kek2014 v2

Towards Understanding QCD Phase Diagram

Lattice and RHIC Experiments

Atsushi Nakamura in Collaboration with K.Nagata

Lattice QCD at finite temperature and density20 Jan. 2014 KEK

1 /38

14年1月20日月曜日

2

QCD Phase Diagram

From Wiki-Pedia “QCD matter”

Very Exciting !Now it’s time for

Lattice QCD.

But how do you reveal it?

Please No Sales-Talk !


Canonical Partition Functions

Experiments

Lattice QCD Simulations

3 / 38


A few years ago,

Nagata and I were looking for a Reduction formula for Wilson fermions in a finite density QCD project:

4

det �̃ = det�

Reduction Matrix Original Fermion Matrix

Z(µ) =�

DU det �(µ)e�SG

Rank(det �̃) < Rank (det �)Nagata and Nakamura Phys. Rev. D82 094027 (arXiv:1009.2149)


5

Obtained formula has the form of the fugacity expansion

� � eµ/T Fugacity

Z(µ) =�

DU�

n

cn�ne�SG

=�

n

Zn�n

Zn : Canonical Partition Function


If

Tr e�(H�µN̂)/T

=�

n

�n|e�(H�µN̂)/T |n�

=�

n

�n|e�H/T |n� eµn/T

=�

n

Zn(T )�n

Z(µ, T ) =

�� eµ/T

�

Fugacity

Z(µ, T )

6 /38

Zn(T )


RHIC(Relativistic Heavy Ion Collider)

7


Multiplicity !Wao,

It is almost

Multiplicity Distribution of RHIC

Interesting !Zn


We assumethe Fireballs created in High Energy Nuclear Collisons are described as

a Statistical System.with (chemical Potential)and T (Temperature)

µ

Grand Canonicalpartition functionZ(µ, T )

All QCD Phase Information is in Z(µ, T )


Partition Function is Sum of the Probabilities with n ...If I know , then I have Zn.�

Experiment

10 /38number


How can we extract from ?Pn = Zn�n

� unknown

Z+n = Z�n

We require (Particle-AntiParticle Symmetry)

Experiment

Zn

Observables in Experiments


12

� = 1.0 � = 1.2

� = 1.4 � = 1.5

� = 1.534Final Value/38

Zn Zn

ZnZn

Z�n

Z�n

Z�n

Z�n

nn

n n

From Experiment

07:12-08:05


1e-07

1e-06

1e-05

0.0001

0.001

0.01

0.1

1

-20 -15 -10 -5 0 5 10 15 20

sqrt(s)=62.4PnP-n

PP n-n

n 1e-10

1e-08

1e-06

0.0001

0.01

1

-20 -15 -10 -5 0 5 10 15 20

Net proton number

ZnZ-n

nZZ-n

sqrt(s)=62.4

�Fit

Experiment

Z+n = Z�nDemand


Fitted are very consistent with those by Freeze-out Analysis.

�

0

2

4

6

8

10

12

0 50 100 150 200

!

sNN1/2

Chemical Freeze-Out

x This work

J.Cleymans, H.Oeschler, K.Redlich and S.WheatonPhys. Rev. C73, 034905 (2006)

Freeze-out

�� eµ/T

�


�s = 19.6GeV

�s = 27GeV

�s = 39GeV

�s = 62.4GeV

�s = 200GeV

from RHIC dataZn

1e-18

1e-16

1e-14

1e-12

1e-10

1e-08

1e-06

0.0001

0.01

-25 -20 -15 -10 -5 0 5 10 15 20 25

'Zn_19.6'

1e-14

1e-12

1e-10

1e-08

1e-06

0.0001

0.01

1

-25 -20 -15 -10 -5 0 5 10 15 20 25

'Zn_27'

1e-14

1e-12

1e-10

1e-08

1e-06

0.0001

0.01

1

-25 -20 -15 -10 -5 0 5 10 15 20 25

'Zn_39'

1e-10

1e-09

1e-08

1e-07

1e-06

1e-05

0.0001

0.001

0.01

0.1

-20 -15 -10 -5 0 5 10 15 20

'Zn_62.4'

1e-07

1e-06

1e-05

0.0001

0.001

0.01

0.1

1

-15 -10 -5 0 5 10 15

'Zn_200'

Experiment

15 /38


Now we have Zn of RHIC data (sqrt(s)= 10.5,19.6, 27, 39, 62.4, 200 GeV)

µ/T

T

Wao ! We can calculate at any density !This includes all QCD Phase information !

(� � eµ/T )14年1月20日月曜日

Do not forget that your n is finite !

I need cooling down


Moments �k

�k ⌘✓T

@

@µ

◆k

logZ


What happens ?

if we increase these points 15%if we dropthese points

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1 1.1 1.2 1.3 1.4 1.5 1.6

freeze-out point

/T

42

=19.6 GeV

0.35

0.4

0.45

0.5

0.55

0.6

0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5/T

Number Susceptibility

freeze-out point

=27 GeV

Number Susceptibilityps = 27GeV

Usually we consider

(only) here.


