56) D n n n1 n4 Z Y Y Y Y The derivative can be replaced either with the forward difference or with the backward difference defined respectively by the equations

φ φ ∂ φ = n+ˆµ − n . (2.57) µ a

φ φ ∂ φ = n − n−µˆ . (2.58) µ a

The µˆ is the unit vector in the direction xµ. The Laplacian on the lattice is deﬁned such that 1 ∂2φ = φ + φ 2φ . (2.59) a2 n+ˆµ n−µˆ − n µ X The free Euclidean action on the lattice is therefore 1 S [φˆ] = d4x φ ∂2 + m2 φ E 2 − Z 1 = φˆ K φˆ . (2.60) 2 n nm m n,m X

K = δ + δ 2δ +m ˆ 2δ . (2.61) nm − n+ˆµ,m n−µ,mˆ − n,m n,m µ X The path integral on the lattice is

−SE [φˆ]+P Jˆnφˆn ZE[J]= dφˆn e n . (2.62) n Z Y The n point functions on the lattice are given by − n 1 δ ZE[J] < 0 T (Φˆ s...Φˆ t) 0 >E = J=0 | | Z[0] δJˆs...δJˆt | ˆ dφˆ φˆ ...φˆ e−SE [φ] = n n s t . (2.63) ˆ −SE [φˆ] R Q ndφn e The path integral of the free scalar ﬁeld on the latticeR Q can be computed in a closed form. We ﬁnd 7 7Exercise: YDRI QFT 15

1 −1 1 −1 −1 P JnKnmJm − P (φˆ−JK )nKnm(φˆ−K J)m ZE[J] = e 2 n,m dφˆn e 2 n,m n − Z 1 J K 1 J Y = e 2 Pn,m n nm m . (2.65) N The 2 point function is therefore given by − 2 1 δ ZE[J] < 0 T (Φˆ sΦˆ t) 0 >E = J=0 | | Z[0] δJˆsδJˆt | −1 = Kst . (2.66) We fourier transform on the lattice as follows π 4 d kˆ ˆ K = Kˆ (k) eik(s−t). (2.67) st (2π)4 Z−π

π 4 d kˆ ˆ K−1 = G(k) eik(s−t). (2.68) st (2π)4 Z−π For Kˆ (k)= G(k) = 1 we obtain the identity, viz

π 4 d kˆ ˆ δ = eik(s−t). (2.69) st (2π)4 Z−π −1 Furthermore we can show that KstKtr = δsr using the equations ˆ (2π)4δ4(kˆ pˆ)= ei(k−pˆ)n. (2.70) − n X

G(k)= Kˆ −1(k) (2.71) Next we compute

K = δ + δ 2δ +m ˆ 2δ nm − n+ˆµ,m n−µ,mˆ − n,m n,m µ X π 4 4 d kˆ ˆ ˆ ˆ d kˆ ˆ = eik(n−m) eikµˆ + e−ikµˆ 2 +m ˆ 2 eik(n−m)(2.72). − (2π)4 − (2π)4 −π µ Z X Z • Perform explicitly the Gaussian integral

N −xiDij xj I = dxi e . (2.64) Z Yi=1 Try to diagonalize the symmetric and invertible matrix D. It is also well advised to adopt an iǫ prescription (i.e. make the replacement D −→ D + iǫ) in order to regularize the integral. • Use the above result to determine the constant of normalization N in equation (2.63). 16 YDRI QFT

Thus kˆ Kˆ (k) = 4 sin2( µ ) +m ˆ 2. (2.73) 2 µ X Hence 1 G(k)= ˆ . (2.74) 2 kµ 2 4 µ sin ( 2 ) +m ˆ The 2 point function is then given byP − π 4ˆ ˆ ˆ d k 1 ikˆ(s−t) < 0 T (ΦsΦt) 0 >E= 4 ˆ e (2.75) | | −π (2π) 2 kµ 2 Z 4 µ sin ( 2 ) +m ˆ In the continuum limit a 0 we scale theP ﬁelds as follows Φˆ = aφ(x), Φˆ = aφ(y) −→ s t where x = as and y = at. The momentum is scaled as kˆ = ak and the mass is scaled as mˆ 2 = a2m2. In this limit the lattice mass mˆ 2 goes to zero and hence the correlation lenght ξˆ = 1/mˆ diverges. In other words the continuum limit is realized at a critical point of a second order phase transition. The physical 2 point function is given by − < 0 T (Φˆ Φˆ ) 0 > < 0 T (Φ(ˆ x)Φ(ˆ y)) 0 > = lim | s t | E | | E a−→0 a2 ∞ d4k 1 = eik(x−y). (2.76) (2π)4 k2 + m2 Z−∞ This is the same result obtain from continuum considerations in the previous section.

2.3 The Eﬀective Action

2.3.1 Formalism We are interested in the φ4 theory on a Minkowski spacetime given by the classical action 1 1 λ S[φ]= d4x ∂ φ∂µφ m2φ2 φ4 . (2.77) 2 µ − 2 − 4! Z The quantum theory is given by the path integral

i i 4 Z[J]= φ e ~ S[φ]+ ~ R d xJ(x)φ(x). (2.78) D Z The functional Z[J] generates all Green functions, viz 1 ~ n δnZ[J] < 0 T (Φ(x )...Φ(x )) 0 > = | 1 n | Z[0] i δJ(x )...δJ(x )|J=0 1 n i S[φ] φ φ(x1)...φ(xn) e ~ = D i . (2.79) φ e ~ S[φ] R D R YDRI QFT 17

The path integral Z[J] generates disconnected as well as connected graphs and it generates reducible as well as irreducible graphs. Clearly the disconnected graphs can be obtained by putting togther connected graphs whereas reducible graphs can be decom- posed into irreducible components. All connected Green functions can be generated from the functional W [J] (vacuum energy) whereas all connected and irreducible Green functions (known also as the 1 particle irreducible) can be generated from the functional − Γ[φc] (eﬀective action). The vacuum energy W [J] is deﬁned through the equation

i Z[J]= e ~ W [J]. (2.80)

In order to define the effective action we introduce the notion of the classical field. This is defined by the equation

δW [J] φ (x)= . (2.81) c δJ(x)

This is a functional of J. It becomes the vacuum expectation value of the ﬁeld operator Φ at J = 0. Indeed we compute

~ 1 δZ[J] φ (x) = =< 0 Φ(x) 0 > . (2.82) c |J=0 i Z[0] δJ(x)|J=0 | |

The eﬀective action Γ[φc] is the Legendre transform of W [J] deﬁned by

Γ[φ ]= W [J] d4xJ(x)φ (x). (2.83) c − c Z This is the quantum analogue of the classical action S[φ]. The eﬀective action generates all the 1 particle irreducible graphs from which the external legs have been removed. − These are the connected, irreducible and amputated graphs. The classical equations of motion are obtained from the principal of least action applied to the classical action S[φ]+ d4xJ(x)φ(x). We obtain

δS[Rφ] = J(x). (2.84) δφ(x) −

Similarly the quantum equations of motion are obtained from the principal of least action applied to the quantum action Γ[φc]. We obtain

δΓ[φ ] c = 0. (2.85) δφc(x)

In the presence of source this generalizes to

δΓ[φ ] c = J(x). (2.86) δφc(x) − 18 YDRI QFT

The proof goes as follows: δΓ δW δJ(y) = d4y φ (y) J δφ (x) δφ (x) − δφ (x) c − c c Z c δW δJ(y) δW = d4y J δφ (x) − δφ (x) δJ(y) − c Z c = J(x) (2.87) − A more explicit form of the quantum equation of motion can be obtained as follows. We start from the identity

~ δ i i 4 0 = φ e ~ S[φ]+ ~ R d xJ(x)φ(x) D i δφ(x) Z δS i i 4 = φ + J e ~ S[φ]+ ~ R d xJ(x)φ(x) D δφ(x) Z δS i = + J e ~ W [J] ~ δφ(x) φ= δ i δJ i δS = e ~ W [J] + J . (2.88) ~ δφ(x) φ= δ + δW i δJ δJ In the last line above we have used the identity

g(x) g(x) F (∂x) e = e F (∂xg + ∂x). (2.89)

We obtain the equation of motion δS δΓ[φ ] = J = c . (2.90) δφ(x) ~ δ δW − δφc φ= i δJ + δJ By the chaine rule we have δ δφ (y) δ = d4y c δJ(x) δJ(x) δφ (y) Z c δ = d4y G(2)(x,y) . (2.91) δφ (y) Z c The G(2)(x,y) is the connected 2 point function in the presence of the source J(x), viz − δφ (y) δ2W [J] G(2)(x,y)= c = . (2.92) δJ(x) δJ(x)δJ(y) The quantum equation of motion becomes δS δΓ[φ ] = J = c . (2.93) ~ 4 (2) δ δφ(x) φ= R d y G (x,y) +φc(x) − δφc i δφc(y) YDRI QFT 19

The connected n point functions and the proper n point vertices are defined as follows. − − The connected n point functions are defined by − n (n) i1...in δ W [J] G (x1, ..., xn)= G = . (2.94) δJ(x1)...δJ(xn) The proper n point vertices are defined by − n (n) δ Γ[φc] Γ (x1, ..., xn)=Γ,i1...in = . (2.95) δφc(x1)...δφc(xn) These are connected 1 particle irreducible n point functions from which the external − − legs are removed (amputated). The proper 2 point vertex Γ(2)(x,y) is the inverse of the connected 2 point function − − G(2)(x,y). Indeed we compute

δφ (z) δ2Γ[φ ] d4z G(2)(x, z)Γ(2)(z,y) = d4z c c δJ(x) δφ (z)δφ (y) Z Z c c δφ (z) δJ(y) = d4z c − δJ(x) δφ (z) Z c = δ4(x y). (2.96) − − We write this as

GikΓ = δi . (2.97) ,kj − j We remark the identities δGi1...in = Gi1...inin+1 . (2.98) δJin+1

k δΓ,i1...in δΓ,i1...in δφc kin+1 = k =Γ,i1...inkG . (2.99) δJin+1 δφc δJin+1

By diﬀerentiating (2.97) with respect to Jl we obatin

ikl ik rl G Γ,kj + G Γ,kjrG = 0. (2.100)

Next by multiplying with Gjs we get the 3 point connected function as − isl ik rl js G = G G G Γ,kjr. (2.101)

Now by diﬀerentiating (2.101) with respect to J we obatin the 4 point connected m − function as

islm ikm rl js ik rlm js ik rl jsm ik rl js nm G = G G G + G G G + G G G Γ,kjr + G G G Γ,kjrnG . (2.102) 20 YDRI QFT

By using again (2.101) we get

′ ′ ′ islm ′ ′ ′ ik kj mr rl js G = Γ,k j r G G G G G Γ,kjr + two permutations ik rl js nm + G G G Γ,kjrnG . (2.103) The diagrammatic representation of (2.94),(2.95),(2.97),(2.101) and (2.103) is shown on ﬁgure 1.

2.3.2 Perturbation Theory In this section we will consider a general scalar ﬁeld theory given by the action 1 1 1 S[φ]= S φi + S φiφj + S φiφjφk + S φiφjφkφl + ... (2.104) i 2! ij 3! ijk 4! ijkl We need the ﬁrst derivative of S[φ] with respect to φi, viz 1 1 1 1 S[φ] = S + S φj + S φjφk + S φjφkφl + S φjφkφlφm + S φjφkφlφmφn + ... ,i i ij 2! ijk 3! ijkl 4! ijklm 5! ijklmn (2.105)

Thus ~ j 1 j jj0 δ k Γ[φ ] = S[φ] ~ ii δ = S + S φ + S φ + G φ c ,i ,i φi=φci+ G 0 i ij c ijk c c | i δφci 2! i δφ 0 cj0 1 ~ δ ~ δ + S φj + Gjj0 φk + Gkk0 φl 3! ijkl c i δφ c i δφ c cj0 ck0 1 ~ δ ~ δ ~ δ + S φj + Gjj0 φk + Gkk0 φl + Gll0 φm 4! ijklm c i δφ c i δφ c i δφ c cj0 ck0 cl0 + ... (2.106)

We ﬁnd upto the ﬁrst order in ~ the result 8

~ ~ 2 1 jk l 1 j m Γ[φ ] = S[φ] ~ ii δ = S[φ ] + G S + S φ + S φ φ + ... + O . c ,i ,i φi=φci+ G 0 c ,i ijk ijkl c ijklm c c | i δφci 2 i 2 i 0 (2.107)

In other words ~ ~ 2 1 jk Γ[φ ] = S[φ] ~ ii δ = S[φ ] + G S[φ ] + O (2.108). c ,i ,i φi=φci+ G 0 c ,i c ,ijk | i δφci 2 i i 0 We expand ~ ~ 2 Γ=Γ + Γ + Γ + ... (2.109) 0 i 1 i 2 8Exercise: Verify this equation. YDRI QFT 21

~ ~ 2 Gij = Gij + Gij + Gij + ... (2.110) 0 i 1 i 2 Immediately we ﬁnd

Γ0[φc],i = S[φc],i. (2.111)

1 Γ [φ ] = GjkS[φ ] . (2.112) 1 c ,i 2 0 c ,ijk Equation (2.111) can be trivially integrated. We obatin

Γ0[φc] = S[φc]. (2.113)

Let us recall the constraint GikΓ = δi . This is equivalent to the constraints ,kj − j GikΓ = δi 0 0,kj − j ik ik G0 Γ1,kj + G1 Γ0,kj = 0 ik ik ik G0 Γ2,kj + G1 Γ1,kj + G2 Γ0,kj = 0 . . (2.114)

ik The ﬁrst constarint gives G0 in terms of Γ0 = S as

Gik = S−1. (2.115) 0 − ,ik ik The second constarint gives G1 in terms of Γ0 and Γ1 as

ij ik jl G1 = G0 G0 Γ1kl. (2.116)

ik The third constraint gives G2 in terms of Γ0, Γ1 and Γ2. Hence the calculation of the 2 point function Gik to all orders in perturbation theory requires the calculation the − eﬀective action to all orders in perturbation theory, viz the calculation of the Γn. In fact the knowledge of the eﬀective action will allow us to calculate all proper n point vertices − to any order in perturbation theory. We are now in a position to integrate equation (2.112). We have

1 jk δS[φc],jk Γ1[φc],i = G0 2 δφci −1 1 jk δ(G0 )jk = G0 −2 δφci 1 δ −1 = ln det G0 . (2.117) −2 δφci 22 YDRI QFT

Thus 1 Γ [φ ] = ln det G−1. (2.118) 1 c −2 0 The effective action upto the 1 loop order is − 1 ~ Γ=Γ + ln det G + ... (2.119) 0 2 i 0 This is represented graphically by the first two diagrams on figure 2.

2.3.3 Analogy with Statistical Mechanics We start by making a Wick rotation. The Euclidean vacuum energy, classical field, classical equation of motion, effective action and quantum equation of motion are defined by

1 − WE[J] ZE[J]= e ~ . (2.120)

δW [J] φ (x) = E =< 0 Φ(x) 0 > . (2.121) c |J=0 − δJ(x) |J=0 | | E

δS [φ] E = J(x). (2.122) δφ(x)

4 ΓE[φc]= WE[J]+ d xJ(x)φc(x). (2.123) Z

δΓ [φ ] E c = J(x). (2.124) δφc(x) Let us now consider the following statistical mechanics problem. We consider a magnetic system consisting of spins s(x). The spin energy density is (s). The system is placed H in a magnetic ﬁeld H. The partition function of the system is deﬁned by

Z[H]= s e−β R dxH(s)+β R dxH(x)s(x). (2.125) D Z The spin s(x), the spin energy density (s) and the magnetic field H(x) play in statistical H mechanics the role played by the scalar field φ(x), the Lagrangian density (φ) and the L source J(x) respectively in field theory. The free energy of the magnetic system is defined through the equation

Z[H]= e−βF [H]. (2.126) YDRI QFT 23

This means that F in statistical mechanics is the analogue of W in ﬁeld theory. The magnetization of the system is deﬁned by

δF 1 = dx s s(x) e−β R dx(H(s)−Hs(x)) −δH |β=ﬁxed Z D Z Z = dx Z = M. (2.127)

Thus the magnetization M in statistical mechanics plays the role of the effective field φ in field theory. In other words φ is the order parameter in the field theory. Finally − c c the Gibbs free energy in statistical mechanics plays the role of the effective action Γ[φc] in field theory. Indeed G is the Legendre transform of F given by

G = F + MH. (2.128)

Furthermore we compute

δG = H. (2.129) δM The thermodynamically most stable state (the ground state) is the minimum of G. Sim- ilarly the quantum mechanically most stable state (the vacuum) is the minimum of Γ. The thermal ﬂuctuations from one side correspond to quantum ﬂuctuations on the other side.

2.4 The O(N) Model

In this section we will consider a generalization of the φ4 model known as the linear sigma model. We are interested in the (φ2)2 theory with O(N) symmetry given by the classical action 1 1 λ S[φ]= d4x ∂ φ ∂µφ m2φ2 (φ2)2 . (2.130) 2 µ i i − 2 i − 4! i Z This classical action is of the general form studied in the previous section, viz

1 1 S[φ]= S φI φJ + S φI φJ φKφL. (2.131) 2! IJ 4! IJKL The index I stands for i and the spacetime index x, i.e I = (i, x), J = (j, y), K = (k, z) and L = (l, w). We have

S = δ (∆ + m2)δ4(x y) IJ − ij − λ S = δ δ4(y x)δ4(z x)δ4(w x) , δ = δ δ + δ δ + δ (2.132)δ . IJKL − 3 ijkl − − − ijkl ij kl ik jl il jk 24 YDRI QFT

The eﬀective action upto the 1 loop order is − 1 ~ Γ[φ]= S[φ]+ ln det G . (2.133) 2 i 0 The proper n point vertex is deﬁned now by setting φ = 0 after taking the n derivatives, − viz δnΓ[φ] Γ(n) (x , ..., x )=Γ = . (2.134) i1...in 1 n ,I1...In φ=0 δφi1 (x1)...δφin (xn)|

2.4.1 The 2 Point and 4 Point Proper Vertices − − The proper 2 point vertex is deﬁned by − 2 (2) δ Γ[φ] Γij (x,y) = φ=0 δφi(x)δφj (y)| 2 2 δ S[φ] ~ δ Γ1[φ] = φ=0 + φ=0 δφi(x)δφj (y)| i δφi(x)δφj (y)| ~ 2 2 4 δ Γ1[φ] = δij(∆ + m )δ (x y)+ φ=0. (2.135) − − i δφi(x)δφj (y)| The one-loop correction can be computed using the result

1 1 Γ [φ] = GmnS[φ] + Gmm0 Gnn0 S[φ] S[φ] . (2.136) 1 ,j0k0 2 0 ,j0k0mn 2 0 0 ,j0mn ,k0m0n0 We get by setting φ = 0 the result 1 Γ [φ] = GmnS[φ] 1 ,IJ 2 0 ,ijmn λ = d4zd4wGmn(z, w) δ δ + δ δ + δ δ δ4(y x)δ4(z x)δ4(w x) − 6 0 ij mn im jn in jm − − − Z λ = δ Gmm(x,y) + 2Gij (x,y) δ4(x y). (2.137) − 6 ij 0 0 − We have

GIJ = S−1 . (2.138) 0 − ,IJ Since S = S and S = δ S(x,y) where S(x,y)=(∆+ m2)δ4(x y) we can write ,IJ IJ IJ − ij − IJ G0 = δijG0(x,y). (2.139) Clearly d4yG (x,y)S(y, z)= δ4(x y). We obtain 0 − R λ Γ [φ] = (N + 2)δ G (x,y)δ4(x y). (2.140) 1 ,IJ − 6 ij 0 − YDRI QFT 25

Now we compute the 4 point proper vertex. Clearly the ﬁrst contribution will be − given precisely by the second equation of (2.132). Indeed we have δ4Γ[φ] Γ(4) (x , ..., x ) = i1...i4 1 4 φ=0 δφi1 (x1)...δφi4 (x4)| 4 4 δ S[φ] ~ δ Γ1[φ] = φ=0 + φ=0 δφi1 (x1)...δφi4 (x4)| i δφi1 (x1)...δφi4 (x4)| 2 ~ δ Γ1[φ] = SI1...I4 + φ=0. (2.141) i δφi1 (x1)...δφi4 (x4)| In order to compute the ﬁrst correction we use the identity mn δG0 mm0 nn0 = G0 G0 S[φc],lm0n0 . (2.142) δφcl We compute 1 1 Γ [φ] = Gmm0 Gnn0 S[φ] S[φ] + Gmm0 Gnn0 S[φ] S[φ] 1 ,j0k0ll0 2 0 0 ,j0k0mn ,ll0m0n0 2 0 0 ,j0lmn ,k0l0m0n0 1 + Gmm0 Gnn0 S[φ] S[φ] . 2 0 0 ,j0l0mn ,k0lm0n0 |φ=0 (2.143)

Thus δ4Γ [φ] 1 λ 2 1 = δ4(x x )δ4(x x ) (N + 2)δ δ + 2δ G (x ,x )2 δφ (x )...δφ (x )|φ=0 2 3 1 − 2 3 − 4 i1i2 i3i4 i1i2i3i4 0 1 3 i1 1 i4 4 1 λ 2 + δ4(x x )δ4(x x ) (N + 2)δ δ + 2δ G (x ,x )2 2 3 1 − 3 2 − 4 i1i3 i2i4 i1i2i3i4 0 1 2 1 λ 2 + δ4(x x )δ4(x x ) (N + 2)δ δ + 2δ G (x ,x )2. 2 3 1 − 4 2 − 3 i1i4 i2i3 i1i2i3i4 0 1 2 (2.144)

2.4.2 Momentum Space Feynman Graphs The proper 2 point vertex upto the 1 loop order is − − ~ λ Γ(2)(x,y) = δ (∆ + m2)δ4(x y) (N + 2)δ G (x,y)δ4(x y). (2.145) ij − ij − − i 6 ij 0 − The proper 2 point vertex in momentum space Γ(2)(p) is deﬁned through the equations − ij

4 4 (2) ipx+iky 4 4 (2) d xd y Γij (x,y) e = (2π) δ (p + k)Γij (p, k) Z = (2π)4δ4(p + k)Γ(2)(p, p) ij − 4 4 (2) = (2π) δ (p + k)Γij (p). (2.146) 26 YDRI QFT

The delta function is due to translational invariance. From the deﬁnition S(x,y)=(∆+ m2)δ4(x y) we have − d4p S(x,y)= ( p2 + m2) eip(x−y). (2.147) (2π)4 − Z Then by using the equation d4y G (x,y)S(y, z)= δ4(x y) we obtain 0 − R d4p 1 G (x,y)= eip(x−y). (2.148) 0 (2π)4 p2 + m2 Z − We get

~ λ d4p 1 Γ(2)(p) = δ ( p2 + m2) (N + 2)δ 1 . (2.149) ij − ij − − i 6 ij (2π)4 p2 + m2 Z − 1 The corresponding Feynman diagrams are shown on ﬁgure 4. The proper 4 point vertex upto the 1 loop order is − − ~ 2 (4) λ 4 4 4 1 λ Γ (x1, ..., x4) = δijδkl + δikδjl + δilδjk δ (y x)δ (z x)δ (w x)+ i1...i4 − 3 − − − 2 i 3 δ4(x x )δ4(x x ) (N + 2)δ δ + 2δ G (x ,x )2 + × 1 − 2 3 − 4 i1i2 i3i4 i1i2i3i4 0 1 3 δ4(x x )δ4(x x ) (N + 2)δ δ + 2δ G (x ,x )2 + 1 − 3 2 − 4 i1i3 i2i4 i1i2i3i4 0 1 2 δ4(x x )δ4(x x ) (N + 2)δ δ + 2δ G (x ,x )2 . 1 − 4 2 − 3 i1i4 i2i3 i1i2i3i4 0 1 2 (2.150)

(4) The proper 4 point vertex in momentum space Γ (p1...p4) is deﬁned through the − i1...i4 equation

d4x ...d4x Γ(4) (x , ..., x ) eip1x1+...+ip4x4 = (2π)4δ4(p + ... + p )Γ(2) (p , ..., p ). 1 4 i1...i4 1 4 1 4 i1...i4 1 4 Z (2.151)

We ﬁnd (with p12 = p1 + p2 and p14 = p1 + p4, etc)

(4) λ Γ (p1, ..., p4) = δi i i i i1...i4 − 3 1 2 3 4 ~ λ 2 1 1 + (N + 2)δ δ + 2δ i 3 2 i1i2 i3i4 i1i2i3i4 ( k2 + m2)( (p k)2 + m2) Zk − − 12 − + 2 permutations . (2.152) The corresponding Feynman diagrams are shown on ﬁgure 5. YDRI QFT 27

2.4.3 Cut-oﬀ Regularization At the one-loop order we have then ~ λ Γ(2)(p) = δ ( p2 + m2) (N + 2)δ I(m2). (2.153) ij − ij − − i 6 ij

~ 2 (4) λ λ 1 2 2 Γ (p1, ..., p4) = δi i i i + (N + 2)δi i δi i + 2δi i i i J(p ,m ) i1...i4 − 3 1 2 3 4 i 3 2 1 2 3 4 1 2 3 4 12 + 2 permutations , (2.154) where 1 d4k d4k ∆(k)= , I(m2)= ∆(k) , J(p2 ,m2)= ∆(k)∆(p k). k2 + m2 (2π)4 12 (2π)4 12 − − Z Z (2.155)

It is not diﬃcult to convince ourselves that the ﬁrst integral I(m2) diverges quadratically 2 2 whereas the second integral J(p12,m ) diverges logarithmically. To see this more carefully it is better we Wick rotate to Euclidean signature. Formally this is done by writing k = ik which is consistent with x0 = ix4. As a consequence we replace k2 = k2 ~k2 0 4 − 0 − with k2 ~k2 = k2. The Euclidean expressions are − 4 − − λ Γ(2)(p) = δ (p2 + m2)+ ~ (N + 2)δ I(m2). (2.156) ij ij 6 ij

2 (4) λ λ 1 2 2 Γ (p1, ..., p4) = δi i i i ~ (N + 2)δi i δi i + 2δi i i i J(p ,m ) i1...i4 3 1 2 3 4 − 3 2 1 2 3 4 1 2 3 4 12 + 2 permutations , (2.157) where now 1 d4k d4k ∆(k)= , I(m2)= ∆(k) , J(p2 ,m2)= ∆(k)∆(p k(2.158)). k2 + m2 (2π)4 12 (2π)4 12 − Z Z Explicitly we have

∞ 4 2 −αm2 d k −αk2 I(m ) = dαe 4 e 0 (2π) Z ∞ Z −αm2 1 3 −αk2 = dαe 2 k dke 0 8π Z Z2 1 ∞ e−αm = dα . (2.159) 16π2 α2 Z0 28 YDRI QFT

To calculate the divergences we need to introduce a cut-oﬀ Λ. In principle we should use the regularized propagator

2 − k e Λ2 ∆(k, Λ) = . (2.160) k2 + m2

Alternatively we can introduce the cut-oﬀ Λ as follows

2 1 ∞ e−αm I(m2, Λ) = dα 16π2 1 α2 Z Λ2 2 1 ∞ e−αm = Λ2 m2 dα 16π2 − 1 α Z Λ2 1 m2 = Λ2 + m2Ei( ) . (2.161) 16π2 − Λ2 This diverges quadratically. The exponential-integral function is deﬁned by

x et Ei(x)= dt. (2.162) t Z−∞ Also by using the same method we compute

2 α1α2 2 4 2 2 −m (α1+α2)− p d k −(α +α )k2 J(p ,m ) = dα dα e α1+α2 12 e 1 2 12 1 2 (2π)4 Z 2 α1Zα2 2 −m (α1+α2)− p 1 e α1+α2 12 = dα dα . (2.163) (4π)2 1 2 (α + α )2 Z 1 2 We introduce the cut-oﬀ Λ as follows

2 α1α2 2 −m (α1+α2)− α +α p12 2 2 1 e 1 2 J(p12,m , Λ) = 2 dα1dα2 2 (4π) 1 (α1 + αa2) Z Λ2 2 2 m xx2 p12 − (x+x2)− 1 e Λ2 x+x2 Λ2 = dxdx . (2.164) (4π)2 2 (x + x )2 Z1 2

The integral can be rewritten as 2 times the integral over the symmetric region x2 >x. We can also perform the change of variables x2 = xy to obtain

2 p2 2 dx dy − m x(1+y)− xy 12 2 2 Λ2 1+y Λ2 J(p12,m , Λ) = 2 2 e (4π) 1 x 1 (1 + y) Z Z 1 2 dx 2 −x a +b(1−ρ) = dρ e ρ . (2.165) (4π)2 x Z1 Z0 YDRI QFT 29

2 2 m p12 In above a = Λ2 and b = Λ2 . We have

1 ∞ 2 2 dx a 2 2 −x ρ +b(1−ρ) J(p12,m , Λ) = 2 dρ e (4π) 0 1 x Z 1 Z 1 2 a = dρ Ei (1 ρ)b . (2.166) −8π2 − ρ − − Z0 The exponential-integral function is such that

a +(1−ρ)b −t a a ρ e 1 Ei (1 ρ)b = C + ln + (1 ρ)b + dt − . (2.167) − ρ − − ρ − t Z0 The last term leads to zero in the limit Λ since a, b 0 in this limit. The −→ ∞ −→ exponential-integral function becomes

a b b b b Ei (1 ρ)b = C + ln a + + √bρ + ln a + + √bρ ln ρ. − − − ρ 4 4 − 4 − 4 − r r r r (2.168)

By using the integral 1 dρ ln(A + Bρ)= 1 (A + B) ln(A + B) A ln A 1 we ﬁnd 0 B − − R 1 a 4a b 1 dρ Ei (1 ρ)b = C + ln a + 1+ ln 1+ + b(b + 4a) + 1. − − − ρ b 2a 2a Z0 r p (2.169)

Hence we have 1 a Λ2 dρ Ei (1 ρ)b = ln a + ... = ln + ... (2.170) − − − − ρ − m2 Z0 Equivalently

1 Λ2 J(p2 ,m2, Λ) = ln + ... (2.171) 12 16π2 2m2 This is the logarithmic divergence. In summary we have found two divergences at one-loop order. A quadratic divergence in the proper 2 point vertex and a logarithmic divergence in the proper 4 point vertex. − − All higher n point vertices are ﬁnite in the limit Λ . − −→ ∞ 2.4.4 Renormalization at 1 Loop − To renormalize the theory, i.e. to remove the above two divergences we will assume that: • 1) The theory comes with a cut-oﬀ Λ so that the propagator of the theory is actually given by (2.160). 30 YDRI QFT

• 2) The parameters of the model m2 and λ which are called from now on bare parameters will be assumed to depend implicitly on the cut-oﬀ Λ.

2 • 3) The renormalized (physical) parameters of the theory mR and λR will be determined from specific conditions imposed on the 2 and 4 proper vertices. − − In the limit Λ the renormalized parameters remain finite while the bare parameters −→ ∞ diverge in such a way that the divergences coming from loop integrals are canceled. In this way the 2 and 4 proper vertices become finite in the large cut-off limit Λ . − − −→ ∞ Since only two vertices are divergent we will only need two conditions to be imposed. 2 We choose the physical mass mR to correspond to the zero momentum value of the proper 2 point vertex, viz −

λ Γ(2)(0) = δ m2 = δ m2 + ~ (N + 2)δ I(m2, Λ). (2.172) ij ij R ij 6 ij

2 We also choose the physical coupling constant λR to correspond to the zero momentum value of the proper 4 point vertex, viz −

2 (4) λR λ λ N + 8 2 Γ (0, ..., 0) = δi i i i = δi i i i ~ δi i i i J(0,m , Λ). i1...i4 3 1 2 3 4 3 1 2 3 4 − 3 2 1 2 3 4 (2.173)

We solve for the bare parameters in terms of the renormalized parameters we ﬁnd

λ m2 = m2 ~ R (N + 2)I(m2 , Λ). (2.174) R − 6 R

λ λ λ 2 N + 8 = R + ~ R J(0,m2 , Λ). (2.175) 3 3 3 2 R The 2 and 4 point vertices in terms of the renormalized parameters are − −

(2) 2 2 Γij (p) = δij(p + mR). (2.176)

2 (4) λR λR 1 2 2 2 Γ (p1, ..., p4) = δi i i i ~ (N + 2)δi i δi i + 2δi i i i J(p ,m , Λ) J(0,m , Λ) i1...i4 3 1 2 3 4 − 3 2 1 2 3 4 1 2 3 4 12 R − R + 2 permutations . (2.177) YDRI QFT 31

2.5 Two-Loop Calculations

2.5.1 The Eﬀective Action at 2 Loop − By extending equation (2.107) to the second order in ~ we get 9

~ 1 ~ 2 δGkl 1 Γ[φ ] = O(1) + O + Gjj0 S + S φm + S φmφn + ... c ,i i 6 i δφ ijkl ijklm c 2 ijklmn c c cj0 3 ~ 3 + S + S φn + ... GjkGlm + O . (2.178) 4 ijklm ijklmn c i Equivalently

~ 1 ~ 2 δGkl 3 ~ 3 Γ[φ ] = O(1) + O + Gjj0 S[φ ] + S[φ ] GjkGlm + O . c ,i i 6 i δφ c ,ijkl 4 c ,ijklm i cj0 (2.179)

We use the identity

kl kl δG δG δJm klm = = G Γ,mj0 δφcj0 δJm δφcj0 − = Gkk0 Gll0 Gmm0 Γ Γ . (2.180) − ,k0l0m0 ,mj0 Thus

~ 1 ~ 2 Γ[φ ] = O(1) + O + Gjj0 Gkk0 Gll0 Gmm0 Γ Γ S[φ ] c ,i i 6 i − ,k0l0m0 ,mj0 c ,ijkl 3 ~ 3 + S[φ ] GjkGlm + O . (2.181) 4 c ,ijklm i By substituting the expansions (2.109) and (2.110) we get at the second order in ~ the equation

1 1 Γ [φ ] = GjkS[φ ] + Gjj0 Gkk0 Gll0 Gmm0 Γ Γ S[φ ] 2 c ,i 2 1 c ,ijk 6 − 0 0 0 0 0,k0l0m0 0,mj0 c ,ijkl 3 + S[φ ] GjkGlm . (2.182) 4 c ,ijklm 0 0 ij Next we compute G1 . Therefore we must determine Γ1kl. By diﬀerentiating equation (2.112) with respect to φcl we get

mn 1 mn 1 δG0 Γ1[φc],kl = G0 S[φc],klmn + S[φc],kmn. (2.183) 2 2 δφcl

9Exercise:Verify this equation. 32 YDRI QFT

By using the identity

mn δG0 mm0 nn0 = G0 G0 S[φc],lm0n0 . (2.184) δφcl

We get

1 1 Γ [φ ] = GmnS[φ ] + Gmm0 Gnn0 S[φ ] S[φ ] . (2.185) 1 c ,j0k0 2 0 c ,j0k0mn 2 0 0 c ,j0mn c ,k0m0n0 Hence

1 1 Gjk = Gjj0 Gkk0 GmnS[φ ] + Gmm0 Gnn0 S[φ ] S[φ ] . (2.186) 1 0 0 2 0 c ,j0k0mn 2 0 0 c ,j0mn c ,k0m0n0 Equation (2.182) becomes

1 1 1 Γ [φ ] = Gjj0 Gkk0 GmnS[φ ] + Gmm0 Gnn0 S[φ ] S[φ ] S[φ ] 2 c ,i 2 0 0 2 0 c ,j0k0mn 2 0 0 c ,j0mn c ,k0m0n0 c ,ijk 1 3 + Gjj0 Gkk0 Gll0 Gmm0 S[φ ] S[φ ] S[φ ] + S[φ ] GjkGlm . 6 − 0 0 0 0 c ,k0l0m0 c ,mj0 c ,ijkl 4 c ,ijklm 0 0 (2.187)

Integration of this equation yields 10

1 1 Γ [φ ] = S[φ ] GijGkl + S[φ ] GijGklGmnS[φ ] . (2.188) 2 c 8 c ,ijkl 0 0 12 c ,ikm 0 0 0 c ,jln The eﬀective action upto the 2 loop order is − 1 ~ ~ 2 1 1 Γ=Γ + ln det G + S[φ ] GijGkl + S[φ ] GijGklGmnS[φ ] + ... 0 2 i 0 i 8 c ,ijkl 0 0 12 c ,ikm 0 0 0 c ,jln (2.189)

This is represented graphically on ﬁgure 2.

2.5.2 The Linear Sigma Model at 2 Loop − The proper 2 point vertex upto 2 loop is given by 11 − − ~ ~ 2 δ2Γ [φ] Γ(2)(x,y) = O(1) + O + 2 . (2.190) ij i i δφ (x)δφ (y)|φ=0 i j 10Exercise:verify this result. 11Exercise: Verify all equations of this section. YDRI QFT 33

The 2 loop correction can be computed using the result − 1 1 1 Γ [φ] = Gjj0 Gkk0 GmnS[φ] + Gmm0 Gnn0 S[φ] S[φ] S[φ] 2 ,i 2 0 0 2 0 ,j0k0mn 2 0 0 ,j0mn ,k0m0n0 ,ijk 1 3 + Gjj0 Gkk0 Gll0 Gmm0 S[φ] S[φ] S[φ] + S[φ] GjkGlm . 6 − 0 0 0 0 ,k0l0m0 ,mj0 ,ijkl 4 ,ijklm 0 0 (2.191)

By setting φ = 0 we obtain

1 1 Γ [φ] = Gi0j0 Gkk0 GmnS[φ] S[φ] Gi0j0 Gkk0 Gll0 Gmm0 S[φ] S[φ] S[φ] 2 ,IJ 4 0 0 0 ,j0k0mn ,ii0kj − 6 0 0 0 0 ,k0l0m0j ,mj0 ,ii0kl 1 λ 2 N + 2 λ 2 = (N + 2)2δ δ4(x y)G (w, w) d4zG (x, z)G (y, z)+ δ G (x,y)3. 4 3 ij − 0 0 0 2 3 ij 0 Z (2.192)

We have then

~ ~ 2 λ 2 N + 2 N + 2 Γ(2)(x,y) = O(1) + O + δ δ4(x y)G (w, w) d4zG (x, z)G (y, z) ij i i 3 2 ij 2 − 0 0 0 Z 3 + G0(x,y) . (2.193) Next we write this result in momentum space. The proper 2 point vertex in momentum (2) − space Γij (p) is deﬁned through the equations

4 4 (2) ipx+iky 4 4 (2) d xd y Γij (x,y) e = (2π) δ (p + k)Γij (p). (2.194) Z We compute immediately

~ ~ 2 λ 2 N + 2 N + 2 d4p d4p 1 Γ(2)(p) = O(1) + O + δ 1 2 ij i i 3 2 ij 2 (2π)4 (2π)4 ( p2 + m2)( p2 + m2)2 Z − 1 − 2 d4p d4p 1 + 1 2 . (2π)4 (2π)4 ( p2 + m2)( p2 + m2)( (p p p )2 + m2) Z − 1 − 2 − − 1 − 2 (2.195)

The corresponding Feynman diagrams are shown on ﬁgure 4. The 4 point proper vertex upto 2 loop is given by − − 2 2 (4) ~ ~ δ Γ2[φ] Γ (x1, ..., x4) = O(1) + O + φ=0. (2.196) i1...i4 i i δφ (x )...δφ (x )| i1 1 i4 4 34 YDRI QFT

We compute

1 Γ [φ] = Gj1n1 Gj0n0 Gk1k0 Gm1m0 S S S + S S + S S 2 ,ijkl|φ=0 2 0 0 0 0 ,j0k0m1m0 ,ilj1k1 ,jkn1n0 ,ikj1k1 ,ljn1n0 ,ijj1k1 ,kln1n0 1 + Gj1j0 Gk1k0 Gm1m0 Gn1n0 S S S + S S + S S 4 0 0 0 0 ,j0k0m1n1 ,ilj1k1 ,jkm0n0 ,ikj1k1 ,jlm0n0 ,ijj1k1 ,klm0n0 1 + Gj1j0 Gk1k0 Gm1m0 Gn1n0 S S S + S S S 2 0 0 0 0 ,ilj1k1 ,jj0m1n1 ,kk0m0n0 ,ikj1k1 ,jj0m1n1 ,lk0m0n0 + S,ijj1k1 S,kj0m1n1 S,lk0m0n0 + S,ij1k1n1 S,jkj0m0 S,lm1k0n0 + S,ij1k1n1 S,klj0m0 S,jm1k0n0

+ S,ij1k1n1 S,jlj0m0 S,km1k0n0 . (2.197)

Thus

δ4Γ [φ] 1 λ 3 2 = (N + 2) (N + 2)δ δ + 2δ δ4(x x )δ4(x x ) δφ (x )...δφ (x )|φ=0 −2 3 i1i4 i2i3 i1i2i3i4 1 − 4 2 − 3 i1 1 i4 4 G (x ,x ) d4zG (z, z)G (x , z)G (x , z) + 2 permutations × 0 1 2 0 0 1 0 2 Z 1 λ 3 (N + 2)(N + 4)δ δ + 4δ δ4(x x )δ4(x x ) − 4 3 i1i4 i2i3 i1i2i3i4 1 − 4 2 − 3 d4zG (x , z)2G (x , z)2 + 2 permutations × 0 1 0 2 Z 1 λ 3 2(N + 2)δ δ + (N + 6)δ δ4(x x ) − 2 3 i1i4 i2i3 i1i2i3i4 1 − 4 G (x ,x )G (x ,x )G (x ,x )2 + 5 permutations . (2.198) × 0 1 2 0 1 3 0 2 3

(4) The proper 4 point vertex in momentum space Γ (p1...p4) is deﬁned through the − i1...i4 equation

d4x ...d4x Γ(4) (x , ..., x ) eip1x1+...+ip4x4 = (2π)4δ4(p + ... + p )Γ(2) (p , ..., p ). 1 4 i1...i4 1 4 1 4 i1...i4 1 4 Z (2.199) YDRI QFT 35

Thus we obtain in momentum space (with p12 = p1 + p2 and p14 = p1 + p4, etc)

(4) ~ Γ (p1, ..., p4) = O(1) + O i1...i4 i ~ 2 N + 2 λ 3 1 (N + 2)δ δ + 2δ − i 2 3 i1i4 i2i3 i1i2i3i4 l2 + m2 Zl − 1 + 2 permutations × ( k2 + m2)2( (p k)2 + m2) Zk − − 14 − ~ 2 1 λ 3 1 (N + 2)(N + 4)δ δ + 4δ − i 4 3 i1i4 i2i3 i1i2i3i4 ( l2 + m2)( (p l)2 + m2) Zl − − 14 − 1 + 2 permutations × ( k2 + m2)( (p k)2 + m2) Zk − − 14 − ~ 2 1 λ 3 1 2(N + 2)δ δ + (N + 6)δ − i 2 3 i1i4 i2i3 i1i2i3i4 ( l2 + m2)( (p l)2 + m2) Zl − − 14 − 1 + 5 permutations . (2.200) × ( k2 + m2)( (l k + p )2 + m2) Zk − − − 2 The corresponding Feynman diagrams are shown on ﬁgure 5.

