relaxation time on the quenched lattice Atsuro Ikeda, Masayuki - - PowerPoint PPT Presentation

Charm quark diffusion coefficient and relaxation time on the quenched lattice

Atsuro Ikeda, Masayuki Asakawa, Masakiyo Kitazawa Osaka University XQCD2016

Anisotropic flow of open charm

• Large elliptic flow of open charm

→charm flow ~ medium flow

• Rapid thermalization of charm

quarks?

• Diffusion coefficient is an

important quantity

Transport coefficient on the lattice

Shear viscosity

Karsch and Wyld 1987, Nakamura and Sakai 2005, Meyer 07, Haas 2013, Borsanyi et al. 2014, etc…

Electric conductivity

Gupta 2004, Aarts et al. 2014, etc…

Quark diffusion coefficient

Ding et al. 2011, Banerjee et al. 2012, Aarts et al. 2015, Francis et al. 2015 etc… Ding et al. arXiv:1504.05274

There is a numerical difficulty, called ill-posed problem, and analyses still have uncertainty.

Measurement of Diffusion coefficient

1. Ansatz for spectral function

 Depend on ansatz  Lattice Euclidean correlator has a lattice artifact

• 2. Maximum entropy method

 Reconstructed spectral function has the strong correlation in whole 𝜕-space  Not sensitive to low energy structure

3. Structure of 𝐻00(𝜐, 𝑞2) (new)

𝜕2𝜍00 𝜕, 𝑞 = 𝑞𝑗𝑞𝑘𝜍𝑗𝑘 𝜕, 𝑞 = 𝑞2𝜍𝑀(𝜕, 𝑞)

𝐸 = 𝜌 3 1 χ lim

𝜕→0 lim Ԧ 𝑞→0

𝜍𝑗𝑗 𝜕, Ԧ 𝑞 𝜕

𝐻𝜈𝜈

𝐹

𝜐, 𝑞 = න 𝑒3𝑦 𝑓𝑗 Ԧ

𝑞 ⋅ Ԧ 𝑦 𝑘𝜈 𝜐, Ԧ

𝑦 𝑘𝜈 † 0, 0 = න

∞

𝑒𝜕 cosh (1/2𝑈 − 𝜐)𝜕 sinh 𝜕/2𝑈 𝜍𝜈𝜈 𝜕, Ԧ 𝑞

Kubo formula

High energy component of 𝜍00(𝜕, 𝑞) is suppressed by 1/𝜕2 comparing with 𝜍𝑗𝑗 𝜕, 𝑞

ill-posed problem Our strategy

for 𝜈 = 0,1,2,3

Linear response theory

Consider the two relaxation process [Kadanoff and Martin 1963]

Classical source ℎ(𝑠) 𝜀 𝑜 𝑠 = 𝜓ℎ(𝑠) Turn off suddenly Perturbative Hamiltonian 𝐼 𝑠, 𝑨 = 𝐼0 𝑠 + 𝜀𝐼 𝑠, 𝑢 𝜀𝐼 𝑠, 𝑢 = 𝑓𝜗𝑢𝜄 −𝑢 ℎ(𝑠) 𝜀 𝑜 𝑠 = −𝑗 න

−∞ 𝑢

𝑒𝑢′ 𝑜 𝑠, 𝑢 , 𝜀𝐼 𝑠′, 𝑢′

𝑓𝑟

compare 𝐸 = 𝜌 3 1 𝜓 lim

𝜕→0 lim 𝑙→0

𝜍𝑗𝑗 𝜕, 𝑙 𝜕

න

Kubo formula 𝜐relax

𝜖2 𝜖𝑢2 + 𝜖 𝜖𝑢 𝑘0 𝑦, 𝑢 = −𝐸𝛼2𝑘0 𝑦, 𝑢

𝜍00

hydro (𝜕, 𝑙)

