アイソスピン化学ポテンシャルにおけるアイソ 化学ポテ シャル おけるクォーク行列の固有値分布と

ダランダム行列理論佐々井祐二 中村 純 高石哲弥 G Akemann佐々井祐二 中村 純 高石哲弥 G.Akemann

津山高専 広島大学 広島経済大学 Brunell University

Contents1 Research situation at μ≠01. Research situation at μ≠0

2. Formulation

3 C i f RMM lt d l tti d t3. Comparison of RMM results and lattice data

4. Pion decay constant Fπ

5. Distribution of the first eigenvalue in RMM

6. Summary

2008年9月3日～5日 熱場の量子論とその応用（京大基研） 1

1. Research situation at μ≠0 RMM LGT

SU(2) Full[1]

[2]SU(3) Quench[2]

SU(3) Ph Q h Thi t lkSU(3) Phase Quench This talk

SU(3) F llSU(3) Full

[1] Osborn Splittorfff & Verbarrschot (2005) Akemann & Bittner (2006)[1] Osborn, Splittorfff & Verbarrschot (2005), Akemann & Bittner (2006)

[2] Akemann & Wettig (2004)

Finite baryon-number density in SU(3) lattice QCDy y ( )

introduces chemical potential μ

quark matrix determinant positive, real for μ=0

2complex for μ≠0

numerical study becomes difficult !

L tti l l ti2. Formulation

Fermion : Kogut-Susskind (Staggered)

Lattice calculationg ( gg )

Even if Quark matrix determinant is complex

one may perform Monte Carlo simulation

Quenching measureg

g

S

Sq

DU OeO

DU e

β

β

−

−

=

∫∫

/ 4

/ 40

det

d

f g

f

N S

N S

DU Δ O eO

DU Δ

β

β

−

−

=

∫∫

Phase quenching measure

∫

/ 40

detf g

N SDU Δ e

β−

∫Phase quenching measure

Nf=2 Phase quench, SU(3), 83×4 lattice, β=5.3, f q , ( ), , β ,

ma=0.05

Calculated eigenvalues:Calculated eigenvalues:

all eigenvalues (NC×N

V=6144) in 980 configurations

3the smallest 100 eigenvalues in 15,000 /10,000 / 5,000

configurations

Random Matrix Model •G.Akemann and G.Vernizzi, 2003

•G.Akemann, 2003

Nf=2 Phase quenched spectral density

G.Akemann, 2003

•J.Osborne, 2004

Nf

2 Phase quenched spectral densityin weak non-Hermiticity limit

⎛ ⎞2*

( 2) ( 0)

* *

( , )( ) ( ) 1

( ) ( )f f

sN NK

K K

ξ ηρ ξ ρ ξ

ξ ξ

= =

⎛ ⎞⎜ ⎟= −⎜ ⎟

where quenched density is given by

( , ) ( , )s s

K Kη η ξ ξ⎜ ⎟⎝ ⎠

where quenched density is given by

( )2

2

2 1Re( )2( 0) *4

1( )

N

K Kξξ −

=

⎛ ⎞⎜ ⎟ ( )22( 0) *4

02 2

1( ) e , .

4 4fN

sK Kα

ξρ ξ ξ ξ ξ

πα α

= ⎜ ⎟=⎜ ⎟⎝ ⎠

( ) ( ) ( )21* 2 *

0 00

,

t

sK dt e I t I t

αξ ξ ξ ξ−

≡ ∫ 0 0( ) ( )I z J iz=

4

Bridge between LGT and RMM

/za V za dξ π= ⋅ Σ = ⋅

rescaled eigenvalue measured eigenvalue

th l tti

/ma V ma dη π= ⋅ Σ = ⋅

on the lattice

/ma V ma dη πΣ

rescaled mass given mass on the lattice

2 2 2( )a F Vπ

α μ= a: lattice spacing

rescaled mass given mass on the lattice

( )π

μ a: lattice spacing

d: mean level spacing

V: lattice volumegiven chemical potentialΣ: chiral condensate

Fπ: pion decay const.

given chemical potential

on the lattice

5

Mean level spacing d is very very important !