2.5.3 The 2 Loop Renormalization of the 2 Point Proper Vertex − − The Euclidean expression of the proper 2 point vertex at 2 loop is given by − − λ λ 2 N + 2 N + 2 Γ(2)(p) = δ (p2 + m2)+ ~ (N + 2)δ I(m2) ~2 δ I(m2)J(0,m2)+ K(p2,m2) . ij ij 6 ij − 3 2 ij 2 (2.201)

d4k d4l K(p2,m2)= ∆(k)∆(l)∆(k + l p). (2.202) (2π)4 (2π)4 − Z We compute

2 α1α2α3 2 4 4 α1α2+α1α3+α2α3 2 2 2 −m (α1+α2+α3)− p d k d l −(α +α )k2 − l K(p ,m ) = dα dα dα e α1α2+α1α3+α2α3 e 1 3 e α1+α3 1 2 3 (2π)4 (2π)4 Z 2 α1α2α3 Z 2 −m (α1+α2+α3)− p 1 e α1α2+α1α3+α2α3 = dα dα dα . (2.203) (4π)4 1 2 3 (α α + α α + α α )2 Z 1 2 1 3 2 3 We have used the result

4 d k 2 1 e−ak = . (2.204) (2π)4 16π2a2 Z 36 YDRI QFT

We introduce the cut-oﬀ Λ as follows

2 α1α2α3 2 −m (α1+α2+α3)− α α +α α +α α p 2 2 1 e 1 2 1 3 2 3 K(p ,m ) = 4 dα1dα2dα3 2 (4π) 1 (α1α2 + α1α3 + α2α3) Z Λ2 2 x1x2x3 p 2 −x1−x2−x3− 2 m e x1x2+x1x3+x2x3 m = 4 dx1dx2dx3 2 (4π) m2 (x1x2 + x1x3 + x2x3) Z Λ2 m2 p2 p2 = (A + B + C( )2 + ...). (2.205) (4π)4 m2 m2 We have e−x1−x2−x3 A = dx1dx2dx3 2 m2 (x1x2 + x1x3 + x2x3) Z Λ2 m2 2 − (x+x2+x3) Λ e Λ2 = dxdx dx . (2.206) m2 2 3 (xx + xx + x x )2 Z1 2 3 2 3 The integrand is symmetric in the three variables x, x2 and x3. The integral can be rewritten as 6 times the integral over the symmetric region x3 > x2 > x. We can also 2 perform the change of variables x2 = xy and x3 = xyz, i.e dx2dx3 = x ydydz to obtain

m2 2 − x(1+y+yz) Λ dx dy e Λ2 A = dz m2 x2 y (1 + z + yz)2 Z1 ∞ dt = 6 e−tψ(t). (2.207) m2 t2 Z Λ2

∞ dy ∞ e−ty(1+z) ψ(t) = dz 2 1 y 1 (1 + z + yz) Z ∞ Z 2 ∞ dy −t y dz −z yt = e y+1 e y+1 y(y + 1) z2 Z1 Zy+2 y(y+2) ∞ 2 −t dy −t y e y+1 yt y(y + 2)t = e y+1 + Ei( . (2.208) y(y + 1) y + 2 y + 1 − y + 1 Z1 The most important contribution in the limit Λ comes from the region t 0. −→ ∞ ∼ Thus near t = 0 we have

∞ −2ty 2 dy e yt −t y y(y + 2) ψ(t) = + e y+1 C + ln t + ln + O(t) y(y + 1) y + 2 y + 1 y + 1 Z1 = ψ0 + ψ1t ln t + ψ2t + ... (2.209)

∞ dy 1 4 ψ = = ln . (2.210) 0 y(y + 1)(y + 2) 2 3 Z1 YDRI QFT 37

∞ dy 1 ψ = = . (2.211) 1 (y + 1)2 2 Z1 ∞ dy ∞ dy ∞ dy y(y + 2) ψ = 2 + C + ln 2 − (y + 1)(y + 2) (y + 1)2 (y + 1)2 y + 1 Z1 Z1 Z1 C ∞ dy y2 1 = 2(ln 2 ln 3) + + ln − − 2 y2 y Z2 C 1 3 1 = 2(ln 2 ln 3) + + ln 3 ln 2 − 2 − 2 2 − 2 1 = (C 1 ln 3 + 3 ln 2). (2.212) 2 − − We have then

dt −t dt −t dt −t A = 6ψ0 e + 6ψ1 e ln t + 6ψ2 e + ... m2 t2 m2 t m2 t Z Λ2 Z Λ2 Z Λ2 2 Λ dt −t dt −t = 6ψ0 + 6ψ1 e ln t + 6(ψ2 ψ0) e + ... m2 m2 t − m2 t Z Λ2 Z Λ2 2 2 2 Λ dt −t m − m t = 6ψ + 6ψ e ln t + 6(ψ ψ ) dt ln t e Λ2 + ... 0 2 1 2 2 0 2 m m t − Λ 1 Z Λ2 Z 2 2 2 2 Λ Λ −t 2 m = 6ψ0 3ψ1 ln + 3ψ1 dt e (ln t) 6(ψ2 ψ0)Ei( )+ ... m2 − m2 m2 − − − Λ2 Z Λ2 Λ2 Λ2 2 Λ2 = 6ψ 3ψ ln + 6(ψ ψ ) ln( )+ ... 0 m2 − 1 m2 2 − 0 m2 (2.213) Now we compute e−x1−x2−x3 B = x1x2x3dx1dx2dx3 3 − m2 (x1x2 + x1x3 + x2x3) Z Λ2 m2 − (x+x2+x3) e Λ2 = xx x dxdx dx − 2 3 2 3 (xx + xx + x x )3 Z1 2 3 2 3 m2 ∞ ∞ ∞ − (x+x2+x3) e Λ2 = 6 xdx x2dx2 x3dx3 3 − 1 x x2 (xx2 + xx3 + x2x3) Z Z Z 2 ∞ ∞ ∞ − m x(1+y+yz) dx e Λ2 = 6 dy zdz − x (1 + z + yz)3 Z1 Z1 Z1 ∞ dt = 6 e−tψ˜(t). (2.214) − m2 t Z Λ2

∞ ∞ e−ty(1+z) ψ˜(t) = dy zdz . (2.215) (1 + z + yz)3 Z1 Z1 38 YDRI QFT

It is not diﬃcult to convince ourselves that only the constant part of ψ˜ leads to a diver- ˜ 1 gence, i.e ψ(0) = 12 . We get

∞ dt Λ2 B = 6 e−tψ˜(0) = 6ψ˜(0) ln . (2.216) − m2 t − m2 Z Λ2 Now we compute

−x1−x2−x3 1 2 e C = (x1x2x3) dx1dx2dx3 4 2 m2 (x1x2 + x1x3 + x2x3) Z Λ2 m2 2 − (x+x2+x3) m e Λ2 = (xx x )2dxdx dx Λ2 2 3 2 3 (xx + xx + x x )4 Z1 2 3 2 3 m2 2 ∞ ∞ ∞ − (x+x2+x3) m e Λ2 = 6 x2dx x2dx x2dx Λ2 2 2 3 3 (xx + xx + x x )4 Z1 Zx Zx2 2 3 2 3 m2 2 ∞ ∞ ∞ − x(1+y+yz) m e Λ2 = 6 dx ydy z2dz Λ2 (1 + z + yz)4 Z1 Z1 Z1 ∞ ∞ ∞ e−ty(1+z) = 6 dt e−t ydy z2dz . (2.217) 2 4 m 1 1 (1 + z + yz) Z Λ2 Z Z This integral is well defined in the limit Λ . Furthermore it is positive definite. −→ ∞ ′ In summay we have found that both K(0,m2) and K (0,m2) are divergent in the limit Λ , i.e K(p2,m2) K(0,m2) is divergent at the two-loop order. This means (2)−→ ∞ (2) 2− 2 that Γij (p) and dΓij (p)/dp are divergent at p = 0 and hence in order to renormlaize the 2 point proper vertex Γ(2)(p) at the two-loop order we must impose two conditions − ij on it. The first condition is the same as before namely we require that the value of the 2 point proper vertex at zero momentum is precisely the physical or renormalized mass. − The second condition is essentially a renormalization of the coefficient of the kinetic term, (2) 2 i.e dΓij (p)/dp . Before we can write these two conditions we introduce a renormalization of the scalar field φ known also as wave function renormalization given by

φ = √ZφR. (2.218)

This induces a renormalization of the n point proper vertices. Indeed the eﬀective action − becomes

1 (n) Γ[φ] = Γ (x1, ..., xn)φi (x1)...φi (xn) n! i1...in 1 n n=0 X 1 (n) = Γ (x1, ..., xn)φi R(x1)...φi R(xn). (2.219) n! i1...inR 1 n nX=0 YDRI QFT 39

The renormalized n point proper vertex Γ(n) is given in terms of the bare n point − i1...inR − proper vertex Γ(n) by i1...in (n) n (n) Γ (x , ..., x )= Z 2 Γ (x , ..., x ). (2.220) i1...inR 1 n i1...in 1 n Thus the renormalized 2 point proper vertex Γ(2) (p) in momentum space is given by − ijR (2) (2) ΓijR(p)= ZΓij (p). (2.221) Now we impose on the renormalized 2 point proper vertex Γ(2) (p) the two conditions − ijR given by Γ(2) (p) = ZΓ(2)(p) = δ m2 . (2.222) ijR |p=0 ij |p=0 ij R d d Γ(2) (p) = Z Γ(2)(p) = δ . (2.223) dp2 ijR |p=0 dp2 ij |p=0 ij The second condition yields immediately 2 1 λ N + 2 ′ Z = =1+ ~2 K (0,m2, Λ). (2.224) 1 ~2 λ 2 N+2 K ′ (0,m2, Λ) 3 2 − 3 2 The ﬁrst condition gives then λ λ 2 N + 2 N + 2 m2 = m2 ~ (N + 2)I(m2, Λ) + ~2 I(m2, Λ)J(0,m2, Λ) + K(0,m2, Λ) R − 6 3 2 2 ′ m2K (0,m2, Λ) − λ λ 2 N + 2 = m2 ~ R (N + 2)I(m2, Λ) + ~2 R 3I(m2 , Λ)J(0,m2 , Λ) + K(0,m2 , Λ) R − 6 3 2 − R R R ′ m2 K (0,m2 , Λ) − R R λ λ 2 N + 2 N + 8 = m2 ~ R (N + 2)I(m2 , Λ) + ~2 R I(m2 , Λ)J(0,m2 , Λ) + K(0,m2 , Λ) R − 6 R 3 2 − 2 R R R ′ m2 K (0,m2 , Λ) . (2.225) − R R In above we have used the relation between the bare coupling constant λ and the renormalized coupling constant λR at one-loop given by equation (2.175). We have also used 2 2 ~ λR 2 2 the relation I(m , Λ) = I(mR, Λ) + 6 (N + 2)I(m , Λ)J(0,mR, Λ) where we have assumed that m2 = m2 ~ λR (N + 2)I(m2, Λ). We get therefore the 2 point proper R − 6 − vertex

2 λ N + 2 ′ Γ(2) (p)= δ (p2 + m2 ) ~2 R δ K(p2,m2 , Λ) K(0,m2 , Λ) p2K (0,m2 , Λ) . ijR ij R − 3 2 ij R − R − R (2.226) 40 YDRI QFT

2.5.4 The 2 Loop Renormalization of the 4 Point Proper Vertex − − The Euclidean expression of the proper 4 point vertex at 2 loop is given by − − 2 (4) λ λ 1 2 2 Γ (p1, ..., p4) = δi i i i ~ (N + 2)δi i δi i + 2δi i i i J(p ,m ) i1...i4 3 1 2 3 4 − 3 2 1 2 3 4 1 2 3 4 12 + 2 permutations λ 3 N + 2 + ~2 (N + 2)δ δ + 2δ I(m2)L(p2 ,m2) 3 2 i1i4 i2i3 i1i2i3i4 14 + 2 permutations λ 3 1 + ~2 (N + 2)(N + 4)δ δ + 4δ J(p2 ,m2)2 3 4 i1i4 i2i3 i1i2i3i4 14 + 2 permutations λ 3 1 + ~2 2(N + 2)δ δ + (N + 6)δ M(p2 ,p2,m2) 3 2 i1i4 i2i3 i1i2i3i4 14 2 + 5 permutations . (2.227)

d4k L(p2 ,m2)= ∆(k)2∆(k p ). (2.228) 14 (2π)4 − 14 Z

d4l d4k M(p2 ,p2,m2)= ∆(l)∆(k)∆(l p )∆(l k + p ). (2.229) 14 2 (2π)4 (2π)4 − 14 − 2 Z For simplicity we will not write explicitly the dependence on the cut-oﬀ Λ in the following. (4) The renormalized 4 point proper vertex Γ (p1,p2,p3,p4) in momentum space is − i1i2i3i4R given by

Γ(4) (p ,p ,p ,p )= Z2Γ(4) (p ,p ,p ,p ). (2.230) i1i2i3i4R 1 2 3 4 i1i2i3i4 1 2 3 4

We will impose the renormalization condition

(4) λR Γ (0, ..., 0) = δi i i i . (2.231) i1...i4R 3 1 2 3 4

We introduce a new renormalization constant Zg deﬁned by

(4) λ ZgΓ (0, ..., 0) = δi i i i . (2.232) i1...i4 3 1 2 3 4 YDRI QFT 41

Equivalently this means

Z g λ = λ. (2.233) Z2 R

The constant Z is already known at two-loop. The constant Zg at two-loop is computed to be

λ λ 2 (N + 2)(N + 8) Z = 1+ ~ (N + 8)J(0,m2) ~2 I(m2)L(0,m2) g 6 − 3 2 (N + 2)(N + 4) + 12 + J(0,m2)2 + (5N + 22)M(0, 0,m2) . (2.234) 4 We compute

Γ(4) (p ,p ,p ,p ) = Z2Γ(4) (p ,p ,p ,p ) i1i2i3i4R 1 2 3 4 i1i2i3i4 1 2 3 4 λR (4) (4) = Zg δi i i i +Γ (p1,p2,p3,p4) 1−loop +Γ (p1,p2,p3,p4) 2−loop. 3 1 2 3 4 i1i2i3i4 | i1i2i3i4 | (2.235)

2 2 2 2 ~ λR 2 2 2 By using the relation J(p12,m )= J(p12,mR)+ 3 (N +2)I(mR)L(p12,mR) we compute

λ λ λ 2 N + 8 λ 3 Z R δ = R δ + ~ R δ J(0,m2 ) ~2 R δ g 3 i1i2i3i4 3 i1i2i3i4 3 2 i1i2i3i4 R − 3 i1i2i3i4 (N + 2)(N + 4) + 12 (N + 8)2 J(0,m2 )2 + (5N + 22)M(0, 0,m2 ) . × 4 − 2 R R (2.236)

2 (4) λR 1 2 2 Γ (p1,p2,p3,p4) 1−loop = ~ (N + 2)δi i δi i + 2δi i i i J(p ,m ) i1i2i3i4 | − 3 2 1 2 3 4 1 2 3 4 12 R + 2 permutations λ 3 N + 8 ~2 R J(0,m2 ) (N + 2)δ δ + 2δ J(p2 ,m2 ) − 3 2 R i1i2 i3i4 i1i2i3i4 12 R + 2 permutations λ 3 N + 2 ~2 R I(m2 ) (N + 2)δ δ + 2δ L(p2 ,m2 ) − 3 2 R i1i2 i3i4 i1i2i3i4 12 R + 2 permutations . (2.237) 42 YDRI QFT

We then ﬁnd

2 (4) λR λR 1 2 2 2 Γ (p1, ..., p4) = δi i i i ~ (N + 2)δi i δi i + 2δi i i i (J(p ,m ) J(0,m )) i1...i4R 3 1 2 3 4 − 3 2 1 2 3 4 1 2 3 4 12 R − R + 2 permutations λ 3 1 + ~2 R (N + 2)(N + 4)δ δ + 4δ (J(p2 ,m2 ) J(0,m2 ))2 3 4 i1i4 i2i3 i1i2i3i4 14 R − R + 2 permutations λ 3 ~2 R 2(N + 2)δ δ + (N + 6)δ J(0,m2 )(J(p2 ,m2 ) J(0,m2 )) − 3 i1i4 i2i3 i1i2i3i4 R 14 R − R + 2 permutations λ 3 1 + ~2 R 2(N + 2)δ δ + (N + 6)δ (M(p2 ,p2,m2 ) M(0, 0,m2 )) 3 2 i1i4 i2i3 i1i2i3i4 14 2 R − R + 5 permutations . (2.238) 2 2 2 2 2 2 2 In the above last equation the combination M(p14,p2,mR) M(0, 0,mR) J(0,mR)(J(p14,mR) 12 − − − J(0,m2 )) must be ﬁnite in the limit Λ . R −→ ∞

2.6 Renormalized Perturbation Theory

The (φ2)2 theory with O(N) symmetry studied in this chapter is given by the action

1 1 λ S = d4x ∂ φ ∂µφ m2φ2 (φ2)2 . (2.239) 2 µ i i − 2 i − 4! i Z 2 This is called a bare action, the ﬁelds φi are the bare ﬁelds and the parameters m and λ are the bare coupling constants of the theory. Let us recall that the free 2 point function < 0 T (φˆ (x)φˆ (y)) 0 > is the prob- − | i,in j,in | ability amplitude for a free scalar particle to propagate from a spacetime point y to a spacetime x. In the interacting theory the 2 point function is < Ω T (φˆ (x)φˆ (y)) Ω > − | i j | where Ω >= 0 > / < 0 0 > is the ground state of the full Hamiltonian Hˆ . On general | | | grounds we can verify that the 2 point function < Ω T (φˆ (x)φˆ (y)) Ω > is given by p − | i j |

iZδ d4xeip(x−y) < Ω T (φˆ (x)φˆ (y)) Ω >= ij + .... (2.240) | i j | p2 m2 + iǫ Z − R 12Exercise:show this result. YDRI QFT 43

2 2 The dots stands for regular terms at p = mR where mR is the physical or renormalized mass. The residue or renormalization constant Z is called the wave function renormalization. Indeed the renormalized 2 point function < Ω T (φˆ (x)φˆ (y)) Ω > is given − | R R | by iδ d4xeip(x−y) < Ω T (φˆ (x)φˆ (y)) Ω >= ij + .... (2.241) | iR jR | p2 m2 + iǫ Z − R The physical or renormalized ﬁeld φR is given by

φ = √ZφR. (2.242) As we have already discussed this induces a renormalization of the n point proper ver- − tices. Indeed the eﬀective action becomes

1 4 4 (n) Γ[φ] = d x1... d xnΓ (x1, ..., xn)φi (x1)...φi (xn) n! i1...in 1 n nX=0 Z Z 1 4 4 (n) = d x1... d xnΓ (x1, ..., xn)φi R(x1)...φi R(xn). (2.243) n! i1...inR 1 n nX=0 Z Z The renormalized n point proper vertex Γ(n) is given in terms of the bare n point − i1...inR − proper vertex Γ(n) by i1...in

(n) n (n) Γ (x , ..., x )= Z 2 Γ (x , ..., x ). (2.244) i1...inR 1 n i1...in 1 n

We introduce a renormalized coupling constant λR and a renormalization constant Zg by

2 ZgλR = Z λ. (2.245) The action takes the form Z Z λZ2 S = d4x ∂ φ ∂µφ m2φ2 (φ2 )2 2 µ iR iR − 2 iR − 4! iR Z = SR + δS. (2.246)

The renormalized action SR is given by 1 1 λ S = d4x ∂ φ ∂µφ m2 φ2 R (φ2 )2 . (2.247) R 2 µ iR iR − 2 R iR − 4! iR Z The action δS is given by δ 1 δ δS = d4x Z ∂ φ ∂µφ δ φ2 λ (φ2 )2 . (2.248) 2 µ iR iR − 2 m iR − 4! iR Z The counterterms δZ , δm and δλ are given by δ = Z 1 , δ = Zm2 m2 , δ = λZ2 λ = (Z 1)λ . (2.249) Z − m − R λ − R g − R 44 YDRI QFT

The new Feynman rules derived from SR and δS are shown on figure 7. The so-called renormalized perturbation theory consists in the following. The renormalized or physical parameters of the theory mR and λR are always assumed to be finite whereas the counterterms δZ , δm and δλ will contain the unobservable infinite shifts between the bare parameters m and λ and the physical parameters mR and λR. The renormalized parameters are determined from imposing renormalization conditions on appropriate proper vertices. In this case we will impose on the 2 point proper vertex − Γ(2) (p) and the 4 point proper vertex Γ(2) (p) the three conditions given by ijR − ijR Γ(2) (p) = δ m2 . (2.250) ijR |p=0 ij R

d Γ(2) (p) = δ . (2.251) dp2 ijR |p=0 ij

(4) λR Γ (0, ..., 0) = δi i i i . (2.252) i1...i4R − 3 1 2 3 4 As an example let us consider the 2 point and 4 point functions upto the 1 loop order. − − − We have immediately the results ~ λ Γ(2) (p)= (p2 m2 ) R (N + 2)I(m2 ) + (δ p2 δ ) δ . (2.253) Rij − R − i 6 R Z − m ij

~ 2 (4) λR λR 1 2 2 Γ (p1, ..., p4) = δi i i i + (N + 2)δi i δi i + 2δi i i i J(p ,m ) Ri1...i4 − 3 1 2 3 4 i 3 2 1 2 3 4 1 2 3 4 12 R δ + 2 permutations λ δ . (2.254) − 3 i1i2i3i4 (2) (4) The ﬁrst two terms in both ΓR and ΓR come from the renormalized action SR and they are identical with the results obtained with the bare action S with the substitutions m m and λ λ . The last terms in Γ(2) and Γ(4) come from the action δS. By −→ R −→ R R R imposing renormalization conditions we get (including a cut-oﬀ Λ)

~ λ ~ λ2 δ = 0 , δ = R (N + 2)I (m2 ) , δ = R (N + 8)J (0,m2 ). (2.255) Z m − i 6 Λ R λ i 6 Λ R In other words

Γ(2) (p) = (p2 m2 )δ . (2.256) Rij − R ij

~ 2 (4) λR λR 1 2 2 2 Γ (p1, ..., p4) = δi i i i + (N + 2)δi i δi i + 2δi i i i (J(p ,m ) J(0,m )) Ri1...i4 − 3 1 2 3 4 i 3 2 1 2 3 4 1 2 3 4 12 R − R + 2 permutations . (2.257) YDRI QFT 45

It is clear that the end result of renormalized perturbation theory upto 1 loop is the − same as the somewhat "direct" renormalization employed in the previous sections to (2) (4) renormalize the perturbative expansion of ΓR and ΓR upto 1 loop. This result extends 13 − also to the 2 loop order . − Let us note at the end of this section that renormalization of higher n point vertices − should proceed along the same lines discussed above for the 2 point and 4 point vertices. − − The detail of this exercise will be omitted at this stage.

2.7 Eﬀective Potential and Dimensional Regularization

Let us go back to our original O(N) action which is given by

1 µ2 g S[φ]= d4x ∂ φ ∂µφ + φ2 (φ2)2 + J φ . (2.258) 2 µ i i 2 i − 4 i i i Z Now we expand the ﬁeld as φi = φci + ηi where φci is the classical ﬁeld. We can always choose φc to point in the N direction, viz φc = (0, ..., 0, φc). By translational invariance we may assume that φci is a constant. The action becomes (where V is the spacetime volume)

µ2 g 1 µ2 S[φ , η] = V φ2 (φ2 )2 + J φ + d4x ∂ η ∂µη + η2 + (µ2 gφ2 )φ η + J η c 2 ci − 4 ci i ci 2 µ i i 2 i − cj ci i i i Z g g φ2 η2 + 2(φ η )2 g(φ η )η2 (η2)2 . (2.259) − 2 ci j ci i − ci i j − 4 i In the spirit of renormalized perturbation theory we will think of the parameters µ2 and g as renormalized parameters and add the counterterms 1 1 1 δS[φ]= d4x δ ∂ φ ∂µφ + δ φ2 δ (φ2)2 + δJ φ . (2.260) 2 Z µ i i 2 µ i − 4 g i i i Z (1) The counterterm δJi is chosen so that the 1 point vertex Γ (x1) is identically zero to all − i1 orders in perturbation theory. This is equivalent to the removal of all tadpole diagrams that contribute to < ηi >. Let us recall the form of the eﬀective action upto 1 loop and the classical 2 point − − function. These are given by 1 ~ Γ= S + ln det G + ... (2.261) 2 i 0

Gij = S−1 . (2.262) 0 − ,ij |φ=φc The effective action can always be rewritten as the spacetime integral of an effective Lagrangian . For slowly varying fields the most important piece in this effective Leff 13Exercise:Try this explicitly. 46 YDRI QFT

Lagrangian is the so-called effective potential which is the term with no dependence on the derivatives of the field. The effective Lagrangian takes the generic form

(φ , ∂φ ,∂∂φ , ...)= V (φ )+ Z(φ )∂ φ ∂µφ + ... (2.263) Leﬀ c c c − c c µ c c For constant classical ﬁeld we have

Γ(φ )= d4xV (φ )= d4x V (φ ). (2.264) c − c − c Z Z We compute immediately

δ2S = ∂2δ + µ2δ g[φ2 δ + 2φ φ ] δ4(x y) δη (x)δη (y)|η=0 − ij ij − ck ij ci cj − i j = ∂2 m2 δ δ4(x y). (2.265) − − i ij −

The masses mi are given by

m2 = gφ2 µ2 , i, j = N and m2 = 3gφ2 µ2 , i = j = N. (2.266) i c − 6 i c − The above result can be put in the form

δ2S ddp = p2 m2 δ eip(x−y). (2.267) δη (x)δη (y)|η=0 (2π)d − i ij i j Z We compute 1 ~ 1 ~ ln det G = ln detG−1 2 i 0 −2 i 0 i~ δ2S = ln det 2 − δη (x)δη (y)|η=0 i j i~ δ2S = T r ln 2 − δη (x)δη (y)|η=0 i j i~ δ2S = d4x 2 | − δη (x)δη (y)|η=0 | Z i j i~ d4p = V ln ( p2 + m2)δ 2 (2π)4 − i ij Z i~ d4p d4p = V (N 1) ln p2 µ2 + gφ2 + ln p2 µ2 + 3gφ2 . 2 − (2π)4 − − c (2π)4 − − c Z Z (2.268)

The basic integral we need to compute is

d4p I(m2)= ln p2 + m2 . (2.269) (2π)4 − Z YDRI QFT 47

This is clearly divergent. We will use here the powerful method of dimensional regularization to calculate this integral. This consists in 1) performing a Wick rotation k0 k4 = ik0 and 2) continuing the number of dimensions from 4 to d = 4. We have −→ − 6 then ddp I(m2)= i E ln p2 + m2 . (2.270) (2π)d E Z We use the identity ∂ x−α = ln x. (2.271) ∂α |α=0 − We get then

∂ ddp 1 I(m2) = i E − ∂α (2π)d (p2 + m2)α |α=0 Z E ∂ Ω pd−1 = i d−1 dp E . (2.272) − ∂α (2π)d E (p2 + m2)α |α=0 Z E d−1 The Ωd−1 is the solid angle in d dimensions, i.e. the area of a sphere S . It is given by 14

d 2π 2 Ωd−1 = d . (2.273) Γ( 2 ) 2 2 2 We make the change of variables x = pE then the change of variables t = m /(x + m ). We get

∞ d −1 ∂ Ω x 2 I(m2) = i d−1 dx − ∂α 2(2π)d (x + m2)α |α=0 Z0 2 d −α 1 ∂ Ωd−1(m ) 2 α−1− d d −1 = i dt t 2 (1 t) 2 . (2.274) − ∂α 2(2π)d − |α=0 Z0 We use the result 1 Γ(α)Γ(β) dt tα−1(1 t)β−1 = . (2.275) − Γ(α + β) Z0 We get then

2 d −α d d ∂ Ω (m ) 2 Γ(α )Γ( ) I(m2) = i d−1 − 2 2 − ∂α 2(2π)d Γ(α) |α=0 d ∂ 1 2 d −α Γ(α 2 ) = i (m ) 2 − . (2.276) − ∂α d Γ(α) |α=0 (4π) 2 14Derive this result. 48 YDRI QFT

Now we use the result that 1 Γ(α) , α 0. (2.277) −→ α −→ Thus

1 d d 2 2 2 I(m ) = i d (m ) Γ( ). (2.278) − (4π) 2 −2

By using this result we have ~ ~ 1 i 1 d 2 2 d 2 2 d ln det G = V i Γ( ) (N 1)( µ + gφ ) 2 + ( µ + 3gφ ) 2 2 i 0 2 − d −2 − − c − c (4π) 2 ~ d Γ( 2 ) 2 2 d 2 2 d = V − (N 1)( µ + gφ ) 2 + ( µ + 3gφ ) 2 . (2.279) 2 d − − c − c (4π) 2 The eﬀective potential including counterterms is given by

2 ~ d µ 2 g 2 2 Γ( 2 ) 2 2 d 2 2 d V (φ ) = φ + (φ ) − (N 1)( µ + gφ ) 2 + ( µ + 3gφ ) 2 c − 2 c 4 c − 2 d − − c − c (4π) 2 1 1 δ φ2 + δ (φ2)2. (2.280) − 2 µ c 4 g c Near d = 4 we use the approximation given by (with ǫ = 4 d and γ = 0.5772 is − Euler-Mascheroni constant)

d 1 ǫ Γ( ) = Γ( ) −2 d ( d 1) 2 2 2 − 1 2 3 = [ γ + + O(ǫ)]. (2.281) 2 ǫ − 2 This divergence can be absorbed by using appropriate renormalization conditions. We 2 remark that the classical minimum is given by φc = v = µ /g. We will demand that the value of the minimum of V remains given by φ = v at the one-loop order by eﬀ c p imposing the condition ∂ V (φc) φc=v = 0. (2.282) ∂φc | As we will see in the next section this is equivalent to saying that the sum of all tadpole diagrams is 0. This condition leads immediately to 15

d 2 Γ(1 2 ) 3 δµ δgv = ~g − . (2.283) d 1− d − (4π) 2 (2µ2) 2

15Exercise: Verify explicitly. YDRI QFT 49

The second renormalization condition is naturally chosen to be given by

∂4 g 4 V (φc) φc=v = 4!. (2.284) ∂φc | 4 This leads to the result 16 Γ(2 d ) ~ 2 2 δg = g (N + 8) − d . (2.285) (4π) 2

As a consequence we obtain 17

Γ(2 d ) ~ 2 2 δµ = gµ (N + 2) − d . (2.286) (4π) 2

After substituting back in the potential we get 18

µ2 g ~ 3 V (φ ) = φ2 + (φ2)2 + (N 1)( µ2 + gφ2)2 ln( µ2 + gφ2) c − 2 c 4 c 4(4π)2 − − c − c − 2 3 + ( µ2 + 3gφ2)2 ln( µ2 + 3gφ2) . (2.287) − c − c − 2 In deriving this result we have used in particular the equation

2 d 4 d (m ) 2 m 2 3 Γ( ) = + ln 4π ln m2 γ + + O(ǫ) . (2.288) −2 d 2(4π)2 ǫ − − 2 (4π) 2 2.8 Spontaneous Symmetry Breaking

2.8.1 Example: The O(N) Model We are still interested in the (φ2)2 theory with O(N) symmetry in d dimensions (d = 4 is of primary importance but other dimensions are important as well) given by the classical action (with the replacements m2 = µ2 and λ/4! = g/4) − 1 1 g S[φ]= ddx ∂ φ ∂µφ + µ2φ2 (φ2)2 . (2.289) 2 µ i i 2 i − 4 i Z This scalar field can be in two different phases depending on the value of m2. The "symmetric phase" characterized by the "order parameter" φ (J = 0) < φ >= 0 ic ≡ i and the "broken phase" with φ = 0. This corresponds to the spontaneous symmetry ic 6 breaking of O(N) down to O(N 1) and the appearance of massless particles called − Goldstone bosons in d 3. For N = 1, it is the Z symmetry φ φ which is broken ≥ 2 −→ − 16Exercise: Verify explicitly. 17Exercise: Verify explicitly. 18Exercise: Verify explicitly. 50 YDRI QFT spontaneously. This is a very concrete instance of Goldstone theorem. In "local" scalar field theory in d 2 there can be no spontaneous symmetry breaking according to the ≤ Wagner-Mermin-Coleman theorem. To illustrate these points we start from the classical potential 1 g V [φ]= ddx µ2φ2 + (φ2)2 . (2.290) − 2 i 4 i Z This has a Mexican-hat shape. The minimum of the system is a configuration which must minimize the potential and also is uniform so that it minimizes also the Hamiltonian. The equation of motion is

φ ( µ2 + gφ2) = 0. (2.291) j − i 2 2 For µ < 0 the minimum is unique given by the vector φi = 0 whereas for µ > 0 we can have as solution either the vector φi < 0 (which in fact is not a minimum) or any vector φi such that µ2 φ2 = . (2.292) i g As one may check any of these vectors is a minimum. In other words we have an inﬁnitely degenerate ground state given by the sphere SN−1. The ground state is conventionally chosen to point in the N direction by adding to the action a symmetry breaking term of the form

d ∆S = ǫ d xφN , ǫ> 0. (2.293) Z

( µ2 + gφ2)φ = ǫδ . (2.294) − i j jN The solution is clearly of the form

φi = vδiN . (2.295) The coeﬃcient v is given by

µ2 ( µ2 + gv2)v = ǫ v = , ǫ 0. (2.296) − ⇒ s g −→ We expand around this solution by writing

φ = π , k = 1, ..., N 1 , φ = v + σ. (2.297) k k − N By expanding the potential around this solution we get 1 1 g g g V [φ]= ddx ( µ2 + gv2)π2 + ( µ2 + 3gv2)σ2 + v( µ2 + gv2)σ + gvσ3 + gvσπ2 + σ2π2 + σ4 + (π2)2 2 − k 2 − − k 2 k 4 4 k Z (2.298) YDRI QFT 51

We have therefore one massive ﬁeld (the σ) and N 1 massless ﬁelds (the pions π ) for − k µ2 > 0. Indeed

m2 = µ2 + gv2 0 , m2 = µ2 + 3gv2 2µ2. (2.299) π − ≡ σ − ≡ For µ2 < 0 we must have v = 0 and thus m2 = m2 = µ2. π σ − It is well known that the O(4) model provides a very good approximation to the dynamics of the real world pions with masses m+ = m− = 139.6 Mev, m0 = 135Mev which are indeed much less than the mass of the 4th particle (the sigma particle) which has mass mσ = 900 Mev. The O(4) model can also be identiﬁed with the Higgs sector of the standard model. The action around the "broken phase" solution is given by

1 1 g g g S[φ]= ddx ∂ π ∂µπ + ∂ σ∂µσ µ2σ2 gvσ3 gvσπ2 σ2π2 σ4 (π2)2 . 2 µ k k 2 µ − − − k − 2 k − 4 − 4 k Z (2.300)

We use the counterterms

1 1 1 δS[φ] = ddx δ ∂ φ ∂µφ + δ φ2 δ (φ2)2 2 Z µ i i 2 µ i − 4 g i Z 1 1 1 1 = ddx δ ∂ π ∂µπ ( δ + δ v2)π2 + δ ∂ σ∂µσ ( δ + 3δ v2)σ2 2 Z µ i i − 2 − µ g i 2 Z µ − 2 − µ g Z 1 1 1 ( δ v + δ v3)σ δ vσπ2 δ vσ3 δ (π2)2 δ σ2π2 δ σ4 . (2.301) − − µ g − g i − g − 4 g i − 2 g i − 4 g We compute the 1 point proper vertex of the sigma ﬁeld. We start from the result −

1 ~ Γ = S + GjkS + ... (2.302) ,i ,i 2 i 0 ,ijk

Gij = S−1 . (2.303) 0 − ,ij |φ=φc We compute immediately

δΓ 1 ~ = 0+ [Gσσ S + Gπiπj S ] δσ |σ=πi=0 2 i 0 ,σσσ 0 ,σπiπj 1 ~ ddk 1 ddk δ = ( 3!gv)+ ij ( 2gvδ ) 2 i (2π)d k2 + 2µ2 − (2π)d k2 + ξ2 − ij Z − Z − ddk 1 ddk 1 = 3igv~ igv(N 1)~ . (2.304) − (2π)d k2 2µ2 − − (2π)d k2 ξ2 Z − Z − 52 YDRI QFT

In the above equation we have added a small mass ξ2 for the pions to control the infrared behavior. We need to compute

ddk 1 ddk 1 = i E (2π)d k2 m2 − (2π)d k2 + m2 Z − Z E d −1 iΩ x 2 = d−1 dx −2(2π)d x + m2 Z 1 iΩd−1 2 d −1 − d d −1 = (m ) 2 t 2 (1 t) 2 dt −2(2π)d − Z0 i d d 2 2 −1 = d (m ) Γ(1 ). (2.305) −(4π) 2 − 2

We get

d δΓ Γ(1 2 ) 3 N 1 σ=π =0 = gv~ − ( + − ). (2.306) i d 1− d 1− d δσ | − (4π) 2 (2µ2) 2 (ξ) 2

By adding the contribution of the counterterms we get

d δΓ 3 Γ(1 2 ) 3 N 1 σ=π =0 = ( δµv + δgv ) gv~ − ( + − ). (2.307) i d 1− d 1− d δσ | − − − (4π) 2 (2µ2) 2 (ξ2) 2

The corresponding Feynman diagrams are shown on ﬁgure 8. We will impose the renormalization condition δΓ = 0. (2.308) δσ This is equivalent to the statement that the sum of all tadpole diagrams giving the 1 point proper vertex for the σ ﬁeld vanishes. In other words we do not allow any − quantum shifts in the vacuum expectation value of φN which is given by < φN >= v. We get then

d 2 Γ(1 2 ) 3 N 1 ( δµ + δgv )= g~ − ( + − ). (2.309) d 1− d 1− d − − (4π) 2 (2µ2) 2 (ξ2) 2

Next we consider the ππ amplitude. We use the result

~ 1 1 Γ = S + GmnS[φ] + Gmm0 Gnn0 S[φ] S[φ] (2.310). ,j0k0 ,j0k0 i 2 0 ,j0k0mn 2 0 0 ,j0mn ,k0m0n0 We compute immediately (including again a small mass ξ2 for the pions)

S = δ (∆ + ξ2)δd(x y). (2.311) ,j0k0 − j0k0 − YDRI QFT 53

1 1 GmnS[φ] = ddzddw[Gσσ (z, w)][ 2gδ δd(x y)δd(x z)δd(x w)] 2 0 ,j0k0mn 2 0 − j0 k0 − − − Z 1 g + ddzddw[δ Gππ(z, w)][ 3! δ δd(x y)δd(x z)δd(x w)] 2 mn 0 − 3 j0k0mn − − − Z = gδ Gσσ(x,x)δd(x y) (N + 1)gδ Gππ(x,x)δd(x y).(2.312) − j0k0 0 − − j0k0 0 − 1 (2) Gmm0 Gnn0 S[φ] S[φ] = ddz ddz ddw ddw [δ Gππ(z, z )][Gσσ (w, w )] 2 0 0 ,j0mn ,k0m0n0 2 0 0 mm0 0 0 0 0 Z Zd Z d Z d d [ 2gvδj0mδ (x z)δ (x w)][ 2gvδk0 m0 δ (y z0)δ (y w0)] × −2 2 ππ −σσ − − − − = 4g v G0 (x,y)G0 (x,y). (2.313) Thus we get ~ Γππ (x,y) = δ (∆ + ξ2)δd(x y)+ gδ Gσσ(x,x)δd(x y) (N + 1)gδ Gππ(x,x)δd(x y) j0k0 − j0k0 − i − j0k0 0 − − j0k0 0 − 2 2 ππ σσ + 4g v δj0k0 G0 (x,y)G0 (x,y) . (2.314) Recall also that ddp 1 ddp 1 Gππ(x,y)= eip(x−y) , Gσσ(x,y)= eip(x−y)(2.315). 0 (2π)d p2 + ξ2 0 (2π)d p2 + 2µ2 Z − Z − The Fourier transform is deﬁned by

d d ππ ipx iky d d ππ d x d yΓj0k0 (x,y)e e = (2π) δ (p + k)Γj0k0 (p). (2.316) Z Z We compute then ~ ddk 1 ddk 1 Γππ (p) = δ (p2 ξ2)+ δ g (N + 1)g j0k0 j0k0 − i j0k0 − (2π)d k2 + 2µ2 − (2π)d k2 + ξ2 Z − Z − d4k 1 1 + 4g2v2 . (2.317) (2π)d k2 + ξ2 (k + p)2 + 2µ2 Z − − By adding the contribution of the counterterms we get ~ ddk 1 ddk 1 Γππ (p) = δ (p2 ξ2)+ δ g (N + 1)g j0k0 j0k0 − i j0k0 − (2π)d k2 + 2µ2 − (2π)d k2 + ξ2 Z − Z − d4k 1 1 + 4g2v2 + (δ p2 + δ δ v2)δ . (2.318) (2π)d k2 + ξ2 (k + p)2 + 2µ2 Z µ − g j0k0 Z − − The corresponding Feynman diagrams are shown on ﬁgure 8. After some calculation we obtain d Γ(1 ) d d ππ 2 2 ~ 2 2 2 −1 2 2 −1 Γj0k0 (p) = δj0k0 (p ξ ) 2 gδj0k0 − d [(ξ ) (2µ ) ] − − (4π) 2 − ~ d4k 1 1 + δ 4g2v2 + δ p2δ (2.319). i j0k0 (2π)d k2 + ξ2 (k + p)2 + 2µ2 Z j0k0 Z − − 54 YDRI QFT

The last integral can be computed using Feynman parameters x1, x2 intrdouced by the identity

1 1 1 1 = dx dx δ(x + x 1). (2.320) A A 1 2 (x A + x A )2 1 2 − 1 2 Z0 Z0 1 1 2 2 We have then (with s = 2, l = k + (1 x )p and M 2 = ξ2x + 2µ2(1 x ) p2x (1 x ) − 1 1 − 1 − 1 − 1 and after a Wick rotation)

ddk 1 1 ddk 1 1 1 = dx1 dx2 δ(x1 + x2 (2π)d k2 + ξ2 (k + p)2 + 2µ2 (2π)d x (k2 ξ2)+ x ((k + p)2 2µ2) s − Z − − Z Z0 Z0 1 − 2 − 1 1 ddl 1 = dx dx δ(x + x 1) 1 2 1 2 − (2π)d (l2 M 2)s Z0 Z0 Z − 1 1 ddl 1 = i( 1)s dx dx δ(x + x 1) E − 1 2 1 2 − (2π)d (l2 + M 2)s Z0 Z0 Z E s 1 1 d−1 i( 1) Ω l dlE = − d−1 dx dx δ(x + x 1) E (2π)d 1 2 1 2 − (l2 + M 2)s Z0 Z0 Z E s 1 1 d −1 i( 1) Ω x 2 dx = − d−1 dx dx δ(x + x 1) 2(2π)d 1 2 1 2 − (x + M 2)s Z0 Z0 Z s 1 1 1 i( 1) Ωd−1 2 d −s d −1 s− d − = − (M ) 2 dx dx δ(x + x 1) (1 t) 2 t 2 2(2π)d 1 2 1 2 − − Z0 Z0 Z0 s 1 1 d d i( 1) Ωd−1 2 d −s Γ(s 2 )Γ( 2 ) = − (M ) 2 dx dx δ(x + x 1) − 2(2π)d 1 2 1 2 − Γ(s) Z0 Z0 1 1 s d i( 1) 2 d −s Γ(s 2 ) = dx dx δ(x + x 1) − (M ) 2 − . (2.321) 1 2 1 2 − d Γ(s) Z0 Z0 (4π) 2 Using this result we have

d Γ(1 ) d d ππ 2 2 ~ 2 2 2 −1 2 2 −1 Γj0k0 (p) = δj0k0 (p ξ ) 2 gδj0k0 − d [(ξ ) (2µ ) ] − − (4π) 2 − d 1 Γ(2 ) d −2 + 4~g2v2δ − 2 dx ξ2x + 2µ2(1 x ) p2x (1 x ) 2 + δ p2δ . j0k0 d 1 1 − 1 − 1 − 1 Z j0k0 (4π) 2 Z0 (2.322)

By studying the amplitudes σσ, σσππ and ππππ we can determine that the countert- 19 erm δZ is ﬁnite at one-loop whereas the counterterm δg is divergent . This means in particular that the divergent part of the above remaining integral does not depend on p.

19Exercise: Show this result explicitly. You need to ﬁgure out and then use two more renormalization conditions. YDRI QFT 55

We simply set p2 = 0 and study d d 1 ππ 2 Γ(1 2 ) 2 d −1 2 d −1 2 2 Γ(2 2 ) Γ (0) = δ ( ξ ) 2~gδ − [(ξ ) 2 (2µ ) 2 ] + 4~g v δ − dx j0k0 j0k0 − − j0k0 d − j0k0 d 1 (4π) 2 (4π) 2 Z0 d −2 ξ2x + 2µ2(1 x ) 2 . (2.323) × 1 − 1 We get (usinggv2 = µ2)

d d 2 ππ 2 Γ(1 2 ) 2 d −1 2 d −1 Γ(1 2 ) 2µ 2 d −1 2 Γ (0) = δ ( ξ ) 2~gδ − [(ξ ) 2 (2µ ) 2 ] + 2~gδ − [(ξ ) 2 (2µ ) j0k0 j0k0 − − j0k0 d − j0k0 d 2µ2 ξ2 − (4π) 2 (4π) 2 − (2.324) This vanishes exactly in the limit ξ 0 and therefore the pions remain massless at 20 −→ one-loop . This is a manifestation of the Goldstone’s theorem which states that there must exist N 1 massless particles associated with the N 1 broken symmetries of the − − breaking pattern O(N) O(N 1). −→ − 2.8.2 Glodstone’s Theorem Spontaneous symmetry breaking of a continuous symmetry leads always to massless particles called Goldstone bosons. The number of massless Goldstone bosons which appear is precisely equal to the number of symmetry generators broken spontaneously. This is a general result known as Goldstone’s theorem. For example in the case of the O(N) model studied in the previous sections the continuous symmetries are precisely O(N) transformations, i.e. rotations in N dimensions which rotate the different components of the scalar field into each other. There are in this case N(N 1)/2 independent − rotations and hence N(N 1)/2 generators of the group O(N). Under the symme- − try breaking pattern O(N) O(N 1) the number of broken symmetries is exactly −→ − N(N 1)/2 (N 1)(N 2)/2= N 1 and hence there must appear N 1 massless − − − − − − Goldstone bosons in the low energy spectrum of the theory which have been already verified explicitly upto the one-loop order. This holds also true at any arbitrary order in perturbation theory. Remark that for N = 1 there is no continuous symmetry and there are no massless Goldstone particles associated to the symmetry breaking pattern φ φ. We sketch now a general proof of Goldstone’s theorem. −→ − A typical Lagrangian density of interest is of the form (φ) = terms with derivatives(φ) V (φ). (2.325) L − The minimum of V is denoted φ0 and satisfies ∂ V (φ) φ=φ0 = 0. (2.326) ∂φa | 20Exercise: Show this result directly. Start by showing that ~ d4k 1 1 ddk 1 4g2v2 = 2i~g[I(ξ2) − I(2µ2)] ,I(m2)= . i Z (2π)d −k2 + ξ2 −k2 + 2µ2 Z (2π)d k2 − m2 56 YDRI QFT

Now we expand V around the minimum φ0 upto the second order in the ﬁelds. We get

1 ∂2 V (φ) = V (φ0)+ (φ φ0)a(φ φ0)b V (φ) φ=φ0 + ... 2 − − ∂φa∂φb | 1 = V (φ )+ (φ φ ) (φ φ ) m2 (φ )+ ... (2.327) 0 2 − 0 a − 0 b ab 0

2 The matrix mab(φ0) called the mass matrix is clearly a symmetric matrix which is also positive since φ0 is assumed to be a minimum configuration. A general continuous symmetry will transform the scalar field φ infinitesimally according to the generic law

′ φ φ = φ + α∆ (φ). (2.328) a −→ a a a

The parameter α is inﬁnitesimal and ∆a are some functions of φ. The invariance of the Lagrangian density is given by the condition terms with derivatives(φ) V (φ) = terms with derivatives(φ + α∆(φ)) V (φ + α∆(φ)). − − (2.329)

For constant ﬁelds this condition reduces to

V (φ)= V (φ + α∆(φ)). (2.330)

Equivalently ∂ ∆a(φ) V (φ) = 0. (2.331) ∂φa

By diﬀerentiating with respect to φb and setting φ = φ0 we get

2 mab(φ0)∆b(φ0) = 0. (2.332)

The symmetry transformations, as we have seen, leave always the Lagrangian density invariant which was actually our starting point. In the case that the above symmetry transformation leaves also the ground state conﬁguration φ0 invariant we must have ∆(φ0) = 0 and thus the above equation becomes trivial. However, in the case that the symmetry transformation does not leave the ground state conﬁguration φ0 invariant, which is precisely the case of a spontaneously broken symmetry, ∆b(φ0) is an eigenstate 2 of the mass matrix mab(φ0) with 0 eigenvalue which is exactly the massless Goldstone particle. 3 Path Integral Quantization of Dirac and Vector Fields

3.1 Free Dirac Field

3.1.1 Canonical Quantization The Dirac field ψ describes particles of spin ~/2. The Dirac field ψ is a 4 component − object which transforms as spinor under the action of the Lorentz group. The classical equation of motion of a free Dirac field is the Dirac equation. This is given by (i~γµ∂ mc)ψ = 0. (3.1) µ − Equivalently the complex conjugate field ψ¯ = ψ+γ0 obeys the equation µ ψ¯(i~γ ←−∂µ + mc) = 0. (3.2) These two equations are the Euler-Lagrange equations derived from the action

S = d4xψ¯(i~cγµ∂ mc2)ψ. (3.3) µ − Z The Dirac matrices γµ satisfy the usual Dirac algebra γµ, γν = 2ηµν . The Dirac { } equation admits positive-energy solutions (associated with particles) denoted by spinors ui(p) and negative-energy solutions (associated with antiparticles) denoted by spinors vi(p). The spinor ﬁeld can be put in the form (with ω(~p)= E/~ = ~p2c2 + m2c4/~) 3 1 d p c i ip ψ(x) = e− ~ pxu(i)(~p)b(~p, i)+ e ~ pxv(i)(~p)d(~p, i)+ . ~ (2π~)3 2ω(~p) Z r Xi (3.4) 58 YDRI QFT

The conjugate field is Π(x)= i~ψ+. In the quantum theory (canonical quantization) the coefficients b(~p, i) and d(~p, i) become operators ˆb(~p, i) and dˆ(~p, i) and as a consequence the spinor field ψ(x) and the conjugate field Π(x) become operators ψˆ(x) and Π(ˆ x) respectively. In order to have a stable ground state the operators ψˆ(x) and Π(ˆ x) must satisfy the anticommutation (rather than commutation) relations given by

ψˆ (x0, ~x), Πˆ (x0, ~y) = i~δ δ3(~x ~y). (3.5) { α β } αβ − Equivalently

ˆb(~p, i), ˆb(~q, j)+ = ~δ (2π~)3δ3(~p ~q) { } ij − dˆ(~p, i)+, dˆ(~q, j) = ~δ (2π~)3δ3(~p ~q) { } ij − ˆb(~p, i), dˆ(~q, j) = dˆ(~q, j)+, ˆb(~p, i) = 0. (3.6) { } { } We find that excited particle states are obtained by acting with ˆb(~p, i)+ on the vacuum 0 > whereas excited antiparticle states are obtained by acting with dˆ(~p, i)+. The vacuum | state 0 > is the eigenstate with energy equal 0 of the Hamiltonian | d3p Hˆ = ω(~p) ˆb(~p, i)+ˆb(~p, i)+ dˆ(~p, i)+dˆ(~p, i) . (3.7) (2π~)3 Z Xi The Feynman propagator for a Dirac spinor field is defined by ¯ (S ) (x y) = < 0 T ψˆ (x)ψˆ (y) 0 > . (3.8) F ab − | a b | The time-ordering operator is defined by

T ψˆ(x)ψˆ(y)=+ψˆ(x)ψˆ(y) , x0 >y0 T ψˆ(x)ψˆ(y)= ψˆ(y)ψˆ(x) , x0

4 µ ~ d p i(γ pµ + mc)ab i (S ) (x y) = e− ~ p(x−y). (3.10) F ab − c (2π~)4 p2 m2c2 + iǫ Z − 3.1.2 Fermionic Path Integral and Grassmann Numbers Let us now expand the spinor ﬁeld as

3 1 d p i ψ(x0, ~x)= χ(x0, ~p)e ~ ~p~x. (3.11) ~ (2π~)3 Z 1There was a serious (well not really serious) error in our computation of the scalar propagator in the ﬁrst semester which propagated to an error in the Dirac propoagtor. This must be corrected there and also in the previous chapter of this current semester in which we did not follow the factors of ~ and c properly. In any case the coeﬃcient ~/c apperaing in front of this propagator is now correct. YDRI QFT 59

The Lagrangian in terms of χ and χ+ is given by

L = d3x L Z = d3xψ¯(i~cγµ∂ mc2)ψ µ − Z c d3p = χ¯(x0, ~p)(i~γ0∂ γipi mc)χ(x0, ~p). (3.12) ~2 (2π~)3 0 − − Z We use the identity ~ω(~p) γ0(γipi + mc)χ(x0, ~p)= χ(x0, ~p). (3.13) c We get then 1 d3p L = χ+(x0, ~p)(i∂ ω(~p))χ(x0, ~p). (3.14) ~ (2π~)3 t − Z Using the box normalization the momenta become discrete and the measure d3~p/(2π~)3 becomes the sum /V . Thus the Lagrangian becomes with θ (t)= χ(x0, ~p)/√~V given ~p p R by P L = θ+(t)(i∂ ω(~p))θ (t). (3.15) p t − p X~p For a single momentum ~p the Lagrangian of the theory simpliﬁes to the single term

L = θ+(t)(i∂ ω(~p))θ (t). (3.16) p p t − p We will simplify further by thinking of θp(t) as a single component ﬁeld. The conjugate + variable is πp(t) = iθp (t). In the quantum theory we replace θp and πp with operators θˆp and πˆp. The canonical commutation relations are

θˆ , πˆ = i~, θˆ , θˆ = πˆ , πˆ = 0. (3.17) { p p} { p p} { p p} There several remarks here: • In the limit ~ 0, the operators reduce to ﬁelds which are anticommuting clas- −→ sical functions. In other words even classical fermion ﬁelds must be represented by anticommuting numbers which are known as Grassmann numbers.

