Quantum Fields and Probability

Home , Cylinder set, Free field, Scalar field theory, Ultraviolet divergence

Antti Kupiainen

September 3, 2018 Contents

1 Introduction5 1.1 Background ...... 5 1.2 Overview ...... 5

2 Quantum Fields7 2.1 Examples ...... 7 2.1.1 Harmonic oscillator ...... 7 2.1.2 Phonon ﬁeld theory - an infrared divergence ...... 10 2.1.3 Vibrating string - an ultraviolet divergence ...... 13 2.2 Second quantization ...... 15 2.2.1 Examples ...... 19 2.2.2 Free scalar ﬁeld ...... 20 2.3 Interacting Fields ...... 22

3 Euclidean Quantum Field Theory 25 3.1 Wiener Measure ...... 26 3.2 Feynman-Kac formula ...... 27 3.3 Harmonic oscillator - Ornstein-Uhlenbeck process ...... 28 3.4 The free ﬁeld ...... 29

4 Random Fields 31 4.1 The Ginzburg-Landau model ...... 31 4.2 Gaussian Integrals ...... 32 4.3 The Gaussian Ginzburg-Landau model ...... 33 4.4 Measures on s0 ...... 35 4.4.1 Bochner’s Theorem ...... 35 4.4.2 Cylinder measures ...... 36 4.4.3 Minlos Theorem ...... 37 4.5 Measures on S0 ...... 41 4.6 Gaussian measures ...... 44 4.6.1 The Euclidean Free Feld ...... 44 4.6.2 Wick Formula ...... 45

5 From Random Fields to Quantum Fields 48 5.1 Reflection positivity ...... 48 5.2 Reflection positive Gaussian measures ...... 50 5.2.1 Example: harmonic oscillator and free field ...... 50 5.3 Reconstruction for Free Field ...... 51 5.3.1 Time zero fields ...... 51 5.3.2 Harmonic oscillator once more ...... 51 5.3.3 Free field ...... 53 5.3.4 Fock space ...... 53

2 6 Perturbation Theory 55 6.1 Regularization ...... 55 6.1.1 Momentum space regularization ...... 55 6.1.2 Scaling limit ...... 56 6.1.3 Lattice Regularization ...... 57 6.2 Gaussian critical point ...... 57 6.2.1 Massless case ...... 58 6.2.2 Massive case ...... 59 6.3 Perturbation Theory for the GL model ...... 60 6.3.1 Measures ...... 60 6.3.2 Perturbation Theory ...... 61 6.3.3 Feynman graphs ...... 61 6.3.4 Momentum space representation ...... 67 6.4 Infrared and Ultraviolet Divergences ...... 69 6.4.1 Infrared divergences ...... 70 6.4.2 Ultraviolet ...... 72

7 The renormalization group 76 7.1 The Block-spin Transformation ...... 76 7.1.1 Gaussian fixed point ...... 78 7.2 Transformations on Measures ...... 78 7.3 Transformations on Hamiltonians ...... 79 7.4 Fixed Points of RL ...... 81 7.5 Linear Analysis around a Fixed Point: Critical Exponents ...... 83 7.6 Effective Field Theories ...... 84 7.7 Scaling limit ...... 86 7.8 Non-Scale Invariant Theories ...... 87 7.9 The Gaussian Fixed Point ...... 89 7.9.1 Block spin fixed point ...... 89 7.9.2 RG in Momentum Space ...... 90

8 Hierarchical model 93 8.1 Local RG transformation ...... 93 8.2 Linearization ...... 94 8.3 Perturbative analysis of the hierarchical RG ...... 95 8.4 d > 4: Gaussian critical point ...... 96 8.5 d = 4: Infrared asymptotic freedom ...... 97 8.5.1 Susceptibility divergence ...... 98 8.6 d < 4: Wilson-Fisher ﬁxed point ...... 99 8.7 d < 4: Superrenormalizable QFT ...... 99 8.7.1 d =2 ...... 101 8.7.2 d =3 ...... 101 8.7.3 Inﬁnite Volume Limit ...... 102 8.7.4 d ↑ 4...... 102 8.7.5 d = 4: Triviality ...... 103 8.8 Nonperturbative analysis ...... 103

9 RG analysis of the Ginzburg Landau Model 107 9.1 Linear RG ...... 107 9.2 The space of local Hamiltonians ...... 109 9.3 Perturbative analysis of RV ...... 112 9.4 Iteration ...... 114 9.4.1 d > 4...... 115 9.4.2 d =4...... 116

3 9.4.3 d = 4 − : anomalous scaling ...... 117 9.4.4 Other fixed points for d < 4...... 119 9.5 Quantum Field Theories ...... 119 9.5.1 Effective Field Theories ...... 119 9.5.2 Polchinski equation ...... 121 9.5.3 Superrenormalizable QFT ...... 122 9.5.4 d < 4: nontrivial fixed points ...... 122 9.5.5 Asymptotic Freedom ...... 123

10 Stochastic PDE’s with Rough Noise 125 10.1 Stochastic Diﬀerential Equations and Stochastic Quantization ...... 125 10.2 Linear case: Gaussian Process ...... 125 10.3 Nonlinear SDE: Local existence ...... 125 10.4 Perturbation Theory ...... 125 10.5 Rough SPDE’s ...... 125 10.5.1 Dimensionless variables ...... 126 10.5.2 Hierarchical SPDE ...... 127 10.5.3 Eﬀective equations ...... 127 10.5.4 Dimensionless formulation ...... 128 10.5.5 RG for SPDE ...... 128 10.5.6 Linearized RG map ...... 131 10.5.7 RG iteration, d =2...... 133

A Some functional analysis 135

B Perron-Frobenius theorem 136

4 Chapter 1

Introduction

1.1 Background

Quantum Field Theory (QFT) has become a universal framework to study physical systems with infinite number of degrees of freedom. Originally developed for high energy physics it lead to the explanation of universality in phase transitions using the Renormalization Group (RG), a powerful method to study scale invariant problems. QFT and RG ideas were thereafter applied to noisy systems, dynamical systems, non-equilibrium systems and many other problems. Early on QFT was also acknowledged as an interesting problem for mathematicians. This lead first to axiomatic characterizations of what sort of mathematical object QFT is and in then to Constructive QFT and the use of probabilistic methods to produce concrete examples of QFT. These techniques were then used in rigorous statistical mechanics and disordered systems. Following the pioneering work of Belavin, Polyakov and Zamoldchikov [?] a beautiful class of QFTs was uncovered by physicists, the two dimensional Conformal Field Theories (CFT). They have inspired a lot of new ideas in mathematics: the Schramm-Loewner Evolution (SLE) and Liouville Quantum Gravity and the theory of random surfaces. In Stochastic Partial Differential Equations (SPDE) physicists had also used QFT to study universality, a notable example being the Kardar-Parisi-Zhang (KPZ) equation where exact non-conventional scaling behavior was uncovered. KPZ and related equations posed hard problems for analysis due to the combination of a very singular noise and nonlinearity. Decisive progress was achieved here by Martin Hairer who introduced QFT ideas on renormalization to their mathematical analysis [?].

1.2 Overview

In quantum mechanics, a system is determined by giving all its possible states, observables and the time evolution. The possible states form a Hilbert space H, while the observables are given by self-adjoint operators A on this space. These operators evolve in time as eitH/~Ae−itH/~ for some positive opertaor H, called the Hamiltonian. In quantum field theory, one considers fields ϕ(t, ~x) that are operators1 on the Hilbert d−1 space H. Here t ∈ R is the time component and ~x ∈ R is a point in space (d = 4 being the physical case). In general, these fields can be spinors, vectors etc., but we will mostly consider scalar fields: ϕ(t, ~x) ∈ R. The fields evolve in time as

ϕ(t, ~x) = eitH/~ϕ(0, ~x)e−itH/~ (1.1)

Additionally, one postulates the existence of a vacuum vector ψ0 ∈ H by Hψ0 = 0. The

1in fact operator valued distribution, see Sect.??

5 basic objects of an axiomatic approach to QFT are then the Wightman functions:

Wn(z1, ..., zn) = (ψ0, ϕ(z1)...ϕ(zn)ψ0) for zi = (ti, ~xi). Finally, H carries a representation of the symmetries of the system: e.g. in a relativistic theory the Poincar´egroup. The positivity of the Hamiltonian and the time evolution (1.1) allow to analytically continue the Wightman functions to ‘imaginary time’. The substitution τ → −iτ gives the so-called Schwinger functions Sn: Sn (τ1, ~x1), ..., (τn, ~xn) = Wn (−iτ1, ~x1), ..., (−iτn, ~xn)

The point of this substitution is that the Schwinger functions can be written as correlation functions: Sn (τ1, ~x1), ..., (τn, ~xn) = E ϕ(τ1, ~x1)...ϕ(τn, ~xn) where ϕ(τ, ~x) is a random field on some probability space Ω. This viewpoint also leads to constructive QFT: construct nontrivial random fields and the reconstruct the quantum fields by analytic continuation. The latter turns out to be possible under a condition on the random fields, so-called reflection positivity, see Section5. The imaginary time formulation of QFT is called Euclidean QFT. Finally, there is a deep connection between Euclidean quantum field theory and statistical physics. In statistical physics, one often considers random variables or ’spins’ φ(x) d located at the vertices of a discrete lattice, e.g. x ∈ Z . The probability of a given configuration φ of spins is given by the Gibbs rule:

−βE(φ) P(φ) ∼ e where β is the inverse temperature and E is the energy of the conﬁguration. The connection with QFT appears when one studies the behaviour of the system around the critical point i.e. a value βc of the inverse temperature where the system exhibits a second order phase transition. Then one can try to construct an Euclidean QFT by a scaling limit where the correlation functions of the random ﬁeld ϕ(x) are constructed as limits

−2n∆ ϕ(x1)...ϕ(x2) = lim φ(x1/)...φ(xn/) = (1.2) E →0 E where the parameter ∆ is called the scaling dimension of the ﬁeld ϕ. The powerful tool to study these limits is the renormalization group which is also useful to study the divergences that appear in interacting quantum ﬁeld theories.

6 Chapter 2

Quantum Fields

2.1 Examples

We start by studying some basic examples of quantum systems. For each of them, we will give the Hamiltonian and solve them ﬁrst classically by solving Hamilton’s equations for the generalised coordinates p and q. Next, we promote p and q to operators on a Hilbert space and quantize the system. Finally we compute the time evolution of the system and the Wightman functions.

2.1.1 Harmonic oscillator

Classical In classical Hamiltonian mechanics, the (one-dimensional) harmonic oscillator is described by the Hamiltonian

1 1 H = p2 + kq2 (2.1) 2m 2

p for real variables p and q. The Hamilton equations give:q ˙ = ∂pH = m andp ˙ = −∂qH = −kq, which can be combined into the Newton equation

k q¨ = − q (2.2) m This equation can easily be solved to give

p(0) q(t) = q(0) cos(ωt) + sin ωt (2.3) mω

2 k where ω = m . For later purposes it is useful to introduce complex coordinates

1 √ i a := √ mωq + √ p (2.4) 2 mω and the coplex conjugatea ¯. Then the Hamiltonian becomes

H = ωaa¯ and the time evolution becomesa ˙ = −iωa and a¯˙ = iωa¯, which gives

a(t) = e−iωta(0) anda ¯(t) = eiωta¯(0).

7 2 Quantum In the quantum mechanical case we consider the Hilbert space H = L (R, dq) and the multiplicative position operator

q : ψ(q) → qψ(q).

The momentum operator is then deﬁned as d p = −i , ~dq , giving the canonical commutation relation

[q, p] = i~. Substituting these operators into the Hamiltonian (2.1) gives the quantum mechanical Hamiltonian

2 d2 1 H = − ~ + kq2 (2.5) 2m dq2 2 1 This is a self-adjoint operator on H. From now on we will set ~ = 1 = m , which allows us to deﬁne, in analogy to (2.4):

1 √ i 1 √ 1 d a := √ ωq + √ p = √ ωq + √ (2.6) 2 ω 2 ω dq which has as adjoint operator 1 √ 1 d a∗ = √ ωq − √ (2.7) 2 ω dq These new operators satisfy the commutation relation

[a, a∗] = 1 (2.8) which is equivalent to the relation for p and q. Applying this relation to the identity 1 ∗ ∗ H = 2 ω(a a + aa ) gives that 1 H = ωa∗a + (2.9) 2 The operators a and a∗ are called the annihilation and creation operator respectively, 1 and can be used to find the spectrum of the Hamiltonian. First note that H ≥ 2 ω since ∗ a a > 0, so H is a positive operator. Next, define the ground state ψ0 by aψ0 = 0. This dψ0 gives the differential equation dq = −ωqψ0 with solution

− ω q2 ψ0 = Ce 2 (2.10)

ω 1/4 1 where C = ( π ) is deﬁned by the normalization condition ||ψ0||2 = 1. Now, Hψ0 = 2 ωψ0 1 so the ground state has energy 2 ω. For the other states, we ﬁrst note that the commutation relation for a and a∗ leads to

[H, a] = −ωa, [H, a∗] = ωa∗ and then [H, (a∗)n] = nω(a∗)n. From this it follows that 1 H(a∗)nψ = nω(a∗)nHψ = (n + )(a∗)nψ , 0 0 2 0 ∗ n so for all n ∈ N,(a ) ψ0 is an eigenvector of H. In other words, acting successively with the creator operator on the ground state gives an inﬁnite sequence of eigenstates of H. In fact, this procedure gives all eigenstates of H, or more precisely: q 1 ~ 0 Make a change of variables q = m q

8 ∗ n (a ) ψ0 ∞ Theorem 1. The set A = ψn = ∗ n is an orthonormal basis of eigenfunctions ||(a ) ψ0||2 n=0 of H in H

Proof. We have already proven that all ψn are eigenfunctions of H, and it is clear from the definition that they have norm 1. The orthogonality can be proven by induction. Notice first that ∗ (a ψ0, ψ0) = (ψ0, aψ0) = 0 by the definition of the vacuum, so (ψ1, ψ0) = 0. Suppose that (ψi, ψj) = 0 for all i < j < n and take m < n. Use the commutation relation to get

∗ n ∗ m n ∗ m n−1 ∗ ∗ m−1 (ψn, ψm) = (a ) ψ0, (a ) ψ0 = ψ0, a (a ) ψ0 = ψ0, a (1 + a a)(a ) ψ0

We get two terms, of which the ﬁrst gives zero by the induction hypothesis such that we are left with only the second. We can repeat this procedure to move one annihilation operator all the way to the right:

n−1 ∗ ∗ m−1 n−1 ∗ m (ψn, ψm) = (ψ0, a a a(a ) ψ0) = ... = (ψ0, a (a ) aψ0) = 0 by the deﬁnition of ψ0. So by induction is (ψn, ψm) = 0 for all m 6= n, which proves the orthogonality. To prove that A is a basis, use (2.7) and (2.10) to get

n √ ∗ n −n/2√ 1 d − ω q2 − ω q2 (a ) ψ = 2 ωq − √ Ce 2 = H ( 2ωq)Ce 2 0 ω dq n for a polynomial Hn of degree n. So the fact that A is a basis is equivalent with the fact 2 that the Hermite polynomials Hn form a basis of L (R). This can be proven as follows: n ﬁrst note that Hn = anq + lower order with an 6= 0 so that {H0,H1, ..., Hn} span the ipq polynomials of order smaller or equal to n. Next, note that B = {e ψ0 | p ∈ R} spans 2 2 L (R). Indeed suppose that ψ(q) ∈ L (R) is orthogonal to all these vectors: Z ipq − 1 ωq2 ipq 0 = (ψ, e ψ0) = C dq ψ(q)e 2 e

− 1 ωq2 for all q. This means that the Fourier transform of ψ(q)e 2 is identically zero, so − 1 ωq2 ψ(q)e 2 = 0 and consequently ψ(q) has to be the zero vector. The only thing left to prove is that the polynomials, and thus the Hermite polynomials, span B. Fix p and expand the exponential: ∞ X ipn eipqψ = qnψ 0 n! 0 n=0 All elements of B can thus be written as sum over polynomials. We just need to show that this sum converges in L2 norm. For this, use the approximation:

n n n 1 ω 2 1 ωq2 2 q2n = q2 n! ≤ e 2 n! n! 2 ω ω From this, we get that

Z n Z 2 2 − 1 ω2 kqnψ k2 = C2 q2ne−ωq dq ≤ C2 n! e 2 q2dq = a2nn! 0 2 ω For some constant a Using this, we ﬁnally arrive at

∞ ∞ ∞ ∞ X ipn X |p|n X |p|n √ X |pa|n k qnψ k ≤ kqnψ k ≤ an n! = √ n! 0 2 n! 0 2 n! n=0 n=0 n=0 n=0 n!

By the ratio test, this sum converges in L2-norm which concludes the proof.

9 Now that we have calculated the spectrum of the Hamiltonian, we want to know the dynamics of the system. The operators evolve in time as follows:

Excercise 1. Prove that

a(t) := eitH a e−itH = e−iωta (2.11) a∗(t) := eitH a∗e−itH = eiωta∗ (2.12) using the commutation relations for the ladder operators and the Hamiltonian.

Next, we can calculate the n = 2 Wightman function by using the relation q(t) = √1 (a + a∗) and the formulae (2.8), (2.10), (2.11) and (2.12): 2ω 1 1 (ψ , q(t )q(t )ψ ) = e−iω(t1−t2)(ψ , aa∗ψ) = e−iω(t1−t2) ≡ G(t − t ) 0 1 2 0 2ω 0 2ω 1 2 The Wick formula gives the other Wightman functions in terms of these functions G:

Excercise 2. Prove Wick’s formula: X Y (ψ0, q(t1)q(t2)...q(tn)ψ0) = G(ti − tj) (2.13)

p∈Pn (i,j)∈p

Here Pn is the set of all pairings of {1, 2, ..., n}, p is one of these pairings and each pair of a pairing is denoted by (i, j) where i < j. It is implied that Wn = 0 if n is odd. We can think about the Harmonic oscillator as follows: a∗ creates a vibration mode or particle with energy ω while a annihilates it. So ψn is a state containing n particles, and 1 the empty state or vacuum ψ0 has energy 2 ω. It gets more interesting when we consider systems of several identical oscillators, which is what we will do in the next examples.

2.1.2 Phonon ﬁeld theory - an infrared divergence Classical Phonons are the vibration modes corresponding to vibrations in a crystal d lattice with a periodic structure. Let us consider a lattice of points x = (x1, x2..., xd) ∈ Z , d and take for now only those points that lie in a ﬁnite box ΛN = [0, 1, ..., N − 1] . At each of the lattice points there is a particle with position ϕ(x) ∈ R and momentum π(x) ∈ R. If these particles undergo harmonic vibrations, the Hamiltonian is of the form:

1 X 2 1 X 2 2 2 H(ϕ, π) = 2 (ϕ(x) − ϕ(y)) + 2 m ϕ(x) + π(x) (2.14) x∼y x∈ΛN

Here we used ∼ to denote the sum over nearest neighbours |x − y| = 1 with periodic d boundary conditions: ϕ(x + nN) = ϕ(x) for all n ∈ Z . Hamilton’s equations give the classical dynamics:

∂H ϕ˙(t, x) = = π(t, x) (2.15) ∂π(x) ∂H π˙ (t, x) = − = (∆ϕ)(t, x) − m2ϕ(t, x) (2.16) ∂ϕ(x) where X (∆ϕ)(x) := (ϕ(x + y) − ϕ(x)) |y|=1 is the discrete Laplacian. Combining these equations gives the discrete Klein-Gordon equation ϕ¨ − ∆ϕ + m2ϕ = 0

10 To solve this equation we make a change of variables from ϕ(x) to the the Fourier transform deﬁned by X ϕˆ(k) = e−ikxϕ(x) (2.17)

x∈ΛN where the product kx is the usual scalar product kx = k1x1 + ... + knxn and where

2π d k ∈ BN := [−π, π] ∩ ( N Z) .

The inverse relation is given by

1 X ϕ(x) = eikxϕˆ(k) (2.18) N d k∈BN

Since ϕ is taken to be real, we needϕ ˆ(k) = ϕˆ(−k). The same goes for the Fourier transform of π. These transforms can be used to write H in a more useful form. Plug the Fourier transforms into the ﬁrst term of H to get:

X 2 1 X X 2 ϕ(x) − ϕ(y) = (eikx − eiky)ϕ ˆ(k) N 2d x∼y x∼y k∈BN 1 1 X X X 2 = eikx(1 − eiku)ϕ ˆ(k) 2 N 2d x∈Z |u|=1 k∈BN 1 1 X X X = ei(k+l)x(1 − eiku)(1 − eilu)ϕ ˆ(k)ϕ ˆ(l) 2 N 2d x∈Z |u|=1 k,l∈BN 1 1 X X = (2 − eiku − e−iku)ϕ ˆ(k)ϕ ˆ(−k) 2 N d k∈BN |u|=1 1 1 X X 1 = 4 sin2 ku |ϕˆ(k)|2 2 N d 2 k∈BN |u|=1

Writing also the other two terms of the Hamiltonian in terms of the fourier transforms gives:

d 1 X X 1 H = ω(k)2|ϕˆ(k)|2 + |πˆ(k)|2 with ω(k)2 = 4 sin2 k + m2 (2.19) 2N d 2 i k∈BN i=1

When we deﬁne the function a similar to (2.4) as

1 1 a(k) = √ pω(k)ϕ ˆ(k) + i πˆ(k) (2.20) 2 pω(k) the Hamiltonian takes ﬁnally the form

1 X H = ω(k)a(k)a(k) (2.21) N d k∈BN which is the Hamiltonian of N d harmonic oscillators.

2 |Λ | Quantum To quantize this system, we take the Hilbert space H = L (R N , Dϕ) where Y Dϕ := dϕ(x) Λ x∈R N

11 |Λ | is the Lebesgue measure in R N . Then H 3 ψ = ψ(ϕ) and the scalar product is Z (ψ, ψ0) = ψ(ϕ)ψ0(ϕ)Dϕ.

Deﬁne the momentum operator as

∂ π(x) = −i ∂ϕ(x) to get the familiar commutation relation

[ϕ(x), π(y)] = iδx,y (2.22)

The quantum Hamiltonian is deﬁned as

1 X 2 1 X 2 2 2 H(ϕ, π) = 2 (ϕ(x) − ϕ(y)) + 2 (m ϕ(x) + π(x) ). x∼y x∈ΛN

As in the classical case, we using the Fourier transform we make a linear change of variables so that (2.19) holds also in the quantum case. Then we deﬁne in analogy to (2.6) and (2.7) the ladder operators:

1 1 a(k) = √ pω(k)ϕ ˆ(k) + i πˆ(k) (2.23) 2 pω(k) 1 1 a∗(k) = √ pω(k)ϕ ˆ(−k) − i πˆ(−k) (2.24) 2 pω(k)

As expected, these operators satisfy the nice commutation relation:

Excercise 3. Prove that (2.22) implies that the ladder operators a and a∗ satisfy the relation ∗ d [a(k), a (l)] = N δk,l Using this, we ﬁnd that the Hamiltonian can be written as

1 X 1 X 1 H = ω(k)a∗(k)a(k) + a(k)a∗(k) = ω(k)a∗(k)a(k) + N d 2N d N d 2 k∈BN k∈BN

Next, we deﬁne the ground state ψ0 to be the vector that satisﬁes

a(k)ψ0 = 0 ∀k ∈ BN (2.25)

Excercise 4. Find the ground state vector ψ0 explicitly as a function of the variables ϕ(x)

1 d For this ground state, we have Hψ0 = 2 N ψ0 which is, as expected, the sum of the ground state energies of N d harmonic oscillators. However, this gives the problem that in the limit N → ∞, the ground state energy goes to infinity. Since this limit corresponds to the limit of long wavelengths, this divergence is called an infrared divergence. This divergence can easily be resolved if we take the attitude that in physics one only measures energy differences, so the absolute ground state energy is meaningless2, The it is natural to subtract this ground state energy before taking the limit. This renormalization gives us the final phonon Hamiltonian

1 X H = ω(k)a∗(k)a(k). N d k∈BN

2In quantum gravity this no more holds and the issue has to be faced

12 Using this Hamiltonian, we get [H, a∗(k)] = ω(k)a∗(k) so the eigenstates are

Y ∗ nk ψn = (a (k)) ψ0

k∈BN with n = {nk}k∈BN and they have the energies X Hψn = ( nkω(k))ψn k

2 Λ Excercise 5. Write ψn explicitly in L (R N ) The ladder operators evolve as

eitH a(k)e−itH = e−iωta(k) eitH a∗(k)e−itH = eiωta∗(k)

To find the dynamics for the fields, we note thatϕ ˆ(k) = √ 1 a(k) + a∗(−k), so 2ω(k) ϕ(x) = 1 P √ 1 eikxa(k) + e−ikxa∗(k). The time-evolved field then becomes: N d k 2ω(k)

1 X 1 ϕ(t, x) = eikx−iω(k)ta(k) + e−ikx+iω(k)ta∗(k) N d p2ω(k) k∈BN The time-evolved ﬁeld satisﬁes the Klein-Gordon equation:

2 2 (∂t − ∆ + m )ϕ = 0 Finally, the Wightman functions give: 1 X 1 (ψ , ϕ(t, x)ϕ(s, y)ψ ) = e−iω(k)(t−s)eik(x−y) ≡ G (t − s, x − y) 0 0 N d 2ω(k) N k∈BN which becomes in the N → ∞ limit: Z 1 1 −iω(k)(t−s) ik(x−y) (ψ0, ϕ(t, x)ϕ(s, y)ψ0) = d e e dk ≡ G(t − s, x − y) (2π) [−π,π]d 2ω(k) Hence all Wightman functions have a limit as N → ∞ because also for phonons we have Wick’s theorem: X Y (ψ0, ϕ(ti, xi)...ϕ(tn, xn)ψ0) = G(ti − tj)

p∈Pn (i,j)∈p

2.1.3 Vibrating string - an ultraviolet divergence Classical Our third example is a harmonically vibrating string: consider a string of length 2π with periodic boundary conditions and variables ϕ(t, x), x ∈ [0, 2π], giving the deviation from the rest position. The classical equation of motion is

2 2 2 (∂t − ∂x + m )ϕ(t, x) = 0 This is an inﬁnite dimensional Hamiltonian system: the equation of motion for the ﬁeld results from the Hamilton’s equations

2 2 ϕ˙ = π π˙ = (∂x − m )ϕ which follow formally from the Hamiltonian Z 2π 1 2 2 2 H(ϕ, π) = 2 [π(x) + (∂xϕ) + m ϕ ]dx 0

13 δH δH usingϕ ˙ = δπ andπ ˙ = − δϕ . Here we also introduced the conjugate momenta π(x). The system is a collection of an inﬁnite number of harmonic oscillators: deﬁne for k ∈ Z the fourier transforms: 1 Z 2π ϕˆ(k) ≡ √ e−ikxϕ(x)dx 2π 0 1 Z 2π πˆ(k) ≡ √ e−ikxπ(x)dx 2π 0

So the inverse fourier transform is ϕ(x) = √1 P eikxϕˆ(k) and since the ﬁelds are real, 2π k∈Z ϕˆ(k) =ϕ ˆ(−k). For the momenta π we get similar expressions. The Hamiltonian then becomes 1 X 2 2 2 2 H = 2 [(k + m )|ϕˆ(k)| + |πˆ(k)| ] k∈Z In analogy to the phonon case, we deﬁne a(k) = √1 ω(k)1/2ϕˆ(k) + iω(k)−1/2πˆ(k), but √ 2 now ω(k) = k2 + m2. We get a familiar expression for the Hamiltonian: X H = ω(k)a(k)a(k) k∈Z meaning that the oscillators oscillate with frequencies ω(k). The dynamics is, as expected, a(k, t) = e−iω(k)a(k, 0) and a(k, t) = eiω(k)ta(k, 0). Using that

1 ϕˆ(k) = a(k) + a(−k) p2ω(k) we ﬁnd that the time-evolved ﬁelds are given by

1 X 1 ϕ(x, t) = √ eikxa(k) + e−ikxa(k) 2ω p2ω(k) k∈Z

Quantum (Heuristic) The Hilbert space H is formally some space of functionals Ψ(ϕ(·)), on which the position operator ϕ acts by multiplication ϕ(x)Ψ and the momentum δ R operator is deﬁned by π = −i δϕ(x) . The latter means that for π(f) = π(x)f(x)dx,

d π(f)Ψ(ϕ) = −i Ψ(ϕ + f). d =0 Formally, we have the commutation relation Z [ϕ(f), π(g)] = i f(x)g(x)dx which can be stated as [ϕ(x), π(y)] = iδ(x − y). With the Fourier transforms deﬁned above, we write the usual ladder operators 1 1 a(k) = √ pω(k)ϕ ˆ(k) + i πˆ(k) 2 pω(k) 1 1 a∗(k) = √ pω(k)ϕ ˆ(−k) − i πˆ(−k) 2 pω(k) that satisfy the commutation relation

∗ [a(k), a (l)] = δk,l

14 and give the Hamiltonian

1 X ∗ ∗ H = 2 ω(k) a (k)a(k) + a(k)a (k) k∈Z

Next, we define the vacuum vector Ψ0 by a(k)ψ0 = 0 ∀k. It is easy to see, using the above commutation relation, that HΨ = 1 P ω(k)Ψ = ∞. Also here, the vacuum 0 2 k∈Z 0 energy is infinite. Contrary to the phonon case, the problem lies in the small wavelengths (large k), so this is an ultraviolet divergence. We renormalize the Hamiltonian to X H = ω(k)a∗(k)a(k). k∈Z Now the vacuum energy is zero, as in the classical case. In both the ultraviolet and infrared case, the divergent vacuum energy stems from the fact that the ground state of a single harmonic oscillator cannot be zero. If it were zero, x and p would simultaneously zero which is in contradiction with the Heisenberg uncertainty principle [x, p] = i. Adding up all the vacuum energies of an infinite number of oscillators will thus always lead to a divergent vacuum energy, so renormalizing the Hamiltonian is necessary to make the results finite. Finally, for the renormalized Hamiltonian, we have the energy eigenstates

Y ∗ nk Ψn = (ak) Ψ0 k where nk is non-zero only for a finite number of k. These states satisfy: X HΨn = ω(k)nkΨn k All of the above discussion was heuristic. In particular, we have not defined the Hilbert space of functionals Ψ(ϕ) precisely. To get a better understanding what this P ∗ space is, define the number operator N = k akak that acts on the energy eigenstates P as NΨn = k nkΨn. As its name suggests, the number operator counts the number of ”particles” in Ψn. The idea is to build the Hilbert space by defining the Hilbert spaces containing n particles X Hn = span{Ψn | nk = n} k and take the dirct sum of all these Hilbert spaces

∞ M F = Hn n=0 The resulting space is called the Fock space, which is exactly the space we were looking for. The next section will be dedicated to making these ideas precise.

2.2 Second quantization

In this section we will study the general formalism for quantum ﬁeld theories and many particle systems called second quantization. Let H be a Hilbert space and deﬁne

Hn := H ⊗ H... ⊗ H as the n-fold tensor product. We will always take H separable, such that there is an orthonormal basis {e }∞ of H. Then {e ⊗ ... ⊗ e }∞ is an orthonormal basis of H . n n=1 i1 in ij =1 n We will also set H0 = C

15 Definition 1. The Fock space Γ(H) over a Hilbert space H is defined as ∞ M Γ(H) = Hn n=0 P∞ So according to this definition, ψ ∈ Γ(H) means that ψ = n=0ψn where each ψn ∈ Hn. The Fock space itself is a Hilbert space with the scalar product ∞ X (ψ, φ) = (ψn, φn) n=0 The grading with n leads to the next definition: Definition 2. The Number operator N is the self-adjoint operator defined by

(Nψ)n ≡ nψn for all vectors in the domain ∞ X 2 D(N) = {ψ ∈ Γ(H)| n (ψn, ψn) < ∞} n=0 We also choose a vacuum vector:

Definition 3. The vacuum vector Ω is the vector with components Ωn = 0 for n 6= 0 and Ω0 = 1 Next, we turn to the action of operators on the Fock space. Let U : D(U) → H be a ⊗n densely defined operator on the Hilbert space. Then U acts on D(U) by Un(ψ1 ⊗ ... ⊗ ψn) = Uψ1 ⊗ ... ⊗ Uψn. U acts on the full Fock space via its second quantization: Definition 4. The second quantization Γ(U) of U is defined on the domain D(Γ(U)) ⊂ ∞ Γ(U) in Fock space F(H) by Γ(U) = ⊕n=0Un. Here D(Γ(U)) consists of vectors Ψn ∈ D(Un) with Ψn 6= 0 for finitely many n. Note that even though U is taken to be bounded, Γ(U) is not always bounded.3 How- ever, if U is unitary, Γ(U) is unitary as well (show!). Let now U(t) be a unitary group on H. Then we can write U(t) = eiHt for some d self-adjoint operator H. From this formula, we get that H = −i dt t=0U(t). Generalizing this gives d d −i Un(t) ψ1 ⊗ ... ⊗ ψn) = −i Un(t) U(t)ψ1 ⊗ ... ⊗ U(t)ψn) dt t=0 dt t=0 = Hψ1 ⊗ ψ2... ⊗ ψn + ψ1 ⊗ Hψ2 ⊗ ... ⊗ ψN + ψ1 ⊗ ... ⊗ Hψn which motivates the following Definition 5. Let A be a self-adjoint operator on H and define dΓ(A) on F(H) as ∞ dΓ(A) = ⊕n=0dΓn(A) where dΓn(A) := A ⊗ 1... ⊗ 1 + 1 ⊗ A ⊗ ... ⊗ 1 + ... + 1 ⊗ ... ⊗ A This operator has the ⊗n domain D = {ψ ∈ F(H) | ψn ∈ D and ψn 6= 0 for a finite number of n} with D(A) the domain of A. It can be proven that dΓ(A) can be extended to a self-adjoint operator on the full Fock space F(H)[35] With this definition we get that Γ(eitH ) = eitdΓ(H) (2.26) We get also the alternative definition for the number operator: N = dΓ(1) 3 P∞ 1 n 2 P∞ 1 Suppose that kUψk = c with kψk = 1 and take F(H) 3 Ψ = n=0 n ⊗i=1 ψ. Then kΨk = n=0 n2 < 2 P∞ cn ∞ whereas kUΨk = n=0 n2 = ∞ if c > 1. Hence if kUk > 1 then Γ(U) is unbounded.

16 The (anti)symmetric Fock space To be able make a distinction between bosons and fermions later, we divide the Fock space into its symmetric and antisymmetric parts:

Deﬁnition 6. The symmetrization operator P : Γ(H) → Γ(H) is deﬁned by

∞ M P = Pn n=0 where 1 X P (f ⊗ ... ⊗ f ) = f ⊗ ... ⊗ f n 1 n n! π(1) π(n) π∈Sn

Here Sn is the permutation group of n elements.

Since it is clear that P ∗ = P and P 2 = P , P is an orthogonal projection operator. Similarly, we deﬁne

Deﬁnition 7. The antisymmetrization operator A : Γ(H) → Γ(H) is deﬁned by

∞ M A = An n=0 where 1 X A (f ⊗ ... ⊗ f ) = (−1)σ(π)f ⊗ ... ⊗ f n 1 n n! π(1) π(n) π∈Sn

Here Sn is the permutation group with n elements and σ(π) is the parity of the permutation π 4

Again A is an orthogonal projection operator. Using A and P , we deﬁne:

Deﬁnition 8. The symmetric or bosonic Fock space is deﬁned as

Γ+(H) = P Γ(H) while the antisymmetric or fermionic Fock space is given by

Γ−(H) = AΓ(H)

Finally, note that dΓ maps the symmetric into the symmetric and the anitsymmetric into the antisymmetric Fock space.

Creation and annihilation operators Having deﬁned the symmetric and antisymmetric Fock spaces, we can deﬁne creation and annihilation operators:

Deﬁnition 9. For f ∈ H, the annihilation operator a(f): Hn → Hn−1 is deﬁned by √ a(f) f1 ⊗ ... ⊗ fn = n(f, f1) f2 ⊗ ... ⊗ fn while the creation operator a∗(f): Hn → Hn+1 is given by √ ∗ a (f) f1 ⊗ ... ⊗ fn = n + 1 f ⊗ f1 ⊗ f2 ⊗ ... ⊗ fn

Note that a(f) is antilinear in f. The notation of the creation and annihilation operators is justiﬁed by the following proposition:

4 Every permutation can be written as composition of transpositions π = t1 ◦ ... ◦ tk. Then the parity of the permutation is given by σ(π) = k mod 2

17 Proposition 1. a(f) and a∗(f) are bounded on Hn, more speciﬁcally: √ ka(f)kHn→Hn−1 ≤ nkfk √ ∗ ka (f)kHn→Hn+1 ≤ n + 1kfk

Moreover, a∗(f): Hn−1 → Hn is the adjoint of a(f): Hn → Hn−1

n Proof. Take a vector f1 ⊗...⊗fn ∈ H and calculate, using the Cauchy-Schwarz inequality: √ ka(f) f1 ⊗ ... ⊗ fn k = k n(f, f1)f2 ⊗ ... ⊗ fnk √ = n|(f, f1)| kf2 ⊗ ... ⊗ fnk √ ≤ nkfk kf1k kf2k...kfnk √ = nkfk kf1 ⊗ ... ⊗ fnk

This proves the ﬁrst inequality, the proof of the second inequality is similar. To prove that a∗ is the adjoint of a, we choose ψ, φ ∈ Hn−1 and f, g ∈ H. Then √ (φ, a(f)g ⊗ ψ) = n(f, g)(φ, ψ) √ √ (a∗(f)φ, g ⊗ ψ) = n(f ⊗ φ, g ⊗ ψ) = n(f, g)(φ, ψ)

So we see that a has indeed adjoint a∗.

The deﬁnitions of the creation and annihilation operators can be extended to the full Fock space, but the proofs are a bit trickier:

Proposition 2. The creation and annihilation operators a∗ and a extend to

D ⊂ Γ(H) → Γ(H) where ∞ X 2 D = {ψ | nkψnk < ∞} n=0 On this domain, we also have that a∗(f) is the adjoint of a(f)

Proof. The ﬁrst statement follows directly from the previous proposition. The proof of the second statement requires some more functional analysis and is given in the appendix

Note that the annihilation operator a maps the symmmetric into the symmetric Fock space and the antisymmetric into the antisymmetric Fock space. The creation operator ∗ ∗ ∗ ∗ ∗ does not, so we deﬁne a+ = P a (f) and a− = Aa (f). We write a±, a±, where the plus signs correspond to the bosonic and the minus signs to the fermionic creation and annihilation operators.

Canonical commutation relations The creation and annihilation operators deﬁned above satisfy the familiar commutation relations for bosons and fermions:

∗ Proposition 3. On the bosonic Fock space F+, the operators a, a satisfy the canonical commutation relations:

∗ [a+(g), a+(f)] = (g, f)I ∗ ∗ [a+(g), a+(f)] = 0 = [a+(g), a+(f)]

On the fermionic Fock space, the same relations hold if the commutator [·, ·] is replaced by the anticommutator {·, ·}

18 With these deﬁnitions, we see that all the examples considered before are bosonic systems.

Proof. Fix a vector f1 ⊗ ... ⊗ fn ∈ Hn and calculate

∗ ∗ a+(g)a (f)P+(f1 ⊗ ... ⊗ fn) = a+(g)P+a (f)P+(f1 ⊗ ... ⊗ fn) + √ X n + 1 = a (g)P f ⊗ f ⊗ ... ⊗ f + + n! π(1) π(n) π∈Sn √ = n + 1 a+(g)P+(f1 ⊗ ... ⊗ fn ⊗ fn+1) X 1 = (n + 1) (g, f )(f ⊗ ... ⊗ f ) (n + 1)! π(1) π(2) π(n+1) π∈Sn+1 n+1 X ˆ = (g, fk)P+(f1 ⊗ ... ⊗ fk ⊗ ... ⊗ fn+1) k=1

Here we have deﬁned fn+1 ≡ f and by writing fˆ we mean that f is omitted. On the other hand,

√ n X a∗ (f)a (g)P (f ⊗ ... ⊗ f ) = a∗ (f) (g, f )(f ⊗ ... ⊗ f ) + + + 1 n + n! π(1) π(2) π(n) π∈Sn n 1 X = a∗ (f) (g, f )P f ⊗ ... ⊗ fˆ ⊗ ... ⊗ f + n k + 1 k n k=1 n X ˆ = (g, fk)P+(f1 ⊗ ... ⊗ fk ⊗ ... ⊗ fn) k=1

Subtracting these expressions yields the desired result. The proof for the fermionic case is a similar (but a bit more diﬃcult, because one has to keep track of all the minus signs) and left as an excercise.

