f giordano collins fragmentation for kaon

September 17th, 2014, TorinoFrancesca Giordano

COLLINS MEASUREMENTS @ BELLE

Francesca Giordano 2

Fragmentation processor how do the hadrons get formed?



Dqh

h

q

Fragmentation function describes the process of hadronization of a parton

D h q (z) is the probability that an hadron h with energy z is generated from a parton q

z Eh

Eq=

Eh

Eb=

2Eh

Q



Strictly related to quark confinement

Dqh

h

q

Fragmentation function describes the process of hadronization of a parton


z Eh

Eq=

Eh

Eb=

2Eh

Q

Francesca Giordano

Dqh

h

q

3

e+

e- q

Fragmentation function describes the process of hadronization of a partonStrictly related to quark confinementCleanest way to access FF is in e+e- qq



z Eh

Eq=

Eh

Eb=

2Eh

Q

Francesca Giordano

Dqh

h

q

3

e+

e- q

Fragmentation function describes the process of hadronization of a partonStrictly related to quark confinementCleanest way to access FF is in e+e- qq


e+e!hX /X

q

e+e!qq(Dhq +Dh

q )


z Eh

Eq=

Eh

Eb=

2Eh

Q

Francesca Giordano

AhUT = " "

" +"

⇒⇒

⇒⇒

/ Dhq

Dqh

h

q

4

e+

e- q

Fragmentation function describes the process of hadronization of a partonStrictly related to quark confinementCleanest way to access FF is in e+e- qqUniversal: can be used to study the nucleon structure when combined with SIDIS and hadronic reactions data


/ H?h1q


z Eh

Eq=

Eh

Eb=

2Eh

Q

Francesca Giordano

Collins Fragmentation

5

Collins mechanism: correlation between the parton transverse spin and the direction of final hadron

left-right asymmetry

H1⊥q

H1⊥ PTq

quark

PT

Francesca Giordano


5



H1⊥q

H1⊥ PTq

quark

PT

strictly related to the outgoing hadron tranverse momentum

TMD!

Francesca Giordano


5



H1⊥q

H1⊥ PTq

quark

PT

strictly related to the outgoing hadron tranverse momentum

TMD!

chiral even

Chiral odd!

chiral oddchiral oddX H1

⊥

Francesca Giordano

KEKB

BELLE @ KEKB


Francesca Giordano

BELLE @ KEKB

Good tracking ϴ [170;1500] and vertex resolution

Good PID: ự(π) ≳ 90%ự(K) ≳ 85%

7

Belle spectrometer:4𝛑 spectrometer optimized for CP violation in B-meson decay

On resonance: √s = 10.58 GeV (e+ e- → Y(4S) → BB)

Off resonance √s = 10.52 GeV (e+ e- → qq (q=u,d,s,c))

Total Luminosity collected:1000 fb-1!!!

KEKB:Asymmetric e+ e- collider

(3.5 / 8 GeV)

Francesca Giordano


8

e+e!hX /X

q

e+e!qq(H?1,q +H?

1,q)h

e-

e+ q

q

e+e ! q q ! hX

Francesca Giordano


8

e+e!hX /X

q

e+e!qq(H?1,q +H?

1,q)h

e-

e+ q

q

e+e ! q q ! hX

chiral even

Chiral odd!


⊥


In e+e- reaction, there is no fixed transverse axis to define azimuthal angles to, and even if there were one the net quark polarization would be 0

Collins Fragmentationh1

h2Back-to-Back jets

q

q

e-

e+


In e+e- reaction, there is no fixed transverse axis to define azimuthal angles to, and even if there were one the net quark polarization would be 0

But if we look at the whole event, even though the q and q spin directions are unknown, they must be parallel

-

e+e ! q q ! h1 h2 Xh = , K

Collins Fragmentationh1

h2Back-to-Back jets

q

q

e-

e+

Francesca Giordano

Reference frames

10

𝜙0 method:hadron 1 azimuthal angle with respect

to hadron 2

reference plane (in blue) given by the e+e- direction and one of the hadron

e+e ! q q ! h1 h2 X h = , K

reference plane (in blue) given by the e+e- direction and the qq axis-

Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥

Figure 3: Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, wherethe angles are formed between the scattering plane and their transverse momenta Phi⊥

around the thrust axis n.

the Collins effect can only be visible in the combination of 2 functions being able to createa single spin asymmetry each. Accordingly the combination of a quark and an antiquarkCollins function in opposing hemispheres gives a product of two sin(φ) modulations forthe two azimuthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). Ine+e− these azimuthal angles are defined as

φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)

where z is just a unit vector in the z-axis defined by the e+e− axis and n is the thrust axis(defined in section 2.2 below), used as a surrogate for the quark-antiquark axis.