• There is no eigenvalues of the operators θˆp and πˆp in the set of complex numbers except 0. The non-zero eigenvalues must be therefore anticommuting Grassmann numbers. • Obviously given two anticommuting Grassmann numbers α and β we have immediately the following fundamental properties

αβ = βα, α2 = β2 = 0. (3.18) − 60 YDRI QFT

The classical equation of motion following from the Lagrangian Lp is i∂tθp = ω(~p)θp. An immediate solution is given by θˆ (t)= ˆb exp( iω(~p)t). (3.19) p p − Thus ˆb , ˆb+ = ~, ˆb , ˆb = ˆb+, ˆb+ = 0. (3.20) { p p } { p p} { p p } The Hilbert space contains two states 0 > (the vacuum) and 1 >= ˆb+ 0 > (the only | | p | exited state). Indeed we clearly have ˆb+ 1 >= 0 and ˆb 1 >= ~ 0 >. We deﬁne the p | p| | (coherent) states at time t = 0 by

ˆ+ + ˆ θ (0) >= ebp θp(0) 0 >, <θ (0) = eθ (0)bp < 0 . (3.21) | p | p | | The number θp(0) must be anticommuting Grassmann number, i.e. it must satisfy 2 + θp(0) = 0 whereas the number θ (0) is the complex conjugate of θp(0) which should be taken as independent and hence (θ+(0))2 = 0 and θ+(0)θ (0) = θ (0)θ+(0). We p p p − p p compute immediately that θˆ (0) θ (0) >= θ (0) θ (0) >, <θ (0) θˆ (0)+ =<θ (0) θ+(0). (3.22) p | p p | p p | p p | p The Feynman propagator for the ﬁeld θp(t) is deﬁned by ′ ′ S(t t )=< 0 T (θˆ (t)θˆ+(t )) 0 > . (3.23) − | p p | We compute immediately (with ǫ> 0) 2

′ 0 ′ ′ dp i i 0 ′ S(t t )= ~e−iω(~p)(t−t ) ~2 e− ~ p (t−t ) , t>t . (3.24) − ≡ 2π~ p0 ~ω(~p)+ iǫ Z −

0 ′ ′ dp i i 0 ′ S(t t ) = 0 ~2 e− ~ p (t−t ) , t

• The integral of f(θ) is therefore

dθf(θ)= dθ(A + Bθ). (3.27) Z Z We demand that this integral is invariant under the shift θ θ + η. This leads −→ immediately to the so-called Berezin integration rules

dθ = 0 , dθθ = 1. (3.28) Z Z 2Exercise: Verify this result using the residue theorem. YDRI QFT 61

• The diﬀerential dθ anticommutes with θ, viz

dθθ = θdθ. (3.29) − • We have immediately

dθdηηθ = 1. (3.30) Z • The most general function of two anticommuting Grassmann numbers θ and θ+ is

f(θ,θ+)= A + Bθ + Cθ+ + Dθ+θ. (3.31)

• Given two anticommuting Grassmann numbers θ and η we have

(θη)+ = η+θ+. = θ+η+. (3.32) − • We compute the integrals

+ dθ+dθe−θ bθ = dθ+dθ(1 θ+bθ)= b. (3.33) − Z Z

+ dθ+dθθθ+e−θ bθ = dθ+dθθθ+(1 θ+bθ) = 1. (3.34) − Z Z It is instructive to compare the ﬁrst integral with the bosonic integral

+ 2π dz+dze−z bz = . (3.35) b Z • We consider now a general integral of the form

+ + dθi dθif(θ ,θ). (3.36) Z Yi ′ + Consider the unitary transformation θi θi = Uijθj where U U = 1. It is rather 3 −→ obvious that

′ dθi = detU dθi. (3.37) Yi Yi ′ + ′ + + Hence i dθi dθi = i dθi dθi since U U = 1. On the other hand, by expanding the function f(θ+,θ) and integrating out we immediately see that the only non-zero Q Q + term will be exactly of the form i θi θi which is also invariant under the unitary transformation U. Hence Q ′+ ′ ′+ ′ + + dθi dθif(θ ,θ )= dθi dθif(θ ,θ). (3.38) Z Yi Z Yi 3Exercise: Verify this fact. 62 YDRI QFT

• Consider the above integral for

+ f(θ+,θ)= e−θ Mθ. (3.39)

M is a Hermitian matrix. By using the invariance under U(N) we can diagonalize the matrix M without changing the value of the integral. The eigenvalues of M are denoted mi. The integral becomes

+ + + −θ Mθ + −θi miθi dθi dθie = dθi dθie = mi = detM. (3.40) Z Yi Z Yi Yi Again it is instructive to compare with the bosonic integral

n + (2π) dz+dz e−z Mz = . (3.41) i i detM Z Yi • We consider now the integral

+ −θ+Mθ−θ+η−η+θ + −(θ++η+M −1)M(θ+M −1η)+η+M −1η dθi dθie = dθi dθie Z i Z i Y Y + −1 = detMeη M η. (3.42)

• Let us consider now the integral

+ + −θ+Mθ δ δ + −θ+Mθ−θ+η−η+θ dθi dθiθkθl e = + dθi dθie δη δη + Z i l k Z i η=η =0 Y −1 Y = detM(M )kl. (3.43)

In the above equation we have always to remember that the order of diﬀerentials and variables is very important since they are anticommuting objects.

• In the above equation we observe that if the matrix M has eigenvalue 0 then the result is 0 since the determinant vanishes in this case.

′ We go back now to our original problem. We want to express the propagator S(t t )=< ′ − 0 T (θˆ (t)θˆ+(t )) 0 > as a path integral over the classical fields θ (t) and θ+(t) which must | p p | p p be complex anticommuting Grassmann numbers. By analogy with what happens in scalar field theory we expect the path integral to be a functional integral of the probability amplitude exp(iSp/~) where Sp is the action Sp = dtLp over the classical fields θp(t) + and θp (t) (which are taken to be complex anticommuting Grassmann numbers instead of R + ordinary complex numbers). In the presence of sources ηp(t) and ηp (t) this path integral reads

i i i Z[η , η+]= θ+ θ exp dtθ+(i∂ ω(~p))θ + dtη+θ + dtθ+η (3.44). p p D p D p ~ p t − p ~ p p ~ p p Z Z Z Z YDRI QFT 63

By using the result (3.42) we know immediately that

+ −1 i i i Z[η , η+] = detMeη M η , M = (i∂ ω(~p)) , η = η , η+ = η+. (3.45) p p −~ t − −~ p −~ p

In other words

′ ′ ′ 1 + −1 + − ~2 R dt R dt ηp (t)M (t,t )ηp(t ) Z[ηp, ηp ] = detMe . (3.46)

From one hand we have

′ θ+ θ θ (t)θ+(t )exp i dtθ+(i∂ ω(~p))θ ~ 2 1 δ2 D p D p p p ~ p t − p ′ + Z = i Z δηp(t )δηp (t) η =η+=0 R + i +R p p θp θ exp dtθp (i∂ ω(~p))θ D D p ~ t − p ′ <θ (tR)θ+(t ) > . R (3.47) ≡ p p From the other hand

~ 2 2 1 δ −1 ′ ′ + Z = M (t,t ). (3.48) i Z δη (t )δηp (t) + p ηp=ηp =0 Therefore

+ ′ −1 ′ <θp(t)θp (t ) >= M (t,t ). (3.49)

We have

′ i ′ M(t,t ) = (i∂ ω(~p))δ(t t ) −~ t − − 0 0 ′ 1 dp p ω(~p) i 0 = − e− ~ p (t−t ). (3.50) ~2 2π~ i Z The inverse is therefore given by

0 ′ ′ ′ dp i i 0 <θ (t)θ+(t ) >= M −1(t,t )= ~2 e− ~ p (t−t ). (3.51) p p 2π~ p0 ω(~p)+ iǫ Z − We conclude therefore that

+ + ′ i + θp θp θp(t)θp (t )exp ~ dtθp (i∂t ω(~p))θp ˆ ˆ+ ′ D D − < 0 T (θp(t)θp (t )) 0 >= . (3.52) | | R + i +R θp θ exp dtθp (i∂ ω(~p))θ D D p ~ t − p R R 64 YDRI QFT

3.1.3 The Electron Propagator We are now ready to state our main punch line. The path integral of a free Dirac ﬁeld in the presence of non-zero sources must be given by the functional integral i i i Z[η, η¯]= ψ¯ ψ exp S [ψ, ψ¯]+ d4xηψ¯ + d4xψη¯ . (3.53) D D ~ 0 ~ ~ Z Z Z

S [ψ, ψ¯]= d4xψ¯(i~cγµ∂ mc2)ψ. (3.54) 0 µ − Z The Dirac spinor ψ and its Dirac conjugate spinor ψ¯ = ψ+γ0 must be treated as independent complex spinors with components which are Grassmann-valued functions of x. Indeed by taking χ (x) to be an orthonormal basis of 4 component Dirac spinors i − (for example it can be constructed out of the ui(p) and vi(p) in an obvious way) we can ¯ ¯ + expand ψ and ψ as ψ = i θiχi(x) and ψi = i θi χ¯i respectively. The coefficients θi and θ+ must then be complex Grassmann numbers. The measure appearing in the above i P P integral is therefore ψ¯ ψ = θ+ θ (3.55) D D D i D i Yi The path integral Z[η, η¯] is the generating functional of all correlation functions of the fields ψ and ψ¯. Indeed we have ¯ ¯ ¯ i ¯ ψ ψ ψα1 (x1)...ψαn (xn)ψβ1 (y1)...ψβn (yn)exp ~ S0[ψ, ψ] < ψα (x1)...ψα (xn)ψ¯β (y1)...ψ¯β (yn) > D D . 1 n 1 n ≡ ψ¯ ψ exp i S [ψ, ψ¯] R D D ~ 0 ~2n δ2nZ[η, η¯] = R . Z[η, η¯] δη¯ (x )...δη¯ (x )δη (y )...δη (y ) α1 1 α1 1 β1 1 β1 1 η=¯η=0 (3.56) For example the 2 point function is given by − ~2 δ2Z[η, η¯] < ψ (x)ψ¯ (y) > . (3.57) α β ≡ Z[η, η¯] δη¯ (x)δη (y) α β η=¯η=0 However by comparing the path integral Z[η, η¯] with the path integral (3.42) we can make the identification i i i M (i~cγµ∂ mc2) δ4(x y) , η η , η+ η¯ . (3.58) ij −→ −~ µ − αβ − i −→ −~ α i −→ −~ α We define i M (x,y) = (i~cγµ∂ mc2) δ4(x y) αβ −~ µ − αβ − 4 ic d p i = (γµp mc) e− ~ p(x−y) − ~ (2π~)4 µ − αβ Z 4 2 2 2 c d p p m c i = − e− ~ p(x−y). (3.59) i~ (2π~)4 γµp + mc Z µ αβ YDRI QFT 65

By using equation (3.42) we can deduce immediately the value of the path integral Z[η, η¯]. We ﬁnd 1 Z[η, η¯] = detM exp d4x d4yη¯ (x)M −1(x,y)η (y) . (3.60) − ~2 α αβ β Z Z Hence the electron propagator is

¯ −1 < ψα(x)ψβ(y) >= Mαβ (x,y). (3.61)

From the form of the Laplacian (3.59) we get immediately the propagator (including also an appropriate Feynman prescription)

4 µ ~ d p (γ pµ + mc)αβ i < ψ (x)ψ¯ (y) > = i e− ~ p(x−y). (3.62) α β c (2π~)4 p2 m2c2 + iǫ Z − 3.2 Free Abelian Vector Field

3.2.1 Maxwell’s Action The electric and magnetic ﬁelds E~ and B~ generated by a charge density ρ and a current density J~ are given by the Maxwell’s equations written in the Heaviside-Lorentz system as

~ E~ = ρ , Gauss′ s Law. (3.63) ∇

~ B~ = 0 , No Magnetic Monopole Law. (3.64) ∇ −

1 ∂B~ ~ E~ = , Faraday′ s Law. (3.65) ∇× − c ∂t

1 ∂E~ ~ B~ = (J~ + ) , Ampere Maxwell′ s Law. (3.66) ∇× c ∂t − The Lorentz force law expresses the force exerted on a charge q moving with a velocity ~u in the presence of an electric and magnetic ﬁelds E~ and B~ . This is given by

1 F~ = q(E~ + ~u B~ ). (3.67) c × The continuity equation expresses local conservation of the electric charge. It reads

∂ρ + ~ J~ = 0. (3.68) ∂t ∇ 66 YDRI QFT

The so-called ﬁeld strength tensor is a second-rank antisymmetric tensor Fµν deﬁned by

0 E E E − x − y − z Ex 0 Bz By F µν =  −  . (3.69) E B 0 B y z − x  E B B 0   z − y x    The dual ﬁeld strength tensor is also a second-rank antisymmetric tensor F˜µν deﬁned by

0 B B B − x − y − z B 0 E E 1 F˜µν =  x z − y  = ǫµναβF . (3.70) B E 0 E 2 αβ y − z x  B E E 0   z y − x    In terms of Fµν and F˜µν Maxwell’s equations will take the form 1 ∂ F µν = J ν , ∂ F˜µν = 0. (3.71) µ c µ The 4 vector current density J µ is given by J µ = (cρ, J , J , J ). The ﬁrst equation − x y z yields Gauss’s and Ampere-Maxwell’s laws whereas the second equation yields Maxwell’s third equation ~ B~ = 0 and Faraday’s law. The continuity equation and the Lorentz ∇ force law respectively can be rewritten in the covariant forms

µ ∂µJ = 0. (3.72)

q dx Kµ = ν F µν . (3.73) c dτ

The electric and magnetic ﬁelds E~ and B~ can be expressed in terms of a scalar potential V and a vector potential A~ as

B~ = ~ A.~ (3.74) ∇×

1 ∂A~ E~ = (~ V + ). (3.75) − c ∇ ∂t We construct the 4 vector potential Aµ as − Aµ = (V/c, A~). (3.76)

The ﬁeld tensor Fµν can be rewritten in terms of Aµ as

F = ∂ A ∂ A . (3.77) µν µ ν − ν µ YDRI QFT 67

This equation is actually equivalent to the two equations (3.74) and (3.75). The ho- µν mogeneous Maxwell’s equation ∂µF˜ = 0 is automatically solved by this ansatz. The µν ν inhomogeneous Maxwell’s equation ∂µF = J /c becomes 1 ∂ ∂µAν ∂ν∂ Aµ = J ν. (3.78) µ − µ c These equations of motion should be derived from a local Lagrangian density , i.e. a L Lagrangian which depends only on the ﬁelds and their ﬁrst derivatives at the point ~x. Indeed it can be easily proven that the above equations of motion are the Euler-Lagrange equations of motion corresponding to the Lagrangian density 1 1 = F F µν J Aµ. (3.79) L −4 µν − c µ The free Maxwell’s action is 1 S [A] = d4xF F µν . (3.80) 0 −4 µν Z The total Maxwell’s action will include a non-zero source and is given by 1 1 S[A] = d4xF F µν d4xJ Aµ. (3.81) −4 µν − c µ Z Z 3.2.2 Gauge Invariance and Canonical Quantization We have a gauge freedom in choosing Aµ given by local gauge transformations of the form (with λ any scalar function)

′ Aµ A µ = Aµ + ∂µΛ. (3.82) −→ Indeed under this transformation we have

′ F µν F µν = F µν . (3.83) −→ These local gauge transformations form a (gauge) group. In this case the group is just the abelian U(1) unitary group. The invariance of the theory under these transformations is termed a gauge invariance. The 4 vector potential Aµ is called a gauge potential or a − gauge field. We make use of the invariance under gauge transformations by working with a gauge potential Aµ which satisfies some extra conditions. This procedure is known as gauge fixing. Some of the gauge conditions so often used are

µ ∂µA = 0 , Lorentz Gauge. (3.84)

i ∂iA = 0 , Coulomb Gauge. (3.85)

A0 = 0 , Temporal Gauge. (3.86) 68 YDRI QFT

A3 = 0 , Axial Gauge. (3.87)

A0 + A1 = 0 , Light Cone Gauge. (3.88)

The form of the equations of motion (3.78) strongly suggest we impose the Lorentz condition. In the Lorentz gauge the equations of motion (3.78) become 1 ∂ ∂µAν = J ν. (3.89) µ c

µ ′µ µ µ µ Clearly we still have a gauge freedom A A = A + ∂ φ where ∂µ∂ φ = 0. In µ −→ µ ′µ other words if A satisfies the Lorentz gauge ∂µA = 0 then A will also satisfy the ′µ µ Lorentz gauge, i.e. ∂µA = 0 iff ∂µ∂ φ = 0. This residual gauge symmetry can be fixed by imposing another condition such as the temporal gauge A0 = 0. We have therefore 2 constraints imposed on the components of the gauge potential Aµ which means that only two of them are really independent. The underlying mechanism for the reduction of the number of degrees of freedom is actually more complicated than this simple counting. We incorporate the Lorentz condition via a Lagrange multiplier ζ, i.e. we add to µ 2 the Maxwell’s Lagrangian density a term proportional to (∂ Aµ) in order to obtain a gauge-fixed Lagrangian density, viz 1 1 1 = F F µν ζ(∂µA )2 J Aµ. (3.90) Lζ −4 µν − 2 µ − c µ The added extra term is known as a gauge-fixing term. The conjugate fields are

δ ζ ζ π0 = L = (∂0A0 ∂iAi). (3.91) δ∂tA0 − c −

δ ζ 1 πi = L = (∂0Ai ∂iA0). (3.92) δ∂tAi c − We remark that in the limit ζ 0 the conjugate ﬁeld π vanishes and as a consequence −→ 0 canonical quantization becomes impossible. The source of the problem is gauge invariance which characterize the limit ζ 0. For ζ = 0 canonical quantization (although a very −→ 6 involved exercise) can be carried out consistently. We will not do this exercise here but only quote the result for the 2 point function. The propagator of the photon ﬁeld in a − general gauge ζ is given by the formula (with ~ = c = 1)

iDµν (x y) = < 0 T Aˆµ (x)Aˆν (y) 0 > F − | in in | d4p i 1 pµpν = ηµν + (1 ) exp( ip(x y)).(3.93) (2π)4 p2 + iǫ − − ζ p2 − − Z In the following we will give a derivation of this fundamental result based on the path integral formalism. YDRI QFT 69

3.2.3 Path Integral Quantization and the Faddeev-Popov Method The starting point is to posit that the path integral of an Abelian vector ﬁeld Aµ in the presence of a source J µ is given by analogy with the scalar ﬁeld by the functional integral (we set ~ = c = 1)

Z[J] = A exp iS[A] D µ µ Z Y i = A exp d4xF F µν i d4xJ Aµ . (3.94) D µ − 4 µν − µ µ Z Y Z Z This is the generating functional of all correlation functions of the ﬁeld Aµ(x). This is clear from the result

µ1 µ2 AµA (x1)....A (x2)exp iS0[A] < Aµ1 (x )....Aµ2 (x ) > µ D 1 2 ≡ A exp iS [A] R Q µ D µ 0 in δnZ[J] = R Q . (3.95) Z[J] δJ (x )...δJ (x ) µ1 1 µn n J=0 The Maxwell’s action can be rewritten as 1 S [A] = d4xF F µν 0 −4 µν Z 1 = d4xA (∂ ∂µηνλ ∂ν∂λ)A . (3.96) 2 ν µ − λ Z We Fourier transform Aµ(x) as

d4k A (x)= A˜ (k)eikx. (3.97) µ (2π)4 µ Z Then 1 d4k S [A] = A˜ (k)( k2ηνλ + kνkλ)A˜ ( k). (3.98) 0 2 (2π)4 ν − λ − Z

We observe that the action is 0 for any configuration of the form A˜µ(k)= kµf(k). Thus we conclude that the so-called pure gauge configurations given by Aµ(x) = ∂µΛ(x) are zero modes of the Laplacian which means in particular that the Laplacian can not be inverted. More importantly this means that in the path integral Z[J] these zero modes (which are equivalent to Aµ = 0) are not damped and thus the path integral is divergent. This happens for any other configuration Aµ. Indeed all gauge equivalent configurations Λ Aµ = Aµ + ∂µΛ have the same probability amplitude and as a consequence the sum of their contributions to the path integral will be proportional to the divergent integral over the Abelian U(1) gauge group which is here the integral over Λ. The problem lies therefore in gauge invariance which must be fixed in the path integral. This entails the 70 YDRI QFT

Λ selection of a single gauge configuration from each gauge orbit Aµ = Aµ + ∂µΛ as a representative and using it to compute the contribution of the orbit to the path integral. In path integral quantization gauge fixing is done in an elegant and efficient way via the method of Faddeev and Popov. Let us say that we want to gauge fix by imposing the Lorentz condition G(A)= ∂ Aµ ω = 0. Clearly G(AΛ)= ∂ Aµ ω + ∂ ∂µΛ and µ − µ − µ thus

Λδ(G(AΛ)) = Λδ(∂ Aµ ω + ∂ ∂ Λ). (3.99) D D µ − µ µ Z Z ′ µ By performing the change of variables Λ Λ = ∂µ∂ Λ and using the fact that ′ ′ −→ Λ = (∂Λ /∂Λ) Λ = det(∂ ∂µ) Λ we get D | |D µ D ′ Λ ′ 1 Λδ(G(AΛ)) = D δ(∂ Aµ ω + Λ )= . (3.100) D det(∂ ∂µ) µ − det(∂ ∂µ) Z Z µ µ This can be put in the form

δG(AΛ) Λδ G(AΛ) det = 1. (3.101) D δΛ Z This is the generalization of

(n) ∂gi daiδ ~g(~a) det = 1. (3.102) ∂aj Z Yi We insert 1 in the form (3.101) in the path integral as follows

δG(AΛ) Z[J] = A Λδ G(AΛ) det exp iS[A] D µ D δΛ µ Z Y Z = det(∂ ∂µ) Λ A δ G(AΛ) exp iS[A] µ D D µ µ Z Z Y = det(∂ ∂µ) Λ AΛδ G(AΛ) exp iS[AΛ]. (3.103) µ D D µ µ Z Z Y Now we shift the integration varaible as AΛ A . We observe immediately that the µ −→ µ integral over the U(1) gauge group decouples, viz

Z[J] = det(∂ ∂µ) Λ A δ G(A) exp iS[A] µ D D µ Z Z µ Y = det(∂ ∂µ) Λ A δ ∂ Aµ ω exp iS[A]. (3.104) µ D D µ µ − Z Z µ Y YDRI QFT 71

Next we want to set ω = 0. We do this in a smooth way by integrating both sides of the above equation against a Gaussian weighting function centered around ω = 0, viz

ω2 ω2 ω exp( i d4x )Z[J] = det(∂ ∂µ) Λ A ω exp( i d4x )δ ∂ Aµ ω exp iS[A D − 2ξ µ D D µ D − 2ξ µ − Z Z Z Z µ Z Z Y (∂ Aµ)2 = det(∂ ∂µ) Λ A exp( i d4x µ )exp iS[A]. (3.105) µ D D µ − 2ξ Z Z µ Z Y Hence (∂ Aµ)2 Z[J] = A exp( i d4x µ )exp iS[A] N D µ − 2ξ µ Z Y Z (∂ Aµ)2 i = A exp i d4x µ d4xF F µν i d4xJ A(3.106)µ . N D µ − 2ξ − 4 µν − µ µ Z Y Z Z Z µ 2 Therefore the end result is the addition of a term proportional to (∂µA ) to the action which ﬁxes gauge invariance to a suﬃcient degree.

3.2.4 The Photon Propagator The above path integral can also be put in the form i 1 Z[J] = A exp d4xA ∂ ∂µηνλ + ( 1)∂ν ∂λ A i d4xJ Aµ . N D µ 2 ν µ ξ − λ − µ µ Z Y Z Z (3.107)

We use the result n n 1 −1 1 −1 1 −1 −ziMij zj −ziji jiM jj −(zi+ jkM )Mij (zj + M jk) dzie = e 4 ij dzie 2 ki 2 jk Z i=1 Z i=1 Y Yn 1 −1 jiM jj −yiMij yj = e 4 ij dyie Z i=1 Yn 1 −1 jiM jj −ximixj = e 4 ij dxie Z i=1 n Y − 1 j M 1j π = e 4 i ij j m i=1 r i − Y 1 j M 1j n − 1 = e 4 i ij j π 2 (detM) 2 . (3.108)

By comparison we have i 1 M ∂ ∂µηνλ + ( 1)∂ν ∂λ δ4(x y) , j iJ . (3.109) ij −→ −2 µ ξ − − i −→ µ 72 YDRI QFT

We deﬁne i 1 M νλ(x,y) = ∂ ∂µηνλ + ( 1)∂ν ∂λ δ4(x y) −2 µ ξ − − i d4k 1 = (k2ηνλ + ( 1)kν kλ)eik(x−y). (3.110) 2 (2π)4 ξ − Z Hence our path integral is actually given by

n − 1 1 4 4 µ −1 ν Z[J] = π 2 (detM) 2 exp d x d yJ (x)M (x,y)J (y) N − 4 µν Z Z ′ 1 = exp d4x d4yJ µ(x)M −1(x,y)J ν (y) . (3.111) N − 4 µν Z Z The inverse of the Laplacian is deﬁned by

d4yM νλ(x,y)M −1(y, z)= ην δ4(x y). (3.112) λµ µ − Z For example the 2 point function is given by − i2 δ2Z[J] < A (x)A (y) > = µ ν Z[J] δJ µ(x)δJ ν (y) J=0 1 = M −1(x,y). (3.113) 2 µν It is not diﬃcult to check that the inverse is given by

d4k 1 k k M −1(x,y) = 2i (η (1 ξ) µ ν )eik(x−y). (3.114) µν − (2π)4 k2 + iǫ µν − − k2 Z Hence the propagator is

d4k i k k < A (x)A (y) > = − (η (1 ξ) µ ν )eik(x−y). (3.115) µ ν (2π)4 k2 + iǫ µν − − k2 Z The most important gauges we will make use of are the Feynman gauge ξ = 1 and the Landau gauge ξ = 0.

3.3 Gauge Interactions

3.3.1 Spinor and Scalar Electrodynamics: Minimal Coupling The actions of a free Dirac ﬁeld and a free Abelian vector ﬁeld in the presence of sources are given by (with ~ = c = 1)

S[ψ, ψ¯]= d4xψ¯(iγµ∂ m)ψ + d4x(ψη¯ +ηψ ¯ ). (3.116) µ − Z Z YDRI QFT 73

1 S[A] = d4xF F µν d4xJ Aµ. (3.117) −4 µν − µ Z Z The action S[A] gives Maxwell’s equations with a vector current source equal to the external vector current J µ. As we have already discussed the Maxwell’s action (J µ = 0) is invariant under the gauge symmetry transformations

A AΛ = A + ∂ Λ. (3.118) µ −→ µ µ µ The action S[A] is also invariant under these gauge transformations provided the vector µ µ current J is conserved, viz ∂µJ = 0. The action describing the interaction of a photon which is described by the Abelian vector ﬁeld Aµ and an electron described by the Dirac ﬁeld ψ must be given by

S[ψ, ψ,¯ A] = S[ψ, ψ¯]+ S[A] d4xj Aµ. (3.119) − µ Z The interaction term j Aµ is dictated by the requirement that this action must also give − µ Maxwell’s equations with a vector current source equal now to the sum of the external vector current J µ and the internal vector current jµ. The internal vector current jµ must clearly depend on the spinor ﬁelds ψ and ψ¯ and furthermore it must be conserved. In order to ensure that jµ is conserved we will identify it with the Noether’s current associated with the local symmetry transformations

ψ ψΛ = exp( ieΛ)ψ , ψ¯ ψ¯Λ = ψ¯ exp(ieΛ). (3.120) −→ − −→ Indeed under these local transformations the Dirac action transforms as

S[ψ, ψ¯] S[ψΛ, ψ¯Λ] = S[ψ, ψ¯] e d4xΛ∂ (ψγ¯ µψ). (3.121) −→ − µ Z The internal current jµ will be identiﬁed with eψγ¯ µψ, viz

jµ = eψγ¯ µψ. (3.122)

This current is clearly invariant under the local transformations (3.120). By performing the local transformations (3.118) and (3.120) simultaneously, i.e. by considering the transformations (3.120) a part of gauge symmetry, we obtain the invariance of the action S[ψ, ψ,¯ A]. The action remains invariant under the combined transformations (3.118) and (3.120) if we also include a conserved external vector current source J µ and Grassmann spinor sources η and η¯ which transform under gauge transformations as the dynamical Dirac spinors, viz η ηΛ = exp( ieΛ)η and η¯ η¯Λ =η ¯ exp(ieΛ). We write this −→ − −→ result as (with S [ψ, ψ,¯ A] S[ψ, ψ,¯ A]) η,η,J¯ ≡

Λ Λ Λ S [ψ, ψ,¯ A] S Λ Λ [ψ , ψ¯ , A ]= S [ψ, ψ,¯ A]. (3.123) η,η,J¯ −→ η ,η¯ ,J η,η,J¯ 74 YDRI QFT

The action S[ψ, ψ,¯ A] with the corresponding path integral deﬁne quantum spinor electrodynamics which is the simplest and most important gauge interaction. The action S[ψ, ψ,¯ A] can also be put in the form

1 S[ψ, ψ,¯ A] = d4xψ¯(iγµ m)ψ d4xF F µν + d4x(ψη¯ +ηψ ¯ ) d4xJ Aµ. ∇µ − − 4 µν − µ Z Z Z Z (3.124)

The derivative operator which is called the covariant derivative is given by ∇µ = ∂ + ieA . (3.125) ∇µ µ µ The action S[ψ, ψ,¯ A] could have been obtained from the free action S[ψ, ψ¯]+ S[A] by making the simple replacement ∂ which is known as the principle of minimal µ −→ ∇µ coupling. In ﬂat Minkowski spacetime this prescription always works and it allows us to obtain the most minimal consistent interaction starting from a free theory. As another example consider complex quartic scalar ﬁeld given by the action

g S[φ, φ+]= d4x ∂ φ+∂µφ m2φ+φ (φ+φ)2 . (3.126) µ − − 4 Z

By applying the principle of minimal coupling we replace the ordinary ∂µ by the covariant derivative = ∂ + ieA and then add the Maxwell’s action. We get immediately the ∇µ µ µ gauge invariant action

g 1 S[φ, φ+, A]= d4x φ+ µφ m2φ+φ (φ+φ)2 d4xF F µν . (3.127) ∇µ ∇ − − 4 − 4 µν Z Z This is indeed invariant under the local gauge symmetry transformations acting on Aµ, φ and φ+ as

A AΛ = A + ∂ Λ , φ exp( ieΛ)φ , φ+ φ+ exp(ieΛ). (3.128) µ −→ µ µ µ −→ − −→ It is not diﬃcult to add vector and scalar sources to the action (3.126) without spoil- ing gauge invariance. The action (3.126) with the corresponding path integral deﬁne quantum scalar electrodynamics which describes the interaction of the photon Aµ with a charged scalar particle φ whose electric charge is q = e. − 3.3.2 The Geometry of U(1) Gauge Invariance The set of all gauge transformations which leave invariant the actions of spinor and scalar electrodynamics form a group called U(1) and as a consequence spinor and scalar electrodynamics are said to be invariant under local U(1) gauge symmetry. The group U(1) is the group of 1 1 unitary matrices given by × U(1) = g = exp( ieΛ), Λ . (3.129) { − ∀ } YDRI QFT 75

In order to be able to generalize the local U(1) gauge symmetry to local gauge symmetries based on other groups we will exhibit in this section the geometrical content of the gauge invariance of spinor electrodynamics. The starting point is the free Dirac action given by

S = d4xψ¯(iγµ∂ m)ψ. (3.130) µ − Z This is invariant under the global transformations

ψ e−ieΛψ , ψ¯ ψe¯ ieΛ. (3.131) −→ −→ We demand next that the theory must be invariant under the local transformations obtained by allowing Λ to be a function of x in the above equations, viz

ψ ψg = g(x)ψ , ψ¯ ψ¯g = ψg¯ +(x). (3.132) −→ −→ The fermion mass term is trivially still invariant under these local U(1) gauge transformations,i.e.

ψψ¯ ψ¯gψg = ψψ.¯ (3.133) −→ The kinetic term is not so easy. The difficulty clearly lies with the derivative of the field which transforms under the local U(1) gauge transformations in a complicated way. To appreciate more this difficulty let us consider the derivative of the field ψ in the direction defined by the vector nµ which is given by

µ ψ(x + ǫn) ψ(x) n ∂µψ = lim − , ǫ 0. (3.134) ǫ −→ The two fields ψ(x+ǫn) and ψ(x) transform under the local U(1) symmetry with different phases given by g(x+ǫn) and g(x) respectively. The point is that the fields ψ(x+ǫn) and ψ(x) since they are evaluated at different spacetime points x + ǫn and x they transform µ independently under the local U(1) symmetry. As a consequence the derivative n ∂µψ has no intrinsic geometrical meaning since it involves the comparison of fields at different spacetime points which transform independently of each other under U(1). In order to be able to compare fields ψ(y) and ψ(x) at different spacetime points y and x we need to introduce a new object which connects the two points y and x and which allows a meaningful comparison between ψ(y) and ψ(x). We introduce a comparator field U(y,x) which connects the points y and x along a particular path with the properties: • The comparator field U(y,x) must be an element of the gauge group U(1) and thus U(y,x) is a pure phase, viz

U(y,x) = exp( ieφ(y,x)) U(1). (3.135) − ∈ • Clearly we must have

U(x,x) = 1 φ(x,x) = 0. (3.136) ⇔ 76 YDRI QFT

• Under the U(1) gauge transformations ψ(x) ψg(x) = g(x)ψ(x) and ψ(y) −→ −→ ψg(y)= g(y)ψ(y) the comparator ﬁeld transforms as

U(y,x) U g(y,x)= g(y)U(y,x)g+(x). (3.137) −→ • We impose the restriction

U(y,x)+ = U(x,y). (3.138)

The third property means that the product U(y,x)ψ(x) transforms as

U(y,x)ψ(x) U g(y,x)ψg(x)= g(y)U(y,x)g+(x)g(x)ψ(x) = g(y)U(y,x)ψ(x)(3.139). −→ Thus U(y,x)ψ(x) transforms under the U(1) gauge group with the same group element as the field ψ(y). This means in particular that the comparison between U(y,x)ψ(x) and ψ(y) is meaningful. We are led therefore to define a new derivative of the field ψ in the direction defined by the vector nµ by

µ ψ(x + ǫn) U(x + ǫn,x)ψ(x) n µψ = lim − , ǫ 0. (3.140) ∇ ǫ −→ This is known as the covariant derivative of ψ in the direction nµ. The second property U(x,x) = 1 allows us to conclude that if the point y is inﬁnites- imally close to the point x then we can expand U(y,x) around 1. We can write for y = x + ǫn the expansion

U(x + ǫn,x) = 1 ieǫn Aµ(x)+ O(ǫ2). (3.141) − µ The coeﬃcient of the displacement vector y x = ǫn is a new vector ﬁeld Aµ which µ − µ µ is precisely, as we will see shortly, the electromagnetic vector potential. The coupling e will, on the other hand, play the role of the electric charge. We compute immediately

ψ = (∂ + ieA )ψ. (3.142) ∇µ µ µ Thus is indeed the covariant derivative introduced in the previous section. ∇µ By using the language of differential geometry we say that the vector field Aµ is a connection on a U(1) fiber bundle over spacetime which defines the parallel transport of the field ψ from x to y. The parallel transported field ψk is defined by

ψk(y)= U(y,x)ψ(x). (3.143)

The third property with a comparator U(y,x) with y inﬁnitesimally close to x, for example y = x + ǫn, reads explicitly

1 ieǫnµA (x) 1 ieǫnµAg (x)= g(y) 1 ieǫnµA (x) g+(x). (3.144) − µ −→ − µ − µ YDRI QFT 77

Equivalently we have

i Ag = gA g+ + ∂ g.g+ Ag = A + ∂ Λ. (3.145) µ µ e µ ⇔ µ µ µ Again we find the gauge field transformation law considered in the previous section. For completeness we find the transformation law of the covariant derivative of the field ψ. We have

ψ = (∂ + ieA )ψ ( ψ)g = (∂ + ieAg )ψg ∇µ µ µ −→ ∇µ µ µ −ieΛ = (∂µ + ieAµ + ie∂µΛ)(e ψ) −ieΛ = e (∂µ + ieAµ)ψ = g(x) ψ. (3.146) ∇µ Thus the covariant derivative of the ﬁeld transforms exactly in the same way as the ﬁeld itself. This means in particular that the combination ψiγ¯ µ ψ is gauge invariant. In ∇µ summary given the free Dirac action we can obtain a gauge invariant Dirac action by the simple substitution ∂ . This is the principle of minimal coupling discussed µ −→ ∇µ in the previous section. The gauge invariant Dirac action is

S = d4xψ¯(iγµ m)ψ. (3.147) ∇µ − Z We need ﬁnally to construct a gauge invariant action which provides a kinetic term for the vector ﬁeld Aµ. This can be done by integrating the comparator U(y,x) along a closed loop. For y = x + ǫn we write U(y,x) up to the order ǫ2 as

U(y,x) = 1 ieǫn Aµ + iǫ2X + O(ǫ3). (3.148) − µ The fourth fundamental property of U(y,x) restricts the comparator so that U(y,x)+ = U(x,y). This leads immediately to the solution X = en n ∂νAµ/2. Thus − µ ν ie U(y,x) = 1 ieǫn Aµ ǫ2n n ∂ν Aµ + O(ǫ3) − µ − 2 µ ν ǫ = 1 ieǫn Aµ(x + n)+ O(ǫ3) − µ 2 ǫ = exp( ieǫn Aµ(x + n)). (3.149) − µ 2 We consider now the group element U(x) given by the product of the four comparators associated with the four sides of a small square in the (1, 2) plane. This is given by − U(x)=trU(x,x + ǫ1)ˆ U(x + ǫ1ˆ,x + ǫ1+ˆ ǫ2)ˆ U(x + ǫ1+ˆ ǫ2ˆ,x + ǫ2)ˆ U(x + ǫ2ˆ,x).(3.150)

This is called the Wilson loop associated with the square in question. The trace tr is of course trivial for a U(1) gauge group. The Wilson loop U(x) is locally invariant under the 78 YDRI QFT gauge group U(1), i.e. under U(1) gauge transformations the Wilson loop U(x) behaves as U(x) U g(x)= U(x). (3.151) −→ The Wilson loop is the phase accumulated if we parallel transport the spinor ﬁeld ψ from the point x around the square and back to the point x. This phase can be computed explicitly. Indeed we have

ǫ ǫ ǫ ǫ U(x) = exp ieǫ A1(x + 1)+ˆ A2(x + ǫ1+ˆ 2)ˆ A1(x + 1+ˆ ǫ2)ˆ A2(x + 2)ˆ 2 2 − 2 − 2 = exp( ieǫ2F ) − 12 e2ǫ4 = 1 ieǫ2F F 2 + ... (3.152) − 12 − 2 12 In the above equation F = ∂ A ∂ A . We conclude that the field strength tensor 12 1 2 − 2 1 F = ∂ A ∂ A is locally gauge invariant under U(1) transformations. This is µν µ ν − ν µ precisely the electromagnetic field strength tensor considered in the previous section. The field strength tensor F = ∂ A ∂ A can also be obtained from the commu- µν µ ν − ν µ tator of the two covariant derivatives and acting on the spinor field ψ. Indeed we ∇µ ∇ν have [ , ]ψ = ie(∂ A ∂ A )ψ. (3.153) ∇µ ∇ν µ ν − ν µ Thus under U(1) gauge transformations we have the behavior [ , ]ψ g(x)[ , ]ψ. (3.154) ∇µ ∇ν −→ ∇µ ∇ν In other words [ , ] is not a differential operator and furthermore it is locally invariant ∇ν ∇ν under U(1) gauge transformations. This shows in a slightly different way that the field strength tensor Fµν is the fundamental structure which is locally invariant under U(1) gauge transformations. The field strength tensor Fµν can be given by the expressions 1 F = [ , ] = (∂ A ∂ A ). (3.155) µν ie ∇µ ∇ν µ ν − ν µ In summary we can conclude that any function of the vector field Aµ which depends on the vector field only through the field strength tensor Fµν will be locally invariant under U(1) gauge transformations and thus can serve as an action functional. By appealing to the requirement of renormalizability the only renormalizable U(1) gauge action in four dimensions (which also preserves P and T symmetries) is Maxwell’s action which is µ quadratic in Fµν and also quadratic in A . We get then the pure gauge action 1 S = d4xF F µν. (3.156) −4 µν Z The total action of spinor electrodynamics is therefore given by 1 S = d4xψ¯(iγµ m)ψ d4xF F µν. (3.157) ∇µ − − 4 µν Z Z YDRI QFT 79

3.3.3 Generalization: SU(N) Yang-Mills Theory We can now immediately generalize the previous construction by replacing the Abelian gauge group U(1) by a diﬀerent gauge group G which will generically be non-Abelian, i.e the generators of the corresponding Lie algebra will not commute. In this chapter we will be interested in the gauge groups G = SU(N). Naturally we will start with the ﬁrst non-trivial non-Abelian gauge group G = SU(2) which is the case considered originally by Yang and Mills. The group SU(2) is the group of 2 2 unitary matrices which have determinant equal × 1. This is given by

SU(2) = u , a, b = 1, ..., 2 : u+u = 1, detu = 1 . (3.158) { ab } The generators of SU(2) are given by Pauli matrices given by 0 1 0 i 1 0 σ1 = , σ2 = − , σ3 = . (3.159) 1 0 i 0 0 1 − Thus any element of SU(2) can be rewritten as σA u = exp( igΛ) , Λ= ΛA . (3.160) − 2 XA The group SU(2) has therefore 3 gauge parameters ΛA in contrast with the group U(1) which has only a single parameter. These 3 gauge parameters correspond to three orthogonal symmetry motions which do not commute with each other. Equivalently the generators of the Lie algebra su(2) of SU(2) (consisting of the Pauli matrices) do not commute which is the reason why we say that the group SU(2) is non-Abelian. The Pauli matrices satisfy the commutation relations σA σB σC [ , ]= if , f = ǫ . (3.161) 2 2 ABC 2 ABC ABC The SU(2) group element u will act on the Dirac spinor ﬁeld ψ. Since u is a 2 2 matrix × the spinor ψ must necessarily be a doublet with components ψa, a = 1, 2. The extra label a will be called the color index. We write ψ1 ψ = . (3.162) ψ2 We say that ψ is in the fundamental representation of the group SU(2). The action of an element u SU(2) is given by ∈ ψa (ψu)a = uabψb. (3.163) −→ XB We start from the free Dirac action

S = d4x ψ¯a(iγµ∂ m)ψa. (3.164) µ − a Z X 80 YDRI QFT

Clearly this is invariant under global SU(2) transformations, i.e. transformations g which do not depend on x. Local SU(2) gauge transformations are obtained by letting g depend on x. Under local SU(2) gauge transformations the mass term remains invariant whereas the kinetic term transforms in a complicated fashion as in the case of local U(1) gauge transformations. Hence as in the U(1) case we appeal to the principle of minimal coupling and replace the ordinary derivative nµ∂ with the covariant derivative nµ µ ∇µ which is deﬁned by

µ ψ(x + ǫn) U(x + ǫn,x)ψ(x) n µψ = lim − , ǫ 0. (3.165) ∇ ǫ −→ Since the spinor ﬁeld ψ is a 2 component object the comparator U(y,x) must be a 2 2 − × matrix which transforms under local SU(2) gauge transformations as

U(y,x) U g(y,x)= u(y)U(y,x)u+(x). (3.166) −→ In fact U(y,x) is an element of SU(2). We must again impose the condition that U(x,x) = 1. Hence for an infinitesimal separation y x = ǫn we can expand U(y,x) as − σA U(x + ǫn,x) = 1 igǫnµAA(x) + O(ǫ2). (3.167) − µ 2 A In other words we have three vector fields Aµ (x). They can be unified in a single object Aµ(x) defined by

σA A (x)= AA(x) . (3.168) µ µ 2 A We will call Aµ(x) the SU(2) gauge field whereas we will refer to Aµ (x) as the components of the SU(2) gauge field. Since Aµ(x) is 2 2 matrix it will carry two color indices × a and b in an obvious way. The components of the SU(2) gauge field in the fundamental µ representation of SU(2) are given by Aab(x). The color index is called the SU(2) funda- A mental index whereas the index A carried by the components Aµ (x) is called the SU(2) A adjoint index. In fact Aµ (x) are called the components of the SU(2) gauge field in the adjoint representation of SU(2). First by inserting the expansion U(x + ǫn,x) = 1 igǫnµAA(x)σA/2+ O(ǫ2) in the − µ definition of the covariant derivative we obtain the result σA ψ = (∂ + igAA )ψ. (3.169) ∇µ µ µ 2 The spinor U(x + ǫn,x)ψ(x) is the parallel transport of the spinor ψ from the point x to the point x + ǫn and thus by construction it must transform under local SU(2) gauge transformations in the same way as the spinor ψ(x+ǫn). Hence under local SU(2) gauge transformations the covariant derivative is indeed covariant, viz

ψ u(x) ψ. (3.170) ∇µ −→ ∇µ YDRI QFT 81

Next by inserting the expansion U(x + ǫn,x) = 1 igǫnµAA(x)σA/2+ O(ǫ2) in the − µ transformation law U(y,x) U g(y,x) = u(y)U(y,x)u+(x) we obtain the transforma- −→ tion law

i A Au = uA u+ + ∂ u.u+. (3.171) µ −→ µ µ g µ For inﬁnitesimal SU(2) transformations we have u = 1 igΛ. We get − A Au = A + ∂ Λ+ ig[A , Λ]. (3.172) µ −→ µ µ µ µ In terms of components we have

σC σC σC σC σA σB AC AuC = AC + ∂ ΛC + ig[AA , ΛB ] µ 2 −→ µ 2 µ 2 µ 2 µ 2 2 σC = AC + ∂ ΛC + igAAΛBif . (3.173) µ µ µ ABC 2 In other words

AuC = AC + ∂ ΛC gf AAΛB. (3.174) µ µ µ − ABC µ The spinor ﬁeld transforms under inﬁnitesimal SU(2) transformations as

ψ ψu = ψ igΛψ. (3.175) −→ − We can now check explicitly that the covariant derivative is indeed covariant, viz

ψ ( ψ)u = ψ igΛ ψ. (3.176) ∇µ −→ ∇µ ∇µ − ∇µ By applying the principle of minimal coupling to the free Dirac action (3.164) we replace the ordinary derivative ∂ ψa by the covariant derivative ( ) ψb. We obtain the µ ∇µ ab interacting action

S = d4x ψ¯a(iγµ( ) mδ )ψb. (3.177) ∇µ ab − ab Z Xa,b Clearly

σA ( ) = ∂ δ + igAA( ) . (3.178) ∇µ ab µ ab µ 2 ab This action is by construction invariant under local SU(2) gauge transformations. It provides obviously the free term for the Dirac field ψ as well as the interaction term between the SU(2) gauge field Aµ and the Dirac field ψ. There remains therefore to find an action which will provide the free term for the SU(2) gauge field Aµ. As opposed to the U(1) case the action which will provide a free term for the SU(2) gauge field Aµ will 82 YDRI QFT also provide extra interaction terms (cubic and quartic) which involve only Aµ. This is another manifestation of the non-Abelian structure of the SU(2) gauge group and it is generic to all other non-Abelian groups. By analogy with the U(1) case a gauge invariant action which depends only on Aµ can µ only depend on A through the field strength tensor Fµν . This in turn can be constructed from the commutator of two covariant derivatives. We have then

1 F = [ , ] µν ig ∇µ ∇ν = ∂ A ∂ A + ig[A , A ]. (3.179) µ ν − ν µ µ ν F is also a 2 2 matrix. In terms of components the above equation reads µν ×

σC σC σC σA σB F C = ∂ AC ∂ AC + ig[AA , AB ] µν 2 µ ν 2 − ν µ 2 µ 2 ν 2 σC = ∂ AC ∂ AC + igAAAB.if . (3.180) µ ν − ν µ µ ν ABC 2 Equivalently

F C = ∂ AC ∂ AC gf AAAB. (3.181) µν µ ν − ν µ − ABC µ ν The last term in the above three formulas is of course absent in the case of U(1) gauge theory. This is the term that will lead to novel cubic and quartic interaction vertices which µ involve only the gauge ﬁeld A . We remark also that although Fµν is the commutator of two covariant derivatives it is not a diﬀerential operator. Since ψ transforms as ∇µ ψ u ψ we conclude that ψ u ψ and hence ∇µ −→ ∇µ ∇µ∇ν −→ ∇µ∇ν F ψ uF ψ. (3.182) µν −→ µν This means in particular that

F F u = uF u+. (3.183) µν −→ µν µν This can be verified explicitly by using the finite and infinitesimal transformation laws A uA u+ + i∂ u.u+/g and A A + ∂ Λ+ ig[A , Λ]4. The infinitesimal form µ −→ µ µ µ −→ µ µ µ of the above transformation law is

F F u = F + ig[F , Λ]. (3.184) µν −→ µν µν µν In terms of components this reads

F C F uC = F C gf F A ΛB. (3.185) µν −→ µν µν − ABC µν 4Exercise: Verify this. YDRI QFT 83

Although the ﬁeld strength tensor Fµν is not gauge invariant its gauge transformation F uF u+ is very simple. Any function of F will therefore transform in the µν −→ µν µν same way as Fµν and as a consequence its trace is gauge invariant under local SU(2) µν transformations. For example trFµν F is clearly gauge invariant. By appealing again to the requirement of renormalizability the only renormalizable SU(2) gauge action in four dimensions (which also preserves P and T symmetries) must be quadratic in Fµν . µν The only candidate is trFµν F . We get then the pure gauge action

1 S = d4xtrF F µν . (3.186) −2 µν Z We note that Pauli matrices satisfy

σA σB 1 tr = δAB. (3.187) 2 2 2

Thus the above pure action becomes 5

1 S = d4xF C F µνC . (3.188) −4 µν Z This action provides as promised the free term for the SU(2) gauge ﬁeld Aµ but also it will provide extra cubic and quartic interaction vertices for the gauge ﬁeld Aµ. In other words this action is not free in contrast with the U(1) case. This interacting pure gauge theory is in fact highly non-trivial and strictly speaking this is what we should call Yang-Mills theory. The total action is the sum of the gauge invariant Dirac action and the Yang-Mills action. This is given by

1 S = d4x ψ¯a(iγµ( ) mδ )ψb d4xF C F µνC . (3.189) ∇µ ab − ab − 4 µν Z Xa,b Z The ﬁnal step is to generalize further to SU(N) gauge theory which is quite straightforward. The group SU(N) is the group of N N unitary matrices which have determinant × equal 1. This is given by

SU(N)= u , a, b = 1, ..., N : u+u = 1, detu = 1 . (3.190) { ab } The generators of SU(N) can be given by the so-called Gell-Mann matrices tA = λA/2. They are traceless Hermitian matrices which generate the Lie algebra su(N) of SU(N). There are N 2 1 generators and hence su(N) is an (N 2 1) dimensional vector space. − − − They satisfy the commutation relations

A B C [t ,t ]= ifABC t . (3.191)

5Exercise: Derive the equations of motion which follow from this action. 84 YDRI QFT

The non-trivial coeﬃcients fABC are called the structure constants. The Gell-Mann generators ta can be chosen such that 1 trtAtB = δAB. (3.192) 2 They also satisfy 1 1 tAtB = δAB + (d + if )tC . (3.193) 2N 2 ABC ABC

The coeﬃcients dABC are symmetric in all indices. They can be given by dABC = 2trtA tB,tC and they satisfy for example { } N 2 4 d d = − δ . (3.194) ABC ABD N CD For example the group SU(3) is generated by the 8 Gell-Mann 3 3 matrices tA = λA/2 × given by

0 1 0 0 i 0 1 0 0 − λ1 = 1 0 0 , λ2 = i 0 0 , λ3 = 0 1 0      −  0 0 0 0 0 0 0 0 0  0 0 1   0 0 i   0 0 0  − λ4 = 0 0 0 , λ5 = 0 0 0 , λ6 = 0 0 1       1 0 0 i 0 0 0 1 0  0 0 0   1 0 0   1 λ7 = 0 0 i , λ8 = 0 1 0 . (3.195)  −  √   0 i 0 3 0 0 2 −     The structure constants fABC and the totally symmetric coeﬃcients dABC are given in the case of the group SU(3) by 1 √3 f = 1 , f = f = f = f = f = f = , f = f = .(3.196) 123 147 − 156 246 257 345 − 367 2 458 678 2

1 d118 = d228 = d338 = d888 = − √3 1 d448 = d558 = d668 = d778 = −2√3 1 d = d = d = d = d = d = d = d = . (3.197) 146 157 − 247 256 344 355 − 366 − 377 2 Thus any ﬁnite element of the group SU(N) can be rewritten in terms of the Gell-Mann matrices tA = λA/2 as λA u = exp( igΛ) , Λ= ΛA . (3.198) − 2 XA YDRI QFT 85

The spinor field ψ will be an N component object. The SU(N) group element u will − act on the Dirac spinor field ψ in the obvious way ψ uψ. We say that the spinor field −→ transforms in the fundamental representation of the SU(N) gauge group. The covariant derivative will be defined by the same formula found in the SU(2) case after making the replacement σA λA, viz ( ) = ∂ δ + igAA(tA) (recall also that the range of −→ ∇µ ab µ ab µ ab the fundamental index a changes from 2 to N). The covariant derivative will transform 2 A covariantly under the SU(N) gauge group. There are clearly N 1 components Aµ of A A − A the SU(N) gauge field, i.e. Aµ = Aµ t . The transformation laws of Aµ and Aµ remain unchanged (only remember that the structure constants differ for different gauge groups). The field strength tensor Fµν will be given, as before, by the commutator of two covariant derivatives. All results concerning Fµν will remain intact with minimal changes involving the replacements σA λA, ǫ f (recall also that the range of the adjoint −→ ABC −→ ABC index changes from 3 to N 2 1). The total action will therefore be given by the same − formula (3.189). We will refer to this theory as quantum chromodynamics (QCD) with SU(N) gauge group whereas we will refer to the pure gauge action as SU(N) Yang-Mills theory.

