Universal properties of Wilson loop operators in large N

Universal properties of Wilson loop operatorsin large N QCDLarge N transition in the 2D SU(N)xSU(N)nonlinear sigma model

• Collaborators: Herbert Neuberger (Rutgers), Ettore Vicari (Pisa)

Rajamani NarayananDepartment of Physics

Florida International University

LATTICE 2008

Williamsburg, July 15

Williamsburg, July 15

LATTICE 2008

Wilson loop operator

• Unitary operator for SU(N) gauge theories.

• A probe of the transition from strong coupling to weak coupling.

• Large (area) Wilson loops are non-perturbative and correspond to strong coupling.

• Small (area) Wilson loops are perturbative and correspond to weak coupling.

Rajamani Narayanan 2

Williamsburg, July 15

LATTICE 2008

Definition of the probeWN(z, b, L) = 〈det(z −W )〉

• W is the Wilson loop operator.

• z is a complex number.

• N is the number of colors.

• b = 1g2N

is the lattice gauge coupling.

• L is the linear size of the square loop.

• 〈· · · 〉 is the average over all gauge fields with the standard gauge action.

Rajamani Narayanan 3

Williamsburg, July 15

LATTICE 2008

Multiplicative matrix model – Janik-Wieczorekmodel

WN(z, b, L) = 〈det(z −W )〉

• W =∏n

i=1 Ui; Uis are the transporters around the individual plaquettes that make up theloop and n = L2 is equal to the area of the loop.

• Two dimensional gauge theory on an infinite lattice can be gauge fixed so that the onlyvariables are the individual plaquettes and these will be independently and identicallydistributed.

• Set Uj = eiεHj and set P (Uj) = N e−N2 Tr H2

j .

• t = ε2n is the dimensionless area which is kept fixed as one takes the continuum limit,n →∞ and ε → 0.

• The parameters b and L get replaced by one parameter, t in the model.

WN(z, b, L) → QN(z, t)

• Note that N can take on any value.

Rajamani Narayanan 4

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LATTICE 2008

Average characteristic polynomial

QN(z, t) =

√

Nτ2π

∫∞−∞ dνe−

N2 τν2 [

z − e−τν−τ2]N

SU(N)√Nt2π

∫∞−∞ dνe−

N2 tν2 [

z − e−tν−τ2]N

U(N)

QN(z, t) =

∑N

k=0

(N

k

)zN−k(−1)ke−

τk(N−k)2N SU(N)

∑Nk=0

(N

k

)zN−k(−1)ke−

tk(N+1−k)2N U(N)

τ = t

(1 +

1

N

)

• Result is exact for the multiplicative matrix model and QCD in two dimensions.

• Both forms are useful in understanding the physics.

Rajamani Narayanan 5

Williamsburg, July 15

LATTICE 2008

Heat-kernel measure

The result for QN(z, t) is consistent with

P (W, τ )dW =∑R

dRχR(W )e−τC2(R)dW

• R denotes the representation.

• dR is the dimension of the representation R.

• C2(R) is the second order Casimir in thr representation R.

QN(z, t) = 〈N∏

j=1

(z − eiθj)〉 =

N∑k=0

zN−k(−1)kMk(t)

Mk(t) = 〈∑

1≤j1<j2<j3....<jk≤N

ei(θj1+θj2+...+θjk)〉 = 〈χk(W )〉 = dke

−τC2(k) =

(N

k

)e−

τk(N−k)2N

Rajamani Narayanan 6

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LATTICE 2008

Zeros of QN (z, t)

We can rewrite QN(z, t) for SU(N) as

ZN(z, t) = QN(z, t)(−1)Ne(N−1)τ

8 (−z)−N2 =

∑σ1,σ2,...σN=±1

2

eln(−z)∑

i σieτN

∑i>j σiσj

• Ferromagnetic interaction for positive τ .

• ln(−z) is a complex external magnetic field.

Conditions for Lee-Yang theorem are fulfilled.

All roots of QN(z, t) lie on the unit circle for SU(N).

This is not the case for U(N).