Transverse polarization: Additionally one still needs an average transverse polarizationof both quarks. Since the e+e− process does not exhibit a well defined polarization axisonly an average transverse polarization can yield this property. In fact the virtual photonemitted has to be spin 1 which in the helicity basis of the incoming leptons can be createdby the combinations +− and −+. In the case of creating a quark-antiquark pair underthe CMS angle of θ = π/2 (see Fig. 3) both lepton helicity combinations would be equallycontributing and transverse polarization of the quarks has to average out. Hence the quark-antiquark pair will have antiparrallel spins on average. Under more general scatteringangles the possibility of antiparallel spins will be proportional to sin2 θ.

2.2. Azimuthal asymmetries

Two different azimuthal asymmetries will become important in the course of the analysis.Therefore the calculation of them will be first described before having a closer look at theslightly different cross sections. The method already mentioned in the previous subsectionjust translates the definition of the Collins function (eq. 3) into the e+e− → qq case. This

7

Ph1

φ0

Ph2 θ2

e−

e+

Ph1⊥

Figure 4: Definition of the azimuthal angle φ0 formed between the plane defined by thelepton momenta and that of one hadron and the second hadron’s transverse momentumPh1⊥ relative to the first hadron.

The dependence on the transverse momentum and on the fractional energy was omittedin the previous formulas for the sake of clarity. The kinematic prefactors are defined as:

A(y) = (12 − y − y2)

CMS=

1

4(1 + cos2 θ) (11)

B(y) = y(1 − y)CMS=

1

4(sin2 θ) , (12)

where y = (1 + cos θ)/2 is a measure of the forwardness of the hard scattering process.Clearly the measurement of the Collins function itself lies hidden in the convolution inte-gral and could at this stage only be obtained under assumptions on the behavior of theintrinsic transverse momentum pT .The second method stays differential in the both azimuthal angles and thus reads[8]:

dσ(e+e− → h1h2X)

dΩdz1dz2dφ1dφ2=

!

q,q

3α2

Q2z21z

22

"

e2q/4(1 + cos2 θ)Dq,[0]

1 (z1)Dq,[0]1 (z2)

+e2q/4 sin2 θ cos(φ1 + φ2)H

⊥,[1],q1 H⊥,[1],q

1

#

, (13)

where the fragmentation functions appear as the zeroth[0] or first[1] moments in theabsolute value of their corresponding transverse momenta:

F [n](z) =

$

d|kT |%

|kT |M

&n

F (z,k2T ) . (14)

Here one is able to access the first kT moment of the Collins function, which can be differ-ent from the preivious convolution seen in equation 9. Nevertheless these cross sections

9

𝜙1+𝜙2 method:hadron azimuthal angles with respect

to the qq axis proxy-

Thrust axis= proxy for the qq axis-

Francesca Giordano

Reference frames

11

e+e ! q q ! h1 h2 X h = , K

Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥




φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)





7



Thrust axis= proxy for the qq axis-

h1

h2 Back-to-Back jets

q

q

e-

e+


Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥




φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)





7

Ph1

φ0

Ph2 θ2

e−

e+

Ph1⊥



A(y) = (12 − y − y2)

CMS=

1

4(1 + cos2 θ) (11)

B(y) = y(1 − y)CMS=

1

4(sin2 θ) , (12)



dΩdz1dz2dφ1dφ2=

!

q,q

3α2

Q2z21z

22

"

e2q/4(1 + cos2 θ)Dq,[0]

1 (z1)Dq,[0]1 (z2)


⊥,[1],q1 H⊥,[1],q

1

#

, (13)


F [n](z) =

$

d|kT |%

|kT |M

&n

F (z,k2T ) . (14)


9

𝜙0 method:hadron 1 azimuthal angle with respect

to hadron 2



Reference frames

F [X] =X

qq

Z[2h · kT1h · kT2 kT1 · kT2]

d2kT1d2kT2

2(kT1 + kT2 qT)X

kTi = zi pTi

M0

1 +

sin

2 21 + cos

2 2cos(20)F

hH?1 (z1) ¯H

?1 (z2)

