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Methods in Theory Contents

1. Motivation 2. Basics—Euclidean quantisation 3. Lattice gauge fields Jonathan Flynn 4. Lattice University of Southampton 5. Lattice QCD 6. Numerical simulations

BUSSTEPP 2002 University of Glasgow

1 2

Why lattice QCD?

evaluate non-perturbative effects in 1 Motivation • physical amplitudes using large scale numerical simulations: observables found directly from QCD lagrangian 1.1 Theoretical long-distance QCD effects in weak processes are • The lattice regularisation of quantum field theories frequently the dominant source of uncertainty in extracting fundamental quantities from experiment is the only known nonperturbative regularisation • admits controllable, quantitative nonperturbative Example: K –K mixing and BK • calculations provides insight into how QFT’s work and enables • study of unsolved problems in QFT’s s W d

0 t 0 1.2 Applications of lattice field theories K t K

QED: ‘triviality’, fixed point structure, . . . d s • W Higgs sector of the SM: bounds on Higgs mass, Since mt ,m Λ can do perturbative analysis at high W  QCD • baryogenesis, . . . scales where QCD is weak and run by renormalisation group down to low scales. Left with: • 2 ν SUSY 0 mt dγν (1 γ5)s dγ (1 γ5)s 0 1 K C αs(µ), − − K µ + O • h | m2 m4 | i m6 QCD: spectrum, strong interaction effects in W W W •    weak decays, confinement, chiral breaking, C( ): calculable perturbative coefficient (as long as • · exotics, finite T and/or density, fundamental µ/ΛQCD not too small) parameters ( , masses) αs evaluate matrix element on a lattice with µ 1/a (a is • lattice spacing) ∼ match lattice result to continuum at scale µ 1/a • ∼

3 4 CKM matrix and unitarity triangle Unitarity triangle

Vud Vus Vub V V ∗ +V V ∗ +V V ∗ = 0 Vcd Vcs Vcb = ud ub cd cb t d t b Vt d Vt s Vt b ! 1 2 3 1 /2 A ( i ) C K M ∆ms & ∆md λ λ λ ρ η f i t t e r ∆m − 2 −2 4 d λ 1 λ /2 Aλ + O(λ ) 0.8 − − Aλ3(1 ρ i η) Aλ2 1 ! − − − 0.6 η Unitarity: VudVu∗b +VcdVc∗b +Vt dVt∗b = 0 0.4 3 7 VudVu∗b = Aλ (ρ + i η) + O(λ ) |ε | |Vub/Vcb| 3 7 0.2 K

VcdVc∗b = Aλ + O(λ ) sin 2β − WA 3 7 0 Vt dVt∗b = Aλ (1 ρ i η) + O(λ ) − − -1 -0.5 0 0.5 1 where ρ = ρ(1 λ2/2) and η = η(1 λ2/2). ρ − − Measurement V other Constraint CKM × b u V 2 (ρ,η) ub 2 2 → ρ + η |ε | b c V K → cb α 2 2 2 2 ∆M V fB BB f (mt ) (1 ρ) + η d | t d | d d − 2 2 ρ + i η 1 ρ i η ∆M V fB BB − − d t d d d (1 )2 + 2 2 ρ η ∆Ms V f B − t s Bs Bs

ε f (A, η,ρ,B ) ∝ η(1 ρ) γ β K K − (CKMfitter Spring 2002: H Hocker¨ et al, hep-ph/0104062; (0,0) (1,0) http://ckmfitter.in2p3.fr/)

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2 Basics: Euclidean quantisation

Lattice embedded in d-dimensional Euclidean

Tat

. . . with sin2β from BaBar and Belle, is in good shape. Errors in the nonperturbative parameters are now the limiting factor in more precise testing to look for effects from New . Las There is also a rich upcoming experimental programme in the next few years which will need or test lattice results: as B-factories: constraining unitarity triangle, rare decays • Tevatron Run II: ∆M , ∆Γ , b-hadron lifetimes, . . . • Bs Bs CLEOc: leptonic and semileptonic D decays, masses of at • quarkonia, hybrids, as ,a lattice spacings LHC: . . . t • Las length in spatial dimension(s)

Tat length in temporal dimension fields live on lattice sites x. Example: scalar field

x = nas , n = 0,...,L 1 φ(x) with j − x = ma , m =,...,T 1 0 t −

7 8 2.1 Lattice as a regulator

Fourier transform of a lattice scalar field in one dimension: x = na, n = 0,...,L 1, with periodic boundary conditions: − L 1 − ipna φ˜(p) = a ∑ e− φ(x) 2.2 Euclidean quantisation on the lattice n=0 φ(x + La) = φ(x) Path integral well-defined in Euclidean space

discretisation implies Wick • Minkowski Euclidean φ˜(p) periodic with period 2π/a rotation • momenta lie in first Brillouin zone • π π iε prescription avoids poles < p − a ≤ a Procedure have introduced a momentum cutoff; • π 1. Continuum classical Euclidean field theory Λ = a 2. Discretisation lattice −→ spatial periodicity implies momentum p quantised in • 3. Quantisation functional integral units of 2π/La −→ gauge invariance and gauge fields, fermions: later • Lattice provides both UV and IR cutoffs. Ultimately want infinite volume (L,T ∞) and continuum (a 0) limits. Most effort devoted to→continuum limit. →

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Step 2: Discretisation

Introduce a hypercubic lattice ΛE with at = as = a. x0 x1,2,3 Step 1 Λ = x aZ4 = 0,...,T 1; = 0,...,L 1 E ∈ a − a − Euclidean fields φ(x) obtained formally from analytic   3 continuation L T lattice sites • 0 finite volume t ix , φ(x,t ) φ(x) • → − → finite number d.o.f. Action: • Lattice action: 4 1 2 SE [φ] = d x (∂µ φ) +V (φ) 2 4 1 Z   SE [φ] = a ∑ ∇µ φ(x)∇µ φ(x) +V (φ) where µ = 0,1,2,3 and x Λ 2 ∈ E   1 λ with forward and backward lattice derivatives V (φ) = m2φ 2 + φ 4 2 4! 1 ∇µ φ(x) φ(x+aµˆ) φ(x) ≡ a − Minkowski Euclidean ←→ 1  Lorentz symmetry O(4) symmetry ∇∗µ φ(x) φ(x) φ(x aµˆ) ≡ a − − t 2 x2 invariant (x 0)2 + x2 invariant −  + + Lattice Laplacian: +  −    + 3 − (x) ( ∗ ) (x)    +  ∆φ ∑ ∇µ ∇µ φ     ≡ µ=0  −    1 3 = (x+a ˆ) + (x a ˆ) 2 (x) 2 ∑ φ µ φ µ φ a µ=0 − − 

11 12 Step 3: Quantisation—functional integral

Lattice action for a free scalar field: SE [φ] ZE D[φ]e− 4 1 ≡ SE [φ] = a ∑ φ(x)∆φ(x) +V (φ) Z x Λ − 2 ∈ E   D[φ] is the measure, eg: D[φ] = ∏ dφ(x) x Λ Remarks ∈ E finite number of integrations Discretisation is not unique. Can use different • • ( ) definitions for ∇µ∗ and/or V (φ) as long as they become Correlation functions the same in the naive continuum limit, a 0. → 1 S [φ] φ(x ) φ(xn) D[φ]φ(x ) φ(xn)e− E Universality: discretisations fall into classes, each h 1 ··· i ≡ Z 1 ··· ∗ member of which has the same continuum limit E Z is shorthand for 0 T 0 , time-ordered vacuum Improvement: optimise choice of lattice action for • h·i h | · | i ∗ a faster approach to the continuum limit expectation value well-defined if S [φ] > 0 O(4) (eventually Lorentz symmetry) is not preserved. • E • particle spectrum implicitly determined by correlation Have cubic symmetry instead; recover O(4) symmetry • as a 0. functions → analytically continue to Minkowski space and get • S-matrix elements (= physics) via LSZ

