Introduction to lattice gauge theories Rainer Sommer DESY, Platanenallee 6, 15738 Zeuthen, Germany

WS 11/12: Di 9-11 NEW 15, 2’101 WS 11/12: Fr 15-17 NEW 15, 2’102

We give an introduction to lattice gauge theories with an emphasis on QCD. Requirements are quantum mechanics and for a better understanding relativistic quantum mechanics and continuum quantum field theory.

These are not lecturenotes written to be easily readable (a script), but my private notes. Still I am of course happy to receive corrections.

References

[1] J. Smit, Introduction to quantum fields on a lattice: A robust mate, Cambridge Lect. Notes Phys. 15 (2002) 1–271.

[2] H. J. Rothe, Lattice gauge theories: An Introduction, World Sci. Lect. Notes Phys. 74 (2005) 1–605.

[3] I. Montvay and G. M¨unster, Quantum fields on a lattice, . Cambridge, UK: Univ. Pr. (1994) 491 p. (Cambridge monographs on mathematical physics).

[4] M. Creutz, QUARKS, GLUONS AND LATTICES, . Cambridge, Uk: Univ. Pr. ( 1983) 169 P. ( Cambridge Monographs On Mathematical Physics).

[5] T. DeGrand and C. E. Detar, Lattice methods for quantum chromodynamics, . New Jersey, USA: World Scientific (2006) 345 p.

[6] C. Gattringer and C. B. Lang, Quantum chromodynamics on the lattice, Lect.Notes Phys. 788 (2010) 1–211.

[7] C. Morningstar, The Monte Carlo method in quantum field theory, hep-lat/0702020. Contents

1 Introduction 2 1.1 Particle physics ...... 2 1.2 Why a lattice? ...... 2 1.3 Where will we go? ...... 2

2 Pathintegral in quantum mechanics 8 2.1 Euclidean Green functions in QM ...... 8 2.2 Quantum field theory ...... 10 2.2.1 Problems with the path integral ...... 11

3 Scalar fields on the lattice 13 3.1 Hypercubic lattice ...... 13 3.2 Momentum space ...... 14 3.3 Green functions of the free field ...... 16 3.4 Transfer matrix ...... 17 3.5 Translation operator, spectral representation ...... 19 3.6 Timeslice correlation function and spectrum of the free thory ...... 22 3.7 Lattice artifacts ...... 23 3.8 Improvement ...... 24 3.9 Universality ...... 25

4 Gauge fields on the lattice 27 4.1 Color, parallel transport, gauge invariance ...... 27 4.2 Group integration ...... 30 4.2.1 Some group integrals (for later) ...... 31 4.3 Pure gauge theory ...... 32 4.3.1 Gauge invariance ...... 32 4.3.2 Transfer Matrix ...... 33

1 1 Introduction

1.1 Particle physics It is about the fundamental forms of matter (quarks = constituents of nucleons and their relatives) and their interactions. About scattering processes (LHC!) but also about the bound states: HOW is a nucleon made up of quarks. The fundamental formulation is a quantum field theory (or string theory, which for energies far below MPlanck is again a quantum field theory). Field theories combine Poincare invariance and quantum mechanics.

1.2 Why a lattice? Field theories have coupling “constants”, e.g. the fine structure constant α in quantum elec- tro dynamics (QED). The standard continuum treatment is an expansion in these coupling constants: “perturbation theory”. QCD is the part of the theory which describes the by far dominant interactions of quarks (up,down,charm,strange,top,bottom). It has a rather large effective coupling constant at dis- tances of the order (0.1 fm to 1 fm).

• There are phenomena, in particular in QCD, which are intrinsically non-perturbative: confinement: quarks are never observed as free particles

mproton = mp  2mu + md (1.1) In fact in the limit of vanishing quark masses:

−const./αs(µ) 2 mp ≈ mp|mu=md=0 ∼ µ e = 0 to all orders in αs : 0 + 0α + 0α + ... (1.2)

• the hadron mass spectrum is non-perturbative and can in principle be computed depend- ing on just a few parameters (neglecting electromagnetism, weak interaction, gravitation):

αs, mu, md, ms, mc, mb, mt

• The lattice formulation is designed to enable such computations

• The lattice formulation is the only known formulation which contains perturbative and non-perturbative “sectors”. It is therefore to be regarded as THE formulation (definition) of a quantum field theory, in particular QCD.

1.3 Where will we go? • Introduce the ingredients - path integral, Euclidean time - lattice (scalar fields) - gauge fields (gluons) - fermions (quarks)

2 • Discuss concepts and computational methods - continuum limit, Symanzik effective theory - strong coupling expansion - MC method - multiscale methods (maybe in part II)

• Discuss some results, e.g. quark — anti-quark potential

Figure 1: Static quark potential in the pure gauge theory. [2001]

3 glueball spectrum pure gauge theory 12 0+−

+− 3−− 10 2 −− *−+ 2 4 2 −− ++ 1 3 0*−+ 3+− 8 2−+ 1+− 3 *++ 0 −+ G 0 6 2++ m (GeV) 0 G r

2 m 4 0++

1 2

0 0 ++ −+ +− −− PC

Figure 2: Glueball spectrum in the pure gauge theory. [1998]

4 hadron mass spectrum

1.4 K input 1.05 φ K input Ξ

1.00 1.2 Σ

[GeV] Λ

0.95 oct m m [GeV]

1.0 K* 0.90 N

0.85 0.8 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 −1 a [GeV ] a [GeV−1]

K input 1.7 Ω

Ξ* 1.5 [GeV] dec

m Σ*

1.3

∆

1.1 0.0 0.2 0.4 0.6 0.8 1.0 1.2 a [GeV−1]

Figure 3: Hadron mass spectrum. [2001]

5 2000

O 1500 X* S* X S D 1000 L K* N M[MeV] r experiment 500 K width input p QCD 0

Figure 4: Hadron mass spectrum. [2009]

6 chiral symmetry breaking decay rates: π → µν

[running of QCD coupling (maybe in part II)]

Figure 5: αs: left: lattice gauge theory [2005], right: experiment + PT [2004].

[interplay with effective theories (maybe in part II) i.e. e.g. expansion in a or 1/mb] [elastic scattering phases (maybe in part II)]

7 2 Pathintegral in quantum mechanics

2.1 Euclidean Green functions in QM We start from a QM Hamiltonian of one degree of freedom pˆ2 Hˆ = + V (ˆq) h/(2π) = 1, c = 1 . (2.1) 2m The statistical mechanics quantum partition function can be written as (T=1/temperature!) Z Z = Tr e−T Hˆ = D[q] e−S[q] (2.2) q(0)=q(T ) (we will “show” that). The Euclidean action

T Z m . 2 Y S[q] = dt ( q +V (q)) , D[q] = “ dq(t)“ (2.3) 2 0 | {z } t = L|t→it A “derivation”: We define the Euclidean time evolution operator (= transfer matrix) by a small (infinitesimal) time unit a, −aHˆ 0 −aHˆ 0 Tˆ = e , T(q, q ) = hq|e |q i , (2.4) such that

N Z = Tr Tˆ ,N = T/a (2.5) Z Z = dq0 dq1 ... dqN−1 T(q0, q1) T(q1, q2) ... T(qN−1, q0) (2.6)