∗ From this proposition, it follows that the antisymmetric ladder operators a−, a− are bounded:

∗ 2 ∗ ∗ ka−(f)ψk = a−(f)ψ, a−(f)ψ ∗ = ψ, a−(f)a−(f)ψ 2 2 2 = −ka−(f)ψk + kfk kψk ≤ kfk2kψ2k

Note the use of the proposition in the third line. The fact that ”fermions are bounded” is a consequence of the Pauli exclusion principle: no two particles can occupy exactly the ∗ ∗ same quantum state, or in our language a−(f)a−(f) = 0. Of course this reasoning is only valid for fermions and it breaks down for bosons: bosons do not obey the Pauli principle and are unbounded. Finally, we see that the vacuum Ω is a cyclic vector:

∗ ∗ Γ(H) = span{a (f1)...a (fn)Ω | n ∈ N, fi ∈ H}

2.2.1 Examples Let us now apply this abstract procedure for some speciﬁc Hilbert spaces. First, we will recover two of the examples introduced above, next we will use the procedure to construct a new quantum ﬁeld theory.

19 ∞ Harmonic oscillator Take H = C. Then Hn = C for all n, so F(H) = ⊕n=1C. Next, deﬁne ladder operators by a∗ = a∗(1), a = a(1) so that we get [a, a∗] = 1. Given the ∗n vacuum vector Ω = 1, we can recover the entire Fock space: Hn = {λ a Ω | λ ∈ C}. Finally, if ω ∈ R acts on the Hilbert space by multiplication, dΓ(ω)n = nω. It is easy to see that this is equivalent to dΓ(ω) = ωa∗a, which is just the (renormalized) Hamiltonian of the harmonic oscillator as discussed earlier.

2 ∼ 2 ˆ ˆ Phonons Take H = ` (ΛN ) = ` (BN ) 3 f = {f(k)}k∈BN . As before, we have the scalar product: X 1 X (f, g) = f(x)g(x) = fˆ(k)ˆg(k) N d x∈ΛN k∈BN and ladder operators

1 X 1 X a∗(f) = a∗fˆ(k) a(f) = a fˆ(k) N d k N d k k∈BN k∈BN

∗ ∗ ikx where ak = a (ek), ak = a(ek) and ek(x) = e , so

∗ X −ikx ilx d [ak, al ] = e e = N δk,l. x

When we ﬁnally deﬁne H = dΓ(ω) where (ωf)ˆ(k) = ω(k)fˆ(k) and

d 2 X 2 1 2 ω(k) = 4 sin 2 kα + m α=1 we recover the phonon theory described earlier. Note that this also works in the in- ﬁnite volume limit: we take H = `2( d) ∼ L2([−π, π]d), (f, g) = P f(x)g(x) = Z = x∈Z R ˆ dk ∗ d [−π,π]d f(k)ˆg(k) (2π)d and [ak, al ] = (2π) δ(k − l).

2.2.2 Free scalar field Now we can finally introduce a relativistic quantum field theory: the free scalar field. We 2 d take H as the Hilbert space L (R , dx) where d is the dimension (the physical dimension is d = 3) and dx is Lebesgue measure. Define Fourier transform fˆ so that Z ˆ ip·x dp f(x) = f(p)e (2π)d and the scalar product is given by Z Z ˆ dp (f, g) = f(x)g(x)dx = f(p)ˆg(p) (2π)d .

By deﬁnition, he ladder operators satisfy [a(f), a∗(f)] = (f, g) and as in the case of phonons ∗ ∗ ip·x we set ap := a (ep), ap = a(ep) and ep(x) = e so that Z Z ˆ dp ∗ ∗ ˆ dp a(f) = ap f(p) (2π)d , a (f) = ap f(p) (2π)d

(recall that a(f) is antilinear in f) and

∗ d [ap, aq] = (2π) δ(p − q).

20 Definition 10. The free quantum field is defined by: 1 ϕ(f) = √ a(ω−1/2f¯) + a∗(ω−1/2f) 2 where the operator ω−1/2 is the Fourier multiplier (ω−1/2f)ˆ(p) = ω(p)−1/2fˆ(p) and p ω(p) = p2 + m2. −1/2 ˆ 2 −1/2 d This is well-defined on Hn if ω f ∈ L i.e. if f ∈ H (R ) More concretely, we have ϕ(f) = R ϕ(x)f(x)dx with (Homework: check the complex conjugations!) Z 1 ip·x −ip·x ∗ ϕ(x) = [e ap + e a ]dp. (2.27) p2ω(p) p We define the Hamiltonian H to be: H = dΓ(ω) As in the other QFT’s we encountered, the Hamiltonian can be rewritten in terms of the ladder operators: Excercise 6. Let A : H → H and take f ∈ H. Show that [a(f), dΓ(A)] = a(A∗f) [a∗(f), dΓ(A)] = −a(Af) Taking A = ω in the above excercise gives [a(f),H] = a(ωf) and [a∗(f),H] = −a∗(ωf) so that ∗ ∗ [ap,H] = ω(p)ap, [ap,H] = −ω(p)ap. Combining these results, we find that formally Z ∗ H = ω(p)apapdp (2.28)

Dynamics Next, we would like to find the dynamics of the system. We begin with the dynamics of the ladder operators: itH −itH itω −itω ap(t) ≡ e ape = Γ(e )apΓ(e ) where we used equation (2.26). Let this act on a vector Ψ in the Fock space and focus on one Ψn = ψ1 ⊗ ... ⊗ ψn. Then we get: itω √ −itω −itω −itω ap(t)Ψn =Γ(e ) n(ep, e ψ1) e ψ2 ⊗ ... ⊗ e ψn √ −itω = n(ep, e ψ1) ψ2 ⊗ ... ⊗ ψn −itω(p) =e apΨn −itω(p) =e apΨn So after a similar calculation for a∗ we find: −itω(p) ∗ itω(p) ap(t) = e ap, ap(t) = e ap. Plugging this into (2.27), we find the time evolution for the field: Z 1 ipx −ipx ∗ ϕ(x, t) = [e ap + e a ]dp p2ω(p) p where we introduced the standard notations for vectors in d + 1-dimensional Minkowski 5 space : p = (p0, p), x = (x0, x), x0 = t, p0 = ω(p) with the Minkowski inner product p 2 2 2 2 ipx px := p·x−x0p0. Since ω(p) = p + m , we see that ∂t −∆+m e = 0 so formally, the field satisfies the Klein-Gordon equation: 2 2 ∂t − ∆ + m ϕ(x, t) = 0. (2.29) 5We use the mostly minus convention for the metric

21 Wightman functions Deﬁne for a nice test function f: ϕ(f, t) := R ϕ(x, t)f(x)dx and compute the Wightman function: Z (Ω, ϕ(f, t)ϕ(g, s)Ω) = e−iω(p)teiω(q)sΩ, a(p)a∗(q)Ω fˆ(p)g ˆ(p) d√p d√q (2π)d ω(p) (2π)d ω(q) Z −iω(q)(t−s) ˆ dp = e f(p)gˆ(p) (2π)dω(p) Z = G(t − s, x − y)f(x)g(y)dxdy

Here we deﬁned the Green’s function G to be6 Z Z −iω(p)t ip·x dp ipx 2 2 dp G(t, x) = e e (2π)dω(p) = 2 e δ(p + m )θ(p0) (2π)d+1

2 2 2 This last line is written in Minkowski space notation, so p = ~p − p0 and we defined dp = dp0dp. Here the step function restricts to positive energies, while the delta function corresponds to the condition that physical particles should be on-shell: p2 = −m2. Furthermore, we can show that the scalar free field is consistent with special relativity, in particular with causality and Lorentz-invariance: Excercise 7. Show that G(t, x) = g(t2 − x2) where g is smooth, except at the origin. How does g blow up here? These points where t2 = x2 are called the light-cone. This implies that the scalar field theory is Lorentz invariant, since the Lorentz transformations are precisely those linear maps (t, x) → (t0, x0) that leave the proper distance t2 − x2 invariant. Excercise 8. Show that [ϕ(x), ϕ(y)] = 0 if (x−y)2 > 0, this is when x and y are spacelike separated. This implies causality: there is no communication between events with spacelike sep- aration. Finally, the Wick formula holds also for free scalar fields: X Y (Ω, ϕ(x1)...ϕ(xn)Ω) = G(xi − xj)

p∈Pn (i,j)∈p

2.3 Interacting Fields

All examples of QFT given before satisfy free-field dynamics: the motion of the particles follows non-interacting dynamics. For example, for the free scalar field, the one-particle state a∗(f)Ω evolves as e−itH a∗(f)Ω = a∗(e−itωf)Ω 2 d −itω(p) so a wave function fˆ ∈ L (R ) evolves after a time t to e fˆ(p). Similarly, an n- particle state evolves as n n Y ∗ Y ∗ −itω a (fi)Ω → a (e fi)Ω i=1 i=1 This means that all the particles move indeed independently. Of course, in nature particles do interact, so a quantum field theory describing real physics should be able to account for these interactions. This means that we have to go back and try to adapt our theory in order to include interacting particles. Recall the Hamiltonian (2.28). If we define the momentum of the field as i π(f) = √ a∗(ωf) − a(ωf) 2

6 P 1 0 For the last line, use δ(g(x)) = 0 δ(x − xi) where xi is a zero of g(x) and g (xi) 6= 0 i |g (xi)|

22 we can rewrite the Hamiltonian in terms of the ﬁelds and the momenta as Z 1 2 2 2 2 H0 = 2 : (∇ϕ(x)) + m ϕ(x) + π(x) ] : dx (2.30) Rd where : : means normal ordering as deﬁned by:

Y # Y ∗ Y : a (fi) := a (fi) a(fj)

i∈I i∈I1 j∈I2

# ∗ where a = a or a and I = I1 ∪ I2. In words: the normal ordering of a product of ladder operators is the product of the same ladder operators, where we put the creation operators ﬁrst and the annihilation operators last. For example, : aa∗ := a∗a. The Klein-Gordon equation (2.29) is just the classical Hamilton equation of this Hamiltonian. The formulation of the free ﬁeld in terms of this Hamiltonian suggests a way to look for new theories that include interacting particles: just add non-linear terms to the Hamil- tonian (2.30) and work out the consequences. Since we want the resulting Hamiltonian to be bounded from below7, the simplest term to add is Z 1 4 Hint = 4 λ ϕ(x) dx Rd where λ denotes the strength of the interaction (which we will take to be small later) and 1 the factor 4 is a convenient normalization. This interaction term gives rise to the new classical equation of motion:

ϕ¨ − ∆ϕ + m2ϕ + λϕ3 = 0

When we want to quantize the system, we ﬁrst need to ask if the new Hamiltian Z 1 4 H = H0 + 4 λ ϕ(x) dx Rd is also well deﬁned in the quantum case. Since we already encoutered infrared and ultraviolet divergences, we anticipate possible problems and regularize the Hamiltonian.

∗ Ultraviolet regularization As before, we have that ϕ(x) = a(fx) + a (fx) where 1 ip·x 2 fˆx(p) = √ e . However, fˆ ∈ / L so ϕ(x) is potentially singular. To regularize it, 2ω(p) ∗ 1 ip·x −(p)2 introduce some > 0 and deﬁne ϕ = a(f,x) + a (f,x) with fˆ,x(p) = √ e e . 2ω(p) Alternatively, we could have taken any function η(p) that satisﬁes η(0) = 1 and decays 4 n rapidly as |p| → ∞. Then ϕ(x) is a bounded operator on H . Later we will be interested in the limit → 0.

Infrared regularization To avoid infrared divergences, turn the interaction oﬀ for large |x|: let BR be a ball of radius R around the origin and set Z 0 4 h,R = λ ϕ(x) dx BR Later we will try to take the limit R → ∞ To see if our new Hamiltonian is well-deﬁned, we try to calculate (Ω, h,RΩ): by Wick’s theorem, we get that

Z 1 2 2 (Ω, ϕ (x)4Ω) = 3!(Ω, ϕ (x)2Ω)2 = 3! e−2 p dp2 2ω(p)

7This way the energies can always taken to be positive

23 But when we try to take the limit → 0, we see that R √ 1 e−22p2 dp ∼ 1−d if d > 1 2 p2+m2 R 1 −22p2 −1 and √ e dp ∼→0 log( ) if d = 1. The vacuum expectation value of the 2 p2+m2 new Hamiltonian diverges in all dimensions! This problem can easily be solved by taking the normal ordering of the interaction Hamiltonian: Z 4 h,R ≡ : ϕ(x) : dx BR

4 4 ∗ 3 ∗2 2 ∗3 ∗4 Here : ϕ : = a +4a a +6a a +4a a+a , so it follows directly that now (Ω, h,RΩ) = 0. We ﬁnally get a ﬁnite result: the vacuum expectation value of the Hamiltonian is zero as expected. 2 There is, however, a much bigger problem: kh,RΩk will diverge in all dimensions d > 1 if we take the limit → 0: Z 2 4 4 kh,RΩk = dxdy(Ω, : ϕ(x) :: ϕ(y) : Ω) BR×BR Z 4 = 4! dxdy(Ω, ϕ(x)ϕ(y)Ω) BR×BR Qn Excercise 9. Prove the second equality above. Can you guess a formula for (Ω, i=1 : 4 ϕ(xi) : Ω)?

R iω(p)(x−y) 1 −22p2 1−d Now, Ω, ϕ(x)ϕ(y)Ω = e 2ω(p) e dp ∼ |x−y| as x → y and d > 1 so the limit → 0 diverges in all dimensions bigger than one. However, in one dimension, the integral above behaves as log |x − y| for x → y, so in one dimension the limit

2 lim kh,RΩk →0

2 exists. Note that the infrared limit limR→∞ kh,RΩk is still divergent: (Ω, ϕ(x)ϕ(y)Ω) depends only on x − y so it is translation invariant. Integrating over an inﬁnite (even one-dimensional) volume will thus always give an inﬁnite quantity. This is as expected: energies should be extensive (proportional to the volume).

Proposition 4. For d = 1, lim→0 h,R is a bounded operator on Hn. Thus it is densely deﬁned in Γ(H).

Proposition 5. hR = lim→0 h,R is self-adjoint on Γ(H) if d = 1. Moreover, it is bounded from below.

Proof. Proven by Nelson in ’66 [?]

For d = 2 we need to add (divergent) counterterms to the interaction Hamiltonian: let Z Z ren 2 h,R ≡ h,R + a(λ) : ϕ(x) : dx + bepsilon(λ) dx BR BR

2 2 −1 3 REN with a(λ) = λ a log and b(λ) = λ b + λ c log . Then h,R is bounded below uniformly as → 0. However, its limit for → 0 does not exist on the Fock space, but has to be defined on a different Hilbert space (see later) For d > 2, no nontrivial limit is expected, as we will explain later. To better understand the behaviour of the interacting scalar field, we will use a different description as given by Euclidean QFT.

24 Chapter 3

Euclidean Quantum Field Theory

To solve the problems encountered in the previous chapter, we will consider a diﬀerent viewpoint of quantum mechanics, due to Richard Feynman. His path integral formalism starts with the idea of ”sum over histories”: the quantum mechanical transition amplitude for a particle moving from point x to point y is a weighted sum of all paths from x to y. Consider for example a free (non-relativistic) quantum mechanical particle (with mass d 1) moving in R . The particle evolves according to the Schr¨odingerequation:

1 i∂tψ(t, x) = − 2 ∆ψ(t, x) = H0ψ(x, t)

−itH The time-evolved state is thus given by ψ(t, x) = e 0 ψ0 (x), where ψ0 is the initial state. For this Hamiltonian, e−itH0 has an integral kernel: Z −itH0 e ψ0 (x) = K0(t, x, y)ψ0(y)dy Rd

2 −d/2 i (x−y) where K0(t, x, y) = (2πit) e 2t Now, divide the time interval t in n equal pieces and −itH −i t H n use e 0 = e n 0 to get Z (i) (f) t t t K0(t, x , x ) = K0( n , x0, x1)K0( n , x1, x2)...K0( n , xn−1, xn)dx1...dxn Z −1 iS(x0,x1,...,xn,t) = Zn e dx1...dxn

(i) (f) Here we defined initial and final positions x0 = x and xn = x , the normalization factor 2 dn/2 Pn 1 (xi−1−xi) Zn = (2πit/n) and S(x0, x1, ..., xn, t) = i=1 2 t/n . So we wrote the evolution of the particle from a fixed initial to final position as the evolution via n − 1 intermediate points xi, we integrated over all the possible intermediate positions and gave each path a certain weight S. This makes the idea of ”sum over histories” exact. Let us clean up the jt t notation: tj ≡ n , x(tj) ≡ xj, ∆x(tj) ≡ x(tj) − x(tj−1) and ∆t = n . With this notation 2 1 Pn ∆x(ti) the weight S becomes S = 2 i=1 ∆t ∆t. Now it is easy to take the formal limit n → ∞, giving Z (i) (t) iS(x) K0(t, x , x ) = e Dx x(0)=x(i)x(t)=x(f) −d/2 Qn 2πit Here the measure is Dx = limn→∞ i=1 n dx(ti) and the weight function S(x) = 1 R t 2 2 0 x˙ (s)ds is the action of the free particle. The classical motion of a particle is the trajectory that extremizes the action: it follows a path x(s) that satisfies x(0) = x(i), (f) d x(t) = x and d S(x + y) = 0 for all admissible paths y (y(0) = 0 = y(t)). In this case, the path has to satisfyx ¨ = 0: the classical free motion is a straight line. The quantum mechanical motion, however, takes into account all possible paths x(i) → x(f).

25 Unfortunately, altough this is conceptually quite nice, the ”measure” eiS(x)Dx giving the appropriate weight to each path does not make any useful mathematical sense. How- ever, if we consider the semigroup e−tH instead of e−itH , we end up with a honest measure. This change corresponds to working in ”imaginary time” by replacing t → −it. Note that this replacement changes the Minkowski space we were working in to the Euclidean space, hence the name Euclidean QFT.1

3.1 Wiener Measure

From now on we will work in imaginary time, and try to make sense of the measure given (x−y)2 −d/2 − −tH0 1 above. Let T0(t, x, y) = (2πt) e 2t be the kernel of e for H0 = − 2 ∆: Z −tH0 e ψ (x) = T0(t, x, y)ψ(y)dy

As above, we then get for all n:

2 n Z n (i) (f) −nd/2 − 1 P ∆x ∆x Y T0(t, x , x ) = (2πt/n) e 2 i=1 ∆t dx(ti) i=1

i f d i f Now, let C(t, x , x ) = {x : [0, t] → R | x(0) = x , x(t) = x , x is continuous }, the set of all continuous paths from a given initial to final position in a fixed time t. Given this set, we define:

i f • A cylinder set in C(t, x , x ) is a set of the form S(t1, ..., tn,I1, ...In) = {x|x(ti) ∈ d Ii} where n ∈ N, ti ∈ (0, t) and Ii is a Borel set in R • The cylinder σ-algebra is the smallest σ-algebra in C(t, xi, xf ) containing all cylinder sets C.

We get the following measure on the cylinder sets:

Z Z n+1 xi,xf xi,xf Y µt : C → R : µt (S({ti},Ii)) = dx1... dxn T0(ti − ti−1, xi−1, xi) I1 In i=1

i f where 0 = t0 < t1 < ...tn < tn+1 = t and x0 = x , xn+1 = x . Since Z T (ti+1 − ti, xi, xi+1)T (ti − ti−1, xi−1, xi)dxi = T (ti+1 − ti−1, xi−1, xi+1) Rd we see that

x,y d µt (S(t1, ..., tn + 1,I1, ..., Ii−1,R ,Ii+1, ..., In−1)) x,y = µt (S(t1, ..., ti−1, ti+1, ..., tn+1,I1, ..., Ii−1,Ii+1, ..., In−1))

So these measures are consistent. x,y Fact: µt extends to a measure on the cylinder σ-algebra and the Borel σ-algebra of C(t, x, y). This is called the conditional Wiener measure on C(t, x, y). d We get a nice formula: let Ai : R → R be bounded ant take t1 < t2 < ... < tn. Then

Z Z n+1 −t1H0 −(t2−t1)H0 −(t−tn)H0 Y x,y e A1e A2...Ane ψ (x) = Ai(x(ti))dµt ψ(y)dy (3.1) d R i=1 1 In particular, this substitution changes the Minkowski scalar product xy = −x0y0 + x · y to the Euclidean one x0y0 + x · y

26 3.2 Feynman-Kac formula

In the discussion before we only treated free particles. Now we add some potential to the 1 Hamiltonian: H = − 2 ∆+V for V (x) continuous and bounded from below. Then we have the Proposition 6. (Feynman-Kac formula) The integral kernel of e−tH is given by Z −tH i f − R t V (x(s))ds xi,xf e (x , x ) = e 0 dµt (x).

Remark. Heuristically, we can write this as Z e−tH (xi, xf ) = e−SE (x)DX (3.2) x(0)=xi,x(t)=xf

R t 1 2 for the Euclidean action functional SE(x) = 0 [ 2 x˙ + V (x(s))]ds. The classical action R t 1 2 functional to our problem is S(x) = 0 [ 2 x˙ −V (x(s))]ds so we see that by formally making the change of variables t = iτ we obtain Feynman’s path integral Z e−iτH (xi, xf ) = eiSE (x)Dx (3.3) x(0)=xi,x(t)=xf . One way to actually prove the Proposition is to use the Trotter product formula: Proposition 7. Let A, B be self-adjoint operators in a (separable) Hilbert space and bounded from below. Take also A + B to be self-adjoint on D(A) ∩ D(B). Then

−t(A+B) − t A − t Bn e = lim e n e n n→∞ where the limit is in the strong topology. Excercise 10. Prove the above proposition for bounded operators. Using this proposition, we get:

−tH − t H − t V n (ϕ, e ψ) = lim (φ, (e n 0 e n ) ψ) n→∞ The last scalar product can be rewritten using (3.1) to

ZZZ n+1 Y − t V x( it )) xixf i f i f e n n dµt ϕ¯(x )ψ(x )dx dx i=1 Since V was taken to be continuous, for each continuous path x:

n+1 X t it Z t V x → V (x(s))ds as n → ∞ n n i=1 0

Qn+1 − t V We also took V to be bounded from below, so there is a constant C such that i=1 e n ≤ C for all n. Hence, by the dominated convergence theorem

n+1 Z Z t Y − t V x( it ) xixf − R V (x(s))ds xixf i f e n n dµt →n→∞ e 0 dµt ≡ R(t, x , x ) i=1

Finally, since R(t, xi, xf ) ≤ C R dµ = Ce−tH0 (xi, xf ): Z (ϕ, e−tH ψ) = φ¯(xi)ψ(xf )R(t, xi, xf )dxidxf from which the formula (3.3) follows.

27 3.3 Harmonic oscillator - Ornstein-Uhlenbeck process

Let us now apply the theory from the previous section for a speciﬁc potential: take d2 H = − 1 + 1 ω2q2 − 1 ω = ωa∗a 2 dq2 2 2

This is our beloved (renormalized) harmonic oscillator in one dimension. Let Ψ0 be the harmonic oscillator ground state2 and deﬁne the measure ν on C[−T,T ] = {q :[−T,T ] → R | q continuous} by

Z Z Z Z 0 R T 1 2 2 1 0 0 y,y − T ( 2 ω q(s) − 2 ωds) F (q)dν(q) = dy dy Ψ0(y)Ψ0(y ) dµ2T (q)e F (q) R R Note that instead of considering intervals [0, t] as above, we consider symmetric time intervals [−T,T ] and we have written y = q(−T ), y0 = q(T ). From (3.1), we get that for −T ≤ t1 < t2 < ... < tn ≤ T : Z n Y −(t1+T )H −(t2−t1)H −(T −tn)H Ai(q(ti))dν(q) = (Ψ0, e A1e A2...Ane Ψ0) i=1 −t1H −(t2−t1)H = (Ψ0, e A1e A2...AnΨ0)

In the last line, we used the deﬁnition of the vacuum: HΨ0 = 0. Clearly, this result does not depend on T , so we can think about ν as a measure on continuous paths R → R. For the harmonic oscillator, we had the Wick formula (2.13), which holds as well if we replace e−itH by e−tH so it is also true in imaginary time. Setting q(t) = etH qe−tH , Wick’s formula gives that

−(t2−t1)H −(tn−tn−1)H (Ψ0, qe q...qe qΨ0) = (Ψ0, q(t1)...q(tn)Ψ0) X Y = G(i(ti − tj))

p∈Pn (i,j)∈p

1 −ωt where we recall that t1 < t2 < . . . and G(it) = 2ω e We can conclude that 2m Z Y X Y q(ti)dν(q) = G(i(ti − tj))

i=1 p∈Pn (i,j)∈p

Since the left-hand side is independent of the ordering of the times ti, our ﬁnal result is: 2m Z Y X Y q(ti)dν(q) = C(ti − tj)

i=1 p∈Pn (i,j)∈p where 1 C(t) = e−ω|t|. 2ω This tells us that ν is a Gaussian measure with covariance C as we will explain in the next chapter. Next, we try to generalize this to a more general Hamiltonian of the type

H = H1 + V where we have renamed the harmonic oscillator Hamiltonian used above to H1. For convenience, choose V to be bounded from below by some V0. We could for example take V = λq4, giving the anharmonic oscillator Hamiltonian. We get: n Z R T −(t1+T )H −(t2−t1)H −(T −tn)H Y − V (q(s))ds (Ψ0, e A1(q)e ...An(q)e Ψ0) = Ai(q(ti))e −T dν(q) i=1

1 2 2 −1/2 − 2 ωq As before: Ψ0 = (πω) e

28 Unlike before, this result does depend on T since the vacuum state Ψ0 is the zero eigenvalue eigenstate of H1 but not of the full Hamiltonian H. If we want to take the limit T → ∞, we have to be more careful than before, but under quite general conditions3, we have that:

• H has a unique ground state Ω with HΩ = E0Ω, the other eigenstates of H have eigenvalues E > E0

• Ω is strictly positive: Ω(q) > 0 ∀q

−t(H−E ) Then, by the spectral theorem, e 0 Ψ0 = PΩ + Rt where PΩ is the orthogonal −δt projector to Ω: PΩψ = (Ω, ψ)Ω and Rt is a remainder with kRtk ≤ e for some δ > 0. Taking the time to inﬁnity, we get:

−t(H−E0) e Ψ0 →t→∞ (Ω, Ψ0)Ω (3.4)

If we now deﬁne a measure ρT by

exp − R T V (q(s))dsdν(q) dρ (q) = −T T R R T exp − −T V (q(s))dq dν(q) we get

n Z −(t1+T )H −(t2−t1)H −(T −tn)H Y (e Ψ0,A1(q)e ...An(q)e Ψ0) A (q(t ))dρ (q) = i i T ke−TH Ψ k2 i=1 0

Now we can take T to inﬁnity using (3.4):

Z n Y −(t2−t1)H lim Ai(q(ti))dρT (q) = (Ω,A1e A2...AnΩ) T →∞ i=1

This result says that the interacting dynamics of e−tH corresponds to the measure dρ = limT →∞ dρT We will later see that the limit indeed exists, and informally ρ is given by

− R ∞ [ 1 q˙2+ 1 ω2q2+V (q)]ds dρ(q) = Z−1e −∞ 2 2 Dq where Z is an (inﬁnite) normalization constant.

3.4 The free ﬁeld

Let us ﬁrst consider the phonon theory of section 2.1.2. Since, as we have seen, this system is a ﬁnite collection of harmonic oscillators, we get a measure dνΛN (ϕ) on the path space ϕ(t, x), t ∈ R, x ∈ ΛN which is formally given by Z ∞ −1 1 X 2 2 2 2 dνΛN = Z exp − 2 (ϕ ˙ + (∇ϕ) + m ϕ )ds Dφ −∞ x∈ΛN where we have the discrete gradient: (∇µϕ)(x) = ϕ(x + eµ) − ϕ(x) with eµ the unit vector in the µ direction. Analoguosly as in the case of the simple harmonic oscillator, we can summarize this by

2n Z Y X Y ϕ(ti, xi)dνΛN = CN (ti − tj, xi − xj) i=1 p∈Pn (i,j)∈p

3For example, one can use the Perron-Frobenius theorem, which is given and proved in the appendix

29 This formula has a limit for N → ∞: CN (t, x) → C(t, x) for Z 1 −ω(k)|t|+ikx dk C(t, x) = e (2π)d [−π,π]d 2ω(k)

Finally, the free quantum ﬁeld gives formally similar formulae but with covariance:

√ Z 1 2 2 C(t, x) = √ e− k +m +ikx dk 2 2 (2π)d Rd 2 k + m Z ipy e dp = 2 2 (2π)d+1 Rd+1 p + m d+1 where in the last line we wrote y = (t, x) ∈ R . The question we want to address next, is what sort of mathematical objects are these measures on function spaces? Note, for example, that for the free ﬁeld: Z Z 2 1 ϕ(t, x) dν = C(0, 0) = 2 2 dp = ∞ Rd+1 p + m so ν will not be supported on functions! In the next chapter, we will study such measures in greater detail.

30 Chapter 4

Random Fields

While trying to construct quantum ﬁeld theories allowing for interactions between the ﬁelds, we found that it is useful to use imaginary time, and hence work in euclidean spacetime instead of Minkowski spacetime. This euclidean QFT gave rise to interpreting QFT dynamics as described by measures on possible paths, but we did not go into detail what these measures mean mathematically. That will be the aim of this chapter, where we will develop a general theory of measures on distribution spaces.

4.1 The Ginzburg-Landau model

We now consider a model that serves many purposes: 1. It will appear as a regularization to the the λφ4 (Euclidean) quantum field theory later. 2. It is a model of second order phase transitions. 3. It also describes our phonon theory viewed as classical statistical mechanics. d We consider a field φ(x) defined in a box Λ ⊂ Z i.e. φ :Λ → R. Define the energy of the field configuration by 1 X r X X H (φ) = (φ − φ )2 + φ2 + λ φ4 Λ 2 x y 2 x x {x,y}∈BΛ x∈Λ x∈Λ where BΛ is the set of nearest neighbors in Λ: |x − y| = 1. In classical statistical mechanics one considers the φ(x) as random variables with probability distribution given by the Gibbs rule, this is we consider the probability measure 1 Y e−βHΛ(φ) dφ(x) ZΛ x∈Λ |Λ| on R . The normalization (the partition function) is given by Z −βHΛ(φ) Y ZΛ = e dφx Λ R x∈Λ This integral exists if λ > 0 or λ = 0 and r > 0: ∞ |Λ| Z 2 4 Z ≤ e−β[rφ +λφ ]dφ < ∞ −∞ Remark The Ising model is a limit of this model: take r r = −2λ so φ2 + λφ4 = λ(φ2 − 1)2 − λ 2 2 x x x

q λ −λ(φ2−1)2 2 and take the limit λ → ∞ (in the sense of distributions). Since π e λ−→→∞ δ(φ −1), the only possible values for φ become ±1 so we indeed recover the Ising model.

31 4.2 Gaussian Integrals

Consider first the case λ = 0. Then our measure is Gaussian. Definition 11. Let A be a real symmetric n×n matrix which is (strictly) positive definite n (so all eigenvalues of A are > 0 or equivalently, (φ, Aφ) > 0 ∀φ ∈ R , φ 6= 0). The n −1 Gaussian measure on R with covariance A and mean 0 is the probability measure 1 −( 1 φ,Aφ) dµ(φ) = e 2 Dφ Z Q P where φ = (φ1, ··· , φn), Dφ = i dφi and (φ, Aφ) = ij φiAijφj. n Thus, each φi : R → R given by φi(φ) = φi is a random variable, with mean 0 and 2 −1 −1 variance hφi i = (A )ii. Also, (φ, f) is a random variable, with variance (f, A f). For Gaussian measures, we can calculate most interesting quantities quite easily: • First, we calculate the partition function Z − 1 (φ,Aφ) Z = e 2 Dφ. Rn The positive matrix A can be diagonalized by an orthogonal matrix S ∈ SO(n): T S AS = diag(λ1, . . . , λn). Since A is orthogonal, | det S| = 1 and we can easily change variables: φ = SΨ we have Dφ = | det S|DΨ = DΨ and thus n n Z Z ∞ 1/2 −1/2 − 1 P λ Ψ2 Y − 1 λ Ψ2 Y 2π h i Z = e 2 i i i DΨ = e 2 i i dΨ = = det(A/2π) i λ i=1 −∞ i=1 i Qk • The correlations functions h α=1 φiα i where iα ∈ {i, . . . , n} (note: several iα may be n the same) can be computed using the generating function : Let f ∈ R . Define Z 1 − 1 (φ,Aφ)+(φ,f) S(f) = he(φ,f)i = e 2 Dφ, Z Pn with (φ, f) = i=1 φifi. Then S(f) is smooth in f and k k ∂ Y Y h φiα i = S(f). ∂f f=0 α=1 α=1 iα But Z 1 − 1 (φ−A−1f,A(φ−A−1f))+ 1 (f,A−1f) S(f) = e 2 2 Dφ Z Z 1 (f,A−1f) 1 − 1 (Ψ,AΨ) = e 2 e 2 DΨ Z 1 (f,A−1f) = e 2 using the change of variables Ψ = φ − A−1f. Hence, say the 2-point function, 2 ∂ 1 (f,A−1f) −1 hφiφji = e 2 = (A )ij. ∂fi∂fj f=0 In general k ( Y 0 k odd h φi i = α Q ∂ 1 −1 m 1 ( ) 2 (f, A f) k = 2m α=1 ∂fiα f=0 m! X Y −1 = (A )iαiβ P {α,β}∈P

where P is a pairing of the set {1, 2, ··· , 2m} i.e. a partition into m sets of size 2. The sum runs through all such pairings, where it is implied that the last line gives zero if k is odd.

32 Example For n = 4 there are three pairings {{1, 2}, {3, 4}}, {{1, 3}, {2, 4}} and (2m)! 1 {{1, 4}, {2, 3}}. In general, the number of pairings is 2mm! . Finally, we arrive at the result:

2m Y X Y h φiα i = hφiα φiβ i α=1 P {α,β}∈P

A general n-point function can be written as a sum of 2-point functions. Note that we have found yet another form of the Wick formula.

4.3 The Gaussian Ginzburg-Landau model

Let us now specialize to the λ = 0 Ginzburg-Landau model with periodic boundary con- d |ΛL| ditions on ΛL = ZL. This gives rise to a gaussian measure on R with density

1 P 2 P 2 −1 − 2 ( {x,y}∈B (φx−φy) +r φx) Z e ΛL

P 2 The quadratic form (φx − φy) deﬁnes a matrix ∆per (lattice-Laplacian with {x,y}∈BΛL periodic boundary conditions):

X 2 (φx − φy) = −(φ, ∆perφ)

{x,y}∈BΛ Concretely X −(∆perφ)x = (φx − φx+u) |u|=1 where the addition is modulo L. For example in d = 1 we have (−∆φ)x = 2φx − φx−1 − d φx+1. This Laplacian is also deﬁned as an operator ∆ in inﬁnite volume i.e. for φ ∈ RZ . 2 d d Then ∆ is self adjoint on ` (Z ) ⊂ RZ

2 d d X 2 ` (Z ) = {φ : Z → C | |φx| < ∞}. x∈Zd P Indeed, ∆ is bounded: k∆k ≤ 2d and in the scalar product (φ, ψ) = d φ¯ ψ we have x∈Z x x (φ, ∆ψ) = (∆φ, ψ).

Excercise 11. Compute the spectrum of ∆, and show as a consequence that ∆ is bounded.

d Z per The periodic Laplacian can be deﬁned in R as well. For this consider the set FL ⊂ d RZ of periodic functions : [we use φx or φ(x) below !]

per d φ ∈ FL : φ(x + Ln) = φ(x) ∀n ∈ Z .

per ΛL Clearly FL can be identiﬁed with R , namely φ is determined by φ|ΛL ,ΛL = {x| |xi ∈ per per {0,...,L−1}. Clearly ∆ : FL → FL so the periodic Laplacian ∆per is the corresponding Λ matrix in R L . d We will now diagonalize ∆per and ∆ by Fourier series. Consider p ∈ [−π, π] and d Z ipx P φp ∈ R φp(x) = e (where px := α pαxα) . Then,

X ip(x+u) ipx X ipu ipx ∆φp(x) = (e − e ) = (e − 1)e = −µ(p)φp(x) |u|=1 |u|=1

1There are (2m)! ways to order m numbers. If you pair the ﬁrst two number, then the next two etc., this gives you (2m)! diﬀerent ways of pairing. However, in each pair, the relative ordering is not important, giving the factor of 2m in the denumerator and the general ordering of the m pairs does not matter iether, giving the factor m!

33 where d X µ(p) = 2 (1 − cos pµ) µ=1 Let 2π B := {p ∈ d | p = n , n ∈ , |n | ≤ L/2}. L R i L i i Z i Then {φp| p ∈ BL} per d is a basis for FL (they are independent and there are L of them). We have

X ip(x−y) d e = L δxy for x, y ∈ ΛL

p∈BL and X i(p−q)x d e = L δpq p, q ∈ BL

x∈ΛL −d/2 so {L φp}p∈BL is an orthonormal basis of eigenvectors of −∆per so we conclude that X 1 (−∆ + r)−1 = L−d eip(x−y) ≡ G (x − y) per xy µ(p) + r L p∈BL

Thus, the correlations of the λ = 0 GL-model are given in terms of the (periodic b.c.) Green’s function GL. 2π d How about L → ∞ ? Before, we had a Riemann sum over cells of size ( L ) . So, in the limit, Z ddp eip(x−y) lim GL(x − y) = d ≡ G(x − y) L→∞ [−π,π]d (2π) µ(p) + r −1 2 d This equals of course the kernel of the operator (−∆ + r) deﬁned on ` (Z ). Indeed, on 2 d d ` (Z ) the operator −∆ has spectrum {µ(p) | p ∈ [−π, π] } and we can diagonalize it by 2 d Fourier series. If f ∈ ` (Z ), let X fˆ(p) = e−ipxf(x) x∈Zd

Then fˆ ∈ L2(B), B = [−π, π]d and

Z d X d p ˆ (f, g) ≡ f(x)g(x) = d f(p)ˆg(p) B (2π) x∈Zd and we have the inverse formula Z d ipx ˆ d p f(x) = e f(p) d B (2π) Thus, Z 1 ddp X ((−∆ + r)−1f)(x) = eipx fˆ(p) = G(x − y)f(y) µ(p) + r (2π)d y −1 so (−∆ + r)xy = G(x − y). We have obtained

Λ −1 Proposition 8. Let µL be the Gaussian measure on R L , with covariance (−∆per + r) . Then Z Y X Y φ(xα) dµL(φ) → G(xα − xβ) as L → ∞ (4.1) P hαβi

34 4.4 Measures on s0

d Can we write the RHS of (4.1) in terms of a measure on RZ ? Let us ﬁrst see how n probability measures in R , n < ∞ can be characterized by their Fourier transforms.

4.4.1 Bochner’s Theorem

n Let µ be a probability (Borel) measure on R . Recall that we deﬁned the generating function of µ as S(f) = R e(φ,f)dµ(φ) if e(φ,f) ∈ L1(µ). It is actually more convenient to n consider S(if), f ∈ R , the characteristic function or the the Fourier transform of µ: Z i(φ,f) n W (f) = e dµ(φ) f ∈ R . which always exists by the dominated convergence theorem. The characteristic function W (f) has some obvious and less obvious properties:

•| W (f)| ≤ R 1dµ = 1.