13 14 Lattice

Relation between W [J] and K : 2.3 Generating functional 1 1 1 W [J] SE [φ] (J,φ) 2 (J,K − J) F e = ∏ dφ(x)e− e = e Scalar product on space of fields φ over ΛE : ZE x Λ Z ∈ E 4 (φ1,φ2) = a ∑ φ1(x)φ2(x) Diagonalise K through Fourier transform: x Λ ∈ E 4 ip y 1 ip x J˜(p) = a e− · J(y), J(x) = e · J˜(p) ∑ 4 3 ∑ Action for free scalar field: y Λ a L T p Λ ∈ E ∈ E∗ 1 2 is the dual lattice (or set of momentum points in the SE [φ] = (φ,K φ), K = ∇∗µ ∇µ +m ΛE∗ 2 − Brillouin zone): K is a linear operator on F . 0 2π 1,2,3 2π Let J(x) be an external field (source) on Λ , J F , and ΛE∗ = p p = n0, p = n1,2,3; E ∈ Ta La define the generating functional W [J] through, 

n = 0,...,T 1, n = 0,...,L 1 eW [J] e(J,φ) 0 − 1,2,3 − ≡ h i  1 S [φ] (J,φ) = ∏ dφ(x)e− E e ZE x Λ Z ∈ E 2π/a aT Correlation functions found by differentiating w.r.t. J(x):

∂ eW [J] = a4 φ(x)e(J,φ) ∂J(x) h i 2 aL ∂ 2π/a eW [J] = (a4)2 φ(x )φ(x ) ∂J(x )∂J(x ) h 1 2 i 1 2 J=0

aL 2π/a ΛE ΛE∗

15 16 Propagator (continued)

Find: 1 4 (K − J)(x) = a G(x y)J(y) Remarks ∑ − y ΛE ∈ As a 0 (and L,T ∞), G(x y) becomes the G(x y) is the Green function for K : • → → − − Euclidean Feynman propagator: ip (x y) 1 e · − 4 ip (x y) G(x y) = a 0 d p e · − 4 3 ∑ 2 2 G(x y) → − a L T pˆ +m 4 2 2 p Λ∗ − −→ (2π) p +m ∈ E Z with using pˆ2 = p2 + O(a2). 3 2 2 apµ pˆ = ∑ pˆµpˆµ , pˆµ = sin Particle masses defined through poles of the µ=0 a 2 • ˆ2 2 1   propagator, here poles of (p +m )− , which is periodic in each component of p with period 2π/a.

1 3 W [J] 8 4 apµ e = exp a J(x)G(x y)J(y) pˆ2 = sin2 2 ∑ − 2 ∑ ( x,y ΛE ) a µ=0 2 ∈   Therefore the propagator is: Unique mass spectrum inside first Brillouin zone, pµ ( π/a,π/a]. 1 ∂ 2 ∈ − φ(x)φ(y) = eW [J] = G(x y) h i a8 ∂J(x)∂J(y) − J=0

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Free Scalar Two-point Correlator in Position Space

3 3 C(x0) = a ∑ φ(x0,x)φ(0) = a ∑G(x0,x) x h i x QCD Lore Create a particle at the origin; propagate it to any • spatial point at time x 0 Confinement: ip x ∑ projects onto zero 3-momentum ( ∑ e · would • x · x · confine and antiquarks into colour project on momentum p). • singlets Can evaluate explicitly for free scalar field (exercise) qq qqq • m x baryons e− | 0| C(x ) = 0 2m(1 +m2a2/4)1/2 gluons confine gluons into glueballs • no free quarks or gluons as asymptotic states with ‘physical mass’ (position of pole in propagator for • p = 0) m, satisfying Lattice QCD: a non-perturbative regulator which preserves gauge symmetry allows us to study these questions am am −→ sinh = 2 2   Fitting exponential decay of a lattice 2-point function • lets you extract masses. Works more generally, see later.

19 20 3 Nonabelian lattice gauge fields

Perform steps 1–3 for Yang-Mills

Step 1

Classical continuum Euclidean Yang-Mills action

1 4 S [A] = d x Tr Fµν (x)Fµν (x) E − 2g 2 Step 2: Discretisation 0 Z  For the discretised gauge field, A (x), x Λ , the g0 is the bare gauge coupling µ E transformation law ∈ SU (N) gauge fields in R4 (N colours): 1 1 A (x) g(x)A (x)g − (x) + g(x)∇ g − (x) R4 µ → µ µ Aµ (x) x , µ = 0,...,3 a a ∈ is inconsistent with group multiplication for nonabelian Aµ = A T Aµ su(N) () µ ∈ groups (Exercise). T a a = 1,...,N 2 1. Generators − Aa real vector field naive discretisation of classical Y-M action fails µ • a b abc c abc [T ,T ] = f T Structure constants f need a different concept to discretise pure gauge • A† = A (antihermitian) theory µ − µ Dµ = ∂µ + Aµ use parallel transport on lattice • Field strength:

F = ∂ A ∂ A + [A ,A ] µν µ ν − ν µ µ ν Gauge transformations:

1 1 A (x) g(x)A (x)g − (x) + g(x)∂ g − (x) µ → µ µ 1 F g(x)F g − (x) µν → µν

21 22 3.1 Defining lattice gauge fields Definition Continuum A lattice gauge field is a set of SU (N) matrices Uµ (x), y x Λ , µ = 0,1,2,3. U (x) is called a link variable. Curve C from y to x parametrised ∈ E µ by: Remarks C z(t ), 0 t 1 ≤ ≤ Where is the gauge potential A (x)? Can define a z(0) = y, z(1) = x • µ x lattice potential Aµ via Parallel transport along C : aAµ (x) Uµ (x) = e

d µ ctm + z˙ Aµ (z) v(t ) = 0 but this is not unique. If A (x) is a given continuum dt µ   gauge potential, one can use a link variable to Solution approximate it for small a: x 1 µ ctm v(1) = POexp dz A (z) v(0) lim Uµ (x) 1 = Aµ (x) µ a 0 a − − y →  Z   C Gauge transformations on the lattice. Let Parallel transporter from y to x along is, • g(x) SU (N) for x Λ . ∈ ∈ E C x µ 1 U (x,y) = POexp dz Aµ (z) ˆ − Uµ (x) g(x)Uµ (x)g − (x + aµ)  Zy  → By inspection, if C is a closed loop of link variables then Lattice C W (C ) = TrU (x,x) Choose C ’s connecting neighbouring lattice sites. is gauge-invariant. This is called a . U (x,x + aµˆ) U (x) SU (N) Approximate locally gauge invariant continuum fields ≡ µ ∈ • by gauge invariant combinations of link variables (see x x a ˆ + µ following example . . . ). 1 1 U (x + aµˆ,x) = U − (x,x + aµˆ) = Uµ− (x)

23 24 3.2 Wilson plaquette action

Return to Step 2: discretise the continuum action:

1 4 S [A] = d x Tr Fµν (x)Fµν (x) E − 2g 2 0 Z  Exercise Consider

Tr F ctm(x)F ctm(x) 1 µν µν S [U ] = Tr 1 Pµν (x) E g 2 ∑ ∑ − ctm 0 x ΛE µ,ν Fµν is a field strength defined in terms of a given ∈ 4  continuum gauge potential Actm. 1 a µ = Tr F (x)F (x) + O(a5) 2 ∑ ∑ µν µν g0 x Λ µ,ν − 2 Consider the plaquette: ∈ E    a 0 1 4 x + aνˆ x + aµˆ + aνˆ → d x Tr Fµν (x)Fµν (x) → − 2g 2 0 Z P (x)  µν ≡ Rewrite as N 1 x x + aµˆ S [U ] = 2 Tr(P + P † ) E 2 ∑ ∑ µν µν g0 x Λ µ,ν − N ∈ E µ<ν   Show that 2N 1 1 1 = 1 ReTrPµν (x) TrPµν (x) = Tr Uµ (x)Uν (x+aµˆ)Uµ− (x+aνˆ)Uν− (x) 2 ∑ ∑ g0 x Λ µ,ν − N 4 ∈ E µ<ν   a 0  a ctm ctm 5 =→ N + Tr Fµν (x)Fµν (x) + O(a ) 1 2 = β ∑ 1 ReTrP2 2 − N   ∑ is sum over all oriented plaquettes • 2 2N is the lattice coupling β 2 • ≡ g0 Last line is the Wilson plaquette action •

25 26

Step 3: Quantisation

Define a functional integral:

3 N 1 † S [U ] = 2 Tr(P + P ) SE [U ] SE [U ] E 2 ∑ ∑ µν µν Z = D[U ]e− = ∏ ∏ dUµ (x)e− g0 x Λ µ,ν − N x Λ µ=0 ∈ E µ<ν   Z Z ∈ E 2N 1 dU (x): invariant group measure for compact , = 1 ReTrP (x) µ 2 ∑ ∑ µν eg SU (N) g0 x Λ µ,ν − N ∈ E µ<ν   g 1 1 Uµ (x) Uµ (x) = g(x)Uµ (x)g − (x + aµˆ) = β ∑ 1 ReTrP2 → 2 − N dU g (x) = dU (x) so that D[U g ] = D[U ]  µ µ ∑ is sum over all oriented plaquettes Measure can be normalised, since SU (N) compact: • 2 no A fields: degrees of freedom are SU (3) matrices • µ dU = 1 2N SU (N) is the lattice coupling Z β 2 • ≡ g0 Not true for dAµ , Aµ su(N) Last line is the Wilson plaquette action ∈ • FunctionalR integral well-defined: finite number of not obligatory to use simple plaquette: all traces of • • closed Wilson loops are proportional to F F as a 0, variables integrated over compact domain allowing other choices for lattice gauge action· → No gauge fixing required in lattice (in • general: but becomes necessary if you want to do a perturbative evaluation of the integral because of zero modes in the quadratic part of the action)

27 28 Group integration (compact Lie groups)

Consider link variable U (x) U SU (N) µ ≡ ∈ U is a complex N N matrix, detU = 1 3.3 Strong coupling expansion × Write Ui j , with matrix (colour) indices i, j Expectation values in lattice Yang-Mills theory: i j = Ui j 1 S [U ] O = D[U ]Oe− E h i Z x x + aµˆ Z U 1 1 β † = i−j SE [U ] = β ∑ 1 Re TrP2 = ∑Tr(P2 +P2)+const i j 2 − N − 2N 2  Group integrals: β = 2N/g 2 is a small parameter for large g 2 • 0 0 evaluate O by expanding exp( S [U ]) in powers of β • h i − E strong coupling expansion (high T , β = 1/T ) dU = 1 • evaluate integrals in group space order by order in β Z • dU Ui j = 0 Z β † 1 1 exp ∑ Tr(P2 + P2) = dU Ui jUk−l = δik δ jl 2N 2 N   Z β 1 1 Tr P P † O 2 dU U U = ε ε ∏ + ( 2 + 2) + (β ) i1 j1 iN jN i1 iN j1 jN 2 2N ··· N! ··· ···   Z

i j 1 = δ δ N ik jl k l

29 30

Area Law O To Show U (y) = 0 Let be a Wilson loop, ie a rectangle of size R T . h ν i ×

1 3 SE [U ] Uν (y) = ∏ ∏ dUµ (x)Uν (y)e− h i Z x Λ µ=0 1 R T Z ∈ E W (R,T ) = TrU × ≡ N with β † SE [U ] = ∑Tr(P2 + P2) − 2N 2

Pick out plaquettes involving Uν (y) Expectation value of W (R,T ) ˆ 1 1 λ W (R,T ) = D[U ] Tr ∏ U h i Z N U C 1 νˆ Z ∈ Uλ− (y) β † 2 1 ∏ 1 + (P2 + P2) + O(β ) contains Tr( U − (y)Uν (y) ) × 2 2N y ··· λ ···   U (y) ν List the contributions to W (R,T ) order by order in β Change variables on other links starting/ending at y. h i Order β 0: only group integrals of type • U (y) U (y)U (y) λ → ν λ dUU = 0 makes S independent of U (y) • E ν Z doesn’t change measure • Order β: consider all plaquettes leaves factor dU (y) U (y) = 0 • • ν ν inside W (R,T ), so that each link of An example of ElitzurR ’s theorem: all gauge non-invariant W (R,T ) pairs up with a link in the combinations of U ’s have vanishing expectation values. opposite direction. This is tiling the Wilson loop with plaquettes.

31 32 Order β contributions

1. Each plaquette inside W (R,T ) contributes Exercise

RT β β Work out the O(β 2) contribution in the strong coupling leading to a factor 2N 2N expansion of the Wilson loop   2. Each group integration contributes

1 1 (R+1)T +(T +1)R leading to a factor N N   3. Each site contributes a factor N colour indices of all links meeting at each site must • be the same (group integration plus trace) leads to N possibilities to choose the colour at each site RT • β β RT (R+1)(T +1) 2N2 4N N leading to a factor N     So that 4. All integrations outside W (R,T ) give 1 RT β β The total contribution is: W (R,T ) = 1 + RT + O(β 2) h i 2N2 4N RT 2RT +R+T RT R T 1 RT     1 β 1 − − − − β   = N 2N N 2N2 Still higher orders involve       evaluation of non-planar Therefore graphs RT β . . . but all successive terms W (R,T ) = + higher orders h i 2N2 depend on the area RT   But RT = A, area of the Wilson loop

33 34

3.4 Area law and linear confinement Relation of C(R,T ) with Wilson loop? Physical interpretation of area law: consider a static quark-antiquark pair separated by distance R:

Q(x0)

C(R,T ) = 0 Q(y 0)U (y 0,y)Q(y)Q(x)U (x,x 0)Q(x0) 0 h | | i R = Γ(x,x0) = Q(x)U (x,x 0)Q(x0) Tr S (x0,y 0)U (y 0,y)S (y,x)U (x,x0) ∼ { Q Q } Tr is over colour and spin • Q(x) Solution for static quark propagator SQ : Static quarks: propagate only in (Euclidean) time 1 + γ 3 0 mQ (y0 x0) SQ (y,x) = δ (y x)U (y,x) e− − , y0 > x0 x = (0,0) y = (0,T ) − 2 † SQ (x0,y 0) = γ5 SQ (y 0,x0) γ5, y00 > x00 Substituting:  

C(R,T ) Tr U (x0,y 0)U (y 0,y)U (y,x)U (x,x 0) x = (R,0) y = (R,T ) ∼ { } 0 0 2 1 + γ0 2m T Tr e− Q Correlation function: × spin 2   †  2m T C(R,T ) = 0 Γ (y,y 0)Γ(x,x0) 0 ∝ W (R,T ) e− Q h | | i h i † = ∑ 0 Γ (y,y 0) n n Γ(x,x 0) 0 So finally: n h | | ih | | i

2 E T (E(R) 2mQ )T V (R)T = 0 Γ n e− n W (R,T ) e− − = e− ∑ h i ∼ n h | | i T ∞ E (R)T V (R) is static quark potential, potential of a QQ pair →∝ e− separated by R E(R): energy of a QQ pair separated by distance R