D?1 (z1)

¯D?1 (z2)

i M12

1 +

sin

2 T1 + cos

2 Tcos(1 + 2)

H?[1]1 (z1) ¯H

?[1]1 (z2)

D[0]1 (z1) ¯D

[0]1 (z2)

F [n](zi) Z

d|kT |2h |kT |Mi

i[n]F (zi, |kT |2)

D. BoerNucl.Phys.B806:23,2009

Francesca Giordano

Product of 2 Collins FFs

13

Like-sign couples

Francesca Giordano


13

Like-sign couples

Fav Ł UnFav Fav Ł UnFav

Francesca Giordano


14

Like-sign couples

Fav Ł UnFav Fav Ł UnFavUnlike-sign couples

All charges couples

Fav Ł Fav UnFav Ł UnFav

Fav + unFav Ł UnFav + Fav


Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥




φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)





7

Ph1

φ0

Ph2 θ2

e−

e+

Ph1⊥



A(y) = (12 − y − y2)

CMS=

1

4(1 + cos2 θ) (11)

B(y) = y(1 − y)CMS=

1

4(sin2 θ) , (12)



dΩdz1dz2dφ1dφ2=

!

q,q

3α2

Q2z21z

22

"

e2q/4(1 + cos2 θ)Dq,[0]

1 (z1)Dq,[0]1 (z2)


⊥,[1],q1 H⊥,[1],q

1

#

, (13)


F [n](z) =

$

d|kT |%

|kT |M

&n

F (z,k2T ) . (14)


9

e+e ! q q ! h1 h2 X h = , K𝜙0 method:

hadron 1 azimuthal angle with respect to hadron 2



Reference frames

R0 =N0(0)

hN0iR12 =

N12(1 + 2)

hN12i


But! Acceptance and radiation effects also contribute to azimuthal

asymmetries!

Double-ratios



asymmetries!

Double-ratios

Simulation: unlike-sign, like-sign



asymmetries!

Double-ratios

Simulation: unlike-sign, like-signData: unlike-sign, like-sign



asymmetries!

Double-ratios

Simulation: unlike-sign, like-signData: unlike-sign, like-sign

To reduce such non-Collins effects:divide the sample of hadron couples in unlike-sign and like-sign (or All-charges),

and extract the asymmetries of the super ratios between these 2 samples:

Unlike-sign couples / Like-sign couples Unlike-sign couples / All charges

Dh1h2ul = RU/RL Dh1h2

uc = RU/RC


Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥




φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)





7

Ph1

φ0

Ph2 θ2

e−

e+

Ph1⊥



A(y) = (12 − y − y2)

CMS=

1

4(1 + cos2 θ) (11)

B(y) = y(1 − y)CMS=

1

4(sin2 θ) , (12)



dΩdz1dz2dφ1dφ2=

!

q,q

3α2

Q2z21z

22

"

e2q/4(1 + cos2 θ)Dq,[0]

1 (z1)Dq,[0]1 (z2)


⊥,[1],q1 H⊥,[1],q

1

#

, (13)


F [n](z) =

$

d|kT |%

|kT |M

&n

F (z,k2T ) . (14)


9

𝜙0 method𝜙1+𝜙2 method

Double-ratios

M0

1 +

sin

2 21 + cos

2 2cos(20)F

hH?1 (z1) ¯H

?1 (z2)

D?1 (z1)

¯D?1 (z2)

i

Fitted by

B12(1 +A12 cos(1 + 2)) B0(1 +A0 cos(20))

M12

1 +

sin

2 T1 + cos

2 Tcos(1 + 2)

H?[1]1 (z1) ¯H

?[1]1 (z2)

D[0]1 (z1) ¯D

[0]1 (z2)

Dh1h2uc = RU/RCDh1h2

ul = RU/RL


Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥




φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)





7

Ph1

φ0

Ph2 θ2

e−

e+

Ph1⊥



A(y) = (12 − y − y2)

CMS=

1

4(1 + cos2 θ) (11)

B(y) = y(1 − y)CMS=

1

4(sin2 θ) , (12)



dΩdz1dz2dφ1dφ2=

!

q,q

3α2

Q2z21z

22

"

e2q/4(1 + cos2 θ)Dq,[0]

1 (z1)Dq,[0]1 (z2)