Use the BCH formula (or explicitly) to show

0 −aV (ˆq)−a 1 pˆ2 0 T(q, q ) = hq|e 2m |q i (2.7) −aV (ˆq)/2 −a 1 pˆ2 −aV (ˆq)/2 0 3 = hq|e e 2m e |q i + O(a ) (2.8) 1 −a V (q) [ hq|e−aV (ˆq)/2|q0i = e 2 δ(q − q0) (2.9) hp|p0i = δ(p − p0) , (2.10) 1 hq|pi = e−ipq (2.11) (2π)1/2 Z −a 1 pˆ2 0 0 −a 1 p2 hq|e 2m |q i = dp hq|pihp|q ie 2m (2.12) Z r 1 −ip(q−q0) −a 1 p2 1 2πm −m 1 (q−q0)2 = dpe e 2m = e 2a (2.13) 2π 2π a ] (2.14) r 1 0 m −a[ (V (q)+V (q0))+ m ( q−q )2] 3 = e 2 2 a + O(a ) (2.15) 2πa And with q(t + a) − q(t) . = q (t + a/2) + O(a2) (2.16) a 8 we have naively shown the equivalence of quantum mechanics and a 1-dimensional Euclidean pathintegral. [remark about unboundedness, therefore a-expansion is not obvious] Z Z = dq0 dq1 ... dqN−1 T(q0, q1) T(q0, q1) ... T(qN−1, q0) (2.17) Y Y [ dq(t) = q(ti) = D[q] (2.18) t i ] (2.19) Z N−1 ! a X qi+1 − qi = const. × D[q] exp − ( )2 + V (q ) , q = q (2.20) 2 a i N 0 i=0 Z = const. × D[q] exp(−S[q]) approximated on a 1-d lattice (2.21) Z a→0 −→ const. × D[q] exp(−Scont[q]) . (2.22)

Euclidean Green functions:

t1 ≥ t2 (2.23) 1 Z G(t , t ) = D[q] e−S[q] q(t ) q(t ) , n = t /a (2.24) 1 2 Z 1 2 i i 1 Z = dq dq ... dq (q , q ) (q , q ) ... (q , q ) q q (2.25) Z 0 1 N−1 T 0 1 T 1 2 T N−1 0 n1 n2 T/a −1 (T −t )/a (t −t )/a t /a = ( Tr Tˆ ) Tr Tˆ 1 qˆTˆ 1 2 qˆTˆ 2 (2.26) ˆ ˆ Tr P |nihn|e−En(T −t1) qˆe−H (t1−t2) qˆe−H t2 = n (2.27) P −En T Tr n |nihn|e ˆ ˆ P hn|e−En(T −t1) qˆe−H (t1−t2) qˆe−H t2 |ni = n (2.28) P −En T n e [ the ground state is non-degenerate ] (2.29) T →∞ ˆ ˆ ˆ ˆ −→ h0|eH t1 qˆe−H t1 eH t2 qˆe−H t2 |0i (2.30)

= h0|qˆ(t1)q ˆ(t2)|0i (2.31) qˆ(t) = eHˆ tqˆe−Hˆ t = Euclidean Heisenberg operators (2.32)

And more generally:

t1 ≥ t2 ≥ t3 ...,T → ∞ (2.33) 1 Z G(t , t , ..., t ) = D[q] e−S[q]q(t ) q(t ) . . . q(t ) . (2.34) 1 2 n Z 1 2 n

= h0|qˆ(t1)q ˆ(t2) ... qˆ(tn)|0i (2.35)

general ti ,T → ∞ (2.36)

G(t1, t2, ..., tn) = h0|T {qˆ(t1)q ˆ(t2) ... qˆ(tn)} |0i (2.37)

In lattice gauges theories, Euclidean Green functions = correlation functions = correlators are the central objects (but for 1+3=4 dimensions). They are mostly computed numerically by a Monte Carlo process. Assume one has the Euclidean Green functions. Real time physics is

9 obtained (in principle) by analytic continuation (see later). But also directly: (T → ∞)

Hˆ t1 −Hˆ t1 Hˆ t2 −Hˆ t2 G(t1, t2) = G(t1 − t2) = h0|e qˆe e qˆe |0i (2.38) ˆ = h0|qˆe−H (t1−t2)qˆ|0ieE0 (t1−t2) (2.39) X = h0|qˆ |nie−En (t1−t2)hn| qˆ|0i eE0 (t1−t2) (2.40) n

X 2 −(En−E0)(t1−t2) = αne , αn = hn|qˆ|0i (2.41) n from the large t2 − t1 behaviour: E1,E2,.... Improvement of precision by different f(ˆq). In various places we dropped O(a2) terms. The result of a lattice path integral is only unique (universal) in the limit a → 0.

2.2 Quantum field theory Let us give a very rough overview what QFT (and therefore QCD) is about. It describes

• particles moving in space and time |pi

• multiparticle states |p1, p2i: q q ˆ 2 2 2 2 H|p1, p2i ≈ ( p1 + m1 + p2 + m2)|p1, p2i (2.42)

≈ because there is always some (small) interaction

• scattering, decays, particle creation

• bound states → one of the main subjects of LGT .

Construction of QFT (scalar particles): Creation and annihilation of particles at any point in space and time: QM operators at any point, quantum fields: ˆ qˆi → φ(x) (2.43)

pˆi → πˆ(x) (2.44) [φˆ(x), φˆ(y)] = 0 = [ˆπ(x), πˆ(y)] (2.45) [φˆ(x), πˆ(y)] = iδ(x − y) (2.46) x not the QM variable, but more an index, labelling dof’s. Free particle:

Z m2 Hˆ = d3x{ 1 πˆ(x)2 + 1 ∂ φˆ(x)∂ φˆ(x) + φˆ2(x)} (2.47) 2 2 j j 2 Interactions: very restricted by general principles:

• unitarity, causality

10 • renormalizability → locality, dimension of fields in the Lagrangian Only λ λ L (x) = 0 φ4(x) → Hˆ = 0 φˆ4(x) (2.48) int 4 int 4 is possible. There is only one parameter: λ0, dimensionless. Let us take two fields, combine them into one complex field, corresponds in the end to a charged particle and its anti-particle.

1 2 1 2 2 λ0 4 φ = √ [φ1 + iφ2] , |φ| = (φ + φ ) , Lint(x) = |φ(x)| (2.49) 2 2 1 2 4 The formal continuum path integral: Z Z = D[φ]D[φ†] e−S[φ] , (2.50) Z λ S[φ] = d4(x) {∂ φ∗(x)∂ φ(x) + m2|φ(x)|2 + 0 |φ(x)|4} (2.51) µ µ 4 Z G(x, y) ≡ hφ(x)φ∗(y) ≡ Z−1 D[φ]D[φ∗] e−S[φ]φ(x)φ∗(y) . (2.52)

2.2.1 Problems with the path integral It is a rather formal object:

• λ0 = 0: gaussian integrals as in QM

• λ0 > 0: perturbation theory (PT) in λ, G(x, y) : (2.53) (0) (1) 2 (2) G = G + λ0G + λ0G + ... (2.54) again gaussian integrals −→Feynman diagrams

• λ0 < 0: singular: ∞ Z 4 e+|λ0|φ (x) → ∞ (2.55) −∞ PT has a zero radius of convergence: asymptotic series