• W (0) = 1.

• W is continuous. Indeed, by the dominated convergence theorem, if fn → f then, i(φ,fn) since |e | ≤ 1,W (fn) → W (f).

n • Let zα ∈ C, fα ∈ R , α = 1, ..., N. Then, Z Z X X i(φ,fα) −i(φ,fβ ) X i(φ,fα) 2 zαz¯βW (fα − fβ) = zαz¯βe e dµ = | αzαe | dµ ≥ 0 α,β α,β

This motivates the following deﬁnition:

n Deﬁnition 12. W : R → C is a function of positive type if conditions a) - d) from above hold.

Theorem 2. (Bochner’s theorem) W (f) is of positive type if and only if there is a n R i(φ,f) Borel probability measure on R such that W (f) = e dµ(φ). We just proved the ﬁrst half of the following theorem:

Proof. ”⇐” is done above. ”⇒”: To prove this, one uses an idea that is useful in other contexts, namely we use W to construct a scalar product. Let

n H = {ψ : R → C | ψ(x) 6= 0 only for ﬁnitely many x}

For φ, ψ ∈ H put X (φ, ψ) = φ¯(x)ψ(y)W (x − y) x,y∈Rn This sum is cleraly well-defined since it is finite. ( , ) satisfies all the requirements for an inner product, except that (φ, φ) = 0 does not imply φ = 0. Thus define H0 = {φ | (φ, φ) = 0}, which is a subspace of H. Put H˜ = H/H0. This is the space of equivalence classes [φ] = [φ + φ0] where φ0 ∈ H0. Then (H˜, (·, ·)) is an inner product space which can be completed into a Hilbert space. n Let for t ∈ R the ”‘ time translation”’ operator Ut : H → H:(Utψ)(x) = ψ(x + t). Clearly, Ut : H0 → H0 so one defines U˜t : H˜ → H˜ by U˜t[φ] = [Utφ]. We then get (U˜tφ, U˜tψ) = (φ, ψ), so U˜t is an isometry. It is also a semigroup since U˜t+s = U˜tU˜s and

35 U˜0 = 1|. Finally, t → U˜t is strongly continuous: ∀ψ ∈ H˜, t → U˜tψ ∈ H˜ is continuous from n R → H˜. Strong continuity follows because W is continuous:

2 kU˜tψ − U˜sψk = 2(ψ, ψ) − (U˜tψ, U˜sψ) − (U˜sψ, U˜tψ)

= 2(ψ, ψ) − (ψ, U˜t−sψ) − (ψ, U˜s−tψ)

This equals X ψ¯(x)ψ(y)2W (x − y) − W (x − y + t − s) − W (x − y + s − t) x,y which tends to 0 as t → s. We have now veriﬁed the assumptions of the following theorem, from which the statement follows.

n Theorem 3. Stone’s Theorem Let R 3 t → Ut unitary in Hilbert space be strongly continuous and UtUs = Ut+s, ∀s, t, U0 = 1|. Then for each φ ∈ H, kφk = 1, there exists a n R i(λ,t) probability Borel measure µφ on R such that (φ, Utφ) = e dµφ(λ). Proof. See Reed-Simon [35], vol. 1, Th. VIII.12. The heuristics behind the proof is that, if H were ﬁnite dimensional, we could easily prove that Ut is diﬀerentiable in t and that ∂U i P t A t = iA for A Hermitian with [A ,A ] = 0 and U = e j j j . Diagonalize all A ∂tj t=0 j j i j t j simultaneously gives  (j)  λ1 0  ..  Aj =  .  (j) 0 λn which gives the measure n ~ X ~ (i) ~ 2 dµφ(λ) = (λ − λ)|φi| i=1

To finish the proof of the first theorem, using Stone’s Theorem, take φ˜ ∈ H˜ ( 1 x = 0 φ˜ = [φ] φ(x) = 0 x 6= 0 Z X ˜ ˜ ˜ i(λ,t) W (t) = φ(x)φ(y + t)W (x − y) = (φ, Utφ) = e dµφ˜(λ) x,y∈Rn k Suppose now we have measures µk on R and want to prove that µk converges as k → ∞. The properties 1,2 and 4 of Wk usually carry to the limit so the main issue to check is 3: continuity. In our case WL however live in spaces whose dimensions tend to infinity as L → infinity. We aim at conditions 1-4 in such a setup and search for a generalization of Bochner’s theorem there.

4.4.2 Cylinder measures

Suppose we had a measure µ on some σ-algebra A of subsets of RN such that sets of the \n n \n form A ⊗ RN ∈ A where A ∈ An = some σ-algebra of subsets of R and RN denotes ∞ N ∞ ⊗i=n+1R (and R ≡ ⊗i=1R). Then we get a measure µn on An by “integrating out” the other variables : N\n µn(A) = µ(A × R ) This resulting set of measures is consistent:

m µn(A) = µn+m(A × R ) (4.2)

36 m ∞ provided A × R ∈ An+m if A ∈ An. Thus, conversely, let {µn}n=1 be a family of n measures, µn a Borel probability measure on R . We say µn are consistent if (4.2) holds n for all Borel sets A ⊂ R . A natural σ-algebra on RN is the σ-algebra generated by the n n cylinder sets. Denote by B Borel sigma-algebra on R and by B the one on R .

N N Deﬁnition 13. A cylinder set on R is a set of the form SB1,··· ,Bn = {x ∈ R |xi ∈ Bi, i = 1, ..., n, Bi ∈ B, n < ∞}. The σ-algebra of subsets of RN generated by cylinder sets is denoted by BN. A measure µ deﬁned on BN is called a cylinder measure.

Note that these deﬁnitions are consistent with the ones given before. We then have the

∞ Theorem 4. Kolmogorov’s extension theorem Let {µn}n=1 be a consistent family of measures on Bn. Then there is a unique cylinder measure µ on BN such that

µ(SB1,··· ,Bn ) = µn(B1 × B2 × · · · × Bn) (4.3)

Proof. The main problem is countable additivity. We can prove this with the Riesz theorem by the following trick. Let R˙ be the one-point compactiﬁcation of R (R˙ = R ∪ {∞} and open sets are open c sets in R and sets of the form A ∪ {∞} where A ⊂ R is open and A ⊂ [−n, n] some n < ∞, R˙ is easily seen to be compact). Let

∞ ˙ ˙ N M = ⊗i=1R = R .

Then M is compact in the product topology. Let C(M) = {f : M → R | f continuous } and C0(M) = {f : M → R | f is continuous and depends on ﬁnitely many xi}. Deﬁne, for f ∈ C0(M), Z `(f) = fdµn

˙ n where n is large enough, and we extend µn to R by putting µn(B1×· · ·×{∞}×· · ·×Bn) = 0. By consistency, ` is n independent if n is large enough. So ` deﬁnes a positive linear functional ` : C0(M) → R, `(1) = 1, |`(f)| ≤ kfk∞. The last inequality implies that ` extends to C(M) since C0(M) is dense in C(M). Hence by Riesz theorem there exists R unique Borel-measure µ on M such that `(f) = fdµ. Recall that M = R˙ N. But

∞ ∞ X X µ({x ∈ X | xn = ∞ some n}) ≤ µ({x | xn = ∞}) = µn({x | xn = ∞}) = 0. n=1 n=1

Hence µ is supported on RN. By construction, this measure satisﬁes (4.3). (BN ⊂ Borel 2 sets of R˙ N) As it turns out, we can embed all interesting function (and distribution) spaces into RN. In the next section we will do this for two subsets.

4.4.3 Minlos Theorem Let ∞ N X 2m 2 2 sm = {x ∈ R | n |xn| ≡ kxkm < ∞} n=1 for m ∈ Z and let \ 0 [ s = sm s = sm m∈Z m∈Z

2 Borel sets of R˙ N = smallest σ-algebra containing open sets and cylinder sets are generated by open sets.

37 Note that sm ⊃ sm+1 ⊃ · · · . s is the set of sequences which decay as faster than any power as n → ∞ while s0 contains the ones with at most polynomial growth. Give s the topology generated by the neighbourhoods of 0 of the form

N(m, ) = {x | kxkm < } so the neighbourhoods of y ∈ s are y + N(m, ). Lemma 1. The space s with the above topology is a complete metrizable space: s has a complete metric d such that d gives the same topology as above. Excercise 12. Prove this lemma: define the metric ∞ X kx − ykm d(x, y) = 2−m 1 + kx − yk m=1 m and show that this distance defines the same topology as above and that s is complete (all Cauchy sequences converge). Remark. s is an example of a Frechet space. This is a complete locally convex metric space which is not a Banach space. Lemma 2.s 0 is the dual space of s: it is the space of continuous linear functionals on s. 0 Proof. Let y ∈ s . Then y defines a continuous linear functional `y on s : ∞ X `y(x) ≡ (y, x) ≡ ynxn. n=1 m m Indeed, this converges since |yn| ≤ Cn some m < ∞ and |xn| ≤ Cmn for all m. To prove continuity, let d(x(k), x) → 0 as k → ∞. Then, by Schwartz’ inequality

∞ (k) X m (k) |`y(x ) − `y(x)| ≤ Cn |xn − xn| n=1 ∞ X 1 = C nm+1|x(k) − x | n n n n=0 ∞ X 1 1/2 ≤ C kx(k) − xk → 0 as k → ∞. n2 m+1 n=0 Conversely, let ` : s → R be a continuous linear map. Then there exist C < ∞, m < ∞ such that |`(z)| ≤ Ckzkm for all z. Indeed, by continuity there exists a neighbourhood U of zero in s such that |`(x)| < 1, for all x ∈ U. On the other hand U contains a set {z | kzkm < δ} for some m, δ. Hence, given a z ∈ s we have δ z ∈ U and so |`(δ z )| < 1 and |`(z)| ≤ δ−1kzk . kzkm kzkm m (k) (k) (n) Let now e ∈ s be given by en = δnk. Put yn = `(e ). By the inequality we just proved, (n) (n) m |yn| = |`(e )| ≤ Cke km = Cn so y ∈ s0. Given x ∈ s, let n (n) X (k) x = e xk. k=1 Then x(n) → x in s (prove !) and so

`(x(n)) → `(x)

(n) Pn But `(x ) = k=1 ykxk. Hence `(x) = (y, x)

38 The reason we introduced s and s0 is the following: let µ be a cylinder (probability) measure on BN. Note that s0 is a µ-measurable set (prove !). Suppose µ(s0) = 1 or µ(BN\s0) = 0 so µ is supported in s0. Then,

0 Excercise 13. Let x ∈ s. Then the function y → (y, x) from s to R is µ-measurable. Deﬁnition 14. Let µ be a cylinder probability measure on s0. The characteristic function W : s → of µ is C Z W (x) = ei(y,x)dµ(y) s0 Lemma 3. The characteristic function W satisﬁes • W (0) = 1, |W (f)| ≤ 1

(i) • W is positive deﬁnite: given zi ∈ C, x ∈ s i = 1, ··· , n we have n X zizjW (xi − xj) ≥ 0 i,j=1

• W : s → R is continuous. Proof. As before, for c) use again the dominated convergence theorem

Finally, we are now equipped to formulate and prove our goal: Theorem 5. Minlos’ Theorem A necessary and suﬃcient condition for a function W on s to be a characteristic function of a probability measure on s’ is that it obeys the three conditions from the lemma above.

n Proof. Let W be given. Let u ∈ R , put x = (u1, ··· , un, 0, 0, ··· ) ∈ s. Then W (x) ≡ n n Wfn(u) is a function of positive type in R so there exists a probability measure µn on R such that Z i(y,u) Wfn(u) = e dµn(u) Rn ∞ {µn}n=1 are consistent since Wfn+m((u1 ··· un, 0, ··· , 0)) = Wfn(u) and Wfn uniquely determines µn (Why ?). Hence there is a measure µ on RN such that Z W (x) = ei(y,x)dµ(y) RN 0 for all x ∈ s with only ﬁnitely many non zero xi. We now need to show that µ(s ) = 1. In that case the integral can be restricted to s0 and the equation above holds for all x ∈ s (because both sides are continuous in x and the set of x with a ﬁnite number of nonzero xi is dense in s). 0 0 ∞ µ(s ) = 1 : Recall that s = ∪−∞sm with

∞ X 2m 2 2 sm = {y| n yn ≡ kykm < ∞}. n=1 We now show that, given > 0, there exists an m such that

µ(sm) ≥ 1 − . (4.4)

0 0 Since µ(s ) ≥ µ(sm) we get µ(s ) ≥ 1 − for all and hence the claim will follow. To prove (4.4), we use ﬁrst the monotone convergence theorem to get a more manage- able expression :

N ! Z α Z α 2 X 2m 2 − kykm lim lim exp − n yn dµ = lim e 2 dµ = µ(sm) α↓0 N→∞ 2 α↓0 n=1

39 where the second equality holds since ( − α kyk2 1 y ∈ sm e 2 m → χ(y ∈ sm) ≡ 0 y 6∈ sm, so if kykm = ∞

N ! Z αX We consider exp − n2my2 dµ. Now choose > 0. Continuity of W (x) then 2 n n=1 implies that there exist m, δ such that

2 |W (x) − 1| ≤ if kxk−m−1 ≤ δ (4.5) recall that W (0) = 1 and note that m here tends to be smaller than zero. Then, for all x ∈ s, 2 Re W (x) ≥ 1 − − kxk2 (4.6) δ −m−1 2 2 since if kxk−m−1 < δ,(4.6) holds by (4.5) and if kxk−m−1 > δ,(4.6) holds because |W (x)| ≤ W (0) = 1 implies Re W (x) ≥ −1. Now write N ! Z αX 2m 2 i(y,x) exp − n yn = e dν(x) 2 N n=1 R N Y x2 dν(x) = (2παn2m)−1/2 exp − n d 2αn2m n n=1 to get

Z N ! Z Z N ! ! αX 2m 2 X exp − n yn dµ = exp i yixi dµ(y) dν(x) 2 N n=1 R i=1 Z Z = W (x)dν(x) = Re W (x)dν(x) RN since the left hand side is real. Hence by (4.6),

Z N ! Z αX 2m 2 2 2 exp − n yn dµ ≥ 1 − − kxk−m−1dν(x). 2 δ RN But Z N Z N ∞ ! 2 X −2m−2 2 X −2m−2 2m X 1 kxk−m−1dν = n xndν(x) = n αn ≤ 2 α. N n R n−1 n=1 n=1 so Z N ! αX 2m 2 cα lim exp − n yn dµ ≥ 1 − − N→∞ 2 δ n=1 From this, (4.4) follows which, as argued above proves the claim.

Example. Gaussian measures. Let C be a matrix Cij i, j = 1, 2, ··· with |Cij| ≤ m m X ai j for some a < ∞, m < ∞ and (x, Cx) = xixjCij > 0 ∀x 6= 0 x ∈ s (note that −n (x, Cx) makes sense since |xi| < bni for all n). Then the function

− 1 (x,Cx) WC (x) = e 2 is of positive type.

40 2 Proof. W is continuous since |(x, Cx)| ≤ Akxkn for some n. Hence ( (k) (k) xi i ≤ k WC (x) = lim WC (x ) xi = k→∞ 0 i > k

But   k Z k ! (k) 1 X X WC (x ) = exp − xixjCij = exp i xiyi dµk(y) 2 k ij=1 R i=1

k where µk is the Gaussian measure on R with covariance matrix Cij, i, j = 1, ··· , k. By k (k) k assumption, this is a positive k × k matrix. Hence, let WC (x) = WC (x ) x ∈ s so WC is k of positive type and WC (x) → WC (x) ∀x ∈ s → WC (x) is of positive type. 0 We call the measure µC corresponding to WC the Gaussian measure in s with covariance C. d d Example. Order Z arbitarily such that Z = {xk|k = 1, 2, ···} and |xk+1| ≥ |xk|. d ˜ ˜ d Then φ : Z → R becomes φ : N → R φi = φxi . Thus, if Cxy x, y ∈ Z is a positive matrix m m with |Cxy| ≤ a(|x|+1) (|y|+1) we can deﬁne the Gaussian measure with covariance C on 0 d d X 2m 2 −1 s (Z ) = {φ : Z → R| (|x|+1) |φx| < ∞ some m ∈ Z}. Thus C = (−∆+r) is an x∈Zd example (it is positive and |Cxy| ≤ constant; this holds for r ≥ 0, d ≥ 3 and r > 0, d = 2).

4.5 Measures on S0

Let n ∞ m d f S(R) = {f : R → R|f is C and sup(1 + |x| ) n | < ∞ for all n, m ≥ 0} (4.7) x∈R dx so the set of functions for which f and its derivatives decay faster than any power. Let us deﬁne the “creation and annihilation operators”: 1 d 1 d a = √ x + , a+ = √ x − 2 dx 2 dx

2 + 1 d 1 2 1 which gives the number operator N = a a = − 2 dx2 + 2 x − 2 . For f ∈ S(R), let Z 1/2 m 2 kfkm = |(N + 1) f| dx R which is ﬁnite due to (4.7)3. We give S the topology where neighbourhoods of 0 are

N(m, ) = {f ∈ S| kfkm < }

Proposition 9. S is homeomorphic to s.

Proof. Consider the functions φn ∈ S :

1 − 1 x2 1 + n φ0 = e 2 , φn = √ (a ) φ0 π1/4 n! These satisfy : Z ∞ (φn, φm) ≡ φnφmdx = δnm −∞

3 See Remark 1 below.

41 + + Furthermore, aφ0 = 0, aa = a a + 1 implies that

1 + n 1 h + n−1 + + n−1 i aφn = √ a(a ) φ0 = √ (a ) φ0 + a a(a ) φ0 n! n! 1 h + n−1 + 2 +n−2 i n + n−1 n n! = √ 2(a ) φ2 + (a ) aa φ0 = ... = √ (a ) φ0 ⇒ a φn = √ φ0 n! n! n!

But also, for φ, ψ ∈ S, (a+φ, ψ) = (φ, aψ) so (φ , φ ) = √1 (φ , anφ ) = (φ , φ ) = 1. n n n! 0 n 0 0 n 6= m gives 0 since aφ0 = 0. 2 The functions φn are thus orthonormal in L (R). They are a basis (they are the Hermite functions). Now our map S → s is f → x = (x1, x2, ··· ), with xn = (f, φn−1). Note that + Nφn = a aφn = nφn

m 0 0 m so (N + 1) φ → x with xn = n xn. Thus

Z ∞ ∞ 2 m 2 X m 2 X 2m 2 2 kfkm = |(N + 1) f| dx = |((N + 1) f, φk)| = n |xn| = kxkm k=0 n=1 i.e. the map f → x maps the neighbourhoods to each other. To conclude, we need to show that this map is a bijection. 2 A) Injection : {φn} is a basis of L .

PN B) Surjection, or the map is onto : Let x ∈ s. Then we need to show that n=1 xnφn−1 PN converges as N → ∞ to an element f of S and xn = (f, φn−1). n=1 xnϕn−1 converges in 2 2 m PN PN m 2 ∞ L to f ∈ L .(N + 1) xnϕn−1 = xnn ϕn−1 converges in L ⇒ f ∈ C . And kfkm = kxkm so f ∈ S.

0 Deﬁnition 15. The space of tempered distributions on R is the set S (R) of continuous linear functionals on S(R).

0 4 Let ϕ ∈ S (R). Put yn = ϕ(φn−1). Then ϕ continuous means : ∃C, m : |ϕ(f)| ≤ m 0 0 Ckfkm ∀f ∈ S ⇒ |yn| = |ϕ(φn−1)| ≤ Ckφn−1km = Cn . Hence y ∈ s . Conversely, y ∈ s ∞ 0 X gives rise to ϕy ∈ S (R) by ϕy(f) = yn(f, φn). Indeed, from above, xn = (f, ϕn) is in n=1 s so |ϕy(f)| ≤ Ckxkm for some m and this = Ckfkm.

2 Remark 1. We see from above that k km is a norm on S : kfkm = 0 ⇒ (f, ϕn) = 0 ∀n ⇒ 2 + m + m + 2m P f = 0. Moreover, since kfkm = (a a + 1) f, (a a + 1) f = f, (a a + 1) f = + k + k d of f, (a a) f and (a a) f = (Polynomial in x and dx ) f, we get

β X α d (m) kfkm ≤ Cm sup |x β f(x)| ≡ Cmkfk x dx α,β≤2m and conversely (show !). So, ϕ ∈ S0 ⇐⇒ ∃C,N : |ϕ(f)| ≤ Ckfk(N). This is in practice useful, see Examples 1,2 below.

4 ϕ continuous : ∀ ∃ N(m, δ): f ∈ N(m, δ) ⇒ |ϕ(f)| < i.e. kfkm < δ ⇒ |ϕ(f)| < . Thus for any f kfkm f ∈ S, |ϕ(f)| = |ϕ δ | ≤ kfkm = Ckfkm. kfkm δ δ

42 Example 1 Let ϕ be a polynomially bounded function : |ϕ(x)| ≤ C(1 + |x|)m. Then it deﬁnes a distribution ϕ ∈ S0 by Z ϕ(f) = ϕ(x)f(x)dx since Z Z | ϕ(x)f(x)dx| ≤ C (1 + |x|)m|f(x)|dx Z 1 ≤ C (1 + |x|)m+2|f(x)| dx [1 + |x|]2 ≤ C0 sup(1 + |x|)m+2|f(x)| x ≤ C00kfk(m+2) so ϕ ∈ S0.

Example 2 ϕ can be a “delta function” or its derivatives : let

dkf ϕ(f) = (x ). dxk 0 Then, k |ϕ(f)| ≤ kD fk∞

0 dk so ϕ ∈ S . We denote ϕ = dxk δ(x = x0). n Remark 2. This works in R too :

n ∞ n α β S(R ) = {f ∈ C (R )| kx D fk∞ ≤ Cαβ ∀α, β}

n n βi α Y αi β Y ∂ where α = (α1, ··· , αn), β = (β1, ··· , βn), x = x ,D = . Basis {ϕk}k∈ n i βi N i=1 i=1 ∂xi n Y (n) n X Y 2m 2 N n ϕk = ϕki (xi) and we deﬁne s = {x ∈ R |x = (xk)k∈N ( ki) |xk| < ∞ ∀m}. i=1 k Do the rest !

0 Deﬁnition 16. The cylinder set σ-algebra in S (R) is the smallest σ-algebra containing all sets 0 A(ϕ, B) = {ϕ ∈ S (R)|ϕ(f) ∈ B} for f ∈ S, B Borel in R. This is just the image of our old cylinder algebra on s0 to S0 by the map constructed above. Indeed, the cylinder σ-algebra C in s0 is the smallest σ-algebra of subsets of s0 that 0 contains the sets σn(B) = {y ∈ s |yn ∈ B} where B is Borel in R. I.e. it is the smallest σ-algebra such that functions

0 pn : s → R , pn(y) = yn

−1 ˜ are measurable (i.e. pn (B) ∈ C ∀B Borel i.e. σn(B) ∈ C ∀B). Let C be the smallest 0 0 σ-algebra in s such that the functions πx : s → R πx(y) = (y, x) are measurable for all ∞ (n) ˜ X x ∈ s. Since pn = π (n) , em = δnm we see that C ⊂ C. But, also, πx = xnpn so πx is en n=1 0 C-measurable. Hence C˜ ⊂ C. Thus C = C˜. Clearly the σ-algebra deﬁned above in S (R) 0 0 is the image of C˜ under the map y ∈ s → ϕy ∈ S . This proves the claim.

43 0 Deﬁnition 17. A cylinder measure on S (R) is a measure on this σ-algebra. Translating from s, s0 to S, S0 we get again:

Theorem 6. Minlos’ theorem A necessary and suﬃcient condition for a function W n 0 n on S(R ) to be the generating function of a cylinder probability measure µ on S (R ), Z W (f) = eiϕ(f)dµ(ϕ) S0 is that W (0) = 1, W be of positive type and continuous.

4.6 Gaussian measures

1 n Here W (f) = exp − 2 C(f, f) where C(f, g) is a continuous bilinear function on S(R ) × n S(R ), with C(f, f) > 0 for f 6= 0. Important examples are given by integral kernels Z Z C(f, g) = dnx dnyC(x, y)f(x)g(y) here R |C(x, y)|[(1 + |x|)(1 + |y|)]mdxdy < ∞ for some m suﬃces. We are interested in translation-invariant C’s i.e. C(x, y) = G(x − y). These are most conveniently expressed n in terms of Fourier transform : If f ∈ S(R ) we put Z fˆ(k) = f(x)e−ikxdnx

n α β ∧ α+β α β and one gets fˆ ∈ S(R ). [This is because (x D f) (k) = i D k fˆ(k), do the details !] Inverse : Z dnk f(x) = eikxfˆ(k) . (2π)n Then we have Z Z dnk |f(x)|2dnx = |fˆ(k)|2 (2π)n Z (f ∗ g)(x) ≡ f(x − y)g(y)dny

(f ∗ g)∧(k) = fˆ(k)ˆg(k)

and Z Z dnk f(x)G(x − y)f(y)dnxdny = |fˆ(k)|2Gˆ(k) (2π)n which holds for f, G ∈ S to start with, but extends to much more general G’s.

4.6.1 The Euclidean Free Feld

The Euclidean Free Feld with mass m is the Gaussian measure µG where 1 Gb(k) = (k ∈ n). k2 + m2 R dnk Clearly S 3 f → R |fˆ(k)|2 is continuous, so (k2 + m2)(2π)n " # Z |fˆ(k)|2 dnk W (f) = exp − 1 2 k2 + m2 (2π)n

44 0 n determines a measure on S (R ), by Minlos’ theorem. Strictly speaking, G(x) is deﬁned as a distribution5, by the Fourier transform Gb but actually is a locally integrable function :

e−m|x| |G(x)| ≤ C . (4.8) |x|n−2

To see this, let us consider a ultraviolet cutoﬀ : let Λ > 0 and deﬁne

1 Λ2 GbΛ(k) = . k2 + m2 k2 + Λ2

This is called Pauli-Villars cutoﬀ. Note that GbΛ(k)Λ−→→∞ Gb(k) pointwise, and as elements 0 0 of S . Note S is given a topology such that ϕi → ϕ as i → ∞ if ϕi(f) → ϕ(f) ∀f ∈ S. 1 n For n < 4, GbΛ ∈ L (R ) and so GΛ(x) is a continuous function (for n ≥ 4 : replace Λ2 Λ2 k by ). p2 + Λ2 p2 + Λ2 So, by rotational symmetry,

Z dnk eik1|x| Λ2 G (x) = Λ (2π)n k2 + m2 k2 + Λ2

So the k1 integral : q q 2 2 2 2 2 2 k + m = (k1 + i ~k + m )(k1 − i ~k + m ), q q 2 2 2 2 2 2 k + Λ = (k1 + i ~k + Λ )(k1 − i ~k + Λ ),

~k = (k2, k3, ··· ).

By Cauchy: √ √ " 2 2 2 2 # Z dn−1~k e−r ~k +m Λ2 e−r ~k +Λ 1 GΛ(x) = n−1 p 2 2 + p 2 2 (2π) 2 ~k2 + m2 m + Λ 2 ~k2 + Λ2 m + Λ which for r = |x|= 6 0 is smooth and converges as Λ → ∞ to √ Z dn−1~k e−r ~k2+m2 G(x) = n−1 p ; (2π) 2 ~k2 + m2 this is a L1 function satisfying (4.8).

4.6.2 Wick Formula Note that, a priori, for ϕ ∈ S0, ϕ(x) is not deﬁned, only ϕ(f), f ∈ S. These are measurable functions in S0 and actually Gaussian random variables on the probability space 0 n (S (R ), µ): − 1 t2(f,Gf) heit(ϕ,f)i = e 2 so the variance of ϕ(f) is (f, Gf). Let fi ∈ S, i = 1, ··· ,N. Then xi ≡ ϕ(fi) are jointly Gaussian : P P − 1 P t t (f ,Gf ) hei tixi i = heiϕ( tifi)i = e 2 ij i j i j

5Fourier transformϕ ˆ of a distribution ϕ ∈ S0 is a distribution,ϕ ˆ ∈ S0, deﬁned by

ϕˆ(fˆ) = (2π)nϕ(f˜) f˜(x) = f(−x)

R ˆ dnk R n which is the usual (see example 1 above) if ϕ ∈ S : ϕˆ(k)f(k) (2π)n = ϕ(x)f(−x)d x.

45 N −1 i.e. x = (x1, ··· , xN ) ∈ R is Gaussian with covariance matrix Aij = (fi, Gfj). Hence e.g. Z Z d d −i[tϕ(f)+sϕ(g)] n n ϕ(f)ϕ(g)dµG(ϕ) = − he i = (f, Gg) = d xd yf(x)G(x − y)g(y) dt 0 ds 0 and, more generally,

2N Z Y X Y ϕ(fi)dµG(ϕ) = (fi, Gfj) i=1 P hiji∈P Z n n X Y = d x1 ··· d x2N f1(x1) ··· fN (xN ) G(xi − xj) P hiji∈P

Hence we may use the notation

Z X Y ϕ(x1) ··· ϕ(x2N )dµG(ϕ) = G(xi − xj) P hiji∈P even if strictly speaking ϕ(x) is not a well deﬁned random variable.

Remark 3. Indeed, let us try to see what ϕ(x)2 would be. Formally Z Z 2 ϕ(x) dµG(ϕ) = lim ϕ(x)ϕ(y)dµG = lim G(x − y) = G(0) = ∞. x→y x→y

This is the familiar ultraviolet divergence in quantum ﬁeld theory. Our G is such that ϕ(x) is not a nice random variable (it is a “distribution valued random variable”). We’ll return to this.

We ﬁnish this discussion of Gaussian measures with the simple and useful

0 d Lemma 4. Suppose µGi , i = 1, 2 are two Gaussian measures on S (R ) and G = G1 + G2. ∞ Then for all F ∈ L (µG), Z Z

F (ϕ)dµG(ϕ) = F (ϕ1 + ϕ2)dµG1 (ϕ1)dµG2 (ϕ2).

Hence vφ is the sum of independent Gaussians ϕ = ϕ1 + ϕ2.

◦ iϕ(f) d R iϕ(f) − 1 (f,Gf) Proof. 1 . The claim holds for F (φ) = e , f ∈ S(R ). Indeed e dµG = e 2 and

Z Z Z 1 1 i(ϕ1(f)+ϕ2(f)) iϕ1(f) iφ2(f) − 2 [(f,G1f)+(f,G2f)] − 2 (f,Gf) e dµG1 dµG2 = e dµG1 e dµG2 = e = e .

◦ 2 . Suﬃces to check (why?) for F = 1C , the indicator of a c cylinder set

0 d C(f1, ··· , fn,A1, ··· ,An) = {ϕ ∈ S (R )|ϕ(fi) ∈ Ai, i = 1, ··· , n}

d where fi ∈ S(R ),Ai Borel sets in R. Now φ(f1), ··· , φ(fn) are Gaussian with covariance

hϕ(fi)ϕ(fj)i = (fi, Gfj) ≡ Aij

n i.e. for g : R → R Borel measurable, Z Z g(ϕ(f1), . . . , ϕ(fn))dµG(φ) = g(x1, . . . , xn)dµA(x) (4.9)

46 and similarity for ϕα(fi) α = 1, 2 : Z

g(ϕ1(f1) + ϕ2(f1), . . . , ϕ1(fn) + ϕ2(fn)dµG1 dµG2 Z

= g(y1 + z1, . . . , yn + zn)dµA1 (y)dµA2 (z) (4.10)

i(λ,x) n But (4.9) = (4.10) if we can show it for g(x) = e for all λ ∈ R i.e. Z Z i(λ,x) i(λ,y+z) e dµA(x) = e dµA1 (y)dµA2 (z) which is a special case of 1◦.

47 Chapter 5

From Random Fields to Quantum Fields

We have seen (at least in the case of the free field) how quantum fields give rise to random fields by going to imaginary time. Now we want to address the converse: under what 0 d conditions a (time translation invariant) measure µ on S (R ) can give rise to a quantum field theory and in particular how to reconstruct the Hilbert space, Hamiltonian and field operators from the measure.

5.1 Reﬂection positivity

d The condition we need turns out to be reﬂection positivity: write (t, x) ∈ R , so d−1 1 d x ∈ R . Let R+ = {(t, x) | t > 0} and deﬁne N X iϕ(fk) ∞ d H+ = cke | ck ∈ C, fk ∈ C0 (R+),N ∈ N k=1

H+ can be thought of as a space of functionals F (ϕ) of ϕ depending on fields restricted to positive times. Next, define the time reflection operator (θf)(t, x) ≡ f(−t, x) and extend it to H+ by N N X iϕ(fk) X −iϕ(θfk) θ cke ≡ c¯ke k=1 k=1 So θF depends on ϕ evaluated at negative times. Definition 18. A measure µ is reflection positive if Z (θF )(ϕ)F (ϕ)µ(dϕ) ≥ 0 for all F ∈ H+.

We get the following sesquilinear form on H+ (F,G) ≡ E[(θF )G] which is a positive form if the measure is reﬂection positive. For this form, let

N := {F ∈ H+ | (F,F ) = 0} 2 0 which is a linear subspace of H+. Using this subspace, define H = H+/N as the set of equivalence classes [F ] = {F + N |,N ∈ N }. Then the form (·, ·) is well defined on H0 since (F + N1,G + N2) = (F,G). Hence it defines a scalar product, since ([F ], [F ]) = 0 ⇒ F ∈ N ⇒ [F ] = 0.

1 d d+1 Note the diﬀerence with the notation before, where we took x ∈ R and thus (x, t) ∈ R . Now the physical dimension is 4 2Use that |(F,G)| ≤ (F,F )1/2(G, G)1/2

48 Definition 19. The physical Hilbert space H is the completion of H0 in the scalar product as constructed above. We have thus found the Hilbert space we were looking for, so next, we try to find the Hamiltonian. Since in physics, the energy or Hamiltonian of a system is a conserved quantity corresponding to time translation invariance, we will search in this direction. We first note that time translation is a natural semigroup that acts on H+: T (τ): H+ → H+ which is defined for all τ ≥ 0 by

T (τ)eiϕ(f) = eiϕ(fτ ) where fτ (t, x) = f(t − τ, x) R d−1 R d−1 so ϕ(fτ ) = ϕ(t, x)f(t − τ, x)d xdt = ϕ(t + τ, x)f(t, x)d xdt. This is indeed a semigroup, since T (t) extends to a map that satisﬁes T (t)T (s) = T (t + s) for all positive t, s, and T (1) = I. Furthermore,

iϕ(f) iϕ(g) −iϕ(θf) iϕ(gτ ) −iϕ((θf) ) iϕ(g) e ,T (τ)e = E e e = E e −τ e iϕ(h) iϕ(h ) by the time translation invariance of the measure: E[e ] = E[e t ] for all t ∈ R, as assumed before. Finally, using (θf)−τ = θfτ we ﬁnd that eiϕ(f),T (τ)eiφ(g) = T (τ)eiϕ(f), eiφ(g) so T (τ) is symmetric on H+. Next we compute, using the notation kF k2 = (F,F ), the seminorm, (T (t)F,T (t)F ) = (F,T (2t)F ) ≤ kF k1/2kT (2t)F k1/2 Iterating this n times gives

n −n kT (t)F k2 ≤ kF k1/2+1/4+...+1/2 kT (2nt)F k2 . Now,

n 2 n+1 n+1 21/2 n+1 21/2 kT (2 t)F k = (F,T (2 t)F ) = E[θF (T (2 t)F )] ≤ E(θF ) E(T (2 t)F ) by Schwartz’ inequality. On the other hand, by time translation invariance:

n+1 2 2 E(T (2 t)f) = EF 2 2 n 2 2 and also E(θF ) = EF so that kT (2 t)F k ≤ EF . Hence we obtained for for all n > 0 2 1−2−n 2 2−n−1 kT (t)F k ≤ kF k (EF ) which implies kT (t)F k2 ≤ kF k. This shows that T (t) is a contraction in seminorm. Furthermore, T (t): N → N since 2 2 kT (t)Nk ≤ kNk = 0. Hence, T (t) extends to a symmetric contraction semgroup on H0 and since H is the completion of H0, the same holds in the phyisical Hilbert space H. Finally, T (t): H → H is strongly continuous (i.e. kT (t)ψk is continuous in t, prove!). We have found that T (t) satisfies all conditions of the Hille-Yoshida theorem: Theorem 7. Let T (t) be a strongly continuous semigroup in a Hilbert space H. Then T (t) = e−tH where H is some positive slef-adjoint operator, defined on D(H) = ψ ∈ H | lim(−1T () − 1)ψ exists} →0 Finally, we have found the QFT Hamiltonian we were looking for. If we then define the vacuum vector by Ω = [1], clearly T (t)Ω = Ω, so HΩ = 0 as we had in all the examples before.

49 5.2 Reﬂection positive Gaussian measures

Gaussian measures are determined by their covariance Their reflection positivity can in fact be described by a similar property of the covariance. To see this suppose the gaussian measure µ with covariance C is reflection positive and consider a test function f supported in positive times. Then C(θf, f) = Eϕ(θf)ϕ(f) ≥ 0. C(θf, f) ≥ 0 is thus a necessary condition for µ to be reflection positive. It is also sufficient: Lemma 5. A gaussian measure with covariance C is reflection positive if and only if for all functions f supported on t ≥ 0: C(θf, f) ≥ 0

Proof. Let f1, ..., fn have support t ≥ 0. We need to show that the matrix

−iϕ(θfa) iϕ(fb) Mab = E e e is positive. We have

−iϕ(θfa−fb) Mab = E e − 1 C(f −θf ,f −θf ) = e 2 b a b a − 1 C(f ,f ) − 1 C(f ,f ) = e 2 a a e 2 b b eC(θfa,fb P n Now, M ≥ 0 ⇔ vavbMab ≥ 0 for all v ∈ R which, by the above, is equivalent to

X c(θfa,fb) wawbe ≥ 0

1 − C(fa,fb) where we deﬁned wa = e 2 va. The claim then follows from the next lemma.

A Lemma 6. Let Aab be a positive n × n matrix. Then Mab = e ab is also positive. P 1 n Proof. Since Mab = n! Aab, all we need to prove is A, B positive implies that Cab = AacBab is positive. To prove this, let D=A ⊗ B, i.e. Dij,kl = AikBjl. Since A ⊗ B is P P positive, ijkl vijDij,klvkl ≥ 0 for all v, taking vij = viδij implies that ik viDii,kkvk = P ik viAikBikvk ≥ 0, from which the claim (and also the ﬁrst lemma) follows.

5.2.1 Example: harmonic oscillator and free field 0 Recall that the harmonic oscillator gives rise to the gaussian measure µ on S (R) with covariance e−ω|t−s| G(t, s) = [q(t)q(s)] = (−∂2 + ω2)−1(s, t) = . E t 2ω We have (θq)(t) = q(−t), so e−ω(t+s) [θq(t)q(s)] = ∀t, s ≥ 0 E 2ω and hence for all f, g supported on [0, ∞): Z ∞ Z ∞ 1 −ωt −ωt E[q(θf)q(g)] = e f(t)dt e g(t)dt ≥ 0. 2ω 0 0 Hence µ is reflection positive. For the free field the argument is similar since it is basically just an infinite collection d−1 of oscillators. Write (t, x) ∈ R × R and Z ˆ ip(x−y) dp E[ϕ(t, x)ϕ(s, y)] = G(t, s, p)e (2π)d−1 Rd−1 with Gˆ(t, s, p) = √ 1 e−ω(p)|t−s|. Thus, 2ω(p) Z Z ∞ Z ∞ dp 1 −ω(p)t −ω(p)s G(θf, g) = (2π)d−1 dtf(t, p)e dsg(s, p)e ≥ 0. 2ω(p) 0 0 So the measure is reflection positive.