𝜕 = 1 𝜌 𝜓𝐸|𝑙|2 𝜕2 + 𝐸|𝑙|2 − 𝜐𝜕2

2

Relaxation process

𝜕2𝜍00 𝜕, 𝑞 = 𝑞𝑗𝑞𝑘𝜍𝑗𝑘 𝜕, 𝑞

Low energy structure of the spectral function

Response lag caused by heavy quark mass

Assumptions

Structure of the spectral function 𝜍00(𝜕, Ԧ 𝑞)=𝜍00

ℎ𝑧𝑒𝑠𝑝 𝜕, Ԧ

𝑞 + 𝜍00

ℎ𝑗𝑕ℎ 𝜕, Ԧ

𝑞 Quark number susceptibility 𝜓 𝑞2 = 𝜓 + 𝜓′ 𝑞 𝑈

2

+ ⋯ 𝜓′ ≪ 𝜓

𝐸𝑞2 𝜕 𝜍00(𝜕) 2𝑛𝑟 hydro part high energy part 𝐻00 𝜐, 0 = 𝜓𝑈

𝜍00

hydro (𝜕, Ԧ

𝑞) 𝜕 = 1 𝜌 𝜓 𝑞2 𝐸| Ԧ 𝑞|2 𝜕2 + 𝐸| Ԧ 𝑞|2 − 𝜐𝜕2 2

𝜍𝜈𝜈

ℎ𝑗𝑕ℎ 𝜕, Ԧ

𝑞 ≥ 0 Scale separation

𝐻00(𝜐, p) around mid-point is the most sensitive to the low energy structure

• f the spectral function 𝜍00(𝜕, 𝑞)

