mass measurement at besiii

June, 14th, 2006 Mo Xiaohu 1

Mass Measurement at BESIII

X.H.MO

Workshop on Future PRC-U.S. Cooperation in High Energy Physics

Beijing, China, Jun 11-18


Content

1.Introduction2.Statistical optimization

of mass measurement 3.Systematic uncertainty

study4.Summary


Introduction


Pseudomass method• ARGUS• CLEO• OPAL• Belle• KEDRThreshold scan• BES

Points : 12 , Lum. : 5 pb1


!,

1 i

Ni

i

n

ii N

ePPLFii

F(x): E.A.Kuraev,V.S.Fadin , Sov.J.Nucl.Phys. 41(1985)466;(s): F.A. Berends et al. , Nucl. Phys. B57 (1973)381.

Ecm (GeV)

B r.c.

obs

BES:PRD53(1995)20


Ecm (GeV)

BES results:the stat. (0.18 0.21 )is compatible withthe syst. (0.25 0.17)M =1776.96 0.18

0.25 M / M = 1.7 10 – 4

0.21 0.17


Statistical optimization of

Mass Measurement


Neglecting all experiment uncertaintiesLuminosity L ; Efficiency =14% ; Branching fraction: Bf =0.1763 • 0.1784 ;[ Bf = B • B e , PDG04]Background BG =0 .

Using Voloshin’s formula for obs

[M.B.Voloshin, PLB556(2003)153.]

Statistical optimization


Statistical optimization for high accurate M measurementAssume : M is known .

To find : 1.What’s the optimal

distribution of data taking point;

2.How many points are needed in scan experiment;

3.How much luminosity is required for certain precision.


Evenly divided :1, for E: E0 + E, E=(Ef–E0)/n2, for lum. : L =Ltot /n= 3pb –1

To eliminate stat. fluctuation, Sampling many times (say, 500)


Ecm (3.545,3.595) GeVLtot = 30 pb –1 Npt : 3 20

1. Sm >> m , using Sm as criterion;

2. Npt =5.

| m

|


(Ecm)

d/dEcm

max. Sm=1.48MeVmin. Sm =0.147MeV

Random sampling 100 times:Ecm (3.545,3.595) GeVLtot =45 pb –1 Npt =5;

1. Points near threshold lead to small Sm ;

2. This corresponds to larger derivative of

The largest derivativepoint may be the optimaldata taking point


(Ecm)

d/dEcm

Scheme I:2 points at region I + Npt(1—20) at region II Scheme II:Only Npt(1—20) at region II

II

I

Scheme I

Scheme II

Only the pointswithin region I are useful for optimal data taking point

L=5 pb –1 for each point


With the region I, one point is enough!

I

Ecm (3.553,3.555) GeVLtot =45 pb –1 Npt = 1—6;

Where should this one point locate?


Ecm = 3553.81 MeV Sm = 0.09559 MeV Ecm = 3554.84 MeV max d/dEcm

Ecm (3.551,3.595) GeVLtot =45 pb –1 Npt = 1; scan

3553.8 MeV

3554.8 MeV


Ltot (pb–1) Sm (MeV)

9 0.285318 0.199027 0.155036 0.138045 0.119954 0.105163 0.097672 0.0913

100 0.07491000 0.0247

10000 0.0079

One pointWith lum.

Ltot


Systematic Uncertainty Study


Study of systematic uncertainty

1.Theoretical accuracy 2.Energy spread E 3.Energy scale 4.Luminosity 5.Efficiency 6.Background analysis


BES:PRD53(1995)20

Accuracy Effect of Theoretical Formula

Energy spread, variation form

s=(Ecm)2

Energy scale, variation form


Ecm = 3554 MeV

Ltot =45 pb –1

m = 1776.99 MeV

Uncertainty due to accuracy of cross section at level of 3 10 – 3 MeV

old [BES, PRD53(1995)20] fit results: m = 1777.028 MeV , m = 0.105 MeVnew [M.B.Voloshin, PLB556(2003)153] fit results: m = 1777.031 MeV , m = 0.094 MeVm = | m (new) – m (old) | < 3 10 – 3 MeV



f(E) ;f(E)=a E+b E2+c E3

a=1; b=0; c=0;a=0; b=1; c=0;a=0; b=0; c=1;a=1; b=1; c=1;m < 1.5 10 – 3 MeV

Ecm (GeV)