50 5.3 Reconstruction for Free Field

5.3.1 Time zero fields To make a connection with the second quantization formalism we need to get a more concrete representation of the abstract Hilbert space of the previous section. Intuitively d−1 we should think of elements in H as functionals of fileds ϕ0(~x) with ~x ∈ R Hence the Hilbert space is a space on time-zero fields. Let us first proceed heuristically. Think of the measure µ as a functional integral Z Z F (ϕ)dµ(ϕ) = e−S(ϕ)F (ϕ)Dϕ with S(ϕ) some local action functional Z S(ϕ) = L(x)dx Rd where L(x) depends on ϕ(x) and the derivatives of ϕ at the point x. An example is

1 2 L(x) = 2 (∇ϕ(x)) + P (ϕ(x)). (5.1)

Given F ∈ H+ let us try to integrate over the fields ϕ(x) for t > 0. Fix a field configuration d−1 ϕ0(x) at the time zero subspace x ∈ R , let Z Z S±(ϕ) = dt dxL(x) R± and define Z −S±(ϕ) F±(ϕ0) = F (ϕ)e Dϕ. ϕ|t=0=ϕ0 Then Z Z −S(ϕ) (F,G) = θF (ϕ)G(ϕ)e Dϕ = (θF )−(ϕ0)G+(ϕ0)Dϕ0

Now suppose we had S− = θS+. (5.2)

Then (θF )− = F+ and we get Z (F,G) = F+(ϕ0)G+(ϕ0)Dϕ0.

2 Hence the map F ∈ H+ → F+ is an isometry from H+ to L (Dϕ0) and reflection positivity follows from positivity of the L2 norm. Superficially (5.2) seems obvious since L(x) is a local functional of the field. However one needs to be careful as L(x) depends not only on ϕ(x) but also on its derivatives. In general the above argument breaks down due to this dependence but it can be saved in the special case where the dependence is as in (5.1). We will show this in the next subsection.

5.3.2 Harmonic oscillator once more e−ω|t−s| Consider the harmonic oscillator measure with covariance G(t, s) = 2ω . H+ is iq(f) R ∞ ∞ spanned by linear combinations of e with q(f) = 0 dtq(t)f(t) for f ∈ C0 (R+). Note also that q(0) is a gaussian random variable with covariance 1 . We will now show that 2ω √ 2 2 √ω −ωq H+ can be identiﬁed as L (R, dµ0) where µ0 is the law of q(0): dµ0(q) = π e dq. We do this by writing q as a sum of independent gaussians:

q(t) = q+(t) + q−(t) + q0(t) (5.3)

51 where the ﬁrst gaussian is nonvanishing only on positive times, the second on negative times and the last one is built out of the time zero random variable. (5.3) follows from a decomposition of the covariance of q to three covariances as follows. Deﬁne G+(t, s) = (G(t, s) − G(t, −s))1t,s≥0 and G−(t, s) = (G(t, s) − G(t, −s))1t,s≤0 for t, s ≤ 0. These are positive operators: G± are the Dirichlet Green functions for the 2 2 operator −∂t + ω on R±. With this notation, the full Green function becomes

G(t, s) = G+(t, s) + G−(t, s) + G0(t, s) where e−ω(|t|+|s|) G (t, s) = 0 2ω Hence (5.3) holds with −ω|t| q0(t) = e x 1 and x ∈ R has the law of µ0 ie it is a Gaussian random variable with variance 2ω . Next, we note that F ∈ H+ is a function of q+ and q0 only: F (q) = F (q+ +q0) and the law of θq+ is the same as that of q−, and the law of θq0 is the same as that of q0. From this we get (F,G)H+ = E[θF G] = E0 E−[θF ]E+[G] = E0 E+[F ]E+[G] = E0UFUG where we deﬁned (UF )(x) = E+[F (q+ + q0)] and E0 is the expectation in the measure µ0. We conclude that

(F,G)H+ = (UF,UG)H which means that U : H+ → H is an isometry. ∞ Since for all f ∈ C0 (R+) 1 R ∞ −ωt iq(f) iq+(f) iq0(f) − (f,G+f) i( f(t)e dt)x Ue = E+[e ]e = e 2 e 0 iαx the range of U contains e for all α ∈ R so the range of U is dense. Thus, U is unitary 2 from H+/N → H and we may identify the physical Hilbert space with L (ψ0(x)dx) where 1 2 p ω − 2 ωx ψ0 = π e is the ground state of the harmonic oscillator. Now, we can compute the Hamitonian via UT (t)F = e−tH UF where it suﬃces to take F (q) = eiq(0)α so that (UF )(x) = eiαx. Then T (t)F = eiq(t)α so that − 1 α2G (t,t) −ωt − 1 α2( 1 − 1 e−2ωt) −ωt UT (t)F = e 2 + eie αx = e 2 2ω 2ω eie αx iαx Thus for eα(x) = e

2 d −tH d d − α (1−e−2ωt) ie−ωtαx Heα = − e eα = − UT (t)eα = − e 4ω e (5.4) dt t=0 dt t=0 dt t=0 d2 d = ( 1 α2 + iωαx)e = (− 1 + ωx )e (5.5) 2 α 2 dx2 dx α This is the Hamiltonian of the harmonic oscillator in disguise. Indeed, previously we 2 2 2 worked in the Hilbert space L (dx) whereas now we have L (ψ0dx). These are related by 1 2 2 2 2 p ω − 2 ωx the unitary map V : L (ψ0dx) → L (dx) given by V f = ψ0f = π e f. Then we recover the old Hamiltonian d2 d d2 VHV −1 = ψ−1 − 1 + ωx ψ = − 1 + 1 ω2x2 − 1 ω. 0 2 dx2 dx 0 2 dx2 2 2

52 5.3.3 Free ﬁeld Starting with the covariance Gˆ(t, s, p) = √ 1 e−ω(p)|t−s| we can repeat the harmonic 2ω(p) oscillator analysis:

G = G+ + G− + G0 with

Gˆ+(t, s, p) = (Gˆ(t, s, p) − Gˆ(t, −s, p))1t≥01s≥0, (5.6)

Gˆ− analogously, and

1 −(|t|+|s|)ω(p) Gˆ0(t, s, p) = e p2ω(p) leading to the corresponding decomposition of the ﬁeld to independents:

ϕ = ϕ+ + ϕ− + ϕ0.

Time zero ﬁeld by ϕ(0, x) has covariance

Z 1 C(x, y) = E[ϕ(0, x)ϕ(0, y)] = eip(x−y) dp . (5.7) 2ω(p) (2π)d−1

We can then write

law −tω ϕ0(t, f) = ψ(e f) where ψ has covariance C and e\−tωf(p) = e−tω(p)fˆ(p). We then get that the map

(UF )(φ) = E+[F (ϕ+ + ϕ0)]

2 0 d−1 is an isometry U : H+ → L (dνC ) where νC is a gaussian measure on S (R ) with covariance C: Z

(F,G)H+ = (UF )(ψ)(UG)(ψ)dνC (ψ)

As before we see that U is onto (show!) and hence we may identify the physical Hilbert 2 space of the free ﬁeld by L (dνC ).

5.3.4 Fock space 0 d 2 We found that the Hilbert space of the free ﬁeld on S (R ) is L (νC ) where νC is a 0 d−1 1 2 −1/2 ˆ 1 gaussian measure on S (R ) with covariance C = 2 (−∆+m ) or C(p) = 2ω(p) where p 2 2 2 ω(p) = p + m . Now, let D ⊂ L (ν0) be the set of polynomials in ψ(f1), . . . , ψ(fn) with d−1 n ≥ 0 and fi ∈ S(R ). Then the multiplication operator ψ(f): D → D . Next, deﬁne δ the operators (f, δψ ): D → D by

n n δ Y X Y (f, ) ψ(f ) = (f, f ) ψ(f ) δψ i i j i=1 i=1 j6=i so formally we can just writre (f, δ ) = R dxf(x) δ . Then we deﬁne operators δψ Rd−1 δψ(x) a(f), a∗(f): D → D by

δ δ a(f) = (C1/2f, ) a∗(f) = ψ(C−1/2f) − ( ,C1/2f) (5.8) δψ δψ

53 where as usual we have (C\±1/2f)(p) = (2ω(p))±1/2fˆ(p). These are the usual ladder operators: they satisfy the commutation relation [a(f), a∗(g)] = (f, g). In this representation the quantum mechanical ﬁeld operator is just the multiplication operator ψ(f), acting on the domain D and it can be written in terms of the creation and annihilation operators by solving from (5.8): ψ(f) = a(C1/2f¯) + a∗(C1/2f) which agrees with Deﬁnition 10. The vacuum vector in this representation is simply Ω = 1 which obviously is annihilated by a(f). Let us still check that the Hamiltonian is what we expect. As in the previous subsection we compute

1 −tω −tH iψ(f) ϕ+(t,f)+ϕ0(t,f)) − (f,C+(t,t)f) iψ(e f) e e = E+[e ] = e 2 e where C+ is deﬁned in (5.6) so that 1 Cˆ (t, t, p)) = (1 − e−2ω(p)t). + 2ω(p) Repeating the computation in (5.4) we get iψ(f) d −tH iψ(f) 1 iψ(f) He = − dt |t=0e e = ( 2 (f, f) + iψ(ωf))e . We leave it as an exercise for the reader to show that this is the same as Z iψ(f) ∗ dp iψ(f) He = ( ω(p)apap (2πd−1 )e so that H agrees with what we had before. The following exercise develops further the 2 connection between the Fock space picture and the L (νG) picture. d−1 Excercise 14. • Deﬁne for f ∈ S(R ) (f real): ψ(f) e ψ(f)− 1 (f,Cg) : eψ(f) := = e 2 E[eψ(f)] Show then that Z ψ(f) ψ(g) (f,Cg) : e :: e : dνC (ψ) = e (5.9)

d−1 • Let f1, ..., fn ∈ S(R ) and deﬁne n d P Y ψ( tifi) : ψ(f1)...ψ(fn) := : e : dt ti=0 i=1 i Using(5.9), prove n Z X Y : ψ(f1)...ψ(fn):: ψ(g1)...ψ(gm) : dνC = δn,m (fi, Cgπ(i)) (5.10) π∈Sn i=1 • Take Ω = 1 and conclude from (5.10) that n Y ∗ 1/2 : ψ(f1)...ψ(fn) : Ω = a (C fi)Ω i=1 ∞ From the above exercise, we can conclude that if {gi}i=1 is an orthonormal basis of 2 d−1 L (R ), then n Y −1/2 {: ϕ(C giα ):Ω | n ≥ 0, iα ∈ N} α1 2 is an orthonormal basis of L (νG).

54 Chapter 6

Perturbation Theory

The theory of measures on distribution spaces as devolped in the previous chapter gives us the tools to finally address the question we posed in the beginning: how can we at- tempt construct a quantum field theory of interacting particles. In physics, the interaction Hamiltonian is often small, so the usual tool is perturbation theory. Here we will develop a theory of interacting systems via non-Gaussian measures as perturbations of Gaussian measures. The first thing to do will be introducing a suitable regularization for ultraviolet divergences.

6.1 Regularization

6.1.1 Momentum space regularization 0 d Let us start with dµG(ϕ) being the Gaussian measure on S (R ) with covariance G = (−∆ + m2)−1. We recall that the integral kernel of G is given by Z eip(x−y) ddp G(x, y) = 2 2 d ≡ G(x − y) Rd p + m (2π) d To make things easier, we start with finite volume Λ ⊂ R . Then we try to add interaction terms that make the measure non-Gaussian. In particular, consider −λ R ϕ(x)4ddx e Λ dµG(ϕ). As before, we immediately run into trouble: ϕ(x)4 is not a well defined random variable. R 4 2 2 Indeed, ϕ(x) dµG = 3G(x, x) = 3G(0) = ∞. In fact, we only know so far that ϕ(f) = R d R n ϕ(x)f(x)dx where f ∈ S(R )are random variables with moments ϕ(f) dµ < ∞, ∀n. d One way to proceed is to regularize the theory. Since G(0) = R Gˆ(p) d p with Gˆ(p) = Rd (2π)d 1 ˆ p2+m2 the divergence is due to insufficient decay of G(p) as |p| → ∞: this divergence is an UV (ultraviolet) divergence.1 Thus, let us multiply by some rapidly decaying function and replace Gˆ(p) by Gˆ(p) = Gˆ(p)χ(p) ≡ Gˆ(p)χ(p) d where χ ∈ S(R ) cuts of large |p|. We demand χ(0) = 1 and χ ≥ 0. For example, −p2 χ(p) = e is a good choice. As → 0, Gˆ(p) → Gˆ(p) pointwise, so we will always keep this limit in mind. Now ϕ(x) is a well defined random variable2: Z (2n)! Z ddp ϕ(x)2ndµ = G (0)n G (0) = Gˆ(p)χ(p) < ∞ G 2nn! (2π)d

1Note that we are working in d > 1: we need at least 1 space and 1 time dimension. The physical dimension is 4. 2 To be pedantic : a priori we know only that ϕ(f) is measurable for f ∈ S, so consider instead ϕ(f,x) (x−y)2 1 − 0 with f (y) = e 2 . This is a measurable function ` : S → : ϕ → ϕ(f ). Thus lim ` ,x (2π)d/2 R ,x →0 R R also is measurable and this is what we call ϕ(x). Formally : ϕ(f,x) = ϕ(y)f,x(y)dy → ϕ(y)δ(x−y)dy = ϕ(x) as → 0.

55 Hence, consider the measure

1 R 2 4 d − Λ[aϕ(x) +λϕ(x) ]d x dν(ϕ) := e dµG (ϕ) (6.1) ZΛ, For later purpose, we added a quadratic term (possibly depending on ) to the action and multiplied by a normalization constant (partition function)

Z R 2 4 d − Λ[aϕ(x) +λϕ(x) ]d x ZΛ, = e dµG (ϕ) to obtain a probability measure. This measure is well deﬁned provided a ∈ R, λ > 0. Indeed, by Jensen’s inequality we get

R R 2 4 d − ( [aϕ(x) +λϕ(x) ]d x)dµG 2 ZΛ, ≥ e Λ = exp −|Λ|(aG(0) + 3λG(0) )

2 a2 a 2 4 +|Λ| 4λ and since aϕ + λϕ ≥ − 4λ , Z ≤ e . We can then pose the problem of interacting QFT by trying to ﬁnd the parameter 3 (mass renormalization) a such that the measures ν converge as → 0, or, equivalently R iϕ(f) d that S(f) = e dν converges to a function of positive type on S(R ). We could also consider the correlation functions

n n Y Z Y G(x1, . . . , xn) = h ϕ(xi)i := ϕ(xi)dν(φ) i=1 i=1 and inquire their limits as → 0.

6.1.2 Scaling limit The parameter used above is a small scale cutoff4: momenta |p| > −1 are suppressed so in x-space the fields are smooth on scales < . It will be useful to make a change of variables so that the smallest scale becomes 1. Define

d−2 φ(x) = 2 ϕ(x)

Then Z d−2 φ(x)φ(y)dµG = G((x − y)) Z ip(x−y) d d−2 e χ(p) d p = 2 2 d (6.2) Rd p + m (2π) Z eip(x−y)χ(p) ddp = 2 2 d (6.3) Rd p + (m) (2π) :=G()(x − y) (6.4)

So we see that φ has covariance with UV cutoﬀ 1 and mass parameter m. The non-gaussian measure becomes

1 R 2 2 4−d 4 d () − −1Λ[ aφ(x) + λφ(x) ]d x dν (φ) := e dµG() (φ) ZΛ,

Note that in these dimensionless variables the coupling constant is 4−dλ and tends to zero as → 0 in d < 4. The volume in turn expands to −1Λ. We have then

(2−d)m () G(x1, . . . , xn) = G (x1/, . . . , xn/) (6.5)

3As we’ll see later in d = 4 other renormalizations are needed 4Recall that small scales mean large momenta, while large scales means low momenta. Hence UV divergences correspond to small scales while IR divergences are due to large scale eﬀects.

56 where 2m Z 2m () Y () Y () G (y1, . . . , yn) = h φ(yi)i = φ(yi)dν (φ) i=1 i=1 Hence the QFT correlation functions are the long distance scaling limits of the unit cutoﬀ theory.

6.1.3 Lattice Regularization This becomes very evident if we use lattice regularization where the ﬁeld ϕ lives on a 0 d lattice of spacing . Consider the Gaussian measure µC on s ((Z) ) with covariance −1 2 d C = (−∆ + r) where the latter is an operator on ` (Z ) i.e. X −2 (−∆ϕ)(x) = [ϕ(x) − ϕ(x + i)] |i|=1

d for ϕ :(Z) → R. Concretely 1 Z eip(x−y) ddp (x, y) = −2 d := C(x − y) −∆ + r π π d µ(p) + r (2π) [− , ]

P 2 pi where µ(p) = i 4 sin 2 , which shows that p is cut-oﬀ to |pi| ≤ π/. This is also intuitively clear: on a lattice, the shortest possible wavelength is twice the lattice spacing k ≥ 2, giving momenta bounded by p ≤ 2π = π . kmax On the lattice, the regularized non-Gaussian measure is then deﬁned as

P d 2 4 1 − d (aφ(x) +λφ(x) ) x∈Λ∩(Z) dν(ϕ) = e dµC (φ) Z˜Λ, R 2 4 where we have used the Riemann sum approximation for the integral Λ(aφ + λφ ). The scaling limit is the defined by introducing the same unit cutoff field

d−2 d φ(x) = 2 ϕ(x) , x ∈ Z so that the lattice correlation function becomes: Z ip(x−y) d () 1 e d p C (x − y) = 2 (x, y) = 2 d −∆1 + (m) [−π,π]d µ(p) + (m) (2π) while the measure becomes 1 P 2 2 4−d 4 () − x∈(−1Λ)∩ d ( aφ(x) + λφ(x) ) dν (φ) = e Z dµC() (φ) Z˜Λ, Thus our UV problem → 0 is the same thing as staying on a ﬁxed lattice, taking 2 distances to inﬁnity (xi/), and going to critical point i.e. zero mass limit ( r → 0). It should now be obvious that both the IR and UV problems are related to Statistical Mechanics at the critical point.

6.2 Gaussian critical point

To make the correspondence between statistical mechanics and QFT precise, we should check that we can recover the continuum correlation functions from the lattice models. Let us now study the Gaussian Ginzburg-Landau-model given by the Gaussian measure d 0 d µC on RZ (or s (Z )) with covariance (we call the mass squared parameter r in what follows) Z Z eip(x−y) ddp C(x − y) = φ(x)φ(y)dµC (φ) = d (6.6) [−π,π]d µ(p) + r (2π) d d where x, y ∈ Z , φ ∈ RZ . First, we see that we have a ﬁnite correlation length ξ

57 Proposition 10. For this model, we have

0 ≤ C(x − y) ≤ Ce−|x−y|/ξ where ξ > 0 and ξ → ∞ as r → 0

Proof. Fix x and y and let R = max |xi−yi|. We may assume that R = |x1−y1| = x1−y1. i=1...d Use the deﬁnition of µ and split the integral as ! Z Z π 1 dp dd−1~p ip1R 1 i~p·(~x−~y) C(x − y) = e e d−1 . −π 2(1 − cos p1) + µ(~p) + r 2π (2π)

Then we can use Cauchy for the p1 integral for the following contour:

ip R The vertical pieces cancel since cos p1, e 1 are periodic ! Z Z π 1 dp dd−1~p −R ip1R 1 i~p(~x−~y) = e e e d−1 −π 2(1 − cos(p1 + i)) + µ(~p) + r 2π (2π)

Choose so that : Re2(1 − cos(p + i)) + r > 0 i.e. 2(1 − cos p cosh ) + r > 0 i.e. 1 1 √ 2 cosh < 2 + r or < cosh−1(1 + r/2). [So, for r small, one can choose ∼ r ]. So, since µ(p) ≥ 0,

Next, we take the long-distance limit L → ∞ and try to recover the continuum quantum ﬁeld theory. We consider the massless and massive case separately.

6.2.1 Massless case Set r = 0, and write the covariance (6.6) in the scaling form:

ip x Z |x| d 2−d e d p 2−d C(x) = |x| 2 d ≡ |x| h(ˆx, |x|) [−π|x|,π|x|]d x µ(p/|x|) (2π) wherex ˆ = x/|x|, provided d > 2.

Excercise 15. Prove that √ 2 Z eip1 ddp Z e− ~p dd−1~p h(ˆx, |x|) −→ = |x|→∞ 2 d p 2 d−1 Rd p (2π) Rd 2 ~p (2π) Z ∞ Ad−1 −r d−3 Γ(d − 2)Ad−1 = d−1 e r d = d−1 ≡ Cd < ∞ (6.7) 2(2π) 0 2(2π)

d−2 for d > 2, where Ad−1 is the volume of S .

We can state this calculation for r = 0 in two ways :

58 • We showed C(x) ∼ G(x) as |x| → ∞: the correlation goes to the 2-point function 1 of the massless free ﬁeld: Gb(p) = p2 . Note that Gb, for d > 2, deﬁnes a continuous − 1 (f,Gf) function e 2 on S : We have Z d d Cd (f, Gg) = d xd y d−2 f(x)g(y) ≤ Ckfkmkgkm Rd×Rd |x − y| for m large enough For d = 2, see below.

• Consider the function

Ld−2C(Lx − Ly) = Ld−2hφ(Lx)φ(Ly)i Z ip(Lx−Ly) d Z ip(x−y) d d−2 e d p e d p = L d = 2 d [−π,π]d µ(p) (2π) [−Lπ,Lπ]d L µ(p/L) (2π)

Formally, this converges

Z eip(x−y) ddp L−→→∞ 2 d = G(x − y), Rd p (2π) since L2µ(p/L) → p2 as L → ∞.

This is indeed the scaling limit we introduced before: if we take the statistical mechan- d−2 ics model at the critical point, look at long distances, scale φ by L 2 as above, we get d−2 quantum ﬁeld theory. Speciﬁcally, if we take xi 6= xj i 6= j, and ϕL(x) = L 2 φ(Lx) (for −1 d x ∈ (L Z) ), then we get in this Gaussian model:

2N 2N Z Y Z Y ϕL(xi)dµC (φ) L−→→∞ ϕ(xi)dµG(ϕ). 0 0 s i=1 S i=1

d Here the LHS is with a measure deﬁned on variables φ(x), x ∈ Z , a lattice model and the d RHS with a measure deﬁned on variables ϕ(x) x ∈ R , a continuum model.

6.2.2 Massive case −1 2 d Let r > 0 and set Cr = (−∆ + r) on ` (Z ). So,

ip x Z |x| d 2−d e d p Cr(x) = |x| 2 2 d . [−|x|π,|x|π]d |x| µ(p/|x|) + |x| r (2π)

In this case, we should divide the integral into two parts. Let r be small, so for |x| < r−1/2, we have 2−d Cr(x) ∼ |x| For |x| > r−1/2 scale diﬀerently substitute p → r1/2p, wich gives

Z ipr1/2x d d−2 e d p d−2 −r1/2|x| C (x) = r 2 ∼ r 2 e r −1 1/2 d π π d r µ(r p) + 1 (2π) [− r , r ] Stated diﬀerently, let us put r = L−2m2 and consider, after the change of variable pm → p,

Z ipx d d−2 e d p L Cm2/L2 (Lx) = 2 2 d [−Lπ,Lπ]d L µ(p/L) + m (2π) Z eipx ddp L−→→∞ 2 2 d ≡ Gm2 (x) Rd p + m (2π)

59 Hence Z 2N Z Y d−2 Y L 2 φ(Lx )dµ (φ) → ϕ(x )dµ (ϕ) i Cm2/L2 i Gm2 0 0 s i=1 S So if we approach the critical point r → 0 in a suitable way and scale the distances and φ accordingly, we get the continuum theory at nonzero m so with correlation length < ∞. Remark All the scalings are completely natural. Formally:

1 − 1 (φ,(−∆+r)φ) Y dµ (φ) = e 2 dφ(x) Cr Z x∈Zd

2−d and φ(x) = L 2 ϕL(x/L) gives X (φ, (−∆ + r)φ)) = φ(x)((−∆ + r)φ)(x)) x∈Zd X h X i = φ(x) (φ(x) − φ(x + u)) + rφ(x) x∈Zd |u|=1 X x h X x x u x i = L2−dϕ ( ) (ϕ ( ) − ϕ ( + )) + rϕ ( ) L L L L L L L L L x∈Zd |u|=1 u X h X ϕL(y) − ϕL(y + ) i = L−dϕ (y) L + L2rϕ (y) L L−2 L 1 d |u|=1 y∈( L Z) which, if ϕL went to a smooth function as L → ∞, Z 2 d −1 −→ ϕ(y)(−∆ + m )ϕ(y)d y = (ϕ, Gm2 ϕ), Rd L → ∞r = m2/L2

6.3 Perturbation Theory for the GL model

6.3.1 Measures Let us consider the λ 6= 0 Ginzburg-Landau model. There are various ﬁnite volume d theories that we can consider : Let Λ ⊂ Z be an L-cube.

• The measures 1 X 4 exp − λ φ(x) dµCΛ (φ) ZeΛ x∈Λ

|Λ| −1 on R where µCΛ is Gaussian with covariance CΛ = (−∆Λ + r) and ∆Λ is the d Laplacian in Z with some boundary conditions on ∂Λ (we have considered periodic and free above; one can consider Dirichlet or Neumann on the lattice as well). ZeΛ normalizes the total measure to 1.

• The measure 1 X 4 exp − λ φ(x) dµC (φ) ZΛ x∈Λ

0 d −1 on s (Z ) when µC is Gaussian with covariance C = (−∆ + r) . Thus µC is in inﬁnite volume and the φ4 perturbation is in ﬁnite volume.

As in Section 6.1.1 using Jensen’s inequality we get a strictly positive lower bound for the partition functions.

60 6.3.2 Perturbation Theory It is possible to prove quite generally, using Ising-model approximations and various corre- d lation inequalities that the Λ → Z limit of the above measures exist. These are, however, special tricks and we want to understand these issues eventually using the renormalization group. But let us for now proceed more heuristically and study the correlation functions perturbatively in λ. Consider e.g. the pair correlation

R X 4 φ(x)φ(y) exp − λ φ(x) dµC (φ) x∈Λ hφ(x)φ(y)iΛ ≡ ≡ G2(x, y) (6.8) R X 4 exp − λ φ(x) dµC (φ) x∈Λ

For |Λ| < ∞, it is easy to show this correlation function is C∞ in λ ∈ [0, ∞) (use the fact P 4 that the random variable V = x∈Λ φ(x) has all moments in the measure µC ). Thus, we have Taylor’s expansion at λ = 0:

N X n G2(x, y) = G2,n(x, y)λ + RN n=0

−N where lim λ RN (λ) = 0. The Taylor coeﬃcients G2,n have a nice graphical representa- λ→0 tion which we now derive.

6.3.3 Feynman graphs

N Denote (6.8) by D and let us start with small n R n=0 We have N = φ(x)φ(y)dµC + O(λ) and D = 1 + O(λ) giving G2,0 = C(x − y). So the zeroth order gives us just our Gaussian covariance. n=1 Z X 4 N = C(x − y) − λ φ(x)φ(y)φ(z) dµC z∈Λ and Z X 4 D = 1 − λ φ(z) dµC . z∈Λ From our rules of Gaussian integrals (in particular the Wick formula) we get: Z 4 2 φ(z) dµC = 3C(0) where the factor 3 comes from the three pairings. The integral Z 4 φ(x)φ(y)φ(z) dµC

6! has 23·3! = 15 pairings and it equals

3C(x − y)C(0)2 + 12C(x − z)C(y − z)C(z − z)

So, X N = C(x − y)(1 − 3λ|Λ|C(0)2) − 12λ C(x − z)C(y − z)C(0) z∈Λ and D = 1 − 3λ|Λ|C(0)2

61 Hence altogether X G2,1 = −12λC(0) C(x − z)C(z − y). z∈Λ To proceed further and get results for higher n, we need to introduce some notation. Clearly, we need to calculate expressions like Z N Z N Y 4 Y 4 φ(x)φ(y) φ(zi) dµC and φ(zi) dµC . i=1 i=1 P Q Both are given as sums of products of pairings P hαβi C(uα − uβ) where {uα} are the points x, y, zi, however such that each zi occurs 4 times. More precisely, let u4(i−1)+j = zi for i = 1,...,N, j = 1, 2, 3, 4 and u4N+1 = x, u4N+2 = y. Then the sum is over pairings of the set 1,..., 4N + 2. The power of Feynnman graphs is that they give an easy diagrammatic expression for each pairing. To a general pairing P we associate a graph G(P ) as follows : G = (V, L) consists of a set V of vertices and a set L of lines. Vertices are the set {x, y, z1, ··· , zN } = V for the ﬁrst case and {z1, ··· , zN } for the second. Lines ` join vertices : ` = {u`1 , u`2 } where u`1 , u`2 ∈ V and u`1 = u`2 is allowed. Let Γ2(V ) be the set of such graphs G such that each vertex zi belongs to four lines (where we count lines {zi, zi} twice) and each vertex x, y belongs to one line. 5 Given a graph G satisfying these conditions, there may be several pairings P such that G(P ) = G. Call the number |{P |G(P ) = G}| ≡ n(G). Then Z N Y 4 X Y φ(zi) φ(x)φ(y)dµC = n(G) C(u`1 − u`2 ) (6.9)

i=1 G∈Γ2(V ) `∈G

4 Example. In hφ(z) φ(x)φ(y)i we had 2 graphs, G1 has lines {z, z}, {z, z}, {x, y} and G2 lines {x, z}, {y, z}, {z, z}. n(G1) = 3, n(G2) = 12.

Now, once we sum (6.9) over the points z1, . . . , zN , the relative order of the points zi will not matter anymore. So we get ﬁnally the following graphical rules which we state for arbitary correlations : Deﬁnition 20. A φ4- graph with 2m external legs, labeled 1, ··· , 2m, and N unlabeled vertices is a graph on N points (vertices) such that each vertex has 4 lines attached (one line can start and end at a vertex and is counted as two here) and 2m lines have one end with no other lines attached.

Some examples of φ4 Feynman graphs are given in ﬁgures 6.1 and 6.2 With this notation, we have

n Z 2m ! d Y X 4 n X φ(xi) exp −λ φ(x) dµC (φ) = (−1) n(G)A(G) dλn λ=0 i=1 x∈Λ G where G runs through all the (2m, n)-graphs, n(G) is the number of pairings giving the same G. The amplitude A(G) corresponding to the graph G is the following expression:

• Label the n vertices of G as z1, ··· , zn

• To each line ` = {u, v} of G put C` = C(u − v) X Y A(G) = C`

z1,··· ,zn∈Λ ` 5 4 Note the fact that each vertex connects to four lines comes from the fact that we have the factor φ(zi) 3 in the Taylor expansion. If our measure would, for example a term φ(zi) , verteces could connect only three lines.

62 2m=2, N=1

2m=2, N=2

2m=2, N=3

2m=4, N=1

2m=4, N=2

Figure 6.1: Examples of Feynman graphs

, ,

Figure 6.2: More examples of Feynman graphs

As shorthand notation, we denote A by .

Example: We get to O(λ):

− λ · 4 · 3 − λ · 3(− ) + O(λ2) G2 = 1 − λ · 3 + O(λ2) = − 12λ − 3λ − + 3λ − + O(λ2)

= C − 12λ + O(λ2) X 2 = C(x − y) − 12λ C(x − z1)C(z1 − z1)C(z1 − y) + O(λ ) z1 X 2 = C(x − y) − 12λC(0) C(x − z1)C(z1 − y) + O(λ )

z1∈Λ

Note the presence of disconnected graphs like that are cancelled upon normalization. For example, there are no disconnected graphs of order 2 or less in the two-point function:

Excercise 16. Prove that

2 3 G2 = − 12λ + λ α + β + γ + O(λ ) and ﬁnd α, β, γ

This statement is true in general: correlation functions only contain connected graphs:

63 Theorem 8.

Z P 4 φ(x)φ(y)e−λ φx(x)dµ dn C n X c n Z = (−) n(G)A(G) dλ λ=0 −λ P φ4(x) e dµC G

X where the sum c runs over connected graphs.

Note that in the numerator the graphs are of the type

· ·

| {z } connected while in the denominator we get

· ·

So connected components have either both x and y or neither.

Note that disconnected graphs are proportional to a power of the volume :

X = C(x − y) C(0)2 = |Λ|C(0)2C(x − y) z∈Λ

= C(x − y)C(0)4|Λ|2 etc

Proof. Introduce the notation :

P 4 R F (φ)e−λ φ dµ hF i = C λ R −λ P φ4 e dµC R so hF i0 = F dµC . Then, d X X hF i = −λ hφ(x)4F i + λ hφ(x)4i hF i dλ λ λ λ λ x x d X so hF i = −λ hφ(x)4; F i where we denote by dλ λ λ x

hF ; Giλ = hFGiλ − hF iλhGiλ (6.10) the so called truncated (or connected) correlation functions. For these we get the result Y Y Lemma 7. Let F = φ(xα),G = φ(xβ). α∈A β∈B X Y Then hF ; Gi0 = C(xγ −xδ) where each pairing P has at least one pair {γ, δ} P {γ,δ}∈P such that γ ∈ A, δ ∈ B.

64 A A 4 2 A 3

Figure 6.3: Example of a connected path

Proof. Obvious, since hF ihGi has those pairings where no such pairs occur and hFGi has all pairings.

We can generalize the above to n d n X 4 4 4 hF iλ = (−1) hF ; φ(x1) ; φ(x2) ; ··· ; φ(xn) i0 dλn λ=0 x1···xn where X |π|−1 Y Y hF1; F2; ··· ; FN i = (−1) (|π| − 1)! h F i0

π α i∈πα |π| where π = {πα}α=1 is a partition of {1, ··· ,N} into |π| subsets.

Example: hF1; F2; F3i = hF1F2F3i−hF1F2ihF3i−hF1F3ihF2i−hF2F3ihF1i+2hF1ihF2ihF3i.

Excercise 17. Deﬁne the free energy FΛ(λ) = − log ZΛ(λ) where ZΛ(λ) is the partition function. Show that dn X F (λ) = (−1)n+1 hφ(x )4; φ(x )4; ··· ; φ(x )4i dλn Λ 1 2 n λ x1···xn Now the thing to check is Y Lemma 8. Let Fi = φ(xiα ). Then α∈Ai X Y hF1; ··· ; FN i0 = C(xγ − xδ) P {γ,δ}∈P

N−1 where for each partition into pairs P , there is a {{γi, δi}}i=1 ⊂ P that forms a connected path, connecting the Ai’s, i.e. there exists a permutation ρ of the Ai’s such that

γ1 ∈ Aρ(1), δ1 ∈ Aρ(2), γ2 ∈ Aρ(2), δ2 ∈ Aρ(3) ··· δN−1 ∈ Aρ(N). An example os such a path is given in ﬁgure 6.3 This lemma yields our theorem since there cannot be a connected graph with only one leg.

The discussion on the two-point functions can easily be generalized to a general N- point function: Q4 Excercise 18. • Let G4(x1 ··· x4) = h i=1 φ(xi)i and deﬁne the connected four point c function G4(x1 ··· x4) = hφ(x1); ... ; φ(x4)i. Prove that

G4(x1 ··· x4) = G2(x1 − x2)G2(x3 − x4) + G2(x1 − x3)G2(x2 − x4) c + G2(x1 − x4)G2(x2 − x3) + G4(x1, ..., x4)

65 z3

z 2

y x z1

Figure 6.4: Example of the tree graph included in a loop graph

c PN n c N+1 c Show that G4 has an expansion n=1 λ G4,n + O(λ ) with G4,n consisting of connected graphs with 4 legs.

• Generalize the previous point to the connected N-point function

c GN (x1 ··· xN ) = hφ(x1); ... ; φ(xN )i

by showing that X Y c GN (x1 ··· xN ) = G|I|(xI ) π I∈π c and that the perturbation expansion of GN consists of connected graphs with N legs.

• . Let Z(J) := he(φ,J)i.

Hence N N Y Y GN (x1, . . . , xN ) := h φ(xi)i = ( ∂/∂J(xi))|J=0Z(J). i=1 i=1 Deﬁne F (J) := log Z(J). Show

N Y ∂ Gc (x , . . . , x ) = ( )| W (J). N 1 N ∂J(x ) J=0 i=1 i

dn • Prove that n log Z is sum of connected vacuum graphs with n vertices. (a vacuum dλ 0 graph is a graph with no legs).

Many interesting physical quantities in quantum ﬁeld theory can be calculated from the amplitudes corresponding to the Feynman graphs we just derived. So it is useful to consider how to calculate the amplitude A(G) of a given graph G in practice. First, let d us observe that in our model G2m,n has a Λ → Z limit : from the previous discussion, it clearly suﬃces to consider connected graphs which are expressions (let say m = 1)

X A(x1, x2, z1 ··· zn)

z1···zn∈Λ

−|x|/ξ and A is a product of C(yi − yj) where the y’s are x’s or z’s. Use |C(x)| ≤ A e where A is some constant, and note that, since G is connected, there exists a connected tree graph in G containing all the vertices and all the legs (a tree graph is a graph without loops). For example, see ﬁgure ?? Now do the z sums using | P C(z − z )| ≤ const (independent on Λ) starting at i zj ∈Λ i j ends of branches (e.g. z3 above).

66 6.3.4 Momentum space representation Actual calculations are easier in Fourier transform space or momentum space. In order d not to worry about analysis let us ﬁrst work in ﬁnite volume Λ = ΛL = ZL. Recall that d 2π d we have the Fourier transform for p ∈ BL = [−π, π] ∩ ( L Z) X fˆ(p) = e−ipxf(x).

x∈ΛL and the inverse formula X Z f(x) = L−d eipxfˆ(p) ≡ dp eipxfˆ(p) L p∈BL R where L dp is a convenient shorthand for the Riemann sum, converging as L → ∞ to R Q dpi [−π,π]d dp where dp is the normalized Lebesgue measure i 2π . We have Z ip(x−y) dp e = δxy for x, y ∈ ΛL L and X i(p−q)x d d e = L δpq ≡ (2π) δL(p − q) p, q ∈ BL.

x∈ΛL where we deﬁned the discrete delta function δL(p − q). Products work out as Z (fg)ˆ(p) = fˆ∗ gˆ := dqfˆ(p − q)ˆg(q) L and (f ∗ g)ˆ = fˆg.ˆ

Example

X 3 3 = C(x − z1)C(z1 − z2) C(z2 − y) = (C ∗ C ∗ C)(x − y) z1z2

Hence using above rules

Z = G(x − y) = dp Gˆ(p)eip(x−y) L with Z Z Gˆ(p) = dq dr Cˆ(q)Cˆ(r)Cˆ(p − q − r) Cˆ(p)2 L L or pictorially q r p p p-q-r

So the amplitude corresponding to this graph is the product of ﬁve factors Cˆ, one corresponding to each line of the graph (including the two external lines), where we integrate over the free momenta in the loop, with momentum conservation at the vertices.

67 A more useful way to derive the same result is to insert for each C its Fourier representation: Z 5 X i[p1(x−z1)+p2(z1−z2)+p3(z1−z2)+p4(z1−z2)+p5(z2−y)] Y = e Cˆ(pi)dpi L z1z2 i=1 Z 5 d d ip1x−ip5y Y = (2π) δ(−p1 + p2 + p3 + p4)(2π) δ(−p2 − p3 − p4 + p5)e Cˆ(pi)dpi i=1

The ﬁrst delta-function is due to the sum over z1 and the second to the sum over z2. Note that also here momentum is conserved at the vertices:

p2 p2 p1 4 4 p X X 3 p3 p5 p1 = pi , p5 = pi i=2 i=2 p p4 4 Solving for the constraints we get again Z ip1(x−y) = dp1dp2dp3Cˆ(p1)Cˆ(p2)Cˆ(p3)Cˆ(p1 − p2 − p3)Cˆ(p1)e L using p5 = p1, p4 = p1 − p2 − p3. These expressions now have the obvious limits as L → ∞ R R with L replaced by .