• Divergences: start with m0, λ0, compute 2 2 2 E = p + mR (2.56) 2 2 2 mR = m0 + λ0 × (divergent integral) + O(λ0)) (2.57)

similar for λR Reason: per unit volume: infinite number of dof

m0, λ0 → mR, λR (2.58) involves infinite (undefined) relations −→regularize, renormalize

11 • Regularize: modify short distances such that theory becomes defined, in particular Feynman diagram integrals become finite, depend on some parameter. Infinities appear when parameter is removed. e.g. a → 0. Most popular versions:

1 Regularize Feynman diagram integrals, e.g. “dimensionally” R d4p → R dnp , n = 4 −  2 Regularize the path integral: lattice with spacing a

• Renormalize: take a limit of some regularization parameter ( → 0 , a → 0)

at fixed mR, λR

observable(mR, λR) = lim {observable(a, m0, λ0)} (2.59) a→0 mR,λR m0(a) , λ0(a) , : lim λ0(a) = ∞ is possible (allowed) (2.60) a→0 more precisely, everything is dimensionless, measure masses etc in units of a

observable(m , λ ) = lim {observable(am , λ )} (2.61) R R 0 0 mR,λR amR→0

am0(amr) , λ0(amr) , : lim λ0(amr) = ∞ is possible (allowed) (2.62) amr→0

bare parameters = parameters in the Lagrangian are not observable, irrelevant; once they are eliminated there are no divergences.

1) Dimensional regularization mixes definition and and approximation

gives asymptotic expansion in λR is tremendously successful (QED!)

2) Lattice regularization non-perturbative definition of the theory

validity, precision of expansion in λR can be checked

12 3 Scalar fields on the lattice

3.1 Hypercubic lattice lattice: 4 Λ = aZ = {xµ = anµ| nµ ∈ Z , µ = 0, 1, 2, 3} (3.1) finite lattice = lattice in finite volume

Λ = {xµ = anµ| nµ = 0, 1,...Lµ/a} (3.2) mostly

L0 = T,L1 = L2 = L3 = L. (3.3) Having the same spacing in all directions enhances the symmetry and is relevant. Integral as for the action Z 4 4 X 4 X S = d x L(x) → a L(x) ≡ a L(an0, an1, an2, an3) with xµ = anµ . (3.4)

x n0,n1,n2,n3

Derivatives: 1 forward: ∂ φ(x) ≡ (∂ φ)(x) = [φ(x + aµˆ) − φ(x)] (3.5) µ µ a 1 backward: ∂∗φ(x) ≡ (∂∗φ)(x) = [φ(x) − φ(x − aµˆ)] (3.6) µ µ a 1 1 symmetric: ∂˜ φ(x) ≡ (∂∗ + ∂ )φ(x) = [φ(x + aµˆ) − φ(x − aµˆ)] (3.7) µ 2 µ µ 2a Partial integration Z Z 4 ∗ 4 ∗ d x (∂µφ (x))(∂µφ(x)) = − d x φ (x)∂µ∂µφ(x) (3.8) is valid

– in finite volume with pbc: φ(x + Lµµˆ) = φ(x) – in infinite space-time and fields which vanish fast enough at infinity Partial summation

4 X 3 X a ψ(x)∂µφ(x) = a ψ(x)[φ(x + aµˆ) − φ(x)] (3.9) x x X X = a3[ ψ(y − aµˆ)φ(y) − ψ(y)φ(y)] (3.10) y y 4 X ∗ = −a (∂µψ(y)) φ(y) (3.11) y ∗ [ ψ(x) = ∂µφ (x) ] (3.12)

4 X ∗ 4 X ∗ ∗ a ∂µφ (x)∂µφ(x) = −a φ (x)∂µ∂µφ(x) . (3.13) x x

13 This gives the action for a scalar field on the lattice

X λ0 S = a4 {∂ φ∗(x)∂ φ(x) + m2|φ|2(x) + |φ|4(x)} (3.14) µ µ 4 x X λ0 = a4 {−φ∗(x)∂∗∂ φ(x) + m2|φ|2(x) + |φ|4(x)} (3.15) µ µ 4 x ∗ † = φnMnmφm = φ Mφ , φn = aφ(na) , n = (n0, n1, n2, n3) , (3.16) M = M † , M > 0 . Check as an exercise (3.17)

Ordered the fields in some way ... Let us study the free field, λ0 = 0. We expect particles, states with a definite momentum. Decoupled (free). So expect that the action is diagonalized in momentum space:

˜† ˜ ˜ ˜ 2 S = φ Mφ , Mnm =µ ˜mδmn . (3.18) Let’s look at momentum space first.

3.2 Momentum space Plane waves are eipx = eipµxµ . (3.19) Expanding a field in plane waves is a Fourrier transformation, Z d4p f(x) = eipxf˜(p) . (3.20) (2π)4 Since eipµxµ = ei(pµxµ+2πnµ) = ei(pµ+2π/a)xµ (3.21) the momenta 2π p ⇔ p + (3.22) µ µ a are equivalent, and we can restrict them to the Brillouin zone, π π − ≤ p < . (3.23) a µ a So we have Z π/a 4 d p ipx ˜ f(x) = 4 e f(p) (3.24) −π/a (2π) in infinite volume. Finite volume

If in addition we put ourselves into a finite volume, there are clearly also a finite number Q of momenta, µ Lµ/a of them. We can then write a field as 1 X f(x) = eipxf˜(p) ,V = L L L L = TL L L . (3.25) V 0 1 2 3 1 2 3 p

14 Usually one chooses periodic boundary conditions (PBC),

iθµ X ipx ipµLµ ˜ X ipx iθµ ˜ f(x +µL ˆ µ) = e f(x) → e e f(p) = e e f(p) , (3.26) p p

(no summation over µ) and one has

ipµLµ iθµ 2π θµ e = e → pµ = kµ + , kµ ∈ {0, 1,...,Lµ/a − 1} (3.27) Lµ Lµ

Lµ Lµ (or a different range: − 2a ≤ pµ < 2a ). Normally one has

θµ = 0 “PBC” (3.28)

θµ = π “APBC” (3.29) but general θµ allows more flexibility. As the volume becomes bigger, our spacings in momentum space 2π µ = (3.30) Lµ shrink and 1 X X 1 f(x) = eipxf˜(p) =  . . .  eipxf˜(p) (3.31) V 0 3 (2π)4 p p Z d4p → eipxf˜(p) . (3.32) (2π)4

15 In the exercise we show a4 X Y eip(x−y) = δ [ x = an , y = am ] (3.33) V nµmµ µ µ µ µ p µ | {z } =a4 δ(x−y)

4 P δ(x − y): lattice delta-function: a x δ(x − y) f(x) = f(y) and we always identify f(x) = f(x + aµLˆ µ). We can take the infinite volume limit Z π/a 4 d p ip(x−y) 4 e = δ(x − y) . (3.34) −π/a (2π) Of course we also have (written in a different way):

a4 X Y Y (2π)4 eix(p−q) = δ = δ(p − q)  = δ(p − q) (3.35) V nµmµ µ V x µ µ

[ pµ = nµ2π/Lµ , qµ = mµ2π/Lµ ] (3.36) X a4 eix(p−q) = (2π)4 δ(p − q) (3.37) x

3.3 Green functions of the free field

X ∗ † S = φnMnmφm = φ Mφ , φn = aφ(na) . (3.38) mn M = M † , M > 0 , (3.39) so there is a unitary matrix U,

