Download presentation
Presentation is loading. Please wait.
Published byDarleen Kennedy Modified over 6 years ago
1
CalculationΒ ofΒ high-orderΒ cumulants Β withΒ canonicalΒ ensembleΒ method Β inΒ latticeΒ QCD
Contents Asobu Suzuki Univ. of Tsukuba for Zn Collaboration S. Oka (Rikkyo Univ.) S. Sakai (Kyoto Univ.) Y. Taniguchi (Univ. of Tsukuba) A. Nakamura (Hiroshima Univ.) R. Fukuda (Tokyo Univ.) Quark number cumulant Sign problem Canonical ensemble method Winding number Numerical results Conclusion
2
Hadronic phase : Hadron Resonance Gas QGP phase : Quark Gluon Gas
Motivation π π QGP phase critical point Hadronic phase QCD phase diagram Hadronic phase : Hadron Resonance Gas QGP phase : Quark Gluon Gas Hadronic phaseγβγQuark-Gluon phase from the viewpoint of quark number cumulant
3
Quark number cumulant H.R. Gas =πππ§π‘ π π π =π =π
π π πΆ β π π π π/π π log π πΊ.πΆ. H.R. Gas π 3 πΆ π 2 πΆ =πππ§π‘ π π π (dotted line) π 4 πΆ π 2 πΆ =π π 3 πΆ π 1 πΆ =π deviation from H.R. Gas STAR Collaboration Phys. Rev. Lett. 112 (2014) something interesting?
4
Canonical ensemble method A.Hasenfratz , D.Toussaint (1992)
π πΊ.πΆ. π,π;π = π π πππ. π;π,π π π ,π= π π π Fugacity expansion Canonical partition function Equivalent Grand canonical partition function Residue theorem π πππ. π;π,π = 1 2ππ πΆ ππ π β π+1 π πΊ.πΆ. π;π,π pure imaginary πΆ:π= π π π π , π π β[βπ,π) = 1 2π βπ π π π π π βπ π π π π πΊ.πΆ. π;ππ,π Fourier transformationοΌ
5
Winding number expansion Li, X. Meng, A. Alexandru,K. F. Liu (2008)
Canonical partition function Grand canonical partition function Fourier trans. π πππ. π;π,π = 1 2π βπ π π π π π βπ π π π π·ππ‘ π· ππ π·ππ‘ π· π sign real , positive calculate π·ππ‘ π·(ππ) at low cost ! πΎ 5 π· π πΎ 5 =π· β π β β π·ππ‘ π· π =π·ππ‘ 1βπ
π π = π ππ log 1βπ
π π
: hopping parameterγγγγ ππ log 1βπ
π =β π π
π π ππ π π = π π π π π π π π ; winding number
6
Numerical results β log π πππ. π π½=1.1 β log π πππ. π π½=2.1
Iwasaki gauge action 2-flavor Wilson Clover Lattice Size : 8 3 Γ4 π· πΏ π»/π» πͺ 0.9 0.137 0.644 1.1 0.133 0.673 1.3 0.706 1.5 0.131 0.813 1.6 0.130 1.7 0.129 1.00 1.8 0.126 1.9 0.125 1.68 2.1 0.122 3.45 β log π πππ. π γπ½=1.1 β log π πππ. π γπ½=2.1 π π = π π π π πππ. π π π π π π π πππ. π π π π π π πΆ π=0 = πππ small π½ β low temperature large π½ β high temperature
7
H.R. gas Q.G. gas π 2 πΆ / π 1 πΆ π½ not depend on π· (solid line)
π 2 πΆ / π 1 πΆ π½ H.R. gas (solid line) π 2 πΆ π πΆ = 3 coth π π΅ π not depend on π· π 2 πΆ π πΆ = π π π΅ 3π π π΅ 3π + 1 π π π΅ 3π 3 Q.G. gas (dotted line)
8
π 2 πΆ / π 1 πΆ π½ consistent with H.R. gas approach to Q.G. gas
π 2 πΆ / π 1 πΆ π½ π π΅ /π=0.12 ,0.24 ,1.08 , π½βΎ1.6 π π΅ /π = π½βΎ , π π΅ /π= π½βΎ1.4 consistent with H.R. gas π½βΌ2.1 approach to Q.G. gas
9
Comparison with previous work
374.7 767.0 222.5 157.1 143.3 T[MeV] Christof Gattringer and Hans-Peter Schadler Phys. Rev. D 91, This work consistent behavior application range of HR gas becomes small as π/π increases measure the H.R. gas / Q.G. gas transition
10
artificial phase transition