But 𝑁0 0 = 𝑈𝜓 and 𝑁2 0 = 0

Study

𝜖𝑁0 𝑞2 𝜖 ෤ 𝑞2

𝑏𝑜𝑒

𝜖𝑁2 𝑞2 𝜖 ෤ 𝑞2

at ෤ 𝑞 → 0

Mid-point expansion of 𝐻00 𝜐, p

𝐻00 𝜐, Ԧ 𝑞 = න

∞

𝑒𝜕 1 + 1 2 1 2 − 𝑈𝜐

2

𝑈2𝜕2 𝜍00 𝜕, Ԧ 𝑞 sinh 𝜕 2𝑈 + 𝑃 1 2 − 𝑈𝜐

4

≡ 𝑁0 𝑞2 +

1 2 1 2 − 𝑈𝜐 2

𝑁2 𝑞2 + ⋯

𝑁𝑜 𝑞2 = 𝑁𝑜

𝑚𝑝𝑥 𝑞2 + 𝑁𝑜 ℎ𝑗𝑕ℎ 𝑞2

𝜐𝑈 𝐻00 𝜐𝑈, Ԧ 𝑞 0.5 𝑁0 𝑞2 M2 𝑞2 ෤ 𝑞 ≡ 𝑞/𝑈

อ 𝜖𝑁0

𝑚𝑝𝑥 𝑞2

𝜖 ෤ 𝑞2

෤ 𝑞2=0

= ℎ0 𝜐relax𝑈 𝜓𝐸𝑈2 + 𝜓′𝑈

𝜖𝑁0 𝑞2 /𝜖 ǁ 𝑞2

ℎ0 𝜐relax𝑈 ≡ lim

෤ 𝑞2→0

𝜖 𝜖 𝐸𝑞2 න

∞

𝑒𝜕 1 sinh 𝜕 2𝑈 𝜍00

ℎ𝑧𝑒𝑠𝑝 𝜕, 𝑞

< 0

𝐸𝑀𝑈 ≡ อ 1 ℎ0 𝜐relax𝑈 𝑈2 𝜓 𝜖 𝜖 ෤ 𝑞2 𝑁0 𝑞2 𝑈3 − 𝜓′ 𝑈2

𝑞2=0

< 𝐸𝑈 with 𝑁0 𝑞2 = 𝑁0

𝑚𝑝𝑥 𝑞2 + 𝑁0 ℎ𝑗𝑕ℎ 𝑞2 and 𝜖 𝜖 ෤ 𝑞2 𝑁0 ℎ𝑗𝑕ℎ 𝑞2 > 0

− 𝑚𝑝𝑕2 𝜌

อ 𝜖𝑁2

𝑚𝑝𝑥 𝑞2

𝜖 ෤ 𝑞2

෤ 𝑞2=0

= ℎ2 𝑈𝜐relax 𝜓𝐸

𝜖𝑁2 𝑞2 /𝜖 ǁ 𝑞2

Opposite sign of h0 < 0 and ℎ2 > 0

ℎ2 𝑈𝜐relax ≡ lim

෤ 𝑞2→0 න

∞

𝑒𝜕 𝑈2 sinh 𝜕 2𝑈 𝜖𝜕2𝜍00 𝜕, 𝑞 𝜖 𝐸𝑞2 > 0

𝐸𝑈 < 𝐸𝑉𝑈 ≡ อ 1 ℎ2(𝑈𝜐𝑠𝑓𝑚𝑏𝑦) 𝜖 𝜖 ෤ 𝑞2 𝑁2 𝑞2 𝜓𝑈

𝑞2=0

𝐸𝑀𝑈 < 𝐸𝑈 < 𝐸𝑉𝑈

with 𝑁2 𝑞2 = 𝑁2

𝑚𝑝𝑥 𝑞2 + 𝑁2 ℎ𝑗𝑕ℎ 𝑞2 , 𝜖 𝜖 ෤ 𝑞2 𝑁2 ℎ𝑗𝑕ℎ 𝑞2 > 0

𝜌 න

Lattice set up

Quenched lattice Wilson Fermion and standard Wilson gauge action 𝛾 = 7.0, 𝛿𝐺 = 3.476

[Asakawa, Hatsuda 2004]

Anisotropic lattice with 𝜊 =

𝑏𝜏 𝑏𝜐 = 4 and 𝑂𝜏 = 128

for high momentum resolution 𝑀𝜏/𝑀𝜐 = 11.5~32

Blue Gene/Q@KEK Iroiro++

Nτ T/Tc 𝑂𝜏 Δp/T Nconf 16 4.68 128 0.196 361 20 3.74 128 0.245 229 24 3.12 128 0.294 240 28 2.67 128 0.344 91 32 2.34 128 0.397 100 32 2.34 64 0.794 304 36 2.08 128 0.442 100 40 1.87 128 0.491 100 44 1.7 128 0.54 89

𝐻00

𝐹 (𝜐, 𝑞)

• 1. Mid-point correlator
• 2. Curvature

𝑂𝜐 = 24 at 𝑞 → 0 Momentum dependence of

𝜖𝑁0(𝑞2)/𝜖 ǁ 𝑞2 and 𝜖𝑁2(𝑞2)/𝜖 ǁ 𝑞2

intercept  Fit with linear function where ෤ 𝑞2 < 1  From 𝑂𝜐 = 32, finite volume dependence is well suppressed 𝜖𝑁0(𝑞2)/𝜖 ෤ 𝑞2 𝜖𝑁2(𝑞2)/𝜖 ෤ 𝑞2

Result: 𝐸𝑀(𝜐𝑠𝑓𝑚𝑏𝑦)𝑈 < 𝐸𝑈 < 𝐸𝑉(𝜐𝑠𝑓𝑚𝑏𝑦)𝑈

Ding et al. arXiv:1504.05274

 Consistent with previous works  High energy contribution become larger for lower T  Information on 𝜐𝑠𝑓𝑚𝑏𝑦 is needed to determine D

Constraint on 𝐸 and 𝜐𝑠𝑓𝑚𝑏𝑦

𝑂𝜐 = 20, 𝑈/𝑈

𝑑 = 3.74

𝐸𝑀𝑈 at 𝜐𝑠𝑓𝑚𝑏𝑦 = 0 is still lower limit.