35 36 Static quark potential

Strong coupling expansion yields V (r) σr • ∼ Expect to see Coulomb part, V (r) 1/r, for small r • ∼ Linear confinement General functional form of V (r): • e Use strong coupling expansion for W (R,T ) to compute h i V (r) = V0 + σr V (R): − r RT σ string tension β ln(β/2N2)RT W (R,T ) = = e e ‘charge’ h i 2N2   Write r = Ra, t = Ta Determine V (r) via numerical simulation by • ‘measuring’ Wilson loops (UKQCD hep-lat/0107021) a 2 ln(β/2N2)rt V (r)t W (R,T ) = e − = e− h i e = π/12 in bosonic string model (Luscher¨ 1981): 2 2 • V (r) = a− ln(β/2N )r σr confirmed numerically (Luscher¨ and Weisz, − ≡ hep-lat/0207003) area law implies linearly rising potential V (r) • need infinite energy to separate Q and Q entirely 2.5 • linear confinement • 1.5 σ is called the string tension • Result suggestive: strong coupling is opposite of 0.5 continuum limit. Should supplement result with numerical studies extrapolated to continuum limit to 5.93 Quenched, 623

[V(r)−V(r0)]*r0 −0.5 5.29, c=1.92, k=0.13400 confirm. Nonetheless, see a characteristic behaviour of 5.26, c=1.95, k=0.13450 strong-coupling gauge theories. 5.20, c=2.02, k=0.13500 5.20, c=2.02, k=0.13550 −1.5 5.20, c=2.02, k=0.13565 Model

−2.5 0 1 2 r/r0

37 38

4 Lattice Fermi fields

Step 1: 3.5 Plaquette-plaquette correlation Classical continuum Euclidean action for free fermions: Correlator of two plaquettes at same spatial position, 4 different times. Smallest linking surface is a 1 1 tube S[ψ,ψ] = d x ψ(x)(γµ ∂µ +m0)ψ(x) joining the plaquettes × Z ψ, ψ Grassmann valued Recall in Minkowski spacetime: γ M,γ M = 2g . Now { µ ν } µν define Euclidean Hermitian γ -matrices by:

t1,x t2,x γ = γ M 0 0 so γ ,γ = 2δ , γ † = γ γ = iγ M { µ ν } µν µ µ j − j  t Step 2: discretisation

mt Tr(U )Tr(U ) e− h 1 2 i ∼ 4 1 S[ψ,ψ] = a ψ(x) γµ (∇µ + ∇∗ ) +m ψ(x) with ∑ 2 µ 0 x ΛE m = 4lnβ + ∈  − ··· = a4 (x)Q (x) Dynamical mass generation in pure Yang-Mills ( ∑ ψ ψ x Λ mass) ∈ E where 1 Q = (∇µ + ∇∗ )γµ +m 2 µ 0 is the ‘naive’ lattice Dirac operator

39 40 Step 3: Quantisation ip x 1 1 e · (Q− )(x) = ˜(p) ξ 4 3 ∑ ξ S[ψ,ψ] a L T p Λ ip/µ +m0 Z D[ψ,ψ]e− ∈ E∗ ≡ ip (x y) Z 1 e · − = a4 (y) Correlation functions: ∑ 4 3 ∑ ξ y Λ a L T p Λ ip/µ +m0 ∈ E ∈ E∗ ! 1 S[ψ,ψ] ψ(x)ψ(y) = D[ψ,ψ]ψ(x)ψ(y)e− 4 h i Z a ∑ SF (x y)ξ(y) Z ≡ y Λ − ∈ E Add Grassmann sources η, ξ to get generating functional Generating functional: 1 eW [η,ξ] = e(η,ψ)+(ψ,ξ) = e(η,Q− ξ) W [η,ξ] 4 Diagonalise via Fourier transform: e = exp a ∑ η(x)SF (x y)ξ(y) x,y Λ − ( ∈ E ) 1 ip x (Qξ)(x) = Qe · ξ˜(p) Two point function: a4L3T ∑ p ΛE∗ ∈ 1 2 1 ∂ W [η,ξ] ip x ˜ ψ(x)ψ(y) = e = (iγµp +m0)e · ξ(p) 8 4 3 ∑ µ h i a ∂η(x)∂ξ(y) η,ξ=0 a L T p Λ ∈ E∗ = SF (x y) 1 − have defined p sin(apµ ) 4 ip (x y) µ a a 0 d p e · − 2 • ≡ → + O(a ) Q acts by multiplication with iγµpµ +m0 −→ (2π)4 ip/ +m • Z 0 now easy to invert . . . •

41 42

4.1 Wilson fermions

Problems with naive discretisation Add extra term to the naive lattice Dirac operator which formally vanishes as a 0: → 1. p/ = p/ + O(a2) 1 S a4 x m x 2. Particle masses are defined through poles of the W [ψ,ψ] = ∑ ψ( ) γµ (∇µ +∇∗µ ) + 0 ψ( ) x Λ 2 1 ∈ E propagator. Here, poles of (ip/ +m )− for m p are 0 0 µ 5  near:  ra 1 ∑ ψ(x)∇∗µ ∇µ ψ(x) − 2 x Λ p/ = 0 or sin(apµ ) = 0 ∈ E a 4 = a ψ(x) QW ψ (x) satisfied for pµ = 0,π/a ∑ • x ΛE corners of Brillouin zone yield additional poles ∈ •   in D = 4 there are 2D = 16 poles and hence a Have defined the Wilson-Dirac operator • 16-fold degeneracy in the spectrum 1 ra Q γµ (∇µ +∇∗ ) +m ∇∗ ∇µ This is the doubling problem W ≡ 2 µ 0 − 2 µ In interacting theory, momenta of order π/a can flip where r is the Wilson parameter, r = O(1) (and usually set you between different doublers: spurious to 1) ‘flavour-changing’ interactions Q acts by multiplication with 3. How to deal with fermion doubling? W ra ignore it: quarks come in sixteen different flavours X ˆ2 • ip/ +m0 + p staggered fermions (Kogut-Susskind): partial lifting 2 • of degeneracy, 16 4. → Wilson propagator: Wilson fermions: complete lifting of degeneracy • ip (x y) but explicit at finite a. 1 e · − S (x y) = W 4 3 ∑ ra 2 − a L T p Λ ip/ +m0 + pˆ ∈ E∗ 2

43 44 Wilson fermion dispersion relation for momentum (k,0,0) Adding the Wilson term, (ra/2)ψ(x)∆ψ(x), modifies the with π < ka π, ma = 0.2 and r = 0,0.2,0.4,0.6,0.8,1. dispersion relation: − − ≤

ra 2 1.2 m0 m0 + pˆ → 2 1 Term proportional to the Wilson parameter r vanishes in 0.8 the classical continuum limit a 0 and we recover the Ea continuum Euclidean fermion pr→opagator. 0.6 After adding the Wilson term, mass terms near corners of 0.4 BZ are: 0.2 0 pµ mass multiplicity π π/2 0 π/2 π − − ka

(0,0,0,0) m0 1 r ( π ,0,0,0) m + 2 4 a 0 a r ( π , π ,0,0) m + 4 6 a a 0 a r ( π , π , π ,0) m + 6 4 a a a 0 a r ( π , π , π , π ) m + 8 1 a a a a 0 a

Choose r = 1: states associated with corners of BZ receive masses of order 1/a, ie of order the cutoff scale these states are removed from the spectrum • one fermion species survives in the continuum limit •