⊥,[1],q1 H⊥,[1],q

1

#

, (13)


F [n](z) =

$

d|kT |%

|kT |M

&n

F (z,k2T ) . (14)


9


Double-ratios

M0

1 +

sin

2 21 + cos

2 2cos(20)F

hH?1 (z1) ¯H

?1 (z2)

D?1 (z1)

¯D?1 (z2)

i

Fitted by

B12(1 +A12 cos(1 + 2)) B0(1 +A0 cos(20))

M12

1 +

sin

2 T1 + cos

2 Tcos(1 + 2)

H?[1]1 (z1) ¯H

?[1]1 (z2)

D[0]1 (z1) ¯D

[0]1 (z2)

A0 =

sin

2

1 + cos

2 FhH?

1 (z1) ¯H?1 (z2)

D?1 (z1)

¯D?1 (z2)

iA12 =

sin

2

1 + cos

2

H?[1]1 (z1) ¯H

?[1]1 (z2)

D[0]1 (z1) ¯D

[0]1 (z2)

Francesca Giordano

Published results:

18

PRD 78, 032011 (2008)PRD 78, 032011 (2008)


𝜋𝜋Belle publications

PRL 96,232002, (2006)PRD 78, 032011 (2008)

http://link.aps.org/abstract/PRD/v78/e032011




http://link.aps.org/abstract/PRL/v96/e232002

http://link.aps.org/abstract/PRL/v96/e232002

Francesca Giordano

Belle vs. Babar

19

Francesca Giordano

SIDIS

20

DF

σFF

chiral even

Chiral odd!


⊥

Francesca Giordano

SIDIS

20

DF

σFF

H?1h1 ⊥

H?1h1

chiral even

Chiral odd!


⊥

Francesca Giordano

[Airapetian et al., Phys. Lett. B 693 (2010) 11-16]

Collins amplitudes in SIDISH?

1h1 AUT /

π+/−

Κ+/−

Deuterium

21

𝜋 Collins FF

Francesca Giordano



1h1 AUT /

π+/−

Κ+/−

Deuterium

21

𝜋 Collins FF

Anselmino et al.Phys.Rev. D75 (2007)

transversity

Francesca Giordano

DY 2011, BNL - May 11th, 2011Gunar Schnell

quark pol.

U L T

nucl

eon

pol.

U f1 h1

L g1L h1L

T f1T g1T h1, h1T

Twist-2 TMDs

significant in size and

opposite in sign for charged

pions

disfavored Collins FF large

and opposite in sign to

favored one

leads to various cancellations

in SSA observables

21

Transversity distribution

(Collins fragmentation)

+

u

[A. Airapetian et al, Phys. Rev. Lett. 94 (2005) 012002]

2005: First evidence from HERMES

SIDIS on proton

Non-zero transversity

Non-zero Collins function

-0.1

0

0.1

0.2

-0.2

-0.1

0

0.1 0.2 0.3

2 !

sin

(" +

"S)#

UT

$

$+

x

$-

z

0.3 0.4 0.5 0.6 0.7

Wednesday, May 11, 2011

H?,unfav1 H?,fav

1



1h1 AUT /

π+/−

Κ+/−

Deuterium

21

𝜋 Collins FF

Anselmino et al.Phys.Rev. D75 (2007)

transversity

Francesca Giordano

Collins amplitudes in SIDIS

22

H?1h1 AUT /

AUU / h?1 H?

1

Francesca Giordano

Collins amplitudes in SIDIS

22

H?1h1 AUT /

K+ amplitudes larger than 𝜋+?

AUU / h?1 H?

1

Francesca Giordano

More recently...

23

z qT sin2𝜭/(1+cos2𝜭) pT

New!

New!

New!

New!New!New!

New! New! New!


New!

New!

New!

New!New!New!

New! New! New!

Duc = RU/RC

Dkuc = RUk/RCk

Dkkuc = RUkk/RCkk

Dkkul = RUkk/RLkk

Dkul = RUk/RLk

Dul = RU/RL

Francesca Giordano

More recently...

23


New!

New!

New!

New!New!New!

New! New! New!


New!

New!

New!

New!New!New!

New! New! New!

Word of caution: this analysis is mainly aimed at kaons, so kinematic cuts and binning are optimized for kaons, and the same values used for pion too.