3.4 Quantization and Renormalization at 1 Loop − 3.4.1 The Fadeev-Popov Gauge Fixing and Ghost Fields We will be interested ﬁrst in the SU(N) Yang-Mills theory given by the action

1 S[A] = d4xtrF F µν −2 µν Z 1 = d4xF A F µνA. (3.199) −4 µν Z The corresponding path integral is given by

Z = AA exp(iS[A]) (3.200) D µ Z Yµ,A This path integral is invariant under ﬁnite SU(N) gauge transformations given explicitly by

i A Au = uA u+ + ∂ u.u+. (3.201) µ −→ µ µ g µ

Also it is invariant under inﬁnitesimal SU(N) gauge transformations given explicitly by

A AΛ = A + ∂ Λ+ ig[A , Λ] A + [ , Λ]. (3.202) µ −→ µ µ µ µ ≡ µ ∇µ 86 YDRI QFT

Equivalently

AC AΛC = AC + ∂ ΛC gf AAΛB AC + [ , Λ]C . (3.203) µ −→ µ µ µ − ABC µ ≡ µ ∇µ As in the case of electromagnetism we must fix the gauge before we can proceed any further since the path integral is ill defined as it stands. We want to gauge fix by imposing the Lorentz condition G(A) = ∂ Aµ ω = 0. Clearly under infinitesimal µ − SU(N) gauge transformations we have G(AΛ)= ∂ Aµ ω + ∂ [ µ, Λ] and thus µ − µ ∇ δG(AΛ) Λδ(G(AΛ))det = Λδ(∂ Aµ ω + ∂ [ µ, Λ])det∂ [ µ,(3.204) ..]. D δΛ D µ − µ ∇ µ ∇ Z Z ′ µ By performing the change of variables Λ Λ = ∂µ[ , Λ] and using the fact that ′ ′ −→ ∇ Λ = (∂Λ /∂Λ) Λ = det(∂ [ µ, ..]) Λ we get D | |D µ ∇ D Λ ′ δG(A ) Λ ′ Λδ(G(AΛ))det = D δ(∂ Aµ ω + Λ )det(∂ [ µ, ..]) = 1. D δΛ det(∂ [ µ, ..]) µ − µ ∇ Z Z µ ∇ (3.205)

This can also be put in the form (with u near the identity)

δG(Au) δG(Au) uδ(G(Au))det = 1 , = ∂ [ µ, ..]. (3.206) D δu δu µ ∇ Z For a given gauge conﬁguration Aµ we deﬁne

∆−1(A)= uδ(G(Au)). (3.207) D Z v + + u uv Under a gauge transformation Aµ Aµ = vAµv + i∂µv.v /g we have Aµ Aµ = + + −→ −→ uvAµ(uv) + i∂µuv.(uv) /g and thus

′ ′ ∆−1(Av)= uδ(G(Auv)) = (uv)δ(G(Auv )) = u δ(G(Au ))=∆−1(A(3.208)). D D D Z Z Z In other words ∆−1 is gauge invariant. Further we can write

1= uδ(G(Au))∆(A). (3.209) D Z As we will see shortly we are interested in conﬁgurations Aµ which lie on the surface G(A) = ∂µA ω = 0. Thus only SU(N) gauge transformations u which are near the µ − identity are relevant in the above integral. Hence we conclude that (with u near the identity)

δG(Au) ∆(A) = det . (3.210) δu YDRI QFT 87

The determinant det(δG(Au)/δu) is gauge invariant and as a consequence is independent of u. The fact that this determinant is independent of u is also obvious from equation (3.206). We insert 1 in the form (3.209) in the path integral as follows

Z = AA uδ(G(Au))∆(A)exp iS[A] D µ D Z Yµ,A Z = u AAδ(G(Au))∆(A)exp iS[A] D D µ Z Z Yµ,A = u AuAδ(G(Au))∆(Au)exp iS[Au]. (3.211) D D µ Z Z Yµ,A Now we shift the integration varaible as Au A . The integral over the SU(N) gauge µ −→ µ group decouples an we end up with

Z = ( u) AAδ(G(A))∆(A)exp iS[A]. (3.212) D D µ Z Z Yµ,A Because of the delta function we are interested in knowing ∆(A) only for conﬁgurations Aµ which lie on the surface G(A) = 0. This means in particular that the gauge transformations u appearing in (3.209) must be close to the identity so that we do not go far from the surface G(A) = 0. As a consequence ∆(A) can be equated with the determinant det(δG(Au)/δu), viz

δG(Au) ∆(A) = det = det∂ [ µ, ..]. (3.213) δu µ ∇ In contrast with the case of U(1) gauge theory, here the determinant det(δG(Au)/δu) actually depends on the SU(N) gauge ﬁeld and hence it can not be taken out of the path integral. We have then the result

Z = ( u) AAδ(∂ Aµ ω)det∂ [ µ, ..]exp iS[A]. (3.214) D D µ µ − µ ∇ Z Z Yµ,A Clearly ω must be an N N matrix since Aµ is an N N matrix. We want to set ω = 0 × × by integrating both sides of the above equation against a Gaussian weighting function centered around ω = 0, viz

ω2 ω2 ω exp( i d4xtr )Z = ( u) AA ω exp( i d4xtr )δ(∂ Aµ ω)det∂ [ µ, ..]exp iS[ D − ξ D D µ D − ξ µ − µ ∇ Z Z Z Z Yµ,A Z Z (∂ Aµ)2 = ( u) AA exp( i d4xtr µ )det∂ [ µ, ..]exp iS[A]. (3.215) D D µ − ξ µ ∇ Z Z Yµ,A Z 88 YDRI QFT

The path integral of SU(N) Yang-Mills theory is therefore given by

(∂ Aµ)2 Z = AA exp( i d4xtr µ )det∂ [ µ, ..]exp iS[A]. (3.216) N D µ − ξ µ ∇ Z Yµ,A Z Let us recall that for Grassmann variables we have the identity

+ + −θi Mij θj detM = dθi dθie . (3.217) Z Yi Thus we can express the determinant det∂ [ µ, ..] as a path integral over Grassmann µ ∇ ﬁelds c¯ and c as follows

det∂ [ µ, ..]= c¯ c exp( i d4x tr¯c∂ [ µ, c]). (3.218) µ ∇ D D − µ ∇ Z Z The fields c¯ and c are clearly scalar under Lorentz transformations (their spin is 0) but they are anti-commuting Grassmann-valued fields and hence they can not describe physical propagating particles (they simply have the wrong relation between spin and statistics). These fields are called Fadeev-Popov ghosts and they clearly carry two SU(N) indices. More precisely since the covariant derivative is acting on them by commutator these fields must be N N matrices and thus they can be rewritten as c = cAtA. We say × that the ghost fields transform in the adjoint representation of the SU(N) gauge group, i.e. as c ucu+ and c¯ ucu¯ + which ensures global invariance. In terms of cA the −→ −→ determinant reads det∂ [ µ, ..]= c¯A cA exp i d4x c¯A ∂ ∂µδAB gf ∂µAC cB (3.219). µ ∇ D D − µ − ABC µ Z YA Z The path integral of SU(N) Yang-Mills theory becomes

(∂ Aµ)2 Z = AA c¯A cA exp( i d4xtr µ ) exp( i d4x tr¯c∂ [ µ, c]) exp iS[A] N D µ D D − ξ − µ ∇ Z Yµ,A Z YA Z Z = AA c¯A cA exp iS [A, c, c¯]. (3.220) N D µ D D FP Z Yµ,A Z YA

(∂ Aµ)2 S [A, c, c¯]= S[A] d4xtr µ d4x tr¯c∂ [ µ, c]. (3.221) FP − ξ − µ ∇ Z Z The second term is called the gauge ﬁxing term whereas the third term is called the Faddev-Popov ghost term. We add sources to obtain the path integral

Z[J, b, ¯b] = AA c¯A cA exp iS [A, c, c¯] i d4xJ AAµA + i d4x(¯bc +cb ¯ ) . N D µ D D FP − µ Z Yµ,A Z YA Z Z (3.222) YDRI QFT 89

In order to compute propagators we drop all interactions terms. We end up with the partition function i 1 Z[J, b, ¯b] = AA c¯A cA exp d4xAA ∂ ∂µηνλ + ( 1)∂ν ∂λ AA i d4xc¯A∂ ∂µcA N D µ D D 2 ν µ ξ − λ − µ Z Yµ,A Z YA Z Z i d4xJ AAµA + i d4x(¯bc +cb ¯ ) . (3.223) − µ Z Z The free SU(N) gauge part is N 2 1 copies of U(1) gauge theory. Thus without any − further computation the SU(N) vector gauge ﬁeld propagator is given by

d4k iδAB k k < AA(x)AB (y) > = − (η (1 ξ) µ ν )eik(x−y). (3.224) µ ν (2π)4 k2 + iǫ µν − − k2 Z The propagator of the ghost ﬁeld can be computed along the same lines used for the propagator of the Dirac ﬁeld. We obtain 6

d4k iδAB < cA(x)¯cB (y) > = eik(x−y). (3.225) (2π)4 k2 + iǫ Z 3.4.2 Perturbative Renormalization and Feynman Rules The QCD action with SU(N) gauge group is given by

SQCD[ψ, ψ,¯ A, c, c¯] = S0[ψ, ψ,¯ A, c, c¯]+ S1[ψ, ψ,¯ A, c, c¯] (3.226)

1 S [ψ, ψ,¯ A, c, c¯] = d4x ψ¯a(iγµ∂ m)ψa d4x(∂ AA ∂ AA)(∂µAνA ∂νAµA) 0 µ − − 4 µ ν − ν µ − a Z X Z 1 d4x(∂ AµA)2 d4x c¯A∂ ∂µcA. (3.227) − 2ξ µ − µ Z Z

S [ψ, ψ,¯ A, c, c¯] = gtA d4x ψ¯aγµψbAA + gf d4x∂µAνC AAAB 1 − ab µ ABC µ ν Z Xa,b Z g2 f f d4xAAABAµDAνE gf d4x(¯cAcB∂µAC +c ¯A∂µcBAc ). − 4 ABC DEC µ ν − ABC µ µ Z Z (3.228)

c We introduce the renormalization constants Z3, Z2 and Z2 by introducing the renor- µ µ malized fields AR, ψR and cR which are defined in terms of the bare fields A , ψ and c respectively by the equations µ µ A ψ c AR = , ψR = , cR = c . (3.229) √Z3 √Z2 Z2 6Exercise: Perform this calculation explicitly. p 90 YDRI QFT

c The renormalization constants Z3, Z2 and Z2 can be expanded in terms of the counter c terms δ3, δ2 and δ2 as c c Z3 =1+ δ3 ,Z2 =1+ δ2 ,Z2 =1+ δ2. (3.230) Furthermore we relate the bare coupling constants g and m to the renormalized coupling gR and mR through the counter terms δ1 and δm by

gZ2 Z3 = gR(1 + δ1) ,Z2m = mR + δm. (3.231) p 3 4 Since we have also AAA, AAAA and ccA vertices we need more counter terms δ1, δ1 c and δ1 which we deﬁne by

3 2 3 2 2 2 4 c c gZ3 = gR(1 + δ1 ) , g Z3 = gR(1 + δ1) , gZ2 Z3 = gR(1 + δ1). (3.232)

We will also define a "renormalized gauge fixing parameter"p ξR by 1 Z = 3 . (3.233) ξR ξ As we will see shortly this is physically equivalent to imposing the gauge fixing condition µ µ on the renormalized gauge field AR instead of the bare gauge field A . The action divides therefore as

S = SR + Scount ter. (3.234)

The action SR is given by the same formula as S with the replacement of all ﬁelds and coupling constants by the renormalized ﬁelds and renormalized coupling constants and also replacing ξ by ξR. The counter term action Scount ter is given explicitly by δ S = δ d4x ψ¯a iγµ∂ ψa δ d4x ψ¯a ψa 3 d4x(∂ AA ∂ AA )(∂µAνA ∂ν AµA) count ter 2 R µ R − m R R − 4 µ νR − ν µR R − R a a Z X Z X Z δc d4x c¯A∂ ∂µcA g δ tA d4x ψ¯a γµψb AA + g δ3f d4x∂µAνC AA AB − 2 R µ R − R 1 ab R R µR R 1 ABC R µR νR Z Z Xa,b Z g2 δ4 R 1 f f d4xAA AB AµDAνE g δcf d4x(¯cAcB∂µAC +c ¯A∂µcBAc ). − 4 ABC DEC µR νR R R − R 1 ABC R R µR R R µR Z Z (3.235)

c c From the above discussion we see that we have eight counter terms δ1, δ2, δ3, δ2, δm, δ1, 4 3 c δ1 and δ1 and five coupling constants g, m, Z2, Z3 and Z2. The counter terms will be determined in terms of the coupling constants and hence there must be only five of them which are completely independent. The fact that only five counter terms are independent means that we need five renormalization conditions to fix them. This also means that the counter term must be related by three independent equations. It is not difficult to discover that these equations are

3 2 gR Z2√Z3 Z3 = 3 . (3.236) g ≡ 1+ δ1 1+ δ1 YDRI QFT 91

g Z √Z Z R 2 3 = 3 . (3.237) g 1+ δ 4 ≡ 1 1+ δ1 p c gR Z2√Z3 Z2√Z3 = c . (3.238) g ≡ 1+ δ1 1+ δ1 c c At the one-loop order we can expand Z3 =1+ δ3, Z2 =1+ δ2 and Z2 =1+ δ2 where c 3 4 c ~ δ3, δ2 and δ2 as well as δ1, δ1, δ1 and δ1 are all of order and hence the above equations become

δ3 = δ + δ δ . (3.239) 1 3 1 − 2

δ4 = δ + 2δ 2δ . (3.240) 1 3 1 − 2

δc = δc + δ δ . (3.241) 1 2 1 − 2 c The independent counter terms are taken to be δ1, δ2, δ3, δ2, δm which correspond c respectively to the coupling constants g, Z2, Z3, Z2 and m. The counter term δ3 will be determined in the following from the gluon self-energy, the counter terms δ2 and δm will be determined from the quark self-energy whereas the counter term δ1 will be determined c 7 from the vertex. The counter term δ2 should be determined from the ghost self-energy . For ease of writing we will drop in the following the subscript R on renormalized quantities and when we need to refer to the bare quantities we will use the subscript 0 to distinguish them from their renormalized counterparts. We write next the corresponding Feynman rules in momentum space. These are shown in ﬁgure 11. In the next two sections we will derive these rules from ﬁrst principle, i.e. starting from the formula (2.136). The Feynman rules corresponding to the bare action are summarized as follows:

• The quark propagator is (γµp + m) < ψb (p)ψ¯a ( p) >= δab µ βα . (3.242) β α − p2 m2 − • The gluon propagator is

1 k k < AA(k)AB ( k) >= δAB η + (ξ 1) µ ν . (3.243) µ ν − k2 µν − k2 • The ghost propagator is 1 < cB(p)¯cA( p) >= δAB . (3.244) − p2

7 c Exercise: Compute δ2 following the same steps taken for the other counter terms. 92 YDRI QFT

• The quartic vertex is

< AAABADAE > = g2 f f (ηρµησν ησµηρν)+ f f (ησρηµν ηρµησν ) µ ν ρ σ − ABC DEC − BDC AEC − + f f (ησµηρν ηµν ησρ) . (3.245) DAC BEC − • The cubic vertex is

< AA(k)AB(p)AC (q) > = igf (2p + k)µηρν (p + 2k)ν ηρµ + (k p)ρηµν , q = p k. µ ν ρ ABC − − − − (3.246)

• The quark-gluon vertex is

A A B A µ < Aµ c¯ (k)c > = g(t )ab(γ )αβ. (3.247)

• The ghost-gluon vertex is

< AC ψ¯a ψb > = igf kµ. (3.248) µ α β − ABC • Impose energy-momentum conservation at all vertices.

• Integrate over internal momenta.

• Symmetry factor. For example if the diagram is invariant under the permutation of two lines we should divide by 1/2.

• Each fermion line must be multiplied by 1. − • All one-loop diagrams should be multiplied by ~/i.

3.4.3 The Gluon Field Self-Energy at 1 Loop − We are interested in computing the proper n point vertices of this theory which are − connected 1 particle irreducible n point functions from which all external legs are am- − − putated. The generating functional of the corresponding Feynman diagrams is of course the eﬀective action. We recall the formal deﬁnition of the proper n point vertices given − by

n (n) δ Γ[φc] Γ (x1, ..., xn)=Γ,i1...in = φ=0. (3.249) δφc(x1)...δφc(xn)| The eﬀective action upto the 1 loop order is − 1 Γ= S + Γ , Γ = ln det G , Gik = S−1. (3.250) i 1 1 0 0 − ,ik YDRI QFT 93

As our first example we consider the proper 2 point vertex of the non-Abelian vector − field Aµ. This is defined by

δ2Γ ΓAB(x,y) = . (3.251) µν δAµA(x)δAνB (y)|A,ψ,c=0

We use the powerful formula (2.136) which we copy here for convenience 1 1 Γ [φ] = GmnS[φ] + Gmm0 Gnn0 S[φ] S[φ] . (3.252) 1 ,j0k0 2 0 ,j0k0mn 2 0 0 ,j0mn ,k0m0n0 We have then immediately four terms contributing to the gluon propagator at 1 loop. − These are given by (with j0 = (x,µ,A) and k0 = (y,ν,B))

2 AB δ S 1 1 AmAn 1 AmAm0 AnAn0 Γ (x,y) = A,ψ,c=0 + G S,A A A A + G G S,A A A S,A A A µν δAµA(x)δAνB (y)| i 2 0 j0 k0 m n 2 0 0 j0 m n k0 m0 n0 ¯ ¯ c¯mcm0 cnc¯n0 ψmψm0 ψnψn0 + ( 1) G0 G0 S,Aj c¯mcn S,Ak cm c¯n + ( 1) G0 G0 S,A ψ¯ ψ S,A ψ ψ¯ . − × 0 0 0 0 − × j0 m n k0 m0 n0 (3.253)

The corresponding Feynman diagrams are shown on figure 9. The minus signs in the last two diagrams are the famous fermion loops minus sign. To see how they actually originate we should go back to the derivation of (3.252) and see what happens if the fields are Grassmann valued. We start from the first derivative of the effective action Γ1 which is given by the unambigous equation (2.112), viz 1 Γ = GmnS . (3.254) 1,j 2 0 ,jmn Taking the second derivative we obtain 1 1 δGmn Γ = GmnS + 0 S . (3.255) 1,ij 2 0 ,ijmn 2 δφi ,jmn

mm0 The ﬁrst term is correct. The second term can be computed using the identity G0 S,m0n = δm which can be rewritten as − n δGmn 0 = Gmm0 S Gn0n. (3.256) δφi 0 ,im0n0 0 We have then 1 1 Γ = GmnS + Gmm0 Gn0nS S 1,ij 2 0 ,ijmn 2 0 0 ,im0n0 ,jmn 1 1 = GmnS + Gm0mGnn0 S S . (3.257) 2 0 ,j0k0mn 2 0 0 ,j0mn ,k0m0n0

m0m Only the propagator G0 has reversed indices compared to (3.252) which is irrelevant for bosonic ﬁelds but reproduces a minus sign for fermionic ﬁelds. 94 YDRI QFT

The classical term in the gluon self-energy is given by δ2S 1 S = = ∂ ∂ρηµν + ( 1)∂µ∂ν δABδ4(x y) ,j0k0 δAµA(x)δAνB (y)|A,ψ,c=0 ρ ξ − − d4k 1 = k2ηµν + ( 1)kµkν δ eik(x−y). − (2π)4 ξ − AB Z (3.258)

We compute d4k 1 kµkν Gj0k0 = δ ηµν + (ξ 1) eik(x−y) 0 AB (2π)4 k2 + iǫ − k2 Z d4k = δ Gµν (k)eik(x−y) AB (2π)4 0 Zµν = δABG0 (x,y). (3.259) The quartic vertex can be put into the fully symmetric form g2 g2 f f d4xAAABAµDAνE = f f d4x d4y d4z d4wAA(x)AB (y)AD(z)AE(w) − 4 ABC DEC µ ν − 8 ABC DEC µ ν ρ σ Z Z Z Z Z δ4(x y)δ4(x z)δ4(x w)(ηρµησν ησµηρν ) × − − − − g2 = d4x d4y d4z d4wAA(x)AB (y)AD(z)AE(w)δ4(x y)δ4 − 4! µ ν ρ σ − Z Z Z Z δ4(x w) f f (ηρµησν ησµηρν )+ f f (ησρηµν ηρµη × − ABC DEC − BDC AEC − + f f (ησµηρν ηµν ησρ) . DAC BEC − In other words (with j0 = (x,µ,A), k0 = (y,ν,B), m = (z, ρ, D) and n = (w, σ, E))

2 4 4 4 ρµ σν σµ ρν σρ µν ρµ σ S,A A A A = g δ (x y)δ (x z)δ (x w) fABC fDEC(η η η η )+ fBDC fAEC(η η η η j0 k0 m n − − − − − − + f f (ησµηρν ηµν ησρ) . (3.261) DAC BEC − We can now compute the ﬁrst one-loop diagram as

2 1 AmAn g DE ρµ σν σµ ρν σρ µν ρµ σν G S,A A A A = δ G (x,x) fABC fDEC(η η η η )+ fBDC fAEC(η η η η ) 2 0 j0 k0 m n − 2 0ρσ − − + f f (ησµηρν ηµν ησρ) δ4(x y) DAC BEC − − g2 = Gρ (x,x)ηµν Gµν (x,x) f f f f δ4(x y) − 2 0ρ − 0 BDC ADC − DAC BDC − = g2 Gρ (x,x)ηµν Gµν (x,x) f f δ4(x y). (3.262) − 0ρ − 0 BDC ADC − YDRI QFT 95

The quantity fBDC fADC is actually the Casimir operator in the adjoint representation of the group. The adjoint representation of SU(N) is (N 2 1) dimensional. The generators A − − in the adjoint representation can be given by (tG)BC = ifABC . Indeed we can easily check A B C that these matrices satisfy the fundamental commutation relations [tG,tG] = ifABC tG. C C We compute then fBDC fADC = (tGtG)BA = C2(G)δBA. These generators must also A B AB 8 satisfy trGtGtG = C(G)δ . For SU(N) we have

fBDC fADC = C2(G)δBA = C(G)δBA = NδBA. (3.263)

Hence

1 AmAn 2 ρ µν µν 4 G S,A A A A = g C2(G)δAB G (x,x)η G (x,x) δ (x y(3.264)). 2 0 j0 k0 m n − 0ρ − 0 − In order to maintain gauge invariance we will use the powerful method of dimensional regularization. The above diagram takes now the form

d µ ν 1 AmAn 2 d p 1 µν p p 4 G S,A A A A = g C2(G)δAB (d + ξ 2)η (ξ 1) δ (x y). 2 0 j0 k0 m n − (2π)d p2 − − − k2 − Z (3.265)

This simpliﬁes further in the Feynman gauge. Indeed for ξ = 1 we get

d 1 AmAn 2 d p 1 µν 4 G S,A A A A = g C2(G)δAB (d 1)η δ (x y) 2 0 j0 k0 m n − (2π)d p2 − − Z ddp ddk (p + k)2 = g2C (G)δ (d 1)ηµν eik(x−y). − 2 AB (2π)d (2π)d p2(p + k)2 − Z Z (3.266)

We use now Feynman parameters, viz

1 1 1 1 = dx dyδ(x + y 1) (p + k)2p2 − 2 2 2 Z0 Z0 x(p + k) + yp 1 1 = dx , l = p +xk , ∆= x(1 x)k2. (3.267) (l2 ∆)2 − − Z0 − We have then (using also rotational invariance)

d 1 d 2 1 AmAn 2 d k µν ik(x−y) d l (l + (1 x)k) G S,A A A A = g C2(G)δAB (d 1)η e dx − 2 0 j0 k0 m n − (2π)d − (2π)d (l2 ∆)2 Z Z0 Z − ddk 1 ddl l2 = g2C (G)δ (d 1)ηµν eik(x−y) dx − 2 AB (2π)d − (2π)d (l2 ∆)2 Z Z0 Z − ddl 1 + (1 x)2k2 . (3.268) − (2π)d (l2 ∆)2 Z − 8Exercise: Derive this result. 96 YDRI QFT

The above two integrals over l are given by (after dimensional regularization and Wick rotation)

ddl l2 ddl 1 = i E (2π)d (l2 ∆)2 − (2π)d (l2 + ∆)2 Z − Z E 1 1 Γ(2 d ) = i − 2 . (3.269) d 1− d 2 − (4π) 2 ∆ 2 1 d −

ddl 1 ddl l2 = i E E (2π)d (l2 ∆)2 (2π)d (l2 + ∆)2 Z − Z E 1 1 d = i Γ(2 ). (3.270) d 2− d (4π) 2 ∆ 2 − 2

We get the ﬁnal result

d 1 1 AmAn i 2 d k µν 2 ik(x−y) dx 1 G S,A A A A = g C2(G)δAB η k e d(d 1)x(1 2 0 j0 k0 m n d (2π)d 2− d − 2 − (4π) 2 Z Z0 x(1 x)k2 2 − − d d Γ(1 ) (d 1)(1 x)2Γ(2 ) . (3.271) × − 2 − − − − 2 We compute now the second diagram. First we write the pure gauge ﬁeld cubic interaction in the totally symmetric form

gf gf d4x∂µAνC AAAB = ABC d4x d4y d4zAA(x)AB(y)AC (z) ηρν ∂µδ4(x z).δ4(x y) ABC µ ν 3! µ ν ρ x − − Z Z Z Z ∂µδ4(x y).δ4(x z) ηρµ ∂ν δ4(y z).δ4(x y) ∂ν δ4(y x).δ4(z y) − x − − − y − − − y − − ηµν ∂ρδ4(z x).δ4(z y) ∂ρδ4(z y).δ4(x z) . (3.272) − z − − − z − −

Thus we compute (with j0 = (x,µ,A), k0 = (y,ν,B) and m = (z, ρ, C))

S = igf Sµνρ(x,y,z). (3.273) ,Aj0 Ak0 Am ABC iSµνρ(x,y,z) = ηρν ∂µδ4(x z).δ4(x y) ∂µδ4(x y).δ4(x z) ηρµ ∂ν δ4(y z).δ4(x y) x − − − x − − − y − − ∂νδ4(x y).δ4(y z) ηµν ∂ρδ4(x z).δ4(y z) ∂ρδ4(y z).δ4(x z) − y − − − z − − − z − − d4k d4p d4l = i Sµνρ(k,p)(2π)4δ4(p + k + l) exp(ikx + ipy + ilz). (2π)4 (2π)4 (2π)4 Z Z Z (3.274) YDRI QFT 97

Sµνρ(k,p) = (2p + k)µηρν (p + 2k)ν ηρµ + (k p)ρηµν . (3.275) − − The second diagram is therefore given by

2 1 AmAm0 AnAn0 g C2(G)δAB 4 4 4 4 µρσ G G S,A A A S,A A A = d zd z0d wd w0G0ρρ (z, z0)G0σσ (w, w0)S (x, 2 0 0 j0 m n k0 m0 n0 − 2 0 0 Z Sνρ0σ0 (y, z , w ) × 0 0 g2C (G)δ d4k d4p = 2 AB G (p)G (k + p)Sµρσ(k,p) − 2 (2π)4 (2π)4 0ρρ0 0σσ0 Z Z Sνρ0σ0 ( k, p)exp ik(x y). (3.276) × − − − In the Feynman gauge this becomes

2 4 4 1 AmAm0 AnAn0 g C2(G)δAB d k d p 1 µρσ G G S,A A A S,A A A = S (k,p) 2 0 0 j0 m n k0 m0 n0 − 2 (2π)4 (2π)4 p2(k + p)2 Z Z Sν ( k, p)exp ik(x y). (3.277) × ρσ − − − We use now Feynman parameters as before. We get

2 4 1 4 1 AmAm0 AnAn0 g C2(G)δAB d k d l 1 G G S,A A A S,A A A = exp ik(x y) dx 2 0 0 j0 m n k0 m0 n0 − 2 (2π)4 − (2π)4 (l2 ∆)2 Z Z0 Z − Sµρσ(k, l xk)Sν ( k, l + xk). (3.278) × − ρσ − − Clearly by rotational symmetry only quadratic and constant terms in lµ in the product µρσ ν S (k, l xk)S ρσ( k, l+xk) give non-zero contribution to the integral over l. These 9 − − − are

2 4 1 1 AmAm0 AnAn0 g C2(G)δAB d k 1 µν G G S,A A A S,A A A = exp ik(x y) dx 6( 1)η 2 0 0 j0 m n k0 m0 n0 − 2 (2π)4 − d − Z Z0 d4l l2 + (2 d)(1 2x)2kµkν + 2(1 + x)(2 x)kµk × (2π)4 (l2 ∆)2 − − − Z − d4l 1 ηµν k2(2 x)2 ηµν k2(1 + x)2 . (3.279) − − − (2π)4 (l2 ∆)2 Z − We now employ dimensional regularization and use the integrals (3.269) and (3.270). We obtain

2 d 1 1 AmAm0 AnAn0 ig C2(G)δAB d k dx G G S,A A A S,A A A = exp ik(x y) 2 0 0 j0 m n k0 m0 n0 − d (2π)d − 2− d 2(4π) 2 0 x(1 x)k2 2 Z Z − − d d 3(d 1)ηµν Γ(1 )x(1 x)k2 + Γ(2 ) (2 d)(1 2x)2kµkν − − − 2 − − 2 − − + 2(1 + x)(2 x)kµkν ηµν k2(2 x)2 ηµν k2(1 + x)2 . (3.280) − − − − 9Exercise: Derive explicitly these terms. 98 YDRI QFT

We go now to the third diagram which involves a ghost loop. We recall ﬁrst the ghost ﬁeld propagator

d4k iδAB < cA(x)¯cB (y) > = eik(x−y). (3.281) (2π)4 k2 + iǫ Z However we will need

4 AB A B d k δ Gc (x)¯c (y) = eik(x−y). (3.282) 0 (2π)4 k2 + iǫ Z The interaction bteween the ghost and vector ﬁleds is given by

gf d4x(¯cAcB∂µAC +c ¯A∂µcBAc )= gf d4x d4y d4zc¯A(x)cB (y)AC (z)∂µ(δ4(x y)δ4(x z)) − ABC µ µ − ABC µ x − − Z Z Z Z (3.283)

In other words (with j0 = (z,C,µ), m = (x, A) and n = (y,B))

µ 4 4 S,A c¯ c = gfABC ∂ (δ (x y)δ (x z)) j0 m n x − − d4k d4p d4l = igf kµ(2π)4δ4(p + k l) exp( ikx ipy + ilz). − ABC (2π)4 (2π)4 (2π)4 − − − Z Z Z (3.284)

We compute the third diagram as follows. We have (with j0 = (z,C,µ), k0 = (w,D,ν), m = (x, A), n = (y,B) and m0 = (x0, A0), n0 = (y0,B0))

4 A0A 4 c¯mcm0 cnc¯n0 4 4 4 4 d k δ ik(x0−x) d p δ G G S,A c¯ c S,A c c¯ = d x d x0 d y d y0 e − 0 0 j0 m n k0 m0 n0 (2π)4 k2 (2π)4 A,AX0,B,B0 Z Z Z Z Z Z eip(y0−y) gf ∂µ(δ4(x y)δ4(x z)) gf ∂ν (δ4(y x )δ4(y × ABC x − − − B0A0D y0 0 − 0 d4k d4p (p + k)µpν = g2f f eik(z−w) ABC ABD (2π)4 (2π)4 (p + k)2p2 Z Z

We use Feynman parameters as before. Also we use rotational invariance to bring the above loop integral to the form

4 1 4 c¯mcm0 cnc¯n0 2 d k ik(z−w) d l µ ν µ ν G G S,A c¯ c S,A c c¯ = g C2(G)δCD e dx l l + x(x 1)k k 0 0 j0 m n k0 m0 n0 (2π)4 (2π)4 − (l2 Z Z0 Z − d4k 1 1 d4l l2 = g2C (G)δ eik(z−w) dx ηµν 2 CD (2π)4 4 (2π)4 (l2 ∆)2 Z Z0 Z − d4l 1 + x(x 1)kµkν . (3.286) − (2π)4 (l2 ∆)2 Z − YDRI QFT 99

Once more we employ dimensional regularization and use the integrals (3.269) and (3.270). Hence we get the loop integral (with C A, D B, z x, w y) −→ −→ −→ −→ d c¯mcm0 cnc¯n0 2 i d k ik(x−y) 1 µν 2 d µ ν G G S,A c¯ c S,A c c¯ = g C2(G) δAB e η k Γ(1 )+ k k Γ(2 0 0 j0 m n k0 m0 n0 − d (2π)d − 2 − 2 − (4π) 2 Z 1 x(1 x) dx − . (3.287) 2− d × 0 x(1 x)k2 2 Z − − By putting equations (3.271), (3.280) and (3.287) together we get

d 1 2 i d k ik(x−y) dx µν 2 (3.271) + (3.280) (3.287) = g C2(G) δAB e η k (d 2)Γ(2 − d (2π)d 2− d − − (4π) 2 0 x(1 x)k2 2 Z Z − − d 1 1 x(1 x)+ ηµν k2Γ(2 ) (d 1)(1 x)2 + (2 x)2 + (1 + x)2 × − − 2 − − − 2 − 2 d d kµkν Γ(2 ) (1 )(1 2x)2 + 2 . (3.288) − − 2 − 2 − The pole at d = 2 cancels exactly since the gamma function Γ(1 d/2) is completely − gone. There remains of course the pole at d = 4. By using the symmetry of the integral over x under x 1 x we can rewrite the above integral as −→ − d 1 2 i d k ik(x−y) dx µν 2 d (3.271) + (3.280) (3.287) = g C2(G) δAB e η k (1 )Γ(2 − d (2π)d 2− d − 2 − (4π) 2 0 x(1 x)k2 2 Z Z − − d d d (1 2x)2 + (1 2x) + ηµν k2Γ(2 ).4x kµkνΓ(2 ) (1 )(1 2x) × − − − 2 − − 2 − 2 −

Again by the symmetry x 1 x we can replace x in every linear term in x by 1/210. −→ − We obtain the ﬁnal result

d d 1 2 iΓ(2 2 ) d k ik(x−y) µν 2 µ ν 2 d −2 dx (3.271) + (3.280) (3.287) = g C (G) − δ e η k k k (k ) 2 − 2 d AB (2π)d − (4π) 2 0 x(1 x Z Z − − d (1 )(1 2x)2 + 2 × − 2 − d d 2 iΓ(2 2 ) d k ik(x−y) µν 2 µ ν 2 d −2 5 = g C (G) − δ e η k k k (k ) 2 + regular terms 2 d AB (2π)d − 3 (4π) 2 Z (3.290)

The gluon field is therefore transverse as it should be for any vector field with an underlying gauge symmetry. Indeed the exhibited the tensor structure ηµν k2 kµkν is consistent − 10Exercise: Why. 100 YDRI QFT with Ward identity. This result does not depend on the gauge fixing parameter although the proportionality factor actually does11. There remains the fourth and final diagram which as it turns out is the only diagram which is independent of the gauge fixing parameter. We recall the Dirac field propagator

d4p (γµp + m) < ψa (x)ψ¯b (y) > = iδab µ αβ e−ip(x−y). (3.291) α β (2π)4 p2 m2 + iǫ Z − However we will need something a little diﬀerent. We have

2 δ S µ y 4 ab S a ¯b A,ψ,c=0 = (iγ ∂µ m)βαδ (y x)δ ,ψα(x)ψβ (y) a ¯b ≡ δψα(x)δψβ (y)| − − d4k = (γµk m) eik(x−y)δab(3.292). (2π)4 µ − βα Z Thus we must have

ψa (x)ψ¯b (y) d4k (γµk + m) G α β = δab µ αβ e−ik(x−y). (3.293) 0 (2π)4 k2 m2 + iǫ Z − Indeed we can check

a0 b ψα (x0)ψ¯ (y) 4 0 β aa0 4 d y S,ψa (x)ψ¯b (y)G0 = δ δαα0 δ (x x0). (3.294) α β − Z Xb,β The interaction bteween the Dirac and vector ﬁleds is given by

gtA d4x ψ¯aγµψbAA = gtA (γµ) d4x d4y d4zψ¯a (x)ψb (y)AA(z)δ4(x y)δ4(x z). − ab µ − ab αβ α β µ − − Z Xa,b Z Z Z (3.295)

In other words (with j0 = (z,A,µ), m = (x, a, α) and n = (y, b, β))

A µ 4 4 S,A ψ¯ ψ = gtab(γ )αβδ (x y)δ (x z). (3.296) j0 m n − −

By using these results we compute the fourth diagram is given by (with j0 = (z,A,µ), k0 = (w,B,ν), m = (x, a, α), n = (y, b, β), m0 = (x0, a0, α0), n0 = (y0, b0, β0) and trγµ = 0, trγµγν = 4ηµν , trγµγνγρ = 0, trγµγνγργσ = 4(ηµν ηρσ ηµρηνσ + ηµσηνρ) ) − ¯ ¯ 4 4 ρ µ ρ ν ψmψm0 ψnψn0 2 A B d p d k tr(γ pρ + m)γ (γ (p + k)ρ + m)γ −ik(z− G0 G0 S,A ψ¯ ψ S,A ψ ψ¯ = g trt t e j0 m n k0 m0 n0 (2π)4 (2π)4 (p2 m2)((p + k)2 m2) Z Z − − d4p d4k pµ(p + k)ν + pν(p + k)µ ηµν (p2 + pk = 4g2trtAtB − − (2π)4 (2π)4 (p2 m2)((p + k)2 m2) Z Z − − e−ik(z−w). (3.297) × 11Exercise: Determine the corresponding factor for an arbitrary value of the gauge ﬁxing parameter ξ. YDRI QFT 101

We use now Feynman parameters in the form

1 1 1 1 2 2 2 2 = dx dyδ(x + y 1) 2 (p m )((p + k) m ) 0 0 − − − Z Z x(p2 m2)+ y((p + k)2 m2) − − 1 1 = dx , l = p + (1 x)k , ∆= m2 x(1 x)k2. (l2 ∆)2 − − − Z0 − (3.298)

By using also rotational invariance we can bring the integral to the form

¯ ¯ 1 4 ψmψm0 ψnψn0 2 A B d k −ik(z−w) G0 G0 S,A ψ¯ ψ S,A ψ ψ¯ = 4g trt t dx e j0 m n k0 m0 n0 (2π)4 Z0 Z d4l 1 2lµlν 2x(1 x)kµkν ηµν l2 x(1 x)k2 m2 × (2π)4 − − − − − − (l2 Z − 1 d4k = 4g2trtAtB dx e−ik(z−w) (2π)4 Z0 Z d4l 1 l2ηµν 2x(1 x)kµkν ηµν l2 x(1 x)k2 m2 × (2π)4 2 − − − − − − (l2 Z 1 d4k = 4g2trtAtB dx e−ik(z−w) x(1 x)(k2ηµν 2kµkν)+ m2 (2π)4 − − Z0 Z d4l 1 1 d4l l2 ηµν × (2π)4 (l2 ∆)2 − 2 (2π)4 (l2 ∆)2 Z − Z − (3.299)

After using the integrals (3.269) and (3.270), the fourth diagram becomes (with z x, −→ w y) −→

¯ ¯ Γ(2 d ) d ψmψm0 ψnψn0 2 2 A B d k 2 µν µ ν −ik(x−y) G0 G0 S,A ψ¯ ψ S,A ψ ψ¯ = 8ig − d trt t (k η k k )e j0 m n k0 m0 n0 (2π)d − (4π) 2 Z 1 2 d −2 x(1 x) (k ) 2 dx − 2 2− d × 0 m x(1 x) 2 Z k2 − − d d 4 2 Γ(2 2 ) d k 2 µν µ ν −ik(x−y) 2 d −2 = g − C(N)δ i(k η k k )e (k ) 2 1 3 d AB (2π)d − (4π) 2 Z + regular terms . (3.300)

a For nf ﬂavors (instead of a single ﬂavor) of fermions in the representation tr (instead of 102 YDRI QFT the fundamental representation t ) we obtain (with also a change k k) a −→ − d d ψ¯mψm ψnψ¯n 4 Γ(2 ) d k d 0 0 2 2 2 µν µ ν ik(x−y) 2 2 −2 G0 G0 S,A ψ¯ ψ S,A ψ ψ¯ = nf g − d C(r)δAB i(k η k k )e (k ) 1 j0 m n k0 m0 n0 3 (2π)d − (4π) 2 Z + regular terms . (3.301)

By putting (3.290) and (3.301) together we get the ﬁnal result

ΓAB(x,y) = (3.290) (3.301) µν − d4k 1 k2ηµν + ( 1)kµkν δ eik(x−y) − (2π)4 ξ − AB Z d d 2 Γ(2 2 ) 5 4 d k 2 µν µ ν ik(x−y) 2 d −2 + g − C (G) n C(r) (k η k k )δ e (k ) 2 1 d 3 2 − 3 f (2π)d − AB (4π) 2 Z + regular terms . (3.302)

The ﬁnal step is to add the contribution of the counter terms. This leads to the one-loop result in the Feynman-t’Hooft gauge given by

ddk 1 ΓAB(x,y) = k2ηµν + ( 1)kµkν δ eik(x−y) µν − (2π)d ξ − AB Z d d 2 Γ(2 2 ) 5 4 d k 2 µν µ ν ik(x−y) 2 d −2 + g − C (G) n C(r) (k η k k )δ e (k ) 2 1 d 3 2 − 3 f (2π)d − AB (4π) 2 Z ddk 1 + regular terms δ k2ηµν + ( 1)kµkν δ eik(x−y). (3.303) − 3 (2π)d ξ − AB Z Equivalently

1 ΓAB(k) = k2ηµν + ( 1)kµkν δ µν − ξ − AB d 2 Γ(2 2 ) 5 4 2 µν µ ν 2 d −2 + g − C (G) n C(r) (k η k k )δ (k ) 2 1 d 3 2 − 3 f − AB (4π) 2 + regular terms δ k2ηµν kµkν δ . (3.304) − 3 − AB Remark that the 1/ξ term in the classical contribution (the ﬁrst term) can be removed by undoing the gauge ﬁxing procedure. In 4 dimensions the coupling constant g2 is dimensionless. YDRI QFT 103

In dimension d = 4 ǫ the coupling constant g is in fact not dimensionless but has − dimension of 1/mass(d/2−2). The dimensionless coupling constant gˆ can therefore be given in terms of an arbitrary mass scale µ by the formula

d −2 2 2 ǫ gˆ = gµ 2 g =g ˆ µ . (3.305) ⇔ We get then

ǫ 1 gˆ2 ǫ 4πµ2 2 5 4 ΓAB(k) = k2ηµν + ( 1)kµkν δ + Γ( ) C (G) n C(r) (k2ηµν kµkν)δ 1 µν − ξ − AB 16π2 2 k2 3 2 − 3 f − AB + regular terms δ k2ηµν kµkν δ − 3 − AB 1 gˆ2 2 k2 5 4 = k2ηµν + ( 1)kµkν δ + + ln 4π γ ln C (G)+ n C(r) (k2ηµν kµ − ξ − AB 16π2 ǫ − − µ2 3 2 3 f − 1 + regular terms δ k2ηµν kµkν δ × − 3 − AB

It is now clear that in order to eliminate the divergent term we need, in the spirit of minimal subtraction, only subtract the logarithmic divergence exhibited here by the the term 2/ǫ which has a pole at ǫ = 0. In other words the counter term δ3 is chosen such that gˆ2 2 5 4 δ = C (G) n C(r) . (3.307) 3 16π2 ǫ 3 2 − 3 f 3.4.4 The Quark Field Self-Energy at 1 Loop − This is deﬁned δ2Γ Γab (x,y) = αβ a ¯b A,ψ,c=0 δψα(x)δψβ (y)| δ2S 1 δ2Γ = + 1 . (3.308) a ¯b A,ψ,c=0 a ¯b A,ψ,c=0 δψα(x)δψβ (y)| i δψα(x)δψβ (y)| The ﬁrst term is given by δ2S = (iγµ∂y m) δ4(y x)δab a ¯b A,ψ,c=0 µ βα δψα(x)δψβ (y)| − − d4k = (γµk m) eik(x−y)δab. (3.309) (2π)4 µ − βα Z Again by using the elegant formula (3.257) we obtain (with j0 = (x, α, a) and k0 = (y, β, b))

¯ ψmψm0 AnAn0 Γ1,j0k0 = G0 G0 S,ψ ψ¯ A S,ψ¯ ψ A . (3.310) − j0 m n k0 m0 n0 104 YDRI QFT

We recall the results 4 µ ν µA νB d k 1 k k GA (x)A (y) = δ ηµν + (ξ 1) eik(x−y). (3.311) 0 AB (2π)4 k2 + iǫ − k2 Z

ψa (x)ψ¯b (y) d4p (γµp + m) G α β = δab µ αβ e−ip(x−y). (3.312) 0 (2π)4 p2 m2 + iǫ Z − A µ 4 4 S,AµA(z)ψ¯a (x)ψb (y) = gtab(γ )αβδ (x y)δ (x z). (3.313) α β − − We compute then d4p d4k 1 k k Γ = g2(tAtA) γµ(γρp + m)γν η + (ξ 1) µ ν ei(k+p)(x−y). 1,j0k0 − ba (2π)4 (2π)4 ρ βα k2(p2 m2) µν − k2 Z Z − (3.314) This is given by the second diagram on ﬁgure 10. In the Feynman- t’Hooft gauge this reduces to (also using γµγργ = (2 ǫ)γρ, γµγ = d and (tAtA) = C (r)δ where µ − − µ ab 2 ab C2(r) is the Casimir in the representation r) d4p d4k 1 Γ = g2C (r)δ γµ(γρ(p + k) + m)γ eip(x−y) 1,j0k0 − 2 ba (2π)4 (2π)4 ρ µ βα k2((p + k)2 m2) Z Z − d4p d4k 1 = g2C (r)δ (2 ǫ)γρ(p + k) + md eip(x−y). − 2 ba (2π)4 (2π)4 − − ρ βα k2((p + k)2 m2) Z Z − (3.315) We employ Feynman parameters in the form 1 1 1 2 2 2 2 2 = dx 2 , l = k + (1 x)p , ∆= x(1 x)p + (1 x)m . k ((p + k) m ) 0 l2 ∆ − − − − − Z − (3.316) We obtain 4 1 4 2 d p ip(x−y) d l 1 ρ Γ1,j0k0 = g C2(r)δba 4 e dx 4 2 (2 ǫ)γ (l + xp)ρ + md βα − (2π) 0 (2π) l2 ∆ − − Z Z Z − 4 1 4 2 d p ip(x−y) d l 1 ρ = g C2(r)δba e dx (2 ǫ)xγ pρ + md . − (2π)4 (2π)4 2 2 − − βα Z Z0 Z l ∆ − (3.317) After Wick rotation and dimensional regularization we can use the integral (3.269). We get d 4 1 iΓ(2 ) d p 1 ǫ dx 2 2 ρ ip(x−y) 2 − 2 Γ1,j0k0 = g C2(r) −d δba (2 ǫ)γ pρ + md e (p ) − (2π)4 − 2 − βα m (4π) 2 Z Z0 x(1 x) + (1 x) − − − d 4 2 iΓ(2 2 ) d p 1 ρ ip(x−y) 2 − ǫ = g C (r) − δ (2 ǫ)γ p + md e (p ) 2 1 + regular terms . − 2 d ba (2π)4 − 2 − ρ βα (4π) 2 Z YDRI QFT 105

The quark ﬁeld self-energy at 1 loop is therefore given by −

d4p Γab (x,y) = (γµp m) eip(x−y)δab αβ (2π)4 µ − βα Z d 4 2 Γ(2 2 ) d p 1 ρ ip(x−y) 2 − ǫ g C (r) − δ (2 ǫ)γ p + md e (p ) 2 1 + regular terms . − 2 d ba (2π)4 − 2 − ρ βα (4π) 2 Z (3.319)

We add the contribution of the counter terms. We obtain

d4p Γab (x,y) = (γµp m) eip(x−y)δab αβ (2π)4 µ − βα Z gˆ2 d4p 2 p2 C (r)δ + ln 4π γ ln γρp + md eip(x−y) 1 + regular terms − 16π2 2 ba (2π)4 ǫ − − µ2 − ρ βα Z d4p + (δ γµp δ ) eip(x−y)δab. (3.320) (2π)4 2 µ − m βα Z

In order to cancel the divergence we must choose the counter terms δ2 and δm to be

gˆ2 2 δ = C (r) . (3.321) 2 −16π2 2 ǫ

gˆ2 8m δ = C (r) . (3.322) m −16π2 2 ǫ These two counter terms allow us to determine the renormalized mass m in terms of the bare mass up to the one-loop order.