45 46

4.2 Chiral symmetry on the lattice Explicit form of the Wilson action: Consider massless free fermions on the lattice with a 1 4 ˆ lattice Dirac operator Q = Q(x y) SW [ψ,ψ] = a ∑ ∑ ψ(x)(γµ r)ψ(x + aµ) − x Λ 2a µ − 4 ∈ E  S[ψ,ψ] = a ψ(x)Q(x y)ψ(y)  ∑ − ψ(x + aµˆ)(γ +r)ψ(x) x,y ΛE − µ ∈ 4r Desirable properties of Q: + m + ψ(x)ψ(x)  0 a  1. Q(x y) is local  − 2 Set r = 1: ‘project out’ components of Dirac spinor 2. Q˜(p) = iγµpµ + O(ap ) through appearance of 1 (1 γ ) to lift the degeneracy. 2  µ 3. Q˜(p) is invertible for p = 0 6 Problem: for m = 0, S [ψ,ψ] is no longer invariant under 0 W 4. Q + Q = 0 chiral transformations γ5 γ5 Nielsen-Ninomiya no-go theorem (1981): 1–4 do not hold ψ(x) eiaγ5 ψ(x) → simultaneously chiral symmetry is broken explicitly by the either left with doublers or chiral symmetry is • regularisation procedure −explicitly→ broken only restored as a 0: chiral and continuum limits ∗ are bound together→for Wilson fermions Ginsparg-Wilson relation lack of chiral symmetry makes operator mixing You can realise exact chiral symmetry on the lattice by ∗ more complicated in lattice case than in continuum replacing 4 with possible to show that explicit chiral symmetry Q Q aQ Q ∗ breaking by Wilson term appears in chiral Ward γ5 + γ5 = γ5 identities and becomes the term as a 0 → (P Ginsparg and KG Wilson PRD 25 (1982) 2649, M Luscher¨ hep-lat/9802011, 1998)

47 48 More on N-N conditions

1. Locality Needed for renormalisability and universality of the continuum limit Range over which fields are coupled in the action is infinitely smaller than any physical distance: compare More on the GW relation x y 1 γ x y γ a x y Q(x,y) | − | e− | − | = e− a | − | ∼ Q Q aQ Q ma x y γ5 + γ5 = γ5 correlation function e− | − | ∼ or where γ = O(1) and m is a physical mass. As a 0 the → 1 1 former is exponentially suppressed with respect to the Q− γ5 + γ5Q− = aγ5 latter. 1 1 Q− is highly non-local, but Q− ,γ5 should be local: the GW relation is highly non-trivial{ } GW relation is expected to imply ‘physical’ chiral symmetry on the lattice. Look at Ward identity for ψ(x)ψ(y) with x y a long distance, using usual chiral | − | (γ5) transformation. Get extra term from variation of the action: Cannot have long-range (non-universal) couplings in ψ(x)ψ(z)(aQγ5Q) ψ(z)ψ(y) the action which would compete with the physical h zz0 i ∼ 1 1 signals arising from universal collective behaviour. (Q− )xz (aQγ5Q) (Q− ) aγ5 zz0 z0y ∼ xy ˜ 2 2. Q(p) = iγµpµ + O(ap ) Want correct continuum limit this is local so negligible at long distances → 3. Q˜(p) invertible for p = 0 No extra poles at non-zero In fact there’s an exact chiral symmetry (Luscher)¨ (see later) momentum: no doublers6 4. Q,γ = 0 Chiral symmetry { 5} Wilson fermions give up entirely on chiral symmetry. Recent breakthrough: modify 4 to get chiral symmetry without doublers.

49 50

Exact chiral symmetry on the lattice

GW relation implies that ψQψ is invariant under flavour singlet chiral transformations: a ψ ψ + iεγ5(1 Q)ψ History → − 2 a ψ ψ + iεψ(1 Q)γ 1982: GW wrote down the relation but no solution was → − 2 5 • found in the interacting case—it was forgotten and non-singlet chiral transformations: 1997 • a realised that the Fixed Point Dirac operator of ψ ψ + iεT γ5(1 Q)ψ • → − 2 ‘classically perfect’ action satisfies GW a followed by observation that Dirac operators for ψ ψ + iεψ(1 Q)γ5T • → − 2 Domain Wall Fermions (Kaplan, Shamir) and overlap formalism (Neuberger) also satisfy GW where T is an SU (Nf ) generator 1998: Luscher¨ demonstrated the chiral symmetry Slightly smeared version of usual chiral transformation. • Looks too good? In fact, singlet chiral transformation alters Led to an explosion of interest. DWF and overlap already the measure used in some numerical studies. δD[ψ,ψ] = Tr(γ Q)D[ψ,ψ] − 5 gives the correct anomalous Ward identity (just like →Fujikawa in the continuum). No anomaly in non-singlet case since TrT = 0

51 52 Anomaly in LGT with GW relation

Expectation value of some fermion operator

1 S O = D[ψ,ψ]Oe− h i Z Z Apply chiral transformation as change of variable, remembering that S is invariant: 1 1 δψ = εγ (1 aQ)ψ δψ = εψ(1 aQ)γ Combining: 5 − 2 − 2 5 1 S O = D[ψ,ψ] 1 εa tr(γ Q) (O + εδO)e− 1 S 1 S h i − 5 O = D[ψ0,ψ0]O0e− = D[ψ,ψ]J(O + εδO)e− Z h i Z Z Z Z Z To order ε  with Jacobian factor J = ∂(ψ0,ψ0) . ∂(ψ,ψ) S D[ψ,ψ] δO a tr(γ Q) e− = 0 − 5 Z ∂ψx0 1 = δxy + εγ5(1 aQxy ) . . . giving the correct anomalous Ward identity for a global ∂ψy − 2 flavour-singlet axial transformation. ∂ψx0 1 = δxy + ε(1 aQxy )γ (Note: tr(γ Q) vanishes in the free case, but it’s non-zero in ∂ψ − 2 5 5 y the presence of gauge fields.)

1 + εγ (1 1 aQ) 0 J = det 5 − 2 0 1 + ε(1 1 aQ)γ  − 2 5  = det(1 + εX )det(1 + εY ) = 1 + ε tr(X +Y ) = 1 εa tr(γ Q) − 5 where X = γ (1 1 aQ), Y = (1 1 aQ)γ ) and used 5 − 2 − 2 5 det = exptrln, trγ5 = 0.

53 54

LH and RH chiral fermions

If have chiral symmetry expect to decompose Neuberger’s operator

ψQψ = ψ+Qψ+ + ψ Qψ An operator Q satisfying the GW relation can be defined as − − follows. Let It’s really possible: 1 ˆ ˆ QW = γµ (∇µ +∇∗µ ) a∇∗µ ∇µ ψ = P ψ ψ+ = P+ψ 2 − − − ψ = ψP+ ψ+ = ψP be the massless free Wilson-Dirac operator. N euberger’s − − 1 operator is defined (in its simplest form) as: where P = (1 γ5) as usual and  2  1 † 1/2 1 Q = 1 A(A A)− ˆ N a − P = (1 γˆ5)  2  where  ˆ γ5 is a ‘smeared’ γ5: A = 1 aQ − W γˆ = γ (1 aQ) 5 5 − Exercise γˆ5γˆ5 = 1 γ Q = Qγˆ Show that Q satisfies the GW relation 5 − 5 N ‘Left’ and ‘right’ become gauge-dependent ideas