Feynman rules The general procedure to calculate the amplitude for an arbitrary graph, known as the Feynman rules, is now obvious : consider a general graph

pn p1

6 P • Put momentum pi to legs , such that pi = 0 The physical reason for this is of course overall momentum conservation. Mathe- matically this is because G(x1, . . . , xn) is translation invariant :

G(x1, ··· , xn) = G(x1 + y, x2 + y, . . . , xn + y) for all y so P X i pixi Gˆ(p1, . . . , pn) = G(x1, x2, . . . , xn)e i x1,...,xn P X i pixi = G(0, x2 − x1, . . . , xn − x1)e i x1,...,xn Pn P X i pixi ix1 pi = G(0, x2, . . . , xn)e i=2 e i x1,...,xn d X =g ˆ(p2, . . . , pn)(2π) δL( pi)

where g(x2, . . . , xn) = G(0, x2 ··· , xn).

6All momenta should point either inwards or outwards. Our convention is that momenta are always directed inwards, outwards directed momenta get a minus sign

68 • Give each line a propagator Gˆ(pi)

• Impose momentum conservation at each vertex (coming from the delta functions).

• Integrate over each independent internal momentum.

Example Consider the diagram from ﬁgure ??. The associated amplitude is given by

3 Z Y Cˆ(p1)Cˆ(−p1 − p2 − p3)Cˆ(p2)Cˆ(p3) dqiCˆ(q1)Cˆ(q2)Cˆ(q3) i=1

× Cˆ(p1 − q1 − q2)Cˆ(q1 + q2 − p1 − p2 − p3)Cˆ(p1 + p2 − q1 − q2 − q3) := Γ(p1, p2, p3) so in x space the graph is

Z 3 Y dpi G(x , ··· , x ) = ei(p1(x1−x4)+p2(x2−x4)+p3(x3−x4))Γ(p , p , p ) 1 4 1 2 3 (2π)d i=1 We have seen so far that

∞ • If |Λ| < ∞, the functions G2m are C in λ

(Λ) d • The Taylor coeﬃcients G2m,n have a Λ → Z limit.

d • It can be shown that if r > 0 and λ is small enough the functions G2m have a Λ → Z ∞ (Λ) limit, they are C in λ and their Taylor coeﬃcients are the lim G2m,n. Λ→Zd Note, however that the expansion is not convergent. Consider for example a lattice consisting of one point giving the integral

Z ∞ −λϕ4− r ϕ2 F (λ) = e 2 dϕ −∞

Then n Z √ d F n 4n − r ϕ2 −2n n (4n)! n αn = = (−) ϕ e 2 dϕ = r (−1) 2π ∼ C (2n)! dλn λ=0 22n(2n)!

P 1 n αn n as n → ∞. Thus the Taylor series n! αnλ has n! ∼ C n! so the series is very badly divergent.

Excercise 19. Prove that F (λ) is analytic in the region C\{λ ∈ R, λ ≤ 0} and has an essential singularity at λ = 0.

6.4 Infrared and Ultraviolet Divergences

In the discussion leading up to this chapter, we hit ultraviolet and infrared divergences every time we tried to consider interacting theories. Now, we finally got finite results (although we found a divergent power series). However, these results were obtained using a (finite) lattice, so by taking the continuum and/orinfinite volume limit, we could still come across divergent results. In this section we will study possible divergences in some more detail.

69 p -p-p-p 1 1 2 3 q1

q2 p-q-q q +q -p-p-p 1 1 2 1 2 1 2 3 q3

p+p-q-q-q 12 1 2 3 p3 p2

Figure 6.5: A complex graph including multiple loops

6.4.1 Infrared divergences Recall that for λ = 0, the critical point of our model is at r = 0. What happens for λ > 0? The following is expected to be true :

• For λ > 0, there exists rc(λ) such that for r > rc(λ)

0 ≤ hφ(x)φ(y)i ≤ Ae−|x|/ξ ξ < ∞

with ξ(r) → ∞ as r ↓ rc(λ).

• For r < rc(λ) there are two phases, where hφ(x)i = ± m 6= 0 and

|hφ(x); φ(y)i |≤ Ae−|x|/ξ ξ < ∞,

with ξ(r) → ∞ as r ↑ rc(λ).

• For r = rc(λ), lim hφ(x)φ(y)i|x − y|d−2+η 6= 0 |x−y|→∞ where η is called the anomalous dimension and it is independent of λ > 0 but depends on the dimension d. For example, η = 0 if d ≥ 4, and for d = 2 and d = 3 it is the same η as in the Ising model.

These claims are proven for small λ and d ≥ 4. The rest remains a conjecture, although with plenty of theoretical and numerical evidence. Let us see how these facts are reﬂected in the behaviour of the perturbation theory when r = 0. The expansion for the two point function G2 starts as (see Homework 16)

2 3 G2 = − 12λ + λ (144 + 48 + 96 ) + O(λ )

Let us consider the momentum space expressions for these graphs at r = 0. Since all integrals are over a bounded region [−π, π]d, the only problem could come from p = 0 Pd 2 4 where µ(p) = 2 i=1(1 − cos pi) = p + O(p ) vanishes. The graphs are of the form

Cˆ(p)2K(p) where C(p)2 comes from the external legs. To lowest order we have from the ”tadpole graph” K(p) = −12λC(0). Since Cˆ(p)2 ∼ (p2)−2 as p → 0 the result is not integrable at 0 if d ≤ 4. The reason for this divergence is that the covariance (−∆ + r)−1 is not −1 −2 smooth in r at r = 0: ∂r(−∆ + r) = −(−∆ + r) which does not have an integrable

70 kernel as r → 0. Recall that we expect there to be a critical point r(λ) which tends to zero as λ → 0. We could thus try to add to the Hamiltonian in (6.8) a ”counterterm” 4 4 2 λφ → λφ + rc(λ)φ with X n rc(λ) = rnλ (6.11) n and determine the coeﬃcients rn so that the divergencies cancel. Lets see what this means to the lowest orders. To O(λ) we take r1 = −6λC(0) which cancels all the tadpole contributions in the graphical expansion. After this the two point function becomes

3 G2 = + 96 + O(λ )

To second order we have left the graph which we write in momentum

2 2 space as λ Cˆ(p) Kˆ2(p). Suppose we had a Taylor expansion

2 Kˆ2(p) = Kˆ2(0) + O(p ). (6.12)

1 2 ˆ ˆ 2 ˆ Then taking the counterterm r2 = 2 λ K2(0) the net eﬀect is that C(p) K2(p) is replaced 2 2 by Cˆ(p) (Kˆ2(p) − Kˆ2(0)) ∼ 1/p which cures our problem. However this approach fails in d < 4. Indeed, Z Z −1 −1 −1 −2 −2 −2 Kˆ2(0) = µ(k + q) µ(k) µ(q) dqdk ∼ (k + q) k q dqdk which diverges for d ≤ 3 (show!). It is instructive to formally consider the dimension taking non-integer values between 3 and 4. Consider then the following graphs q

p p–q .

This graph is a chain consisting of n links, each of which has amplitude

q Z ddq 1 1 = d ≡ I(p) [−π,π]d (2π) µ(q) µ(p − q) p-q

This is integrable in d > 2 if p 6= 0 and behaves as  |p|d−4 d < 4  I(p) ∼ log |p| d = 4 (6.13) constant d > 4

Hence the above graph has amplitude

Z I(q)n ddq µ(p − q) (2π)d which is ill-deﬁned for d < 4 and n large. Hence the perturbative approach fails formally for all d < 4.

71 At d = 4, Kˆ2(0) is ﬁnite and its second derivative at p = 0 diverges only mildly (show!):

2 ∆pKˆ2(p) ∼ C log p as p → 0.

In higher orders (like the chains above) powers of logarithms appear. In d > 4 (6.12) holds and, indeed, our program can be carried out to all orders in λ, that is, the coefficients rn in (6.11) can be chosen so that the resulting Taylor coefficients in λ for all correlation functions are finite. d = 4 is a special borderline case as we will see later. As a clue for what comes out easily from the renormalization group analysis let us consider the four point function and the following graphs including the bubble graphs

At d = 4 we can formally sum the series:

∞ X λ (−2I(p))nλn+1 = := λ(p) 1 + 2λI(p) n=0 and call the sum the strength of the interaction at momentum scale p. In particular it is a contribution to the connected four point function and thereby measures the deviation of the four point function from the gaussian expression (sum of products of two-point functions). We note that λ(p) → 0 as p → 0 which indicates that in large length scales (small momenta) our theory becomes gaussian! This ﬁshy argument actually has a grain of truth in it and indeed in d = 4 the theory becomes gaussian in large scales as we’ll see using the renormalization group. It turns out that for d < 4 the theory will be strongly non gaussian in large scales invalidating completely the perturbative approach.

6.4.2 Ultraviolet Consider now (continuum) ﬁeld theory and in particular the cutoﬀ measure (6.1)

1 R 2 4 d − Λ[aϕ(x) +λϕ(x) ]d x dν(ϕ) := e dµG (ϕ) ZΛ,

For > 0 the correlation functions are again C∞ in λ and a, and we have the expansion

N X n p N+1 N+1 G2k = λ a G2k,n,p + RN RN = O(λ , a ) n,p=0

2 d given in terms of graphs. The coeﬃcients, G2k,n,p for m > 0, have a |Λ| → R limit but not an → 0 limit, as we saw above. Hence, how to proceed? We have seen already earlier that one can try to add counterterms to the action to cancel divergencies. Hence consider the computation of G2 in Homework 16:

2 3 G2 = − 12λ + λ (144 + 48 + 96 ) + O(λ )

72 The ﬁrst loop graph in this sequence is called the tadpole graph G(0) and diverges as 2−d if → 0 in d > 2, and it diverges as log −1 in d = 2. To try to solve this divergence, let us take a = −6λG(0). Then, as in the previous section, we end up with

3 G2 = + 96 + O(λ )

2 The remaining graph (in momentum space) is Cˆ(p) J(p) with Z J(p) = Cˆ(p − k − l)Cˆ(k)Cˆ(l)dkdl.

−1 −1 J(p) converges in d = 2 whereas it diverges as log in d = 3 and as in d = 4. Next we note that if d = 3 the following limit exists poitnwise in p: Z (p − k − l)2 − (k + l)2 lim(J(p) − J(0)) = dkdl. →0 ((p − k − l)2 + m2)((k + l)2 + m2)(k2 + m2)(l2 + m2) (6.14)

Excercise 20. Show that this limit indeed exists in d = 3 and that is of the form

2 lim(J(p) − J(0)) = p K(p) →0 with K a bounded function. If we hence take 2 a = −6λG(0) + 48λ J(0) 2 we find that G2 has a → 0 limit to order λ . Hence we needed to add another diverging mass counterterm to the cutoff action in order to have a well defined (perturbative) limit for the correlation function up to order two. In fact one has the following general results:

2 Theorem 9. Let m > 0. There exist constants bi such that by setting a = λb1 log in −1 2 d = 2 and a = λb2 + λ b3 log in d = 3, the perturbation series has a limit order by order in λ as → 0 The proof of this is not very hard but the following is harder: Theorem 10. With the assumptions of the previous theorem the limit

lim lim Gn(x1, . . . , xn) Λ↑R3 →0

iϕ(f) exists as well as the limit limΛ↑R3 lim→0he i which deﬁnes a function of positive type 0 3 and thereby a measure on S (R ). 2 Actually the last theorem holds for all m ∈ R and all λ > 0. These two theorems basically solve the problem in two and three dimensions (we already discussed one dimension while studying Euclidean ﬁeld theory). However, we still do not have any results for the physical dimension d = 4. Now the limit (6.14) does not exist but diverges logarithmically in : we get that

2 lim J(p) − J(0) − p b log →0 exists. We thus need yet another counterterm, so called wave function renormalization. Indeed, this divergence is cancelled with the second order couterterm added to the action: Z 1 2 2 − 2 bλ log (∇ϕ(x)) dx. (6.15) Λ There is more to come however. Also the four point function diverges in the second order: the graph

73 diverges. This is the graph we encountered in the infrared problem (with r = 0 and UV cutoﬀ 1). The integral is given by Z L(p) = Cˆ(p − k)Cˆ(k)dk.

One gets lim(L(p) − L(0)) →0 exists where L(0) = d log + O(1). The divergence is then cancelled by (show!) adding the counterterm Z −18dλ2 log ϕ(x)4dx. Λ

Observe ﬁnally that the reference Gaussian measure dµG is formally given by

− 1 R ((∇ϕ(x))2+m2ϕ(x)2)dx dµG(ϕ) = e 2 Dϕ so that we can formally absorb the counterterm (6.15) to a rescaling of the ﬁeld:

− 1 ϕ → (1 − 2b log λ2) 2 ϕ. These considerations lead us to formulate the following Question: Find three functions Z(), a(), λ() such that m m Y lim lim Z() h ϕ(xi)i,Λ (6.16) Λ↑ d →0 R i=1 0 md m Qm d exists in S (R ) or in other words Z() h i=1 ϕ(fi)i, converge for all fi ∈ S(R )). Here h i is the expectation w.r.t. the limit of the measure

1 R 2 4 d − Λ[a()ϕ(x) +λ()ϕ(x) ]d x dν,Λ(ϕ) := e dµG (ϕ). ZΛ, These three functions are called the ﬁeld strength renormalization, bare mass and bare coupling constant respectively. Obviously λ() = 0, a() = 0,Z = 1 is a solution. The real question we ask is if there are any limits that are non-Gaussian. We have argued so far that

• For d = 2,Z = 1, a = −6λG(0), λ() = λ works.

2 • For d = 3,Z = 1, a = −6λG(0) + αλ log (α explicit), λ() = λ works. • For d = 4 the answer is more complex. The perturbative renormalization theory developed in the end of 40’s by Feynman, Schwinger, Tomonaga and Dyson leads to the following result: Theorem 11. There exist formal power series ∞ ∞ ∞ X n X n X n Z() = Zn()λ , a() = an()λ , λ() = λn()λ n=0 n=0 n=0 such that the coeﬃcients Gm,n() of resulting formal power series m ∞ m Y X n Z() h ϕ(xi)i,R4 = Gm,n()λ i=1 n=0 have nontrivial limits as → 0.

74 This should be contrasted with the following amazing result (Aizenman, Duminil-Copin 2018):

Theorem 12. Let Z() > 0, a() ∈ R and λ() > 0. Then the only possible limits (6.16) are Gaussian. This holds as well in d > 4.

Our next objective will be to understand these claims, but before that let us pause to comment on the origins of the counterterms. The question of continuum limit will not arise in statistical mechanics where the lattice spacing is a fixed nonzero quantity. There the subtle question is that of phase transitions: in the Ginzburg Landau model one looks for the existence of the critical point r(λ) where the correlation length diverges, and studies nontrivial scaling properties of the correlation functions at the critical point. However, the continuum limit question does arise in quantum field theory. In this case there is no fundamental length and one would like to have a quantum theory of continuum fields. The problem of divergences of the coefficients of perturbation theory arose in the 1930’s when physicists were studying Quantum Electrodynamics (QED), the quantum theory of electromagnetic field interacting with electrically charged matter. In this theory there are two sorts of fields, the electromagnetic field A(x) which is resembles our ϕ(x) but instead 4 takes values in A(x) ∈ R . A(x) is the four dimensional electromagnetic vector potential. It is a Gaussian field with zero mass, thus resembling our m = 0 free field. The second field describes electrons and has non-zero mass. The interaction term in the Hamiltonian makes the theory non-Gaussian; the analogue of our parameter λ is played by the electron charge e. One can then proceed to derive a perturbation series as above in powers of e and as above the individual terms are ill defined, diverging due to small scale (large momentum) behavior of the integrands. One would like to think about the electron charge (and its mass) as given physical constants that enter the action functional describing the theory. The divergence of the perturbation theory indicates that this point of view is incorrect. Rather one should think about these parameters depending on scale. Thus physicists introduced in the 50’s the idea of renormalization. We should think about the parameters entering -cutoff theory describing the charge and mass of the electron in that scale. These can very well be different from the ones we measure that are interpreted as the charge and mass of the electron in the measurement scale. Once we have introduced the renormalization group we will make this picture more precise. For the time being we just remark that from this point of view the -dependence of the parameters a and λ and Z is less surprising.

75 Chapter 7

The renormalization group

In the discussion about non-gaussian measures, we have encountered two problems where perturbation theory around a Gaussian measure is problematic: statistical mechanics (lattice) models at the critical point and the continuum limit of QFT. We also noted that the problem in both cases is related to scale invariance at the critical point. The tool to study this scale invariance is the renormalization group.

7.1 The Block-spin Transformation

Suppose that we have a Statistical Mechanics model at its critical point such that A B hφ(x)φ(y)i − ≤ (7.1) |x − y|a |x − y|a+ as |x − y| → ∞) for some > 0. We are after a theory that explains the leading term in this asymptotics and the fact that this term (or the exponent a) is universal: it turns out that it stays the same when the Hamiltonian is changed (at least under some changes). Thus suppose there is a class of Hamiltonians that satisfy (7.1). Then the details of these Hamiltonians affect only the sub leading asymptotics in the RHS of (7.1). We can get a special theory where the scale invariance satisfied by the leading term is exact by taking the scaling limit as we considered in section 6.1.2. Recall the relations (6.4) and (6.5) between cutoff quantum field theory and statistical mechanics. Let, for L > 0 a ϕL(x) = L 2 φ(x/L). (7.2) Then if (7.1) is satisfied we get

A lim hϕL(x)ϕL(y)i = . L→∞ |x − y|a

−1 d On the other hand ϕL can be seen as a cutoff 1/L field as it depends on x ∈ (L Z) . d Hence if the LHS is given by an expectation hϕ(x)ϕ(y)i the field ϕ lives on R . Thus we expect the limits Y lim h ϕL(xi)i L→∞ i if they exist, to be the correlation functions of a field theory. This field theory is called the scaling limit of our model. By definition it is scale invariant, for example:

Lahϕ(Lx)ϕ(Ly)i = hϕ(x)ϕ(y)i.

Universality then would mean that several Hamiltonians give rise to the same scaling limit. The above formulation has two serious drawbacks. First, the scaling limit is a diﬀerent object than the one we started with. The model we started with is a ﬁxed (unit) lattice

76 model whereas the scaling limit is a continuum object. More importantly, we have not given any constructive approach to the study of the scaling limit: we still need to understand the statistical mechanics at the critical point, in particular we need to show (7.1). The Renormalization group addresses these two problems. First, it supplements scaling by another operation, coarse graining, that allows one to stay in the category of fixed lattice spacing theories. Second, it provides actually a tool to study the critical theory. To develop this new theory, we replace the scale transformation (7.2) by a average φL(x) = L 2 φ (Lx) (7.3) where φaverage(Lx) is the average of φ in a L-sided cube centered at Lx: X φaverage(Lx) = L−d φ(Lx + y) (7.4) L |yi|< 2 d d (7.3) and (7.4) define a map φ ∈ RZ → φL ∈ RZ , i.e. the scaled field is also defined on d Z . We take for convenience L > 1 odd integer. For example, take L = 3. Then we have to average over •••

•• Lx •

••• Lx + y

We see that φ → φL involves: • “Coarse graining” : average over the details of φ for scales ≤ L.

a/2 • Scaling : φL(x) depends on φ near Lx and we multiply φ by L so it ﬁrst averages out the details and then rescales to same level of detail as before. φ → φL is called the Block-spin transformation.(7.3) and (7.4) deﬁne a linear map d d CL : RZ → RZ , concretely, ( a −d 1 |(Lx − Ly)i |< L/2 ∀i (CL)xy = L 2 (7.5) 0 otherwise Note that 2 CL = CL2 . Thus, the iteration of (7.4) n times is the same as doing it once with Ln : n φLn (x) = (CLn φ)(x) = (CLφ)(x) n( a −d) X n = L 2 φ(L x + y) n |yi|

Proposition 11. Suppose that (7.1) holds. Then, for maxi |xi − yj| > 1, n n ∗ h(CLφ)(x)(CLφ)(y)i−→n→∞AG (x, y) with Z Z G∗(x, y) = du dv|x − y + u − v|−a

77 7.1.1 Gaussian ﬁxed point

d Consider the (massless) Gaussian Ginzburg-Landau model on Z , d > 2. Its is given by the 0 d 1 gaussian measure on s ( ) with covariance C = where ∆ d is the lattice Laplacean. Z −∆ d Z Z In (6.7) we saw that

C 1 hφ(x)φ(y)i = C(x − y) = d + o( ). |x − y|d−2 |x − y|d−2

Hence we need to take a = d − 2 and get the Gaussian ﬁxed point covariance for the block spin transformation:

Z Z C C∗(x − y) = du dv d . |x − y + u − v|d−2 It is instructive to generalize a bit and consider a lattice covariance

Z d ip(x−y) 1 d p D(x − y) = e d . [−π,π]d ρ(p) (2π)

P 2 pi For GL model we had ρ(p) = µ(p) where µ(p)) = 4 sin ( 2 ). We compute

Z d ip(Lnx−Lny+u−v) n n n(d−2) −2nd X d p e hCLφ(x)CLφ(y)i = L L d d (2π) ρ(p) u,v [−π,π] Z d ip(x−y+L−n(u−v)) −2nd X d p e = L d 2n −n n n d (2π) L ρ(L p) u,v [−L π,L π]

Suppose now that ρ(p) = p2 + o(p2) as p → 0 (µ satisﬁes this). Then L2nρ(L−np) → p2, as n → ∞. Recalling that

Z d 1 ip(x−y) 1 d p Cd (x − y) = e 2 d = d−2 . −∆Rd Rd p (2π) |x − y|

2 d where −∆Rd is the usual −∆ on L (R ) we conclude Z Z n n 1 ∗ hCLφ(x)CLφ(y)i−→n→∞ du dv (x − y + u − v) = C (x − y) −∆Rd as well. This is an example of universality: details of the covariance are erased upon coarse graining and scaling. n n 1 n T Note that in this example, CLφ(x) are Gaussian, with covariance CL −∆ (CL) where T denotes the transpose.

7.2 Transformations on Measures

Now we turn transformations for general measures. Suppose µ is a cylinder measure in d RZ and consider the random variables CLφ. Their generating function is

Z Z T i(CLφ,f) i(φ,C f) WL(f) = e dµ(φ) = e L dµ(φ)

T = W (CL f) which is obviously of positive type, so there is a measure µL such that Z i(Ψ,f) WL(f) = e dµL(Ψ).

78 d d So µL is the probability distribution of the block spins CLφ. Thus CL : RZ → RZ induces a map ∗ CL : B → B 0 d where B are the cylinder probability measures on s (Z ). ∗ CL is called the Renormalization Group Transformation on measures. It is a linear map in B. It turns out that a we need to express it as a nonlinear map on Hamiltonians in order to be able to draw useful consequences.

7.3 Transformations on Hamiltonians

N Concretely, let us work in a ﬁnite volume ΛN which is taken to be a box of side L Λ Λ centered at the origin. With the previous deﬁnitions, we have φ ∈ R N ,CLφ ∈ R N−1 and n ΛN−n CLφ ∈ R . Let µ be a probability measure

dµ(φ) = e−H(φ)dΛN φ

Λ −H where H is some function H : R N → R such that e is integrable (we include the normalization constant, i.e. the partition function to H). Then,

0 −H (ψ) ΛN−1 dµL(ψ) = e d ψ with 0 Z −H (ψ) −H(φ) ΛN e = e δ(ψ − CLφ)d φ where Y δ(ψ − CLφ) := δ(ψ(x) − (CLφ)(x)).

x∈ΛN−1

More explicitely, recall that the block spins CLφ(x) live on the unit lattice and box ΛN−1 a and equal L 2 times the average of the original spins φ in the L-cube centered at Lx. Thus, if we fix CLφ(x) to ψ(x), for all x ∈ ΛN−1, we need to integrate out all fluctuations around this average. In other words, let φ0 be a configuration that is constant in each L-cube, and equals the average of φ there:

− a − a φ0(Lx + y) = L 2 CLφ(x) = L 2 ψ(x) for all x ∈ ΛN−1, all y with |yi| < L/2. Using our previous notation this equals

− a − a +d T d−a T φ0 = L 2 L 2 C ψ = L C ψ

T a −d (recall, Cxy = L 2 if x ∈ L-cube at Ly, and equals 0 otherwise). Thus φ = φ0 + z with z having 0 average over L-cubes i.e. CLz = 0. Thus

0 Z d−a T −H (ψ) −H(L C ψ+z) ΛN e = e δ(CLz)d z

Λ \LΛ d Even more concretely, take ζ ∈ R N N−1 i.e. ζ(x) ∈ R, ∀x ∈ ΛN where x 6= Lw, w ∈ Z . Put  ζ(x) x ∈ ΛN \LΛN−1  X z(x) = − ζ(Lw + y) x = Lw  |yi|

Then CLζ = 0 and so

0 Z d−a T e−H (ψ) = e−H(L C Ψ+z)dΛN \LΛN−1 ζ

79 a concrete integral over “ﬂuctuation variables” ζ(x). 0 The map RL : H → H is called the RG Transformation on Hamiltonians, i.e. Z −H(Ld−aCT ψ+z) RLH = − log e dζ (7.6)

Λ Λ Here H : R N → R and RLH : R N−1 → R. Note how the volume contracted from an LN -box to an LN−1-box. Remarks:

∗ • It was trivial to define the RG on measures and in infinite volume : CLµ is auto- matically a measure. On Hamiltonians, we need to work in finite volume and make some assumptions on H so that RLH in (7.6) is well defined.

n n ∗ n ∗ n • RG is called a “group” because CL = CL :(CL) = CLn and (RL) = RLn . −1 Actually, it is a semigroup : R1 is the identity but RL is not deﬁned.

0 d Our Proposition 11 says that (here µ is in inﬁnite volume i.e. on s (Z )) Z n n ∗ (CLφ)(x)(CLφ)(y)dµ(φ)n−→→∞ G (x, y) where we absorbed the constant A in the deﬁnition of G∗. So Z ∗ ψ(x)ψ(y)dµLn (ψ) → G (x, y)

∗ Thus, we might hope that the measures µLn converge to some measure µ such that Z ψ(x)ψ(y)dµ∗(Ψ) = G∗(x, y)

∗ ∗ and µ is a ﬁxed point of CL : ∗ ∗ ∗ CLµ = µ Equivalently, provided we make sense of H in inﬁnite volume, we might expect

n ∗ ∗ ∗ (RL) Hn −→→∞ H ,RLH = H in some topology in a space of H’s. However, this is not as easy as it seems:

• It is actually quite diﬃcult to set up nice spaces of Hamiltonians of unbounded spins (like our φ(x) ∈ R) where RL would act. This due to the fact that we need H to tend to +∞ on large values of |φ(x)|’s since we need the φ(x) integral to converge at ±∞. This large ﬁeld behaviour is quite non universal and plays little role in the qualitative analysis of RL. However it is quite a headache for rigorous proofs! We will mostly ignore this aspect in the heuristic discussion below, see however Section ?? where we adress this question in the case of the hierarchical model.

• It might seem that the large ﬁeld problem does not occur if φ were a bounded variable (a bounded spin model) like in the Ising model where φ(x) ∈ {−1, 1}. However even in this case setting up R near the critical temperature is nontrivial. Moreover, for bounded spins φ the RG, as deﬁned above, will in general change the range of values that φ takes : for example for Ising spins φ(x) ∈ {±1}, CLσ takes values in the set a −d a {φ ∈ L 2 Z | |φ| ≤ L 2 } so for a > 0, it can take larger values. Upon iteration, n ( a −d)n n an CLφ(x) ∈ L 2 Z with |CLφ(x)| ≤ L 2 . Thus, for example in d = 3, where we expect a = d−2+η with η ≈ 0.1, the block spins become unbounded and continuous as n → ∞!

80 • For unbounded spins, RL and its iterations have been rigorously studied for the Ginzburg-Landau model. It turns out that one can, to a certain extent set up a space of Hamiltonians where RL acts; however, one has to supplement this with a diﬀerent representation of µ for large values of φ’s. For the rest of the time, we will not discuss these problems. Rather we

a) Pretend that RL is deﬁned in some space of H’s and see what implications this could have n b) Carry out a perturbative analysis of RLH for H = Ginzburg-Landau model and try to understand the IR and UV problem this way.

7.4 Fixed Points of RL

We assume the following setup

• We have some space K of “Hamiltonians” H such that H ∈ K determines a unique d R 1 −H measure µH on Z . We denote hF iH = F dµH. Here µH is the limit of e Λ Dφ R ZΛ d as Λ % Z and HΛ is some ﬁnite volume version of H (or more properly, H = {HΛ}Λ⊂Zd ).

• . The RG is deﬁned in K, so RL : K → K:

∗ CLµH = µRLH

At the moment we do not ﬁx the scaling exponent a in the deﬁnition of CL:

a −d X (CLφ)(x) = L 2 L φ(Lx + y). L |yi|< 2

• We also suppose that there is a metric in K and if Hn → H, then µHn → µH. Let us see what kind of picture of critical phenomena emerges from such assumptions.

Deﬁnition 21. H ∈ K is non-critical if there exist A > 0, ξ < ∞ such that

−|x−y|/ξ |hφ(x)φ(y)iH| ≤ Ae for all x, y. If this does not hold then we say H is critical.

∗ ∗ ∗ Let us start by considering ﬁrst a ﬁxed point H of RL so RLH = H . Then

∗ ∗ Proposition 12. Suppose RLH = H .

∗ • If H is non-critical then a = d and hφ(x)φ(y)iH∗ = 0 for x 6= y.

∗ −α −α • If H is critical and C1(|x − y| + 1) ≤ hφ(x)φ(y)iH∗ ≤ C2(|x − y| + 1) then necessarily α < d and a = α.

∗ Proof. First, take H non-critical. Suppose hφ(x)φ(y)iH∗ 6= 0 for some x 6= y. Then since n ∗ ∗ RLH = H

n n ∗ |hφ(x)φ(y)iH∗ | = |hφ(x)φ(y)iRnH∗ | = |hCLφ(x)CLφ(y)iH| n(a−2d) X X n n ∗ = |L hφ(L x + u)φ(L y + v)iH| n n |ui|

81 If |x − y| > 1 then the RHS tends to zero as n → ∞ since |Ln(x − y) + u − v| ≥ Ln. Thus hφ(x)φ(y)iH∗ = 0 for |x − y| > 1. By translation invariance hφ(x)φ(x)iH∗ = A and hφ(x)φ(y)iH∗ = B for |x − y| = 1 where A ≥ 0. By Schwartz inequality |B| ≤ A so that we may suppose A > 0. Then from the previous calculation we obtain for |x − y| = 1:

n(a−2d) n(d−1) B = hφ(x)φ(y)iH∗ = L L B Hence B = 0 unless a = d + 1. For x = y we get

n(a−2d) nd nd n(d−1) A = hφ(x)φ(x)iH∗ = L (L A + 2dL b − (2d − 1)L B) The second B term is due to the boundary of the Ln cube. Since this holds for all n we get a = d and hence B = 0 from which the ﬁrst claim follows. For the second claim, let

X −α Ad(L) = (|u − v| + 1) . 1 |ui|,|vi|< 2 L Then we have n(a−2d) 2 n(a−2d) C1L Ad(L) ≤ hφ(0) iH∗ ≤ C2L Ad(L). d d For α > d we have A1L ≤ Aα(L) ≤ A2L . This implies a = d. Then as above (do it!) we get hφ(x)φ(y)iH∗ = 0 for |x − y| > 1, a contradiction. d d For α = d we have A1L log L ≤ Aα(L) ≤ A2L log L. Thus no a is possible. 2d−α 2d−α For α < d we have A1L ≤ Aα(L) ≤ A2L . This implies a = α. The claim follows.

This result is consistent with the statement that the possible ﬁxed points are (1) the one where φ(x) and φ(y) being independent random variables for x 6= y and (2) the one where the correlations decay with rate < d. In the former case the exponent a = d corresponds to the central limit scaling. Indeed, for independent spins the averages

− d n X n φLn (x) = L 2 φ(L x + y) converge to Gaussians by central limit theorem and the same holds true if φ(x) and φ(y) are weakly correlated. Such a non-critical fixed point is called a high temperature fixed point. The latter case a = α < d is the one we are now going to study. n ∗ Suppose now that R H → H . Then, hφ(x)φ(y)i n → hφ(x)φ(y)i ∗ and we conclude L RLH H that if H∗ is high temperature fixed point, the central limit theorem holds for the block spins. For the critical case we define: Definition 22. The stable manifold of H∗ in K is

n ∗ Ms = {H ∈ K | RLH → H }

What can we say about the decay of correlations of H ∈ Ms? Identical arguments as above show that it can not be the case that H ∈ Ms and

−a0 hφ(x)φ(y)iH ≤ A|x − y|

0 with a > a. Thus it is reasonable to expect that all H ∈ Ms have the same critical ∗ exponent a. We call Ms the critical surface of the ﬁxed point H . Next, consider Mξ = {H | H has correlation length ξ} For this we assume that all H ∈ K have a well deﬁned ξ, i.e. that the limit 1 lim − loghφ(0)φ(x)i = ξ−1 |x|→∞ |x| exists. Then we have

82 Figure 7.1: Graphical representation of K

•M s = M∞

• RL : Mξ → Mξ/L

The second statement follows since RL scales by L:

α X −2d −L|x|/ξ hφ(0)φ(x)iRLH = L hφ(u)φ(Lx + v)iL ∼ e . u,v

Thus, we have a nice picture of K as given in ﬁgure 7.1. ∗ From this we see that RL takes critical H to H upon iteration while taking noncritical H away from H∗ upon iteration.

7.5 Linear Analysis around a Fixed Point: Critical Expo- nents

∗ Let us now study RL near H . We assume that the usual dynamical systems ideas are 1 applicable: we assume that RL is, say, a smooth map in K. Thus, let L = DRH∗ be the derivative of R at H∗. This means that, for any H ∈ K,

R(H∗ + H) = H∗ + LH + O(2).

The spectrum of L plays now a crucial rule in the analysis of R near H∗. In the known examples, the spectrum of L has the following structure :

• It is discrete, consisting of, real, positive, eigenvalues λi of ﬁnite multiplicity.

• There is a ﬁnite number of λi with λi ≥ 1.

Let Hi be an eigenvector : LHi = λiHi. We say that :

•H i is relevant if λi > 1

•H i is marginal if λi = 1

•H i is irrelevant if λi < 1

n n n ∗ ∗ Note that DRH∗ = D(R ◦ R ◦ R ··· R)H∗ = (DRH∗ ) = L (since R(H ) = H ), so

n ∗ ∗ n R (H + Hi) = H + λi Hi + ...

1 If K is a Banach space, the derivative of R at H is deﬁned as usual as the linear map DRH : K → K : 1 DRHH1 = lim [R(H + H1) − R(H)]. →0

83 This means that the Hi perturbation increases (λi > 1) or decreases (λi < 1) exponentially in n or stays constant (λi = 1) upon iteration. Let us interpret this. Consider ﬁrst the case where there is only one λ > 1, with multiplicity 1, say λ1. Then for |x − y| >> 1

n n −na n n ∗ ∗ hφ(L x)φ(L y)iH +H1 ≈ L hCLφ(x)CLφ(y)iH +H1 −na = L hφ(x)φ(y)iRn(H∗+H) −na = L hφ(x)φ(y)i ∗ n n 2 H +λ1 H1+O((λ1 ) )

n Take n such that λ1 = O(1). Recall that we had one direction in K where we depart from H∗ upon iteration, namely the one parametrized by ξ. Thus, it is natural to assume ∗ ∗ that H + H1 has ξ < ∞ and H + O(1)H1 has ξ = O(1). From above, then

∗ n n ∗ n ∗ ξ(H + H1) =L ξ(R (H + H1)) = L ξ(H + O(1)H1) =LnO(1)

α −nα ∗ −1/α Write λ1 = L (so α > 0). Then ∼ L ⇒ ξ(H + H1) ∼ . ∗ Let us identify = 0 with T = Tc, the critical temperature. Thus, our Hamiltonian H ∗ ∗ could be say the Ising model βcHIsing and H + H1 would be βHIsing with (1 + )β = β or ∼ T − Tc. More generally, we could start with Ising and consider some eﬀective long distance model (see below for precise deﬁnitions) for its correlations. Then again ∼ T − Tc. So then

−1/α ξ ∼ (T − Tc) and −1/α is the so called correlation length critical exponent describing how ξ diverges as T → Tc.

Universality

∗ Consider now H = H + Hi, λi < 0. As before,

na n n n n L hφ(L x)φ(L y)i ≈ hC φ(x)C φ(y)i = hφ(x)φ(y)i ∗ n H L L H H +λi Hi n = hφ(x)φ(y)iH∗ + O(λi )

Take |x − y| = O(1) and set u = Lnx, v = Lny so that |u − v| ≈ Ln. Then we got

−a −a−αi hφ(u)φ(v)iH = A|u − v| + O(|u − v| )

−α n −α n −α where we wrote λi = L i , αi > 0. Then, since |u−v| ∼ L , we get that L i ∼ |u−v| i . Thus these directions give subleading corrections to the decay determined by H∗. We have universality : we expect that the leading asymptotics of the correlation functions QN n QN n h i=1 φ(L xi)iH coincide with those of h i=1 φ(L xi)iH∗ if H ∈ Ms as n → ∞ and xi are apart from each other. The details of H are only seen in the O(L−αin) corrections. We say that all H ∈ Ms have the same critical behaviour. Also, all the exponents αi are determined by H∗.

7.6 Eﬀective Field Theories

We ﬁnally consider the QFT problem from the RG point of view. Suppose we have a scalar 0 d n −n QFT, described by a measure ν on S (R ). Let σx be be the cube of side L centered at −n d x for x ∈ (L Z) . Deﬁne Z nd ϕn(x) := L ϕ(u)du n σx

84 2 n so ϕn is the average of ϕ in the cube σx . Let νn be the law of ϕn i.e. νn is the measure 0 −n d on s ((L Z) ) deﬁned through the characteristic function Z Z iϕn(f) iϕ(fn) e dνn(ϕ) = e dν(ϕ). where we denote X −nd ϕn(f) := f(x)ϕn(x)L x∈(L−nZ)d

P n and fn := x f(x)1σx . In other words, νn gives the joint distribution of the variables ϕn(x) and correlation functions of νn are the averages of those of ν: m Z Z m DY E mdn DY E ϕn(xi) = L du1··· dum ϕ(ui) . νn σn σn ν i=1 x1 xm i=1

−n We call νn the eﬀective QFT on scale L corresponding to ν. νn describes the behaviour of the QFT on scales larger than L−n: we have ”integrated out” the information on smaller scales. Suppose now that the measure νn is Gibbsian in the following sense. Let −n d Λ ⊂ (L Z) be a box and let νn,Λ be the law of ϕn restricted to Λ so νn,Λ is deﬁned through the characteristic function Z Z iϕn(f) iϕn(f) e dνn,Λ(ϕ) = e dνn(ϕ)

−n d where f ∈ s((L Z) ) has support in Λ. Suppose νn,Λ is absolutely continuous with respect to the Lebesgue measure Y Dn,Λϕ = dϕ(x) x∈Λ∩(L−nZ)d) (we expect this to be true) and write the Radon-Nikodym derivative as

−Sn,Λ(ϕ) dνn,Λ(ϕ) = e Dn,Λϕ.

−n We call Sn,Λ(ϕ) the Effective Action of the QFT on scale L (and in box Λ). d It is convenient to relate all of these effective theories to a fixed lattice Z . For this, d −an −n d define a Z field φn(x) = L ϕn(L x) for x ∈ Z where a is a parameter to be fixed later, and let µn be the distribution of φn, i.e.