3.4.5 The Vertex at 1 Loop − The quark-gluon vertex at one-loop is given by

δ3Γ ΓabA (x,y,z) = αβµ ¯a b A A,ψ,c=0 δψα(x)δψβ (y)δAµ (z)| δ3S 1 δ3Γ = + 1 ¯a b A A,ψ,c=0 ¯a b A A,ψ,c=0 δψα(x)δψβ (y)δAµ (z)| i δψα(x)δψβ (y)δAµ (z)| 1 δ3Γ = g(tA) (γµ) δ4(x y)δ4(x z)+ 1 . ab αβ ¯a b A A,ψ,c=0 − − i δψα(x)δψβ (y)δAµ (z)| (3.323) 106 YDRI QFT

In this section we compute the one-loop correction using Feynman rules directly. We write 1 d4x d4y d4ze−ikx−ipy−ilzΓabA (x,y,z) = g(tA) (γµ) (2π)4δ4(k + p + l)+ d4x d4y d4ze−ikx αβµ ab αβ i Z Z Z 3 Z Z Z δ Γ1 ¯a b A A,ψ,c=0 × δψα(x)δψβ (y)δAµ (z)| 1 = g(tA) (γµ) + Feynman diagrams (2π)4δ4(k + p + l) ab αβ i It is not difficult to convince ourselves that there are only two possible Feynman diagrams contributing to this 3 point proper vertex which we will only evaluate their leading − divergent part in the Feynman-’t Hooft gauge. The first diagram on figure 12 is given explicitly by ddk 12a = ig3f (tDtC ) ( k + p 2p )ρηλµ (k + 2p p )ληρµ + (2k + p + p )µηλρ − CDA ab (2π)d − 1 − 2 − 1 − 2 1 2 Z γλ(γ.k + m)γρ αβ × (k2 m2)(k + p )2(k + p )2 − 1 2 g3C (G) ddk = 2 (tA) ( k + p 2p )ρηλµ (k + 2p p )ληρµ + (2k + p + p )µηλρ − 2 ab (2π)d − 1 − 2 − 1 − 2 1 2 Z γλ(γ.k + m)γρ αβ . (3.325) × (k2 m2)(k + p )2(k + p )2 − 1 2 D C D C E In the second line we have used the fact that fCDAt t = fCDA[t ,t ]/2= ifCDAfDCEt /2. We make now the approximation of neglecting the quark mass and all external momenta since the divergence is actually independent of both 12. The result reduces to g3C (G) ddk 12a = 2 (tA) kρηλµ kληρµ + 2kµηλρ − 2 ab (2π)d − − Z γλ(γ.k)γρ αβ 2 3 × (k ) g3C (G) ddk 1 = 2 (tA) 2(γµ) k2 2(2 ǫ)kµkν (γ ) − 2 ab (2π)d − αβ − − ν αβ (k2)3 Z g3C (G) ddk 2(2 ǫ) 1 = 2 (tA) 2(γµ) k2 − k2(γµ) − 2 ab (2π)d − αβ − d αβ (k2)3 Z g3C (G) 4(d 1) ddk 1 = 2 (tA) − (γµ) 2 ab d αβ (2π)d (k2)2 Z 3ig3C (G) d = 2 (tA) (γµ) Γ(2 ). (3.326) 2(4π)2 ab αβ − 2

12Exercise: Compute this integral without making these approximations and show that the divergence is indeed independent of the quark mass and all external momenta. YDRI QFT 107

The second diagram on ﬁgure 12 is given explicitly by µ λ ddk γλ( γ.(k + p2)+ m)γ ( γ.(k + p1)+ m)γ 12b = g3(tC tAtC ) − − αβ . ab (2π)d k2((k + p )2 m2)((k + p )2 m2) Z 1 − 2 − (3.327) We compute tC tAtC = tC tC tA + tC [tA,tC ] A C B = C2(N)t + ifACBt t i = C (N)tA + f [tC ,tB] 2 2 ACB 1 = C (N) C (G) tA. (3.328) 2 − 2 2 We get then µ λ 1 ddk γλ( γ.(k + p2)+ m)γ ( γ.(k + p1)+ m)γ 12b = g3 C (N) C (G) (tA) − − αβ . 2 − 2 2 ab (2π)d k2((k + p )2 m2)((k + p )2 m2) Z 1 − 2 − (3.329) Again as before we are only interested at this stage in the leading divergent part and thus we can make the approximation of dropping the quark mass and all external momenta13. We obtain thus µ λ 1 ddk γλ( γ.k)γ ( γ.k)γ 12b = g3 C (N) C (G) (tA) − − αβ 2 − 2 2 ab (2π)d (k2)3 Z µ λ ρ σ 1 ddk γλγργ γσγ k k = g3 C (N) C (G) (tA) αβ 2 − 2 2 ab (2π)d (k2)3 Z µ ρ λ 1 1 ddk γλγργ γ γ = g3 C (N) C (G) (tA) αβ 2 − 2 2 ab d (2π)d (k2)2 Z 1 (2 ǫ)2 ddk 1 = g3 C (N) C (G) (tA) − (γµ) 2 − 2 2 ab d αβ (2π)d (k2)2 Z ig3 1 d = C (N) C (G) (tA) (γµ) Γ(2 ). (3.330) (4π)2 2 − 2 2 ab αβ − 2 By putting the two results 12a and 12b together we obtain ig3 d 12a + 12b = C (N)+ C (G) (tA) (γµ) Γ(2 ). (3.331) (4π)2 2 2 ab αβ − 2 a Again if the quarks are in the representation tr instead of the fundamental representation ta we would have obtained ig3 d 12a + 12b = C (r)+ C (G) (tA) (γµ) Γ(2 ). (3.332) (4π)2 2 2 ab αβ − 2 13 Exercise: Compute this integral without making these approximations. 108 YDRI QFT

The dressed quark-gluon vertex at one-loop is therefore given by g3 d d4x d4y d4ze−ikx−ipy−ilzΓabA (x,y,z) = g(tA) (γµ) + C (r)+ C (G) (tA) (γµ) Γ(2 αβµ ab αβ (4π)2 2 2 ab αβ − 2 Z Z Z (2π)4δ4(k + p + l) × g3 2 = g(tA) (γµ) + C (r)+ C (G) (tA) (γµ) + ... ab αβ (4π)2 2 2 ab αβ ǫ (2π)4δ4(k + p + l) × (3.333)

Adding the contribution of the counter terms is trivial since the relevant counter term is of the same form as the bare vertex. We get g3 2 d4x d4y d4ze−ikx−ipy−ilzΓabA (x,y,z) = g(tA) (γµ) + C (r)+ C (G) (tA) (γµ) + ... αβµ ab αβ (4π)2 2 2 ab αβ ǫ Z Z Z A µ 4 4 + δ1g(t )ab(γ )αβ (2π) δ (k + p + l). (3.334) We conclude that, in order to subtract the logarithmic divergence in the vertex, the counter term δ1 must be chosen such that g2 2 δ = C (r)+ C (G) . (3.335) 1 −(4π)2 2 2 ǫ In a more careful treatment we should get 14 gˆ2 2 δ = C (r)+ C (G) . (3.336) 1 −(4π)2 2 2 ǫ We recall that the renormalized coupling g is related to the bare coupling g0 by the relation g Z √Z = 2 3 g0 1+ δ1 1 = 1 δ + δ + δ − 1 2 2 3 gˆ2 1 11 4 = 1+ C (G) n C(r) 16π2 ǫ 3 2 − 3 f g2 1 11 4 = 1+ µ−ǫ C (G) n C(r) . (3.337) 16π2 ǫ 3 2 − 3 f This is equivalent to g3 1 11 4 g = g + µ−ǫ 0 C (G) n C(r) . (3.338) 0 16π2 ǫ 3 2 − 3 f 14This should become apparent if you solve the previous two exercises. YDRI QFT 109

We compute then

∂g g3 11 4 µ = µ−ǫ 0 C (G) n C(r) ∂µ − 16π2 3 2 − 3 f g3 11 4 = C (G) n C(r) . (3.339) −16π2 3 2 − 3 f 110 YDRI QFT

A, µ B, ν

δAB k k = η + (ξ 1) µ ν . k2 µν − k2

¯a b ψα ψβ

δab(γ.p + m) = βα . p2 m2 −

c¯A cB

δAB = . p2

¯a ψα

µ, A

A µ = g(t )ab(γ )αβ. b ψβ

ν, B, p

ρ, C, q

= igf (k + 2p)µηρν (p + 2k)ν ηρµ + (k p)ρηµν . ABC − − µ, A, k YDRI QFT 111 ρ, D σ, E

= g2 f f (ηρµησν ησµηρν )+ f f (ησρηµν ηρµησν ) − ABC DEC − BDC AEC − µ, A ν, B + f f (ησµηρν ηµν ησρ) . DAC BEC − c¯A

µ, C

µ = igfABC k . cB −

. 112 YDRI QFT

. 4 The Renormalization Group

4.1 Critical Phenomena and The φ4 Theory

4.1.1 Critical Line and Continuum Limit We are interested in the critical properties of systems which are ergodic at finite volume, i.e. they can access all regions of their phase space with non zero probability. In the infinite volume limit these systems may become non ergodic and as a consequence the phase space decomposes into disjoint sets corresponding to different phases. The ther- modynamical limit is related to the largest eigenvalue of the so-called transfer matrix. If the system remains ergodic then the largest eigenvalue of the transfer matrix is non degenerate while it becomes degenerate if the system becomes non ergodic. The boundary between the different phases is demarcated by a critical line or a second order phase transition which is defined by the requirement that the correlation length, which is the inverse of the smallest decay rate of correlation functions or equivalent the smallest physical mass, diverges at the transition point. The properties of these systems near the transition line are universal and are described by the renormalization group equations of Euclidean scalar field theory. The requirement of locality in field theory is equivalent to short range forces in second order phase transitions. The property of universality is intimately related to the property of renormalizability of the field theory. More precisely universality in second order phase transitions emerges in the regime in which the correlation length is much larger than the macroscopic scale which corresponds, on the field theory side, to the fact that renormalizable local field theory is insensitive to short distance physics in the sense that we obtain a unique renormalized Lagrangian in the limit in which all masses and momenta are much smaller than the UV cutoff Λ. The Euclidean O(N) φ4 action is given by (with some change of notation compared 114 YDRI QFT to previous chapters and sections)

1 1 λ S[φ]= ddx (∂ φi)2 + m2φiφi + (φiφi)2 . (4.1) − 2 µ 2 4 Z d d i i We will employ lattice regularization in which x = an, d x = a n, φ (x) = φn and ∂ φi = (φi φi )/a. The lattice action reads µ n+ˆµ − n R P ad−2 adλ S[φ] = ad−2 φi φi (m2a2 + 2d)φi φi (φi φi )2 n n+ˆµ − 2 n n − 4 n n n µ X X = 2κ Φi Φi Φi Φi g(Φi Φi 1)2 . (4.2) n n+ˆµ − n n − n n − n µ X X The mass parameter m2 is replaced by the so-called hopping parameter κ and the coupling constant λ is replaced by the coupling constant g where 1 2g λ g m2a2 = − 2d , = . (4.3) κ − ad−4 κ2 i i The ﬁelds φn and Φn are related by

i 2κ i φn = d−2 Φn. (4.4) ra

1 2g λ g m2a2 = − 2d , = . (4.5) κ − ad−4 κ2 The partition function is given by

i S[φ] Z = dΦn e Z Yn,i 2κ Φi Φi = dµ(Φ) e Pn Pµ n n+ˆµ . (4.6) Z The measure dµ(φ) is given by

i i i i 2 i − Pn ΦnΦn+g(ΦnΦn−1) dµ(Φ) = dΦn e Yn,i N −Φ~ 2 −g(Φ~ 2 −1)2 = d Φ~ n e n n n Y dµ(Φ ). (4.7) ≡ n n Y This is a generalized Ising model. Indeed in the limit g the dominant conﬁgura- 2 2 −→ ∞ N−1 tions are such that Φ1 + ... +ΦN = 1, i.e. points on the sphere S . Hence YDRI QFT 115

dµ(Φ )f(Φ~ ) dΩ f(Φ~ ) n n = N−1 n , g . (4.8) dµ(Φ ) dΩ R n R N−1 −→ ∞ For N = 1 we obtain R R dµ(Φ )f(Φ~ ) 1 n n = (f(+1) + f( 1)) , g . (4.9) dµ(Φ ) 2 − −→ ∞ R n Thus the limit g R of the O(1) model is precisely the Ising model in d dimen- −→ ∞ sions. The limit g of the O(3) model corresponds to the Heisenberg model in d −→ ∞ dimensions. The O(N) models on the lattice are thus intimately related to spin models. There are two phases in this model. A disordered (paramagnetic) phase characterized by < Φi >= 0 and an ordered (ferromagnetic) phase characterized by < Φi >= v = 0. n n i 6 This can be seen in various ways. The easiest way is to look for the minima of the classical potential 1 λ V [φ]= ddx m2φiφi + (φiφi)2 . (4.10) − 2 4 Z The equation of motion reads λ [m2 + φjφj]φi = 0. (4.11) 2 For m2 > 0 there is a unique solution φi = 0 whereas for m2 < 0 there is a second solution given by φjφj = 2m2/λ. − i A more precise calculation is as follows. Let us compute the expectation value < Φn > on the lattice which is deﬁned by

2κ Φi Φi dµ(Φ) Φi e Pn Pµ n n+ˆµ < φi > = n n 2κ P P Φi Φi R dµ(Φ) e n µ n n+ˆµ κ Φi (Φi +Φi ) Rdµ(Φ) Φi e Pn n Pµ n+ˆµ n−µˆ = n . (4.12) κ P Φi P P (Φi +Φi ) R dµ(Φ) e n n n µ n+ˆµ n−µˆ i i Now we approximate the spins ΦRn at the 2d nearest neighbors of each spin Φn by the i i average v =< Φn >, viz (Φi +Φi ) µ n+ˆµ n−µˆ = vi. (4.13) P 2d This is a crude form of the mean ﬁeld approximation. Equation (4.12) becomes

i 4κd Φi vi dµ(Φ) Φ e Pn n vi = n 4κd P Φi vi R dµ(Φ) e n n i i dµ(Φ )Φi e4κdΦnv = R n n . (4.14) i 4κdΦi vi R dµ(Φn) e n R 116 YDRI QFT

The extra factor of 2 in the exponents comes from the fact the coupling between any two nearest neighbor spins on the lattice occurs twice. We write the above equation as

i ∂ v = ln Z[J] i i . (4.15) ∂J i |J =4κdv

Φi Ji Z[J] = dµ(Φn) e n Z i i i i 2 i i N i −ΦnΦn−g(ΦnΦn−1) +ΦnJ = d Φn e . (4.16) Z The limit g 0: In this case we have −→ i i i i i i J J N i −ΦnΦn+ΦnJ Z[J] = d Φn e = Z[0] e 4 . (4.17) Z In other words 1 vi = 2κ dvi κ = . (4.18) c ⇒ c 2d

The limit g : In this case we have −→ ∞

i i Z[J] = dN Φi δ(Φi Φi 1) eΦnJ N n n n − Z 1 = dN Φi δ(Φi Φi 1) 1+Φi J i + Φi Φj J iJ j + ... . (4.19) N n n n − n 2 n n Z By using rotational invariance in N dimensions we obtain

dN Φi δ(Φi Φi 1) Φi = 0. (4.20) n n n − n Z

δij δij Z[0] dN Φi δ(Φi Φi 1) Φi Φj = dN Φi δ(Φi Φi 1) Φk Φk = . (4.21) n n n − n n N n n n − n n N Z Z N Hence

J iJ i Z[J] = Z[0] 1+ + ... . (4.22) 2N Thus

J i 4κ dvi N vi = = c κ = . (4.23) N N ⇒ c 4d YDRI QFT 117

The limit of The Ising Model: In this case we have N = 1 , g . (4.24) −→ ∞ We compute then

Z[J] = dΦ δ(Φ2 1) eΦnJ N n n − Z = Z[0] cosh J. (4.25)

Thus

v = tanh 4κdv. (4.26)

A graphical sketch of the solutions of this equation is shown on ﬁgure 17. Clearly for κ near κc the solution v is near 0 and thus we can expand the above equation as 1 v = 4κdv (4κd)3v2 + .... (4.27) − 3 The solution is 1 (4d)2κ3v2 = κ κ . (4.28) 3 − c

Thus only for κ > κc there is a non zero solution. In summary we have the two phases

κ > κc : broken, ordered, ferromagnetic (4.29)

κ < κc : symmetric, disordered, paramagnetic. (4.30) The critical line κ = κ (g) interpolates in the κ g plane between the two lines given c c − by N κ = , g . (4.31) c 4d −→ ∞

1 κ = , g 0. (4.32) c 2d −→ See figure 18. For d = 4 the critical value at g = 0 is κc = 1/8 for all N. This critical value can be derived in a different way as follows. From equation (2.172) we know that the renormalized mass at one-loop order in the continuum φ4 with O(N) symmetry is given by the equation (with λ 6λ) −→ 2 2 2 mR = m + (N + 2)λI(m , Λ) (N + 2)λ (N + 2)λ m2 (N + 2)λ = m2 + Λ2 + m2 ln + m2C + finite terms. 16π2 16π2 Λ2 16π2 (4.33) 118 YDRI QFT

This equation reads in terms of dimensionless quantities as follows

(N + 2)λ (N + 2)λ (N + 2)λ a2m2 = am2 + + a2m2 ln a2m2 + a2m2C + a2 ﬁnite terms. R 16π2 16π2 16π2 × (4.34)

The lattice space a is formally identified with the inverse cut off 1/Λ, viz 1 a = . (4.35) Λ Thus we obtain in the continuum limit a 0 the result −→ (N + 2)λ (N + 2)λ (N + 2)λ a2m2 + a2m2 ln a2m2 + a2m2C + a2 finite terms. −→ − 16π2 16π2 16π2 × (4.36)

2 In other words (with r0 = (N + 2)/8π ) r a2m2 a2m2 = 0 λ + O(λ2). (4.37) −→ c − 2 This is the critical line for small values of the coupling constant as we will now show. Expressing this equation in terms of κ and g we obtain 1 2g r g − 8 0 + O(λ2). (4.38) κ − −→ − 2 κ2 This can be brought to the form

1 2 1 κ (1 2g) 1 + 16r g 4g + O(g2/κ2). (4.39) − 16 − −→ 256 0 − We get the result 1 r 1 κ κ = + ( 0 )g + O(g2). (4.40) −→ c 8 2 − 4 This result is of fundamental importance. The continuum limit a 0 corresponds −→ precisely to the limit in which the mass approaches its critical value. This happens for every value of the coupling constant and hence the continuum limit a 0 is the limit −→ in which we approach the critical line. The continuum limit is therefore a second order phase transition.

4.1.2 Mean Field Theory We start from the partition function of an O(1) model given by

P ΦnVnmΦm+P JnΦn Z(J) = dµ(Φn) e n,m n . (4.41) n Z Y YDRI QFT 119

The positive matrix Vnm (for the case of ferromagnetic interactions with κ> 0) is deﬁned by

Vnm = κ δm,n+ˆµ + δm,n−µˆ . (4.42) µˆ X The measure is deﬁned by

−Φ2 −g(Φ2 −1)2 dµ(Φn)= dΦn e n n . (4.43)

We introduce the Hubbard transformation

1 −1 − P XnVnm Xm+P ΦnXn P ΦnVnmΦm dXn e 4 n,m n = Ke n,m . (4.44) n Z Y We obtain

− 1 − 1 X V 1X (X +J )Φ Z(J) = dX e 4 Pn,m n nm m dµ(Φ ) ePn n n n K n n Z n Z n Y − Y 1 − 1 X V 1X − A(X +J ) = dX e 4 Pn,m n nm m Pn n n . (4.45) K n n Z Y The function A is deﬁned by

A(X + J )= ln z(X + J ) , z(X + J )= dµ(Φ ) e(Xn+Jn)Φn . (4.46) n n − n n n n n Z In the case of the Ising model we have explicitly

′ 1 ′ z(X + J )= dµ(Φ ) e(Xn+Jn)Φn = K e(Xn+Jn) + e−(Xn+Jn) = K cosh(X + J ). n n n 2 n n Z (4.47)

We introduce a new variable φn as follows

φn = Xn + Jn. (4.48)

The partition function becomes (using also the fact that V and V −1 are symmetric matrices)

− − − 1 − 1 φ V 1φ + 1 φ V 1J − 1 J V 1J − A(φ ) Z(J) = dφ e 4 Pn,m n nm m 2 Pn,m n nm m 4 Pn,m n nm m Pn n . K n n Z Y (4.49)

ˆ L l We replace Vij by Wij = Vij/L and we replace every spin Φn by Φn = l=1 Φn, i.e. by l the sum of L spins Φn which are assumed to be distributed with the same probability l P dµ(Φn). We get the partition function 120 YDRI QFT

ˆ ˆ ˆ l Pn,m ΦnWnmΦm+Pn JnΦn Z(J) = dµ(Φn) e Z Yn,l − 1 − 1 X W 1 X (X +J )Φ L = dX e 4 Pn,m n nm m dµ(Φ ) ePn n n n K n n Z n Z n Y Y − − − −L 1 φ V 1φ − 1 φ V 1J + 1 J V 1J + A(φ ) 1 4 Pn,m n nm m 2 Pn,m n nm m 4 Pn,m n nm m Pn n = dφ e K n n Z Y 1 dφ e−LV (φn). (4.50) ≡ K n n Z Y In the limit L we can apply the saddle point method. The partition function is −→ ∞ dominated by the configuration which solves the equation of motion dV dA = 0 φ J + 2 V = 0. (4.51) dφ ⇔ n − n nm dφ n m m X In other words we replace the field at each site by the best equivalent magnetic field. This approximation performs better at higher dimensions. Clearly steepest descent allows an expansion in powers of 1/L. We see that mean field is the tree level approximation of the field theory obtained from (4.50) by neglecting the quadratic term in Jn and redefining redefined −1 the current Jn as Jn = m Vnm Jm/2. The partition function becomes (up to a multiplicative constant factor) P − − − −L 1 φ V 1φ − 1 φ V 1J + 1 J V 1J + A(φ ) Z(J) = e 4 Pn,m n nm m 2 Pn,m n nm m 4 Pn,m n nm m Pn n . |saddle point (4.52) The vacuum energy (which plays the role of the thermodynamic free energy) is then given by 1 W (J) = ln Z[J] L 1 1 1 = φ V −1φ φ V −1J + J V −1J + A(φ ) . − 4 n nm m − 2 n nm m 4 n nm m n |saddle point n,m n,m n,m n X X X X (4.53)

The order parameter is the magnetization which is conjugate to the magnetic ﬁeld Jn. It is deﬁned by

∂W Mm = ∂Jm 1 = (φ J )V −1 2 n − n nm n X dA = . (4.54) −dφm YDRI QFT 121

The eﬀective action (which plays the role of the thermodynamic energy) is the Legendre transform of W (J) deﬁned by

Γ(M) = M J W (J) n n − n X = MnJn + MnVnmMm + A(φn) n n,m n X X X = M V M + B(M ). (4.55) − n nm m n n,m n X X The function B(Mn) is the Legendre transform of A(φn) given by

B(Mn)= Mnφn + A(φn). (4.56) For the Ising model we compute (up to an additive constant) A(φ ) = ln cosh φ n − n 1+ e−2φn = φ ln . (4.57) − n − 2 The magnetization in the Ising model is given by

1 e−2φn 1 1 Mn = − φn = ln(1 + Mn) ln(1 Mn). (4.58) 1+ e−2φn ⇔ 2 − 2 − Thus 1 1 A(φ )= ln(1 + M )+ ln(1 M ). (4.59) n 2 n 2 − n

1 1 B(M )= (1 + M ) ln(1 + M )+ (1 M ) ln(1 M ). (4.60) n 2 n n 2 − n − n From the definition of the effective potential we get the equation of motion ∂Γ ∂B = 2 V M + ∂M − nm m ∂M n m n X = (J φ )+ φ n − n n = Jn. (4.61) Thus for zero magnetic field the magnetization is given by an extremum of the effective potential. On the other hand the partition function for zero magnetic field is given by Z = exp( LΓ) and hence the saddle point configurations which dominate the partition − function correspond to extrema of the effective potential. In systems where translation is a symmetry of the physics we can assume that the magnetization is uniform, i.e. Mn = M = constant and as a consequence the effective potential per degree of freedom is given by Γ(M) = vM 2 + B(M). (4.62) − N 122 YDRI QFT

The number is the total number of degrees of freedom, viz = 1. The positive N N n parameter v is ﬁnite for short range forces and plays the role of the inverse temperature P β = 1/T . It is given explicitly by

Vnm v = n,m . (4.63) P N It is a famous exact result of statistical mechanics that the effective potential Γ(M) is a convex function of M, i.e. for M, M and M such that M = xM + (1 x)M with 1 2 1 − 2 0 )2 > . (4.65) dφ2 − − Thus d2A/dφ2 > 0 and as a consequence A is a convex function of φ 1. From the − definition of the partition function z(φ) and the explicit form of the measure dµ(Φ) we can see that Φ 0 for φ and hence we obtain the condition −→ −→ ±∞ d2A 0 , φ . (4.66) dφ2 −→ −→ ∞ Since M =< Φ > this condition also means that M 2 < Φ2 > 0 for φ . Now − −→ −→ ±∞ by differentiating Mn = ∂W/∂Jn with respect to Mn and using the result ∂Jn/∂Mn = 2 2 ∂ Γ/∂Mn we obtain ∂2W ∂2Γ 1= 2 2 . (4.67) ∂Jn ∂Mn 2 2 2 2 We compute (using Vnn = 0) the result ∂ Γ/∂Mn = d B/dMn. By recalling that φ = X + J we also compute (using V −1 = 0) the result ∂2W/∂J 2 = d2A/dφ2 . n n n nn n − n Hence we obtain d2B d2A 1= 2 2 . (4.68) − dMn dφn Thus the function B is also convex in the variable M. Furthermore the condition (4.65) leads to the condition that the function B goes to infinity faster than M 2 for M 2 −→ ±∞ (or else that M is bounded as in the case of the Ising model). | | The last important remark is to note that the functions A(φ) and B(M) are both even in their respective variables. There are two possible scenario we now consider:

1Exercise: Verify this explicitly. 2Exercise: Verify this explicitly. YDRI QFT 123

• First Order Phase Transition: For high temperature (small value of v) the effective action is dominated by the second term B(M) which is a convex function. The minimum of Γ(M) is M = 0. We start decreasing the temperature by increasing v. At some T = Tc (equivalently v = vc) new minima of Γ(M) appear which are degenerate with M = 0. For T < Tc the new minima become absolute minima and as a consequence the magnetization jumps discontinuously from 0 to a finite value corresponding to these new minima. See figure 14. ′′ In this case the second derivative of the effective potential at the minimum Γ (0) is always strictly positive and as a consequence the correlation length, which is ′′ inversely proportional to the square root of Γ (), is always finite.

• Second Order Phase Transition: The more interesting possibility occurs when the minimum at the origin M = 0 becomes at some critical temperature T = Tc a maximum and simultaneously new minima appear which start moving away from the origin as we decreasing the temperature. The critical temperature Tc is deﬁned ′′ by the condition Γ (0) = 0 or equivalently

′′ 2vc = B (0). (4.69)

Above Tc we have only the solution M = 0 whereas below Tc we have two minima moving continuously away from the origin. In this case the magnetization remains continuous at v = vc and as a consequence the transition is also termed continuous. Clearly the correlation length diverges at T = Tc. See ﬁgure 14.

4.1.3 Critical Exponents in Mean Field In the following we will only consider the second scenario. Thus we assume that we have a second order phase transition at some temperature T = Tc (equivalently v = vc). We are interested in the thermodynamic of the system for temperatures T near Tc. The transition is continuous and thus we can assume that the magnetization M is small near T = Tc and as a consequence we can expand the eﬀective action (thermodynamic energy) in powers of M. We write then

Γ(M) = M V M + B(M ) − n nm m n n,m n X X a b = M V M + M 2 + M 4 + ... . (4.70) − n nm m 2! n 4! n n,m n X X The function B(Mn) is the Legendre transform of A(φn), i.e.

B(Mn)= Mnφn + A(φn). (4.71)

We expand A(φn) in powers of φn as

′ ′ a b A(φ )= φ2 + φ4 + ... (4.72) n 2! n 4! n 124 YDRI QFT

Thus

′ dA ′ b 3 Mn = = a φn φn + ... (4.73) −dφn − − 6 We compute

2 2 d A d B −1 2 = [ 2 ] dφn − dMn 1 b = + M 2 + ... −a 2a2 n 1 b ′ = + a 2φ2 + ... (4.74) −a 2a2 n By integration this equation we obtain

1 b ′ A = φ2 + a 2φ4 + ... (4.75) −2a n 4!a2 n Hence

′ 1 ′ b a = , b = . (4.76) −a a4

′′ The critical temperature is given by the condition Γ (0) = 0 (where Γ here denotes the eﬀective potential Γ(M)= ( vM 2 + B(M))). This is equivalent to the condition ′′ N − B (0) = 2vc which gives the value (recall that the coeﬃcient a is positive since B is convex) a v = . (4.77) c 2

′ ′ The equation of motion Γ (0) = 0 gives the condition B (M) = 2vM. For v < vc we have no spontaneous magnetization whereas for v > vc we have a non zero spontaneous magnetization given by

12 1/2 M = (v vc) . (4.78) r b −

The magnetization is associated with the critical exponent β deﬁned for T near Tc from below by

M (T T )β. (4.79) ∼ c − We have clearly

1 β = . (4.80) 2 YDRI QFT 125

The inverse of the (magnetic) susceptibility is defined by (with J being the magnetic field) ∂M χ−1 = ∂J δ2Γ = δM 2 b = ( 2v + a + M 2). (4.81) N − 2 We have the 2 cases − v < v , M = 0 χ−1 = 2(v v) c ⇒ c − 12 1/2 −1 v > vc , M = (v vc) χ = 4(v vc). (4.82) r b − ⇒ − The susceptibility is associated with the critical exponent γ defined by

χ T T −γ. (4.83) ∼ | − c| Clearly we have

γ = 1. (4.84)

The quantum equation of motion (equation of state) relates the source (external magnetic ﬁeld), the temperature and the spontaneous magnetization. It is given by ∂Γ J = ∂M b = (2(v v)M + M 3) N c − 6 b = N M 3. (4.85) 3 The equation of state is associated with the critical exponent δ deﬁned by

J M δ. (4.86) ∼ Clearly we have

δ = 3. (4.87)

Let us derive the 2 point correlation function given by − δ2Γ −1 G(2) = nm δM δM n m b −1 = 2V + aδ + M 2δ . (4.88) − nm nm 2 n nm 126 YDRI QFT

Define b Γ(2) = 2V + aδ + M 2δ . (4.89) nm − nm nm 2 n nm (2) (2) The two functions Gnm and Γnm can only depend on the difference n m due to invariance − under translation. Thus Fourier transform is and its inverse are defined by

π ddk K = K˜ (k) eik(n−m) , K˜ (k)= K e−ik(n−m). (4.90) nm (2π)d nm −π n Z X For simplicity we assume a uniform magnetization, viz M = Mn. Thus ˜(2) (2) −ik(n−m) Γ (k) = Γnme n X b = 2V˜ (k)+ a + M 2. (4.91) − 2 Hence π ddk 1 G(2) = eik(n−m). (4.92) nm d b 2 −π (2π) 2V˜ (k)+ a + M Z − 2 The function V˜ (k) is given explicitly by

−ik(n−m) V˜ (k)= Vnme . (4.93) n X We assume a short range interaction which means that the potential Vnm decays exponentially with the distance n m . In other words we must have | − | −κ|n−m| Vnm < Me , κ> 0. (4.94)

This condition implies that the Fourier transform V˜ (k) is analytic for Im k < κ 3. | | Furthermore positivity of the potential Vnm and its invariance under translation gives the requirement

V˜ (k) V = V˜ (0) = v. (4.95) | |≤ nm n X For small momenta k we can then expand V˜ (k) as

V˜ (k)= v(1 ρ2k2 + O(k4)). (4.96) − The 2 point function admits therefore the expansion − π ddk G˜(2)(0) G(2) = eik(n−m). (4.97) nm (2π)d 1+ ξ2k2 + O(k4) Z−π 3Exercise: Construct an explicit argument. YDRI QFT 127

1 G˜(2)(0) = . (4.98) 2(v v)+ b M 2 c − 2

2vρ2 ξ2 = . (4.99) 2(v v)+ b M 2 c − 2 The length scale ξ is precisely the so-called correlation length which measures the exponential decay of the 2 point function. Indeed we can write the 2 point function − − as π d d k 2 2 G(2) = G˜(2)(0)e−ξ k eik(n−m). (4.100) nm (2π)d Z−π More generally it is not difficult to show that the denominator 2V˜ (k)+ a + bM 2/2 is − strictly positive for v > v and hence the 2 point function decays exponentially which c − indicates that the correlation length is finite. We have the two cases vρ2 v < v : M = 0 ξ2 = c ⇒ v v c − 12 vρ2 v > v : M = (v v )1/2 ξ2 = . (4.101) c b − c ⇒ 2(v v ) r − c The correlation length ξ is associated with the critical exponent ν defined by

ξ T T −ν. (4.102) ∼ | − c| Clearly we have 1 ν = . (4.103) 2

The correlation length thus diverges at the critical temperature T = Tc. A more robust calculation which shows this fundamental result is easily done in the continuum. In the continuum limit the 2 point function (4.97) becomes − ddk 1 G(2)(x,y) = eik(x−y). (4.104) (2π)d m2 + k2 Z The squared mass parameter is given by

b 2 1 2(vc v)+ M m2 = = − 2 v v [T T . (4.105) ξ2 2vρ2 ∼ | − c|∼ − c| We compute 4 2 2m G(2)(x,y) = ( )d/2−1K (mr). (4.106) (4π)d/2 r 1−d/2

4Exercise: Do this important integral. 128 YDRI QFT

For large distances we obtain 5 1 m e−mr G(2)(x,y) = ( )(d−1)/2 , r . (4.107) 2m 2π r(d−1)/2 −→ ∞ The last crucial critical exponent is the anomalous dimension η. This is related to the behavior of the 2 point function at T = T . At T = T we have v = v and M = 0 and − c c c hence the 2 point function becomes − π ddk 1 G(2) = eik(n−m). (4.108) nm (2π)d 2v (ρ2k2 O(k4)) Z−π c − Thus the denominator vanishes only at k = 0 which is consistent with the fact that the correlation length is inﬁnite at T = Tc. This also leads to algebraic decay. This can be checked more easily in the continuum limit where the 2 point function becomes − ddk 1 G(2)(x,y) = eik(x−y). (4.109) (2π)d k2 Z We compute 6

2d−2 1 G(2)(x,y) = Γ(d/2 1) . (4.110) (4π)d/2 − rd−2 The critical exponent η is deﬁned by the behavior 1 G(2)(x,y) . (4.111) ∼ rd−2+η The mean ﬁeld prediction is therefore given by

η = 0. (4.112)

In this section we have not used any particular form for the potential Vmn. It will be an interesting exercise to compute directly all the critical exponents β, γ, δ, ν and η for the case of the O(1) model corresponding to the nearest-neighbor interaction (4.42) 7. This of course includes the Ising model as a special case.

4.2 The Callan-Symanzik Renormalization Group Equation

4.2.1 Power Counting Theorems We consider a φr theory in d dimensions given by the action 1 µ2 g S[φ]= ddx ∂ φ∂µφ + φ2 φr . (4.113) 2 µ 2 − 4 Z 5Exercise: Check this limit 6Exercise: Do this important integral. 7Exercise: Compute the exponents β, γ, δ, ν and η for the potential (4.42). YDRI QFT 129

The case of interest is of course d = 4 and r = 4. In natural units where ~ = c = 1 the action is dimensionless, viz [S] = 1. In these units time and length has the same dimension whereas mass, energy and momentum has the same dimension. We take the fundamental dimension to be that of length or equivalently that of mass. We have clearly (for example from Heisenberg uncertainty principle)

1 L = . (4.114) M

[t] = [x]= L = M −1 , [m] = [E] = [p]= M. (4.115)

It is clear that the Lagrangian density is of mass dimension M d and as a consequence the ﬁeld is of mass dimension M (d−2)/2 and the coupling constant g is of mass dimension M d−rd/2+r (use the fact that [∂]= M). We write

d−2 [φ]= M 2 . (4.116)

− d−r d 2 δ d 2 [g]= M 2 M r , δ = d r − . (4.117) ≡ r − 2 r The main result of power counting states that φ theory is renormalizable only in dc dimension where dc is given by the condition 2r δ = 0 d = . (4.118) r ⇔ c r 2 − The eﬀective action is given by

∞ 1 Γ[φ ]= ddx ... ddx Γ(n)(x , ..., x )φ (x )...φ (x ). (4.119) c n! 1 n 1 n c 1 c n n=0 X Z Z Since the eﬀective action is dimensionless the n point proper vertices Γ(n)(x , ..., x ) − 1 n have mass dimension such that

− 1 (n) n d 2 (n) nd +n 1= [Γ (x , ..., x )]M 2 [Γ (x , ..., x )] = M 2 . (4.120) M nd 1 n ⇔ 1 n The Fourier transform is deﬁned as usual by

d d (n) ip1x1+...+ipnxn d d (n) d x1... d xnΓ (x1, ..., xn) e = (2π) δ (p1 + ... + pn)Γ˜ (p1, ..., p(4.121)n). Z Z From the fact that ddpδd(p) = 1 we conclude that [δd(p)] = M −d and hence

R (n) d−n( d −1) [Γ˜ (p1, ..., pn)] = M 2 . (4.122) 130 YDRI QFT

The n point function G(n)(x , ..., x ) is the expectation value of the product of n ﬁelds − 1 n and hence it has mass dimension

− (n) n d 2 [G (x1, ..., xn)] = M 2 . (4.123)

The Fourier transform is deﬁned by

d d (n) ip1x1+...+ipnxn d d (n) d x1... d xnG (x1, ..., xn) e = (2π) δ (p1 + ... + pn)G˜ (p1, ..., p(4.124)n). Z Z Hence

(n) d−n( d +1) [G˜ (p1, ..., pn)] = M 2 . (4.125)

We consider now an arbitrary Feynman diagram in a φr theory in d dimensions. This diagram is contributing to some n point proper vertex Γ˜(n)(p , ..., p ) and it can be − 1 n characterized by the following: • L=number of loops.

• V =number of vertices.

• P =number of propagators (internal lines).

• n=number of external lines (not to be considered propagators). We remark that each propagator is associated with a momentum variable. In other words we have P momenta which must be constrained by the V delta functions associated with the V vertices and hence there can only be P V momentum integrals in this diagram. − However, only one delta function (which enforces energy-momentum conservation) sur- vives after integration and thus only V 1 delta functions are actually used. The number − of loops L must be therefore given by

L = P (V 1). (4.126) − − Since we have r lines coming into a vertex the total number of lines coming to V vertices is rV . Some of these lines are propagators and some are external lines. Clearly among the rV lines we have precisely n external lines. Since each propagator connects two vertices it must be counted twice. We have then

rV = n + 2P. (4.127)

(n) V It is clear that Γ˜ (p1, ..., pn) must be proportional to g , viz

(n) V Γ˜ (p1, ..., pn)= g f(p1, ..., pn). (4.128)

We have clearly d [f(p , ..., p )] = M δ , δ = V δ + d n( 1). (4.129) 1 n − r − 2 − YDRI QFT 131

The index δ is called the superficial degree of divergence of the Feynman graph. The physical significance of δ can be unraveled as follows. Schematically the function f is of the form Λ Λ 1 1 f(p , ..., p ) ddk ... ddk ... δd( p k) V −1. (4.130) 1 n ∼ 1 P k2 µ2 k2 µ2 − Z0 Z0 1 P − − X X If we neglect, in a first step,the delta functions than we can see immediately that the P (d−2) asymptotic behavior of the integral f(p1, ..., pn) is Λ . This can be found by factoring out the dependence of f on Λ via the rescaling k Λk. By taking the delta functions −→ into considerations we see immediately that the number of independent variables reduces and hence the asymptotic behavior of f(p1, ..., pn) becomes

f(p , ..., p ) ΛP (d−2)−d(V −1). (4.131) 1 n ∼ By using P = (rV n)/2 we arrive at the result − −V δ +d−n( d −1) f(p , ..., p ) Λ r 2 1 n ∼ Λδ. (4.132) ∼ The index δ controls therefore the ultraviolet behavior of the graph. From the last two equations it is obvious that δ is the diﬀerence between the power of k in the numerator and the power of k in the denominator, viz

δ = (power of k in numerator) (power of k in denominator) (4.133) − Clearly a negative index δ corresponds to convergence whereas a positive index δ corresponds to divergence. Since δ is only a superﬁcial degree of divergence there are ex- ceptions to this simple rule. More precisely we have the following ﬁrst power counting theorem:

• For δ > 0 the diagram diverges as Λd. However symmetries (if present) can reduce/eliminate divergences in this case.

• For δ = 0 the diagram diverges as ln Λ. An exception is the trivial diagram (P = L = 0).

• For δ < 0 the diagram converges absolutely if it contains no divergent subdiagrams. In other words a diagram with δ < 0 which contains divergent subdiagrams is generically divergent.

As an example let us consider φ4 in 4 dimensions. In this case

δ = 4 n. (4.134) − Clearly only the 2 point and the 4 point proper vertices are superficially divergent, i.e. − − they have δ 0. In particular for n = 4 we have δ = 0 indicating possible logarithmic ≥ 132 YDRI QFT divergence which is what we had already observed in actual calculations. For n = 6 we observe that δ = 2 < 0 which indicates that the 6 point proper vertex is superficially − − convergent. In other words the diagrams contributing to the 6 point proper vertex − may or may not be convergent depending on whether or not they contain divergent subdiagrams. For example the one-loop diagram on figure 13 is convergent whereas the two-loop diagrams are divergent. The third rule of the first power counting theorem can be restated as follows: • A Feynman diagram is absolutely convergent if and only if it has a negative superficial degree of divergence and all its subdiagrams have negative superficial degree of divergence. The φ4 theory in d = 4 is an example of a renormalizable field theory. In a renormalizable field theory only a finite number of amplitudes are superficially divergent. As we have already seen, the divergent amplitudes in the case of the φ4 theory in d = 4 theory, are the 2 point and the 4 point amplitudes. All other amplitudes may diverge only − − if they contain divergent subdiagrams corresponding to the 2 point and the 4 point − − amplitudes. Another class of field theories is non-renormalizable field theories. An example is φ4 in D = 6. In this case

δ = 2 , δ = 2V + 6 2n. (4.135) r − − The formula for δ depends now on the order of perturbation theory as opposed to what happens in the case of D = 4. Thus for a fixed n the superficial degree of divergence increases by increasing the order of perturbation theory, i.e. by increasing V . In other words at a sufficiently high order of perturbation theory all amplitudes are divergent. In a renormalizable field theory divergences occur generally at each order in perturbation theory. For φ4 theory in d = 4 all divergences can be removed order by order in perturbation theory by redefining the mass, the coupling constant and the wave function renormalization. This can be achieved by imposing three renormalization conditions on (2) (2) 2 (4) Γ˜ (p), dΓ˜ (p)/dp and Γ˜ (p1, ..., p4) at 0 external momenta corresponding to three distinct experiments. In contrast we will require an infinite number of renormalization conditions in order to remove the divergences occurring at a sufficiently high order in a non-renormalizable field theory since all amplitudes are divergent in this case. This corresponds to an infinite number of distinct experiments and as a consequence the theory has no predictive power. From the formula for the superficial degree of divergence δ = δ V + d n(d/2 1) − r − − we see that δr, the mass dimension of the coupling constant, plays a central role. For 4 3 δr = 0 (such as φ in d = 4 and φ in d = 6 we see that the index δ is independent of the order of perturbation theory which is a special behavior of renormalizable theory. 4 For δr < 0 (such as φ in d > 4) we see that δ depends on V in such a way that it increases as V increases and hence we obtain more divergencies at each higher order of perturbation theory. Thus δr < 0 defines the class of non-renormalizable field theories as δr = 0 defines the class of renormalizable field theories. YDRI QFT 133

Another class of field theories is super-renormalizable field theories for which δr > 0 (such as φ3 in D = 4). In this case the superficial degree of divergence δ decreases with increasing order of perturbation theory and as a consequence only a finite number of Feynman diagrams are superficially divergent. In this case no amplitude diverges. The second (main) power counting theorem can be summarized as follows: • Super-Renormalizable Theories: The coupling constant g has positive mass dimension. There are no divergent amplitudes and only a finite number of Feynman diagrams superficially diverge.