55 56 4.3 Domain Wall Fermions

∂ 2 +V (s) u(s) = m2u(s) Dirac fermion in 5 dimensions − s   γ5 = +1 V (s) Assume eigenfunctions have D = γ ∂ + γ ∂ φ(s) 5 µ µ 5 s − definite chirality since 2 ∂s +V (s) commutes with γ5 = γ0γ1γ2γ3, µ = 0,1,2,3 − s − γ . Three cases: γ5 = 1 s: extra spatial coordinate 5 − φ(s) φ is a given potential 1. Continuous spectrum M representing a domain wall s ∞ V (s) | |→ M 2 leads to eigenvalues with m2 M 2 with height and width set by −→ ≥ a scale M, e.g. s 2. Discrete spectrum φ(s) = M tanh(Ms), but eigenfunctions with m2 < M 2 decay exponentially exact form not needed. discrete spectrum. All non-zero masses are of order−M→ 1/M (only scale). No negative masses since ∂ 2 +V (s) = ( γ ∂ + φ)†( γ ∂ + φ). Planewave solutions − s − 5 s − 5 s 3. Massless modes ip x D5χ(x,s) = 0 with χ(x,s) = e · u(s) ( γ5∂s + φ)u(s) = 0, γµpµu(s) = 0 p = (iE,p) physical 4-momentum − with solutions m2 = E 2 p2 mass of the mode − s P v = v Allowed m2 determined from: u(s) = exp φ(t )dt v,   γµpµv = 0  Z0  γ ∂ φ(s) u(s) = iγ p u(s) n 5 s − − µ µ u(s) Only LH solution is Multiply on left by iγ p  µ µ normalisable. Massless ∂ 2 +V (s) u(s) = m2u(s) mode bound to the wall s − s 2 with V (s) = γ5∂sφ(s) + φ (s).

57 58

on the boundary you find: •

G(x,0;y,0) = 2MP S(x,y)P+ Summary − where S(x,y) is the 4-dimensional propagator of the all but one mode have mass m O(M) operator • ≥ massless mode: left-handed and bound to domain wall 1/2 • D M + (D M) 1 (D /M)2 − at energies E M, theory describes a left-handed ≡ 4 − − 4 • fermion in 4-dimensions = D (1 D /2M + ) 4 − 4  ···  Domain Wall Fermions D describes a massless 4-dim fermion, reduces to D4 as M . → ∞ Mechanism is stable against changes in setup: D satisfies a Ginsparg-Wilson relation • domain wall Dirichlet boundary condition • −→ 1 γ D + Dγ = Dγ D Dirac fields χ(x,s) in s 0 with 5 5 M 5 • ≥ D = D + γ ∂s M (Kaplan 1992) The construction also works 5 4 5 − • in the presence of gauge fields (no s-dependence) satisfying ∗ and on the lattice: M 1/a, D Q (massless ∗ → 4 → W D χ(x,s) = 0, P+χ(x,s) = 0 Wilson-Dirac) 5 |s=0 1 1/2 massless mode as before D = 1 (1 aQ ) (1 aQ )†(1 aQ ) − • −→ a − − W − W − W 5-dim fermion propagator satisfies • 1  † 1/2   = 1 A(A A)− D G(x,s;y,t ) = (x y) (s t ) a − 5 s,t 0 δ δ ≥ − −  P G(x,s;y,t ) = 0 where A = 1 aQW + s=0 − use a finite 5th-dimension: can have one chirality • exponentially bound to one wall, other chirality on other wall

59 60 5.1 Fermion action in LQCD

5 Lattice QCD 1 3 ˆ SF [U ,ψ,ψ] = ∑ ∑ ψ(x)(1 γµ )Uµ (x)ψ(x+µ) x Λ − 2 µ=0 − ∈ E Formulate a lattice theory of quarks and gluons.  ˆ † Lattice action: + ψ(x+µ)(1+γµ )Uµ ψ(x)  SQCD [U ,ψ,ψ] = SG [U ] + SF [U ,ψ,ψ] + ψ(x) m0 + 4 ψ(x)   SG [U ] Wilson plaquette action Rescale ψ and ψ by SF [U ,ψ,ψ] Wilson fermion action ψ(x) √2κ ψ(x), ψ(x) ψ(x)√2κ Define a covariant derivative: → →

1 and fix κ by requiring (m0 + 4)2κ = 1 Dµ ψ(x) = Uµ (x)ψ(x + aµˆ) ψ(x) a − Lattice action for QCD with Wilson fermions becomes: 1 †  D∗ ψ(x) = ψ(x) U (x aµˆ)ψ(x aµˆ) 1 µ a − µ − − SQCD[U ,ψ,ψ] = β ∑ 1 ReTrP2 2 − 3 For the Wilson-Dirac oper ator:  3  1 ra + ∑ κ ∑ ψ(x)(1 γµ )Uµ (x)ψ(x+µˆ) γµ (∇µ +∇∗ ) +m ∇∗ ∇µ x Λ − µ=0 − 2 µ 0 − 2 µ ∈ E  † 1 ra + ψ(x+µˆ)(1+γµ )Uµ ψ(x) γµ (Dµ +D∗ ) +m D∗ Dµ → 2 µ 0 − 2 µ + ψ(x)ψ(x)  Set: r = 1  We have traded parameters: (g ,m ) (β,κ), with: a = 1 express all quantities in units of a 0 0 7→ 6 1 = , = (hopping parameter) β 2 κ g0 2m0 + 8

61 62

Introduce the effective gauge action, using 5.2 Effective gauge action detX = elogdetX = eTr log X Rewrite fermionic piece of LQCD action: so that S [U ,ψ,ψ] = ψ(x)[Q ψ](x) F ∑ W S [U ] x Λ Z = D[U ]e− eff , S [U ] S [U ] Tr logQ[U ] ∈ E eff ≡ G − Z ∑ ψ(x)Qxy ψ(y) ≡ x,y Λ Quark propagator: ∈ E 1 Qxy is Wilson-Dirac operator in matrix notation (‘quark 1 S [U ] ψ(y)ψ(x) = D[U ]Q− [U ]e− eff matrix’) h i Z yx Z 3 Now examine the fermionic contribution to S [U ] in Q 1 U x eff xy = δxy κ ∑ δy,x+µˆ ( γµ ) µ ( ) − µ=0 − greater detail. Split: 1 U † x (0) + δy,x µˆ ( +γµ ) µ ( ) Q[U ] = Q V [U ] − − Functional integral: Q(0) describes free Wilson fermions:

S [U ] S [U ,ψ,ψ] 3 Z = D[U ,ψ,ψ]e− G − F Q(0) 1 1 xy = δxy κ ∑ δy,x+µˆ ( γµ ) + δy,x µˆ ( +γµ ) Z − µ=0 − − Integrate out fermions:   1 (0)− (0) Q SW (free Wilson propagator) SG[U ] ≡ Z = D[U ]e− detQ[U ] while V is the interaction term: Z 3 V U 1 U x 1 xy [ ] = κ ∑ δy,x+µˆ ( γµ ) µ ( ) Exercise µ=0 − − h  1 U † y 1 Show that detQ[U ] is real. + δy,x µˆ ( +γµ ) µ ( ) − − i

63 64 5.3 The

There is phenomenological evidence that quark loops have only small effects on hadronic physics. Zweig’s (OZI) rule: φ 3π is suppressed relative to + → Now write φ K K − → 0 (0) (0) 1 (0) (0) 1 π Q[U ] = Q Q − Q[U ] = Q 1 Q − V [U ] d − s K − d The effective gauge action becomes  s u s u φ π − φ S [U ] = S [U ] logdetQ[U ] u eff G − s d s (0) 1 = S [U ] Tr log 1 Q − V [U ] + const d s + G − − u K ∞ 1 j π + = S [U ] + Tr S(0)V [U ]  G ∑ j W j=1 This motivates the quenched approximation which  Trace here is over all quark indices: Dirac, colour, site corresponds to setting • each term is a closed loop of j free quark detQ[U ] = 1, ie Seff[U ] = SG [U ] • and j vertices detQ[U ] = 1 corresponds to setting κ = 0 for internal the sum contributes closed quark loops to the effective • • action quarks (in loops)