0.35

0.4

0.45

0.5

0.55

0.6

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7µ/T

Number Susceptibility, sNN1/2=19.6

freeze-out point

0.35

0.4

0.45

0.5

0.55

0.6

0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5µ/T

Number Susceptibility, sNN1/2=27

freeze-out point

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4µ/T


freeze-out point

0.38 0.4

0.42 0.44 0.46 0.48

0.5 0.52 0.54 0.56 0.58

0.6

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1µ/T

Number Susceptibility, sNN1/2=62.4

freeze-out point

0.4

0.45

0.5

0.55

0.6

0.65

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1µ/T


freeze-out point

Susceptivility


-1

-0.5

0

0.5

1

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4µ/T

R42, sNN1/2=19.6

freeze-out point

0.6 0.7 0.8 0.9

1 1.1 1.2 1.3 1.4 1.5

0.86 0.88 0.9 0.92 0.94 0.96 0.98 1µ/T

R42, sNN1/2=27

freeze-out point

0.4 0.6 0.8

1 1.2 1.4 1.6 1.8

2 2.2

0.5 0.6 0.7 0.8 0.9 1µ/T

R42, sNN1/2=39

freeze-out point

0.5

0.6

0.7

0.8

0.9

1

1.1

0.35 0.4 0.45 0.5 0.55 0.6 0.65µ/T

R42, sNN1/2=62.4

freeze-out point

0.2

0.4

0.6

0.8

1

1.2

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45µ/T

R42, sNN1/2=200

freeze-out point

Kurtosis

Usually we consider (only) here.


22

Multiplicity tells us

Not only Freeze-out points

Information of wider regions

Nmax →largeWider


Lee-Yang Zeros

23


Lee-Yang ZerosZeros of in Complex Fugacity Plane.

Z(↵k) = 0

Great Idea to investigate a Statistical System

�xxxxxx

Phase Transition

Z(�)

24 /38

(1952)


25

Lee-Yang Zeros

Non-trivial to obtain.But once they are got, it is easy to figure out the Free-energy

Lee-Yang zeros

F: Free-energy

: zeros�k

2-d Electro-Magnetic

F: Potential

:Point charge�k

Z(�, T ) = e�F/T


26

-1-0.5

0 0.5

1 -1-0.5

0 0.5

1

-10

-5

0

5

10

15

20

’out’

F (�) = ��

k

log(� � �k)


cut Baum-Kuchen (cBK) Algorithm

1

1

　Number ofZeros in

Contour C

A Coutour is cut into four pieces

if there are zeros inside.50 - 100 number

of significant digits

i ( )

and

27/38


28

Is this myOriginal ?

I donot think so.

Let us waituntil someone

claims.


28

Is this myOriginal ?

I donot think so.

Let us waituntil someone

claims.

Riemann (1826 - 1866)

It’s me !


Lee-Yang Zeros: RHIC Experiments

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=19.6

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=27

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=39

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=62.4

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=200

Experiment

29/40

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

cos(t), cos(t),sin(t)’plotXY11.5’


30

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=200


Lee-Yang Zeros Lattice QCD

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[!]

Re[!]

T/Tc=0.99

� = 1.85

T/Tc � 0.99

Z(�) =�

m

Z3m � �3m

Zn = 0 n �= 3munless

� = ei�

� =2�

3Periodicity

( )

31 /38


-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[!]

Re[!]

T/Tc=0.99

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

'./plotXY_B1870'cos(t), sin(t)

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

'./plotXY_B1890'cos(t), sin(t)

� = 1.85

� = 1.87

� = 1.89

T/Tc � 0.99

T/Tc � 1.01

T/Tc � 1.04

32 /38


Lee-Yang Zeros Lattice QCD

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[!]

Re[!]

T/Tc=1.20⇠ = eµ/T

The Unit Cirle in

Imaginary

⇠µ

Roberge-WeiseTransition !

T/Tc � 1.20

33 /38

(� � eµ/T )


-1 -0.5 0 0.5 1

Im[ q

]

Re[ ]

-1

-0.5

0

0.5

1

q

q

10 43

8 43

T/Tc=1.08

-1 -0.5 0 0.5 1Re[ B]

B

10 43

�q �B = �q3

34 /38


A Message to Experimentalists

In the Lee-Yang Diagram constructed from your multiplicity,

Zeroshere �q�B

No Roberge-Weise Transition T � TRW� 1.2Tc

Your Temperature

35 /38


Lee-Yang Zeros: RHIC Experiments

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=19.6

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=27

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=39

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=62.4

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Im[! B

]

Re[!B]

sNN1/2=200

36/38


Effects of NmaxKim’s Model

37

Z(µq) = I0 + (⇠3q + ⇠�3q )I1 +(⇠6q + ⇠�6

q )I2 + · · ·Ik :Modified Bessel

In Confinement

Lesson from the Model

Nmax Large

Lee Yang ZerosLarge regions|µ|

It should be so!14年1月20日月曜日

SummaryGrand-Partition functions, , provide us the QCD phase information, which can be constructed from .

Lattice QCD can calculate But we need much more works to obtain reliable

Experiments provide us the multiplicitiesWe can calculate from them.