DY E DY −an −n E φn(xα) = L ϕn(L xα) . µ ν α n α n

How are the diﬀerent φn’s related ? Easy : Z −a(n−1) −n+1 −a(n−1) (n−1)d φn−1(x) = L ϕn−1(L x) = L L ϕ(u)du n−1 σx Z −a(n−1) −d X nd a−d X = L L L ϕ(u)du = L φn(Lx + y) σn |yα|

∗ µn−1 = CLµn.

d Let V be a box in Z and µn,V be the restriction of µn to φ|V . Then we get

−Hn,V (φ) dµn,V (φ) = e DV φ

2ϕ(f) is a priori deﬁned only for test functions f ∈ S0. We assume here that it is also deﬁned on characteristic functions of cubes. This assumption holds in the known cases.

85 where −S (ϕ) e−Hn,V (φ) = Lan|V |e n,L−nV

−an −n |V | where φ and ϕ are related by φ(x) = L ϕn(L x) and DV φ is Lebesgue on R . We −n call Hn,V the effective Hamiltonian of the QFT on scale L (and in box V ). Note that all these effective Hamiltonians are defined on a fixed lattice (spacing 1) whereas the effective actions are defined on lattices with varying spacing. It should now be obvious how the effective Hamiltonians are related to each other:

Hn−1,V = RLHn,LV

In the spirit of the previous section we denote the collection of Hn,V with various boxes V by Hn and then write m Hn−m = RL Hn ∀n, m.

7.7 Scaling limit

We saw above that a QFT gives rise to Hamiltonians related to each other by the renormalization group transformation RL. Note that this was just a change of variables and in particular the scaling exponent a in the definition of RL was arbitrary. Of course we expect only one choise of a to give rise to nice Hn. In order to understand this and to understand how the RG can be used to study the continuum limit of cutoff QFTs we now ∗ suppose we have the setup RL : K → K and a fixed point H . How would we get a QFT from these data? d d The most obvious way is to consider the scaling limit. Given φ : Z → R set for x ∈ R

ϕ(N)(x) = LaN φ([LN x]) (7.7)

d d where for y ∈ R we deﬁne [y] to be the closest point on Z to y (and say largest in some d (N) (N) 0 d ordering of Z in case there are several such points). Let ν be the law of ϕ on S (R ) d i.e. for all f ∈ S(R ) Z iϕ(N)(f) iϕ(f) (N) he iH∗ = e dν (ϕ).

The scaling limit is then deﬁned as

ν = lim ν(N) N→∞ provided the limit exists which we assume. Then we expect that also the correlation functions converge for xi 6= xj i 6= j:

m Z m NamDY N E Y lim L φ([L xi]) = ϕ(xi)dν(ϕ) := G(x1, . . . , xm). N→∞ H∗ i=1 i=1

1 md We assume this convergence takes place in Lloc(R ). G(x1, . . . , xm) are correlations of a QFT. What are the eﬀective ﬁeld theories νn or µn or Hn of this QFT ?

∗ Answer All Hn are just H !

Proof. This is mainly bookkeeping. By deﬁnition

m Z DY E n(d−a)m n φ(xα) = L dy1 . . . dym1(yα ∈ σL−nx )G(y1, ··· , ym) µ α α=1 n

86 n −n −n 1 md where σ −n = L cube at L x . Since (7.7) holds in L ( ) we obtain L xα α loc R m Z m DY E Nam n(d−a)m n DY N E φ(xα) = lim L L dy1 . . . dym1(yα ∈ σx ) φ([L yi]) µ N→∞ α H∗ α=1 n i=1 m Nam n(d−a)m −Nmd X N−n DY N−n E = lim L L L 1(|uα,i| < L ) φ(L xα + uα) N→∞ H∗ d u1,...um∈Z α=1 m m m DY N−n E DY E DY E = lim (CL φ)(xα) = lim φ(xα) = φ(xα) N→∞ H∗ N→∞ RN−nH∗ H∗ α=1 α=1 L α=1

The scaling limit is a scale-invariant, massless, ﬁeld theory :

−Na G(λx1, . . . , λxm) = λ G(x1 . . . xm).

This is reﬂected in the fact that all eﬀective Hamiltonians are the same.

7.8 Non-Scale Invariant Theories

∗ How to get a QFT with effective theories Hn 6= H ? For example, we would like to get a massive theory. We would like to pose this constructively : find (lattice) cutoff-L−N theories that converge as N → ∞ to a QFT. The way to do this is obvious from the calculation in the previous section. Pick a Hamiltonian H(N) and define the measure ν(N) 0 d on S (R ) by Z iϕ(N)(f) iϕ(f) (N) he iH(N) = e dν (ϕ). where as in (7.7) ϕ(N)(x) = LaN φ([LN x]). ν(N) has a cutoff L−N as it really lives on fields −N d −n defined on the lattice (L Z) . We can define the effective field theories on scales L (N) (N) with n < N determined by ν . Call this measure νn . Then as above (check it!) we obtain Z iϕ(f) (N) iϕ(n)(f) e dνn (ϕ) = he iRN−nH(N) .

Thus the problem of continuum limit can be stated as follows. We want to ﬁnd the Hamiltonian H(N) s.t. the limit (N) lim νn := νn N→∞ exists. This is equivalent to the existence of the limit

N−n (N) lim R H := Hn. (7.8) N→∞

Then (we assume R is continuous) the effective theories are mapped to each other by the RG: m R Hn := Hn−m. The easiest way to satisfy (7.8) is to take H(N) = H∗. This was discussed in the previous (N) c c section. A modification of this is to take H = H where H ∈ Ms is critical. In this ∗ case too the limit is the scale invariant QFT with Hn = H for all n. A more interesting case is obtained by recalling the definition of the unstable manifold:

∗ ∗ ∗ Deﬁnition 23. Let RH = H . The unstable manifold Mu of H is deﬁned by

n (k) k (k) (k) ∗ o Mu = H ∈ K ∃ H ∈ K s.t. R H = H and lim H = H , . k→∞

87 Note that if R were invertible then Mu is the stable manifold of the inverse of R −k ∗ i.e. the set of H s.t. limk→∞ R H = H . Mu is an an invariant manifold for R i.e R : Mu → Mu.

(k) −N Let now H ∈ Mu and H as in the definition. Taking the Hamiltonian of our L - (N) (n) cutoff QFT H we get a continuum limit with effective Hamiltonians Hn = H . In (N) this example the cutoff theory H ∈ Mu. This is not necessary. It suffices to take the cutoff theories so that (N) lim H ∈ Ms N→∞ as N → ∞ and (7.8) holds. I.e. if we let the cutoff Hamiltonians tend to the critical surface in a suitable way the the continuum limit exists. In all these cases the effective theories lie on the unstable manifold of the RG fixed point. Hence we have obtained a nice picture: critical phenomena lie on the stable manifold Ms, and quantum field theories on the unstable manifold Mu!

Example. (a) From QFT to eﬀective Hamiltonians. Let ν = Gaussian measure, covari- 1 −n d ance Gb(p) = p2+m2 (free ﬁeld, mass m). Then νn has covariance (x, y ∈ (L Z) )

Z d ip(x−y) d p Gn(x, y) = e Gb(p)ρn(p) d . Rd (2π) where Z 2 nd ipx ρn(p) = L e dx n σ0 d−2 Taking a = 2 , µn has covariance Cn :

d Z −n 1 d p C (x, y) = L(2−d)n eipL (x−y) ρ (p) n p2 + m2 n (2π)d Z 1 ddp = eip(x−y) ρ (Lnp) p2 + L−2nm2 n (2π)d

88 where d Z 2 2 n ipx Y 2 sin pµ/2 ρn(L p) = e dx = ≡ ρ(p) 0 pµ σ0 µ=1

d is n-independent. Since now x, y ∈ Z we get

Z d ip(x−y) −n d p Cn(x, y) = e C(p, L m) d [−π,π]d (2π) where X ρ(p + 2πk) C(p, m) = (p + 2πk)2 + m2 k∈Zd C(p, m), m ≥ 0 are the (Fourier transforms of) covariances that parametrize the Gaussian part of the unstable manifold of the Gaussian ﬁxed point C(p, 0) of the block spin RG. They have correlation length ∼ m−1.

(b) Scaling limit of lattice GL model. Take h·iHN) be given by the Gaussian measure −2N 2 −1 P 2 pi with (Fourier transform of) covariance (µ(p) + L m ) where µ(p) = 4 sin ( 2 ). Then RN−nH(N) deﬁnes a Gaussian measure with (Fourier transform of) covariance

X rN−n(p + 2πk) Cˆ (p, L−nm) = N−n L2(N−n)µ(L−N+n(p + 2πk)) + L−2nm2 k∈Zd∩[−LN−n,LN−n]d where d 2 Y 2 sin pµ/2 r (p) = . N−n LN−n sin(L−N+np ) µ=1 µ

−n −nn Clearly limN→∞ C˜N−n(p, L m) = C(p, L m).

To get less trivial examples of scaling limits we need to study the RG ﬂow near the Gaussian ﬁxed point.

7.9 The Gaussian Fixed Point

7.9.1 Block spin ﬁxed point

As we saw in Section 7.1.1, the RG drives upon iteration the Gaussian measure µC where 1 d C = −∆ (on Z ) to the ﬁxed point

Z Z Z d ∗ ip(x−y) ip(u−v) 1 d p d C (x − y) = du dv e e 2 d x, y ∈ Z 1 1 d 1 1 d d p (2π) [− 2 , 2 ] [− 2 , 2 ] R i.e. n n ∗ lim hC φ(x)C φ(y)iµ = C (x − y). n→∞ L L C Since Z d Z 1/2 d ipu Y ipµuµ Y 2 sin pµ/2 due = e duµ = , 1 1 d pµ [− 2 , 2 ] µ=1 −1/2 µ=1 we have, Z d 2 d ∗ ip(x−y) 1 Y 2 sin pµ/2 d p C (x − y) = e 2 d d p p (2π) R µ=1 µ

89 (note that this converges absolutely) and writing

Z X Z dpµf(pµ) = dpµf(pµ + 2πn) [−π,π] R n∈Z we have Z d ∗ ip(x−y) ˆ∗ d p C (x − y) = e C (p) d [−π,π]d (2π) d 2 ˆ∗ X 1 Y 2 sin pµ/2 C (p) = 2 . (p + 2πn) pµ + 2πnµ n∈Zd µ=1 This is the Gaussian ﬁxed point of the block spin RG. We should now like to perform the linear analysis near C∗. We will do this in a slightly diﬀerent model, where it is less messy (everything can be done in the case of block spin with identical conclusions, see [22]).

7.9.2 RG in Momentum Space d d We will consider, instead of a spin model on Z (lattice cutoff), a “spin model” on R where we use a momentum space cutoff. A priori this is as good from the physics point 0 d of view. Thus, let dµG be the Gaussian measure on S (R ) with covariance Z d ip(x−y) χ(p) d p G(x − y) = e 2 d Rd p (2π) where d > 2, χ(p) is a cutoff, say 2 χ(p) = e−p d but anything with χ(0) = 1 and fast decay at ∞ will do (say χ ∈ S(R )). 1 Remark 4. In d = 2, p2 χ(p) does not define a covariance (it is not integrable at p = 0) and we should introduce an infrared cutoff. A convenient way to do this from the RG point of view is to take χ(p) χ(Rp) GbR(p) = − . p2 p2 −p2 −c|x|/R Then GR(x) is smooth and decays at scale R; e.g. for χ(p) = e , |GR(x)| ≤ Ce . 2 For test functions f ∈ S(R ) with fˆ(0) = 0 the readom variables φ(f) are well defined in the limit R → ∞. The Ginzburg-Landau measures defined below also have a R → ∞ limit as long as λ > 0.

d A priori we know that only smeared fields φ(f), f ∈ S(R ) are integrable w.r.t. µG and Z iφ(f) − 1 (f,Gf) e dµG = e 2 so that e.g. hφ(f)φ(g)i = (f, Gg) = R f(x)g(y)G(x − y)dxdy. However, now φ(x) itself −d/2 − 1 (x−y)2 makes sense (defined as lim φ(f) where lim f = δx (e.g. f(y) = (2π) e 2 ) →0 →0 R and we have φ(x)φ(y)dµG = G(x − y) well defined, ∀x, y. (Technically, µG is supported on distributions that are C∞ functions, see Homework). Thus, let us consider the finite volume measures 1 −VΛ(φ) dνΛ(φ) = e dµG(φ) ZΛ R R where ZΛ normalizes dνΛ = 1,VΛ is a function of φ’s localized in Λ, e.g., VΛ = Λ dxP (φ(x)) P e.g. a polynomial.

90 We now describe the RG in this setup. We need to split φ into “local” and “global” or “high momentum” and “low momentum” parts as in the block spin case. We do this by ﬁrst splitting G : write, for L > 1, 1 1 1 Gˆ(p) = χ(p) = χ(Lp) + (χ(p) − χ(Lp)). p2 p2 p2 In position space this reads

Z d Z ip(x−y) 1 d p ip(x−y) 1 G(x − y) = e 2 χ(Lp) d + e 2 (χ(p) − χ(Lp)) Rd p (2π) Rd p d Z x−y 1 d p Z 1 2−d ip L ip(x−y) = L e 2 χ(p) d + e 2 (χ(p) − χ(Lp)) Rd p (2π) Rd p x − y = L2−dG( ) + Γ(x − y) (7.9) L Here Γ(x − y) is the “high momentum” or “ﬂuctuation within blocks” part : χ(p) − −p2 −L2p2 2 1 χ(Lp) = e − e = O(p ) for p ≈ 0 and it gets its main contribution if |p| ∈ [ L , 1]. In particular, for χ analytic (prove !)

|Γ(x − y)| ≤ Ae−|x−y|/L i.e. it has correlation length < ∞.

2−d Remark 5. In d = 2 we get GR(x) = L GR/L(x/L) + Γ(x) which means that the IR cutoﬀ does not enter in the RG transformation as Γ does not depend on R.

d−2 − 2 x Let us apply the Lemma4 to (7.9). We note that the covariance of L φ( L ) is 2−d x−y L G( L ) if φ has covariance G. Hence : Z Z − d−2 · F (φ)dµ (φ) = F (L 2 φ( ) + Z)dµ (φ)dµ (Z). G L G Γ This gives our RG :

Z R F (φ)e−VΛ(φ)dµ (φ) hF i = F (φ)dν (φ) = G νΛ Λ R −V (φ) e Λ dµG(φ) d−2 d−2 − 2 · R − 2 · −VΛ(L φ( L )+Z) F (L φ( L ) + Z)e dµG(φ)dµΓ(Z) = d−2 = hFeiν˜ − · R −VΛ(L 2 φ( )+Z) e L dµGdµΓ with 1 Z dν˜(φ) = e−Ve (φ)dµ (φ) Z = e−Ve dµ Z G G d−2 Z − · −Ve (φ) −VΛ(L 2 φ( )+Z) e = e L dµΓ(Z)

d−2 d−2 − 2 · R − 2 · −VΛ(L φ( L )+Z) F (L φ( L ) + Z)e dµΓ(Z) Fe(φ) = d−2 . − · R −VΛ(L 2 φ( )+Z) e L dµΓ(Z) Hence, we got a renormalized measure ν˜ which is again of same form as ν and we can deﬁne the RG as a map of V ’s by

Z d−2 −V (L− 2 φ( · )+Z) RLV (φ) = Ve(φ) = − log e L dµΓ(Z). (7.10)

Compare with block-spin ! The latter can actually, with a suitable deﬁnition of φ on the RHS, be written exactly as (7.10), see [22].

91 R Remark that if V = Λ P (φ(x))dx then Z − d−2 · − d−2 x V (L 2 φ( ) + Z) = P (L 2 φ( ) + Z(x))dx L Λ L Z d − d−2 = L P (L 2 φ(y) + Z(Ly))dy L−1Λ

−1 so it depends on φ in L Λ. We may call Ve by VeL−1Λ if we wish. Of course this will not d matter in the limit Λ % Z . We have now a ﬁxed point V = 0, i.e. dµG → dµG under our RG. Call this the Gaussian ﬁxed point. For later purpose let us note that the decomposition

− d−2 0 x φ(x) = L 2 φ ( ) + Z(x) (7.11) L leads to the following decompostion of the GFF

∞ X − d−2 n (n) x φ(x) = L 2 z ( ) (7.12) Ln n=0 where z(n) are i.i.d.

92 Chapter 8

Hierarchical model

8.1 Local RG transformation

The RG map (7.10) is local in the following sense. The fluctuation field Z is weakly correlated: as we saw above, it has correlation length O(1/L). We consider L to be here a fixed number (e.g. L = 2) so that we expect that if we start with a local VΛ like e.g., R VΛ = Λ dxP (φ(x)) the resulting RLV is (aproximately) local as well. This is Wilson’s original intuition when he developed the RG. A single application of RL is a nice map and the singularities of critical phenomena come only once we iterate RL sufficiently (infinitely) many times. The hierarchical model is a modification of the lattice Ginzburg-Landau model where the RG map is completely local, mapping a local V back to a local RLV . It was introduced by Dyson [?, ?, ?, ?, ?] and also by Wilson [?, ?, ?, ?] who considered it a an ”approximate recursion relation”, an approximation to the RG map. We can arrive at the hierarchical model by modifying the recursion (7.10) as follows. d First, let φ(x) be a lattice field i.e. x ∈ Z . Second, replace Z by a lattice field z whose covariance decouples across blocks of size L. More explicitely, let us take L ∈ N odd and consider the gaussian variables with covariance

z(x)z(y) = Γxyδ x y . (8.1) E [ L ],[ L ] where [·] denotes the integer part and for all x, y, z

Γx+Lz y+Lz = Γxy.. (8.2) Γ can be taken arbitrary, but the one motivated by GFF is

−d Γxy = δxy − L . (8.3)

d y d Let x ∈ Z and BL(x) = {y :[ L ] = x} be the L-block in Z centered at Lx.(8.1) implies that z|BL(x) and z|BL(y) are independent if x 6= y and (8.2) implies they are identically distributed. Finally (8.3) implies P z(y) = 0 almost surely. This condition is y∈BL(x) shared by the corresponding z of the GFF. Let us now consider the recursion (7.10) with this Γ for a local potential X V (φ) = v(φ(x)). x

Since z|BL(x) and z|BL(y) are independent and identically distributed if x 6= y we conclude that RV is also local X RV (φ) = Rv(φ(x)) x with d−2 0 Z P − 2 0 −Rv(φ ) − x∈B (0) v(L φ +z(x)) e = e L dµΓ(z). (8.4)

93 Ld Note that this is a ﬁnite dimensional integral: z ∈ R . (8.4) is in fact a block spin RG transformation for a lattice model where the covariance of the GFF is replaced by a hierarchical covariance. Let d 1 M 1 M ΛM := {x ∈ Z : − 2 L < xi < 2 L , i = 1, . . . , d}.

We deﬁne the hierarchical GFF φ(x), x ∈ ΛM by (compare with (7.12)) M X − d−2 n x φ(x) = L 2 z ([ ]). (8.5) n Ln n=0

Here the ﬁelds zn deﬁned on ΛM−n are independent. For n < M they have the covariance (8.1) and for n = M we take zM = zM (0) to be unit normal variable (Λ0 consists of one point, the origin). The decomposition (7.11) becomes then in the hierarchical model

− d−2 0 x φ(x) = L 2 φ ([ ]) + z (x) (8.6) L 0 0 where φ is the hierarchical ﬁeld on ΛM−1. This leads to the RG map (8.4).

8.2 Linearization

The Gaussian fixed point corresponds to v = 0. Let us study the linearization of R at this d fixed point. By definition Lv = d |=0R(v) so that Z 0 X − d−2 0 (Lv)(φ ) = v(L 2 φ + z(x))dµΓ(z)

x∈BL(0) Note that if P (φ) = φn + ... is a polynomial with degree n where ... are lower degree terms then d−n d−2 n (LP )(φ) = L 2 φ + .... d−n d−2 Hence L has eigenvalues L 2 . If d > 2 we may also diagonalize L in the space of polynomials. For this consider d−2 d−2 αφ+β L− 2 αφ+β X αz(x) d L− 2 αφ+α2Γ +β (Le )(φ) = e Ee = L e xx x∈Λ1

Γxx Let G = 1−L2−d and αφ− 1 Gα2 eα(φ, G) := e 2 . Then d Leα := L e d−2 . L− 2 α and deﬁning the Hermite polynomial dn H (φ, G) = | e (φ, G). (8.7) n dαn α=0 α we get d−n d−2 LHn = L 2 Hn d = 2 is a degenerate case: all eigenvalues are L2 and the matrix of L is triangular, non-diagonalizable. Since R and L preserve the space of even functions v let us study RG in this space. For QFT we are interested in the number of relevant eigenvalues. This depends on dimension: • d > 4. Only the mass term φ2 is relevant. • d = 4. φ2 is relevant and φ4 is marginal. • d = 3. φ2 and φ4 are relevant and φ6 is marginal. Hence in d < 4 we expect to construct a nongaussian QFT around the Gaussian ﬁxed point.

94 8.3 Perturbative analysis of the hierarchical RG

Let us now study perturbatively the RG ﬂow near the Gaussian (v = 0) ﬁxed point. Consider a perturbation v(φ) = rφ2 + λφ4 of the GL type where we suppose λ is small and (as will become clear later) r = O(λ). Then we will compute

Z 0 − P v(ψ(x)) Rv(φ ) = − log e x dµΓ(z)

d where the sum is over the L points x ∈ BL(0) and

− d−2 0 ψ(x) := L 2 φ + z(x).

Let us expand this in powers of v (i.e. λ) to second order:

0 X 1 X 3 Rv(φ ) = hv(ψ(x))iΓ − 2 hv(ψ(x)); v(ψ(y))iΓ + O(λ ). (8.8) x xy where h · iΓ denotes expectation in µΓ and the second term on the RHS is the truncated expectation (6.10). We ignore here the fact that φ0 may be arbitrary large in which case v is not small and the O(λ3) is dubious. This is the large ﬁeld problem and is discussed in section 8.8. The ﬁrst term on the RHS of (8.8) is the linearized RG i.e.

X 2 02 4−d 04 hv(ψ(x))iΓ = (L r + γ1λ)φ + L λφ + const.. (8.9) x

2−d P 2 −d with γ1 = L x Γxx = L (1 − L ) and the second term becomes

X X 2d−4 2 2 2 04 d−2 3 3 2 02 hv(ψ(x)); v(ψ(y))iΓ = (36L λ hz(x) ; z(y) iΓφ + 16L hz(x) ; z(y) iΓλ φ xy xy d−2 2 4 2 02 d−2 2 2 02 + 12L hz(x) ; z(y) iΓλ φ + 12L hz(x) ; z(y) iΓrλφ ) + const.

P where x z(x) = 0 was used repeatedly as well as vanishing of odd moments of z. The 2 2 2 remaining expectations are evaluated by Wick theorem, e.g. hz(x) ; z(y) iΓ = 2Γxx and we conclude Rv(φ0) = const. + r0φ02 + λ0φ04 + w(φ0). (8.10) where

0 2 2 3 r = L r + γ1λ + γ2λ + γ3rλ + O(λ ) (8.11) λ0 = L4−dλ + bλ2 + O(λ3). (8.12) and w(φ0) collects higher powers in φ0: w(φ0) = g0φ06 + ... . In particular with g0 = O(λ3). Note that under subsequent iterations we have

g0 = L6−2dg + O(λ3).

Let us now see what the consequences for such a recursion in various dimensions.

95 8.4 d > 4: Gaussian critical point

If d > 4 then only the mass parameter r is a relevant variable and we expect (for small λ) that the critical point is Gaussian. To determine the critical point we take

v = rφ2 + λφ4 and try to ﬁnd r(λ) such that v ∈ Ms(0). Hence, setting

n 2 4 R v := vn = rnφ + λnφ + wn(φ) the problem is to ﬁnd r0 = r(λ) such that

rn, λn, wn → 0 as n → ∞.

From the previous section it is reasonable to postulate that the recursion is as follows1

2 rn+1 = L rn + γ1λn + ρn 4−d 2 λn+1 = L λn + bλn + ln wn+1 = Lwn + νn where L is the linearization of R and we suppose that in some norm

−α kLwnk ≤ L kwnk where α > 0 if d > 3 (recall that the exponent for the least irrelevant term φ6 is 6 − 2d). Furthermore, anticipating that rn = O(λn) we suppose the remainders satisfy

2 |ρn| ≤ A(λn + kwnk) 3 |ln| ≤ A(λn + kwnk) 3 kνnk ≤ Aλn for some constant A. The construction of r(λ) is a standard probelm of dynamical systems and it is done inductively in n. Given λ (small), take

r0 ∈ [−γ1λ, γ1λ] := I0.

Then, for λ small 2 |r1 − L r0| ≤ 2γ1λ

Consider r1 as a function of r0. r1 maps our interval

2 2 r1(I0) ⊃ [−(L − 2)γ1λ, (L − 2)γ1λ]

4−d 2 i.e., since λ1 ≤ L λ + O(λ ) < λ for λ small we have r1(I0) ⊃ [−γ1λ1, γ1λ1] (provided L2 > 3 which holds). Thus by continuity (which we assume here), ﬁnd can ﬁnd an interval

I1 ⊂ I0 such that r1(I1) = [−γ1λ1, γ1λ1]. Furthermore we have, for λ small, that

−β λ1 ≤ L λ 3 3 kw1k ≤ Aλ < Bλ1,

1For the rigorous analysis see section ??

96 where β > 0 for d > 4 and B = L3βA. Now, keep on iterating. We ﬁnd closed intervals In ⊂ In−1 ⊂ · · · I0 such that rn as a function of r0 satisﬁes

2 2 rn(In) = [−γ1λn, γ1λn]

−nβ where λn ≤ L λ and −α 3 kwn+1k ≤ L kwnk + Aλn. Thus the inductively the bound 3 kwnk ≤ Bλn iterates with a suitable B (indeed, with B = (L−3β − L−α)−1A if we take 3β < α). Since In closed ⊂ In−1 we have

∞ ∩n=0In 6= ∅

∞ and, for r0 ∈ ∩n=0In, we have

wn, λn, rn → 0 as n → ∞.

Actually it is readily seen that |I | −→ 0 so ∩∞ I consists of one point := r(λ), the n n→∞ n=0 n critical r value. Clearly r(λ) = O(λ). This is the desired critical point.

8.5 d = 4: Infrared asymptotic freedom

2 In d = 4 the coupling constant is marginal λn+1 = λn to leading order so the O(λ ) term will determine what happens. We have

2 3 λn+1 = bλn + O(λn). (8.13)

−d with (using Γxy = δxy − L )

2d−4 X 2 2 2d−4 X 2 2d−4 d b = −18L hz(x) ; z(y) iΓ = −36L Γxy = −36L (L − 1) (8.14) xy xy which is negative. Hence λn decreases upon iteration. To see how, consider ﬁrst the diﬀerential equation dλ = −bλ2 dn which is solved by λ λ = → 0 as n → ∞ n 1 + nbλ 1 i.e. for large n, λn ∼ nb . 1 log n Excercise 21. Prove that (8.13) ⇒ λn = nb + O( n2 ). Excercise 22. Near critical theory. Study what happens if we take

r = r(λ) + µ for small ν as follows. Let us replace the discrete iteration by a continuous one by replacing Ln by es and considering diﬀerential equation

λ˙ = β(λ, r) r˙ = 2r + γ(λ, r). with 2 3 2 2 β(λ, r) = β2λ + O(λ , λ r) γ(λ, r) = O(λ , λr).

97 We have seen it has the critical point solution rs, λs → 0 as s → ∞ with initial condition r0 = r(λ), λ0 = λ. Take now initial condition r0 = r(λ) + µ, λ0 = λ. Call the solution rs(µ), λs(µ) Derive a linear diﬀerential equation to d (as, bs) = dµ |µ=0(rs(µ), λs(µ)) and deduce from it the asymptotics α 2s as ∼ s e µ as s → ∞ and compute α. Argue that this is the leading asymptotics for our hierarchical model iteration.

8.5.1 Susceptibility divergence As an application of the last exercise let us study the susceptibility of our model. To deﬁne this let us consider the hierarchical model with an external magnetic ﬁeld. In the box ΛM this is given by the measure P 2 4 −1 − x∈Λ (hφ(x)+rφ(x) +λφ(x) ) νh(dφ) = Z e M µ(dφ) where µ is the Gaussian hierarchical measure. Let X φ¯ := L−Md φ(x)

x∈ΛM Then the magnetization m is deﬁned as ¯ m = hφih and the susceptibility by dm Md ¯ ¯ T χ := dh |h=0 = L hφ; φi0 . Note that φ¯ is the M:th block spin up to a multiplicative factor:

2−d M M φ¯ = L 2 C φ. Hence 2M χ = L hz, ziM where the expectation is in the measure

1 2 −1 −vM (z)− z h·iM = N e 2 dz 2M Hence χ = L (1 + O(λM )) blows up with the volume. How about near the critical point? Take r0 = r(λ) + µ. Let us iterate the RG to the scale n such that the eﬀective mass rn = O(1). From the exercise we get L2nnαµ = O(1).

The measure h·in is then close to gaussian with covariance −∆h + O(1) in the box ΛM−n so that 2n 2n ˜ ˜ T χ = L hz, zin = L hφ; φin

− d (M−n) where φ˜ = L 2 P φ(x). By Central limit theorem x∈ΛM−n χ = L2nO(1). Since L2n ∼ µ−1(log µ)−α we conclude that the susceptibility diverges as µ → 0 as χ ∼ µ−1(log µ−1)−α. If d > 4 the logarithmic factor is absent and if d < 4 the exponent −1 will change.

98 8.6 d < 4: Wilson-Fisher ﬁxed point

4−d 2 Now λn = L λn + O(λn) increases, so sooner or later λn is big enough so we can not trust our perturbative analysis that holds only if λn is small enough. It is very instructive to pretend that d is a continuous variable and take d = 4 − , small. Then 2 3 λn+1 = L λn − b(L)λn + O(λn). ∗ 2 2 log L Let us look for a ﬁxed point λ = a + O( ). We get 1 = L − ba + O( ) or a = b where b is given by (8.14) at d = 4.. Thus a ﬁxed point would be

λ∗ = a + O(2).

We conclude that the RG has a non-Gaussian ﬁxed point

v∗(φ) = r∗φ2 + λ∗φ4 + w∗(φ). where r∗ = O() and w∗ = O(2). This ﬁxed point is called (in the context of the GL model) Wilson-Fisher ﬁxed point. We’ll return to it in section ??.

8.7 d < 4: Superrenormalizable QFT

Let us now inquire about the continuum limit for d < 4 which we expect to be non- Gaussian since the unstable manifold of the Gaussian ﬁxed point is two dimensional including the φ4 direction. Let us call the UV cutoﬀ = L−N . Recall the connection between the statistical mechanics (dimensionless, unit lattice) φ and QFT ϕ variables:

d−2 N N ϕ(x) = L 2 φ(L x)

−N d Using (8.5) we obtain a decomposition for the hierarchical ﬁeld ϕ(x) for x ∈ L Z ∩ ΛK K where ΛK is the box of side L centred at origin (take M = N + K in (8.5) ):

N+K d−2 N N X d−2 (N−n) N−n ϕ(x) := L 2 φ(L x) = L 2 zn([L x]). (8.15) n=0

N−n −N d −(N−n) d The field zn([L x]) is constant on cubes in L Z which have centres on L Z −(N−n) and side length L . Let GN,K be the covariance of ϕ. Then the QFT measure with UV cutoff L−N and IR cutoff LM is given as

R 2 4 d 1 − [aN ϕ(x) +λϕ(x) ]d x ΛK dνN,K (ϕ) := e dµGN,K (ϕ). (8.16) ZN where we have anticipated a cutoﬀ dependence (counterterm) in the mass parameter.

Remark 6. It is natural to introduce the analogue of Laplacean and its Green function in n d the hierarchical setup. For each n ∈ Z let Bn be the set of L -adic cubes in R i.e. b ∈ Bn n n d is a cube with side L and center at x ∈ (L Z) . We denote by Bn(x) the cube b in Bn s.t. x ∈ b i.e. b has center Ln[L−nx] where [·] is integer part. Deﬁne the orthogonal projector 2 d Pn in L (R ) Z −nd (Pnψ)(x) = L ψ(y)dy. Bn(x) n Pn projects to functions that are constant on L -cubes. Deﬁne

Qn = Pn − Pn+1

99 Since PnPn+1 = Pn+1 we have that Qn are mutually orthogonal projectors. Qn projects to functions that are constant on Ln cubes and zero average on Ln+1 cubes. We have the 2 d decomposition of the identity in L (R ): ∞ X 1 = Qn n=−∞ 2 In finite volume ΛK we have in L (ΛK ): K−1 X 1 = Qn + PK . n=−∞ where the last factor PK takes care of the constant functions. With this notation we define the hierarchical Laplacean as the self adjoint operator in 2 d L (R ) (h) X −2n − ∆ = L Qn. (8.17) n∈Z The hierarchical GFF is then defined as the Gaussian field ϕ whose covariance is given by the inverse of −∆(h):

(h) −1 X 2n Eϕ(x)ϕ(y) = (−∆ ) (x, y) = L Qn(x, y) n∈Z −nd where the kernel is given by Qn(x, y) = L 1y∈Bn(x). This field is a S0( d) valued random field if d ≥ 3. In d = 2 only ϕ(f) with R f = 0 R Rd are well defined random variables. Indeed, e.g. for f = 1B0 (0) we get 2 X 2n −dn −d(n+1)) Eϕ(f) = L (L − L ) n≥0 which diverges in d = 2. We define the UV cutoff L−N and IR cutoff LM field by the covariance K−1 X 2n 2K GN,K = L Qn + L PK . n=−N This field is constant on L−N cubes and may be identified by the field (8.15). In dimensionless variables (8.16) gives rise to the potential (N) −2N 2 −(4−d)N 4 v (φ) = L aN φ + L λφ .

Our problem then is to determine the mass parameter aN such that eﬀective action ( in dimensionless variables) at scale L−n (N) N−n (N) vn := R v has a limit as the UV cutoﬀ is removed (N) vn → vn as N → ∞. Note that the parameters in v(N) are very small so that we expect our perturbative RG analysis to be applicable. We write (N) (N) 2 (N) 4 (N) vn (φ) = rn φ + λn φ + wn (φ). From (8.11) and (8.12) we get, dropping the superscript 2 2 2 rn−1 = L rn + γ1λn + γ2λn + γ3rnλn + O(λn(rn + λn) ) 4−d λn−1 = L λn + O(λn(rn + λn)). 3 We also dropped the irrelevant term wn as it is O(λn) and will not contribute to the counterterms (verify this!) in d = 2, 3.

100 8.7.1 d = 2 −2n −2n Write rn = L ρn, λn = L gn. Then

−2n ρn−1 = ρn + γ1gn + O(L gn(gn + ρn)) −2n gn−1 = gn(1 + O(L (gn + ρn)).

Note ρn increases so that rn would be proportional to N − n and hence unbounded in the cutoﬀ N unless we choose the initial condition appropriately. Obviously, we need to take

ρN = −Nγ1λ i.e. −2N rN = −NL γ1λ (8.18) Then it is easy to prove inductively

2 −2n ρn = −nγ1λ + O(λ L n) −2n gn = λ(1 + O(λL n)) and then for the running mass and coupling

−2n 2 −4n rn = −nL γ1λ + O(λ L n) −2n −2n λn = L λ(1 + O(λL n)).

In terms of the QFT variables the parameter 8.18 translates to a diverging mass parameter

aN = −Nγ1λ i.e. it diverges as log . In the dimensionless variables nothing is diverging: vn is small throughout the iteration as its should be for the argument to be valid.

8.7.2 d = 3 −n −n Now we write rn = L ρn, λn = L gn.

−n 2 −2n 2 ρn−1 = Lρn + γ1gn + L (γ2gn + γ3ρngn) + O(L gn(gn + ρn) ) −n gn−1 = gn(1 + O(L (gn + ρn)).

Now the divergence of ρn is more severe. Anticipating that gn equals λ up to exponentially small correction in n we set

−1 −n ρn = (1 − L) γ1gn + L sn.

Then

2 −n 2 sn−1 = sn + γ4gn + +O(L gn(gn + ρn) ) −n gn−1 = gn(1 + O(L gn)).

−1 where γ4 = L(γ2 + (1 − L) γ1γ3). Note that now the second order perturbative contri- 2 bution γ4gn = O(λ) to sn causes sn to increase so that sn would be proportional to N − n and hence unbounded in the cutoﬀ N unless we choose the initial condition appropriately as in d = 2 we had to do in ﬁrst order. We set

2 sN = −Nγ4λ so that the dimensionless bare mass parameter is

−1 −N −2N 2 rN = (1 − L) γ1L λ − NL γ4λ (8.19)

101 We then prove inductively 2 −2n 3 sn = −nγ4λ + O(L nλ ) −n gn = λ(1 + O(λL n)) and then for the running mass and coupling −1 −n −2n 2 −3n 3 rn = (1 − L) γ1L λ − γ4nL λ + O(λL nλ ) −n −n λn = L λ(1 + O(λL n)). In terms of the QFT variables we have now two diverging counterterms −1 N 2 aN = −(1 − L) L γ1λ − Nγ4λ i.e. we have a −1 and log divergences. In the dimensionless variables again nothing is diverging and vn is small throughout the iteration.

8.7.3 Infinite Volume Limit The perturbative analysis sketched above is valid in dimensions 2 and 3 as long as the dimensionless coupling λn is small, say λn < λ¯. In order to construct the QFT at infinite volume we then need to control the infinite volume limit M → ∞ of the unit lattice theory in box ΛM defined by the measure P − x∈Λ vn(φ(x)) dνn(φ) = e M dµM (φ). In general this is a hard problem since we know that this theory should have a non- perurbative non-gaussian fixed point. The only question admitting perturbative analysis is the massive QFT. For this we take the initial condition, say in d = 3: 0 −2N 2 rN = rN + +L m 0 where the counterterm rN is given by (8.19). Then, for λ/m small the infinite volume limit can be controlled. E.g. for m = 1 and λ small the measure ν0 is P 2 4 − x∈Λ (φ(x) +λφ(x) +... ) dν0(φ) = e M dµM (φ) which is a small perturbation of Gaussian theory with unit correlation length. This can be studied with expansion methods (cluster expansion).

8.7.4 d ↑ 4 It is instructive to study the continuum limit for non-integer dimension in the spirit of section 8.6. We consider the iteration 2 rn−1 = L rn + γ(λn, rn, wn) 4−d λn−1 = L λn + β(λn, rn, wn)

wn−1 = Lwn + η(λn, rn, wn). n(d−4) k Since we expect λn ∼ L we expect that we have to ﬁne tune λn contributions to 2 rn−1 for all t k such that k(4 − d) ≤ 2. Hence we need [ 4−d ] counterterms for r:

2 [ 4−d ] X k rN = tiλN +r ˜N k=1 −N(d−4) where λN = L λ. In the QFT variables, if k(4 − d) 6= 0 k X (d−4)i−2 i a = ti λ i=1 and if if k(4 − d) = 0 there is a log correction to the tk. We leave it to the reader to ponder how the ti are found in practise!

102 8.7.5 d = 4: Triviality As we saw in section 8.5 λ decreases logarithmically as we increase the scale:

2 3 λn−1 = −bλn + O(λn) so that λN λn ∼ . 1 + b(N − n)λN

If λN (0) stays bounded in N, as we have to assume in our perturbative analysis, then (N) λn → 0 as N → ∞. Thus the only perturbative theory (= the one where the coupling constant remains small on all scales) is the Gaussian. Same holds for d > 4. One might hope that, by taking λN large enough, things might change. For example, there might be another ﬁxed point λ∗ for the RG, i.e. the ﬂow in λ would be

λ = 0 λ∗

Unfortunately, this is unlikely to happen. There is strong numerical evidence that no other ﬁxed points exist in d ≥ 4. We’ll return to this issue in section 9.5.2.

8.8 Nonperturbative analysis

In this section we illustrate in the simplest case how the RG iteration can be rigorously set up and controlled. We carry out the analysis for the d = 2 hierarchical QFT. Let us normalize the R slightly diﬀerently. Given a bounded positive function g on R deﬁne Z Y − d−2 (T g)(φ) = g(L 2 φ + zx)µ(dz)

x∈Λ1

The RG map R is then deﬁned by

(Rg)(φ) = (T g)(φ)/(T g)(0).

