アイソスピン化学ポテポテ シャル おけるンシャル...
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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
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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∫
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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.
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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)
19