Fisher’s zeros for SU 3 with N ﬂavors and RG ﬂows · 2013. 8. 5. · 1212 1616 2016 2020 Plaq vs beta; Nf=12; m=0.02 Figure: Plaquette discontinuities for various lattices

Fisher’s zeros for SU(3) with Nf flavors and RG flows

Yuzhi Liu

The University of Iowa

Fermilab

Lattice 2013 MainzJuly 30, 2013

In collaboration with Z. Gelzer, Y. Meurice and D. K. Sinclair (ANL).

Yuzhi Liu (U. of Iowa) Fisher’s zeros and RG flows Lattice 2013 Mainz July 30, 2013 1 / 32

Content of the talk

Global Aspects of RG flows for SU(3) with Nf flavors

Fisher’s zeros and finite size scaling

Two lattice matching and renormalization group flows

Fisher’s zeros for SU(3) with Nf flavors

Conclusions


Attempt to describe globally the RG flows of SU(3)with Nf flavors

Generic descriptions of the the global properties of the RG flows:

β (inverse bare coupling) direction

bare mass direction

"other" directions (induced by blocking or coarse graining the bare theory)

The bare theory is a point in the β-mass plane

To start with, this could describe the situation of SU(3) with Nf flavors ofunimproved staggered fermions at zero temperature.

Deformations will be discussed later


The bare plane

Figure: The set of initial theoriesYuzhi Liu (U. of Iowa) Fisher’s zeros and RG flows Lattice 2013 Mainz July 30, 2013 4 / 32

SU(3) with Nf flavors: pure gauge plane

Figure: The pure gauge (m → ∞) planeYuzhi Liu (U. of Iowa) Fisher’s zeros and RG flows Lattice 2013 Mainz July 30, 2013 5 / 32

SU(3) with Nf flavors: the massless plane

Figure: The massless planeYuzhi Liu (U. of Iowa) Fisher’s zeros and RG flows Lattice 2013 Mainz July 30, 2013 6 / 32

SU(3) with Nf flavors: types of flows

Figure: Schematic flows


SU(3) with Nf flavors: bulk transitions

Figure: Surface of bulk transitions?


SU(3) with Nf flavors

Figure: Schematic flows (a better picture).


SU(3) with Nf flavors: IR fixed point

Figure: β∗ from " the least irrelevant direction".


βbulk and β∗ as a function of Nf

Figure: βbulk and β∗ (see de Forcrand et al. arXiv:1211.3374, TomboulisarXiv:1211.4842). Could the solid line and the dotted line merge at Nc

f ?


SU(3) with Nf flavors: Non-integer Nf ?

4.6 4.7 4.8 4.9 5β

0.48

0.5

0.52

0.54

0.56

0.58

0.6

0.62

<Pl

aq>

Nf = 4

Nf = 4.5

Nf = 5

<Plaq> for different Nf; V = 4

4; m = 0.02.

Figure: Average Plaq for integer and non-integer Nf ; V = 44; m = 0.02.


Finite T

Figure: What is the lowest value of Nf for which the finite T transition turns into a bulktransition?


SU(3) with Nf = 12, Finite T transition turning into abulk transition

3.8 3.9 4 4.1 4.2 4.3 4.4beta

0.5

0.55

0.6

0.65

Pla

q44841241646686126881212161620162020

Plaq vs beta; Nf=12; m=0.02

Figure: Plaquette discontinuities for various lattices. 8 4 means a 83× 4 lattice.


SU(3) with Nf = 12: Finite T transition turning into abulk transition

3.8 3.9 4 4.1 4.2 4.3 4.4beta

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Pbp

44841241646686126881212161620162020

Pbp vs beta; Nf=12; m=0.02

Figure: < ψ̄ψ > discontinuities for various lattices. 8 4 means a 83× 4 lattice.


Nf = 12 Finite T transition turning into a bulk transition

3.8 3.9 4 4.1 4.2 4.3 4.4beta

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Wl

44841241646686126881212161620162020

Wl vs beta; Nf=12; m=0.02

4.05 4.1 4.15 4.2 4.25 4.3beta

0

0.02

0.04

0.06

0.08

Wl

44841241646686126881212161620162020

Wl vs beta; Nf=12; m=0.02

Figure: Polyakov’s loop (time-periodic Wilson’s line) for various lattices. 8 4 means83

× 4 lattice.