Rajamani Narayanan 7

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LATTICE 2008

Weak coupling vs strong coupling

QN(z, t) =

N∑k=0

(Nk

)zN−k(−1)ke−

t(1+ 1N )k(N−k)

2N

• Weak coupling; small area; t = 0

QN(z, t) = (z − 1)N

All roots at z = 1 on the unit circle.

• Strong coupling; large area; t = ∞

QN(z, t) = zN + (−1)N

Roots uniformly distributed on the unit circle.

QN(z, t) is analytic in z for all t at finite N . This is not the case as N →∞.

Rajamani Narayanan 8

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LATTICE 2008

Phase transition in an observable –Durhuus-Olesen transition

There is a critical area, t = 4, such that the distribution of zeros of Q∞(z, t) on the unit circlehas a gap around z = −1 for t < 4 and has no gap for t > 4.

The integral

QN(z, t) =

√Nτ

2π

∫ ∞

−∞dνe−

N2 τν2 [

z − e−τν−τ2]N

is domimated by the saddle point, ν = λ(t, z), given by

λ = λ(t, z) =1

zet(λ+12) − 1

With z = eiθ and w = 2λ + 1, ρ(θ) = − 14πRe w gives the distribution of the eigenvalues of W

on the unit circle.

The saddle point equation at z = −1 is

w = tanht

4w

showing that w admits a non-zero solution for t > 4.

Rajamani Narayanan 9

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LATTICE 2008

Double scaling limit

t =4

1 + α√3N

; z = −e(4

3N )34ξ

limN→∞

(4N

3

)14

(−1)Ne(N−1)τ

8 (−z)−N2 QN(z, t) =

∫ ∞

−∞due−u4−αu2+ξu ≡ ζ(ξ, α)

Claim

The behavior in the double scaling limit is universal and should be seen in the large N limit of3D QCD, 4D QCD, 2D PCM ....

The modified Airy function, ζ(ξ, α), is a universal scaling function.

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LATTICE 2008

Large N universality hypothesis

Let C be a closed non-intersecting loop: xµ(s), s ∈ [0, 1].

Let C(m) be a whole family of loops obtained by dialation: xµ(s, m) = 1mxµ(s),with m > 0.

Let W (m, C(∗)) = W (C(m)) be the family of operators associated with the family of loopsdenoted by C(∗) where m labels one member in the family.

DefineON(y, m, C(∗)) = 〈det(e

y2 + e−

y2W (m, C(∗))〉

Then our hypothesis is

limN→∞

N (N, b, C(∗))ON

(y =

(4

3N 3

)14 ξ

a1(C(∗)), m = mc

[1 +

α√3Na2(C(∗))

])= ζ(ξ, α)

Rajamani Narayanan 11

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LATTICE 2008

Numerical test of the universality hypothesis –3D large N QCD• Use standard Wilson gauge action

• The lattice coupling b = 1g2N

has dimensions of length.

• Use square Wilson loops and use the linear length, L, to label C(∗).

• Change b to generate a family of square loops labelled by L.

• Need to keep b > bB = 0.43 to be in the continuum phase.

• Need to keep b < b1 where b1 depends on the lattice size in order to be in the confinedphase.

• Need to use smeared links in the construction of the Wilson loop operator to avoid cornerand perimeter divergences.

• Need to obtain bc(L), a1(L) and a2(L) such that

limN→∞

N (b, N)ON

(y =

(4

3N 3

)14 ξ

a1(L), b = bc(L)

[1 +

α√3Na2(L)

])= ζ(ξ, α)

Rajamani Narayanan 12

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LATTICE 2008

Numerical test of the universality hypothesis –3D large N QCD

• Fix N and L.

• Obtain estimates for bc(L, N), a1(L, N) and a2(L, N).

• Check that there is a well defined limit as N →∞.

• Check that bc(L), a1(L) and a2(L) have proper continuum limits as L →∞.

Rajamani Narayanan 13

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LATTICE 2008

Binder cumulant

WithON(y, b) = C0(b, N) + C1(b, N)y2 + C2(b, N)y4 + · · ·

define

Ω(b, N) =C0(b, N)C2(b, N)

C21(b, N)

.