𝜋𝜋 results cannot be compared directly to published results

Duc = RU/RC

Dkuc = RUk/RCk

Dkkuc = RUkk/RCkk

Dkkul = RUkk/RLkk

Dkul = RUk/RLk

Dul = RU/RL


Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥




φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)





7

Ph1

φ0

Ph2 θ2

e−

e+

Ph1⊥



A(y) = (12 − y − y2)

CMS=

1

4(1 + cos2 θ) (11)

B(y) = y(1 − y)CMS=

1

4(sin2 θ) , (12)



dΩdz1dz2dφ1dφ2=

!

q,q

3α2

Q2z21z

22

"

e2q/4(1 + cos2 θ)Dq,[0]

1 (z1)Dq,[0]1 (z2)


⊥,[1],q1 H⊥,[1],q

1

#

, (13)


F [n](z) =

$

d|kT |%

|kT |M

&n

F (z,k2T ) . (14)


9


M0

1 +

sin

2 21 + cos

2 2cos(20)F

hH?1 (z1) ¯H

?1 (z2)

D?1 (z1)

¯D?1 (z2)

i M12

1 +

sin

2 T1 + cos

2 Tcos(1 + 2)

H?[1]1 (z1) ¯H

?[1]1 (z2)

D[0]1 (z1) ¯D

[0]1 (z2)

More recently...


Ph1

φ1

Ph2

φ2 − πθ

Thrust axis

e−

e+

Ph1⊥

Ph2⊥




φ1,2 =n

|n|·!

z × n

|z||n|×

n × Ph1,2

|n||Ph1,2|

"

acos

!

z × n

|z||n|·n × Ph1,2

|n||Ph1,2|

"

, (4)





7

Ph1

φ0

Ph2 θ2

e−

e+

Ph1⊥



A(y) = (12 − y − y2)

CMS=

1

4(1 + cos2 θ) (11)

B(y) = y(1 − y)CMS=

1

4(sin2 θ) , (12)



dΩdz1dz2dφ1dφ2=

!

q,q

3α2

Q2z21z

22

"

e2q/4(1 + cos2 θ)Dq,[0]

1 (z1)Dq,[0]1 (z2)


⊥,[1],q1 H⊥,[1],q

1

#

, (13)


F [n](z) =

$

d|kT |%

|kT |M

&n

F (z,k2T ) . (14)


9


M0

1 +

sin

2 21 + cos

2 2cos(20)F

hH?1 (z1) ¯H

?1 (z2)

D?1 (z1)

¯D?1 (z2)

i M12

1 +

sin

2 T1 + cos

2 Tcos(1 + 2)

H?[1]1 (z1) ¯H

?[1]1 (z2)

D[0]1 (z1) ¯D

[0]1 (z2)

Both interesting: different integration of FFs in pTi, might provide information on the Collins pT

dependence

Technically more complicated: require the determination of a qq proxy (Thrust axis)-

More recently...

Francesca Giordano

N j,raw = PijNi

i = ,K

25

PID correction

Francesca Giordano

Perfect PID j = i

N j,raw = PijNi

j = e, µ,,K, p

i = ,K

25

PID correction

Francesca Giordano

Perfect PID j = i

ự(π) ≳ 90% ự(K) ≳ 85%

N j,raw = PijNi

BUT!!

j = e, µ,,K, p

i = ,K

25

PID correction

Francesca Giordano

Perfect PID j = i

ự(π) ≳ 90% ự(K) ≳ 85%

90%

N j,raw = PijNi

BUT!!

j = e, µ,,K, p

i = ,K

25

PID correction

Francesca Giordano

Perfect PID j = i

ự(π) ≳ 90% ự(K) ≳ 85%

e, µ,K, p

90%

10%

Pij =

0

BBBB@

Pe!e Pe!µ Pe! Pe!K Pe!p

Pµ!e Pµ!µ Pµ! Pµ!K Pµ!p

P!e P!µ P! P!K P!p

PK!e PK!µ PK! PK!K PK!p

Pp!e Pp!mu Pp! Pp!K Pp!p

1

CCCCA

N j,raw = PijNi

BUT!!

j = e, µ,,K, p

i = ,K

25

PID correction

Francesca Giordano

Perfect PID j = i

ự(π) ≳ 90% ự(K) ≳ 85%

e, µ,K, p

90%

10%

Pij =

0

BBBB@



P!e P!µ P! P!K P!p



1

CCCCA

N j,raw = PijNi

BUT!!

j = e, µ,,K, p

i = ,K

N i = P1ij N j,raw

25

PID correction

Francesca Giordano

ijHow to determine the Pij?