• Renormalizable Theories: The coupling constant g is dimensionless. There is a finite number of superficially divergent amplitudes. However since divergences occur at each order in perturbation theory there is an infinite number of Feynman diagrams which are superficially divergent.

• Non-Renormalizable Theories: The coupling constant g has negative mass dimension. All amplitudes are superﬁcially divergent at a suﬃciently high order in perturbation theory.

4.2.2 Renormalization Constants and Renormalization Conditions We write the φ4 action in d = 4 as 1 1 λ S[φ] = ddx ∂ φ∂µφ m2φ2 (φ2)2 . (4.136) 2 µ − 2 − 4! Z The bare ﬁeld φ, the bare coupling constant λ and the bare mass m2 are given in terms of the renormalized ﬁeld φR, the renormalized coupling constant λR and the renormalized 2 mass mR respectively by the relations

φ = √ZφR. (4.137)

2 λ = Zg/Z λR. (4.138)

2 2 m = (mR + δm)/Z. (4.139) The renormalization constant Z is called wave function renormalization constant (or 2 equivalently ﬁeld amplitude renormalization constant) whereas Zg/Z is the coupling constant renormalization constant. The action S given by equation (4.136) can be split as follows

S = SR + δS. (4.140)

The renormalized action SR is given by 1 1 λ S [φ ] = ddx ∂ φ ∂µφ m2 φ2 R (φ2 )2 . (4.141) R R 2 µ R R − 2 R R − 4! R Z 134 YDRI QFT

The counter-term action δS is given by δ 1 δ δS[φ ] = d4x Z ∂ φ ∂µφ δ φ2 λ (φ2 )2 . (4.142) R 2 µ R R − 2 m R − 4! R Z The counterterms δZ , δm and δλ are given by

δ = Z 1 , δ = Zm2 m2 , δ = λZ2 λ = (Z 1)λ . (4.143) Z − m − R λ − R g − R The renormalized n point proper vertex Γ(n) is given in terms of the bare n point − R − proper vertex Γ(n) by

(n) n 2 (n) ΓR (x1, ..., xn)= Z Γ (x1, ..., xn). (4.144) The effective action is given by (where φ denotes here the classical field) 1 Γ [φ ] = Γ(n)(x , ..., x )φ (x )...φ (x ). (4.145) R R n! R 1 n R 1 R n nX=0 We assume a momentum cutoff regularization. The renormalization constants Z and Zg and the counterterm δm are expected to be of the form

2 δm = a1(Λ)λr + a2(Λ)λr + ... 2 Z =1+ b1(Λ)λr + b2(Λ)λr + ... 2 Zg =1+ c1(Λ)λr + c2(Λ)λr + ... (4.146)

All other quantities can be determined in terms of Z and Zg and the counterterm δm. We can state our third theorem as follows: • Renormalizability of the φ4 theory in d = 4 means precisely that we can choose the constants ai, bi and ci such that all correlation functions have a ﬁnite limit order by order in λ when Λ . R −→ ∞ We can eliminate the divergences by imposing appropriate renormalization conditions at zero external momentum. For example we can choose to impose conditions consistent with the tree level action, i.e.

(2) 2 Γ˜ (p) 2 = m R |p =0 R d (2) Γ˜ (p) 2 = 1 dp2 |p =0 ˜(4) Γ (p1, ..., p4) p2=0 = λR. (4.147) | i This will determine the superﬁcially divergent amplitudes completely and removes divergences at all orders in perturbation theory 8.

8Exercise: YDRI QFT 135

It is well established that a far superior regularization method,than the simple cutoff used above, is dimensional regularization in which case we use, instead of renormalization conditions, the so-called minimal subtraction (MS) and modified minimal subtraction (MMS) schemes to renormalize the theory. In minimal subtraction scheme we subtract only the pole term and nothing else. In dimension d = 4 the coupling constant λ is not dimensionless. The dimensionless 6 coupling constant in this case is given by g defined by

g = µ−ǫλ, ǫ = 4 d. (4.148) − The bare action can then be put in the form

Z Z m2 µǫg Z S = ddx ∂ φ ∂µφ m R φ2 R g (φ2 )2 . (4.149) 2 µ R R − 2 R − 4! R Z The new renormalization condition Zm is deﬁned through the equation

Z m2 = m2 m . (4.150) R Z The mass µ2 is an arbitrary mass scale parameter which plays a central role in dimensional regularization and minimal subtraction. The mass µ2 will deﬁne the subtraction point. In other words the mass scale at which we impose renormalization conditions in the form

(2) 2 Γ˜ (p) 2 = m R |p =0 R d (2) Γ˜ (p) 2 2 = 1 dp2 |p =µ Γ˜(4)(p , ..., p ) = µǫg . (4.151) 1 4 |SP R The symmetric point SP is deﬁned by p .p = µ2(4δ 1)/3. For massive theories we i j ij − can simply choose µ = mR. According to Weinberg’s theorem (and other considerations) the only correlation functions of massless φ4 which admit a zero momentum limit is the 2 point function. This means in particular that the second and third renormalization − conditions (4.147) do not make sense in the massless limit m2 0 and should be re- R −→ placed by the second and third renormalization conditions (4.151). This is also the reason why we have kept the ﬁrst renormalization condition unchanged. The renormalization conditions (4.151) are therefore better behaved. As pointed above the renormalization prescription known as minimal subtraction is far superior than the above prescription of imposing renormalization conditions since it

• Show that the loopwise expansion is equivalent to an expansion in powers of λ. • Write down the one-loop eﬀective action of the φ4 theory in d = 4. Use a Gaussian cutoﬀ.

• Compute a1, b1 and c1 at the one-loop order of perturbation theory. • Consider one-loop renormalization of φ3 in d = 6. 136 YDRI QFT is intimately tied to dimensional regularization. In this prescription the mass scale µ2 appears only via (4.148). We will keep calling µ2 the subtraction point since minimal subtraction must be physically equivalent to imposing the renormalization conditions (4.151) although the technical detail is generically different in the two cases. ˜(n) The renormalized proper vertices ΓR depend on the momenta p1,...,pn but also on 2 the renormalized mass mR, the renormalized coupling constant gR and the cutoff Λ. In the case of dimensional regularization the cutoff is ǫ = 4 d whereas in the case of − ˜(n) lattice regularization the cutoff is the inverse lattice spacing. The proper vertices ΓR 2 2 will also depend on the mass scale µ explicitly and implicitly through mR and gR. The renormalized proper vertices are related to the bare proper vertices as

(n) n ˜ 2 2 2 ˜(n) 2 ΓR (pi,µ ; mR, gR, Λ) = Z Γ (pi; m , g, Λ). (4.152)

The renormalization constant Z (and also other renormalization constants Zg, Zm and counterterms δZ , δm and δλ) will only depend on the dimensionless parameters gR, 2 2 2 2 2 2 mR/µ , Λ /µ , mR/Λ as well as on Λ, viz m2 Λ2 m2 Z = Z(g , R , , R , Λ). (4.153) R µ2 µ2 Λ2 In dimensional regularization we have

2 (n) 2 2 n mR (n) 2 Γ˜ (p ,µ ; m , g ,ǫ)= Z 2 (g , ,ǫ)Γ˜ (p ; m ,g,ǫ). (4.154) R i R R R µ2 i

Renormalizability of the φ4 theory in d = 4 via renormalization conditions (the fourth theorem) can be stated as follows:

˜(n) 2 2 2 2 • The renormalized proper vertices ΓR (pi,µ ; mR, gR, Λ) at ﬁxed pi, µ , gR, mR ˜(n) 2 2 have a large cut-oﬀ limit ΓR (pi,µ ; mR, gR) which are precisely the physical proper vertices, viz

(ln Λ)L Γ˜(n)(p ,µ2; m2 , g , Λ) = Γ˜(n)(p ,µ2; m2 , g )+ O . (4.155) R i R R R i R R Λ2 ˜(n) 2 2 The renormalized physical proper vertices ΓR (pi,µ ; mR, gR) are universal in the sense that they do not depend on the speciﬁc cut-oﬀ procedure as long as the renormalization conditions (4.151) are kept unchanged. In the above equation L is the number of loops.

4.2.3 Renormalization Group Functions and Minimal Subtraction The bare mass m2 and the bare coupling constant λ are related to the renormalized mass 2 mR and renormalized coupling constant λR by the relations Z m2 = m2 m . (4.156) R Z YDRI QFT 137

Z Z λ = g λ = g µǫg . (4.157) Z2 R Z2 R In dimensional regularization the renormalization constants will only depend on the di- 2 2 mensionless parameters gR, mR/µ as well as on ǫ. We may choose the subtraction mass 2 2 2 scale µ = mR. Clearly the bare quantities m and λ are independent of the mass scale µ. Thus by diﬀerentiating both sides of the above second equation with respect to µ2 keeping m2 and λ ﬁxed we obtain

∂λ ∂ 2 0= µ β = ǫgR µ ln Zg/Z gR. (4.158) ∂µ 2 ⇒ − − ∂µ 2 λ,m λ,m The so-called renormalization group beta function β (also called the Gell-Mann Law function) is defined by 2 mR ∂gR β = β(gR, 2 )= µ (4.159) µ ∂µ 2 λ,m Let us define the new dimensionless coupling constant Z G = g g . (4.160) Z2 R Alternatively by differentiating both sides of equation (4.157) with respect to µ keeping m2 and λ fixed we obtain ∂λ ∂ ∂ ∂ 0= µ 0= ǫG + β G + µ mR G. (4.161) ∂µ 2 ⇒ ∂g ∂µ 2 ∂m λ,m R λ,m R The last term is absent when µ = mR. Next by differentiating both sides of equation (4.156) with respect to µ keeping m2 and λ fixed we obtain 2 2 ∂m ∂mR 2 ∂ 0= µ 0= µ + mR µ ln Zm/Z . (4.162) ∂µ 2 ⇒ ∂µ 2 ∂µ 2 λ,m λ,m λ,m We define the renormalization group function γ by 2 mR ∂ 2 γm = γm(gR, 2 ) = µ ln mR µ ∂µ 2 λ,m ∂ = µ ln Zm/Z . (4.163) − ∂µ 2 λ,m In the minimal subtraction scheme the renormalization constants will only depend on the dimensionless parameters gR and as a consequence the renormalization group functions will only depend on gR. In this case we find d Z −1 β(g ) = ǫg 1+ g ln g R − R R dg Z R d −1 = ǫ ln G(g ) . (4.164) − dg R R 138 YDRI QFT

d Zm γm(gR) = β(gR) ln . (4.165) − dgR Z

We go back to the renormalized proper vertices (in minimal subtraction) given by

˜(n) 2 2 n/2 ˜(n) 2 ΓR (pi,µ ; mR, gR,ǫ)= Z (gR,ǫ)Γ (pi; m ,g,ǫ). (4.166)

Again the bare proper vertices must be independent of the subtraction mass scale, viz

∂ 0= µ Γ˜(n) . (4.167) ∂µ 2 λ,m By diﬀerentiating both sides of equation (4.166) with respect to µ keeping m2 and λ ﬁxed we obtain

∂ ˜(n) n ∂ ˜(n) µ ΓR = µ ln Z ΓR . (4.168) ∂µ 2 2 ∂µ 2 λ,m λ,m Equivalently we have

2 ∂ ∂mR ∂ ∂gR ∂ ˜(n) n ∂ ˜(n) µ + µ 2 + µ ΓR = µ ln Z ΓR . (4.169) ∂µ ∂µ ∂m ∂µ ∂g 2 2 ∂µ 2 R R λ,m λ,m We get ﬁnally

∂ ∂ ∂ n µ + m2 γ + β η Γ˜(n) = 0. (4.170) ∂µ R m ∂m2 ∂g − 2 R R R This is our first renormalization group equation. The new renormalization group function η (also called the anomalous dimension of the field operator) is defined by

∂ η(gR) = µ ln Z ∂µ 2 λ,m d = β(gR) ln Z. (4.171) dgR

Renormalizability of the φ4 theory in d = 4 via minimal subtraction (the ﬁfth theorem) can be stated as follows:

˜(n) 2 2 • The renormalized proper vertices ΓR (pi,µ ; mR, gR,ǫ) and the renormalization group functions β(g ), γ(g ) and η(g ) have a ﬁnite limit when ǫ 0. R R R −→

By using the above the theorem and the fact that G(gR) = gR + .... we conclude that the beta function must be of the form

β(g )= ǫg + β (ǫ)g2 + β (ǫ)g3 + ... (4.172) R − R 2 R 3 R YDRI QFT 139

The functions β (ǫ) are regular in the limit ǫ 0. By using the result (4.164) we ﬁnd i −→ ′ G ǫg g = R R G − β β (ǫ) β2(ǫ) β (ǫ) = 1+ 2 g + ( 2 + 3 )g2 + ... (4.173) ǫ R ǫ2 ǫ R

The most singular term in ǫ is captured by the function β2(ǫ). By integrating this equation we obtain

β (0) −1 G(g )= g 1 2 g + less singular terms. (4.174) R R − ǫ R

The function G(gR) can then be expanded as

n ˜ G(gR)= gR + gRGn(ǫ). (4.175) n=2 X

The functions G˜n(ǫ) behave as

βn−1(0) G˜ (ǫ)= 2 + less singular terms. (4.176) n ǫn−1 Alternatively we can expand G as

G (g ) G(g )= g + n R + regular terms , G (g )= O(gn+1). (4.177) R R ǫn n R R n=1 X This is equivalent to

Z H (g ) g =1+ n R + regular terms ,H (g )= O(gn ). (4.178) Z2 ǫn n R R n=1 X We compute the beta function

′ G (g ) G (g ) −1 β(g ) = ǫ g + n R 1+ n R R − R ǫn ǫn n=1 n=1 X ′ X ′ ′ G G (G )2 G = ǫ g + 1 + ... 1 1 + 1 2 + ... − R ǫ − ǫ ǫ2 − ǫ2 ′ b (g ) = ǫg G (g )+ g G (g )+ n R . (4.179) − R − 1 R R 1 R ǫn nX=1 The beta function is ﬁnite in the limit ǫ 0 and as a consequence we must have −→ bn(gR) = 0 for all n. The beta function must therefore be of the form

′ β(g ) = ǫg G (g )+ g G (g ). (4.180) R − R − 1 R R 1 R 140 YDRI QFT

The beta function β is completely determined by the residue of the simple pole of G, i.e. by G . In fact all the functions G with n 2 are determined uniquely by G (from the 1 n ≥ 1 condition bn = 0). Similarly from the ﬁniteness of η in the limit ǫ 0 we conclude that the renormal- −→ ization constant Z is of the form α (g ) Z(g )=1+ n R + regular terms , α (g )= O(gn+1). (4.181) R ǫn n R R Xn=1 We compute the anomalous dimension

d η = β(gR) ln Z(gR) dgR ′ 1 ′ = ǫg G (g )+ g G (g ) α + ... . (4.182) − R − 1 R R 1 R ǫ 1 Since η is ﬁnite in the limit ǫ 0 we must have −→ ′ η = g α . (4.183) − R 1 4.2.4 CS Renormalization Group Equation in φ4 Theory We will assume d = 4 in this section although much of what we will say is also valid in other dimensions. We will also use a cutoﬀ regularization throughout.

Inhomogeneous CS RG Equation: Let us consider now φ4 theory with φ2 insertions. We add to the action (4.136) a source term of the form ddxK(x)φ2(x)/2, i.e. R 1 1 λ 1 S[φ, K] = ddx ∂ φ∂µφ m2φ2 (φ2)2 + Kφ2 . (4.184) 2 µ − 2 − 4! 2 Z Then we consider the path integral

Z[J, K] = φ exp(iS[φ, K]+ i ddxJφ). (4.185) D Z Z It is clear that diﬀerentiation with respect to K(x) generates insertions of the operator φ2/2. The corresponding renormalized ﬁeld theory will be given by the path integral −

Z [J, K] = φ exp(iS[φ , K]+ i ddxJφ ). (4.186) R D R R R Z Z

1 1 λ Z S[φ , K] = ddx ∂ φ ∂µφ m2 φ2 R (φ2 )2 + 2 Kφ2 + δS.(4.187) R 2 µ R R − 2 R R − 4! R 2 R Z YDRI QFT 141

d 2 Z2 is a new renormalization constant associated with the operator d xK(x)φ (x)/2. We have clearly the relations R J Z2 WR[J, K]= W [ , K]. (4.188) √Z Z

Z Γ [φ , K]=Γ[√Zφ , 2 K]. (4.189) R c c Z The renormalized (l,n) point proper vertex Γ(l,n) is given in terms of the bare (l,n) point − R − proper vertex Γ(l,n) by

(l,n) n 2 −l l (l,n) ΓR (y1, ..., yl; x1, ..., xn)= Z Z2Γ (y1, ..., yl; x1, ..., xn). (4.190) (1,2) The proper vertex Γ (y; x1,x2) is a new superﬁcially divergent proper vertex which requires a new counterterm and a new renormalization condition. For consistency with the tree level action we choose the renormalization condition

Γ˜(1,2)(q; p ,p ) = 1. (4.191) R 1 2 |q=pi=0 Let us remark that correlation functions with one operator insertion iφ2(y)/2 are deﬁned by

i 2 1 1 δ δ δ < φ (y)φ(x1)...φ(xn) >= n ... Z[J, K] J=K=0. (4.192) 2 i Z[J, K] δK(y) δJ(x1) δJ(xn) | This can be generalized easily to

l i 2 2 1 1 δ δ δ δ < l φ (y1)...φ (yl)φ(x1)...φ(xn) >= n ...... Z[J, K] J=K=0. 2 i Z[J, K] δK(y1) δK(yl) δJ(x1) δJ(xn) | (4.193)

From this formula we see that the generating functional of correlation functions with l operator insertions iφ2(y)/2 is defined by δ δ Z[y1, ..., yl; J]= ... Z[J, K] K=0. (4.194) δK(y1) δK(yl) | The generating functional of the connected correlation functions with l operator insertions iφ2(y)/2 is then defined by δ δ W [y1, ..., yl; J]= ... W [J, K] K=0. (4.195) δK(y1) δK(yl) | We write the effective action as 1 Γ[φ , K] = ddy ... ddx Γ(l,n)(y , ..., y ; x , ..., x )K(y )...K(y )φ (x )...φ (x ). c l!n! 1 n 1 l 1 n 1 l c 1 c n l,nX=0 Z Z (4.196) 142 YDRI QFT

2 The generating functional of 1PI correlation functions with l operator insertions iφc (y)/2 is deﬁned by

δlΓ[φ , K] 1 c = ddx ... ddx Γ(l,n)(y , ..., y ; x , ..., x )φ (x )...φ (x ). δK(y )...δK(y ) n! 1 n 1 l 1 n c 1 c n 1 l n=0 X Z Z (4.197)

Clearly

l+n δ Γ[φc, K] (l,n) = Γ (y1, ..., yl; x1, ..., xn). (4.198) δK(y1)...δK(yl)δφc(x1)...δφc(xn)

We also write

1 ddq ddp Γ[φ , K] = 1 ... n Γ˜(l,n)(q , ..., q ; p , ..., p )K˜ (q )...K˜ (q )φ˜ (p )...φ˜ (p ). c l!n! (2π)d (2π)d 1 l 1 n 1 l c 1 c n l,nX=0 Z Z (4.199)

We have deﬁned

d d (l,n) iq1y1 iqlyl ip1x1 ipnxn (l,n) d y1... d xnΓ (y1, ..., yl; x1, ..., xn) e ...e e ...e = Γ˜ (q1, ..., ql; p1, ..., pn). Z Z (4.200)

(l,n) The deﬁnition of the proper vertex Γ˜ (q1, ..., ql; p1, ..., pn) in this equation includes a delta function. We recall that δW [J, K] Γ[φ , K]= W [J, K] ddxJ(x)φ (x) , φ (x)= . (4.201) c − c c δJ(x) Z We calculate immediately

∂W δW ∂Γ δΓ = ddz = ddz . (4.202) ∂m2 |λ,Λ − δK(z) ⇒ ∂m2 |λ,Λ − δK(z) Z Z As a consequence

∂Γ(l,n)(y , ..., y ; x , ..., x ) δΓ(l,n)(y , ..., y ; x , ..., x ) 1 l 1 n = ddz 1 l 1 n ∂m2 |λ,Λ − δK(z) Z = ddzΓ(l+1,n)(z,y , ..., y ; x , ..., x ) (4.203) − 1 l 1 n Z Fourier transform then gives

∂Γ˜(l,n)(q , ..., q ; p , ..., p ) 1 l 1 n = Γ˜(l+1,n)(0,q , ..., q ; p , ..., p ). (4.204) ∂m2 |λ,Λ − 1 l 1 n YDRI QFT 143

By using equation (4.190) to convert bare proper vertices into renormalized proper vertices we obtain ∂ n ∂ ln Z ∂ ln Z /Z l 2 Γ˜(l,n) = ZZ−1Γ˜(l+1,n). (4.205) ∂m2 − 2 ∂m2 − ∂m2 R − 2 R λ,Λ The factor of 1 multiplying the right hand side of this equation will be absent in the − 9 ˜(l,n) Euclidean rotation of the theory . The renormalized proper vertices ΓR depend on 2 the momenta q1,...,ql,p1,...,pn but also on the renormalized mass mR, the renormalized coupling constant λR and the cutoﬀ Λ. They also depend on the subtraction mass scale 2 2 2 2 µ . We will either assume that µ = 0 or µ = mR. We have then ∂m ∂ ∂λ ∂ n ∂ ln Z ∂ ln Z /Z R + R l 2 Γ˜(l,n) = ZZ−1Γ˜(l+1,n). ∂m2 ∂m ∂m2 ∂λ − 2 ∂m2 − ∂m2 R − 2 R R R λ,Λ (4.206)

We write this as ∂m ∂ ∂m2 ∂λ ∂ n ∂m2 ∂ ln Z ∂m2 ∂ ln Z /Z R m + m R m lm 2 Γ˜(l,n) = ∂m2 R ∂m R ∂m ∂m2 ∂λ − 2 R ∂m ∂m2 − R ∂m ∂m2 R R R R R R λ,Λ m ZZ−1Γ˜(l+1,n)(4.207). − R 2 R We deﬁne m ∂m2 ∂λ β(λ , R ) = m R R Λ R ∂m ∂m2 R λ,Λ ∂λ = m R . (4.208) R ∂m R λ,Λ

m ∂m2 ∂ ln Z η(λ , R ) = m R Λ R ∂m ∂m2 R λ,Λ ∂ ∂ = m ln Z + β ln Z . (4.209) R ∂m ∂λ R R λ,Λ

m ∂m2 ∂ ln Z /Z η (λ , R ) = m 2 2 R Λ R ∂m ∂m2 R λ,Λ ∂ ∂ = m ln Z /Z + β ln Z /Z . (4.210) R ∂m 2 ∂λ 2 R R λ,Λ

m ∂m2 m2 σ(λ , R ) = ZZ−1 m . (4.211) R R Λ 2 R ∂m R λ,Λ 9Exercise: Check this. 144 YDRI QFT

The above differential equation becomes with these definitions ∂ ∂ n m + β η lη Γ˜(l,n) = m2 σΓ˜(l+1,n). (4.212) R ∂m ∂λ − 2 − 2 R − R R R R This is the original Callan-Symanzik equation. This equation represents only the response of the proper vertices to rescaling (φ φ ) and to reparametrization (m2 m2 , −→ R −→ R λ λ ). We still need to impose on the Callan-Symanzik equation the renormaliza- −→ R tion conditions in order to determine the renormalization constants and show that the renormalized proper vertices have a finite limit when Λ . The functions β, η, η −→ ∞ 2 and σ can be expressed in terms of renormalized proper vertices and as such they have an infinite cutoff limit. The Callan-Symanzik equation (4.212) can be used to provide an inductive proof of renormalizability of φ4 theory in 4 dimensions. We will not go through this involved exercise at this stage.

Homogeneous CS RG Equation-Massless Theory: The renormalization conditions for a massless φ4 theory in d = 4 are given by

(2) Γ˜ (p) 2 = 0 R |p =0 d (2) Γ˜ (p) 2 2 = 1 dp2 |p =µ Γ˜(4)(p , ..., p ) = λ . (4.213) 1 4 |SP R ˜(n) The renormalized proper vertices ΓR depend on the momenta p1,...,pn, the mass scale 2 µ , the renormalized coupling constant λR and the cutoff Λ. The bare proper vertices (n) Γ˜ depend on the momenta p1,...,pn, the bare coupling constant λ and the cutoff Λ. The bare mass is fixed by the condition that the renormalized mass is 0. We have then

2 (n) 2 n Λ (n) Γ˜ (p ,µ ; λ , Λ) = Z 2 (λ, , Λ)Γ˜ (p ; λ, Λ). (4.214) R i R µ2 i The bare theory is obviously independent of the mass scale µ2. This is expressed by the condition ∂ µ Γ˜(n)(p ; λ, Λ) = 0. (4.215) ∂µ i λ,Λ We diﬀerentiate equation (4.214) with respect to µ2 keeping λ and Λ ﬁxed. We get

∂ (n) ∂λR ∂ (n) n ∂ ln Z n (n) Γ˜ + Γ˜ = Z 2 Γ˜ . (4.216) ∂µ R ∂µ ∂λ R 2 ∂µ λ,Λ R λ,Λ We obtain immediately the diﬀerential equation ∂ ∂ n µ + β(λ ) η(λ ) Γ˜(n) = 0. (4.217) ∂µ R ∂λ − 2 R R R YDRI QFT 145

∂λ ∂ ln Z β(λ )= µ R , η(λ )= µ . (4.218) R ∂µ R ∂µ λ,Λ λ,Λ This is the Callan-Symanzik equation for the massless theory. The functions β and η do not depend on Λ/µ since they can be expressed in terms of renormalized proper vertices and as such they have an infinite cutoff limit. For the massless theory with φ2 insertions we need, as in the massive case, an extra renormalization constant Z2 and an extra renormalization condition to fix it given by ˜(1,2) Γ (q; p1,p2) q2=p2=µ2 = 1. (4.219) R | i We will also need an extra RG function given by ∂ η (λ )= µ ln Z /Z . (4.220) 2 R ∂µ 2 λ,Λ The Callan-Symanzik equation for the massless theory with φ2 insertions is then given (by the almost obvious) equation

∂ ∂ n µ + β(λ ) η(λ ) lη (λ ) Γ˜(l,n) = 0. (4.221) ∂µ R ∂λ − 2 R − 2 R R R Homogeneous CS RG Equation-Massive Theory: We consider again a massless φ4 theory in d = 4 dimensions with φ2 insertions. The action is given by the massless limit of the action (4.187), namely

1 1 λ Z Z S[φ , K] = ddx Z∂ φ ∂µφ δ φ2 R g (φ2 )2 + 2 Kφ2 . (4.222) R 2 µ R R − 2 m R − 4! R 2 R Z The eﬀective action is still given by 1 Γ[φ , K] = ddy ... ddx Γ(l,n)(y , ..., y ; x , ..., x )K(y )...K(y )φ (x )...φ (x ). c l!n! 1 n 1 l 1 n 1 l c 1 c n l,nX=0 Z Z (4.223)

˜(n) An arbitrary proper vertex ΓR (p1, ..., pn; K) can be expanded in terms of the proper (l,n) vertices ΓR (q1, ..., ql; p1, ..., pn) as follows 1 ddq ddq Γ˜(n)(p , ..., p ; K)= 1 ... l Γ˜(l,n)(q , ..., q ; p , ..., p )K˜ (q )...K˜ (q ). R 1 n l! (2π)d (2π)d R 1 l 1 n 1 l Xl=0 Z Z (4.224)

We consider the diﬀerential operator

∂ ∂ n d δ D = µ + β(λR) η(λR) η2(λR) d qK˜ (q) . (4.225) ∂µ ∂λR − 2 − δK˜ (q) Z 146 YDRI QFT

We compute

d δ (n) 1 d d (l,n) d qK˜ (q) Γ˜ (p1, ..., pn; K) = d q1... d qlΓ˜ (q1, ..., ql; p1, ..., pn) δK˜ (q) R l! R Z Xl=0 Z Z d δ d qK˜ (q) K˜ (q1)...K˜ (ql) × δK˜ (q) Z 1 = ddq ... ddq lΓ˜(l,n)(q , ..., q ; p , ..., p ) l! 1 l R 1 l 1 n Xl=0 Z Z K˜ (q )...K˜ (q ). (4.226) × 1 l By using now the Callan-Symanzik equation (4.221) we get

∂ ∂ n d ˜ δ ˜(n) µ + β η η2 d qK(q) ΓR (p1, ..., pn; K) = 0. (4.227) ∂µ ∂λR − 2 − δK˜ (q) Z A massive theory can be obtained by setting the source K(x) equal to a constant which 2 − will play the role of the renormalized mass mR. We will then set K(x)= m2 K˜ (q)= m2 (2π)dδd(q). (4.228) − R ⇔ − R We obtain therefore the Callan-Symanzik equation ∂ ∂ n ∂ µ + β η η m2 Γ˜(n)(p , ..., p ; m2 ) = 0. (4.229) ∂µ ∂λ − 2 − 2 R ∂m2 R 1 n R R R This needs to be compared with the renormalization group equation (4.170) and as a consequence the renormalization function η2 must be compared with the renormaliza- − ˜(n) tion constant γm. The renormalized proper vertices ΓR will also depend on the coupling constant λR, the subtraction mass scale µ and the cutoﬀ Λ.

4.2.5 Summary We end this section by summarizing our main results so far. The bare action and with φ2 insertion is

1 1 λ 1 S[φ, K] = ddx ∂ φ∂µφ m2φ2 (φ2)2 + Kφ2 . (4.230) 2 µ − 2 − 4! 2 Z The renormalized action is 1 1 λ 1 S [φ , K] = ddx ∂ φ ∂µφ m2 φ2 R (φ2)2 + Z Kφ2 . (4.231) R R 2 µ R R − 2 R R − 4! 2 2 R Z The dimensionless coupling gR and the renormalization constants Z, Zg and Zm are deﬁned by the equations

g = µ−ǫλ , ǫ = 4 d. (4.232) R R − YDRI QFT 147

φ = √ZφR Z λ = λ g R Z2 Z m2 = m2 m . (4.233) R Z The arbitrary mass scale µ deﬁnes the renormalization scale. For example renormalization conditions must be imposed at the scale µ as follows

(2) 2 Γ˜ (p) 2 = m R |p =0 R d (2) Γ˜ (p) 2 2 = 1 dp2 |p =µ Γ˜(4)(p , ..., p ) = µǫg 1 4 |SP R (1,2) Γ˜ (q; p ,p ) 2 = 1. (4.234) R 1 2 |q=pi=µ However we will use in the following minimal subtraction to renormalize the theory instead of renormalization conditions. In minimal subtraction, which is due to ’t Hooft, the renormalization functions β, γm, η and η2 depend only on the coupling constant gR and they are deﬁned by

−1 ∂gR d Zg β(gR)= µ = ǫ ln G(gR) , G = 2 gR. (4.235) ∂µ 2 − dg Z λ,m R

∂ 2 d Zm γm(gR)= µ ln mR = β(gR) ln . (4.236) ∂µ 2 − dg Z λ,m R

∂ d η(gR)= µ ln Z = β(gR) ln Z. (4.237) ∂µ 2 dg λ,m R

∂ Z2 d Z2 η2(gR)= µ ln = β(gR) ln . (4.238) ∂µ Z 2 dg Z λ,m R We may also use the renormalization function γ deﬁned simply by η(g ) γ(g )= R . (4.239) R 2 The renormalized proper vertices are given by

˜(n) 2 2 n/2 ˜(n) 2 ΓR (pi,µ ; mR, gR)= Z (gR,ǫ)Γ (pi; m , λ, ǫ). (4.240)

(l,n) n ˜ 2 2 2 −l l ˜(l,n) 2 ΓR (qi; pi; µ ; mR, gR)= Z Z2Γ (qi; pi; m , λ, ǫ). (4.241) 148 YDRI QFT

They satisfy the renormalization group equations

∂ ∂ ∂ n µ + γ m2 + β η Γ˜(n) = 0. (4.242) ∂µ m R ∂m2 ∂g − 2 R R R

∂ ∂ n m + β η lη Γ˜(l,n) = m2 σΓ˜(l+1,n). (4.243) R ∂m ∂λ − 2 − 2 R − R R R R In the ﬁrst equation we have set K = 0 and in the second equation the renormalization scale is µ = mR. The renormalization function σ is given by

Z 1 ∂m2 σ(g ) = m R Z m2 R ∂m 2 R R λ Z d Z = m 2+ β(g ) ln m . (4.244) Z R dg Z 2 R ˜(n) An alternative renormalization group equation satisﬁed by the proper vertices ΓR can be obtained by starting from a massless theory, i.e. m = m = 0 with K = 0 and then R 6 setting K = m2 at the end. We obtain − R ∂ ∂ n ∂ µ + β η η m2 Γ˜(n) = 0. (4.245) ∂µ ∂λ − 2 − 2 R ∂m2 R R R In this form the massless limit is accessible. As we can see from (4.242) and (4.245) the renormalization functions γ (g ) and η (g ) are essentially the same object. Indeed m R − 2 R since the two equations describe the same theory one must have

η (g )= γ (g ). (4.246) 2 R − m R

Alternatively we see from equation (4.244) that the renormalization constant Z2 is not an independent renormalization constant since σ is ﬁnite. In accordance with (4.246) we choose

Z2 = Zm. (4.247)

Because Z2 = Zm equation (4.244) becomes

∂ σ(g ) = m ln m2 R R ∂m R λ = 2 γ . (4.248) − m YDRI QFT 149

4.3 Renormalization Constants and Renormalization Func- tions at Two-Loop

4.3.1 The Divergent Part of the Eﬀective Action The 2 and 4 Point Proper Vertices: Now we will renormalize the O(N) sigma − model at the two-loop order using dimensional regularization and (modiﬁed) minimal subtraction. The main divergences in this theory occur in the 2 point proper vertex − (quadratic) and the 4 point proper vertex (logarithmic). Indeed all other divergences − in this theory stem from these two functions. Furthermore only the divergence in the 2 point proper vertex is momentum dependent. − The 2 point and 4 point (at zero momentum) proper vertices of the O(N) sigma − − model at the two-loop order in Euclidean signature are given by equations (2.201) and (2.227), viz

1 N + 2 λ2 N + 2 λ2 N + 2 Γ(2)(p) = δ p2 + m2 + λ (a) 2(b) (c) .(4.249) ij ij 2 3 − 4 3 − 6 3

δ 3 N + 8 3 (N + 2)(N + 8) Γ(4) (0, 0, 0, 0) = i1i2i3i4 λ λ2(d)+ λ3 (g) i1...i4 3 − 2 9 2 27 3 (N + 2)(N + 4) + 12 5N + 22 + λ3 (e) + 3λ3 (f) . (4.250) 4 27 27 The Feynman diagrams corresponding to (a), (b), (c), (d), (g), (e) and (f) are shown on ﬁgure 16. Explicitly we have

ddk 1 (a)= I(m2)= . (4.251) (2π)d k2 + m2 Z

ddk 1 (d)= J(0,m2)= . (4.252) (2π)d (k2 + m2)2 Z

(b)= I(m2)J(0,m2) = (a)(d). (4.253)

ddk ddl 1 (c)= K(p2,m)= . (4.254) (2π)d (2π)d (l2 + m2)(k2 + m2)((l + k + p)2 + m2) Z

ddk 1 ddl 1 (g)= I(m2)L(0,m2)= . (4.255) (2π)d k2 + m2 (2π)d (l2 + m2)3 Z Z

(e)= J(0,m2)2 = (d)2. (4.256) 150 YDRI QFT

ddl ddk 1 (f)= M(0, 0,m2)= . (4.257) (2π)d (2π)d (l2 + m2)(k2 + m2)((l + k)2 + m2) Z We remark that the two-loop graph (g) is a superposition of the one-loop graphs (a) and (d) and thus it will be made ﬁnite once (a) and (d) are renormalized. At the two-loop order only the diagram (c) is momentum dependent. We introduce the notation ∂ (c)=Σ(2)(p) = Σ(2)(0) + p2 Σ(2)(0) + ... ∂p2 2d−6 2 2d−8 = m I2 + p m I3 + ... (4.258) We will also introduce the notation

d−2 (a)= m I1. (4.259) 10 All other integrals can be expressed in terms of I1 and I2. Indeed we can show ∂ d (d)= (a) = (1 )md−4I . (4.260) −∂m2 − 2 1

d (b) = (1 )m2d−6I2. (4.261) − 2 1

d (e) = (1 )2m2d−8I2. (4.262) − 2 1

1 ∂ 1 (f)= Σ(2)(0) = (d 3)m2d−8I . (4.263) −3 ∂m2 −3 − 2

1 ∂ 1 d d (g)= (a) (d)= (1 )(2 )m2d−8I2. (4.264) −2 ∂m2 2 − 2 − 2 1

Calculation of The Poles: We have already met the integral I1 before. We compute ddk 1 I = 1 (2π)d k2 + 1 Z 2 kd−1dk = (4π)d/2Γ(d/2) k2 + 1 Z 2 1 xd/2−1dx = . (4.265) (4π)d/2Γ(d/2) 2 x + 1 Z We use the formula uαdu Γ(α + 1)Γ(β α 1) = aα+1−β − − . (4.266) (u + a)β Γ(β) Z 10Exercise: Derive these results. YDRI QFT 151

Thus (with d = 4 ǫ) − m2 (m2)−ǫ/2 (a) = Γ( 1+ ǫ/2) (4.267) 16π2 (4π)−ǫ/2 − We use the result 2 Γ( 1+ ǫ/2) = 1+ γ + O(ǫ). (4.268) − − ǫ − Hence we obtain m2 2 m2 (a) = 1+ γ + ln + O(ǫ) . (4.269) 16π2 − ǫ − 4π The ﬁrst Feynman graph is then given by m2 2 m2 λ(a) = g 1+ γ ln 4π + ln + O(ǫ) . (4.270) 16π2 − ǫ − − µ2 In minimal subtraction (MS) we subtract only the pole term 2/ǫ whereas in modiﬁed − minimal subtraction (MMS) we subtract also any other extra constant such as the term 1+ γ ln 4π. − − We introduce 2 N = . (4.271) d (4π)d/2Γ(d/2) We compute N I = d Γ(d/2)Γ(1 d/2) 1 2 − N 2 = d ( + O(ǫ)). (4.272) 2 − ǫ Then

ǫ 2 d/2−1 λ(a) = gµ (m ) I1 µ = gm2( )ǫI m 1 N 2 m2 = d [ + ln + O(ǫ)]. (4.273) 2 − ǫ µ2 From this formula it is now obvious that subtracting N /ǫ is precisely the above mod- − d iﬁed minimal subtraction. We have also met the integral Σ(2)(p) before (see (2.202)). By following the same steps that led to equation (2.205) we obtain

2 x1x2x3 2 −m (x1+x2+x3)− p 2 2 1 e ∆ K(p ,m ) = dx1dx2dx3 , ∆= x1x2 + x1x3 + x2x3. (4π)d ∆d/2 Z (4.274) 152 YDRI QFT

Thus

I2 = K(0, 1) 1 e−(x1+x2+x3) = dx1dx2dx3 . (4.275) (4π)d ∆d/2 Z

∂ 2 I = K(p , 1) 2 3 ∂p2 |p =0 −(x1+x2+x3) 1 x1x2x3e = dx1dx2dx3 . (4.276) −(4π)d ∆1+d/2 Z We perform the change of variables x = stu, x = st(1 u) and x = s(1 t). Thus 1 2 − 3 − x + x + x = s, dx dx dx = s2tdsdtdu and ∆ = s2t(1 t + ut(1 u)). The above 1 2 3 1 2 3 − − integrals become

∞ 1 1 1−d/2 1 −s 2−d t I2 = dse s du dt (4π)d (1 t + ut(1 u))d/2 Z0 Z0 Z0 − − Γ(3 d) 1 1 t1−d/2 = − du dt . (4.277) (4π)d (1 t + ut(1 u))d/2 Z0 Z0 − −

∞ 1 1 2−d/2 1 −s 3−d t (1 t) I3 = dse s duu(1 u) dt − −(4π)d − (1 t + ut(1 u))1+d/2 Z0 Z0 Z0 − − Γ(4 d) 1 1 t2−d/2(1 t) = − duu(1 u) dt − . (4.278) − (4π)d − (1 t + ut(1 u))1+d/2 Z0 Z0 − − We want to evaluate the integral

1 1 J = du dt t1−d/2(1 t + ut(1 u))−d/2 0 0 − − Z 1 Z 1 = du dt t1−d/2 + (1 t + ut(1 u))−d/2 1 − − − Z0 Z0 + (t1−d/2 1) (1 t + ut(1 u))−d/2 1 . (4.279) − − − − The first term gives the first contribution to the pole term. The last term is finite at d = 4. Indeed with the change of variable x = t u(1 u) 1 + 1 we calculate − − 1 1 1 du dt(t−1 1) (1 t + ut(1 u))−2 1 = du ln u(1 u) − − − − − − Z0 Z0 Z0 1 = 2 du ln u − Z0 = 2. (4.280) YDRI QFT 153

We have then 2 1 1 J = 1+ du dt(1 t + ut(1 u))−d/2 +(2+ O (ǫ)) ǫ − − − 1 Z0 Z0 2 2 1 1 = 1+ du (u(1 u))1−d/2 1 +(2+ O (ǫ)) ǫ − ǫ 2 u(1 u) 1 − − 1 − Z0 − − 2 2 1 2 1 u(1 u) = 1 du (u(1 u))1−d/2 1 + du − ǫ − − ǫ 2 − − ǫ 2 u(1 u) 1 − Z0 − Z0 − − (u(1 u))1−d/2 1 +(2+ O (ǫ)) × − − 1 2 2 2 1 2 = 1+ du(u(1 u))1−d/2 + ( 1+ O (ǫ)) + (2 + O (ǫ)) ǫ − ǫ 2 − ǫ 2 − ǫ 2 − 2 1 − − Z0 − 2 2 2 Γ2(2 d/2) 2 = 1+ − + ( 1+ O (ǫ)) + (2 + O (ǫ)) (4.281) ǫ − ǫ 2 − ǫ 2 Γ(4 d) ǫ 2 − 2 1 − − − − In the last line we have used the result (2.275) whereas in the third line we have used the fact that the second remaining integral is ﬁnite at d = 4. From this result we deduce that 6 J = +3+ O(ǫ). (4.282) ǫ

Γ(3 d) I = − J 2 (4π)d Γ( 1+ ǫ) = − J (4π)d 1 6 9 6γ = + + O(1) . (4.283) (4π)d − ǫ2 − ǫ ǫ We compute Γ(d/2)=1+ γǫ/2 ǫ/2+ O(ǫ2) and hence Γ2(d/2) =1+ γǫ ǫ + O(ǫ2). − − Thus the above result can be rewritten as 3 ǫ I = N 2 (1 + )+ O(1). (4.284) 2 − d 2ǫ2 2

Now we compute the integral I3. The only divergence has already been exhibited by the term Γ(4 d). Thus we have − Γ(ǫ) 1 1 1 t I = duu(1 u) dt − + O(1) 3 −(4π)d − (1 t + ut(1 u))3 Z0 Z0 − − Γ(ǫ) 1 1 = du + O(1) −(4π)d 2 Z0 N 2Γ2(d/2) = d Γ(ǫ)+ O(1) − 8 N 2 = d + O(1). (4.285) − 8ǫ 154 YDRI QFT

4.3.2 Renormalization Constants One-Loop Renormalization: We prefer to go back to the original expressions

1 N + 2 Γ(2)(p) = δ p2 + m2 + λ (a) ij ij 2 3 1 N + 2 = δ p2 + m2 + λ I(m2) . (4.286) ij 2 3

δ 3 N + 8 Γ(4) (0, 0, 0, 0) = i1i2i3i4 λ λ2(d) i1...i4 3 − 2 9 δ 3 N + 8 = i1i2i3i4 λ λ2J(0,m2) . (4.287) 3 − 2 9 The renormalized mass and the renormalized coupling constant are given by

Z Z m2 = m2 m , λ = λ g . (4.288) R Z R Z2 We will expand the renormalization constants as

2 (2) Z =1+ λRZ . (4.289)

(1) 2 (2) Zm =1+ λRZm + λRZm . (4.290)

(1) 2 (2) Zg =1+ λRZg + λRZg . (4.291)

(2) (2) (2) At one-loop of course Z = Zm = Zg = 0. We will also deﬁne the massless coupling constant by

g = λm−ǫ. (4.292)

The renormalized 2 point and 4 point proper vertices are given by − − (2) (2) (ΓR)ij (p) = ZΓij (p) 1 N + 2 = δ p2 + m2 + λ m2 Z(1) + I(m2 ) . (4.293) ij R R R m 2 3 R

(Γ )(4) (0, 0, 0, 0) = Z2Γ(4) (0, 0, 0, 0) R i1...i4 i1...i4 δ 3 N + 8 = i1i2i3i4 λ + λ2 Z(1) J(0,m2 )) . (4.294) 3 R R g − 2 9 R YDRI QFT 155

Minimal subtraction gives immediately

2 (1) N + 2 I(mR) N + 2 −ǫ Nd Zm = 2 = mR . (4.295) − 6 mR 6 ǫ

N + 8 N + 8 N Z(1) = J(0,m2 )= m−ǫ d . (4.296) g 6 R 6 R ǫ The renormalized mass and the renormalized coupling constant at one-loop order are given by

N + 2 N + 8 m2 = m2 λ I(m2 ) , λ = λ + λ2 J(0,m2 ). (4.297) R − 6 R R R 6 R R

Two-Loop Renormalization of The 2 Point Proper Vertex: The original ex- − pression of the 2 point vertex reads − 1 N + 2 λ2 N + 2 λ2 N + 2 Γ(2)(p) = δ p2 + m2 + λ (a) 2(b) (c) .(4.298) ij ij 2 3 − 4 3 − 6 3 We use the result

N + 2 I(m2)= I(m2 )+ λ I(m2 )J(0,m2 )+ O(λ2 ). (4.299) R 6 R R R R By using the one-loop results we ﬁnd the renormalized 2 point proper vertex to be given − by

(2) (2) (ΓR)ij (p) = ZΓij (p) N + 2 (N + 2)2 = δ p2 + m2 + Z(2)λ2 p2 + Z(2)λ2 m2 + Z(1)λ2 I(m2 )+ λ2 I(m2 )J(0,m2 ) ij R R m R R 6 g R R 36 R R R λ2 N + 2 λ2 N + 2 R 2(b) R (c) − 4 3 R − 6 3 R N + 2 λ2 N + 2 = δ p2 + m2 + Z(2)λ2 p2 + Z(2)λ2 m2 + Z(1)λ2 I(m2 ) R (m2d−6I ij R R m R R 6 g R R − 6 3 R 2 2 2d−8 + p mR I3) . (4.300) 2 2 2d−6 In the last equation we have used the results (b)R = I(mR)J(0,mR) and (c)R = mR I2+ 2 2d−8 p mR I3. By requiring ﬁniteness of the kinetic term we obtain the result