As N →∞, Ω(b,∞) is a step function with

• Strong coupling; b < bc(L); Ω = 16.

• Weak coupling; b > bc(L); Ω = 12.

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LATTICE 2008

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2b=4/t

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Ω(b

,N)

Multiplicative matrix model

Rajamani Narayanan 15

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LATTICE 2008

-30 -20 -10 0 10 20 30α

0.2

0.3

0.4

0.5

Ω(α

)

Scaling limit of the multiplicative matrix modelΩ(0)=0.364739936

Rajamani Narayanan 16

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LATTICE 2008

Extraction of bc(L, N), a2(L, N) and a1(L, N)

We useΩ(bc(L, N), N) = 0.364739936

to extract bc(L, N).

We usedΩ(b, N)

dα

∣∣∣∣α=0

=1

a2(L, N)√

3N

dΩ

db

∣∣∣∣b=bc(L,N)

= 0.0464609668

to extract a2(L, N).

We use √4

3N 3

1

a21(L, N)

C1(bc(L, N), N)

C0(bc(L, N), N)= 0.16899456

to extract a1(L, N).

Rajamani Narayanan 17

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LATTICE 2008

0 0.01 0.02 0.03 0.04 0.05 0.061/N

0.17

0.18

0.19

0.2

b c(L,N

)/L

Linear fit using ΩFitted values0.174(3)Nonlinear simultaneous fitFitted values0.173

L=6, 83 lattice

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LATTICE 2008

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24

1/N0.5

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6a 2(N

)

Linear fit using ΩFitted values2.46(25)Nonlinear simultaneous fitFitted values2.27

L=6, 83 lattice

Rajamani Narayanan 19

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LATTICE 2008

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24

1/N0.5

0.8

0.82

0.84

0.86

0.88

0.9

0.92

0.94

0.96

0.98

1

1.02

1.04

a 1(L,N

)

Linear fit using ΩResults from the linear fit0.99(4)Nonlinear simultaenous fitFitted values0.955

L=6, 83 lattice

Rajamani Narayanan 20

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LATTICE 2008

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.51/L

0.1

0.12

0.14

0.16

0.18

0.2

0.22

0.24b c/L

bb

I

fit without 2-by-2 continuum value is 0.113(5), χ2

/d.o.f=0.102

fit with 2-by-2 continuum value is 0.126(3), χ2

/d.o.f=4.63

fit without 2-by-2 continuum value is 0.112(5), χ2

/d.o.f=0.078

fit with 2-by-2 continuum value is 0.124(3), χ2

/d.o.f=3.72

Rajamani Narayanan 21

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LATTICE 2008

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.51/L

0

1

2

3

4

5

6

7

8

a 2

bb

I

fit without 2-by-2 continuum value is 6.2(2.0), χ2

/d.o.f=0.065

fit with 2-by-2 continuum value is 6.3(1.4), χ2

/d.o.f=0.034

fit without 2-by-2 continuum value is 5.1(1.6), χ2

/d.o.f=0.116

fit with 2-by-2 continuum value is 4.5(9), χ2

/d.o.f=0.171

Rajamani Narayanan 22

Williamsburg, July 15

LATTICE 2008

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.51/L

0.85

0.86

0.87

0.88

0.89

0.9

0.91

0.92

0.93

0.94

0.95

0.96

0.97

0.98

0.99

1

1.01

a 1

datafit without 2-by-2 continuum value is 0.97(6), χ2

/d.o.f=0.456

fit with 2-by-2 continuum value is 0.96(4), χ2

/d.o.f=0.301

Rajamani Narayanan 23

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LATTICE 2008

Two dimensional SU(N) X SU(N) principal chiralmodel

• Similar to four dimensional SU(N) gauge theory in many respects.

•S =

N

T

∫d2xTr∂µg(x)∂µg

†(x)

g(x) ∈ SU(N).

• The global symmetry group SU(N)L× SU(N)R reduces down to a single SU(N) “diagonalsubgroup” if we make a translation breaking “gauge choice”, g(0) = 1.