Function of π π΅ /π We truncate the fugacity expansion deviate from HR gas approach to QG gas π πΊ.πΆ. π π /π βΌ π=β π ππ’π‘ π ππ’π‘ π πππ. 3π π 3π π π π π πππ. 3π =0 , π > π ππ’π‘ deviate from QG gas not depend on π΅ πππ π 2 πΆ / π 1 πΆ π π΅ /π ,π½=0.9 HR gas(blue line) QG gas(red line) π΅ πππ creates artificial phase transition
11
π 2 πΆ / π 1 πΆ π π΅ /π at low temperature
HR gas(blue line) QG gas(red line) π·=π.π π·=π.π consistent small π π© /π» βΌ H.R. gas π π© /π» βΌπ for π·=π.π π π© /π» βΌπ.π for π·=π.π large π π© /π» βΌ Q.G. gas Application range of HR gas becomes small as π½ increases. circumstantial evidence of phase transition
12
π 2 πΆ / π 1 πΆ π π΅ /π at high temperature
π·=π.π HR gas(blue line) QG gas(red line) π·=π.π approach to QG gas as π· increases π·=π.π volume dependence π π Γπ ππ π Γπ approach to QG gas as volume increases interaction remains
13
Skewness (fig. above) Kurtosis (fig. bottom)
Skewness , Kurtosis π£π . π½ π 3 πΆ π 1 πΆ π 4 πΆ π 2 πΆ solid line : H.R. gas dotted line : Q.G. gas Skewness (fig. above) Kurtosis (fig. bottom) π π΅ /π=0.12 ,1.08 , π½βΎ1.6 π π© /π»=π,ππ ,π·=π.π deviate from H.R. gas negative value consistent with H.R. gas as same as π΅ π πͺ / π΅ πͺ
14
circumstantial evidence of phase transition
Kurtosis π£π . π π΅ /π ,π½=1.6 consistent with H.R. gas approach to Q.G. gas singularity circumstantial evidence of phase transition
15
more high-order cumulant π½=1.6
π 5 πΆ / π 1 πΆ π π΅ /π π 6 πΆ / π 2 πΆ π π΅ /π β ππ π π© /π»~π β ππ π π© /π»~π π 5 πΆ π πΆ = π 6 πΆ π 2 πΆ =81 H.R. gas (blue line) 2nd ,3rd ,4th : consistent with H.R. gas 5th ,6th : inconsistent with H.R. gas (suggestion)
16
Conclusion Canonical ensemble method Hadron Resonance gas model
Sign problem become mild. We can see as a function of π π΅ /π . π ππ’π‘ makes an artificial phase transition! Hadron Resonance gas model Less than 4th-order β HR gas is good model. It may be difficult to adapt HR gas for 5th and 6th. Quark Gluon gas model large π½ or π/π β approach to Q.G. gas Circumstantial evidence of phase transition We measured singular behavior for π½=1.6 . We saw HR gas / QG gas transition.
17
back up
18
Grand Canonical ensemble
Quark number cumulant π π πΆ β ππΊ π π π π π=0 πΊ π β log β«ππ π ππ π(π) Distribution π(π) characterize π 1 πΆ =π π 2 πΆ = π 2 π π πΆ =0 , πππ π>2 π π = 1 2π π π β πβπ π 2 2 π 2 ex. Gaussian π non-Gaussian Grand Canonical ensemble π π β π π π π π πππ. (π) π πΊ.πΆ. (π) system heat bath π π πΆ = π π π π/π π log π πΊ.πΆ. energy particle
19
around phase transition line finite density lattice QCD
Quark number cumulant π π πΆ = π π π π/π π log π πΊ.πΆ. π 2 πΆ = πΏ π 2 π 3 πΆ = πΏ π 3 π 4 πΆ = πΏ π 4 β3 πΏ π 2 2 2nd 4th 3rd π 3 πΆ π 2 πΆ around phase transition line π 4 πΆ π 2 πΆ π π π π πΆ expectation value π 3 πΆ π 1 πΆ sign problem finite density lattice QCD STAR Collaboration Phys. Rev. Lett. 112 (2014)
20
Volume dependence 8 3 Γ4 π£π . 12 3 Γ4 π 2 πΆ / π 1 πΆ π π΅ /π π½=0.9
8 3 Γ4 π£π Γ4 π 2 πΆ / π 1 πΆ π π΅ /π π½=0.9 π 2 πΆ / π 1 πΆ π π΅ /π π½=1.5
21
How to determine π ππ’π‘ ? dβAlembert π π π lim πββ π π+1 π π <1
π πΊ.πΆ. (π)= π π πππ. π π ππ/π π πππ. π+1 π πππ. π π π π <1
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.