𝐸𝑉(𝜐𝑠𝑓𝑚𝑏𝑦)𝑈 𝐸𝑀(𝜐𝑠𝑓𝑚𝑏𝑦)𝑈

𝜐𝑠𝑓𝑚𝑏𝑦 from Langevin dynamics

𝑂𝜐 = 20, 𝑈/𝑈

𝑑 = 3.74

[Petreczky, Teany 2006 Caron-Huot et al. 2009]

𝜐/𝐸𝑉(𝜐𝑠𝑓𝑚𝑏𝑦) 𝜐/𝐸𝑀(𝜐𝑠𝑓𝑚𝑏𝑦)

kinetic 𝑛𝑑 on the lattice?

We need 𝜐𝑠𝑓𝑚𝑏𝑦 or 𝜐𝑠𝑓𝑚𝑏𝑦/𝐸 on the lattice

𝜐𝑠𝑓𝑚𝑏𝑦 𝐸 𝑈 = 𝑛𝑑

𝜐𝑠𝑓𝑚𝑏𝑦 𝐸 𝑈 = 𝑛𝑑

from Langevin dynamics or heavy quark limit

Conclusion

Constraint on 𝐸 and 𝜐𝑠𝑓𝑚𝑏𝑦 in (𝐸, 𝜐𝑠𝑓𝑚𝑏𝑦)-plane from the p-dependence

• f the mid-point correlator 𝐻00

𝐹 1 2𝑈 , 𝑞 on the lattice with basic

assumptions for the spectral function 𝜍00(𝜕, Ԧ 𝑞) . We obtain 𝜖𝑁0(𝑞2)/𝜖 ෤ 𝑞2 and 𝜖𝑁2(𝑞2)/𝜖 ෤ 𝑞2 with good statistics. Spatial volume dependence was well suppressed even with

𝑀𝜏 𝑀𝜐 = 8.

Future work

Can we measure 𝜓′ = ฬ

𝜖𝜓 𝑞2 𝜖 𝑞2 𝑞2=0

• n the lattice?

Other information on D and 𝜐relax? Estimate of high energy contribution of 𝜍𝜈𝜈 𝜕, Ԧ 𝑞 (MEM, ansatz for spectral function): 𝐸𝑀(𝜐𝑠𝑓𝑚𝑏𝑦)𝑈 < 𝐸𝑈 < 𝐸𝑉(𝜐𝑠𝑓𝑚𝑏𝑦)𝑈 ֜ equality

ℎ0 𝜐𝑈 and ℎ2(𝜐𝑈)

ℎ0 𝜐relax𝑈 = −

𝑚𝑝𝑕2 𝜌 + 𝑈𝜐relax 1 − 𝐺 1 𝑈𝜐relax

< 0 ℎ2 𝑈𝜐relax =

1 𝑈𝜐relax 𝐺 1 𝑈𝜐relax >0

𝐺 𝑏 ≡ 𝑏 𝜌 න

∞

𝑦 𝑦2 + 1 1 sinh 𝑏 2 𝑦 = −1 + alog2 𝜌 − a 𝜌 Ψ 𝑏 4𝜌 − Ψ 1 2𝜌 Ψ 𝑨 ≡ 𝑒 𝑒𝑨 𝑚𝑝𝑕Γ(𝑨)