I d 1 di i l 2 di i l i ?

0 B k C h f l

Is d 1-dimensional or 2-dimensional spacing?

It seems that we should think of d as 1-dimensional spacing.

μ=0 Banks –Casher formula

Σ = = − = − ∝(0) 1

V Vd d

π ρ πψψ

y

V Vd dψψ

Measure the mean level spacing d between neighbor eigenvaluesbetween neighbor eigenvalues.

O x

yμ≠0

yfor the smallest 7eigenvalues

O x

y y

Calculate the mean Project

eigenvalues

i

level spacing d on

y-axis

O xO xon y-axis

6

3. Comparison of RMM results3. Comparison of RMM results

and lattice data

Our purpose again

Eigen-value distribution

function of Lattice

Spectral density of RMM2

*( )K ξ η⎛ ⎞

p p g

function of Lattice

∫

( , )x yρ

( 2) ( 0)

* *

( , )( ) ( ) 1

( , ) ( , )f f

sN N

s s

K

K K

ξ ηρ ξ ρ ξ

η η ξ ξ

= =

⎛ ⎞⎜ ⎟= −⎜ ⎟⎝ ⎠

3

( , )

3 8 4 6144

x y dxdy Nρ =

= × × =

∫We want to determine parameters

hi h d th l tti d t3 8 4 6144× ×

which reproduce the lattice data.

7

83×4 lattice, Nf=2, β=5.3, ma=0.05, μa=0.10

① Calculate mean level-spacing d, and rescale lattice data by it. µ a=0.1

(15,000 configurations,

−

= ×3

2.775 10d

ξ /d

the smallest 7

eigenvalues)ξ π= ⋅ /za d

rescaled eigenvalue measured eigenvalue

on the latticeTh i l i i

② Obtain the rescaled mass η by dThe aerial view is

obtained from 980

configurations.

/ 57.6ma dη π= ⋅ =

These values are determined

uniquely8

uniquely.

③ Put η and choose α suitably in RMM

☆ Choose α in order to match those

distribution latitudes, peaks and plateaus , p p

on the real and imaginary axes.

☆ Then α =1 68 is obtained µ a=0 1☆ Then α =1.68 is obtained. µ a 0.1

LGT 84×4 lattice, Nf=2, β=5.3，

0 05 0 10RMM N

f=2, α=1.68, η=57.6

ma=0.05, μa=0.10

ξ π= ⋅ /za d

f

9

G G

μa=0.10 15000 configurations

histogram LGT phase quench

RMM phase quench

RMM quench

histogram LGT phase quench

RMM phase quench

RMM quench

� Charts coincide without tuning of those normalizations.g

� Because the effect of is small, it is difficult to

know which of RMM graphs corresponds to LGT graph

detΔ

know which of RMM graphs corresponds to LGT graph.

� Free parameter is α only.

10

1st eigenvalue

Distribution of the first 3 eigenvalues in LGT1 eigenvalue

2nd eigenvalue

3rd eigenvalue

Tuning of parameter α

α= 1.58

α= 1 68α= 1.68

α= 1.78

11

μa=0.00−

= ×3

2.284 10d2 2 2( ) 0.0a F Vα μ= =( ) 0.0a F V

πα μ

No free parameter !

Spectral density of RMM

( ) ( )21( 0) 2 *

0 00

( )2

fN tydt e I t I tαρ ξ ξ ξ

= ′−

= ∫ ξ = +x iy

Spectral density of RMM

5000 configurations

histogram LGT full

RMM full

RMM quench

0

� This statistics are not so rich. The first three peaks of LGT full are

very well in agreement with the one of RMM full.12

μa=0.004773 This aerial view is the almost same one

at µa=0 0 The close-up near the origin−

= ×3

2.661 10d

at µa=0.0. The close-up near the origin

has very narrow distribution width.

15,000 configurations

RMM Nf=2，

1st peak y=1.635

histogram LGT phase quench

RMM h h

f

α=0.08，η=59.4

RMM phase quench

RMM quench

� This statistics are not so poor. It seems that only the first peak of p y p

LGT is in agreement with RMM. 13

µa=0.20 10,000 configurations

−

= ×3

4.341 10d

histogram LGT phase quenchLeft aerial view is histogram LGT phase quench

RMM phase quench

RMM quench

obtained from 580

configurations.