This normalization constant will not enter in expectations. We want to study iteration of R starting with the function 2 4 g(φ) = e−rφ −λφ where we suppose r and λ are small. We obtain

2 2 4−d 4 (T g)(φ) = e−L rφ −L λφ −ν(φ) (8.20) where Z e−ν(φ) = e−σ(φ,z)µ(dz) and 2−d 2 X 2 − d−2 X 3 X 4 2 σ(φ, z) = 6λL φ zx + 4λL 2 φ zx + (λzx + rzx) (8.21) x∈Λ1 x∈Λ1 x∈Λ1 P 2 We used zx = 0. If φ is not too big, namely |φ| < ρ such that λρ is small we expect to be able to determine ν perturbatively in λ and then write

L2rφ2 + L4−dλφ4 + ν(φ) = r0φ2 + λ0φ4 + w(φ) (8.22) with suitable r0, λ0 and where w is a small perturbation. If |φ| ≥ ρ it turns out only a crude bound for w suﬃces. We will do this in the easiest case, d = 2.

103 Proposition 13. The following holds for λ, δ suﬃciently small. Let for |φ| > ρ := λ−δ

− 1 λφ4 |g(φ)| ≤ e 2 (8.23) and for |φ| ≤ ρ assume g(φ) = e−v(φ) (8.24) with v(φ) = rφ2 + λφ4 + w(φ) (8.25) where2 |r| ≤ λρ and sup |w(φ)| ≤ λ3/2 (8.26) |φ|≤ρ Then g0 := R(g) satisﬁes (8.23)-(8.26)

λ0 = L2λ, µ0 = L2µ + γλ + O(λ3/2) (8.27)

2 where γ = 6L Γxx and sup |w0(φ)| ≤ λ03/2 (8.28) |φ|≤ρ0 where ρ0 := λ0−δ Here and in what follows O(λk) means bounded by C(L)λk.

L is ﬁxed and λ < λ(L) is taken small enough. Generic constants C, c do not depend on −c(L)ρ2 n 3 L. We use e = O(λ ) for all n. The factor 2 in (8.26) is for convenience, anything less than 2 is ok by making δ small enough.

Proof. We start with bounding v0 so let |φ| ≤ ρ0. Let χ(z) be the indicator function

0 χ(z) := 1(∀x ∈ Λ1 : {|zx| ≤ ρ − ρ })

(recall that ρ0 = L−δρ) and deﬁne Y gχ(φ) := E(χ(z) g(φx)) x∈Λ1 where φx := φ + zx and E denotes expectation in dµ(z). We deﬁne g1−χ in the same way so that T (g) = gχ + g1−χ. Note that on the support of χ we have |φx| ≤ ρ for all x. Hence we may use (8.24) and (8.25) to write

P 0 4 2 2 P − v(φx) −λ φ −L rφ −σ(φ,z)− w(φx) gχ(φ) = E(e x χ(z)) = e E(e x χ(z)). (8.29) where σ is given by (8.21). Since

z4 + 4ψz3 + 6ψ2z2 = z2((z + 2ψ)2 + 2ψ2) ≥ 2z2ψ2 we get the bound 2 X 2 X 2 σ(φ, z) ≥ 2λφ zx + r zx. (8.30) x∈Λ1 x∈Λ1 Hence we can write

P −σ(φ,z)− w(φx) −σ(φ,z) Ee x χ(z) = Ee χ(z) + R1(φ) (8.31) where by the mean value theorem

P 1−δ 1−δ 2 3/2 −σ(φ,z)−t x w(φx) C(L)λ 2 3/2 C(L)λ −1 03/2 1 03/2 |R1(φ)| ≤ L λ sup Ee χ(z) ≤ e L λ = e L λ ≤ 4 λ . t∈[0,1] (8.32)

2It turns out we need to consider later r ∼ λ log λ so this is covered by our assumption on r

104 using (8.30) and (8.26). Next we Taylor expand the exponential

−σ(φ,z) Ee χ(z) = E(1 − σ(φ, z))χ(z) + R2(φ) (8.33) where by Taylor’s theorem

1 2 −tσ(φ,z) |R2(φ)| ≤ 2 sup E|σ(φ, z))| e χ(z). t∈[0,1]

We use the simple bound

2 X 4 |σ(φ, z))| ≤ C(L)λ(ρ + zx). (8.34) x whereby

2 C(L)λ 2 4 1 03/2 |R2(φ)| ≤ C(L)λ e E(ρ + zx) ≤ 4 λ (8.35) (8.36)

Finally E(1 − σ(φ, z))χ(z) = E(1 − σ(φ, z)) + E(1 − σ(φ, z))(1 − χ(z)) (8.37) and 2 2 2 2 E(1 − σ(φ, z)) = 1 − 6λL φ Γxx − L (6λΓxx + rΓxx) (8.38) Using (8.34)

X 2 4 |E(1 − σ(φ, z))(1 − χ(z))| ≤ C(L)λ E(ρ + |zx| )(1 − χ(z)). (8.39) x The last expectation is very small since on the support of (1 − χ) at least for one x we 0 − 1 δ have |zx| > ρ − ρ = (1 − L 2 )ρ:

2 4 2 4 2 1 1 −c(L)ρ2 E(ρ + |zx| )(1 − χ(z)) ≤ (E(ρ + |zx| ) ) 2 (E(1 − χ(z))) 2 ≤ C(L)e (8.40) Combining (8.31), (8.32), (8.36), (8.39) and (8.40) we conclude for (8.29)

−λ0φ4−L2rφ2 2 gχ(φ) = e (1 + E − γλφ + R3(φ)) (8.41)

3 03/2 with |E| ≤ Cλρ and |R3(φ)| ≤ 4 λ . To conclude the induction for v(φ) we need to bound Y g1−χ(φ) = E((1 − χ(z)) g(φx)) (8.42) x∈Λ1

Since kgk∞ < C we get

−c(L)ρ2 g1−χ(φ) ≤ C(L)E(1 − χ(z)) ≤ e . (8.43) Combining this with (8.41) and using |λ0φ4 + L2rφ2| ≤ C(L)λρ4 we obtain

−λ0φ4−L2rφ2 2 (T g)(φ) = e (1 + E − γλφ + R4(φ)) (8.44)

7 03/2 with |R4(φ)| ≤ 8 λ . From this we conclude 2 (T g)(0) = 1 + E + R4(0) ≥ 1 − Cλρ (8.45) and 0 2 2 1 03/2 |w (φ) − R4(φ))| ≤ C(L)λ ρ ≤ 8 λ for δ small. The bounds for v0(φ) now follow.

105 0 0 Let now |φ| ≥ ρ . Recall that χ(z) be the indicator of the event |zx| ≤ ρ − ρ for all x. Under χ we then have

0 −δ −δ |φx| ≥ 2ρ − ρ = (2L − 1)ρ > c(L)ρ = c(L)λ provided we take δ small depending on L. Thus, under χ, if φx < ρ we still have v(φx) ≥ 1 4 2 λφx and so we can bound

1 P 4 − 2 λ x∈Λ φx gχ(φ) ≤ Ee 1 χ(z). Using (8.30) we get by an easy Gaussian integral estimate

− 1 λ0φ4 −c(L)λρ2 P z2 − 1 λ0φ4−c(L)λρ2 gχ(φ) ≤ e 2 Ee x x ≤ e 2 . (8.46)

Finally consider g1−χ. Under 1 − χ φx may be small so we use the bound

1 4 − λφx g(φx) ≤ Ce 2 valid for all φx. Using (8.30) we obtain

− 1 λ0φ4 − 1 λ0φ4 −c(L)ρ2 g1−χ(φ) ≤ C(L)e 2 E(1 − χ) ≤ e 2 e . (8.47)

0 Since T (g) = gχ + g1−χ, and g = T (g)/T (g)(0) using (8.45) we get

− 1 λ0φ4 2 2 − 1 λ0φ4 g0(φ) ≤ e 2 eO(λρ)(e−c(L)λρ + e−c(L)ρ ) ≤ e 2 . (8.48) as claimed.

(N) (N) −vN We now take gN = e with

(N) (N) 2 (N) 4 vN (φ) = rN φ + λN φ .

(N) −N (N) (N) with λN = L λ. We want to determine the coeﬃcient rN so that limN→∞ gn exists. This is easily done given the recursion (8.27):

(N) 2 (N) −2n −3n 3/2 rn−1 = L rn + γL λ + O(L λ )

−2n Substituting rn = L ρn we obtain

(N) (N) −n 3/2 ρn−1 = ρn + γλ + O(L λ ).

(N) This shows that ρn stays bounded only if we take

(N) ρN = −γλN + ρ. Then (N) −n 3/2 ρn = −γλn + ρ + O(L λ ). so that (N) −2n −2n −3n 3/2 rn = −γλnL + ρL + O(L λ ). Note that this dimensionless mass term indeed is small for all N, n whereas in the QFT units the bare mass is (N) µN = −γλN + ρ. i.e. the counterterm −γλN blows up logarithmically in the UV cutoﬀ L−N . We have (N) proved that with this counterterm gn satisﬁes the assumptions of the Proposition for all (N) N. One can also easily prove along similar lines that limN→∞ gn exists.

106 Chapter 9

RG analysis of the Ginzburg Landau Model

9.1 Linear RG

We will now start analysing the RG transformation (7.10) generalizing what we learnt in the hierarchical model. From (7.10) we get immediately for the linearization of R:

Z d−2 d−2 − 2 · − 2 · (LV )(φ) = V (L φ( L ) + Z)dµΓ(Z) := hV (L φ( L ) + Z)iΓ.

We will next ﬁnd some eigenfunctions of this linear map. Let us ﬁrst see some examples. R 2 Example 1. Let V = Λ φ(x) dx and assume d > 2. Then

Z d−2 Z Z − 2 x 2 2−d x 2 2 LV (φ) = h (L φ( L ) + Z(x) dxiΓ = L φ( L ) dx + dxhZ(x) iΓ Λ Λ Λ Z Z = L2 φ(x)2dx + |Λ|Γ(0) = L2 (φ(x)2 + Ld−2Γ(0))dx. Λ/L Λ/L

2 2−d where we used hZ(x) iΓ = Γ(0). Hence, since, from (7.9), G(0) = L G(0) + Γ(0), we get Ld−2Γ(0) = Ld−2G(0) − G(0), and Z Z L (φ(x)2 − G(0))dx = L2 (φ(x)2 − G(0))dx Λ Λ/L

d R 2 2 i.e. we found an eigenfunction (as Λ → R ) (φ(x) −G(0))dx with eigenvalue L i.e. with exponent α = 2. Note that this corresponds to a mass term. In the lattice cutoﬀ case we have

2 1 − m P φ(x)2 lim e 2 x∈Λ dµ (φ) = dµ (φ) G Gm2 Λ→Rd ZΛ Z ip(x−y) −p2 1 G 2 (x − y) = e e . m µ(p) + m2

(Can you ﬁgure out what happens in the momentum space cutoﬀ case?). This is a relevant eigenvector.

Remark 7. This argument fails in d = 2. As we saw already in the hierarchical model L is not diagonalizable in d = 2. We rather obtain Z Z L φ(x)2dx = L2 (φ(x)2 + Γ(0))dx. Λ Λ/L

107 2 R x Example 2. Consider V (φ) = Λ ∇φ(x) dx. In the same way, using ∇(φ( L )) = L−1(∇φ)( x ) we get : L Z LV = ∇φ(x))2dx − |Λ|∇2Γ(0) Λ/L and Z Z L [(∇φ(x))2 + ∇2G(0)])dx = [(∇φ(x))2 + ∇2G(0)]dx Λ Λ/L so this is a marginal eigenvector. Note that it corresponds to a change of the ﬁxed point which is formally given by H = R (∇φ)2. In fact we have a one-parameter family of Gaussian ﬁxed points dµzG z > 0. Note also that this eigenvector exists also at d = 2. R 4 2m+1 Example 3. Consider V (φ) = Λ φ(x) dx. Now (using hZ iΓ = 0) we get Z Z LV = L4−d φ(x)4dx + 6L2Γ(0) φ(x)2dx + 3Γ(0)2|Λ| Λ/L Λ/L so we get a familiar eigenvector Z 4 2 2 V4 = (φ(x) − 6G(0)φ(x) + 3G(0) )dx Λ which is relevant if d < 4, marginal if d = 4 and irrelevant if d > 4, of course exactly as in the hierarchical model. At d = 2 again there is no eigenvector and we just have Z Z L φ(x)4dx = L2 (φ(x)4 + 6Γ(0)φ(x)2 + 3Γ(0)2)dx. Λ Λ/L where the φ(x)2 term is the source of the diverging mass counterterm for the QFT.

It is obvious now that we have eigenvectors Vn, polynomial of degree n, with eigenvalues d n 2−d L L 2 . We can get the formula as we did in the hierarchical model by considering

2−d φ(f) L 2 R φ( x )f(x)dx Z(f) φ(f˜)+ 1 (f,Γf) Le = e L he iΓ = e 2

2−d d 2+d where f˜(x) = L 2 L f(Lx) = L 2 f(Lx). Since Z Z ˜ ˜ 2+d 2−d x−y (f, Gf) = L f(Lx)G(x − y)f(Ly)dxdy = L f(x)G( L )f(y)dxdy we have from (7.9)(f, Γf) = (f, Gf) − (f,˜ Gf˜) and so

φ(f)− 1 (f,Gf) φ(f˜)− 1 (f,G˜ f˜) Le 2 = e 2 .

We can generate the eigenvectors from this. Recall the Hermite polynomials (8.7) and let

n d −(f,Gf)λ2+λφ(f) Vn(φ, f) = Hn φ(f), (f, Gf) = e . dλn λ=0 Then LVn(φ, f) = Vn(φ, f˜). Taking the limit φ(f) → φ(x) (i.e. letting f tend to a delta function), (f, Gf) → G(0) and Vn(φ, f) becomes a polynomial Pn(φ(x) in φ(x),

n n−2 Pn(φ(x)) = Hn(φ(x),G(0)) = φ(x) + aφ(x) G(0) + ··· and Z Z d+n 2−d L Pn(φ(x))dx = L 2 Pn(φ(x))dx. Λ Λ/L

108 A local V is an integral of a polynomial in φ(x) and its derivatives. Clearly we get eigenvectors ` Z Y nk φ(x)n ∇kφ(x) dx + lower order polynomial Λ k=1 with eigenvalue d 2−d (n+P n ) −(P kn ) L L 2 k L k i.e. the more there are φ’s or derivatives, the more V is irrelevant. In the class of even local V ’s the relevant ones are R φ2, R φ4 for d < 4 and the marginal ones are R (∇φ)2, R φ6 at d = 3 and R φ4 at d = 4. R d+2 In case of odd V , we have φ with eigenvalue L 2 always relevant (this corresponds R 3 6−d R 5 to a magnetic ﬁeld), φ with eigenvalue L 2 relevant for d < 6, φ with eigenvalue 5− 3d R R 2 L 2 , relevant 3d < 10 (note that ∇φ, ∇φφ are boundary terms).

9.2 The space of local Hamiltonians

We have seen that the linearized RG L preserves the set of local Hamiltonians. However, unlike in the hierarchical model R does not. We will now set up a space K of semi local Hamiltonians that is preserved by L, and, as we’ll see also by R in the perturbative analysis. We consider V ’s of the form

n Z Y dx1 ··· dxnK(x1, ··· , xn) φ(xi) i=1 and specify the K’s. For this, generalize a bit what we found above. Deﬁne the normal ordered product :

n Y X |I| Y Y : φ(xi) := (−1) 2 φ(xj)h φ(xi)i. i=1 I⊂{1,··· ,n} j6∈I i∈I

Excercise 23. Show that Z n n n Y Y Y ∂ h 1 i : φ(xi): f(xi)dxi = exp − (f, Gf) + φ(f) ∂λ λ =0 2 i=1 i=1 i=1 i i

Pn d where f = i=1 λifi, fi ∈ S(R ). This implies : n n Y 2−d Y 2 n xi L : φ(xi) := L : φ L : . i=1 i=1

m Q mα d More generally, let ∇ φ = α ∂α φ for m = (m1, . . . , md) ∈ N . Then our space K consists of ﬁnite sums of the form

Z N N X Y mi ni Y VΛ(φ) = Kmn(x1, . . . , xN ): (∇ φ(xi)) : 1Λ(xi)dxi (9.1) mn i=1 i=1

d where m = (m1, . . . , mN ) with mi ∈ N and n = (n1, . . . , nN ). We assume further that the kernels are translation invariant and rotation covariant so that for φ0(x) = φ(Ox + y), d O ∈ O(d), x ∈ R 0 VΛ0 (φ ) = V (φ)

109 where Λ0 = OΛ + y. Furthermore we suppose they are exponentially decaying in the sense Z γL(x) |Kmn(0, x2, . . . , xN )|e dx2 . . . dxN := kKmnkγ < ∞

d where γ > 0 and L(x) is the length of the shortest connected path Γ in R such that 0 and all xi belong to Γ. Then LV has kernels K0:

2−d PN P 0 i=1ni − ni|mi| Nd Kmn(x) = L 2 L L Kmn(Lx).

Since Z Z γL(x) −(N−1)d γL(L−1x) |Kmn(0,Lx)e dx = L |Kmn(0, x)|e dx and obviously L(L−1x) = L−1L(x) we obtain

2−d PN P 0 i=1ni − ni|mi| d kKmnkLγ = L 2 L L kKmnkγ.

Hence LV ∈ KLγ and thus also LV ∈ Kγ. We conclude that an even V in Kγ is irrelevant (the norm contracts under L) except for the following cases: Z Z Z K(x − y)φ(x)φ(y), Kαβ(x − y)∇αφ(x)∇βφ(y), K(x − y)φ(x)∆φ(y) Z Z 2 2 K(x1 ··· x4)φ(x1) ··· φ(x4), K(x − y)φ(x) φ(y) Z Z 2 3 K(x1x2x3)φ(x1)φ(x2)φ(x3) , K(x − y)φ(x)φ(y) . (9.2)

Our last task is to express these in the ”basis” R φ2, R φ4, R (∇φ)2 modulo irrelevant terms. We’ll do this to the ﬁrst expression in (9.2), the others being similar. These can be split into relevant/marginal and irrelevant as follows. Consider e.g. the ﬁrst term in (9.2).

R γ|x| Lemma 9. Let K(x) be rotation invariant with kKkγ = |K(x)|e dx < ∞. Then for d all φ ∈ S(R ) Z Z K(x − y)φ(x)φ(y)dxdy = (αφ(x)2 + β(∇φ(x))2)dx + Ve(φ)

where |α|, |β| ≤ CkKkγ and Ve ∈ Kγ/2 is irrelevant with kernel ≤ CkKkγ.

Proof. Since R |K(x)|eγ|x|dx < ∞ we get that the Fourier transform Z Z |Kb(p)| = | eipxK(x)dx| ≤ e|Imp||x||K(x)|dx is analytic in |Imp| < γ. Taylor expand

Kb(p) = Kb(0) + βp2 + Rb(p) (9.3)

(we used the rotation symmetry to get the second term of the form βp2) where

Z 1 4 1 3 d Rb(p) = dt(1 − t) 4 Kb(tp) 6 0 dt We get |α| = |Kb(0)| ≤ kKkγ

110 and Z Z 1 1 2 −2 γ|x| 2 −2 |β| = 2 |∆Kb(0)| = 2 | K(x)x dx| ≤ γ e K(x)x dx = γ kKkγ. It remains to study the remainder. We have

d4 X Kb(tp) = pαpβpγpδ(∂α∂β∂γ∂δKb)(tp) dt4 αβγδ and so X Rb(p) = pαpβpγpδHbαβγδ(p). αβγδ with Z 1 1 3 Hbαβγδ(p) = 6 dt(1 − t) (xαxβ\xγxδK(x))(tp). 0 Taking inverse Fourier transform and changing variables we get

Z 1 Z 1 3 −d ipx/t Hαβγδ(x) = 6 dt(1 − t) t e (xαx\βxγK(x))(p)dp 0 Z 1 1 2 −d−3 = 6 dt(1 − t) t xαxβxγK(x/t). 0 Hence Z 1 Z 1 −d 3 γ|x|/2 kHαβγkγ/2 ≤ 6 dtt dx(|x|/t) |K(x/t)|e 0 Z 1 Z 1 3 γt|x|/2 = 6 dt dx|x| |K(x)|e . 0

1 3 8 γ|x|/2 Since 6 |x| ≤ γ3 e we get ﬁnally 8 kH k ≤ kKk . αβγδ γ/2 γ3 γ Thus X Z V˜ = Hαβγδ(x − y)∂α∂βφ(x)∂γ∂δφ(y)dxdy is irrelevant and satisﬁes the claim.

The other non-irrelevant Hamiltonians can be analyzed similarly, e.g. Z Z Y 4 K(x1, . . . , x4) φ(xi) = a φ dx + Ve

R where a = K(0, x2, x3, x4)dx2dx3dx4 and Ve is irrelevant. Thus every V ∈ Kγ can be written as Z Z Z 2 2 4 V = r : φ : dx + z :(∇φ) : dx + λ : φ : dx + V,e Ve ∈ Kγ/2 and then Z Z Z LV = L2r : φ2 : dx + z :(∇φ)2 : dx + L4−dλ : φ4 : dx + Ve 0

0 0 −2 with Ve ∈ KLγ/2, kVe kLγ/2 ≤ L kVek. These are our “coordinates” in K.

111 9.3 Perturbative analysis of RV

Consider now the full R:

Z h 2−d · i RV (φ) = − log dµ (Z) exp −V (L 2 φ( ) + Z) . (9.4) Λ Γ Λ L Suppose V is in K as above. As discussed in the end of section 7.3 to deﬁne RV (φ) one needs to address the convergence of (9.4) as Z becomes large. This is the large ﬁeld R 4 problem. Say for VΛ = λ Λ φ we need at least Re λ > 0. For a V as in the previous section, this positivity property is not easy to state. With a lot of work, RV can actually be analyzed rigorously. One does the following:

a) For |φ(x)| < C one uses the representation in terms of Kmn. b) For |φ(x)| > C one uses another representation where positivity is explicit. The problem is to combine a) and b): φ(x) can be small and φ(y) large nearby. One uses expansion ideas from statistical mechanics to decouple these problems. Instead of trying to address the stability problem, we will limit ourselves to calculating RV perturbatively in V . R 4 Thus, let us start with VΛ(φ) = λ Λ φ(x) dx and see what RVΛ looks like. Expand in powers of λ: N X n (RVΛ)(φ) = λ vn(φ) + remainder n=0 with n n 2−d 1 d 1 d −V (L 2 φ( · +Z) vn(φ) = (RVΛ)(φ) = − loghe L iΓ n! dλn λ=0 n! dλn λ=0 Z (−1)n+1 4 4 Y = n! hψ(x1) ; ... ; ψ(xn) iΓ dxi n Λ i where we used Exercise 17 (recall that ”;” refers to truncated correlation) and denoted

2−d 2 x ψ(x) = L φ( L ) + Z(x). 2−d 2 x As an example consider v2. Denote L φ( L ) := Ψ(x). So using Wick theorem 4 4 4 4 4 4 hψ(x) ; ψ(y) iΓ = hψ(x) ψ(y) iΓ − hψ(x) ihψ(y) iΓ = 72Ψ(x)2Ψ(y)2Γ(x − y)2 + 48 Ψ(x)Ψ(y)3 + Ψ(x)3Ψ(y) Γ(x − y)Γ(0) + 4 ! Γ(x − y)4 + 16Γ(x − y)2Γ(0)2 + 96Ψ(x)Ψ(y)Γ(x − y)3 + 6 · 4 · 3(Ψ(x)2 + Ψ(y)2)Γ(x − y)2Γ(0) + 16Ψ(x)3Ψ(y)3Γ(x − y) + 144Ψ(x)Ψ(y)Γ(x − y)Γ(0)2 or graphically 1:

+ + + +

+ + +

1There is one line missing on the right of graph 6

112 The conventions for these graphs are as follows: - The edges (x, y) carry Γ(x − y) 2−d 2 x - External vertices x carry Ψ(x) = L φ( L ). Thus we end up with the representation X vn(φ) = WG(φ) G where the sum runs through all connected graphs G with n+m vertices x1, . . . , xn, y1, . . . , ym. Vertices yi have four edges ({z, z} allowed) attached and vertices xj have 4 − nj edges attached where nj > 0. The amplitude of the graph equals Z n m n Y 0 Y ni Y Y WG(φ) = n(G) Γ(z − z ) Ψ(xi) dyj dxi m+n Λ {z,z0}∈e(G) i=1 i=1 i=1 Z n Y ni := KG(x1, . . . , xn, Λ) Ψ(xi) dxi n Λ i=1

Recalling that |Γ(x − y)| ≤ Ce−α|x−y|, for some α and the fact that the graph is connected we get

Excercise 24. Show that kKGkγ < ∞, for some γ > 0 and uniformly in Λ. Thus PN n n=0 λ vn(φ) ∈ Kγ. R 4 Note that if we consider instead V (φ) = λ Λ : φ(x) : dx then there are no ”tadpole” graphs. Thus e.g. for v2 we get the graphs + +

From our analysis in the previous section, we get, to 3rd order in λ Z Z 4−d 2 3 4 2 3 2 (RVΛ)(φ) = (L λ + aλ + O(λ )) : φ(x) : +(bλ + O(λ )) : φ : L−1Λ L−1Λ Z + (cλ2 + O(λ3)) (∇φ)2 + Ve L−1Λ where Ve = O(λ2) is irrelevant. Remark 8. Strictly speaking this analysis was done in the previous section for translation d and rotation invariant kernels whereas these properties hold for KG(x, Λ) only in the ΛR limit. Since this limit exists this problem is easily resolved and we work here in that limit.

Here the constant a comes from the graph and after performing the

Wick ordering form the graph :

Z Z 2 2−d x y a = −36 Γ(x) dx + 72 Γ(x − y)L G( L − L ) b, c come from and and

(see computation for d = 4 below).

113 9.4 Iteration

R 4 R 2 R 2 Suppose now V = λ : φ : +r : φ : +z :(∇φ) : +Ve where Ve ∈ K is irrelevant, −α kLVekγ ≤ L kVekγ for α > 0. Calculate perturbatively RV :

N X n+1 (RV )(φ) = (−) hV ; V ; ··· ; V iΓ + Remainder n=1 N X = (RV)n(φ) + Remainder n=0

d−2 − 2 · where V = V (L φ( L ) + Z). The point is now that (RV )n ∈ Kγ again. Indeed, we get graphs as before and also from r, z and Ve. Pictorially, the Ve ones are

where represent the kernels Kmn and the lines are Γ(x − y). It is not hard to show that the exponential decay of K and Γ give rise to exponential decay of these graphs. ∼n m∼ Thus the claim.

4 DR Y 2E Example K(x1, ··· , x4) φ(xi) gives for instance i=1

Z 4 Z 4 Y 2−d xi Y = dx L 2 φ dy K(x , x , y , y )Γ(y − y )Γ(y − y )K(y , y , x , x ) i L i 1 2 1 2 1 3 2 4 3 4 3 4 i=1 i=1 Z Y := Ke(x1, ··· , x4) φ(xi)dxi where

Z 4 4−2d 4d Y Ke(x) = L L dyiK(Lx1, Lx2, y1, y2)Γ(y1 − y3)Γ(y2 − y4)K(y3, y4, Lx3, Lx4) i=1

114 and Z γL(0,x2,x3,x4) e |Ke(0, ··· , x4)|dx2 ··· dx4

Z −1 −3d L γL(0,x2,x3,x4) −1 −1 −1 = L e |Ke(0,L x2,L x3,L x4)|dx2 ··· dx4 (9.5)

Use L(0, x2, x3, x4) ≤ L(0, x2, y1, y2) + |y1 − y3| + |y2 − y4| + L(y3, y4, x3, x4) (since the RHS is the length of a particular path Γ such that 0, x2, x3, x4 ∈ Γ). So,

Z 4 4 4−d Y Y (9.5) ≤ L dyi dxiK¯ (0, x2, y1y2)Γ(¯ y1 − y3)Γ(¯ y2 − y4)K¯ (y3, ··· , x4) i=1 i=2

L−1γL L−1γ|x| with K¯ = e K, Γ(¯ x) = e Γ(x). Use then |Γ(¯ y2 − y4)| ≤ C, integrate over y4, x3, x4 :

Z 3 4−d Y ¯ ¯ (9.5) ≤ L dyidx2|K(0, x2, y1, y2)Γ(y1 − y3)|kKkL−1γ. i=1

R 0 Integrate y3 : |Γ(¯ y1 − y3)|dy3 < C and then the rest :

4−d 2 4−d 2 (9.5) ≤ CL kKkL−1γ ≤ CL kKkγ.

Summary Perturbatively, the RnV retains the form : Z Z Z n 2 2 4 Vn ≡ R V (φ) = zn :(∇φ) : +rn : φ : +λn : φ : +Ven L−nΛ L−nΛ L−nΛ

−α where Ven ∈ K and kLVenk ≤ L kVnk, α > 0. We get a recursion of the same form as in the hierarchical model the only new aspect being the wave function renormalization z which iterates as

2 zn+1 = zn + O(λn)

Let us brieﬂy discuss the various cases we considered in the hierarchical model.

9.4.1 d > 4

The determination of the critical point r(λ) goes exactly as before so consider next zn. 2 Since zn+1 = zn + O(λn) and z0 = 0 we have

n−1 n−1 X 2 X (4−d)m 2 2 zn = O(λm) = O((L λ) ) → z∞ = O(λ ) as n → ∞. m=0 m=0

d So we expect (formally in the limit Λ → R ) Z RnV −→ z :(∇φ)2 : . n→∞ ∞

Our RG has a one-parameter family of Gaussian fixed points given by µαG with α > 0 −1 and the GL model flows to α = (1+2z∞) . The Gaussian line has one unstable direction : φ2 : and, once that is fixed to the critical value, we flow to the line.

115 9.4.2 d = 4

As in the hierarchical model λn+1 = λn to leading order and we have

2 3 λn+1 = λn + aλn + O(λn).

Let us compute the coeﬃcient a from the second order from the term Z 1 2−d 2−d y 2 x 4 2 4 − 2 h:(L φ( L ) + Z(x)) ::(L φ( L ) + Z(y)) : iΓ dxdy. (9.6)

This gives graph a) which contributes directly and b) which contributes indirectly. a) equals Z 4−2d x 2 y 2 2 −36L : φ( L ) :: φ( L ) : Γ(x − y) dxdy Z = −36L4 : φ(x)2φ(y)2 : Γ(L(x − y)2dxdy + P (φ) where P (φ) is quadratic in φ. b) equals Z −8L6−d : φ(x)3 : Γ(L(x − y)) : φ(y)3 : dxdy.

Use

: φ(x)3 :: φ(y)3 :=: φ(x)3φ(y)3 : +9G(x − y): φ(x)2φ(y)2 : +P (φ) where P is quadratic in φ (and so contributes to r and z). Altogether the quartic term is Z : φ(x)2φ(y)2 : K(x − y)dxdy with

K(x) = −36L4(Γ(Lx)2 + 2L2−dΓ(Lx)G(x)) = −36L4(G(Lx)2 − L−4G(x)2) where we used Γ(Ly)2 + 2L2−dG(y)Γ(Ly) = G(Ly)2 − (L2−dG(y))2 and set d = 4. Finally we have Z Z : φ(x)2φ(y)2 : K(x − y)dxdy = a : φ(x)4 : dx + irrelevant with Z Z a = K(x)dx = −36 dy[G(y)2 − L−4G(y/L)2] (9.7)

Proposition 14. a is universal i.e. independent of the cutoﬀ function χ(p) in the covariance Gˆ(p) = p−2χ(p) and given by

a = β2 log L

−2 with β2 = −9π log L.

116 Proof. We have Z Z log L d a = 36 d4y ds (e−4sG(e−s|y|)2) 0 ds where we recall that G(x ) depends on |x| := r.Thus,

Z log L Z ∞ d a = −36 · 4π2 ds dr r3 4 + e−4sr G(e−sr)2 0 0 dr Z log L Z ∞ d = −144π2 ds dr r4e−4sG(e−sr)2 0 0 dr Z log L = −144π2 dse−4s lim r4G(e−sr)2. 0 r→∞ By the residue theorem √ 2 Z 4 ip1r Z 3 − p r Z ∞ d p e −p2 d p e 1 −ρ 1 G(r) = 4 2 e = 3 p = 2 ρe dρ = 2 (2π) p (2π) 2 p2 (2πr) 0 (2πr)

So e−4s lim r4G(e−sr)2 = (2π)−4 r→∞ and the claim follows for the cutoff function χ(p) = e−p2 . d ˆ ˆ 2 Let now χ(p) ∈ S(R ) be another rotation invariant cutoff. Then Gχ(p)−G(p) = O(p ) d 2 as p → 0. Hence G − Gχ ∈ S(R ) and so limr→∞ r G(r) = 0. Finally, the z parameter flows

2 zn+1 = zn + O(λn) so that ∞ X 2 zn → z∞ = O(λn) < ∞ n=0 1 since λn ∼ n .

9.4.3 d = 4 − : anomalous scaling Let us ﬁnally see how the presence of the z-parameter changes the hierarchical model analysis of the Wilson-Fisher ﬁxed point. The λ iteration formally becomes for d = 4 − :

2 3 λn+1 = L λn + bλn + O(λn). where b = β2 log L + O(). As in the hierarchical model we ﬁnd a ﬁxed point

λ∗ = − + O(2). β2 For z we have 2 3 ∗2 zn+1 = zn + γλn + O(λn) = zn + O(λ ) ∗ so zn increases linearly in n. What went wrong ? We have indeed a new ﬁxed point λ 6= 0 and we need a diﬀerent scaling of φ in our RG :

− d−2 − d−2+η L 2 → L 2

− η i.e. we need to rescale the ﬁeld φ → L 2 φ. Let us proceed formally and write

− 1 (φ,G−1φ)−V − 1 R (∇φ)2−V e−HDφ = e 2 Dφ ≈ e 2 Dφ

117 Then Z Z Z −η 1 2 2 −η 2 −2η 4 RH = L ( 2 + γλ ) (∇φ) + L r1 : φ : +L λ1 : φ : +V.e

2 1 We fix η so that at the fixed point the coefficient of (∇φ) is 2 . Hence

η = 2γλ∗2/ log L + O(3). The coeﬃcient γ is computed from the second order graphs along the same lines as we did for β2. We ﬁrst need to extract from (9.6) the quadratic term. Similar calculation as for the quartic term gives Z Z 2−d x y 3 −3 x−y 3 −48L φ( L )φ( L )(G(x − y) − L G( L ) )dxdy = φ(x)φ(y)K(x − y)dxdy with K(x) = −48L2+d(G(L(x − y))3 − L−6G(x − y)3). Then we write Z Z φ(x)φ(y)K(x − y)dxdy = (aφ(x)2 + γ(∇φ(x))2)dx + irrelevant where Z Z 1 2 2 3 −6 3 γ = − 2 x K(x)dx = 24 x (G(x) − L G(x/L) )dx.

We write this as Z Z log L d γ = −24 dx x2 ds (e−2sG(e−s|x|))3 0 ds where we recall that G(x ) depends on |x| := r.Thus, Z log L Z ∞ d γ = 24 · 4π2 ds dr r5 6 + r (e−2sG(e−sr))3 0 0 dr Z log L Z ∞ d = 96π2 dse−6s dr r6(G(e−sr)3 0 0 dr Z log L 3 2 6 −2s −s 3 log L = 96π ds lim r e G(e r) = 4 . 0 r→∞ 4π Hence we get 1 η = 2 + O(3) 54 a celebrated result of Wilson and Fisher. Note that the leading term is very small even for = 1! The value of η for the three dimensional Ising model is η = 0.036.. and by universality this value should be the one for GL model too. The above value is not that far oﬀ!

Summary d = 4 − We have a new ﬁxed point Z Z Z ∗ 1 2 ∗ 2 ∗ 4 ∗ H = 2 (∇φ) + r : φ : +λ : φ : +Ve

∗ −1 2 ∗ 2 ∗ 2 with λ = −β2 + O( ), r = O( ), Ve = O( ). The two point function has anomalous decay 1 hφ(x)φ(y)i ∗ ∼ H |x − y|d−2+η

1 2 3 ∗ with η = 54 + O( ). H has only one relevant direction, namely the correlation length and no marginal ones ((∇φ)2 is irrelevant under the new scaling).

118 9.4.4 Other fixed points for d < 4 Tri-critical point At d = 3 we get a new marginal eigenvector at the Gaussian fixed point namely : φ6 : whose scaling exponent is 6 − 3d. It becomes relevant for d < 3. The tricritical point at d = 3 is the interaction Hamiltonian X V = (r : φ(x):2 +λ : φ(x)) :4 +u : φ(x)) :6) x where we fix the two relevant parameters r and λ to critical values r(u) and λ(u) such n that R V tends to the Gaussian fixed point. Excercise 25. Compute the flow of u and show it indeed is infrared asymptotically free:

2 3 un+1 = ζun + O(un) with ζ < 0. Calculate ζ and show it is universal.

For d < 3 the coupling u is relevant and at d = 3 − we have a new ﬁxed point with

u∗ = α, r, λ = O(2).

Excercise 26. Compute α and the anomalous dimension η.

Two dimensional minimal models

2n 2n The dimension of : φ is (n + 1)d − 2n. Hence it becomes relevant for d < dn = n+1 . dn is an ”n-critical point” where n − 1 parameters have to be fine tuned. At d = 2 this leads us to expect that there are an infinite number of critical theories constructed from (even) polynomials of φ and thus an infinite number of RG fixed points. The scaling limits at these fixed points are two dimensional scale invariant quantum field theories. In 2d such theories posses a rich symmetry, local conformal invariance; they are conformal field theories (CFTs). In [5] Belavin, Polyakov and Zamolodchikov found such a discrete series of CFTs and it was later argued by A.B. Zamolodchikov [45] that these are the same CFT as the ones coming from even polynomials in φ.

9.5 Quantum Field Theories

9.5.1 Effective Field Theories The conclusions from our RG analysis to the problem of constructing Quantum Field Theories are the same as in the context of the hierarchical model and we will be rather brief in this section. We consider the regularized QFT measure (we write this for momentum cutoff but identical conclusions hold for lattice cutoff)

1 R 2 R 2 R 4 −z Λ:(∇ϕ) :−ρ Λ:ϕ :−g Λ:ϕ : −W(ϕ) dν(ϕ) = e dµG := e dµG . Z where 1 Gb(p) = χ(p) p2

d −p2 and χ ∈ S(R ) is a positive cutoﬀ function with χ(0) = 1, for instance χ(p) = e . Remark 9. As discussed in section 7.9.2 we use in two dimensions the IR cutoﬀ covari- 1 ance Gb(p) = p2 (χ(p) − χ(Rp)). As we saw there it will not enter the RG so we ignore it from now on.