− 𝑚𝑝𝑕2 𝜌

𝜌

Low energy structure of 𝜍00 𝜕, Ԧ 𝑞

𝐸 = 𝜌 3 1 𝜓 lim

𝜕→0 lim Ԧ 𝑞→0

𝜍𝑗𝑗 𝜕, Ԧ 𝑞 𝜕

න

Kubo formula 𝜐relax

𝜖2 𝜖𝑢2 + 𝜖 𝜖𝑢 𝑘0 𝑦, 𝑢 = 𝐸𝛼2𝑘0 𝑦, 𝑢

𝜍00

hydro (𝜕, Ԧ

𝑞) 𝜕 = 1 𝜌 𝜓( Ԧ 𝑞)𝐸| Ԧ 𝑞|2 𝜕2 + 𝐸| Ԧ 𝑞|2 − 𝜐𝜕2 2

Consider the diffusion eq. as the relaxation process

𝜕2𝜍00 𝜕, 𝑞 = 𝑞𝑗𝑞𝑘𝜍𝑗𝑘 𝜕, 𝑞

Low energy structure of the spectral function

[Kadanoff and Martin 1963] Response lag caused by heavy quark mass

𝜖𝑁0

𝑚𝑝𝑥 𝑞2

𝜖 ෤ 𝑞2 = න

∞

𝑒𝜕 1 sinh 𝜕 2𝑈 𝜖 𝜖 𝐸𝑞2 𝜍00

ℎ𝑧𝑒𝑠𝑝 𝜕, 𝑞 𝜓𝐸𝑈2 + 𝜓′𝑈

𝜖𝑁0 𝑞2 /𝜖 ǁ 𝑞2

𝑞2→0 ℎ0 𝜐relax𝑈 = − 𝑚𝑝𝑕2

𝜌 + 𝑈𝜐relax 1 − 𝐺 1 𝑈𝜐relax < 0

𝐸𝑀𝑈 ≡ อ 1 ℎ0 𝜐relax𝑈 𝑈2 𝜓 𝜖 𝜖 ෤ 𝑞2 𝑁0 𝑞2 𝑈3 − 𝜓′ 𝑈2

𝑞2=0

< 𝐸𝑈

with 𝑁0 𝑞2 = 𝑁0

𝑚𝑝𝑥 𝑞2 + 𝑁0 ℎ𝑗𝑕ℎ 𝑞2 and 𝜖 𝜖 ෤ 𝑞2 𝑁0 ℎ𝑗𝑕ℎ 𝑞2 > 0

𝐺 𝑏 ≡ 𝑏 𝜌 න

∞

𝑦 𝑦2 + 1 1 sinh 𝑏 2 𝑦 = −1 + alog2 𝜌 − a 𝜌 Ψ 𝑏 4𝜌 − Ψ 1 2𝜌 Ψ 𝑨 ≡ 𝑒 𝑒𝑨 𝑚𝑝𝑕Γ(𝑨)

− 𝑚𝑝𝑕2 𝜌

𝜖 𝜖 ෤ 𝑞2 𝑁2

𝑚𝑝𝑥 𝑞2 = න ∞

𝑒𝜕 1 sinh 𝜕 2𝑈 𝜖 𝜖 𝐸𝑞2 𝜕2𝑈2𝜍00 𝜕, 𝑞 𝜓𝐸 + 𝜓′ 𝜓 𝑁2

𝑚𝑝𝑥(𝑞2)

𝑁2 𝑞2 : Momentum dependence

Opposite sign of h0 < 0 and ℎ2 > 0

𝑞2→0 ℎ2 𝑈𝜐relax = 1 𝑈𝜐relax 𝐺 1 𝑈𝜐relax >0

𝐸𝑈 < 𝐸𝑉𝑈 ≡ อ 1 ℎ2(𝑈𝜐𝑠𝑓𝑚𝑏𝑦) 𝜖 𝜖 ෤ 𝑞2 𝑁2 𝑞2 𝜓𝑈

𝑞2=0

𝐸𝑀𝑈 < 𝐸𝑈 < 𝐸𝑉𝑈

with 𝑁2 𝑞2 = 𝑁2

𝑚𝑝𝑥 𝑞2 + 𝑁2 ℎ𝑗𝑕ℎ 𝑞2 , 𝜖 𝜖 ෤ 𝑞2 𝑁2 ℎ𝑗𝑕ℎ 𝑞2 > 0

𝜌