κ = 0 mq = ∞ 1 1 ⇔ + + + 2 3 ··· infinitely heavy quarks in loops contributing to the effectiv−→ e interaction quenching is an enormous simplification for • numerical simulations: cost of full QCD > 10000 cost of quenched QCD

65 66

6 Numerical simulations 6.1 Monte Carlo integration

Return to the problem of computing observables in QCD Integrand is strongly peaked around configurations C with (restrict to SU (3) gauge group) large values of S [C ] S [U ] 1 S [U ] W (C ) e− eff = e− eff O = D[U ]Oe− eff ≡ h i Z Z C Boltzmann factor or statistical 1 3 W ( ) : C O Seff[U ] weight of configuration = ∏ ∏ dUµ (x) e− Z x Λ µ=0 Z ∈ E Monte Carlo procedure strong coupling expansions have a small radius of • convergence generate a sample or ensemble of gauge field • C weak coupling expansion is asymptotic configurations, i , i = 1,...,Ncfg, with statistical • weights W (C ) . . . and the two don’t overlap i • sample comprises predominantly configurations with O • exact evaluation of or Z on a computer is not large W (C ) • practical (although hpossiblei in principle) i importance sampling: design an algorithm which instead use stochastic methods to evaluate O or Z • C C • h i generates a configuration with likelihood W ( ) Monte Carlo integration: evaluate the observable on a common algorithms • finite number of ‘typical’ field configurations • Metropolis • Field configuration heat bath (for SU (N) gauge theory, scalar field • theories, spin systems) Assignment of an SU (3) matrix Uµ (x) to every link (x, µ) cluster algorithms (Swendsen-Wang, Wolff) (for • on the lattice: spin systems, O(N) models, not gauge theories) hybrid Monte Carlo (HMC) or multiboson C = U (x) x Λ , µ = 0,1,2,3 , C = U • µ ∈ E { } algorithms (for QCD with dynamical fermions) 

67 68 6.2 Hadronic correlation functions S [C ] S [U ] quenched QCD: use W (C ) = e− G = e− G as • probability measure Recall spectral decomposition of two-point function: (E E )(x y ) C SG [U ] n 0 0 0 ‘full’ QCD: use W ( ) = detQ[U ]e− as measure A(x)B(y) = ∑ 0 A(0,x) n e− − − n B(0,y) 0 • h i n,p h | | i h | | i detQ[U ] is real but not positive definite n • † S use det(Q Q)e− G , corresponding to two flavours of Now consider the pion two-point function: • dynamical quark † hard to simulate odd numbers of fermions Cπ (t ) ∑ 0 P(t ,x)P (0) 0 • ≡ x h | | i evaluate observables on each configuration in the • ensemble, O[C ], i = 1,...,N , giving N P(x) = ψ(x)γ ψ(x) is an interpolating operator i cfg cfg • 5 between the pion state and the vacuum. P † = P ‘measurements’ − sample average of observable ∑ projects onto zero momentum: states n at rest • • x | i

Ncfg states n in sum have same quantum numbers as O 1 O C • |P i = ∑ [ i] pion, J = 0− Ncfg i=1 0 P(0) n n P †(0) 0 expectation value p=0 p=0 M (π)t • Cπ (t ) = h | | ih | | ie− n ∑ 2M (π) O O n n = lim 2 h i Ncfg ∞ → 0 P n M (π)t = h | | i e− n ∑ 2M (π) results from Monte Carlo integration have statistical n n • error ∝ 1/ Ncfg For large Euclidean times t the state with the lowest • mass dominates, call it M p π

69 70

Numerical calculation of 2pt correlator

Let Calculating propagators P(x) = ψ1(x)γ5ψ2(x) S (x,y) has site, spin and colour indices and P †(x) = ψ (x)γ ψ (x) • W ab,µν − 2 5 1 depends on the gauge field and the quark mass (κ) 1,2 label distinct quark flavours: limits the contractions on a given configuration the propagator for quark type • appearing; different choices of 1,2 let you study mπ ,mK ,... j (with mass fixed by κ ) solves The correlator: j C C † Q i S i (x,y) = δx,y Cπ (t ) = ∑ 0 TP(t ,x)P (0) 0 zx W, j x h | | i suppressing colour and spin indices = 0 Tψ (x)γ ψ (x)ψ (0)γ ψ (0) 0 ∑−h | 1 5 2 2 5 1 | i impractical to solve for the whole matrix: instead, fix x • 1 y = 0: Seff[U ] 1 1 C C = ∑ ∏dUµ (x)e− Tr γ5Q2− [U ]x0γ5Q1− [U ]0x Q i S i (x,0) = δ x Z x Λ zx W, j x,0 ∈ E Zµ=0,...,3  solve matrix equation Q X = b for vector X 1 1 = Tr γ Q− [U ] γ Q− [U ] ∗ · ∑ 5 2 x0 5 1 0x repeat for each configuration i x ∗  gives propagator from 0 to any x 2 −→ = γ5 γ5 ∑ t ,x 1 0 correlation function also contains S (0,x). x • W † † since Q = γ5Q γ5, then SW = γ5SW γ5 Sample average ∗ get S (0,x) from S (x,0) ∗ −→ W W N cfg have all propagators needed 1 C C i i −→ Cπ (t ) = Tr γ S (x,0)γ S (0,x) N ∑ ∑ 5 W,2 5 W,1 now just evaluate the trace with γ ’s using propagators cfg i=1 x • 5  evaluated on each gauge configuration C where S i (y,x) is propagator for quark type j from x to y W, j C on the ith configuration i .

71 72 Effective mass plot On a finite lattice with T t • − periodic temporal Plot boundary conditions 2 2 C (t ) is symmetric for 0 Cπ (t ) + C (t ) C (T /2) 0 t T π eff π π   Mπ (t ) = ln − Mπ t T t C (t +1) + C 2(t +1) C 2(T /2) ≈ → − " π p π − π # t p

1 2 0 t T Mπ t Mπ (T t ) Cπ (t )   0 P π (e− + e− − ) 0.60 → 2Mπ h | | i Quenched β = 6.0, κ = 0.1337 1 2 M T /2 = 0 P π e− π cosh M (T /2 t ) h | | i π − Mπ 0.50  Obtain Mπ and the matrix element Z = 0 P π ∝ fπ by a M = 0.3609+0.0012−0.0013 • h | | i P fitting Cπ (t ) to the above cosh formula ZP = 0.1553+0.0039−0.0041 1 0.40 mass

0.01 0.30 fitted curve Cπ (t ) 4 10− 0.20 0 10 20 30 t 6 10− (D Lin, APE data) 0 5 10 15 20 25 30 t Mπ t Simpler: if T ∞, then Cπ (t ) ∝ e− and plot Example: Quenched, β = 6, κ = 0.1337, 323 64 lattice. → × Fitted curve has aM = 0.3609+12, Z = 0.1553+39. (D Lin, ln C (t )/C (t +1) M π 13 41 π π ≈ π APE data) − − Differs only at righthand end of above plot

73 74

6.3 Elimination of bare parameters

Hadron masses obtained from correlation functions depend implicitly on the input bare parameters, β and κ. Moreover, you determine dimensionless quantities, like aMπ and have to fix a afterwards. Eliminate bare parameters by matching lattice hadron masses to experiment. Can study quark mass dependence of on the

lattice by computing aMhad for several values of κ at fixed By choosing appropriate interpolating operators for: • β. From leading order chiral perturbation theory:

vector mesons ρ,K ∗,φ,... 2 Mπ = B(mu +md) octet baryons N,Σ,Λ,Ξ,... 2 M = B(m +m ) K  u s decuplet baryons ∆,Σ∗,Ξ∗,Ω Mρ = A +C(mu +md ) one can extract the hadron spectrum from fits to the M = A +C(mu +ms ) K ∗ correlation functions information on quark mass dependence resides in −par→ameters A, B and C Motivates ansatz for quark mass dependence of lattice data:

(aM )2 = (aB)(am + am ) PS q1 q2 aM = (aA) +C(am + am ) V q1 q2 C = (aA) + (aM )2 (aB) PS

75 76 To first approximation, assume mu = md = 0, so that one 2 2 2 expects Mπ = 0 (in real life: Mπ = 0.018GeV compared to 2 2 Mρ = 0.59GeV ). To compute masses of strange hadrons, one has to 2 Compute aMρ by plotting aMV versus (aMPS) and determine the value of κ which corresponds to the strange 2 extrapolating to (aMPS) = 0. quark mass: κs

0.4 Fix κs at the point where

(aM )2 M 2 (494MeV)2 PS = K  = = 0.4116 2 2 2 (aMρ ) Mρ (770MeV) aMV 0.35 Use similar procedure for κc , κb Summary

parameter fixed through 0.3 0 0.02 0.04 0.06 0.08 2 2 κu = κd (aMPS) = 0 (aMPS) a aMV = aMρ at κ = κu Example: Quenched, β = 6.2 (UKQCD PRD 62 054506,2000) 2 2 κs (aMPS) /(aMρ ) = 0.4116 Then use experimental value to ‘calibrate’ lattice spacing: . . . . M 1 ρ,phys a− = (aM ) ρ latt Mπ , Mρ and MK are used to eliminate β, κu, κd . This is called a hadronic renormalisation scheme. The fix all other masses in terms of M • ρ dependence of lattice estimates on β and κ has been have traded a hadronic quantity, M , for a bare • ρ eliminated by matching to the observed hadron spectrum. parameter, β could use other physical (dimensionful) quantities, • such as fπ , to fix a

77 78

6.4 Systematic errors

Lattice computations are truly first principles. Errors can Light hadron spectrum in quenched LQCD be systematically reduced.

1.8

1.6 aL Ξ Ω a 1.4 Σ Ξ* φ Λ 1.2 Σ* N K* m (GeV) 1.0 ∆ We want 0.8 K input 1 aL 1fm and a− Λ φ input   QCD K experiment 0.6 Computer power limits the number of lattice points which can be used and hence the precision of the calculation. 0.4 Typically, full QCD simulations use about 24 points in each spatial direction (O(50) in quenched simulations) so Errors shown are statistical and sum of statistical and compromises have to be made. systematic. Statistical errors Functional integral is evaluated by (CP-PACS collaboration hep-lat/0206009) importance sampling. Statistical error estimated from fluctuations of computed quantities within different clusters of configurations

79 80 Discretisation errors Current simulations typically have

a 0.05 to 0.1fm ∼

Errors with Wilson fermions are O(aΛQCD), O(amq) and can be particularly severe in heavy quark physics, although we are helped by: guidance from heavy quark symmetry • Finite volume effects Pion is light (pseudo Goldstone use of discretised effective theories boson of chiral symmetry breaking): it can propagate • Efforts to reduce discretisation errors: over large distances. Simulations are performed with heavier pions (ie using quarks around the strange Use several lattice spacings a and extrapolate a 0 • → mass) and results are extrapolated to the chiral limit. Improvement (Symanzik) Adjust the discretisation • so that errors are formally reduced. Simple eg: Typically impose mπ aL > 4 f (x+a) f (x) Quenching Repeated evaluation of fermion determinant f 0(x) = − + O(a) to generate unquenched gauge configurations very a expensive. More and more simulations now use compared to dynamical quarks, although typically have two flavours of degenerate sea-quarks a bit below the strange mass. f (x+a) f (x a) 2 f 0(x) = − − + O(a ) 2a Renormalisation Need to relate bare lattice operators to standard renormalised ones (eg MS): introduces Relatively easy to reduce errors from O(a) to uncertainties. O(αsa). Also possible, though more involved, to use nonperturbative improvement to get to O(a2). Perfect actions: apply renormalisation group to • continuum action to construct (classical) action with no discretisation errors. Truncations are necessary in practice: not used in large-scale QCD simulations to date

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6.5 Continuum Limit

Consider calculating a physical mass Mphys lattice mass vanishes in the continuum limit Dependence of bare coupling g0 • lattice gives on cutoff a • dimensionless g m = M a B(g ) = a ∂ 0 phys 0 − ∂a = β g 3 β g 5 + M should not − 0 0 − 1 0 ··· • phys depend on a (at least find g 0 as a 0 • 0 → → as a 0), hence m calculate B(g ) in lattice PT → • 0 depends on g0(a): corresponding correlation length ξ = 1/m diverges ∗ (in lattice units) dMphys = 0 continuum limit is a critical point da ∗ ∂m once ξ a the system ‘forgets’ the fine details of m + B(g ) = 0 . . . or nonperturbatively ∗  0 • the original lattice universality ∂g0 (ALPHA) −→ mass ratios should be pure numbers, independent of • Solve to find g0, a: M = C Λ 1 m physi i latt m = C exp − 100MeV 50MeV 5MeV 2 (lattice 2β0g Λlatt says how ‘strong’ the strong interaction is  0  mass) ∗ it’s strongly-dependent on the details of with a different C for ∗ regularisation: Λ /Λ = 28.8 for SU (3) YM each physical mass: MS latt finding C is the hard part (where all the 2 g0 ‘physics’ lies).

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Typically (ignoring operator mixing):

ren latt O (µ) = ZO (µa,g)O (a)

Scaling 1 if a− Λ and µ Λ can use PT to relate •  QCD  QCD ZO depends on short-distance physics calibrate a from mρ , fπ , σ, . . . • • ren,latt further calculations yield mass ratios m /m IR physics common to matrix elements of O • i 0 • if close enough to ctm limit, m /m is constant as • i 0 Example: axial vector current in Wilson LQCD β % this is scaling latt • Aµ = ψ(x)γµ γ5ψ(x) Use this in a 2-point correlation function: Asymptotic Scaling latt latt† C(t ) = ∑ 0 TA0 (x,t )A0 (0) 0 2 2 x h | | i PT in g0 should work for large enough β = 2N/g0 • latt† 2 observe scaling according to the β-function (1-loop) large t >0 π(p = 0) A0 (0) 0 m t • = h | | i e− π 1 2mπ M a exp − phys ∝ 2 2β0g0 But   this is asymptotic scaling Aren = Z Alatt and π(p = 0) Aren†(0) 0 = f m • µ A µ h | 0 | i π π so that latt† ZA π A0 0 fπ = h | | i mπ

. . . you need ZA to get the physical fπ .

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Perturbative renormalisation

Calculate matrix element of quark bilinear O between, say, Nonperturbative renormalisation the same quark states and fix ZO by demanding agreement Impose a physical condition to fix ZO (ZO is a property of O so use any convenient states). Example 1: local Wilson vector current •

Vµ = ψ(x)γµ ψ(x) ctm ZO = not conserved Z = 1 −→ V 6 Possible to define a conserved lattice vector current

V C, which has Z = 1. Hence, fix Z using latt µ V

αs π(p) V C(0) π(p) ZO = 1 + γ ln(µa) + c + h | µ | i 4π ··· ZV = π(p) Vµ (0) π(p) For axial current with µa =1  h | | i Example 2: Use Ward Identities to relate Z’s of different αs Z = 1 15.8 C • operators. For example, impose continuum axial A − 4 F π current WID 15.8 is a large coefficient. . . ∂ A O = 2m PO h µ µ i h i αMS/αlatt 2.7: αlatt is a poor expansion parameter with O arbitrary operator • s s ≈ s related to tadpoles: extra vertices in lattice PT from m renormalised quark mass • expanding exp aAµ (x) P pseudoscalar density turn to nonperturbative renormalisation. . . • 

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