26

Francesca Giordano

ijHow to determine the Pij?From data!

26

Francesca Giordano

ijHow to determine the Pij?D D0

+slow

From data!

26

Francesca Giordano

ijHow to determine the Pij?D D0 K

+slow

+fastFrom data!

26

Francesca Giordano


+slow

+fast

mD m0

D

Negative hadron = .

(no PID likelihood used)K

From data!

26

Francesca Giordano


+slow

+fast

mD m0

D

Negative hadron = .


Negative hadron identified as .

mD m0

D

From data!

26

Francesca Giordano


+slow

+fast

mD m0

D

Negative hadron = .



mD m0

D PK!

From data!

26

Francesca Giordano


+slow

+fast

mD m0

D

Negative hadron = .



mD m0

D PK!

K

PK!K

From data!

26

Francesca Giordano


+slow

+fast

mD m0

D

Negative hadron = .



mD m0

D PK!

K

PK!K

PK!p

p

From data!

26

Francesca Giordano


+slow

+fast

mD m0

D

Negative hadron = .



mD m0

D PK!

µ

PK!µ

K

PK!K

PK!p

p

From data!

26

Francesca Giordano


+slow

+fast

mD m0

D

Negative hadron = .



mD m0

D PK!

eµ

PK!e

PK!µ

K

PK!K

PK!p

p

From data!

26

Francesca Giordano

2D correctionDetector performance depends on momentum

and scattering angle!

27

Francesca Giordano


and scattering angle!

Pij Pij(p, )

27

Francesca Giordano


and scattering angle!Bin

#plab [GeV/c] bin ranges

0 [0.5,0.65)1 [0.65,0.8)2 [0.8,1.0)3 [1.0,1.2)4 [1.2,1.4)5 [1.4,1.6)6 [1.6,1.8)7 [1.8,2.0)8 [2.0,2.2)9 [2.2,2.4)

10 [2.4,2.6)11 [2.6,2.8)12 [2.8,3.0)13 [3.0,3.5)14 [3.5,4.0)15 [4.0,5.0)16 [5.0,8.0)

Bin #

cos𝜽lab

bin ranges0 [-0.511,-0.300)1 [-0.300,-0.152)2 [-0.512,0.017)3 [0.017,0.209)4 [0.209,0.355)5 [0.355,0.435)6 [0.435,0.542)7 [0.542,0.692)8 [0.692,0.842)

Pij Pij(p, )

27

Francesca Giordano

2D correction

no PID cut e/μ PID cut π PID cut

K PID cut p PID cut Unsel.

K from D* decay for plab in [1.4,1.6) and cos𝜽lab in [0.209,0.355)

0

BBBB@



P!e P!µ P! P!K P!p



1

CCCCA

28

Francesca Giordano

2D correction

no PID cut e/μ PID cut π PID cut

K PID cut p PID cut Unsel.

K from D* decay for plab in [1.4,1.6) and cos𝜽lab in [0.209,0.355)

0

BBBB@



P!e P!µ P! P!K P!p



1

CCCCA

pπ, K -> j from D* decay pπ, p -> j from ờ decay

pe, µ -> j from J/ψ decay

28

Francesca Giordano

2D correction0

BBBB@



P!e P!µ P! P!K P!p



1

CCCCA

pπ, K -> j from D* decay pπ, p -> j from ờ decay

pe, µ -> j from J/ψ decay

29

Francesca Giordano

PID correction

K

< 15%

30% >30

Francesca Giordano

uds-charm-bottom-tau contributions

31

𝜋𝜋 couples

Published 𝜋𝜋 studied a charm enhanced data and found charm contribute only

as dilution=> charm contribution corrected out


𝜋K couples



KK couples



KK couples


For the moment charm contributionis not being corrected out

in any of the samples (𝜋𝜋, 𝜋K, KK)

Francesca Giordano

yCollins asymmetries

34


𝜋𝜋 => non-zero asymmetries, increase with z1, z2

𝜋K => asymmetries compatible with zero

KK => non-zero asymmetries, increase with z1,z2

similar size of pion-pion

𝜙0 yasymmetries


yasymmetries𝜙0𝜋𝜋 => non-zero asymmetries,

increase with z1, z2

𝜋K => asymmetries compatible with zero

KK => non-zero asymmetries, increase with z1,z2

similar size of pion-pion

But! charm have different contributions, we need to account for it!