† † 2 U U = 1 , (U MU)nm = δmnµm , µm ∈ R. (3.40) To solve the free theory, we introduce a generating functional (J, J¯ = sources)

Z ∗ −S[φ]+a4 P [J¯(x)φ(x)+φ∗(x)J(x)] Z(J, J¯) = D[φ]D[φ ] e x , (3.41)

∗ −1 −8 ∂ ∂ ¯ hφ(x)φ (y)i = Z(0, 0) a ¯ Z(J, J) . (3.42) ∂J(x) ∂J(y) J=J¯=0 It is evaluated as Z P ˜ 2 2 ¯˜ ˜ ˜∗ ˜ ¯ ∗ m[−|φ|mµm+Jmφm+φmJm] ˜ † ∗ Z(J, J) = D[φ]D[φ ] e [ φm = Umnφn = Unmφn ,... (3.43) D[φ˜]D[φ˜∗] = D[φ]D[φ∗] det U det U † = D[φ]D[φ∗] ] (3.44) Z ˜ 2 2 ¯˜ ˜ ˜∗ ˜ Y ˜ ˜∗ −|φm| µm+Jmφm+φmJm = dφmdφm e (3.45) m Y π ¯˜ ˜ −2 ¯T −1 = eJmJmµm = (det(M/π))−1 eJ M J (3.46) µ2 m m −1 a8 P J¯(x)G(x,y)J(y) = (det(M/π)) e x,y (3.47)

16 with

3 −1 2 Jn = a J(na),Mnm = a G(an, am) (3.48) with

∗ 2 [−∂µ∂µ + m ]G(x, y) = δ(x − y) (3.49)

From this we get the two-point function ∂ ∂ Z(0, 0)−1 Z(J, J¯) (3.50) ∂J¯(x) ∂J(y)

∂ 8 X a8 P J¯(z)G(z,z0)J(z) = a [ G(x, z)J(z)] × e z,z0 (3.51) ∂J(y) z 8 a8 P J¯(z)G(z,z0)J(z) = a G(x, y) e z,z0 (3.52) + terms that vanish when we set J = J¯ = 0 (3.53) → hφ(x)φ∗(y)i = G(x, y) . (3.54)

The matrix U which diagonalizes M is the matrix formed from the fourrier transformation. This is due to translation invariance (periodic boundary conditions or infinite volume) and the fact that M just contains finite differences as non-trivial terms. So we have

1 X G(x, y) = G(x − y, 0) = eip(x−y) G˜(p) a graph (3.55) V p 1 1 G˜(p) = ∼ + O(a2 p2 ) exercise (3.56) pˆ2 + m2 p2 + m2 µ

Exercise

What are the higher point functions, such as G(u, v, w),G(u, v, w, x)?

3.4 Transfer matrix We treat here a real scalar field. The complex one is basically two copies of the real one. The simple scalar action allows for the explicit derivation of a transfer matrix just like in quantum mechanics. We set a = 1 (all dimensionful quantities in units of a until eq. (3.79)). The action is

2 X m λ0 S = { 1 ∂ φ(x)∂ φ(x) + φ2(x) + φ4(x)} (3.57) 2 µ µ 2 4 x The transfer matrix acts between timeslices. Therefore we collect all variables in a timeslice:

Φ(x0) = {φ(x0, x) , x ∈ Λspace} (3.58)

17 The action is rewritten as

X X 1 2 S = 2 [φ(x0 + 1, x) − φ(x0, x)] + V (Φ(x0)) (3.59) x0 x 2 X m λ0 V (Φ(x )) = { 1 ∂ φ(x)∂ φ(x) + φ2(x) + φ4(x)} . (3.60) 0 2 j j 2 4 x We want to show as in QM Z Z = Tr e−T Hˆ = D[φ] e−S . (3.61)

First we need a Hilbert space and operators. Introduce φ(x) as operator φˆ(x) and canonical conjugateπ ˆ(x) (Schr¨odingerpicture): [φˆ(x), φˆ(y)] = 0 = [ˆπ(x), πˆ(y)] (3.62) ˆ Y [φ(x), πˆ(y)] = iδ(x − y) ≡ δxiyi (3.63) i and just as usually (think of φ(x) = q, π(x) = p) hφ0(x)|φ(x)i = δ(φ0(x) − φ(x)) (3.64) hπ0(x)|π(x)i = δ(π0(x) − π(x)) (3.65) hφ0(x)|F (φˆ(x))|φ(x)i = δ(φ0(x) − φ(x))F (φ(x)) (3.66) 1 hφ(x)|π(x)i = e−iφ(x)π(x) . (3.67) (2π)1/2 The total Hilbert space is the direct product of Hilbert spaces at each x Y |Φi = |φ(x)i = |φ((0, 0, 0))i|φ(a, 0, 0)i ... |φ((L − 1,L − 1,L − 1))i . (3.68) x An explicit representation is the Schr¨odingerrepresentation: Z Y ψ[u] = ψ[{u(x)}] , hψ|ψi = du(x) |ψ[u]|2 (3.69) x φˆ(x) ψ[u] = u(x) ψ[u] (3.70) ∂ πˆ(x) ψ[u] = −i ψ[u] . (3.71) ∂u(x) We factorize the “Boltzmann factor” 1 1 1 Y − V (Φ(x +1)) − P [φ(x +1,x)−φ(x ,x)]2 − V (Φ(x )) e−S = e 2 0 e 2 x 0 0 e 2 0 (3.72)

x0 Now consider the difficult term (as in quantum mechanics):

1 1 − P [φ(x +1,x)−φ(x ,x)]2 Y − [φ(x +1,x)−φ(x ,x)]2 e 2 x 0 0 = e 2 0 0 (3.73) x 1 Y − πˆ(x)2 = hφ(x0 + 1, x)|e 2 |φ(x0, x)i (3.74) x 1 P 2 − x πˆ(x) = hΦ(x0 + 1)|e 2 |Φ(x0)i (3.75)

18 and then

1 1 1 − V (Φ(x +1)) − P [φ(x +1,x)−φ(x ,x)]2 − V (Φ(x )) e 2 0 e 2 x 0 0 e 2 0 (3.76) 1 ˆ 1 P 2 1 ˆ − V (Φ) − x πˆ(x) − V (Φ) = hΦ(x0 + 1)|e 2 e 2 e 2 |Φ(x0)i . (3.77)

The TM,

1 1 1 − V (Φ)ˆ − P πˆ(x)2 − V (Φ)ˆ Tˆ = e 2 e 2 x e 2 . (3.78) is hermitian and positive. Therefore (restoring a) writing

−aHˆ Tˆ = e , (3.79) makes sense. It defines the lattice hamiltonian Hˆ , a hermitian operator.

Hˆ = Hˆ † , Hˆ ≥ 0 . (3.80)

A formal expansion in a gives:

2 X m λ0 Hˆ = { 1 πˆ(x)2 + 1 ∂ φˆ(x)∂ φˆ(x) + φˆ2(x) + φˆ4(x)} + O(a2) . (3.81) 2 2 j j 2 4 x Remarks:

• The exisitence of a positive, hermitian operator is called positivity. It corresponds to unitarity in Minkowsky space, the conservation of probablity, very important.

• An action with just 2nd order derivatives was important in deriving Tˆ. • We emphasize the huge Hilbert space. A QM degree of freedom at each space-point.