119 0 We ask : are there functions of z, ρ, g such that ν converges to a measure ν on S as → 0 ? We know that λ = 0 is a solution, and ask if a non-Gaussian ν is possible. The effective QFT measures were defined in Section 7.6 for the lattice cutoff. In the present setup this goes via splitting the covariance to small and large momentum pieces. Write, for ` >

1 1 1 Gb(p) = χ(p) = χ(`p) + χ(p) − χ(`p) p2 p2 p2

:= Gb`(p) + Γb,`(p) and deﬁne the eﬀective QFT measure on scales ≥ ` as

Z () () −W(ϕ+z) −W` (ϕ) dν` (ϕ) = e dµΓ,` (z) dµG` (ϕ) := e dµG`

() which has cutoff `. In the Block spin case ν` was the law of the average of the original random field ϕ over ` cubes. In the present case the connection is not as sharp but if we d take the cutoff χ to have compact support, say χ(p) = 0 for |p| ≥ 1 then if f ∈ S(R ) is a test function s.t. fˆ(p) has support in the ball |p| < `−1 then

Z Z iϕ(f) iϕ(f) () e dν(ϕ) = e dν` (ϕ)

() −1 so that ν` gives the law ofϕ ˆ(p), |p| < ` . To connect to Statistical Mechanics and RG, we need to scale as in (7.6):

(d−2) φ(x) = 2 ϕ(x)

d−2 (we take a = 2 to anticipate what comes later). Then ν becomes µ:

2−d 2 · −W( φ( )) −V(φ) dµ(φ) = e dµC (φ) ≡ e dµC

1 2 2 where Cb(p) = p2 χ(p), has UV-cutoff 1, and is Gaussian. (Note the IR cutoff m ). Here Z Z Z 2 2 2 4−d 4 V(φ) = z :(∇ϕ) : + ρ : ϕ : + g : ϕ : note the powers of (so called canonical dimensions of mass, coupling). Then we define the effective dimensionless action

() V` := R`/V where RL is as before :

Z 2−d 2 · −V (L φ( L )+Z) RLV = − log e dµΓL (Z) with 1 ΓbL(p) = (χ(p) − χ(Lp)). p2 The eﬀective QFT action is then given by a change of variables

() () d−2 2 W` (ϕ) = V` (` ϕ(`·)).

120 9.5.2 Polchinski equation Unlike in the block spin case where L has to take integer values here we can consider arbitrary L > 1 and in particular let it tend to 1. The result will be a diﬀerential equation version of the RG. Thus given a Hamiltonian V let V (s) = Res V for s ≥ 0. Thus V (0) = V . Let us compute the derivative ∂sV (s). Since V (s + σ) = Reσ V (s)we get

Z 2−d −V (e 2 σφ(e−σ·)+Z) ∂ V (s, φ) = ∂ R σ V = −∂ log e s dµ (Z). s σ σ=0 e s σ σ=0 Γeσ

0 d Excercise 27. Let µs be a Gaussian measure on S (R ) with covariance Gs depending d 2 smoothly on a parameter s. Denote gs := ds Gs. Then, for C functions F (ϕ) d Z Z F (ϕ)dµ (ϕ) = 1 ( δ , g δ )F (ϕ)dµ (ϕ) ds s 2 δϕ s δϕ s where formally Z δ2F ( δ , g δ )F (ϕ) = g (x, y) dxdy. δϕ s δϕ s δϕ(x)δϕ(y) ϕ(f) δ δ Concretely, for F (ϕ) = e we have ( δϕ , gs δϕ )F = (f, gsf)F . Applying to our case we get a second order PDE

1 δV δV ∂sV = AV − 2 ( δϕ , γ δϕ ) (9.8) where Z δV Z δ2V AV (φ) = (( 2−d − x · ∇ )φ(x)) dx + 1 γ(x − y) dxdy 2 x δφ(x) 2 δφ(x)δφ(y)

dΓes and γ(x) = ds s=0 or in Fourier space p · ∇ χ(p) γ(p) = − p . b p2

Of course we have L = e(log L)A. The differential equation (9.8) is called the Pochinski equation. It was used in [?] to give a simple proof of perturbative renormalizability of the d = 4 theory. While (9.8) looks quite simple differential equation with quadratic nonlinearity in practise for a rigorous nonperturbative analysis it has been more convenient to work with the discrete ”time” step iteration. Let us apply it to derive once more the perturbative RG equations for our model. Let t s () = e and ` = e , s > t. We are interested in t → −∞ and s fixed. Denote V` by vs and write Z 2 2 4 vs(φ) = (zs :(∇φ(x)) : +rs : φ(x) : +λs : φ(x) :)dx + ws(φ)

2t (4−d)t where ws is irrelevant and we take initial conditions zt = 0, rt = e ρet , λt = e get and wt = 0. We obtain

2 3 z˙ = γ1λ + O(λ ) 2 3 r˙ = 2r + γ2λ + O(λ ) 2 3 λ˙ = (4 − d)λ + β2λ + O(λ ) w˙ = Aw + O(λ2) where we anticipated that r = O(λ2).

Excercise 28. Compute in this formalism γ1 and β2 and check that they agree with our earlier results.

121 9.5.3 Superrenormalizable QFT As an example let us consider again the d = 3 case. It is convenient to go to the dimensional −s −2s parameters: let gs = e λs and ρs = e rs. Then

2 s 3 ρ˙s = γ2gs + O(e gs ) s 2 g˙s = O(e gs )

s 2 The ”initial” condition is gt = g with t → −∞. Hence gs = g + O(e g ) stays almost constant as long as ges < O(1). We get

2 2 ρs = ρt + (s − t)γ2g + o(g )

2 Thus, we need to take ρt = tγ2g i.e. the bare mass and coupling are

2 g = g, ρ = −γ2g log this is a “mass counterterm” of O(g2), coming from the graph that diverges logarithmically in UV. Remark 10. This perturbative analysis remains valid as long as the dimensionless cou- s¯ pling λs is small, say λs < λ¯. At the corresponding scale e we then have an effective d Hamiltonian vs¯ that will define Gibbs measures for arbitrary volumes Λ ⊂ Z . Note in particular that this statement holds for arbitrarily large coupling constant g. For large g we just have to stop the iteration at a smaller scale s¯. As in the case of the hierarchical model (see Section 8.7.3) the question of infinite volume limit is hard. The case that can be done 2 with expansions is to add a mass term m to ρ and assume g/m small. Using correlation inequalities the infinite volume limit for d = 2 and d = 3 QFT:s can be constructed for 2 all values of m ∈ R and g > 0. However the properties of the limit are hard to study except for the perturbative region. In particular we believe there exists a unique critical value m2(g) where the correlation length is infinite and for m2 < m2(g) the model should exhibit a first order phase transition with symmetry breaking. The latter property has been proven for m2 large negative.

9.5.4 d < 4: nontrivial ﬁxed points . Consider the d = 3 QFT discussed above. This was parametrized by the mass squared parameter m2 and coupling constant parameter g in the expressions for the cutoﬀ ”bare” parameters:

2 2 ρ = m − γ2g log , g = g.

Above we could have taken m = 0 and started with r, λ to produce an effective Hamiltonian on scale 1 of the form Z Z Z z : ∇ϕ2 : +r : ϕ2 : +λ : ϕ4 : +Ve with various values of r. To solve for the IR we would need to keep iterating the RG. Now λ increases and eventually, if we really choose r = r(λ) = critical point, we would tend to the nontrivial fixed point H∗. Thus, UV is controlled by the Gaussian fixed point, IR by the non-Gaussian. We could construct a scaling limit at H∗ : this would be a scale-invariant QFT with non- Gaussian IR and UV. H∗ has (at least) one relevant direction : the correlation length. We could construct, as above, a massive QFT corresponding to the unstable manifold of ∗ ∗ H . We expect the unstable manifold of the Gaussian fixed point H0 to be 2-dimensional

122 Figure 9.1: A typical ﬂow around a critical point

(r, λ), and the unstable manifold of H∗ to be one dimensional. So the flow is as given in figure 6.2 Thus, a “typical” QFT is a typical point of the r, λ plane with Gaussian UV and finite ∗ correlation length. The ones on Ms(H ) have non-Gaussian IR, Gaussian UV and the ∗ ones on Mu(H ) have non-Gaussian UV, ξ < ∞. (We should really include also the r = ∞ (ξ = 0) “high-temperature” fixed point).

9.5.5 Asymptotic Freedom

What would have happened if β2 above had been negative at d = 4 ? Now

dλ 2 3 λ(s) = |β2|λ + O(λ ) ⇒ λ(0) = ds 1 + |β2|sλ(s)

−t i.e. if = e , s = t, and we want to ﬁx λe−t (t) = g, then the “bare coupling” λ is g λ = −1 −→ 0 1 + |β2| log g →0 i.e. to have a non-Gaussian theory, we need to let λ → 0 as → 0. This is obvious, since λ is now unstable. Actually, it is instructive to calculate exactly λ : we need the λ-equation to O(λ3) (i.e. again, we need to consider z, r, Ve too, to O(λ3)). So consider dλ = β(λ) = β λ2 + β λ3 + β˜(λ), β˜ = O(λ4). ds 2 3 So Z g dλ = log −1 λ β(λ) Z g 1 1 Z g 1 β = dλ = dλ − 3 + O(1) 2 β3 2 2 2 λ β2λ 1 + λ + O(λ ) λ β2λ β λ β2 2 1 1 β3 g −1 = − − 2 log + O(g) = log β2λ β2g β2 λ g ⇒ λ = . 1 + β g log −1 − gβ3 log log −1 + O(1) 2 β2 This is how the bare coupling should be chosen to have a continuum limit.

123 Remark We have seen that β2 is universal (independent on regularization). It can be shown that β3 is also. (β4 is not). They are intrinsic, depend only on the continuum limit (which does not depend on the regularization). An example of asympotically free theory is the Ginzburg-Landau model with λ negative. This is not stable, but can be rigorously dg 2 deﬁned by analytic continuation. Since λ = −g, g > 0 we get ds = β2g + ··· β2 > 0. More interesting one is the SU(N) gauge theory. There one has two couplings z and λ, both marginal, with dz = γλ2 + ··· ds dλ = β λ3 + ··· ds 3

2 g2 −t 2 2 R t 2 γ so λ(s) ≈ 2 where = e , λ (t) = g and z(t) − z(0) = γ λ(s) ∼ log t 1+β3g (t−s) 0 β3 so z (0) must diverge as − γ log log −1. β3

Remarks

1. Thus, we expect to find QFT’s at fixed points and on their unstable manifolds. For Ginzburg-Landau model, d < 4 we find several non-Gaussian ones. For d ≥ 4 the fixed points and unstable manifolds are Gaussian.

2. Asymptotically free theories have Gaussian ﬁxed points with marginally unstable non-Gaussian direction. Here we have a non-Gaussian continuum limit.

Example : Non Abelian gauge theories with not too many fermions (QCD is one). Non-linear σ-models in d = 2 (e.g. O(N), N > 2 spin systems we discussed earlier). Certain d = 2 fermionic models (Gross-Neveu model).

3. IR behaviour of asymptotically free models is interesting. Now λ(s) % as s increases. E.g. in non-Abelian gauge and σ-models. These theories have no relevant variable like r in Ginzburg-Landau model that would provide a mass (correlation length). Nevertheless, they are massive (ξ < ∞): ξ with be ∼ e−const/λ2 (so called dimensional transmutation). There is no other critical point than λ = 0. (λ2 is the temperature in Heisenberg model).

4. Theories that are expected to be Gaussian (trivial) are : Ginzburg-Landau model, QED, Standard Model of weak and EM interactions. This sounds paradoxical since QED has a very successful renormalized perturbation theory and is the most accurate known physical theory. The point is that, as long as we do not take to 0, we can keep λ > 0 and so λ(s) > 0. Now if one calculates a correlation function in perturbation R theory, the graphs that are convergent integrals of the form |p|<−1 I(p)dp depend λ on the cutoﬀ like to some power. But we may take (recall, λ = −1 has 1−β2 log λ to be small so ((log −1)−1 ∼ λ)

∼ e−const/λ.

−const/λ 1 Hence, the dependence on cutoff is ∼ e , very small for λ small ( λ = 137 for QED). We say that all the above theories are good effective theories for distances >> (in practice, energies << 1015 GeV). For smaller distances one needs to find a new theory (say string theory).

124 Chapter 10

Stochastic PDE’s with Rough Noise

10.1 Stochastic Diﬀerential Equations and Stochastic Quan- tization

10.2 Linear case: Gaussian Process

10.3 Nonlinear SDE: Local existence

10.4 Perturbation Theory

10.5 Rough SPDE’s

We motivated the study of the SPDE (??) by stochastic quantization. In this case the time d variable is non-physical. The space variable x ∈ R is the Euclidean space time variable i.e. d = 4 is the physical case. However it is of great physical interest to study nonlinear parabolic PDE’s driven by a space time decorrelated noise. They are of the form

∂tϕ = ∆ϕ + V (ϕ) + ξ (10.1)

d where ϕ(t, x) is deﬁned on Λ ⊂ R , V (ϕ) is a function of ϕ and possibly its derivatives which can also be non-local and ξ is white noise on R × Λ, formally 0 0 0 0 E ξ(t , x )ξ(t, x) = 2δ(t − t)δ(x − x). Examples are the KPZ equation with d = 1 and

2 V (ϕ) = (∂xϕ) (KPZ) describing random deposition in surface growth and the Ginzburg-Landau model

V (ϕ) = −λϕ3 (GL) describing stochastic dynamics of spin systems and, as we saw, also stochastic quantization of the φ4 theory. Usually in these problems one is interested in the behavior of solutions in large time and/or long distances in space. In particular one is interested in stationary states and their scaling properties. These can be studied with regularized versions of the equations where the noise is replaced by a molliﬁed version that is smooth in small scales. Often one expects that the large scale behavior is insensitive to such regularization. From the mathematical point of view and sometimes also from the physical one it is of interest to inquire about the short distance properties i.e. about the well-posedness

125 of the equations without regularizations. Then one is encountering the problem which we saw above namely that the solutions are expected to have very weak regularity, they are distributions, and it is not immediately clear how to set up the solution theory for the nonlinear equations in distribution spaces. Indeed, the Picard iteration fails here: denoting η(t, x) the solution of the linear equation as above (i.e. th GFF) then for the 2 KPZ equation V (η(s)) = (∂xη(s, x)) is not defined as η has the regularity of Brownian motion in x. For the GL equation V (η(s)) = η(s, x)3 and as we have seen is not defined as a random field. From the QFT experience we expect that in general the equations (10.1) need to be renormalized in order to be well posed. Then the question arises how do we find what kind of counterterms are needed. To understand this it is natural to try apply the idea of effective equations related to each other by a RG in the same way as effective actions were studied in QFT.

10.5.1 Dimensionless variables Consider now the regularization of equation (10.1). We can do this several ways, e.g. by regularizing the white noise by Z ξ(t, x) = χ((x − y)/)ξ(t, y)dy or taking the lattice regularization discussed in Section ??. Then the equation (10.1) is replaced by

∂tϕ = ∆ϕ + V (ϕ) + ξ where in the lattice case ∆ is the discrete Laplacean. As in the QFT it will be useful to introduce dimensionless variables in terms of which the cutoﬀ = 1. Deﬁne a space time scaling operation sL by

− d−2 −2 −1 (sLϕ)(t, x) := L 2 ϕ(L t, L x).

This scaling preserves the linear equationϕ ˙ = ∆ϕ + ξ. Indeed, by a change of variables 0 0 0 0 ϕ := sLϕ satisﬁesϕ ˙ = ∆ϕ + ξ with

0 − d−2 −2 −2 −1 ξ (t, x) = L 2 ξ(L t, L x) so that

0 0 0 0 −2−d −2 0 −1 0 0 0 Eξ (t, x)ξ (t , x ) = L δ(L (t − t ))δ(L (x − x )) = δ(t − t )δ(x − x ) We will now set φ := sϕ. Then the KPZ and GL nonlinear terms 2 (∂xϕ) KPZ V(ϕ) = 3 ϕ + rϕ GL become ( 2−d 2 () 2 (∂xφ) KPZ v (φ) = 4−d 3 2 φ + rφ GL In this dimensionless formulation the equation has cutoﬀ on unit scale (instead of scale ) and the nonlinearity is small if d < 2 (KPZ), d < 4 (GL). These are the subcritical cases. However, φ is now deﬁned on [0, −2T ] × −1Λ i.e. we need to control arbitrary large times and volumes as → 0 (we denote the noise in (??) again by ξ: it equals in law the space time white noise on [0, −2T ] × −1Λ).

126 10.5.2 Hierarchical SPDE

We will illustrate the RG analysis in the context of hierarchical SPDE’s

(h) ∂tϕ = ∆ ϕ + V (ϕ) + ξ (10.2)

d (h) where ξ is the white noise on R × R and ∆ is the hierarchical Laplacean (8.17)

(h) X −2n − ∆ = L Qn. (10.3) n∈Z

The UV cutoﬀ L−N version is deﬁned by

∞ (h) X −2n −∆N = L Qn. n=−N

Similarly we can discretize the noise by setting

Z Nd ξN (t, x) = L ξ(t, y)dy BN (x)

The regularized hierarchical GL equation is deﬁned by

(h) 3 ∂tϕ = ∆N ϕ − λϕ − aN ϕ + ξN (10.4) where we look for solutions that are constant on L−N -cubes so that this equation is really −N d deﬁned on (L Z) . Restricting it to cube Λ = BM (0) the equation (10.4) has an invariant measure

− R ( λ ϕ(x)4+ aN ϕ(x)2)dx dνN,M (ϕ) = e Λ 4 2 dµN,M (ϕ). (10.5) where µN,M is the Gaussian measure with covariance

M X −2n Eϕ(x)ϕ(y) = L Qn(x, y). n=−N

10.5.3 Eﬀective equations

Recall that we studied the limit of the measure (10.5) as N → ∞ by introducing the eﬀective theories at scales L−n > L−N i.e. the law of P ϕ(x) = Lnd R ϕ(y)dy given as n Bn(x)

(N) R (N) − Λ Sn (ϕ(x))dx dνn,M (ϕ) = e dµ0 (ϕ). (10.6)

(N) and we studied the limit of effective actions limN→∞ Sn := Sn. For the SPDE we want to find the effective equation for ϕn = PnϕN . It turns out to be of the form

(h) (N) ∂tϕn = ∆n ϕn + Vn (ϕn) + ξn. (10.7)

(N) and we will next explain how to determine the function Vn .

127 10.5.4 Dimensionless formulation As in QFT it will be useful to introduce the dimensionless formulation where the UV cutoff is 1. For simplicity of notation we set the IR cutoff M = 0 i.e. we work in the unit 1 1 d box [ 2 , 2 ] . We may identify ϕN with a function φ defined on R × ΛN where

1 N 1 N d d ΛN = [− 2 L , 2 L ] ∩ Z by setting −N d−2 −2N −N φ(t, x) = L 2 ϕN (L t, L x) Then ˙ 3 φ = ∆N φ − λN φ − ρN φ + ξN . (10.8) −(4−d)N −2N where λN = L λ, ρN = L r and

N−1 X −2n −∆N = QnL n=0 where Qn = Pn − Pn+1 with

−nd X (Pnφ)(x) = L φ(y) [L−ny]=[L−nx]

n n is the averaging in L lattice cube. Note that P0 = 1 and Pnφ is constant on L lattice cubes. The noise is EξN (t, x)ξN (s, y) = δ(t − s)δxy.

Remark. This equation has a stationary measure

− P ( λN φ4 + rN φ2 ) x∈ΛN 4 x 2 x e dµGN (φ) where µGN is the Gaussian measure with covariance

N−1 X 2n GN = QnL . (10.9) n=0

10.5.5 RG for SPDE Let us now study the equation (10.8):

˙ 3 φ = ∆N φ − λN φ − rN φ + ξN . (10.10)

We decompose φ = Pφ + Qφ (10.11) where P := P1 and Q = 1 − P. Then z(t) := Qφ(t) has zero average on L-blocks and Pφ is constant on them. Thus we may identify Pφ with a ﬁeld deﬁned on ΛN−1. We set

2−d 0 −2 −1 0 (Pφ)(t, x) = L 2 φ (L t, [L x]) := (sLφ )(t, x).

Thus we have a decomposition 0 φ = sLφ + z. (10.12) Note that φ0 is just the block spin ﬁeld: extending the deﬁnition of the block spin map CL to the space time setting we have

0 − 2+d X 2 φ (t, x) = (CLφ)(t, x) := L 2 φ(L t, Lx + u) u [ L ]=0

128 Remark. Recall that in the QFT problem we arrived at this decomposition and noted 0 that if φ is the Gaussian hierarchical ﬁeld with covariance GN given by (10.9) then φ has −d covariance GN−1 and z is independent with covariance Γxy = δxy − L δ x y . [ L ][ L ] Let us now plug the decomposition (10.11), (10.12) into the equation (10.10). We have QP = (1 − P)P = 0 and QnP = Qn for n > 0. This leads to

0 −2 0 ∆N (sLφ ) = L sL(∆N−1φ ).

Then, since QQn = δn0Q we get ∆N z = −z.

Consider ﬁrst the linear version of the equation (10.10) i.e. λN = 0 = rN . It becomes a pair of equations

˙0 0 0 φ = ∆N−1φ + ξ

z˙ = −z + ξ⊥

0 where the noises ξ , ξ⊥ are as follows. First,

0 2− 2−d 2 ξ (t, x) = L 2 (PξN )(L t, Lx) and we get

0 0 2+d −2d X 2 Eξ (t, x)ξ (s, y) = L L EξN (L t, u)ξN (s, v) = δ(t − s)δxy. u v [ L ]=x,[ L ]=y

0 law Hence ξ (t, x) = ξN−1, the white noise on R × ΛN−1. 0 Next, ξ⊥ = QξN = ξN − PξN . This noise is independent of ξ and furthermore ξ⊥(x) and ξ⊥(y) are independent if x, y are in diﬀerent L-cubes. The solution of the z equation is a Gaussian process (an Ornstein-Uhlenbeck process)

Z t −t s−t z(t, y) = e z(0, y) + e ξ⊥(s, u)ds (10.13) 0 with covariance

−|t−s| −t−s −t−s Ez(t, x)z(s, y) = (e − e )Qxy + e Ez(0, x)z(0, y) where we allowed for a possibly random initial condition. Let us for simplicity consider the original equation (10.10) with the random initial condition given by the invariant measure of the gaussian hierarchical ﬁeld i.e. φ(0, x) has covariance GN . Then z(0, x) is gaussian with covariance Qxy and z(t, y) = η(t, y), the process with covariance

−|t−s| Eη(t, x)η(s, y) = e Qxy

Hence, for the linear equation the eﬀective equation is simply the same with just N replaced by N − 1:

˙0 0 φ = ∆N−1φ + ξN−1

The z and φ0 variables decouple. Let us consider now the nonlinearity in equation (10.10). It will be useful to think about it as a functional in φ taking values in functions on R × ΛN :

(N) 3 v (t, x; φ) := λN φ(t, x) + rN φ(t, x).

129 Then (10.10) becomes the pair

˙0 0 0 φ = ∆N−1φ − w + ξ

z˙ = −z − u + ξ⊥ where

2−d X (N) 2 0 w(t, x) = L 2 v (L t, y; sLφ + z) y [ L ]=x (N) 0 (N) 0 −d X (N) 0 u(t, x) = Qv (sLφ + z) (t, x) = v (t, x, sLφ + z) − L v (t, x, sLφ + z) y x [ L ]=[ L ]

Consider ﬁrst the z equation. We can sove it separately in each L cube bL(x) = y 0 {y | [ L ] = x}, x ∈ ΛN−1. Indeed, u(t, y) for y ∈ bL(x) depends only on z(t, y ) for 0 y ∈ bL(x). Moreover, the noises ξ⊥|bL(x), x ∈ ΛN−1 are i.i.d for diﬀerent x’s. Hence the solution (which we of course need to prove exists)

0 z(t, y) = z(t, y, φ (·, x)), y ∈ bL(x)

0 −2 is a function of the noise in the cube bL(x) and of the ﬁeld φ (s, x) with s ≤ L t. Plugging this solution into the φ0-equation we get a renormalized SPDE for φ0:

˙0 0 (N) 0 0 φ = ∆N−1φ − vN−1(φ ) + ξ where

(N) 2−d X 0 2 2 0 vN−1(t, x; φ ) = L vN (L t, y; sLφ + z). y [ L ]=x

0 0 vN−1(t, x; φ ) for x ∈ ΛN−1 is a random function of the φ (s, x) with s ≤ t. It depends on 2 y the noise ξN (s, y) for s ≤ L t and [ L ] = x i.e. in the L-block centered at Lx. In particular 0 {vN−1(t, x; φ )}x∈ΛN−1 are i.i.d. Remark. Note the close analogy to the RG map of the QFT problem

(N) 0 Z P y 0 −V (φ ) − [ ]=x VN (sLφ +z(x)) e N−1 = e L dµ(z)

We can obviously iterate this procedure to derive the eﬀective equation for φn = N−n CL φ: ˙ (N) (N) φn = ∆nφn − vn (φn) + ξn

(N) where vn (t, x; φ) are i.i.d. random functions of {φ(s, x)}s≤t, independent of the noise

(N) 2(N−n) N−n ξn = L CL ξN which equals in law the space time white noise ξn on R × Λn. (N) (N) 0 (N) vn are constructed inductively. Set v = vn and v = vn−1. Then

0 2−d X 2 v (t, x; φ) = L 2 v(L t, y, sLφ + z). y [ L ]=x where {z(t, y)}y∈b(x) solve the SDE

(N) z˙(t, y) = −z(t, y) + Q(−v(sLφ + z) + ξn )(t, u) (10.14)

130 It is convenient to rewrite this in the integral form

(N) z(t, y) = ηn (t, y) + ζ(t, y) (10.15)

(N) where ηn is the OU process

(N) (N) −|t−s| Eηn (t, x)ηn (s, y) = e Qxy and Z t X s−t (N) ζ(t, y) = − Qyu e v(s, u; sLφ + ηn + ζ)ds := F(t, y, ζ). (10.16) u 0 Given a realization of the original white noise ξ our task is to construct the functions (N) (N) (N) vn (t, x; φ) and study the limit limN→∞ vn (t, x; φ). Note that vn is a function of the noise PN ξ so increasing N means adding more noise.

10.5.6 Linearized RG map v0 is non-linear in v since the solution z is non-linear in v. Let us compute the linearization of this map i.e. we approximate z(t, y) by η(t, y). Let us denote the resulting iteration by (N) un i.e.

(N) 2−d X (N) 2 (N) un−1(t, x; φ) = L 2 un (L t, y; sLφ + ηn ) y with the initial condition

(N) 3 uN (t, x; φ) = λN φ(t, x) + rN φ(t, x) P (N) Using y ηn (s, y) = 0 we get [ L ]=x

(N) 3 (N) (N) un (t, x; φ) = λnφ(t, x) + an (t, x)φ(t, x) + bn (t, x).

(N) (N) The coeﬃcients an and bn are random processes. They satisfy the iteration

(N) 2−d X (N) 2 2−d X (N) 2 2 an−1(t, x) = L an (L t, y) + 3L λn ηn (L t, y) y y [ L ]=x [ L ]=x (N) 2−d X (N) 2 (N) 2 (N) 2 (N) 2 3 bn−1(t, x) = L 2 (bn (L t, y) + an (L t, y)ηn (L t, y) + λnηn (L t, y) ) y [ L ]=x

(N) (N) (N) (N) with initial conditions aN (t, x) = rN , bN (t, x) = 0. Let us deﬁne An and Bn by

(N) (N) (N) λnAn (t, x) =an (t, x) − Ean (t, x) (N) (N) λnBn (t, x) =bn (t, x). Then we obtain

(N) 3 (N) (N) (N) un (t, x; φ) = λn(φ(t, x) + An (t, x)φ(t, x) + Bn (t, x)) + rn φ(t, x) with

(N) −2 X (N) 2 (N) 2 2 (N) 2 2 An−1(t, x) = L (An (L t, y) + 3(ηn (L t, y) − Eηn (L t, y) )) y [ L ]=x (N) d −3 X (N) 2 (N) 2 (N) 2 (N) 2 3 Bn−1(t, x) = L 2 (Bn (L t, y) + An (L t, y)ηn (L t, y) + ηn (L t, y) ) y [ L ]=x (N) 2 (N) rn−1 = L rn + λnρ

131 2 2 3 2 −d 2 (N) where ρ = 3L Eη = 4 L (1 − L ) . Thus rn blows up just as in the QFT case and we have to ﬁne tune the initial condition by taking 4−d 2 −1 −2 rN = (L − L ) ρλN , d > 2, rN = L ρNλN , d = 2. (N) With this choise we have rn = rn for all n ≤ N. (N) (N) The random variables {An (t, y)}y are i.i.d. and {Bn (t, y)}y are i.i.d. and since η (N) (N) (N) is stationary the processes An and Bn are also stationary in time. The process Bn (N) has EBn = 0. Consider the variances. Let (N) (N) (N) E(An (t, x)An (s, x)) = Gn (t − s). Then using independence

(N) d−4 (N) 2 −2L2t Gn−1(t) = L Gn (L t) + ge −4 P 2 with g = 9L xy(Qxy) . Thus we get that

N−n (N) X (d−4)m −2L2mt −2L2t Gn (t) = g L e ≤ Ce m=1 uniformly in N and for t > s

N−n (N) (N) 2 (N) (N) X (d−4)k −2L2k(t−s) 4−d E(An (t)−An (s)) = 2(Gn (0)−Gn (t−s)) = g L (1−e ) ≤ C|t−s| 2 . k=1 For (N) (N) (N) EBn (t, x)Bn (s, x) = Hn (t − s) we get

(N) 2(d−3) (N) 2 d−6 (N) 2 X (N) 2 2 X (N) 2 3 2 Hn−1(t) =L Hn (L t) + L (Gn (L t)E( ηn (L t, y)) + E( ηn (L t, y) ) y y 2(d−3) (N) 2 (N) =L Hn (L t) + hn (t) where (N) (N) 2 −L2t −3L2t −3L2t hn (t) = g1Gn (L t)e + g2e ≤ Ce for positive constants gi. Hence N−n (N) X 2(d−3)m (N) 2m 3−d −3L2t Hn (t) = g L hn (L t) ≤ Ct e m=1 for d > 3 and C log t−1e−3L2t for d = 3. For d < 3 we get

N−n (N) (N) 2 (N) (N) X 2(d−3)k −2L2k(t−s) 3−d E(Bn (t)−Bn (s)) = 2(Hn (0)−Hn (t−s)) ≤ C L (1−e ) ≤ C|t−s| . k=1 (N) Proposition Almost surely the process An is H¨oldercontinuous uniformly in n. The (N) R t −(t−s) (N) same holds for Bn if d < 3 and if d = 3 for the process 0 e Bn (s)ds. (N) Proof. An belongs to the Wiener chaos of order 2 of the noise ξ. Hence one has the hypercontractivity estimate

(N) (N) 2p (N) (N) 2 p p p 4−d E(An (t) − An (s)) ≤ Cp(E(An (t) − An (s)) ) ≤ Cp |t − s| 2 for all p. From this we infer the estimates: ˜(N) −p P{ sup sup kAn ≥ R} ≤ CpTR N≥n t≤T for al p. Similar estimate holds for the H¨oldernorm.

132 10.5.7 RG iteration, d = 2

0 2n Denote the sup norm in C ([0,L ] × Λn) by k · kn i.e.

kφkn = sup sup |φ(t, x)| t∈[0,L2n] x∈Λn

Let En be the event

˜(N) (N) En = {∀m ≥ n, ∀N ≥ m : kAm km, kBm km ≤ Rm}

−δ where Rn = λn where δ is taken small enough.

Proposition. Almost surely En holds for some n < ∞. 0 2n Pick now a realization of the noise ξ s.t. En holds. Let Bn ⊂ C ([0,L ] × Λn) be the ball of radius 3Rn. Then, for φ ∈ Bn we have

1−3δ 1−2δ kun(φ)kn ≤ Cλn , kun(φ1) − un(φ2)kn ≤ Cλn kφ1 − φ2kn

1−2δ i.e. un is a Lipschitz map from Bn to Bn with small Lipschitz constant λn . The same holds for vn as well: (N) Proposition Under En the following holds. The functions vn (t, φ) are deﬁned on the 0 2n (N) ball φ ∈ Bn ⊂ C ([0,L ] × Λn) of radius 3Rn. φ → vn (·, φ) is a Lipschitz map from Bn 1−2δ to Bn with Lipschitz constant λn . (N) (N) 0 Proof We proceed by induction. Denote vn by v and vn−1 by v . We have

0 X 2 v (t, x; φ) = v(L t, y; sLφ + η + ζ) y [ L ]=x where ζ is the solution to (10.16). We solve (10.16) by contraction mapping in the ball 1−3δ kζkn ≤ Aλn for suitable A. Indeed,

−2δ ksLφ + ηy + ζykn ≤ kφkn−1 + Rn + kζkn ≤ (L + 2)Rn ≤ 3Rn

1−3δ so that F(t, y, ζ) is deﬁned. Then, by the induction assumption kF(ζ)kn ≤ C(L)λn and F is a contraction. Further, the solution is Lipschitz in φ with

1−2δ kζ(φ1) − ζ(φ2)kn ≤ C(L)λn kφ1 − φ2kn−1.

Write v(t, x; φ) = u(t, x; φ) + w(t, x; φ) We get

0 X 2 2 2 w (t, x; φ) = (u(L t, y : sLφ+η+ζ)−u(L t, y; sLφ+η)+w(L t, y; sLφ+η+ζ)) (10.17) y [ L ]=x

3 2 Assume inductively supψ∈Bn kw(ψ)kn ≤ λn . Then from (10.18) we infer

3 3 0 2 1−2δ 2−5δ 2 2 2 kw (φ)kn ≤ L (Cλn kζkn + kw(sLφ + η + ζ)kn) ≤ C(L)λn + L λn ≤ λn−1 (10.18) if δ < 1/10. 2 (N) Thus under En, supφ∈Bm kvm (φ)km are uniformly bounded in N ≥ m ≥ n. Conver- gence as N → ∞ follows by iterating the bound

(N+1) (N) −(N−m)α 1−3δ sup kvm (φ) − vm (φ)km ≤ L λm φ∈Bm

133 for small enough α. To do this we need to modify the event En a bit by adding to it the condition

˜(N+1) ˜(N) (N+1) (N) −(N−m)α ∀m ≥ n, ∀N ≥ m : kAm − Am km, kBm − Bm km ≤ L Rm

Summary. Let us summarize the intuition behind the proof. The scaling part of the linearized RG where the z variable is ignored is given by

2−d d 2−d SLf(φ) = L 2 L f(L 2 φ).

Thus SL has eigenfunctions k (k+1) 2−d d k SLφ = L 2 L φ . d+2 These expand if k < d−2 . This is interesting for critical phenomena i.e. the problem of large scale behaviour with N → ∞ and λN ﬁxed. However, we are interested in N → ∞ (d−4)N and λN = L and λ ﬁxed. Hence the relevant calculation is

k 3−k (d−2) k SL(λnφ ) = L 2 λn−1φ .

Thus these monomials contract unless k ≤ 3. Taking into account the randomness z in the k RG we get random coefficients An and Bn for φ , k = 1, 0. The expectation of these thus expand under RG. However, by symmetry EBn = 0 so only expectation EAn has to be dealt with by cancelling it with a counterterm. For variances we gain an extra L−d factor and in d = 2 they contract. Only the variance of Bn(t) increases for d ≥ 3. However, R t the relevant object for solving the effective equation is 0 B(s)ds whose variance remains bounded. The RG generates terms in second (and higher) orders in λn. In second order there is d−4 an extra λn and we gain a L factor: in d = 2 everything contracts whereas in d = 3 this becomes 1 2 k 2 (1−k) 2 k SL(λnφ ) = L λn−1φ so only k = 0 and k = 1 require care. We thus expect a mass counterterm in the second order in d = 3. In the hierarchical model setup one can modify the iteration so that d may be taken arbitrary real number. Then one finds that one needs mass counterterms k 2 up to λn with k ≤ 4−d .

134 Appendix A

Some functional analysis

We begin with some basic deﬁnitions: • An operator A on a Hilbert space H is a linear map A : D(A) → H where D(A) (called the domain of A) is a linear subspace of H

• A bounded operator A has D(a) = H and kAψk ≤ kAkkψk where kAk = supkψk=1 kAψk is the norm of A.

• An operator A with domain D(A) is closed if whenever ψn ∈ D(A), φn → φ and Aφn → ψ, then φ ∈ D(A) and ψ = Aφ.

• A1 is an extension of A if D(A1) ⊃ D(A) and A1ψ = Aψ for all ψ ∈ D(A) • A is closable if it has a closed extension. The smallest closed extension is the closure of A, denoted by A The definition of the adjoint operator A∗ requires more care. First, let A be bounded. Then for a given φ ∈ H, the map ψ → (Aψ, φ) is bounded since k(Aψ, φ)k ≤ kAkkφkkψk The Riesz lemma states that there is a unique vector η ∈ H such that (Aψ, φ) = (ψ, η). Denote the map φ → η by η = A∗φ, so (Aψ, φ) = (ψ, A∗φ) for all ψ, φ ∈ H. This map A∗ is the adjoint operator. If A is not bounded, we can still define an adjoint operator if D(A) is dense in H. In that case, define D(A∗) to be the set of φ ∈ H such that there is a vector η ∈ H for which (Aψ, φ) = (ψ, η)∀ ψ ∈ D(A) We then define the adjoint A∗ by A∗φ = η, since (Aψ, φ) = (ψ, A∗φ) ∀ψ ∈ D(A), ∀φ ∈ D(A∗) Proposition 15. A∗ is closed. A is closeable ⇐⇒ D(A∗) is dense Proof. Given in [35], page 253

We will now apply this to the situation of proposition2 and prove that a∗(f) = a(f)∗. We have already shown that (a(f)ψ, φ) = (ψ, a∗(f)φ) for all ψ, φ ∈ D. Hence, D ⊂ D(a(f)∗) and a∗φ = a(f)∗ for all φ ∈ D, so a(f)∗ is an extension of a∗(f). To ∗ ∗ P∞ show that they are equal, let φ ∈ D(a(f) ) and set η = a(f) φ. Write φ = n=0φnand P∞ n n η = n=0ηn with φn, ηn ∈ H . We have for ALL φn ∈ H that

(a(f)φn, φ) = (φn, η) = (φn, ηn) since φ ∈ D(a(f)∗). On the other hand,

∗ (a(f)φn, φ) = (a(f)φn, φn−1) = (φn, a (f)φn−1)

∗ ∗ Combining these results, we see that ηn = a (f)φn−1. Since we took φ ∈ D(a(f) , P∞ ∗ ∗ ∗ ∗ n=0ka (f)φn−1k < ∞ so also φ ∈ D(a (f)). We have proven that D(a (f)) = D(a(f) ) and a∗(f)φ = η = a(f)∗φ, which proves proposition2. Moreover proposition 15 implies that a∗(f) is closed and a(f) is closable.

135 Appendix B

Perron-Frobenius theorem

2 n 2 n Theorem 13. Let A = L (R ) → L (R ) have integral kernel Z (Aψ)(x) = A(x, y)ψ(y)dy Rn where A(x, y) > 0 ∀x, y. Suppose that A has an eigenvalues λ = kAk. Then λ has multiplicity one and eigenfunction ψλ(x) > 0 ∀x

Proof. Since A(x, y) ∈ R, A Re ψλ = Re Aψλ so we may assume that ψλ is real. Since A(x, y) > 0,

2 2 2 λkψλk = (Aψλ, ψλ) ≤ (A|ψλ|, |ψλ|) ≤ kAkkψλk = λkψλk so we get (Aψλ, ψλ) = (A|ψλ|, |ψλ|). Writing this in terms of its positive and negative parts ψλ = ψ+ − ψ− gives that

(Aψ+, ψ+) + (Aψ−, ψ−) − (Aψ−, ψ+) − (Aψ+, ψ−) = (Aψ+, ψ+) + (Aψ−, ψ−) so (Aψ−, ψ+) + (Aψ+, ψ−) = 0. This means that either ψ+ = 0 or ψ− = 0 almost everywhere since A(x, y) > 0. We thus may assume that ψλ ≥ 0. Now there is only left to prove that λ has multiplicity one. To do this, take a positive vector θ 6= 0. Then Aθ ≥ 0, Aθ 6= 0, so 0 < (Aθ, ψλ) = (θ, Aψλ) = λ(θ, ψλ). Since θ was arbitrary, this means that ψλ > 0 almost everywhere. Suppose now that A has two 1 2 1 2 independent eigenvectors ψλ, ψλ. We may take ψλ ⊥ ψλ, but this is impossible since both are strictly positive almost everywhere. There can only be one independent eigenvector corresponding to λ, so its multiplicity is one, which concludes the proof.

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