N + 2 N + 2 N 2 Z(2) = m2d−8I = m−2ǫ d . (4.301) 18 R 3 − 144 R ǫ 156 YDRI QFT

Cancellation of the remaining divergences gives

2 (2) N + 2 (1) I(mR) N + 2 2d−8 Zm = Zg 2 + mR I2 − 6 mR 18 (N + 2)(N + 8) N 2 N + 2 N 2 N + 2 N 2 = m−2ǫ d m−2ǫ d m−2ǫ d 36 R ǫ2 − 12 R ǫ2 − 24 R ǫ (N + 2)(N + 5) N 2 N + 2 N 2 = m−2ǫ d m−2ǫ d . (4.302) 36 R ǫ2 − 24 R ǫ

Two-Loop Renormalization of The 4 Point Proper Vertex: The original ex- − pression of the 4 point vertex reads − δ 3 N + 8 3 (N + 2)(N + 8) Γ(4) (0, 0, 0, 0) = i1i2i3i4 λ λ2(d)+ λ3 (g) i1...i4 3 − 2 9 2 27 3 (N + 2)(N + 4) + 12 5N + 22 + λ3 (e) + 3λ3 (f) . (4.303) 4 27 27 We use the result N + 2 J(0,m2)= J(0,m2 )+ λ I(m2 )L(0,m2 )+ O(λ2 ). (4.304) R 3 R R R R By using the one-loop results we ﬁnd the renormalized 4 point proper vertex to be given − by

(Γ )(4) (0, 0, 0, 0) = Z2Γ(4) (0, 0, 0, 0) R i1...i4 i1...i4 δ N + 8 (N + 8)(N + 2) = i1i2i3i4 λ + λ3 Z(2) λ3 Z(1)(d) λ3 I(m2 )L(0,m2 ) 3 R R g − 3 R g R − 18 R R R 3 (N + 2)(N + 8) 3 (N + 2)(N + 4) + 12 5N + 22 + λ3 (g) + λ3 (e) + 3λ3 (f) 2 R 27 R 4 R 27 R R 27 R δ N + 8 3 (N + 2)(N + 4) + 12 = i1i2i3i4 λ + λ3 Z(2) λ3 Z(1)(d) + λ3 (e) 3 R R g − 3 R g R 4 R 27 R 5N + 22 + 3λ3 (f) . (4.305) R 27 R Cancellation of the remaining divergences gives N + 8 3 (N + 2)(N + 4) + 12 5N + 22 Z(2) = Z(1)(d) (e) 3 (f) g 3 g R − 4 27 R − 27 R N + 8 ǫ 3 (N + 2)(N + 4) + 12 ǫ 5N + 22 = Z(1)(1 )m−ǫI (1 )2m−2ǫI2 + (1 ǫ)m−2ǫI − 3 g − 2 R 1 − 4 27 − 2 R 1 27 − R 2 (N + 8)2 N 2 N 2 N 2 + 6N + 20 N 2 N 2 5N + 22 N 2 N 2 = m−2ǫ( d d ) m−2ǫ( d d ) m−2ǫ( d d ) 18 R ǫ2 − 2ǫ − 36 R ǫ2 − ǫ − 18 R ǫ2 − 2ǫ (N + 8)2 N 2 5N + 22 N 2 = m−2ǫ d m−2ǫ d . (4.306) 36 R ǫ2 − 36 R ǫ YDRI QFT 157

4.3.3 Renormalization Functions The renormalization constants up to two-loop order are given by

N + 2 N 2 Z = 1 g2 d . (4.307) − R 144 ǫ

N + 8 N (N + 8)2 N 2 5N + 22 N 2 Z = 1+ g d + g2 d d . (4.308) g 6 R ǫ R 36 ǫ2 − 36 ǫ

N + 2 N (N + 2)(N + 5) N 2 N + 2 N 2 Z = 1+ g d + g2 d d . (4.309) m 6 R ǫ R 36 ǫ2 − 24 ǫ The beta function is given by

ǫgR β(gR)= d d . (4.310) −1+ gR ln Zg 2gR ln Z dgR − dgR We compute

d N + 8 N (N + 8)2 N 2 5N + 22 N 2 g ln Z = g d + g2 d d . (4.311) R dg g 6 R ǫ R 36 ǫ2 − 18 ǫ R

2 2 d N + 2 Nd d N + 2 2 Nd ln Z = gR 2gR ln Z = gR . (4.312) dgR − 72 ǫ ⇒− dgR 36 ǫ We get then the fundamental result

N + 8 3N + 14 β(g )= ǫg + g2 N g3 N 2. (4.313) R − R 6 R d − 12 R d The second most important renormalization function is η. It is deﬁned by

d η(gR) = β(gR) ln Z dgR N + 8 3N + 14 N + 2 N 2 = ǫg + g2 N g3 N 2 g d − R 6 R d − 12 R d − 72 R ǫ N + 2 = g2 N 2. (4.314) 72 R d

The renormalization constant Zm is associated with the renormalization function γm deﬁned by

d Zm γm(gR) = β(gR) ln . (4.315) − dgR Z 158 YDRI QFT

We compute d N + 2 N (N + 2)(N + 8) N 2 N + 2 N 2 ln Z = d + g d d . (4.316) dg m 6 ǫ R 36 ǫ2 − 12 ǫ R The renormalization function γm at the two-loop order is then found to be given by N + 2 5(N + 2) γ = N g g2 N 2. (4.317) m 6 d R − 72 R d From this result we conclude immediately that N + 2 5(N + 2) η = γ = N g + g2 N 2. (4.318) 2 − m − 6 d R 72 R d 4.4 Critical Exponents

4.4.1 Critical Theory and Fixed Points We will postulate that quantum scalar field theory, in particular φ4, describes the critical domain of second order phase transitions which includes the critical line T = Tc where the correlation length ξ diverges and the scaling region with T near Tc where the correlation length ξ is large but finite. This is confirmed for example by mean field calculations. From now on we will work in Euclidean signature. We write the action in the form 1 1 Λǫg S[φ]= β (φ)= ddx (∂ φ)2 + m2φ2 (φ2)2 . (4.319) H 2 µ 2 − 4! Z In the above action the cutoff Λ reflects the original lattice structure, i.e. Λ = 1/a. The cutoff procedure is irrelevant to the physics and as a consequence we will switch back and forth between cutoff regularization and dimensional regularization as needed. The critical domain is defined by the conditions m2 m2 << Λ2 | − c| momenta << Λ d −1 < φ(x) ><< Λ 2 . (4.320) 2 2 In above mc is the value of the mass parameter m at the critical temperature Tc where 2 2 mR = 0 or the correlation length ξ diverges. Clearly mc is essentially mass renormalization. We will set

2 2 T Tc m = mc + t, t − . (4.321) ∝ Tc The critical theory should be renormalized at a scale µ in such a way that the renormalized mass remains massless, viz (2) Γ˜ (p; µ, g , Λ) 2 = 0 R R |p =0 d (2) Γ˜ (p; µ, g , Λ) 2 2 = 1 dp2 R |p =µ Γ˜(4)(p , ..., p ; µ, g , Λ) = µǫg . (4.322) 1 4 R |SP R YDRI QFT 159

The renormalized proper vertices are given by Λ Γ˜(n)(p ; µ, g )= Zn/2(g, )Γ˜(n)(p ; g, Λ). (4.323) R i R µ i

The bare proper vertices Γ˜(n) are precisley the proper vertices of statistical mechanics. ˜(n) Now since the renormalized proper vertices ΓR are independent of Λ we should have ∂ Λ Zn/2Γ˜(n) = 0. (4.324) ∂Λ µ,gR We obtain the renormalization group equation 11 ∂ ∂ n Λ + β η Γ˜(n) = 0. (4.325) ∂Λ ∂g − 2 The renormalization functions are now given by ∂g β(g)= Λ . (4.326) ∂Λ gR,µ

∂ η(g)= Λ ln Z . (4.327) − ∂Λ gR,µ Clearly the functions β and η can not depend on the ratio Λ/µ since Γ˜(n) does not depend on µ. We state the (almost) obvious theorem

• The renormalization group equation (4.325) is a direct consequence of the existence of a renormalized ﬁeld theory. Conversely the existence of a solution to this renormalization group equation implies the existence of a renormalized theory.

The ﬁxed point g = g∗ and the critical exponent ω: The renormalization group equation (4.325) can be solved using the method of characteristics. We introduce a dilatation parameter λ, a running coupling constant g(λ) and an auxiliary renormalization function Z(λ) such that

d λ Z−n/2(λ)Γ˜(n)(p ; g(λ), λΛ) = 0. (4.328) dλ i (n) We can verify that proper vertices Γ˜ (pi; g(λ), λΛ) solves the renormalization group equation (4.325) provided that β and η solves the ﬁrst order diﬀerential equations

d β(g(λ)) = λ g(λ) , g(1) = g. (4.329) dλ 11Exercise: Show this result. 160 YDRI QFT

d η(g(λ)) = λ ln Z(λ) ,Z(1) = Z. (4.330) dλ We have the identiﬁcation

(n) −n/2 (n) Γ˜ (pi; g, Λ) = Z (λ)Γ˜ (pi; g(λ), λΛ). (4.331)

Equivalently Λ Γ˜(n)(p ; g, )= Z−n/2(λ)Γ˜(n)(p ; g(λ), Λ). (4.332) i λ i The limit Λ is equivalent to the limit λ 0. The functions β and η are assumed −→ ∞ −→ to be regular functions for g 0. ≥ The integration of (4.329) and (4.330) yields the integrated renormalization group equations

g(λ) dx ln λ = . (4.333) β(x) Zg

λ dx ln Z(λ)= η(g(x)). (4.334) x Z1

The zeros g = g∗ of the beta function β which satisfy β(g∗) = 0 are of central importance to quantum ﬁeld theory and critical phenomena. Let us assume that the a zero g = g∗ of the beta function does indeed exist. We observe then that any value of the running coupling constant g(λ) near g will run into g in the limit λ 0 regardless of the ∗ ∗ −→ initial value g = g(1) which can be either above or below g∗. This can be made precise as follows. We expand β(g) about the zero as follows

β(g)= β(g ) + (g g )ω + ... (4.335) ∗ − ∗

′ β(g∗) = 0 , ω = β (g∗). (4.336)

We compute

1 g(λ) g ln λ = − ∗ g(λ) g λω , λ 0. (4.337) ω g g ⇒ − ∗ ∼ −→ − ∗ If ω > 0 then g(λ) g when λ 0. The point g = g is then called an attractive −→ ∗ −→ ∗ or stable infrared (since the limit λ 0 is equivalent to the massless limit λΛ 0) −→ −→ fixed point (since dng(λ)/dλn = 0). If ω < 0 then the point g = g is called a repulsive |g∗ ∗ infrared fixed point or equivalently a stable ultraviolet fixed point since g(λ) g when −→ ∗ λ . −→ ∞ YDRI QFT 161

′ The slope ω = β (g∗) is our ﬁrst critical exponent which controls leading corrections to scaling laws. As an example let us consider the beta function

3 β(g)= ǫg + bg2 , b = . (4.338) − 16π2

There are in this case two ﬁxed points the origin and g∗ = ǫ/b with critical exponents ω = ǫ < 0 (infrared repulsive) and ω = +ǫ (infrared attractive) respectively. We − compute immediately

1 1/g(λ) dx g ln λ = g(λ)= ∗ . (4.339) ǫ x 1/g ⇒ 1+ λǫ(g /g 1) Z1/g − ∗ ∗ − Since ǫ = 4 d> 0, g(λ) g when λ 0 and as a consequence g = ǫ/b is a stable − −→ ∗ −→ ∗ infrared fixed point known as the non trivial (interacting) Wilson-Fisher fixed point. In the limit λ we see that g(λ) 0, i.e. the origin is a stable ultraviolet fixed −→ ∞ −→ point which is the famous trivial (free) Gaussian fixed point. See figure 15. The fact that the origin is a repulsive (unstable) infrared fixed point is the source of the strong infrared divergence found in dimensions < 4 since perturbation theory in this case is an expansion around the wrong fixed point. Remark that for d > 4 the origin becomes an attractive (stable) infrared fixed point while g = g∗ becomes repulsive.

The critical exponent η: Now we solve the second integrated renormalization group ′ equation (4.330). We expand η as η(g(λ)) = η(g ) + (g(λ) g )η (g )+ .... In the limit ∗ − ∗ ∗ λ 0 we obtain −→ ln Z(λ)= η ln λ + ... Z(λ)= λη. (4.340) ⇒ The critical exponent η is deﬁned by

η = η(g∗). (4.341)

The proper vertex (4.332) becomes

(n) Λ − n η (n) Γ˜ (p ; g, )= λ 2 Γ˜ (p ; g(λ), Λ). (4.342) i λ i

˜(n) Λ However from dimensional considerations we know the mass dimension of Γ (pi; g, λ ) d−n(d/2−1) (n) d−n(d/2−1) to be M and hence the mass dimension of Γ˜ (λpi; g, Λ) is (λM) . We get therefore

(n) Λ −d+n( d −1) (n) Γ˜ (p ; g, )= λ 2 Γ˜ (λp ; g, Λ). (4.343) i λ i By combining these last two equations we obtain the crucial result 162 YDRI QFT

(n) −d+ n (d−2+η) (n) Γ˜ (λp ; g, Λ) = λ 2 Γ˜ (p ; g , Λ) , λ 0. (4.344) i i ∗ −→ The critical proper vertices have a power law behavior for small momenta which is independent of the original value g of the φ4 coupling constant. This in turn is a manifestation of the universality of the critical behavior. The mass dimension of the ﬁeld φ has also changed from the canonical (classical) value (d 2)/2 to the anomalous (quantum) value −

1 d = (d 2+ η). (4.345) φ 2 − In the particular case n = 2 we have the behavior

Γ˜(2)(λp; g, Λ) = λη−2Γ˜(2)(p; g , Λ) , λ 0. (4.346) ∗ −→ Hence the 2 point function must behave as − 1 G˜(2)(p) , p 0. (4.347) ∼ p2−η −→

The critical exponent ν: The full renormalized conditions of the massless (critical) theory when K = 0 are (4.322) plus the two extra conditions 6 ˜(1,2) ΓR (q; p1,p2; µ, gR, Λ) q2=p2=µ2,p p =− 1 µ2 = 1. | i 1 2 3 (4.348)

˜(2,0) ΓR (q; q; µ, gR, Λ) q2= 4 µ2 = 0. (4.349) − | 3

The first condition fixes the renormalization constant Z2 while the second condition provides a renormalization of the < φ2φ2 > correlation function. The renormalized proper vertices are defined by (with l + n> 2)

Λ Λ Γ˜(l,n)(q ; p ; µ, g )= Zn/2−l(g, )Zl (g, )Γ˜(l,n)(q ; p ; g, Λ). (4.350) R i i R µ 2 µ i i We have clearly the condition

∂ Λ Zn/2−lZl Γ˜(l,n) = 0. (4.351) ∂Λ 2 µ,gR We obtain the renormalization group equation

∂ ∂ n Λ + β η lη Γ˜(l,n) = 0. (4.352) ∂Λ ∂g − 2 − 2 YDRI QFT 163

The renormalization functions β and η are still given by equations (4.326) and (4.327) while the renormalization function η2 is deﬁned by

∂ Z η (g)= Λ ln 2 . (4.353) 2 − ∂Λ Z gR,µ As before we solve the above renormalization group equation (4.352) by the method of characteristics. We introduce a dilatation parameter λ, a running coupling constant g(λ) and auxiliary renormalization functions Z(λ) and ζ2(λ) such that

d λ Z−n/2(λ)ζ−l(λ)Γ˜(l,n)(q ; p ; g(λ), λΛ) = 0. (4.354) dλ 2 i i We can verify that proper vertices Γ˜(l,n) solves the above renormalization group equation (4.352) provided that β, η solve the first order differential equations (4.329) and (4.330) and η2 solves the first order differential equation

d η (g(λ)) = λ ln ζ (λ) , ζ (1) = ζ . (4.355) 2 dλ 2 2 2 We have the identiﬁcation

˜(l,n) −n/2 −l ˜(l,n) Γ (qi; pi; g, Λ) = Z (λ)ζ2 (λ)Γ (qi; pi; g(λ), λΛ). (4.356)

Equivalently

Λ Γ˜(l,n)(q ; p ; g, )= Z−n/2(λ)ζ−l(λ)Γ˜(l,n)(q ; p ; g(λ), Λ). (4.357) i i λ 2 i i The corresponding integrated renormalization group equation is

λ dx ln ζ = η (g(x)). (4.358) 2 x 2 Z1 We obtain in the limit λ 0 the behavior −→

η2 ζ2(λ)= λ . (4.359)

The new critical exponent η2 is deﬁned by

η2 = η2(g∗). (4.360)

We introduce the mass critical exponent ν by the relation 1 ν = ν(g∗) , ν(g)= . (4.361) 2+ η2(g) 164 YDRI QFT

We have then the infrared behavior of the proper vertices given by

(l,n) Λ − n η−lη (l,n) Γ˜ (q ; p ; g, )= λ 2 2 Γ˜ (q ; p ; g(λ), Λ). (4.362) i i λ i i (l,n) From dimensional considerations the mass dimension of the proper vertex Γ˜ (qi; pi; g, Λ/λ) is M d−n(d−2)/2−2l and hence

(l,n) Λ −d+ n (d−2)+2l (l,n) Γ˜ (q ; p ; g, )= λ 2 Γ˜ (λq ; λp ; g(λ), Λ). (4.363) i i λ i i By combining the above two equations we obtain

(l,n) d− n (d−2+η)− l (l,n) Γ˜ (λq ; λp ; g, Λ) = λ 2 ν Γ˜ (q ; p ; g , Λ) , λ 0. (4.364) i i i i ∗ −→

4.4.2 Scaling Domain (T > Tc)

In this section we will expand around the critical theory. The proper vertices for T > Tc can be calculated in terms of the critical proper vertices with φ2 insertions.

The correlation length: In order to allow a large but ﬁnite correlation length (non zero renormalized mass) in this massless theory without generating infrared divergences we consider the action 1 1 Λǫg S[φ]= β (φ)= ddx (∂ φ)2 + (m2 + K(x))φ2 (φ2)2 . (4.365) H 2 µ 2 c − 4! Z We want to set at the end T T K(x)= t − c . (4.366) ∝ Tc The n point proper vertices are given by − 1 ddq ddq Γ˜(n)(p ; K,g, Λ) = 1 ... l Γ˜(l,n)(q ; p ; g, Λ)K˜ (q )...K˜ (q ). (4.367) i l! (2π)d (2π)d i i 1 l Xl=0 Z Z We consider the diﬀerential operator

∂ ∂ n d δ D = Λ + β η η2 d qK˜ (q) . (4.368) ∂Λ ∂g − 2 − δK˜ (q) Z We compute

d δ (n) 1 d d (l,n) d qK˜ (q) Γ˜ (pi; K,g, Λ) = d q1... d qlΓ˜ (qi; pi; g, Λ) δK˜ (q) l! Z Xl=0 Z Z d δ d qK˜ (q) K˜ (q1)...K˜ (ql) × δK˜ (q) Z 1 = ddq ... ddq (lΓ˜(l,n)(q ; p ; g, Λ))K˜ (q )...K˜ (q ). l! 1 l i i 1 l Xl=0 Z Z (4.369) YDRI QFT 165

By using now the Callan-Symanzik equation (4.352) we get

∂ ∂ n d δ (n) Λ + β η η2 d qK˜ (q) Γ˜ (pi; K,g, Λ) = 0. (4.370) ∂Λ ∂g − 2 − δK˜ (q) Z We now set K(x)= t or equivalently K˜ (q)= t(2π)dδd(q) to obtain

∂ ∂ n ∂ Λ + β η η t Γ˜(n)(p ; t,g, Λ) = 0. (4.371) ∂Λ ∂g − 2 − 2 ∂t i We employ again the method of characteristics in order to solve this renormalization group equation. We introduce a dilatation parameter λ, a running coupling constant g(λ), a running mass t(λ) and an auxiliary renormalization functions Z(λ) such that

d λ Z−n/2(λ)Γ˜(n)(p ; t(λ), g(λ), λΛ) = 0. (4.372) dλ i (n) Then Γ˜ (pi; t(λ), g(λ), λΛ) will solve the renormalization group equation (4.371) provided the renormalization functions β, η and η2 satisfy d β(g(λ)) = λ g(λ) , g(1) = g. (4.373) dλ

d η(g(λ)) = λ ln Z(λ) ,Z(1) = Z. (4.374) dλ

d η (g(λ)) = λ ln t(λ) , t(1) = t. (4.375) 2 − dλ

The new definition of η2 given in the last equation is very similar to the definition of γm given in equation (4.246). We make the identification

(n) −n/2 (n) Γ˜ (pi; t,g, Λ) = Z (λ)Γ˜ (pi; t(λ), g(λ), λΛ). (4.376)

From dimensional considerations we have

(n) d− n (d−2) (n) pi t(λ) Γ˜ (p ; t(λ), g(λ), λΛ) = (λΛ) 2 Γ˜ ( ; , g(λ), 1). (4.377) i λΛ λ2Λ2 Thus

(n) −n/2 d− n (d−2) (n) pi t(λ) Γ˜ (p ; t,g, Λ) = Z (λ)m 2 Γ˜ ( ; , g(λ), 1). (4.378) i m m2 We have used the notation m = λΛ. We use the freedom of choice of λ to choose

t(λ)= m2 = λ2Λ2. (4.379) 166 YDRI QFT

The theory at scale λ is therefore not critical since the critical regime is deﬁned by the requirement t << Λ2. The integrated form of the renormalization group equation (4.375) is given by

λ dx t(λ)= t exp η (g(x)). (4.380) − x 2 Z1 This can be rewritten as

tλ2 λ dx 1 ln = . (4.381) t(λ) x ν(g(x)) Z1 Equivalently

t λ dx 1 ln = . (4.382) Λ2 x ν(g(x)) Z1 This is an equation for λ. In the critical regime ln t/Λ2 . For ν(g) > 0 this means −→ −∞ that λ 0 and hence g(λ) g . By expanding around the ﬁxed point g(λ)= g we −→ −→ ∗ ∗ obtain in the limit λ 0 the result −→ t λ = ν. (4.383) Λ2 (n) By using this result in (4.378) we conclude that the proper vertices Γ˜ (pi; t,g, Λ = 1) must have the infrared (λ 0) scaling −→ (n) d− n (d−2+η) (n) pi Γ˜ (p ; t,g, Λ=1)= m 2 Γ˜ ( ; 1, g , 1) , t<< 1 , p<< 1. (4.384) i m ∗ The mass m is thus the physical mass ξ−1 where ξ is the correlation length. We have then

m(Λ = 1) = ξ−1 = tν. (4.385)

Clearly ξ when t 0 or equivalently T T since ν > 0. −→ ∞ −→ −→ c The critical exponents α and γ: At zero momentum the above proper vertices are ﬁnite because of the non zero mass t. They have the infrared scaling

(n) d− n (d−2+η) Γ˜ (0; t,g, Λ=1) = m 2 ν(d− n (d−2+η)) = t 2 , t<< 1 , p<< 1. (4.386)

The case n = 2 is of particular interest since it is related to the inverse susceptibility, viz

χ−1 = Γ˜(2)(0; t,g, Λ = 1) = tγ. (4.387) YDRI QFT 167

The critical exponent γ is given by

γ = ν(2 η). (4.388) − The obvious generalization of the renormalization group equation (4.371) is ∂ ∂ n ∂ Λ + β η η (l + t ) Γ˜(l,n)(q ; p ; t,g, Λ) = 0. (4.389) ∂Λ ∂g − 2 − 2 ∂t i i This is valid for all n + l > 2. The case l = 2, n = 0 is special because of the non multiplicative nature of the renormalization required in this case and as a consequence the corresponding renormalization group equation will be inhomogeneous. However we will not pay attention to this difference since the above renormalization group equation is sufficient to reproduce the leading infrared behavior, and as a consequence the relevant critical exponent, of the proper vertex with l = 2 and n = 0. We find after some calculation, similar to the calculation used for the case l = 0, the leading infrared (λ 0) behavior −→ (l,n) − l +d− n (d−2+η) (l,n) qi pi Γ˜ (q ; p ; t,g, Λ=1)= m ν 2 Γ˜ ( ; ; 1, g , 1) , t<< 1 , q,p<< 1. i i m m ∗ (4.390)

By applying this formula naively to the case l = 2, n = 0 we get the desired leading infrared behavior of Γ˜(2,0) which corresponds to the most infrared singular part of the energy-energy correlation function. We obtain

(2,0) − 2 +d (2,0) q Γ˜ (q; t,g, Λ=1)= m ν Γ˜ ( ; 1, g , 1) , t<< 1 , q<< 1. (4.391) m ∗ By substituting K˜ (q)= t(2π)dδd(q) in (4.367) we obtain

tl Γ˜(n)(p ; K,g, Λ) = Γ˜(l,n)(0; p ; g, Λ). (4.392) i l! i Xl=0 Hence ∂2Γ(K,g, Λ) = Γ˜(2,0)(0; g, Λ). (4.393) ∂t2 |t=0 In other words Γ˜(2,0) at zero momentum is the specific heat since t is the temperature and Γ is the thermodynamic energy (effective action). The infrared behavior of the specific heat is therefore given by

(2,0) Cv = Γ˜ (0; t,g, Λ = 1) −α (2,0) = t Γ˜ (0; 1, g∗, 1) , t<< 1 , q<< 1. (4.394)

The new critical exponent α is deﬁned by

α = 2 νd. (4.395) − 168 YDRI QFT

4.4.3 Scaling Below Tc

In order to describe in a continuous way the ordered phase corresponding to T < Tc starting from the disordered phase (T > Tc) we introduce a magnetic ﬁeld B, i.e. a source J = B. The corresponding magnetization M is precisely the classical ﬁeld φc =< φ(x) >, viz

M(x)=< φ(x) > . (4.396)

The Helmholtz free energy (vacuum energy) will depend on the magnetic field B, viz W = W (B) = ln Z(B). We know that the magnetization and the magnetic field − are conjugate variables, i.e. M(x) = ∂W (B)/∂B(x). The Gibbs free energy or thermodynamic energy (effective action) is the Legendre transform of W (B), viz Γ(M) = ddxM(x)B(x) W (B). We compute then − R ∂Γ(M) B(x)= . (4.397) ∂M(x) The effective action can be expanded as

1 ddp ddp Γ[M,t,g, Λ] = 1 ... n Γ˜(n)(p ; t,g, Λ)M(p )...M(p ).(4.398) n! (2π)d (2π)d i 1 n n=0 X Z Z Thus 1 ddp ddp B(p) = 1 ... n Γ˜(n+1)(p ,p; t,g, Λ)M(p )...M(p ). (4.399) n! (2π)d (2π)d i 1 n n=0 X Z Z By assuming that the magnetization is uniform we obtain M n Γ[M,t,g, Λ] = Γ˜(n)(p = 0; t,g, Λ). (4.400) n! i n=0 X

M n B[M,t,g, Λ] = Γ˜(n+1)(p = 0; t,g, Λ). (4.401) n! i n=0 X By employing the renormalization group equation (4.371) we get

∂ ∂ n + 1 ∂ M n ∂ ∂ n + 1 ∂ Λ + β η η t B = Λ + β η η t Γ˜(n+1) ∂Λ ∂g − 2 − 2 ∂t n! ∂Λ ∂g − 2 − 2 ∂t n=0 X = 0. (4.402)

Clearly ∂ M n M B = n Γ˜(n+1)(p = 0; t,g, Λ). (4.403) ∂M n! i nX=0 YDRI QFT 169

Hence the magnetic ﬁeld obeys the renormalization group equation

∂ ∂ 1 ∂ ∂ Λ + β (1 + M )η η t B = 0. (4.404) ∂Λ ∂g − 2 ∂M − 2 ∂t By using the method of characteristics we introduce as before a running coupling constant g(λ), a running mass t(λ) and an auxiliary renormalization functions Z(λ) such as equations (4.373), (4.374) and (4.375) are satisﬁed. However in this case we need also to introduce a running magnetization M(λ) such that

d 1 λ ln M(λ)= η[g(λ)]. (4.405) dλ −2 By comparing (4.374) and (4.405) we obtain

− 1 M(λ)= MZ 2 (λ) (4.406)

We must impose

d λ Z−1/2(λ)B(M(λ),t(λ), g(λ), λΛ) = 0. (4.407) dλ In other words we make the identiﬁcation

B(M,t,g, Λ) = Z−1/2(λ)B(M(λ),t(λ), g(λ), λΛ). (4.408)

From dimensional analysis we know that [Γ˜(n)] = M d−n(d−2)/2 and [M] = M (d−2)/2 = M 1−ǫ/2 and hence [B]= M (d+2)/2 = M 3−ǫ/2. Hence

M t B(M,t,g, Λ) = Λ3−ǫ/2B( , , g, 1). (4.409) Λ1−ǫ/2 Λ2 By combining the above two equations we get

M(λ) t(λ) B(M,t,g, Λ) = Z−1/2(λ)(λΛ)3−ǫ/2B( , , g(λ), 1). (4.410) (λΛ)1−ǫ/2 λ2Λ2

Again we use the arbitrariness of λ to make the theory non critical and as a consequence avoid infrared divergences. We choose λ such that

M(λ) = 1. (4.411) (λΛ)1−ǫ/2

The solution of equation (4.405) then reads

M(λ) 1 λ dx M 1 λ dx ln = η(g(x)) ln = [d 2+ η(g(x))]. (4.412) M −2 x ⇒ Λ1−ǫ/2 2 x − Z1 Z1 170 YDRI QFT

The critical domain is deﬁned obviously by M << Λ1−ǫ/2. For d 2+ η positive we − conclude that λ must be small and thus g(λ) is close to the ﬁxed point g∗. This equation then leads to the infrared behavior M d−2+η = λ 2 . (4.413) Λ1−ǫ/2 From equation (4.381) we get the infrared behavior

t(λ) − 1 = λ ν . (4.414) tλ2 We know also the infrared behavior Z(λ)= λη. (4.415) The infrared behavior of equation (4.410) is therefore given by

− 2+d η 3−ǫ/2 t −1/ν B(M,t,g, Λ) = λ 2 Λ B(1, λ , g , 1). (4.416) Λ2 ∗ This can also be rewritten as − 1 B(M,t,g, 1) = M δf(tM β ). (4.417) This is the equation of state. The two new critical exponents β and δ are deﬁned by ν β = (d 2+ η). (4.418) 2 −

d + 2 η δ = − . (4.419) d 2+ η − From equations (4.413) and (4.414) we observe that λ2 M = tβ( )β. (4.420) t(λ)

For negative t (T < Tc) the appearance of a spontaneous magnetization M = 0 at B = 0 −1/β 6 means that the function f(x), where x = tM , admits a negative zero x0. Indeed the condition B = 0, M = 0 around x = x reads explicitly 6 0 ′ 0= f(x ) + (x x )f (x )+ .. (4.421) 0 − 0 0 This is equivalent to M = x −β( t)β. (4.422) | 0| − We state without proof that correlation functions below Tc have the same scaling behavior as above Tc. In particular the critical exponents ν, γ and α below Tc are the same as those deﬁned earlier above Tc. We only remark that in the presence of a magnetic ﬁeld B we have two mass scales tν (as before) and m = M ν/β where M is the magnetization which is the correct choice in this phase. In the limit B 0 (with T < T ) the magnetization −→ c becomes spontaneous and m becomes the physical mass YDRI QFT 171

4.4.4 Critical Exponents from 2 Loop and Comparison with Experi- ment − The most important critical exponents are the mass critical exponent ν and the anomalous dimension η. As we have shown these two critical exponents deﬁne the infrared behavior of proper vertices. At T = Tc we ﬁnd the scaling

(l,n) d− n (d−2+η)− l (l,n) Γ˜ (λq ; λp ; g, Λ) = λ 2 ν Γ˜ (q ; p ; g , Λ) , λ 0. (4.423) i i i i ∗ −→ The critical exponent η provides the quantum mass dimension of the ﬁeld operator, viz 1 [φ]= M dφ , d = (d 2+ η). (4.424) φ 2 − The scaling of the wave function renormalization is also determined by the anomalous dimension, viz

Z(λ) λη. (4.425) ≃ The 2 point function at T = T behaves therefore as − c 1 1 G(2)(p)= G(2)(r)= . (4.426) p2−η ⇔ rd−2+η

The critical exponent ν determines the scaling behavior of the correlation length. For T > Tc we ﬁnd the scaling

(n) d− n (d−2+η) (n) pi T Tc Γ˜ (pi; t,g, Λ=1)= m 2 F ( ) , t = − << 1 , pi << 1. (4.427) m Tc The mass m is proportional to the mass scale tν. From this equation we see that m is the physical mass ξ−1 where ξ is the correlation length ξ. We have then

m = ξ−1 tν. (4.428) ∼ The 2 point function for T > T behaves therefore as 12 − c 1 G(2)(r)= exp( r/ξ). (4.429) rd−2+η −

The scaling behavior of correlation functions for T < Tc is the same as for T > Tc except that there exists a non zero spontaneous magnetization M in this regime which sets an extra mass scale given by M 1/β besides tν. The exponent β is another critical exponent associated with the magnetization M given by the scaling law ν β = (d 2+ η). (4.430) 2 − 12Exercise: Give an explicit proof. 172 YDRI QFT

In other words for T close to Tc from below we must have M ( t)β (4.431) ∼ − For T < Tc the physical mass m is given by m = ξ−1 M ν/β ( t)ν. (4.432) ∼ ∼ − There are three more critical exponents α (associated with the speciﬁc heat), γ (associated with the susceptibility) and δ (associated with the equation of state) which are not independent but given by the scaling laws

α = 2 νd. (4.433) −

γ = ν(2 η). (4.434) − d + 2 η δ = − . (4.435) d 2+ η − The last critical exponent of interest is ω which is given by the slope of the beta function at the ﬁxed point and measures the approach to scaling. The beta function at two-loop order of the O(N) sigma model is given by N + 8 3N + 14 β(g )= ǫg + g2 N g3 N 2. (4.436) R − R 6 R d − 12 R d

The ﬁxed point g∗ is deﬁned by 3N + 14 N + 8 β(g ) = 0 g2 N 2 g N + ǫ = 0. (4.437) R∗ ⇒ 12 R∗ d − 6 R∗ d The solution must be of the form

2 gR∗Nd = aǫ + bǫ + ... (4.438) We ﬁnd the solution 6 18(3N + 14) a = , b = . (4.439) N + 8 (N + 8)3 Thus 6 18(3N + 14) g N = ǫ + ǫ2 + ... (4.440) R∗ d N + 8 (N + 8)3 The critical exponent ω is given by

′ ω = β (gR∗) 3(3N + 14) = ǫ ǫ2 + ... (4.441) − (N + 8)2 YDRI QFT 173

The critical exponent η is given by

η = η(gR∗). (4.442)

The renormalization function η(g) is given by N + 2 η(g ) = g2 N 2. (4.443) R 72 R d We substitute now the value of the ﬁxed point. We obtain immediately N + 2 η = ǫ2. (4.444) 2(N + 8)2 The critical exponent ν is given by 1 ν = . (4.445) 2+ η2

ν = ν(gR∗) , η2 = η2(gR∗). (4.446)

The renormalization function η2(g) is given by N + 2 5(N + 2) η = N g + g2 N 2. (4.447) 2 − 6 d R 72 R d By substituting the value of the ﬁxed point we compute immediately N + 2 (N + 2)(13N + 44) η = ǫ ǫ2 + .... (4.448) 2 −N + 8 − 2(N + 8)3

1 N + 2 (N + 2)(N 2 + 23N + 60) ν = + ǫ + ǫ2 + .... (4.449) 2 4(N + 8) 8(N + 8)3 All critical exponents can be determined in terms of ν and η. They only depend on the dimension of space d and on the dimension of the symmetry space N which is precisely the statement of universality. The epsilon expansion is divergent for all ǫ and as a consequence a resummation is required before we can coherently compare with experiments. This is a technical exercise which we will not delve into here and content ourselves by using what we have already established and also by quoting some results. The most important predictions (in our view) correspond to d = 3 (ǫ = 1) and N = 1, 2, 3. • The case N = 1 describes Ising-like systems such as the liquid-vapor transitions in classical ﬂuids. Experimentally we observe

ν = 0.625 0.006 ± γ = 1.23 1.25. (4.450) − 174 YDRI QFT

The theoretical calculation gives

1 1 7 203 ν = + + + ... = + ... = 0.6265 0.0432 2 12 162 324 ± 1 η = + .... = 0.019 γ = ν(2 η) = 1.241. (4.451) 54 ⇔ −

The agreement for ν and η up to order ǫ2 is very reasonable and is a consequence of the asymptotic convergence of the ǫ series. The error is estimated by the last term available in the epsilon expansion.

• The case N = 2 corresponds to the Helium superﬂuid transition. This system allows precise measurement near Tc of ν and α given by

ν = 0.672 0.001 ± α = 0.013 0.003. (4.452) − ± The theoretical calculation gives

1 1 11 131 ν = + + + ... = + ... = 0.6550 0.0550 2 10 200 200 ± α = 2 νd = 0.035. (4.453) − − Here the agreement up to order ǫ2 is not very good. After proper resummation of the ǫ expansion we ﬁnd excellent agreement with the experimental values. We quote the improved theoretical predictions

ν = 0.664 0.671 − α = (0.008 0.013). (4.454) − −

• The case N = 3 corresponds to magnetic systems. The experimental values are

ν = 0.7 0.725 − γ = 1.36 1.42. (4.455) − The theoretical calculation gives

1 5 345 903 ν = + + + ... = + ... = 0.6874 0.0648 2 44 5324 1331 ± 5 η = = 0.021 γ = 1.36. (4.456) 242 ⇔

There is a very good agreement. YDRI QFT 175

4.5 The Wilson Approximate Recursion Formulas

4.5.1 Kadanoff-Wilson Phase Space Analysis We start by describing a particular phase space cell decomposition due to Wilson which is largely motivated by Kadanoff block spins. We assume a hard cutoff 2Λ. Thus if φ(x) is the field (spin) variable and φ˜(k) is its Fourier transform we will assume that φ˜(k) is zero for k> 2Λ. We expand the field as

∞ φ(x)= ψ~ml(x)φ~ml. (4.457) X~m Xl=0 The wave functions ψ~ml(x) satisfy the orthonormality condition

d ∗ d xψ~m1l1 (x)ψ~m2l2 (x)= δ~m1 ~m2 δl1l2 . (4.458) Z ˜ The Fourier transform ψ~ml(k) is deﬁned by

˜ d ikx ψ~ml(k)= d xψ~ml(x) e . (4.459) Z The interpretation of l and ~m is as follows. We decompose momentum space into thin spherical shells, i.e. logarithmically as 1 k 1 | | . (4.460) 2l ≤ Λ ≤ 2l−1 The functions ψ˜ (k) for a ﬁxed l are non zero only inside the shell l, i.e. for 1/2l ~ml ≤ k /Λ 1/2l−1. We will assume furthermore that the functions ψ˜ (k) are constant | | ≤ ~ml within its shell and satisfy the normalization condition

ddk ψ˜ (k) 2 = 1. (4.461) (2π)d | ~ml | Z ˜ The functions ψ~ml(k) and ψ~ml(x) for a ﬁxed l and a ﬁxed ~m should be thought of as ˜ minimal wave packets, i.e. if ∆k is the width of ψ~ml(k) and ∆x is the width of ψ~ml(x) then one must have by the uncertainty principle the requirement ∆x∆k = (2π)d. Thus for each shell we divide position space into blocks of equal size each with volume inversely proportional to the volume of the corresponding shell. The volume of the lth momentum shell is proportional to Rd where R = 1/2l, viz ∆k = (2π)d2−ldw where w is a constant. Hence the volume of the corresponding position space box is ∆x = 2ldw−1. The functions ψ~ml(x) are non zero (constant) only inside this box by construction. This position space box is characterized by the index ~m as is obvious from the normalization condition

ddx ψ (x) 2 = ddx ψ (x) 2 = 1. (4.462) | ~ml | | ~ml | Z Z~x∈box ~m 176 YDRI QFT

In other words

ddx = ddx. (4.463) ~x∈box ~m Z X~m Z The normalization conditions in momentum and position spaces lead to the relations

ψ˜ (k) = 2ld/2w−1/2. (4.464) | ~ml |

ψ (x) = 2−ld/2w1/2. (4.465) | ~ml | Obviously ψ˜ (0) = 0. Thus ddxψ (x) = ψ˜ (0) = 0 and as a consequence we will | ~ml | ~ml ~ml assume that ψ (x) is equal to +2−ld/2w1/2 in one half of the box and 2−ld/2w1/2 in ~ml R − the other half. The meaning of the index ~m which labels the position space boxes can be clariﬁed further by the following argument. By an appropriate scale transformation in momentum space we can scale the momenta such that the lth shell becomes the largest shell l = 0. Clearly the correct scale transformation is k 2lk since 1 2lk /Λ 2. This −→ ≤ | | ≤ corresponds to a scale transformation in position space of the form x x/2l. We l −→ obtain therefore the relation ψ~ml(x) = ψ~m0(x/2 ). Next we perform an appropriate translation in position space to bring the box ~m to the box ~0. This is clearly given by the translation ~x ~x a ~m. We obtain therefore the relation −→ − 0 ψ (x)= ψ (x/2l a ~m). (4.466) ~ml ~00 − 0 ˜ The functions ψ~ml(k) and ψ~ml(x) correspond to a single degree of freedom in phase space occupying a volume (2π)d, i.e. a single cell in phase space is characterized by l and ~m. Each momentum shell l corresponds to a lattice in the position space with a lattice spacing given by

1/d l −1/d al = (∆x) = 2 a0 , a0 = w . (4.467)

The largest shell l = 0 correspond to a lattice spacing a0 and each time l is increased by 1 the lattice spacing gets doubled which is the original spin blocking idea of Kadanoﬀ. We are interested in integrating out only the l = 0 modes. We write then

φ(x)= ψ~m0(x)φ~m0 + φ1(x). (4.468) X~m

∞ φ1(x)= ψ~ml(x)φ~ml. (4.469) X~m Xl=1 From the normalization (4.465) and the scaling law (4.466) we have

−d/2 ψ~ml(x) = 2 ψ~ml−1(x/2). (4.470) YDRI QFT 177

′ We deﬁne φ (x/2) by

−d/2 ′ φ1(x) = 2 α0φ (x/2). (4.471)

In other words

∞ ′ ′ ′ −1 φ (x/2) = ψ~ml−1(x/2)φ ~ml−1 , φ~ml−1 = α0 φ~ml. (4.472) X~m Xl=1 We have then

−d/2 ′ φ(x)= ψ~m0(x)φ~m0 + 2 α0φ (x/2). (4.473) X~m

4.5.2 Recursion Formulas

We will be interested in actions of the form

K S (φ(x)) = ddxR (φ(x))∂ φ(x)∂µφ(x)+ ddxP (φ(x)). (4.474) 0 2 0 µ 0 Z Z

We will assume that P0 and R0 are even polynomials of the ﬁeld and that dR0/dφ is much smaller than P0 for all relevant conﬁgurations. The partition function is

Z = φ(x) e−S0(φ(x)) 0 D Z ′ = φ (x/2) dφ e−S0(φ(x)). (4.475) D ~m0 Z Z Y~m

′ The degrees of freedom contained in the ﬂuctuation φ (x/2) correspond to momentum shells l 1 and thus correspond to position space wave packets larger than the box ~m ≥ ′ by at least a factor of 2. We can thus assume that φ (x/2) is almost constant over the ′ box ~m. If x0 is the center of the box ~m we can expand φ (x/2) as

′ ′ ′ 1 ′ φ (x/2) = φ (x /2) + (x x )µ∂ φ (x /2) + (x x )µ(x x )ν ∂ ∂ φ (x /2) + ... 0 − 0 µ 0 2 − 0 − 0 µ ν 0 (4.476)

The partition function becomes

′ Z = φ (x /2) dφ e−S0(φ(x)). (4.477) 0 D 0 ~m0 Z Z Y~m 178 YDRI QFT

The Kinetic Term: We compute

d µ d µ ′ ′ d xR0∂µφ(x)∂ φ(x) = d xR0 ∂µψ~m0(x)∂ ψ~m 0(x).φ~m0φ~m 0 ′ Z Z ~m,~mX 1−d/2 ′ −d 2 ′ µ ′ + 2 α0 φ~m0∂µψ~m0(x)∂µφ (x/2) + 2 α0∂µφ (x/2)∂ φ (x/2) . X~m (4.478)

d The integral d x is ~m ~x∈box ~m. In the box ~m the function R0(φ(x)) can be replaced ′ by the function R (ψ (x)φ + 2−d/2α φ (x/2)). In this box ψ (x) is approximated R 0 P~m0 R ~m0 0 ~m0 by +w1/2 in one half of the box and by w1/2 in the other half, i.e. by a step function. − Thus the third term in the above equation can be approximated by (dropping also higher −d/2 ′ derivative corrections and deﬁning u~m = 2 α0φ (x0/2))

′ ′ 2−d−1α2w−1 R (w1/2φ + u )+ R ( w1/2φ + u ) ∂ φ (x /2)∂µφ (x /2). 0 0 ~m0 ~m 0 − ~m0 ~m µ 0 0 X~m (4.479)

The contribution of ∂µψ~m0(x) is however only appreciable when ψ~m0(x) = 0, i.e. at the center of the box. In the first and second terms of the above equation we can then replace R0(φ(x)) by R0(u~m). The second term in the above equation (4.478) vanishes by conservation of momentum. In the first term we can neglect all the coupling terms ′ φ φ ′ with ~m = ~m since ψ ′ (x) is zero inside the box ~m. We define the integral ~m0 ~m 0 6 ~m 0

d µ ρ = d x∂µψ~m0(x)∂ ψ~m0(x). (4.480) Z This is independent of ~m because of the relation (4.466). The ﬁrst term becomes therefore 2 ρ ~m R0(u~m)φ~m0. Equation (4.478) becomes

Pd µ 2 d xR0∂µφ(x)∂ φ(x) = ρ R0(u~m)φ~m0 Z X~m + 2−d−1α2w−1 R (w1/2φ + u )+ R ( w1/2φ + u ) 0 0 ~m0 ~m 0 − ~m0 ~m ~m ′ X′ ∂ φ (x /2)∂µφ (x /2). (4.481) × µ 0 0 The Interaction Term: Next we want to compute

d (m) d −d/2 ′ d xP0(φ (x)) = d xP0(ψ~m0(x)φ~m0 + 2 α0φ (x/2)). (4.482) Z~x∈box ~m Z~x∈box ~m We note again that within the box ~m the ﬁeld is given by

(m) −d/2 ′ φ (x)= ψ~m0(x)φ~m0 + 2 α0φ (x/2). (4.483) YDRI QFT 179

Introduce z0 = ψ~m0(x)φ~m0 + u~m. We have then

1 ddxP (φ(m)(x)) = ddx P (z )+ (x x )µ∂ u + (x x )µ(x x )ν 0 0 0 − 0 µ ~m 2 − 0 − 0 Z~x∈box ~m Z~x∈box ~m dP 1 d2P ∂ ∂ u 0 + (x x )µ(x x )ν ∂ u ∂ u 0 z + ... . × µ ν ~m dz |z0 2 − 0 − 0 µ ~m ν ~m dz2 | 0 (4.484)

As stated earlier ψ (x) is approximated by +w1/2 in one half of the box and by w1/2 ~m0 − in the other half. Also recall that the volume of the box is w−1. The ﬁrst term can be approximated by

w−1 ddxP (z )= P (w1/2φ + u )+ P ( w1/2φ + u ) . (4.485) 0 0 2 0 ~m0 ~m 0 − ~m0 ~m Z~x∈box ~m We compute now the third term in (4.484). We start from the obvious identity

1 1 1 ddx (x x )µ(x x )ν = ddx (x x )µ(x x )ν + ddx (x x )µ(x x )ν . 2 − 0 − 0 2 − 0 − 0 2 − 0 − 0 Zbox Zbox+ Zbox− (4.486)

We will think of the box ~m as a sphere of volume w−1. Thus

1 1 ddx (x x )µ(x x )ν = V ηµν . (4.487) 2 − 0 − 0 2 Zbox We have the deﬁnitions 1 V = r2ddx , w−1 = ddx. (4.488) d Zbox Zbox Explicitly we have

1 dw−1 (d+2)/d Ω V = d−1 . (4.489) d + 2 Ω d d−1 The above identity becomes then

1 1 1 V ηµν = ddx (x x )µ(x x )ν + ddx (x x )µ(x x )ν . (4.490) 2 2 − 0 − 0 2 − 0 − 0 Zbox+ Zbox− The sum of these two integrals is rotational invariant. As stated before we think of the box as a sphere divided into two regions of equal volume. The ﬁrst region (the ﬁrst half of the box) is a concentric smaller sphere whereas the second region (the second half of the box) is a thin spherical shell. Both regions are spherically symmetric and thus we can assume that 180 YDRI QFT