3.5 Translation operator, spectral representation Let us introduce the spatial translation operator by

Uˆ(x)|ψi = |ψ0i , (3.82) 0 ˆ 0 ˆ hψ1|φ(y)|ψ2i = hψ1|φ(y − x)|ψ2i . (3.83)

This means that ˆ † ˆ ˆ ˆ hψ1|U(x) φ(y)U(x)|ψ2i = hψ1|φ(y − x)|ψ2i (3.84) and therefore the operators transform as

Uˆ(x)†φˆ(y)Uˆ(x) = φˆ(y − x) . (3.85)

Clearly our Hamiltonian is chosen invariant

Uˆ(x)†Hˆ Uˆ(x) = H.ˆ (3.86)

19 Another property that we need is that Uˆ(x) is unitary. We look at the Schr¨odingerrepresen- tation Z ˆ Y ∗ hψ1|φ(y − x)|ψ2i = dv(z) ψ1[v]v(y − x)ψ2[v] (3.87) z Z Y 0 ∗ 0 0 0 0 = dv (z) ψ1[v ] v (y) ψ2[v ] , [ v (y) = v(y + x) ] (3.88) z Z Y 0 ∗ 0 0 ˆ 0 = dv(z)(ψ1) [v] v(y) ψ2[v] = hψ1|φ(y)|ψ2i (3.89) z 0 ˆ 0 0 ψi[v] = U(x)ψi[v] = ψi[v ] , v (y) = v(y + x) (3.90) From this we see immediately that

0 0 ˆ † ˆ hψ1|ψ2i = hψ1|U(x) U(x)|ψ2i = hψ1|ψ2i . (3.91) There are simultaneous eigenstates of Hˆ and Uˆ. For eigenstates of Uˆ we have

Uˆ(x)|λ(x)i = λ(x)|λ(x)i , |λ(x)|2 = 1 , λ(x) = eiα(x) (3.92) α(x) + α(y) = α(x + y) → α(x) = −xp with some p . (3.93)

The last equation is just because of the linearity seen before. So we have with |λ(x)i = |p, ni, n for other quantum numbers,

Uˆ(x)|p, ni = e−ixp|p, ni , (3.94) ˆ H|p, ni = (E(p, n) + E0)|p, ni , (3.95)

E0 : the ground state energy (3.96) The physical interpretation is that p is the momentum of the state. And on a finite, periodic lattice we have the restrictions for p as before. As normalization we choose

0 0 3 0 hp, n|p , n i = 2 E(p, n) L δ(p − p ) δnn0 (3.97) Let us now look at the 2-point function and use translation invariance to derive the spectral representation.

T → ∞ ,L finite (3.98) G(x − y) = hφ(x) φ(y)i (3.99) ˆ = h0|φˆ(x)e−H |x0−y0|φˆ(y)|0i eE0 (x0−y0) (3.100) 1 X 1 = e−E(p,n) |x0−y0|h0|φˆ(x)|p, nihp, n|φˆ(y)|0i L3 2 E(p, n) p,n and, assuming the translation invariance of the ground state, Uˆ †(x)|0i = |0i (for a finite system it can be proven that the ground state is translation invariant)

h0|φˆ(x)|p, ni = h0|Uˆ(x)φˆ(0)Uˆ †(x)|p, ni (3.101) = eipxh0|φˆ(0)|p, ni (3.102) h0|φˆ(y)|p, nihp, n|φˆ(x)|0i = eip(x−y)|h0|φˆ(0)|p, ni|2 (3.103)

20 This gives

1 X |h0|φˆ(0)|p, ni|2 G(x − y) = eE(p,n) |x0−y0| eip(x−y) (3.104) L3 2E(p, n) p,n 1 Z X = dω ρ (ω, p) e−ω |x0−y0| eip(x−y) , (3.105) L3 L p X |h0|φˆ(0)|p, ni|2 ρ (ω, p) = δ(ω − (E(p, n))) . (3.106) L 2E(p, n) n And in infinite volume Z d3p Z G(x − y) → dω ρ(ω, p) e−ω |x0−y0| eip(x−y) , (3.107) (2π)3

ρ(ω, p) = lim ρL(ω, p) = spectral density. (3.108) L→∞ From our derivation we have seen that the spectral density is nothing but the coupling of a field-operator to states of definite momentum: |h0|φˆ(0)|p, ni|2 . The x-dependence of the two- point function follows from space and time translations, in terms of two dynamical quantities, E(p, n) , ρ(ω, p).

21 3.6 Timeslice correlation function and spectrum of the free thory The free propagator is

Z d4p Z d3p G(x) = hφ(x)φ(0)i = eipxG˜(p) = eipxG(x ; p) (3.109) (2π)4 (2π)3 0 Z π/a dp 0 −ip0x0 G(x0; p) = e G(p) . (3.110) −π/a (2π) From the above use of translation invariance we have Z −ω|x0| X 2 −E(p,n)|x0| G(x0; p) = dω ρ(ω, p) e = cn(p) e . (3.111) n We now evaluate this explicitly.

Z π/a ip0x0 dp0 e G(x0; p) = 2 2 (3.112) −π/a (2π) pˆ + m

for x0 ≤ 0 [ φ = ap˜0 , n0 = −x0/a ] (3.113) Z π dφ 1 iφn0 = a e 2 2 2 2 (3.114) −π (2π) a m + a pˆ + 2(1 − cos φ) Z π dφ 1 = a eiφn0 [ A = 2 + a2m2 + a2pˆ2 ] (3.115) −π (2π) A − 2 cos φ [ z = eiφ , dz = izdφ , 2 cos φ = (z + z−1) ] (3.116) 1 I 1 n0 X = a dz z −1 = Residues (3.117) 2πi |z|=1 z[A − (z + z )] (3.118)

The poles are at (ω > 0)

−1 −aω aω D = z[A − (z + z )] = 0 , z1 = e , z2 = e , (3.119) A = 2 cosh(aω) ω > 0 , (3.120) → D = −(z − e−aω)(z − eaω) . (3.121)

Only z1 is inside the circle. Its residue is

e−n0aω . (3.122) 2 sinh(aω) So we have

e−|x0|ω G(x ; p) = , (also for x > 0) (3.123) 0 2 sinh(aω)/a 0 → E(p) = ω(p) , 2[cosh(aE(p)) − 1] = a2m2 + a2pˆ2 , (3.124) 1 spectral density: ρ(ω, p) = aδ(ω − E(p)) (3.125) 2 sinh(aω) We observe

22 • There is only one intermediate state per p; the spectral density is a single δ-function. Such a state is created from the vacuum by

ˆ 3 X ˆ ipx O1(p) = Ca φ(x)e . (3.126) x Namely we had

† −|x0−y0|Hˆ G(x0 − y0; p) ∝ hO1(x0, −p)O1(y0, p)i ∝ h0|O1(p) e O1(p)|0i (3.127)

2 2 2 2 • The energy momentum relation is (when a pi  1 , a m  1): E2 = p2 + m2 + O(a2) (3.128)

⇒ a free particle with relativistic energy momentum relation.