versus sin2𝜭/(1+cos2𝜭)

A0 =

sin

2

1 + cos

2 FhH?

1 (z1) ¯H?1 (z2)

D?1 (z1)

¯D?1 (z2)

i

linear in sin2𝜭/(1+cos2𝜭),

go to 0 for sin2𝜭/(1+cos2𝜭) 0

fit form: p0 + p1 sin2𝜭/(1+cos2𝜭)

𝜋𝜋

𝜋K

KK

QCD test?


Collins fragmentation: u-d-s contributions(vs 0.1, 27th Sept, 2012)

1 Definition and assumptions

u, d → π (ud, ud)

Dfav = Dπ+

u = Dπ−

d = Dπ−

u = Dπ+

d(1)

Ddis = Dπ−

u = Dπ+

d = Dπ+

u = Dπ−

d(2)

s → π (ud, ud)

Ddiss→π = Dπ+

s = Dπ−

s = Dπ+

s = Dπ−

s (3)

u, d → K (us, us)

Dfavu→K = DK+

u = DK−

u (4)

Ddisu,d→K = DK−

u = DK+

u = DK+

d = DK−

d= DK−

d = DK+

d(5)

s → K (us, us)

Dfavs→K = DK−

s = DK+

s (6)

Ddiss→K = DK+

s = DK−

s (7)

In the end we are left with 7 possible fragmentation functions:

Dfav , Ddis, Ddiss→π, D

favu→K , Ddis

u,d→K , Dfavs→K , Ddis

s→K (8)

2 Pion-Pion

e+e− → π±π∓ (unLike sign)

NUππ ∝

!

q

e2q(Dfav1 D

fav2 +Ddis

1 Ddis2 ) (9)

e+e− → π±π± (Like sign)

NLππ ∝

!

q

e2q(Dfav1 Ddis

2 +Ddis1 D

fav2 ) (10)

3 Pion-Kaon

e+e− → π±K∓ (unLike sign)

NUπK ∝

!

q

e2q (11)

e+e− → π±K± (Like sign)

NLπK ∝

!

q

e2q (12)

1

Assuming charm contribute only as a dilution

Fragmentation contributions


For pion-pion couples:

For pion-Kaon couples:

For Kaon-Kaon couples:



For pion-pion couples:

For pion-Kaon couples:

For Kaon-Kaon couples:

Not so easy! A full phenomenological study needed!


Francesca Giordano

ySummary & outlook

40

𝜙0 asymmetries present similar features for 𝜋𝜋 and KK couplesvery small/compatible with zero for 𝜋K couplesfor 𝜋𝜋 and 𝜋K the sin2𝜭/(1+cos2𝜭) dependence of asymmetries are not inconsistent with a linear dependence going to zeroKK show a more convoluted sin2𝜭/(1+cos2𝜭) dependence

Francesca Giordano

ySummary & outlook

40


𝜙12 asymmetries with Thrust axis in progress study using jet algorithm instead of Thrust in progress

Francesca Giordano

ySummary & outlook

40


𝜙12 asymmetries with Thrust axis in progress study using jet algorithm instead of Thrust in progress

Stay tuned!

Francesca Giordano

http://abstrusegoose.com/342

41




Backups

Francesca Giordano

Kinematic variables

43

hadron energy fraction with respect to parton

qT component of virtual photon momentum transverse to the h1h2 axis in the frame

where h1 and h2 are back-to-back

z 0.2 0.25 0.3 0.42 1

pT12 0 0.13 0.3 0.5 3

pT0 0 0.13 0.25 0.4 0.5 0.6 0.75 1 3

qT 0 0.5 1 1.25 1.5 1.75 2 2.25 2.5

sin2𝜭/(1+cos2𝜭) 0.4 0.45 0.5 0.6 0.7 0.8 0.9 0.97 1

z1, z2

pT component of hadron momentum transverse to reference direction1. 𝜙1+𝜙2 method: the thrust axis

2. 𝜙0 method: hadron 2

pT1, pT2

pT0


𝜙0 yasymmetries

Francesca Giordano

yMore asymmetries

46

𝜙0

f giordano collins fragmentation for kaon

Science