Present data are those of net-proton, which are not conserved quantities.

Either correction, or ask experimentalists to measure net-baryon

Charge multiplicity is a conserved quantity, and another probe.

Large Nmax are wanted, but even finite Nmax data give us the lower bound.

Lee-Yang zeros provide us a new tool of the QCD phase study.

They are sensitive to the data, i.e., they teach us which regions are important.

38

Zn

Zn

Zn

Z(µ, T )


Backup Slide

39


BES(Beem Energy Scan)


Find Rooms where No Criminal.

The Target is in other Room.

Not here ! Then, ..

Hunting the QCD Phase Transition Regions

Lower Bound14年1月20日月曜日

0.15

0.16

0.17Te

mpe

ratu

re (G

eV)

0 0.1 0.2 0.3 0.4 0.5Chemical Potential (GeV)!

200

62.439

27

19.6

Freeze-out PointLower bound determined by susceptivility

Phase Transition Regions estimated byLee-Yang Zero distribution

Lower bound determined by negative Kurtosis


Other Messages

Net proton multiplicity is Not a conserved quantity.

Baryon multiplicity is perfect

Can you estimate Baryon multiplicity from that of Proton ?

Another conserved quantity is the Charge multiplicity. It should work as well.

43


44 /40

You have a Big Chance to find QCD phase Transition !


Canonical Partition Functionsis a Bridge

between Two Approachesto Study QCD Phase

Lattice QCD Simulaitions

Experiments

45 /40


Lattice QCD Canonical Approach

Miller and RedlichPhys. Rev. D35 (1987) 2524

A.Hasenfratz and ToussaintNucl. Phys.B371 (1992) 539

Barbour and BellNucl. Phys. B372 (1992) 385

Engels, Kaczmarek, Karsch and LaermannNucl.Phys. B558 (1999) 307

deForcrand and KratochvilaNucl. Phys. B (P.S.) 153 (2006) 62 (hep-lat/0602024)

A.Li, Meng, Alexandru, K-F. LiuPoS LAT2008:032 and 178

Phys.Rev. D82(2010) 054502, D84 (2011) 071503

Danzer and GattringerarXiv:1204-1020 Europian Journal

Lattice

46 /40


Lattice: How to Calculate

Fugacity ExpansionNagata and A. Nakamura, Phys. Rev. D82 (2010) 094027

Lattice

47 /40

Alexandru and Wenger, Phys.Rev.D83 (2011) 034502


Four Excuses why not Baryon Multiplicities

1. This is a formulation. Let’s wait until Experimentalists measure Baryon multiplicities

2. After the Freeze-out, the proton number is essentially constant.

3. Expect the proton multiplicity is similar to the baryon multiplicity

4. By some event generators or models, let us calculate the proton and baryon multiplicity.From that data, we can estimate the baryon multiplicity.

48


ZGC(µ) =�

Zn�n

�DU(

�an�n)(det�(0))Nf e�SG

=�

DU(�

an�n)(det�(0))Nf e�SG

=�

n

�n

�DUan(det�(0))Nf e�SG

Lattice

49

=

/40


from lattice QCDZn

1e-160

1e-140

1e-120

1e-100

1e-80

1e-60

1e-40

1e-20

1

-60 -40 -20 0 20 40 60m=3n

·=Q��RULJ·

1e-60

1e-50

1e-40

1e-30

1e-20

1e-10

1

-40 -30 -20 -10 0 10 20 30 40

m=3n

'Zn1950-orig'

1e-90

1e-80

1e-70

1e-60

1e-50

1e-40

1e-30

1e-20

1e-10

1

-50 -40 -30 -20 -10 0 10 20 30 40 50

m=3n

'Zn1850-orig'

1e-40

1e-35

1e-30

1e-25

1e-20

1e-15

1e-10

1e-05

1

-30 -20 -10 0 10 20 30

m=3n

'Zn2000-orig'

1e-40

1e-35

1e-30

1e-25

1e-20

1e-15

1e-10

1e-05

1

-30 -20 -10 0 10 20 30

m=3n

'Zn1900-orig'

1e-20

1e-19

1e-18

1e-17

1e-16

15 16 17 18 19 20m=3n

'Zn1850-orig'

Im(Zn) are used as an error

Lattice

50 /40


A Strange Fact

There are Lee-Yang Zeros on the unit circle.

Theoretically, a bit annoying.

Phenomenologically, very natural

51 /40


52

Z(µ) =� �

f

det �(mf , µf )e�SG

is REAL det �(m,µ)

if is pure Imaginary.µ

On the unit circle in complex plane.�

(� = eµ/T )


53

det �(m,µ) is REAL and Positive,

if is pure Imaginary and m is sufficiently large.

µ

Z(µN ) =�

det �(Nucleon)e�SG > 0

Lee-Yang zeros on the unit circle tell usthat Nucleon is a composite.


54

Current lattice QCD simulations assumes

mu = md

Z(µN ) =�

(det�(mq, µq))2 � det �(ms, µs) · · · e�SG

Z(µN ) can not take zero.


55

µp = 2µu + µd

mu > md

µpPure imaginary does not mean

µu andµd are pure imaginary.


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