• 2-particle states are simply created by ˆ ˆ ˆ ˆ ˆ O2(p, q) = O1(p) O1(q) = O1(q) O1(p) (3.129) A correlation function is

12 X −i(pz+qw−px−qy) G4 = a e hφ(t, z) φ(t, w)φ(0, x)φ(0, y)i (3.130) x,y,z,w this gives: a graph

−|t|(E(p)+E(q)) time dependence C1(p, q) e + C2(p, q) δ(p + q) (3.131) So

E = E(p) + E(q) ≥ 2m a graph (3.132)

For later use we note that above we have shown that Z dp0 sinh(aω) e−|x0|ω = eip0x0 . (3.133) 2π cosh(aω) − cos(ap0)

3.7 Lattice artifacts Let us include the O(a2) effects: 2 2 [cosh(aE(p)) − 1] = E2 + a2E4 + O(a4) (3.134) a2 4! (m2 + p2)2a2 = E2 + + O(a4) (3.135) 12 2 p4a2 pˆ2 = (1 − cos p a) = p2 − j + O(a4) , (3.136) j a2 j j 12 a2 X E2(p) = m2 + p2 − [(m2 + p2)2 + p4] +O(a4) . (3.137) 12 j j | {z } ≈10% when aE(p)=1

23 Now we have to be careful, however. This includes the mass in the Lagrangian, not an observ- able. We should renormalize first, even at tree level! Not unique, but very natural:

mass = energy at rest : renormalization condition (3.138) a2 m2 = E2(p = 0) = m2 (1 − m2 + ...) (3.139) R 12 a2 m2 = m2 (1 + m2 ) + O(a4) (3.140) R 12 R → (3.141) a2 X E2(p) = m2 + p2 − [2m2 p2 + (p2)2 + p4] +O(a4) . (3.142) R 12 R j j | {z } ≈10% when aE(p)=1

See Fig. 5, left. Can this be improved? Better discretization?

3.8 Improvement In general there is Symanzik improvement:

4 X X dO−4 S → Simpr = S + δS , δS = a cia Oi(x) , (3.143) x i Here we have

3 X a2 X c adO−4O (x) = c [∂ ∂ φ(x)]2 (3.144) i i 1 2 µ µ i µ=0 3 X 2 6= [ ∂µ∂µφ(x)] (3.145) µ=0

2 2 2 2 1 2 2 2 X 4 4 → E (p) = m + p + a [c − 12 ][(m + p ) + pj ] + O(a ) (3.146) j

Remarks (here not really explained):

• In general in a scalar theory in 4 dimensions: Oi(x) local fields with mass dimension 6 Even more generally: local fields with mass dimension ≥ 5

ci = ci(λ0) (3.147)

• On-shell improvement: terms which vanish by the e.o.m can be dropped (see exercise for the e.o.m.)

• Also fields in correlation functions have to be improved (correction terms). We here looked only at energies (which are on-shell).

24 1.00 1.09

1.08

0.95 1.07

1.06

1.05 0.90 1.04

1.03

0.85 1.02

1.01

0.80 1.00 0 1 2 3 0 1 2 3

2 2 2 2 Figure 6: E (p)/E (p)cont for p = (p, 0, 0), m = 0, against the square lattice spacing a E . Left side: c = 0, right side: c = 1/12.

3.9 Universality

Z Z = D[φ] e−S[φ] , (3.148) Z hφ(x)φ(y)i = Z−1 D[φ] e−S[φ]φ(x)φ(y) . (3.149)

X x →∞ a3 hφ(x)φ(0)i 0∼ e−E(p=0)x0 = e−mRx0 = e−n0/ξ (3.150) x A change of notation

S → H/(kTtemp) (3.151) d = 3 + 1 → d = 4 (+0 : static) (3.152) −1 (amR) → ξ (3.153) a → 0 → ξ → ∞ (3.154) shows that a lattice field theory in 3 + 1 dimensions is a statistical model in 4 dimensions. The continuum limit is reached at a critical point. Statistical models are known to have universality there. Universality means that a change of the Hamilton function which does not change the symmetries (axis permutations, φ → −φ etc.) gives the same correlation functions. In the QFT this means the continuum limit is unique; it does not depend on the details 2 of the discretization (adding e.g. (∂µ∂µφ) term). More precisely: the continuum limit only depends on the coefficients of the renormalizable 6 interaction terms: [O(φ(x), ∂µφ(x)] ≤ 4 . Eg. the addition of φ changes only the cutoff effects. In terms of continuum field theory best looked at in the corresponding action: X m2 S = a4 { 1 ∂ φ(x)∂ φ(x) + φ2(x)} + S (3.155) 2 µ µ 2 int x

25 1.09 1.09 1.08 1.08

1.07 1.07

1.06 1.06

1.05 1.05

1.04 1.04

1.03 1.03

1.02 1.02

1.01 1.01

1.00 1.00 0.0 0.5 1.0 1.5 2.0 2.5 0.0 0.5 1.0 1.5 2.0 2.5

2 2 2 2 Figure 7: E (p)/E (p)cont against the square lattice spacing a E . Improved case: c = 1/12. Left side: for p = (p, p, 0), m = 0, right side: p = (p, 0, 0), mR = p. then

4 P −(n+4k−4) n k • assume locality: Sint = a x const. M φ (x)(∂µφ(x)∂µφ(x)) + ... • renormalizability when (we just state this):

4 X Sint = a Lint(x) , [Lint(x)] ≤ 4 (3.156) x .[Φ] = mass dimension: [∂µ] = 1 , [φ(x)] = 1 , (3.157)

and no odd powers of φ and Euclidean invariance λ λ L (x) = 0 φ4(x) → Hˆ = 0 φˆ4(x) (3.158) int 4 int 4

λ0 is the only possible coupling constant, free parameter QFT’s are very predictive

• how well is this established (proven)?

to all order in λ0 → λR (T. Reisz, lattice power counting theorem) 6 2 −2 non-renormalizability of eg. φ (x)/m6 in the same way to all orders in m6 . non-perturbatively: numerical investigations are in agreement with this.

Remark: do not confuse higher dimensional operators in Symanzik improvement and a theory with higher dimensional operators in the continuum limt. In one case the coefficients are proportional to adO−4, in the other case they are proportional to 1/mdO−4.

26 4 Gauge fields on the lattice

4.1 Color, parallel transport, gauge invariance Quarks carry color, A, B = 1 ... 3 (1 ...N in an SU(N) gauge theory):   ψ1(x) 3 ψ(x) = ψ2(x) ∈ C . (4.1) ψ3(x)

The basic principle is gauge invariance: different colors are completely equivalent. One can rotate the fields and physics does not change.

Λ(x) ∈ SU(3) : ψ(x) → ψΛ(x) = Λ(x) ψ(x) . (4.2)

(There is also the possibility to rotate by a phase ( U(1) ), but this corresponds to electrody- namics, a separate issue). Λ(x) ∈ SU(3) is an arbitrary function of x. So it makes no sense to compare ψ(x) and ψ(y). Before comparing we have to parallel-transport ψ(y), such that it transforms as ψ(x). Parallel transporter:

P (x ← y): P (x ← y) → P Λ(x ← y) = Λ(x) P (x ← y)Λ−1(y) (4.3) P (x ← y) ψ(y) → Λ(x) P (x ← y) ψ(y) . (4.4)

The parallel transporter will in general depend on the path from y to x. Think of a straight path for definiteness. For the definition of a (continuum derivative) we need the transporter by an infinitessimal distance:

Dµψ(x) = (Dµψ)(x) = lim[P (x ← x + µˆ) ψ(x + µˆ) − ψ(x)] . (4.5) →0 Every SU(N) matrix can be written in the form