Questions

What is the lowest value of Nf for which the finite T transition turns into abulk transition?How does the end point in (β,m) for 12 flavors (Jin and MawhinneyarXiv:1203.5855) and 8 flavors (Christ and Mawhinney, Phys.Rev. D46(1992)) disappear? (Could it shrink to zero mass?)

3.8 4 4.2 4.4 4.6 4.8β

0.45

0.5

0.55

0.6

0.65

0.7

<Pl

aq>

Nf = 4

Nf = 5

Nf = 7

Nf = 9

Nf = 12

<Plaq> for different Nf and V = 4

3 x 4

Figure: Average Plaquette for different Nf (preliminary).


Fisher’s zeros and Finite Size Scaling

Decomposition of the partition function (Niemeijer and van Leeuwen)

Z = Zsing.eGbounded

Zsing. = e−LD fsing.

RG transformation: the lattice spacing a increases by a scale factor b

a → ba

L → L/b

fsing. → bD fsing.

Zsing. → Zsing.

Important Conclusion (Itzykson et al. 83)The zeros of the partition functions are RG invariant.

Fisher’s zeros: zeros of the partition function in the complex β plane.



We consider discrete RG transformations.

Lattice size in lattice spacing unit: L −→ Lb

Scaling variables: ui −→ λiui = byi ui

Relevant variables: λi = b1/νi > 1.Irrelevant variables: λj = b−ωj < 1.

Singular part of the partition function: Zs = Q({uiL1/νi }, {ujL−ωj})

If we only keep one relevant variable u ≃ β − βc , Fisher’s zeros have thefollowing relation

uL1/ν = A + BL−ω +O(L−2ω),

and the lowest Fisher’s zeros

Re(β1(L))− βc = Re(A)L−1/ν + Re(B)L−1/ν−ω ,

Im(β1(L)) = Im(A)L−1/ν + Im(B)L−1/ν−ω .



0

1

2

3

0 2 4 6 8 10 12

nmax

Ising HM, Re and Im of lowest zeros, D=3

βc

Lowest Re(β)Lowest Im(β)

Fits



Calculate both the real and imaginary part of the partition function.

The “logarithmic residue” method

12πi

∮

Cβn Z ′(β)

Z (β)=

k∑

i=1

βni ,

where βi are all the zeros in the integration region C.

Single point reweighting → multiple point reweighting.

Approximate the density of state via Chebyshev polynomial.

Different methods allow us to crosscheck the final results.



Two lattice matching is a way to measure the running of the barecouplings and to construct the approximate RG flows. (A. HasenfratzPRD 80 034505)

One of the two lattice matching observables:

R(β,V/aD) ≡

⟨

(∑

x∈B1~φx )(

∑

y∈B2~φy )

⟩

β⟨

(∑

x∈B1~φx )(

∑

y∈B1~φy ))

⟩

β

The matching condition can be chosen as

R(β, LD) = R(β′, (L/b)D)

β = β0 =⇒ β′

1

β = β′

1 =⇒ β′′

1

...



-4

-2

0

2

4

0 2 4 6

Im

β

Reβ

HM matching n=4 and n=5

-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


-4

-2

0

2

4

0 2 4 6

Im

β

Reβ


RG flowsZeros n=4Zeros n=5

Figure: Fisher’s zeros seperate complex RG flows (arrows) (PRL 104 25160).


Fisher’s zeros of the 2D O(2) model: TRG and MC

Figure: Fisher’s zeros of XY model with L = 4, 8, 16, 32, 64 for 30 states compared toMC (Alan Denbleyker and Haiyuan Zou).



Zeros of SU(3) with 3 flavors with reweighting method.

0.50 0.52 0.54

50

100

150

0

0.005

0.01

0.015

5.115 5.12 5.125 5.13 5.135Im

�

Re�

SU(3) 3 flavors 4 x 123

ReZ=0ImZ=0

Figure: Fisher’s zeros obtained by reweighting for 3 light quarks on a 4 × 123 lattice.Left: plaquette distributions for several values of β. The high-statistics (veryexpensive!) double peak distribution at β =5.124 is superimposed. L−3 scaling for theImaginary part of lowest zero for 4 × L3 lattices.