1 1 ddx (x x )µ(x x )ν = V ηµν . (4.491) 2 − 0 − 0 2 ± Zbox±

Clearly V = V+ + V−. The integral of interest is

d 1 µ ν dP0 dP0 d 1 µ ν d x (x x0) (x x0) ∂µ∂νu~m z = 1/2 ∂µ∂ν u~m d x (x x0) (x x0) 2 − − dz | 0 dz |w φ ~m0+u ~m 2 − − Z~x∈box Zbox+ dP0 d 1 µ ν + 1/2 ∂µ∂ν u~m d x (x x0) (x x0) dz |−w φ ~m0+u ~m 2 − − Zbox− V+ dP0 µ V− dP0 µ = 1/2 ∂µ∂ u~m + 1/2 ∂µ∂ u~m 2 dz |w φ ~m0+u ~m 2 dz |−w φ ~m0+u ~m −1−d/2 dP0 µ ′ = 2 α0V+ 1/2 ∂µ∂ φ (x0/2) dz |w φ ~m0+u ~m −1−d/2 dP0 µ ′ + 2 α0V− 1/2 ∂µ∂ φ (x0/2). (4.492) dz |−w φ ~m0+u ~m Similarly the fourth term in (4.484) is computed as follows. We have

2 2 d 1 µ ν d P0 d P0 d 1 µ d x (x x ) (x x ) ∂ u ∂ u z = 1/2 ∂ u ∂ u d x (x x ) 0 0 µ ~m ν ~m 2 0 2 w φ ~m0+u ~m µ ~m ν ~m 0 ~x∈box ~m 2 − − dz | dz | box+ 2 − Z ν Z (x x0) × 2 − d P0 d 1 µ + 1/2 ∂ u ∂ u d x (x x ) 2 −w φ ~m0+u ~m µ ~m ν ~m 0 dz | box− 2 − ν Z (x x0) × − 2 V+ d P0 µ = 1/2 ∂µu~m∂ u~m 2 dz2 |w φ ~m0+u ~m 2 V− d P0 µ + 1/2 ∂µu~m∂ u~m 2 dz2 |−w φ ~m0+u ~m 2 −1−d 2 d P0 ′ µ ′ = 2 α V+ 1/2 ∂µφ (x0/2)∂ φ (x0/2) 0 dz2 |w φ ~m0+u ~m 2 −1−d 2 d P0 ′ µ ′ + 2 α V− 1/2 ∂µφ (x0/2)∂ φ (x0/2). 0 dz2 |−w φ ~m0+u ~m (4.493)

Finally we compute the ﬁrst term in (4.484). we have

d µ dP0 dP0 d µ d x(x x0) ∂µu~m z = 1/2 ∂µu~m d x(x x0) − dz | 0 dz |w φ ~m0+u ~m − Z~x∈box ~m Zbox+ dP0 d µ + 1/2 ∂µu~m d x(x x0) . dz |−w φ ~m0+u ~m − Zbox− (4.494) YDRI QFT 181

Clearly we must have

ddx(x x )µ + ddx(x x )µ = 0. (4.495) − 0 − 0 Zbox+ Zbox− We will assume that both integrals vanish again by the same previous argument. The linear term is therefore 0, viz

dP ddx(x x )µ∂ u 0 = 0. (4.496) − 0 µ ~m dz |z0 Z~x∈box ~m The ﬁnal result is w−1 ddxP (φ(m)(x)) = P (w1/2φ + u )+ P ( w1/2φ + u ) 0 2 0 ~m0 ~m 0 − ~m0 ~m Z~x∈box ~m −1−d/2 dP0 dP0 µ ′ + 2 α0 V+ 1/2 + V− 1/2 ∂µ∂ φ (x0/2) dz |w φ ~m0+u ~m dz |−w φ ~m0+u ~m 2 2 −1−d 2 d P0 d P0 ′ µ ′ + 2 α V+ 1/2 + V− 1/2 ∂µφ (x0/2)∂ φ (x0/2). 0 dz2 |w φ ~m0+u ~m dz2 |−w φ ~m0+u ~m (4.497)

The Action: By putting all the previous results together we obtain the expansion of the action. We get K S (φ(x)) = ddxR (φ(x))∂ φ(x)∂µφ(x)+ ddxP (φ(x)) 0 2 0 µ 0 Z Z Kρ = R (u )φ2 2 0 ~m ~m0 X~m ′ ′ + 2−d−2Kα2w−1 R (w1/2φ + u )+ R ( w1/2φ + u ) ∂ φ (x /2)∂µφ (x /2) 0 0 ~m0 ~m 0 − ~m0 ~m µ 0 0 X~m w−1 + P (w1/2φ + u )+ P ( w1/2φ + u ) 2 0 ~m0 ~m 0 − ~m0 ~m X~m −1−d/2 dP0 dP0 µ ′ + 2 α0 V+ 1/2 + V− 1/2 ∂µ∂ φ (x0/2) dz |w φ ~m0+u ~m dz |−w φ ~m0+u ~m ~m X 2 2 −1−d 2 d P0 d P0 ′ µ ′ + 2 α V+ 1/2 + V− 1/2 ∂µφ (x0/2)∂ φ (x0/2). 0 dz2 |w φ ~m0+u ~m dz2 |−w φ ~m0+u ~m X~m (4.498)

The Path Integral: We need now to evaluate the path integral

−S0(φ(x)) dφ~m0 e . (4.499) Z Y~m 182 YDRI QFT

We introduce the variables Kρ u z = 1/2u . (4.500) ~m −→ ~m 2w ~m Kρ φ y = 1/2φ . (4.501) ~m0 −→ ~m 2 ~m0 2w R W (x)= R ( 1/2x). (4.502) 0 −→ 0 0 Kρ 2w P Q (x)= w−1P ( 1/2x). (4.503) 0 −→ 0 0 Kρ We compute then

2 1/2 1 1 dφ e−S0(φ(x)) = dy exp y2 W (z ) Q (y + z ) Q ( y + z ) ~m0 Kρ ~m − ~m 0 ~m − 2 0 ~m ~m − 2 0 − ~m ~m Z ~m ~m Z Y Y dQ dQ ′ 2−(3+d)/2α (Kρw)1/2 V 0 (y + z )+ V 0 ( y + z ) ∂ ∂µφ (x /2) − 0 + dy ~m ~m − dy − ~m ~m µ 0 ~m ~m 2 2 d Q d Q ′ ′ 2−d−2α2Kρ V 0 (y + z )+ V 0 ( y + z ) ∂ φ (x /2)∂µφ (x /2) − 0 + dy2 ~m ~m − dy2 − ~m ~m µ 0 0 ~m ~m ′ ′ 2−d−2α2Kw−1 W (y + z )+ W ( y + z ) ∂ φ (x /2)∂µφ (x /2) . − 0 0 ~m ~m 0 − ~m ~m µ 0 0 (4.504)

Deﬁne 1 1 M (z ) = dy exp y2 W (z ) Q (y + z ) Q ( y + z ) 0 ~m ~m − ~m 0 ~m − 2 0 ~m ~m − 2 0 − ~m ~m Z dQ dQ ′ 2−(3+d)/2α (Kρw)1/2 V 0 (y + z )+ V 0 ( y + z ) ∂ ∂µφ (x /2) − 0 + dy ~m ~m − dy − ~m ~m µ 0 ~m ~m 2 2 d Q d Q ′ ′ 2−d−2α2Kρ V 0 (y + z )+ V 0 ( y + z ) ∂ φ (x /2)∂µφ (x /2) − 0 + dy2 ~m ~m − dy2 − ~m ~m µ 0 0 ~m ~m ′ ′ 2−d−2α2Kw−1 W (y + z )+ W ( y + z ) ∂ φ (x /2)∂µφ (x /2) . − 0 0 ~m ~m 0 − ~m ~m µ 0 0 (4.505)

′ In this equation φ (x0/2) is given in terms of z~m by

d/2 ′ 2 2w 1/2 φ (x0/2) = z~m. (4.506) α0 Kρ YDRI QFT 183

The remaining dependence on the box ~m is only through the center of the box x0. Then

2 1/2 dφ e−S0(φ(x)) = M (z ) ~m0 Kρ 0 ~m Z Y~m Y~m 2 = 1/2 exp(ln M (z )) Kρ 0 ~m ~m Y M0(z~m) 2 1/2 = exp ln I0(0). (4.507) I0(0) Kρ ~m ~m X Y The function I0(z) is deﬁned by

1 1 I (z)= dy exp y2W (z) Q (y + z) Q ( y + z) . (4.508) 0 − 0 − 2 0 − 2 0 − Z

In order to compute M0 we will assume that the derivative terms are small and expand the exponential around the ultra local approximation. We compute

dQ ′ M (z ) = I (z ) 1 2−(3+d)/2α (Kρw)1/2V < 0 (y + z ) > ∂ ∂µφ (x /2) 0 ~m 0 ~m − 0 dy ~m ~m µ 0 ~m 2 ρV d Q ′ ′ 2−d−1α2K < 0 (y + z ) > +w−1 ∂ φ (x /2)∂µφ (x /2) + ... . − 0 2 dy2 ~m ~m 0 ~m ~m µ 0 0 ~m (4.509)

The path integral (4.499) becomes

−S0(φ(x)) 2 1/2 I0(z~m) −(3+d)/2 1/2 dφ~m0 e = I0(0) exp ln 2 α0(Kρw) V Kρ × I0(0) − Z ~m ~m ~m ~m Y Y X 2 X dQ0 µ ′ −d−1 2 ρV d Q0 < (y~m + z~m) > ∂µ∂ φ (x0/2) 2 α0K < 2 (y~m + z~m) > × dy~m − 2 dy~m X~m −1 ′ µ ′ + w ∂µφ (x0/2)∂ φ (x0/2) . (4.510) We make the change of variable x /2 x (this means that the position space wave 0 −→ packet with l = 1 corresponding to the highest not integrated momentum will now ﬁt into the box) to obtain

1/2 −S0(φ(x)) 2 1/2 I0(z~m) −(3+d)/2 α0(Kρw) V dφ~m0 e = I0(0) exp ln 2 Kρ × I0(0) − 4 Z ~m ~m ~m ~m Y Y X 2 2 X dQ0 µ ′ −d−1 α0K ρV d Q0 < (y~m + z~m) > ∂µ∂ φ (x) 2 < 2 (y~m + z~m) > × dy~m − 4 2 dy~m X~m −1 ′ µ ′ + w ∂µφ (x)∂ φ (x) . (4.511) 184 YDRI QFT

1/2 −d/2 ′ Now z~m is given by z~m = (Kρ/2w) 2 α0φ (x). It is clear that dQ0(y~m +z~m)/dy~m = −1 dQ0(y~m+z~m)/dz~m, etc. Recall that the volume of the box ~m is w and thus we can make the identification w ddx = . However we have also made the rescaling x 2x ~m 0 −→ and hence we must make instead the identification 2dw ddx = . We obtain R P ~m 1/2 −S0(φ(x)) 2 1/2 d R d PI0(z) −(3+d)/2 α0(Kρw) V d dφ~m0 e = I0(0) exp 2 w d x ln 2 2 w Kρ × I0(0) − 4 Z ~m ~m Z Y Y 2 2 dQ ′ α K ρV d Q ddx< 0 (z) > ∂ ∂µφ (x) 2−d−1 0 2dw ddx < 0 (z) > × dz µ − 4 2 dz2 Z Z −1 ′ µ ′ + w ∂µφ (x)∂ φ (x) . (4.512) 1/2 −d/2 ′ n Now z is given by z = (Kρ/2w) 2 α0φ (x). The expectation values < O (z) > are defined by 1 O(y + z)+ O( y + z) n 1 1 < On(z) >= dy − exp y2W (z) Q (y + z) Q ( y + z) . I(z) 2 − 0 − 2 0 − 2 0 − Z (4.513) Next we derive in a straightforward way the formula d dQ d2Q dQ dQ < 0 (z) > = < 0 (z) > < 0 2(z) > + < 0 (z) >2 dz dz dz2 − dz dz dW dQ 1 1 dQ + 0 < 0 (z) > dyy2e... dy 0 (y + z)y2e... . dz dz I (z) − I (z) dz 0 Z 0 Z (4.514) By integrating by part the second term in (4.512) we can see that the first term in (4.514) cancels the 3rd term in (4.512). The last term can be neglected if we assume that dW0/dz is much smaller than Q0. We obtain then

−S0(φ(x)) 2 1/2 d d I0(z) dφ~m0 e = I0(0) exp 2 w d x ln Kρ × I0(0) Z ~m ~m Z Y Y α2Kw ρV dQ dQ 0 ddx < 0 2(z) > < 0 (z) >2 + w−1 − 8 2 dz − dz 0 Z ′ ′ ∂ φ (x)∂µφ (x) . (4.515) × µ The Recursion Formulas: The full path integral is therefore given by

′ Z = φ (x /2) dφ e−S0(φ(x)) 0 D 0 ~m0 Z Z ~m Y 2 ′ I (z) α Kw ρV dQ dQ φ (x) exp 2dw ddx ln 0 0 ddx < 0 2(z) > < 0 (z) >2 ∝ D I (0) − 8 2 dz − dz Z Z Z 0 Z ′ ′ −1 µ + w ∂µφ (x)∂ φ (x) . (4.516) YDRI QFT 185

We write this as

′ ′ Z = φ (x) e−S1(φ (x)). (4.517) 1 D Z The new action S1 has the same form as the action S0, viz

′ K ′ ′ ′ ′ S (φ (x)) = ddxR (φ (x))∂ φ (x)∂µφ (x)+ ddxP (φ (x)). (4.518) 1 2 1 µ 1 Z Z

The new polynomials P1 and R1 (or equivalently Q1 and W1) are given in terms of the old ones P0 and R0 (or equivalently Q0 and W0) by the relations

d/2 −1 ′ W1(2 α0 z) = R1(φ (x)) α2C dQ dQ α2 = 0 d < 0 2(z) > < 0 (z) >2 + 0 . 8 dz − dz 4 0 (4.519)

d/2 −1 −1 ′ Q1(2 α0 z) = w P1(φ (x)) I (z) = 2d ln 0 . (4.520) − I0(0)

The constant Cd is given by

Cd = wρV. (4.521)

Before we write down the recursion formulas we also introduce the notation

′ φ(0) = φ , φ(1) = φ . (4.522)

The above procedure can be repeated to integrate out the momentum shell l = 1 and get from S to S . The modes with l 2 will involve a new constant α and instead of P , 1 2 ≥ 1 0 R0, S0 and I0 we will have P1, R1, S1 and I1. Aside from this trivial relabeling everything else will be the same including the constants w, V and Cd since l = 1 can be mapped to l = 0 due to the scaling x 2x (see equations (4.470) and (4.472)). This whole 0 −→ process can be repeated an arbitrary number of times to get a renormalization group flow of the action given explicitly by the sequences P P P ... P 0 −→ 1 −→ 2 −→ −→ i and R R R ... R . By assuming that dR /dφ(i) is sufficiently small 0 −→ 1 −→ 2 −→ −→ i i compared to Pi for all i the recursion formulas which relates the different operators at the renormalization group steps i and i + 1 are obviously given by

d/2 −1 (i+1) Wi+1(2 αi z) = Ri+1(φ (x)) α2C dQ dQ α2 = i d 2 + i . 8 dz − dz 4 i (4.523) 186 YDRI QFT

d/2 −1 −1 (i+1) Qi+1(2 αi z) = w Pi+1(φ (x)) I (z) = 2d ln i . (4.524) − Ii(0) The function I (z) is given by the same formula (4.508) with the substitutions I I , i 0 −→ i W0 Wi and Q0 Qi. −→ (i+1) −→ 1/2 −d/2 (i+1) The ﬁeld φ and the variable z are related by z = (Kρ/2w) 2 αiφ . The full action at the renormalization group step i is K S (φ(i)(x)) = ddxR (φ(i)(x))∂ φ(i)(x)∂µφ(i)(x)+ ddxP (φ(i)(x)). (4.525) i 2 i µ i Z Z The constants αi will be determined from the normalization condition

Wi+1(0) = 1. (4.526) Since Q is even this normalization condition is equivalent to α2 i = 1. (4.527) 4 i |z=0

The Ultra Local Recursion Formula: This corresponds to keeping in the expansion (4.476) only the ﬁrst term. The resulting recursion formula is obtained from the above recursion formulas by dropping from equation (4.523) the ﬂuctuation term dQ dQ 2 . (4.528) dz − dz The recursion formula (4.523) becomes

α2 W (2d/2α−1z) = i . (4.529) i+1 i 4 i The solution of this equation together with the normalization condition (4.527) is given by

Wi = 1 , αi = 2. (4.530) The remaining recursion formula is given by

d/2 −1 d Ii(z) Qi+1(2 2 z) = 2 ln . (4.531) − Ii(0) We state without proof that the use of this recursion formula is completely equivalent to the use in perturbation theory of the Polyakov-Wilson rules given by the approximations:

2 2 2 2 • We replace every internal propagator 1/(k + r0) by 1/(Λ + r0).

Λ d d • We replace every momentum integral Λ/2 d p/(2π) by the volume c/4 where c = 4Ω Λd(1 2−d)/(d(2π)d). d−1 − R YDRI QFT 187

4.5.3 The Wilson-Fisher Fixed Point Let us start with a φ4 action given by

1 1 S [φ ]= ddx (∂ φ )2 + r φ2 + u φ4 . (4.532) 0 0 2 µ 0 2 0 0 0 0 Z The Fourier transform of the ﬁeld is given by

Λ ddp φ (x)= φ˜ (p) eipx. (4.533) 0 (2π)d 0 Z0 We will decompose the ﬁeld as

′ φ0(x)= φ1(x)+Φ(x). (4.534)

′ The background field φ (x) corresponds to the low frequency modes φ˜ (p) where 0 1 0 ≤ p Λ/2 whereas the fluctuation field Φ(x) corresponds to high frequency modes φ˜ (p) ≤ 0 where Λ/2

Z = φ e−S0[φ0] 0 D 0 Z ′ ′ = φ Φ e−S0[φ1+Φ] D 1 D Z Z ′ ′ −S [φ ] = φ e 1 1 . (4.536) D 1 Z ′ The ﬁrst goal is to determine the action S1[φ1]. We have

′ ′ e−S1[φ1] = dΦ e−S0[φ1+Φ]

Z ′ ′ ′ ′ −S [φ ] −S [Φ] −u ddx[4Φ3φ +4Φφ 3+6Φ2φ 2] = e 0 1 dΦ e 0 e 0 R 1 1 1 . (4.537) Z ′ We will expand in the ﬁeld φ1 up to the fourth power. We deﬁne expectation values with respect to the partition function

Z = dΦ e−S0[Φ]. (4.538) Z 188 YDRI QFT

Let us also introduce

′ V = 4u ddxΦ3φ 1 − 0 1 Z ′ V = 6u ddxΦ2φ 2 2 − 0 1 Z ′ V = 4u ddxΦφ 3. (4.539) 3 − 0 1 Z Then we compute

′ ′ −S [φ ] −S [φ ] 1 2 2 1 3 2 1 4 e 1 1 = Ze 0 1 < 1+ V + V + V + (V + 2V V + V + 2V V )+ (V + 3V V )+ V > . 1 2 3 2 1 1 2 2 1 3 6 1 1 2 24 1 (4.540)

By using the symmetry Φ Φ we obtain −→ − ′ ′ −S [φ ] −S [φ ] 1 2 2 1 2 1 4 e 1 1 = Ze 0 1 < 1+ V + (V + V + 2V V )+ (3V V )+ V > . 2 2 1 2 1 3 6 1 2 24 1 (4.541)

The term < V1V3 > vanishes by momentum conservation. We rewrite the diﬀerent expectation values in terms of connected functions. We have

< V2 >=< V2 >co 2 2 < V1 >=< V1 >co 2 2 2 < V2 >=< V2 >co + < V2 >co 2 2 2 < V1 V2 >=< V1 V2 >co + < V1 >co< V2 >co 4 4 2 2 < V1 >=< V1 >co +3 < V1 >co . (4.542) By using these results the partition function becomes

′ ′ −S [φ ] −S [φ ] + 1 ( + )+ 1 + 1 e 1 1 = Ze 0 1 e 2 co 2 1 co 2 co 2 1 2 co 24 1 co . (4.543)

In other words the partition function is expressible only in terms of irreducible connected ′ functions. This is sometimes known as the cumulant expansion. The action S1[φ1] is given by

′ ′ 1 1 1 S [φ ] = S [φ ] < V > (< V 2 > + < V 2 > ) < V 2V > < V 4 > . 1 1 0 1 − 2 co −2 1 co 2 co − 2 1 2 co −24 1 co (4.544)

We need the propagator

< Φ(x)Φ(y) > = < Φ(x)Φ(y) > 12u ddz < Φ(x)Φ(z) > < Φ(y)Φ(z) > < Φ(z)Φ(z) > +O(u2). co 0 − 0 0 0 0 0 Z (4.545) YDRI QFT 189

The free propagator is obviously given by

Λ ddp 1 < Φ(x)Φ(y) > = eip(x−y). (4.546) 0 (2π)d p2 + r ZΛ/2 0 Thus

ddp 1 ddp ddp 1 < Φ(x)Φ(y) > = 1 12u 1 2 + O(u2) eip1(x−y). co (2π)d p2 + r − 0 (2π)d (2π)d (p2 + r )2(p2 + r ) 0 Z 1 0 Z 1 0 2 0 (4.547)

We can now compute

′ < V > = 6u ddxφ 2(x) < Φ2(x) > − 2 co 0 1 co Z d d d d d p ′ d p 1 d p d p 1 = 6u φ˜ (p) 2 1 12u 1 2 + O(u2) . 0 (2π)d | 1 | (2π)d p2 + r − 0 (2π)d (2π)d (p2 + r )2(p2 + r ) 0 Z Z 1 0 Z 1 0 2 0 (4.548)

1 ′ ′ < V 2 > = 8u2 ddx ddx φ (x )φ (x ) < Φ3(x )Φ3(x ) > −2 1 co − 0 1 2 1 1 1 2 1 2 co Z ′ ′ = 8u2 ddx ddx φ (x )φ (x ) 6 < Φ(x )Φ(x ) >3 +3 < Φ(x )Φ(x ) > < Φ(x )Φ(x ) > − 0 1 2 1 1 1 2 1 2 0 1 2 0 1 1 0 Z < Φ(x )Φ(x ) > × 2 2 0 d d d d p ′ d p d p 1 = 8u2 φ˜ (p) 2 6 1 2 + O(u ) . − 0 (2π)d | 1 | (2π)d (2π)d (p2 + r )(p2 + r )((p + p + p )2 + r ) 0 Z Z 1 0 2 0 1 2 0 (4.549)

The second term of the second line of the above equation did not contribute because of momentum conservation. Next we compute

1 ′ ′ < V 2 > = 18u2 ddx ddx φ 2(x )φ 2(x ) < Φ2(x )Φ2(x ) > −2 2 co − 0 1 2 1 1 1 2 1 2 co Z ′ ′ = 36u2 ddx ddx φ 2(x )φ 2(x ) < Φ(x )Φ(x ) >2 − 0 1 2 1 1 1 2 1 2 0 Z d d d d p d p ′ ′ ′ d k 1 = 12u2 1 ... 3 φ˜ (p )...φ˜ (p )φ˜ ( p p p ) − 0 (2π)d (2π)d 1 1 1 3 1 − 1 − 2 − 3 (2π)d k2 + r Z Z 0 1 + 2 permutations + O(u ) . (4.550) × (k + p + p )2 + r 0 1 2 0 190 YDRI QFT

The last two terms of the cumulant expansion are of order u3 and u4 respectively which ′ 0 0 we are not computing. The action S1[φ1] reads then explicitly

Λ/2 d d d d ′ 1 d p ′ d k 1 d k d k 1 S [φ ] = φ˜ (p) 2 p2 + r + 12u 1 144u2 1 2 1 1 2 (2π)d | 1 | 0 0 (2π)d k2 + r − 0 (2π)d (2π)d (k2 + r )2(k2 + r ) Z0 Z 1 0 Z 1 0 2 0 ddk ddk 1 96u2 1 2 + O(u3) − 0 (2π)d (2π)d (k2 + r )(k2 + r )((p + k + k )2 + r ) 0 Z 1 0 2 0 1 2 0 Λ/2 d d d d p d p ′ ′ ′ d k 1 + 1 ... 3 φ˜ (p )...φ˜ (p )φ˜ ( p p p ) u 12u2 (2π)d (2π)d 1 1 1 3 1 − 1 − 2 − 3 0 − 0 (2π)d k2 + r Z0 Z 0 1 + 2 permutations + O(u3) . (4.551) × (k + p + p )2 + r 0 1 2 0 ′ The Fourier mode φ˜ (p) is of course equal φ˜ (p) for 0 p Λ/2 and 0 otherwise. We 1 0 ≤ ≤ scale now the ﬁeld as ˜′ ˜ φ1(p)= α0φ1(2p). (4.552) The action becomes 1 Λ ddp p2 ddk 1 ddk ddk S [φ ] = α22−d φ˜ (p) 2 + r + 12u 1 144u2 1 2 1 1 2 0 (2π)d | 1 | 4 0 0 (2π)d k2 + r − 0 (2π)d (2π)d Z0 Z 1 0 Z 1 ddk ddk 1 96u2 1 2 + O(u3) × (k2 + r )2(k2 + r ) − 0 (2π)d (2π)d (k2 + r )(k2 + r )(( 1 p + k + k )2 + r ) 0 1 0 2 0 Z 1 0 2 0 2 1 2 0 Λ ddp ddp ddk 1 + α42−3d 1 ... 3 φ˜ (p )...φ˜ (p )φ˜ ( p p p ) u 12u2 0 (2π)d (2π)d 1 1 1 3 1 − 1 − 2 − 3 0 − 0 (2π)d k2 + r Z0 Z 0 1 + 2 permutations + O(u3) . (4.553) × (k + 1 p + 1 p )2 + r 0 2 1 2 2 0

In the above equation the internal momenta ki are still unscaled in the interval [Λ/2, Λ]. The one-loop truncation of this result is given by

1 Λ ddp p2 ddk 1 S [φ ] = α22−d φ˜ (p) 2 + r + 12u 1 + O(u2) 1 1 2 0 (2π)d | 1 | 4 0 0 (2π)d k2 + r 0 Z0 Z 1 0 Λ ddp ddp ddk 1 + α42−3d 1 ... 3 φ˜ (p )...φ˜ (p )φ˜ ( p p p ) u 12u2 0 (2π)d (2π)d 1 1 1 3 1 − 1 − 2 − 3 0 − 0 (2π)d k2 + r Z0 Z 0 1 + 2 permutations + O(u3) . (4.554) × (k + 1 p + 1 p )2 + r 0 2 1 2 2 0

We bring the kinetic term to the canonical form by choose α0 as

1+d/2 α0 = 2 . (4.555)

Furthermore we truncate the interaction term in the action by setting the external momenta to zero since we are only interested in the renormalization group ﬂow of the YDRI QFT 191 operators present in the original action. We get then

1 Λ ddp Λ ddp ddp S [φ ] = φ˜ (p) 2(p2 + r )+ u 1 ... 3 φ˜ (p )...φ˜ (p )φ˜ ( p p p ). 1 1 2 (2π)d | 1 | 1 1 (2π)d (2π)d 1 1 1 3 1 − 1 − 2 − 3 Z0 Z0 (4.556)

The new mass parameter r1 and the new coupling constant u1 are given by

ddk 1 r = 4r + 48u 1 + O(u2). (4.557) 1 0 0 (2π)d k2 + r 0 Z 1 0

ddk 1 u = 24−d u 36u2 + O(u3) . (4.558) 1 0 − 0 (2π)d (k2 + r )2 0 Z 0 Now we employ the Wilson-Polyakov rules corresponding to the ultra local Wilson recursion formula (4.531) consisting of making the following approximations:

2 2 2 2 • We replace every internal propagator 1/(k + r0) by 1/(Λ + r0).

Λ d d • We replace every momentum integral Λ/2 d p/(2π) by the volume c/4 where c = 4Ω Λd(1 2−d)/(d(2π)d). d−1 − R

The mass parameter r1 and the coupling constant u1 become

u r = 4 r + 3c 0 + O(u2) . (4.559) 1 0 Λ2 + r 0 0

u2 u = 24−d u 9c 0 + O(u3) . (4.560) 1 0 − (Λ2 + r )2 0 0

This is the result of our first renormalization group step. Since the action S1[φ1] is of the same form as the action S0[φ0] the renormalization group calculation can be repeated without any change to go from r1 and u1 to a new mass parameter r2 and a new coupling constant u2. This whole process can evidently be iterated an arbitrary number of times to define a renormalization group flow (r , u ) (r , u ) ....(r , u ) (r , u )..... 0 0 −→ 1 1 −→ l l −→ l+1 l+1 The renormalization group recursion equations relating (rl+1, ul+1) to (rl, ul) are given precisely by the above equations, viz

u r = 4 r + 3c l . (4.561) l+1 l Λ2 + r l

u2 u = 24−d u 9c l . (4.562) l+1 l − (Λ2 + r )2 l 192 YDRI QFT

The ﬁxed points of the renormalization group equations is deﬁne obviously by

u r = 4 r + 3c ∗ . (4.563) ∗ ∗ Λ2 + r ∗

u2 u = 24−d u 9c ∗ . (4.564) ∗ ∗ − (Λ2 + r )2 ∗ We ﬁnd the solutions

Gaussian ﬁxed point : r∗ = 0 , u∗ = 0, (4.565) and (by assuming that u∗ is suﬃciently small)

4cu Λ4 Wilson Fisher fixed point : r = ∗ , u = (1 2d−4). (4.566) − ∗ − Λ2 ∗ 9c − For ǫ = 4 d small the non trivial (interacting) Wilson-Fisher fixed point approaches − the trivial (free) Gaussian fixed point as

4 Λ4 r = Λ2ǫ ln 2 , u = ǫ ln 2. (4.567) ∗ −9 ∗ 9c

The value u∗ controls the strength of the interaction of the low energy (infrared) physics of the system.

4.5.4 The Critical Exponents ν

In the Gaussian model the recursion formula reads simply rl+1 = 4rl and hence we have two possible solutions. At T = Tc the mass parameter r0 must be zero and hence rl = 0 for all l, i.e. r = 0 is a fixed point. For T = T the mass parameter r is non zero and 0 6 c 0 hence r = 4lr for l (r = is the second fixed point). For T near T l 0 −→ ∞ −→ ∞ 0 ∞ c the mass parameter r is linear in T T . 0 − c In the φ4 model the situation is naturally more complicated. We can be at the critical temperature T = Tc without having the parameters r0 and u0 at their fixed point values. Indeed, as we have already seen, for any value u0 there will be a critical value r = r (u ) of r corresponding to T = T . At T = T we have r r and u u 0c 0c 0 0 c c l −→ ∗ l −→ ∗ for l . For T = T we will have in general a different limit for large l. −→ ∞ 6 c The critical exponent ν can be calculated by studying the behavior of the theory only for T near T . As stated above r (T ) r and u (T ) u for l . c l c −→ ∗ l c −→ ∗ −→ ∞ From the analytic property of the recursion formulas we conclude that rl(T ) and ul(T ) are analytic functions of the temperature and hence near Tc we should have rl(T ) = ′ ′ r (T )+(T T )r (T )+... and u (T )= u (T )+(T T )u (T )+... and as a consequence l c − c l c l l c − c l c rl(T ) and ul(T ) are close to the fixed point values for sufficiently large l and sufficiently small T T . We are thus led in a natural way to studying the recursion formulas only − c around the fixed point, i.e. to studying the linearized recursion formulas. YDRI QFT 193

The Linearized Recursion Formulas: Now we linearize the recursion formulas around the ﬁxed point. We ﬁnd without any approximation

12cu 12c 1 1 r r = 4 ∗ (r r )+ (u u ) + 12cu . l+1 − ∗ − (Λ2 + r )(Λ2 + r ) l − ∗ Λ2 + r l − ∗ ∗ Λ2 + r − Λ2 + r l ∗ ∗ l ∗ (4.568)

9cu2(2Λ2 + r + r ) 9c u u = 24−d ∗ ∗ l (r r ) + 24−d 1 (u + u ) (u u ). l+1 − ∗ (Λ2 + r )2(Λ2 + r )2 l − ∗ − (Λ2 + r )2 l ∗ l − ∗ l ∗ l (4.569)

Keeping only linear terms we ﬁnd

12cu 12c r r = 4 ∗ (r r )+ (u u ). (4.570) l+1 − ∗ − (Λ2 + r )2 l − ∗ Λ2 + r l − ∗ ∗ ∗

18cu2 18cu u u = 24−d ∗ (r r ) + 24−d 1 ∗ (u u ). (4.571) l+1 − ∗ (Λ2 + r )3 l − ∗ − (Λ2 + r )2 l − ∗ ∗ ∗ This can be put into the matrix form

r r r r l+1 − ∗ = M l − ∗ . (4.572) u u u u l+1 − ∗ l − ∗ The matrix M is given by

12cu∗ 12c 4 2 2 2 4 12c 4 (Λ +r∗) Λ +r∗ 4 3 ǫ ln 2 Λ2 (1 + 9 ǫ ln 2) M = − 2 = − . 24−d 18cu∗ 24−d(1 18cu∗ ) 0 1 ǫ ln 2 (Λ2+r∗)3 − (Λ2+r∗)2 ! − (4.573)

After n steps of the renormalization group we will have

r r r r l+n − ∗ = M n l − ∗ . (4.574) u u u u l+n − ∗ l − ∗ In other words for large n the matrix M n is completely dominated by the largest eigenvalue of M. Let λ1 and λ2 be the eigenvalues of M with eigenvectors w1 and w2 respectively such that λ1 > λ2. Clearly for u∗ = 0 we have

1 λ = 4 , w = 1 1 0 0 λ = 1 , w = . (4.575) 2 2 1 194 YDRI QFT

The matrix M is not symmetric and thus diagonalization is achieved by an invertible (and not an orthogonal) matrix U. We write

M = UDU −1. (4.576)

The eigenvalues λ1 and λ2 can be determined from the trace and determinant which are given by

λ + λ = M + M , λ λ = M M M M . (4.577) 1 2 11 22 1 2 11 22 − 12 21 We obtain immediately 4 λ = 4 ǫ ln 2 , λ = 1 ǫ ln 2. (4.578) 1 − 3 2 − The corresponding eigenvectors are

4c 5 1 Λ2 (1 + 9 ǫ ln 2) w1 = , w2 = − . (4.579) 0 1

We write the equation Mwk = λkwk as (with (wk)j = wjk)

Mijwjk = λkwik. (4.580)

The identity M = λ λ >< λ can be rewritten as k k| k k| P Mij = λkwikvkj = λ1wi1v1j + λ2wi2v2j. (4.581) Xk T The vectors vk are the eigenvectors of M with eigenvalues λk respectively, viz (with (vk)j = vkj)

T Mij vkj = vkjMji = λkvki. (4.582)

We ﬁnd explicitly

1 0 v = , v = . (4.583) 1 4c (1 + 5 ǫ ln 2) 2 1 Λ2 9 The orthonormality condition is then

vkjwjl = δkl. (4.584) Xj From the result (4.581) we deduce immediately that

n n n Mij = λ1 wi1v1j + λ2 wi2v2j λnw v . (4.585) ≃ 1 i1 1j YDRI QFT 195

The linearized recursion formulas take then the form r r λnw v (r r )+ v (u u ) . (4.586) l+n − ∗ ≃ 1 11 11 l − ∗ 12 l − ∗ u u λnw v (r r )+ v (u u ) . (4.587) l+n − ∗ ≃ 1 21 11 l − ∗ 12 l − ∗ Since rl = rl(T ) and ul = ul(T ) are close to the fixed point values for sufficiently large l and sufficiently small T T we conclude that r r and u u are both linear in − c l − ∗ l − ∗ T T and as a consequence − c v (r r )+ v (u u )= c (T T ). (4.588) 11 l − ∗ 12 l − ∗ l − c The linearized recursion formulas become r r c λnw (T T ). (4.589) l+n − ∗ ≃ l 1 11 − c u u c λnw (T T ). (4.590) l+n − ∗ ≃ l 1 21 − c The Critical Exponent ν: The correlation length corresponding to the initial action is given by

ξ0(T )= X(r0(T ), u0(T )). (4.591) After l + n renormalization group steps the correlation length becomes

ξl+n(T )= X(rl+n(T ), ul+n(T )). (4.592) At each renormalization group step we scale the momenta as p 2p which corresponds −→ to scaling the distances as x x/2. The correlation length is a measure of distance −→ and thus one must have −l−n X(rl+n, ul+n) = 2 X(r0, u0). (4.593) From equations (4.589) and (4.590) we have (r r ) = (r r ) , (u u ) = (u u ) . l+n+1 − ∗ |T −Tc=τ/λ1 l+n − ∗ |T −Tc=τ l+n+1 − ∗ |T −Tc=τ/λ1 l+n − ∗ |T −Tc=τ (4.594) Hence X(r , u ) = X(r , u ) . (4.595) l+n+1 l+n+1 |T =Tc+τ/λ1 l+n l+n |T =Tc+τ By using the two results (4.593) and (4.595) we obtain −l−n−1 −l−n 2 ξ0(Tc + τ/λ1) = 2 ξ0(Tc + τ). (4.596) We expect ξ (T + τ) τ −ν. (4.597) 0 c ∝ In other words

1 τ −ν −ν ν ln 2 ( ) = τ λ1 = 2 ν = . (4.598) 2 λ1 ⇔ ⇔ ln λ1 196 YDRI QFT

4.5.5 The Critical Exponent η The ultra local recursion formula (4.531) used so far do not lead to a wave function renormalization since all momentum dependence of Feynman diagrams has been dropped and as a consequence the value of the anomalous dimension η within this approximation is 0. This can also be seen from the ﬁeld scaling (4.552) with the choice (4.555) which are made at every renormalization group step and hence the wave function renormalization is independent of the momentum. In any case we can see from equation (4.554) that the wave function renormalization at the ﬁrst renormalization group step is given by

α2 Z = 0 . (4.599) 22+d

From the other hand we have already established that the scaling behavior of Z(λ) for small λ (the limit in which we approach the infrared stable ﬁxed point) is λη. In our case λ = 1/2 and hence we must have

Z = 2−η. (4.600)

Let α∗ be the ﬁxed value of the sequence αi. Then from the above two equations we obtain the formula ln Z 2 ln α η = = d + 2 ∗ . (4.601) − ln 2 − ln 2

1+d/2 As discussed above since α∗ = 2 for the ultra local recursion formula (4.531) we get immediately η = 0. To incorporate a non zero value of the critical exponent η we must go to the more accurate yet more complicated recursion formulas (4.523) and (4.524). The field scaling at each renormalization group step is a different number αi. These numbers are determined from the normalization condition (4.527). Recall that integrating out the momenta 1 k /Λ 2 resulted in the field φ1(x)= ∞ ≤ | | ≤ ′ (1) ~m l=1 ψ~ml(x)φ~ml which was expressed in terms of the field φ (x)= φ which appears −d/2 ′ in the final action as φ1(x) = 2 α0φ (x/2). After n renormalization group steps P P 1−n we integrate out the momenta 2 k /Λ 2 which results in the field φn(x) = ∞ ≤ | | ≤ ′ (n) ~m l=n ψ~ml(x)φ~ml. However the action will be expressed in terms of the field φ = φ defined by P P −nd/2 (n) n φn(x) = 2 α0α1....αn−1φ (x/2 ). (4.602)

We are interested in the 2 point function − ∞ 1 −S [φ] < φ φ ′ ′ > = dφ φ φ ′ ′ e 0 . (4.603) ~ml ~m l Z ~m1l1 ~ml ~m l Z lY1=0 Y~m1 YDRI QFT 197

′ Let us concentrate on the integral with l1 = l and ~m1 = ~m and assume that l > l. We have then the integral

l−1 ′ −S0[φ] −Sl[φ ] ... dφ~mlφ~ml dφ~m1l1 e = ... dφ~mlφ~ml e . (4.604) Z Z lY1=0 Y~m1 Z ′ (l) (l) l−1 We have φ = φ where φ contains the momenta k /Λ 1/2 . Since φ~ml is −1/2 | | ≤ not integrated we have φ~ml = α0...αl−1(Kρ/2) y~m which is the generalization of ′ φ~ml = α0φ~ml−1. We want now to further integrate φ~ml. The ﬁnal result is similar to (4.504) except that we have an extra factor of y~m and z~m contains all the modes with l1 > l. The integral thus clearly vanishes because it is odd under y~m ~y~m. ′ ′ −→ − We conclude that we must have l = l and ~m = ~m otherwise the above 2 point − function vanishes. After few more calculations we obtain

Kρ l =l+1 ~m dφ ~m1l1 ~m Ml(z~m1 ).Rl(z~m) ′ ′ ′ 2 2 −1 1 1 1 < φ~mlφ~m l > = δll δ ~′ α0...αl−1( ) . ~mm 2 dφ Ml(z ) R Q l1=Ql+1 ~m1 Q~m1l1 ~m1 ~m1 R Q Q Q (4.605) The function M is given by the same formula (4.505) with the substitutions M M , l 0 −→ l W W and Q Q . The variable z is given explicitly by 0 −→ l 0 −→ l ~m Kρ z = 1/22−d/2α φ(l+1)(x /2). (4.606) ~m 2w l 0 The function Rl(z) is deﬁned by

−1 2 Rl(z)= Ml (z) dyy exp ... . (4.607) Z The exponent is given by the same exponent of equation (4.505) with the substitutions M M , W W and Q Q . 0 −→ l 0 −→ l 0 −→ l An order of magnitude formula for the 2 point function can be obtained by replacing − the function Rl(z) by Rl(0). We obtain then

2 2 Kρ −1 < φ φ ′ ′ > = δ ′ δ ′ α ...α ( ) Rl(0). (4.608) ~ml ~m l ll ~mm~ 0 l−1 2 At a ﬁxed point of the recursion formulas we must have

W W , Q Q R R , (4.609) l −→ ∗ l −→ ∗ ⇔ l −→ ∗ and

α α . (4.610) l −→ ∗ The 2 point function is therefore given by − 2l Kρ −1 < φ φ ′ ′ > δ ′ δ ′ α ( ) R∗(0). (4.611) ~ml ~m l ∝ ll ~mm~ ∗ 2 198 YDRI QFT

The modes < φ > correspond to the momentum shell 2−l k /Λ 21−l, i.e. k ~ml ≤ | | ≤ ∼ Λ2−l. From the other hand the 2 point function is expected to behave as − 1 < φ φ ′ ′ > δ ′ δ ′ , (4.612) ~ml ~m l ∝ ll ~mm~ k2−η where η is precisely the anomalous dimension. By substituting k Λ2−l in this last ∼ formula we obtain

η−2 l(2−η) < φ φ ′ ′ > δ ′ δ ′ Λ 2 . (4.613) ~ml ~m l ∝ ll ~mm~ By comparing the l dependent bits in (4.611) and (4.613) we ﬁnd that the anomalous − dimension is given by 2 ln α 22−η = α2 η = 2 ∗ . (4.614) ∗ ⇒ − ln 2 A Exercises

Power Counting Theorems for Dirac and Vector Fields

• Derive power counting theorems for theories involving scalar as well as Dirac and vector ﬁelds by analogy with what we have done for pure scalar ﬁeld theories.

• What are renormalizable ﬁeld theories in d = 4 dimensions involving spin 0, 1/2 and 1 particles.

• Discuss the case of QED.

Renormalization Group Analysis for The Eﬀective Action

• In order to study the system in the borken phase we must perform a renormalization group analysis of the eﬀective action and study its behavior as a function of the mass parameter. Carry out explicitly this program. 200 YDRI QFT Examination QFT Master 2 2012-2013 2 h

Solve 3 exercises out of 6 as follows:

• Choose between 1 and 2.

• Choose between 3 and 4.

• Choose between 5 and 6.

Exercise 1: We consider the two Euclidean integrals

d4k 1 I(m2)= . (2π)4 k2 + m2 Z

d4k 1 1 J(p2,m2)= . (2π)4 k2 + m2 (p k)2 + m2 Z − • Determine in each case the divergent behavior of the integral.

• Use dimensional regularization to compute the above integrals. Determine in each case the divergent part of the integral. In the case of J(p2,m2) assume for simplicity zero external momentum p = 0.

Exercise 2: The two integrals in exercise 1 can also be regularized using a cutoﬀ Λ. First we perform Laplace transform as follows

∞ 1 2 2 = dαe−α(k +m ). k2 + m2 Z0 • Do the integral over k in I(m2) and J(p2,m2). In the case of J(p2,m2) assume for simplicity zero external momentum p = 0.

• The remaining integral over α is regularized by replacing the lower bound α = 0 by α = 1/Λ2. Perform the integral over α explicitly. Determine the divergent part in each case. Hint: Use the exponential-integral function

−x et x e−t 1 Ei( x)= dt = C + ln x + dt − . − t t Z−∞ Z0 YDRI QFT 201

Exercise 3: Let zi be a set of complex numbers, θi be a set of anticommuting Grass- mann numbers and let M be a hermitian matrix. Perform the following integrals

+ + + + −Mij zi zj −zi ji−ji zi dzi dzie . Z Yi

+ + + + −Mij θi θi−θi ηi−ηi θi dθi dθie . Z Yi Exercise 4: Let S(r, θ) be an action dependent on two degrees of freedom r and θ which is invariant under 2 dimensional rotations, i.e. ~r = (r, θ). We propose to gauge − ﬁx the following 2 dimensional path integral − W = eiS(~r)d2~r. Z We will impose the gauge condition

g(r, θ) = 0.

• Show that ∂g(r, θ) dφδ g(r, θ + φ) = 1. | ∂θ |g=0 Z • Use the above identity to gauge ﬁx the path integral W .

Exercise 5: The gauge ﬁxed path integral of quantum electrodynamics is given by

(∂ Aµ)2 i Z[J] = A exp i d4x µ d4xF F µν i d4xJ Aµ . D µ − 2ξ − 4 µν − µ µ Z Y Z Z Z • Derive the equations of motion.

• Compute Z[J] in a closed form.

• Derive the photon propagator.

Exercise 6: We consider phi-four interaction in 4 dimensions. The action is given by 1 1 λ S[φ]= d4x ∂ φ∂µφ m2φ2 (φ2)2 . 2 µ − 2 − 4! Z • Write down Feynman rules in momentum space.

• Use Feynman rules to derive the 2 point proper vertex Γ2(p) up to the one-loop − order. Draw the corresponding Feynman diagrams. 202 YDRI QFT

• Use Feynman rules to derive the 4 point proper vertex Γ4(p ,p ,p ,p ) up to the − 1 2 3 4 one-loop order. Draw the corresponding Feynman diagrams.

• By assuming that the momentum loop integrals are regularized perform one-loop renormalization of the theory. Impose the two conditions

2 2 4 Γ (0) = mR , Γ (0, 0, 0, 0) = λR.

Determine the bare coupling constants m2 and λ in terms of the renormalized 2 coupling constants mR and λR. 2 4 • Determine Γ (p) and Γ (p1,p2,p3,p4) in terms of the renormalized coupling constants. YDRI QFT 203 Final Examination QFT Master 2 2012-2013 2 h

Exercise 1: We consider the two Euclidean integrals

d4k 1 I(m2)= . (2π)4 k2 + m2 Z

d4k 1 1 J(p2,m2)= . (2π)4 k2 + m2 (p k)2 + m2 Z − • Determine in each case the divergent behavior of the integral.

Exercise 2: The gauge ﬁxed path integral of quantum electrodynamics is given by

(∂ Aµ)2 i Z[J] = A exp i d4x µ d4xF F µν i d4xJ Aµ . D µ − 2ξ − 4 µν − µ µ Z Y Z Z Z • Derive the equations of motion.

• Compute Z[J] in a closed form.

• Derive the photon propagator. 204 YDRI QFT B References 206 YDRI QFT Bibliography

[1] M. E. Peskin and D. V. Schroeder, “An Introduction to quantum ﬁeld theory,” Reading, USA: Addison-Wesley (1995) 842 p

[2] J. Zinn-Justin, “Quantum ﬁeld theory and critical phenomena,” Int. Ser. Monogr. Phys. 113, 1 (2002).

[3] J. Smit, “Introduction to quantum ﬁelds on a lattice: A robust mate,” Cambridge Lect. Notes Phys. 15, 1 (2002).

[4] H. Kleinert, “Critical Poperties of Phi4 Theories,” World Scientiﬁc, Singapore, 2004

[5] K. G. Wilson and J. B. Kogut, “The Renormalization group and the epsilon expansion,” Phys. Rept. 12, 75 (1974).

[6] K. G. Wilson, “Renormalization group and critical phenomena. 2. Phase space cell analysis of critical behavior,” Phys. Rev. B 4, 3184 (1971).

[7] G. R. Golner, “Calculation of the Critical Exponent eta via Renormalization-Group Recursion Formulas,” Phys. Rev. B 8, 339 (1973).