N 2−1 X P = eB ,B = −B† = BaT a (4.6) a=1 a a a † a a b 1 a b abc c B ∈ R ,T = −(T ) , tr T = 0 , tr T T = − 2 δab , [T ,T ] = f T . (4.7) For example in SU(2): 1 T a = τ a , Pauli matrices . (4.8) 2i So we can write for an infinitessimal path

Aµ(x) P (x ← x + µˆ) = e → Dµ = ∂µ + Aµ . (4.9)

Aµ has mass dimension one. If quarks were scalars, gauge invariance would force the kinetic term in the Lagrangian to look like

X 2 |DµψA| . (4.10) A

27 It contains a coupling to the field Aµ (gluon field), an interaction term. What about a kinetic term for Aµ? Gauge invariance is the basic principle. First in the continuum: The basic object is Dµ because it is gauge covariant,

Λ −1 Dµ = Λ(x)DµΛ (x) . (4.11)

We want something Euclidean invariant and gauge invariant:

tr DµDµ = tr (∂µ∂µ + ...) ∂µ acting on what? (4.12)

tr Fµν Fµν , (4.13)

Fµν = [Dµ,Dν] = ∂µAν − ∂νAµ + [Aµ,Aν] (4.14) a a a abc b c = T {∂µAν − ∂νAµ + f AµAν} (4.15) a a = T Fµν . (4.16)

Other terms have higher dimension, are not renormalizable. So the action is Z Z 1 4 1 4 2 SG = − 2 d x tr Fµν(x)Fµν(x) = 2 d x F . (4.17) 2g0 2g0 |{z} convention Unlike QED, there are interaction terms already in here even without the quark fields. On the lattice, the quark fields are at the lattice points x, but the gluon field has to provide the parallel transporter from point to point, it is sitting on a link

P (x ← x + aµˆ) = U(x, µ): (4.18) P (x + aµˆ ← x) = U −1(x, µ): . (4.19)

Gauge covariant derivative: 1 forward: D ψ(x) = [U(x, µ)ψ(x + aµˆ) − ψ(x)] (4.20) µ a 1 backward: D∗ψ(x) = [ψ(x) − U −1(x − aµ,ˆ µ)ψ(x − aµˆ)] (4.21) µ a With these we could build a kinetic term for scalar quarks

4 X X 2 4 X X ∗ ∗ 4 X † ∗ Sφ = a |DµψA| = a ψA(−DµDµψ)A = a ψ (−DµDµ)ψ . (4.22) x A x A x For fermions we do that later. For the kinetic term we need a local object, gauge invariant. The most local one is the parallel-transporter around a plaquette (elementary square):

x + a ν

−1 −1 x x + a µ Oµν(x) = tr U(x, µ)U(x + aµ,ˆ ν)U (x + aν,ˆ µ)U (x, ν): . (4.23)

28 Assume that we have a smooth classical field, so the parallel transporters are close to one, we can then write

U(x, µ) = eaAµ(x) (4.24)

Then we can expand in a (classical continuum limit)

−1 −1 Oµν(x) = tr U(x, µ)U(x + aµ,ˆ ν)U (x + aν,ˆ µ)U (x, ν) (4.25) = tr eaAµ(x)eaAν (x+aµˆ) e−aAµ(x+aνˆ)e−aAν (x) , (4.26) | {z } | {z } eB eC 2 1 3 B = a(Aµ(x) + Aν(x + aµˆ)) + a 2 [Aµ(x) ,Aν(x + aµˆ)] + O(a ) , (4.27) 2 1 3 C = a(−Aµ(x + aνˆ) − Aν(x) + a 2 [Aµ(x + aνˆ) ,Aν(x)] + O(a ) , (4.28) 1 B+C+ [B,C]+... eBeC = e 2 = eD , (4.29) 2 2 2 3 D = −a ∂νAµ(x) + a ∂µAν(x) + a [Aµ(x) ,Aν(x)] + O(a ) (4.30) 2 3 = a Fµν(x) + O(a ) (4.31) We note that

D = DaT a , → tr D = 0 . (4.32)

This is true for the O(a2) terms but also for the higher ones: they are formed from derivatives or from commutators. Commutators always can be written again as CaT a . Therefore we can write

D 1 2 5 Oµν(x) = tr e = N + tr D + 2 tr D + O(a ) (4.33) 1 4 2 5 = N + 2 a tr (Fµν(x)) + O(a ) (4.34) x + a ν

x x + a µ = tr U(p) , p = (x, µ, ν) . (4.35)

And finally have the Wilson plaquette action 1 X 1 X X S [U] = tr {1 − U(p)} = tr P (x, µ, ν) , (4.36) G g2 g2 0 p 0 x µ,ν P (x, µ, ν) = 1 − U(x, µ) U(x + aµ,ˆ ν) U(x + aν,ˆ µ)−1 U(x, ν)−1 . (4.37)

Of course other forms are possible. In fact, take any small loop on the lattice (a graph), sum over all orientations. The result will be

X ˜ 4 X 2 2 Oµν(x) = c1 + c2a tr (Fµν(x)) + O(a ) , (4.38) x,µ,ν x,µ,ν because there is no other gauge invariant field of dimension ≤ 4. It was therefore not really necessary to do the above calculation. By the same logics, there is no dimension five axis- permutation invariant field. This is why the next term is dimension 6, giving O(a2) for the lattice spacing corrections.

29 4.2 Group integration We remain in the pure gauge theory. In order to fully define the path integral we have to specify the integration measure. The variables are in the group SU(3), so we want to know dU =? ,U ∈ SU(N) . (4.39) The basic principle is gauge invariance, so we want the measure to be gauge invariant. Parametrize the SU(N) matrices by W (ω) = exp(ωaT a) . (4.40) A gauge transformation gives U(x, µ) = W (ω) → U 0(x, µ) = Λ(x) U(x, µ) Λ(x + aµˆ)−1 = W (ω0) (4.41) W (ω) → W (ω0(ω)) , (4.42) we want dU 0(x, µ) = dU(x, µ) . (4.43)

QN 2−1 a A naive expectation would be that dU ∝ a=1 dω , but it is not quite correct because SU(N) is a curved manifold. Eg. for SU(2) a parametrization is the following.

0 k k µ µ µ W = w + iw τ , w ∈ R , w w = 1 . (4.44) So the manifold is a 3-sphere. It has a curvature. We may choose (these ω are not the ω of exp(ωaT a) √ W = 1 − ωaωa + iωaτ a , ωa ∈ [0, 1] . (4.45) To account for the curvature we define a metric tensor on the manifold ∂W ∂W † G = tr ( ) = G ,G ≥ 0 , ← exercise (4.46) ab ∂ωa ∂ωb ba In a transformation ω → ω0 it changes to ∂W ∂W † ∂W ∂W † ∂ωc ∂ωd ∂ωc ∂ωd G (ω0) = Tr ( ) = Tr ( ) = G (ω) . (4.47) ab ∂ω0a ∂ω0b ∂ωc ∂ωd ∂ω0a ∂ω0b cd ∂ω0a ∂ω0b As a remark (we do not need it): an invariant line element is ∂ωa ∂ωb ds2 = G dωadωb = G (ω) dω0c dω0d = G (ω0)dω0cdω0d . (4.48) ab ab ∂ω0c ∂ω0d cd Now take N 2−1 Y dW (ω) = pdet(G) dωa . (4.49) a=1 The pieces transform as eq. (4.47) ∂ωa det(G0) = det(G)(det( ))2 (4.50) ∂ω0b N 2−1 N 2−1 Y ∂ω0a Y dω0a = det( ) dωc (4.51) ∂ωb a=1 c=1 → dW (ω0) = dW (ω) . (4.52)

30 So we choose

N 2−1 Y √ dU(x, µ) = C dωa det G (4.53) a=1 Z dU = 1 fixes C. (4.54)

This is the Haar measure, the invariant measure on the group.