• Model is asymptotically free and there are N − 1 particle states with masses

MR = Msin(Rπ

N )

sin( πN )

, 1 ≤ R ≤ N − 1.

The states corresponding to the R-th mass are a multiplet transforming as an R componentantisymmetric tensor of the diagonal symmetry group.

Rajamani Narayanan 24

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LATTICE 2008

Connection to multiplicative matrix model• W = g(0)g†(x) plays the role of Wilson loop with the separation x playing the role of area.

• One expects

GR(x) = 〈χR(g(0)g†(x))〉 ∼ CR

(N

R

)e−MR|x|

where χR is the trace in the R-antisymmetric representation.

• Comparison with the multiplicative matrix model suggests that M |x| plays the role of thedimensionless area.

• Numerical measurement of the correlation length using the lattice action

SL = −2Nb∑x,µ

<Tr[g(x)g†(x + µ)]

and

ξ2G =

1

4

∑x x2G1(x)∑x G1(x)

yields the following continuum result:

MξG = 0.991(1)

Rajamani Narayanan 25

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LATTICE 2008

Setting the scale

• ξG will be used to set the scale and it is well described by

ξG = 0.991

[e

2−π4

16π

]√

E exp(π

E

)in the range 11 ≤ ξG ≤ 20 with

E = 1− 1

N<〈Tr[g(0)g†(1)]〉 =

1

8b+

1

256b2+

0.000545

b3− 0.00095

b4+

0.00043

b5

The above equations will be used to find a b for a given ξ.

Rajamani Narayanan 26

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LATTICE 2008

Smeared SU(N) matrices

One needs to smear to defined well defined operators.

• Start with g(x) ≡ g0(x).

• One smearing step takes us from gt(x) to gt+1(x).

• Define Zt+1(x) by:

Zt+1(x) =∑±µ

[g†t (x)gt(x + µ)− 1]

• Construct antihermitian traceless SU(N) matrices At+1(x)

At+1(x) = Zt+1(x)− Z†t+1(x)− 1

NTr(Zt+1(x)− Z†t+1(x)) ≡ −A†

t+1(x)

• SetLt+1(x) = exp[fAt+1(x)]

• gt+1(x) is defined in terms of Lt+1(x) by:

gt+1(x) = gt(x)Lt+1(x)

Rajamani Narayanan 27

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LATTICE 2008

Numerical details

• We need L/ξG > 7 to minimize finite volume effects.

• Since we want 11 ≤ ξG ≤ 20, we chose L = 150.

• We used a combination of Metropolis and over-relaxation at east site x for our updates. Thefull SU(N) group was explored.

• 200-250 passes of the whole lattices was sufficient to thermalize starting from g(x) ≡ 1.

• 50 passes were enough to equilibriate if ξG was increased in steps of 1.

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LATTICE 2008

Test of the universality hypothesis

The test of the universality hypothesis proceeds in the same manner as for three D large Ngauge theory.

Given an N and a ξ, we find the the dc the makes the Binder cumulantΩ(dc, N) = 0.364739936.

We look at dc as a function of ξ for a given N . This gives us the continuum value of dc/ξ forthat N .

We then take the large N limit and it gives us

dc

ξG|N=∞ = 0.885(3)

Rajamani Narayanan 29

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LATTICE 2008

2 3 4 5 6 7 8

x 10−3

0.65

0.7

0.75

0.8

1/ξ2

d c/ξ

N=30

d

c/ξ at finite ξ

fitted 2−nd deg polynomial

fit: dc/ξ =1543/ξ4−26.10/ξ2+0.7682

With error estimation: (d/ξ)critical, continuum

=0.768(2)

Rajamani Narayanan 30

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LATTICE 2008

0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07

0.65

0.7

0.75

0.8

0.85

1/N

(d/ξ

G) cr

it co

nt

Extrapolation of continuum critical dc/ξ

G to N=∞

d

c/ξ

G data

linear fit

dc (N) /ξ

G = − 3.535/N + 0.885

dc/ξ

G |

N=∞ = 0.885(3)

Rajamani Narayanan 31