Fisher zeros for SU(3) with Nf = 4 and 12

0

0.01

0.02

0.03

0.04

0.05

3.6 3.8 4 4.2 4.4 4.6 4.8 5 5.2 5.4 5.6 5.8 6

Imβ

Reβ

Fisher’s zeros for Nf=4 and Nf=12, m=0.02

Nf=4, V=44

Nf=4, V=84

Nf=4, V=124

Nf=12, V=44

Nf=12, V=64

Nf=12, V=84

Nf=12, V=164

Nf=12, V=204

Figure: Zeros for Nf = 4 and Nf = 12 for L4 lattices (L = 16, 20 preliminary). 12flavors seems to pinch the real axis. Higher volumes in progress.


Fisher zeros scaling for SU(3) with Nf = 12

1e-07

1e-06

1e-05

0.0001

0.001

0.01

4 8 16 32

Imβ

L

Imaginary part of Fisher’s zeros for Nf=12, m=0.02

Nf=12, V=44

Nf=12, V=64

Nf=12, V=84

Nf=12, V=164

Nf=12, V=204

fit function 8.51892*L-4.0432

Figure: Lowest zeros scale like L−4 (bulk). (L = 16, 20 preliminary) .


Questions

Will the Fisher’s zero pinch the real axis like L−2 (ν=1/2, mean field for afree scalar) instead of L−4 near the end point (for m = mc)?

Is it possible to find a hint of the IR fixed point from the behavior of thezeros over a broader β interval as a function of m?

Are the two questions related?


Conclusion

The RG flows can be understood from the Fisher’s zeros point of view.

There is a clear first order phase transition for Nf = 12 from the scaling ofthe zeros. (L−4)

We are looking for the smallest Nf for which the finite temperaturetransition turns into a bulk transition. It may be possible to makeconnection between βbulk and β∗.

Understanding the mass dependence of the transition and calculating themass anomalous dimension γm from the zeros is in progress.

Adding improvement terms to the fermion action may change the phasestructure (S4b phase, Zech Gelzer, in progress).

New methods to perform the RG blocking is possible. (YM’s talk onTensor RG.)


Backup Slides


Finite T transition turning into a bulk transition?

The two loop β function for arbitrary Nc and Nf reads

β(g) = −a∂g∂a

= −β0g3 − β1g5 + · · ·

where

β0 =1

16π2 (113

Nc −23

Nf )

and

β1 =1

(16π2)2 (34N2

c

3−

10NcNf

3−

(−1 + N2c )Nf

Nc)

The solution of the above differential equation is

a =1ΛL

exp(−1

2β0g2 )(β0g2)−

β12β2

0

Since the physical temperature T is give by aT = N−1t and β = 2Nc/g2, we

can get

aTΛL/T = exp(−β

4Ncβ0)(

β

2Ncβ0)

β12β2

0


Finite T transition turning into a bulk transition?

ln(Nt ) = ln(ΛL

T) +

β

4Ncβ0−

β1

2β20

ln(β

2Ncβ0)

For Nf = 12 and Nc = 3, we will have

ln(Nt ) = ln(ΛL

T) +

4π2β

9+

259

ln(8π2β

9)

at 2-loop level.Assuming the physical critical temperature Tc and ΛL is fixed, then we willhave

ln(Nt2

Nt1) =

4π2(βc2 − βc1)

9+

259

ln(βc2

βc1).

The δβc = βc2 − βc1 is almost proportional to ln(Nt2Nt1

) since βc2βc1

≈ 1 for largeadjacent Nt .

ln(1612

) =4π2(βc2 − βc1)

9+

259

ln(βc2

βc1)

For Nt2 = 16 and Nt1 = 12, δβc = βc2 − βc1 ≈ 0.058.


Documents

Fisher’s zeros for SU 3 with N ﬂavors and RG ﬂows · 2013. 8. 5. · 1212 1616 2016 2020 Plaq vs beta; Nf=12; m=0.02 Figure: Plaquette discontinuities for various lattices