Exercise

Show that for our SU(2)-parametrization ωaωb + (1 − ωcωc) δ G = ab , (4.55) ab (1 − ωcωc) Y 1 dU(x, µ) = const. × (1 − ωcωc)−1/2 dωa = δ(1 − wµwµ)d4w . (4.56) π2 a So the parameter space of the group is a 3-sphere.

4.2.1 Some group integrals (for later) What about Z Z Z hUi ≡ dUU = d(Λ−1U) U = dU 0 ΛU 0 = ΛhUi (4.57) for all Λ ∈ SU(N) . In particular for Λ = exp(i 2πn/N) ≡ exp(i 2πn/N)1 , n = 0 ...N − 1 , the center of SU(N) (4.58) then N−1 1 X hUi = exp(i 2πn/N)hUi = exp(i 2πn/N)hUi = 0 (4.59) N n=0 hU †i = 0 . (4.60) A non-trivial group integral: Z Z † ∗ fijkl = dUUij(U )kl = dUUij(Ulk) (4.61) Z Z −1 † ˜ −1 † = d(Λ U) Uij(U )kl = d(UΛ ) Uij(U )kl (4.62) Z Z † † ˜ ˜ −1 † = dU (ΛU)ij(U Λ )kl = dU (UΛ)ij(Λ U )kl. (4.63)

For:

−1 fixed j, k: Fil = fijkl , ΛF Λ = F ⇒ F = c 1 (4.64) | {z } exercise ˜ ˜ −1 0 fixed i, l: Gjk = fijkl , ΛGΛ = G ⇒ G = c 1 (4.65)

⇒ fijkl = c δilδjk (4.66)

31 It remains to determine c: X Z X fijjl = dU δil = c δilδjj (4.67) j j → c = 1/N . (4.68)

Exercise

F is an N × N matrix. Prove that if ΛF Λ−1 = F holds for all Λ ∈ SU(N) then F = c 1. Hints: Start with N = 2. Find two special SU(2) matrices which allow to show F = c 1. Embed SU(2) in SU(N) and use the N = 2 property to show it for all N.

4.3 Pure gauge theory The path integral is now

Z Y Z = D[U] e−SG[U] , D[U] = dU(x, µ) , (4.69) x,µ 1 Z hO[U]i = D[U] O[U] e−SG[U] (4.70) Z

U(x + Lνν,ˆ µ) = U(x, µ) , PBC (4.71)

Λ(x + Lννˆ) = Λ(x) , for the gauge transformations (4.72) where S[U] = SG[U] when matter fields are neglected. This is the pure gauge theory. We will always first work with the finite volume theory and then discuss the limit of large volume.

4.3.1 Gauge invariance Consider some observable, O[U], any polynomial of the fields U(x, µ). Its expectation value is the same as the expectation value of any gauge transform of it:

1 Z Y hO[U]i = dU(x, µ) O[U] e−S[U] (4.73) Z x,µ Z 1 Y −1 = dU Λ (x, µ) O[U] e−S[U] (4.74) Z x,µ 1 Z Y = dU(x, µ) O[U Λ] e−S[U] = hO[U Λ]i . (4.75) Z x,µ

We may define a projector onto the gauge invariant part of O[U]: Z Y Λ P0O[U] = dΛ(x) O[U ] . (4.76) x Then integrating above over Λ(x) we see that

hP0O[U]i = hO[U]i , (4.77)

32 so one needs to consider only the gauge invariant part of any observable. Consider in particular an observable (e.g. a parallel transporter from y to x) which transforms as

O[U Λ] = Λ(x)O[U]Λ−1(y) → hO[U]i = 0 . (4.78)

Open parallel transporters (in contrast to closed loops) have no gauge invariant part. Their expectation values vanish. Such a local symmetry can’t break spontaneously (Elitzur theorem).

4.3.2 Transfer Matrix

Hilbert space

Wave functionals of the spatial link variables V (x, k) with scalar product

Z Y hψ0|ψi = D[V ](ψ0[V ])∗ ψ[V ] , D[V ] = dV (x, k) . (4.79) x,k

We will see that the physical Hilbert space consists of the gauge invariant wave functionals only:

Λ ψphys[V ] = ψphys[V ] , Λ(x) ∈ SU(3) (4.80) a projector onto gauge invariant wave functionals is Z ˆ Y Λ P0 ψ[V ] = dΛ(x) ψ[V ] , (4.81) x ˆ P0 ψphys[V ] = ψphys[V ] . (4.82)

Transfer matrix

Let us first remember quantum mechanics: the path integral with a finite time-spacing a is

N Z = Tr Tˆ ,N = T/a (4.83) Z 0 −∆S(q,q0) 0 Tˆ ψ(q) = dq e ψ(q ) (4.84) 1 q − q0 ∆S(q, q0) = 1 aV (q) + S (q, q0) + 1 aV (q0) ,S (q, q0) = ( )2 . (4.85) 2 k 2 k 2m a So this is an integral operator on wave functions ψ(q). We can write the pure gauge theory

33 path integral in the same way Z 0 0 0 Tˆ ψ[V ] = DV K[V,V ] ψ[V ] (4.86) Z Y 0 K[V,V 0] = dW (x)e−∆S[V,W,V ] , (4.87) x 0 1 0 1 0 ∆S[V, W, V ] = 2 Sp[V ] + Sk[V, W, V ] + 2 Sp[V ] , (4.88) 3 1 X X S [V ] = tr P (x, k, l) . (4.89) p g2 0 x k,l=1 3 1 X X S [V, W, V 0] = tr [P (x, k, 0) + P (x, k, 0)†] , (4.90) k g2 0 x k=1 P (x, k, 0) = 1 − V 0(x, k) W (x + akˆ) V (x, k)−1 W (x)−1 . (4.91)

The kernel is real and symmetric, so the transfer matrix is hermitean. It is also bounded. Therefore the spectrum is discrete. (General theorem in functional analysis). Now perform a gauge transformation Λ(x) just on the layer V

∆S[V, W, V 0] = ∆S[V Λ,W Λ,V 0] , (4.92) Z Y Λ −1 0 K[V,V 0] = dW (x)e−∆S[V ,W Λ ,V ] , choose Λ = W (4.93) x Z Y W 0 = dW (x)e−∆S[V ,1,V ] (4.94) x ˆ ˆ ˆ → T = P0T0 (4.95) 0 −∆S[V,1,V 0] K0[V,V ] = e , (4.96) and finally (perform a gauge trafo on both layers):

ˆ ˆ ˆ ˆ T = P0T0P0 . (4.97)

We have learnt that the timelike gauge links in the path integral do not really corrspond to QM degrees of freedom, but represent a projector onto the gauge invariant subspace of the Hilbert space. The physical transfer matrix is defined in that space. Physical states are gauge invariant! Positivity

Tˆ > 0 (4.98) can be shown rigorously; ≥ 0 is a rather simple exercise. The rest requires some mathematics...

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