RMM Nf=2，

α=2.38，η=36.2

� There is the effect of at μa=0 2 It seems that statisticsdetΔ� There is the effect of at μa 0.2. It seems that statistics

are still insufficient in order to know whether the phase quenched

graph of LGT corresponds to the same graph of RMM14

detΔ

５．Distribution of the first eigenvalue in

RMM (Preliminary)For more precise α fitting we consider the comparison ofFor more precise α fitting, we consider the comparison of

the distribution of the first eigenvalue of LGT and RMM.

RMM 1st eigenvalue

(up to 1st 2 terms)

μa=0.10

1st eigenvalue

1 1 2 3

1( ) ( ) ( , ) ( , , )

2J J

p d d dξ ρ ξ ξ ρ ξ ξ ξ ξ ρ ξ ξ ξ′ ′ ′ ′′ ′ ′′= − + +∫ ∫ �

g

15ρi : i-point density correlation function

2 2( ) ( , )

J

p dξ ξ ρ ξ ξ′ ′= +∫ �2nd eigenvalue

where the integration is 2

0 0

( , )R

J

d drr dπ

ξ ρ ξ ξ φ′ ′ =∫ ∫ ∫Joint probability distribution （μ=0）

*( ) ( )Kξ ξ ξ( 0)

( )N =

（This corresponds to ）1( ) ( , )

sKρ ξ ξ ξ=

* *

2 1 1 1 2 1 2 2 1( ) ( ) ( ) ( , ) ( , )K Kρ ξ ρ ξ ρ ξ ξ ξ ξ ξ= −

( 0)( )fN

ρ ξ=

2 1 1 1 2 1 2 2 1( ) ( ) ( ) ( , ) ( , )

s sK Kρ ξ ρ ξ ρ ξ ξ ξ ξ ξ

Kernel （ν =0）

( )2 2

1 22 2

2 2 1 11 1Re( ) Re( )2 1 2* 8 82 2

1 2 1 2 0 02 2 2

1, ( ) ( ) e e

4 4 4s

K K Kξ ξ

α αξ ξ

ξ ξ ξ ξπα α α

− −

=

( ) ( )212 *

0 00

4 4 4

tdtte I t I t

α

πα α α

ξ ξ−∫ ( )0∫

16

５. Pion decay constant Fπ

/

β = 5.30a F Vπ

α μ =π

μa αfit αfit /μa

0 0 confinement none none0.0 confinement none none

0.004773 (β < βC= 0.08 16.8(β β

C

5.3197(9)) confinement

0 1 (β < β =5 314(1)) 1 68 16 80.1 (β < βC=5.314(1))

confinement

1.68 16.8

0.2 (β > βC=5.298(2))

deconfinement

2.38 11.9deconfinement

� βC

is from Kogut and Sinclear (2004).

� It th t F β i d fi t h i� It seems that Fπ

on βC

or in deconfinement phase is

smaller than Fπ

in confinement phase. 17

６. SummaryyA) 位相クェンチconfigurationの固有値分布を解析するために RMMの計算と比較したするために，RMMの計算と比較した．

B) µa=0.00の場合，フリーパラメータはない．LGT )fullの最初の3つのピークはRMM fullと非常によく一致している．く 致し る

C) µa=0.004773, 0.1, 0.2の場合, α パラメータの調節のみで RMMグラフはLGTグラフに 致す調節のみで，RMM グラフはLGTグラフに一致する．

D) βCに近づくとFπの値は下がるようである．

18

E) 今後は 0 125 0 150 0 175 （β β ）におE) 今後は，µa=0.125，0.150，0.175 （β<βC）においてFπ の変化を吟味したい．（固有値計算はほぼ

π

ぼ終了）

F) RMMにおける第１ 第２固有値分布を計算しF) RMMにおける第１，第２固有値分布を計算しLGTの結果と比較したい．このことにより， α パラメ タやF をより精密に吟味したい （マシンパラメータやFπをより精密に吟味したい．（マシンパワーが必要）

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