Cross section (n

b)

'

J/

/

/

/

/

J

J

J

J

EfEfEfEf

(1.51MeV)

J/

(1.06MeV)

3 m < 6 10 – 3 MeV

2

20

32

GJ

GC

Eq


f(E) ;f(E)=a E+b E2+c E3

a=1; b=0; c=0;a=0; b=1; c=0;a=0; b=0; c=1;a=1; b=1; c=1;

Ecm (GeV)

Cross section (n

b)

'

J/

E

EJ/

/

/

/

/

J

J

J

J

MMME

EEEW

W=E+ (E=M+ ); ～ 10– 4

m < 8 10 – 3 MeV


BES:PRD53(1995)20

Luminosity L : 2% m < 1.4 10 – 2 MeVEfficiency : 2% m < 1.4 10 – 2 MeV Branching fraction: Bf : 0.5% m < 3.5 10 – 3 MeV[ Bf = B • B e , PDG04]Background BG : 10% m < 1.7 10 – 3 MeV [ BG = 0.024 pb –1: PLR68(1992)3021 ]

Total : m < 2.02 10 – 2 MeV


Term m

(10 – 3 MeV) m / m

(10 – 6)Theoretical accuracy 3 1.7

Energy spread 6 3.4

Energy scale 8 4.5

Luminosity 14 7.9

Efficiency 14 7.9

Branching Fraction 3.5 2.0

Background 1.7 1.0

Total 22.7 12.7

Summary:systematic


KEDR Collab. , depolarization method:Single energy scale at level of 0.8 keV, or 10 –4 MeVTotal systematic error at level of 9 keV, or 10 – 3 MeV

Absolute calibration of energy scale

Fix, stable, regular, eliminate and controllable

UNSTABLE and IRREGULAR, uncontrollable

BESI: E=0.2MeV

Bottleneck


BKG.study

Eventselection

Optimalpoint

Data taking design

>100 pb –1 , 50 pb –1 , >100 pb –1


Statistical and systematic uncertainties have been studied based on BESI performance experience. Monte Carlo simulation and sampling technique are adopted to obtain optimal data taking point for high accurate mass measurement. We found: optimal position is located at large derivative of cross section near threshold ; one point is enough, and 45 pb–1 is sufficient for accuracy up to 0.1 MeV .

Many factors have been taken into account to estimate possible systematic uncertainties, the total relative error is at the level of 1.3 10 – 5. However the absolute calibration of energy scale may be a key issue for further improvement of accuracy of mass.

Summary

Thanks!


Backup


Evenly divided :1,for E: E0 + E, E=(Ef–E0)/n2, for lum. : L =Ltot /n= 3pb –1

To eliminate stat. fluctuation, Sampling many times (say, 500)

The point below threshold Have no effect for fit results

M=1777.0367 MeV Sm =0.4273 MeV


Optimization study shows that: optimal position is locate at large derivation of cross section near threshold ; one point is enough , and 45 pb–1 is sufficient for accuracy up to 0.1 MeV .

Summary:statistical

1. What’s the distribution of data taking point ;

2. How many points are needed in scan experiment ;

3. How much luminosity is required for certain precision.


smB ,Improved the previous calculation, accuracy close to 0.1%M.B.Voloshin, PLB556(2003)153.

NRQCD, NNLO, accuracy better that 0.1%P.Ruiz-Femenia and A.Pich, PRD64(2001)053001.

v

h(v)

Fc(v)10–3

S(v)/ 10–3

h(v)

new


Ecm = 3554 MeV

Ltot =45 pb –1

m = 1776.99 MeV

Uncertainty due to accuracy of cross section at level of 3 10 – 3 MeV

old fit results: m = 1777.028 MeV m = 0.105 MeVnew fit results: m = 1777.031 MeV m = 0.094 MeV

m = | m (new) – m (old) | < 3 10 – 3 MeV

± 10 – 4 m < 10 – 4 MeV ± 2 10 – 4

m < 10 – 4 MeV


mass measurement at besiii

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