Unravelling the mystery of quark confinement: how transversely polarized quarks fragment into Kaon hadrons

1. September 17th, 2014,Torino
Francesca Giordano
COLLINS
MEASUREMENTS @ BELLE

2. Francesca Giordano 2
Fragmentation process
or how do the hadrons get formed?

3. Francesca Giordano 2
Fragmentation process
or how do the hadrons get formed?
Dq
h
h
q
Fragmentation function describes the
process of hadronization of a parton
Dh
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

4. Francesca Giordano 2
Fragmentation process
or how do the hadrons get formed?
Strictly related to quark confinement
Dq
h
h
q
Fragmentation function describes the
process of hadronization of a parton
Dh
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

5. Francesca Giordano
Dq
h
h
q
3
e+
e- q
Fragmentation function describes the
process of hadronization of a parton
Strictly related to quark confinement
Cleanest way to access FF is in e+e- ➞ qq
Fragmentation process
or how do the hadrons get formed?
Dh
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

6. Francesca Giordano
Dq
h
h
q
3
e+
e- q
Fragmentation function describes the
process of hadronization of a parton
Strictly related to quark confinement
Cleanest way to access FF is in e+e- ➞ qq
Fragmentation process
or how do the hadrons get formed?
e+
e !hX
/
X
q
e+
e !q¯q
⇥(Dh
q + Dh
¯q )
Dh
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

7. Francesca Giordano
Ah
UT =
" "
" + "
⇒⇒
⇒⇒
/ Dh
q
Dq
h
h
q
4
e+
e- q
Fragmentation function describes the
process of hadronization of a parton
Strictly related to quark confinement
Cleanest way to access FF is in e+e- ➞ qq
Universal: can be used to study the nucleon
structure when combined with SIDIS and
hadronic reactions data
Fragmentation process
or how do the hadrons get formed?
/ H?h
1q
Dh
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

8. 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
⊥ PT
q
quark
PT

9. 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
⊥ PT
q
quark
PT
strictly related to the outgoing
hadron tranverse momentum
TMD!

10. 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
⊥ PT
q
quark
PT
strictly related to the outgoing
hadron tranverse momentum
TMD! }chiral even
Chiral odd!
⇥
chiral oddchiral odd
X H1
⊥

11. Francesca Giordano
KEKB
BELLE @ KEKB
Francesca Giordano 6

12. 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)

13. Francesca Giordano
Collins Fragmentation
8
e+
e !hX
/
X
q
e+
e !q¯q
⇥(H?
1,q + H?
1,¯q)
h
e-
e+
q
¯q
e+
e ! q ¯q ! hX

14. Francesca Giordano
Collins Fragmentation
8
e+
e !hX
/
X
q
e+
e !q¯q
⇥(H?
1,q + H?
1,¯q)
h
e-
e+
q
¯q
e+
e ! q ¯q ! hX
}chiral even
Chiral odd!
⇥
chiral oddchiral odd
X H1
⊥

15. Francesca Giordano 9
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 Fragmentation
h1
h2
Back-to-Back jets
¯q
q
e-
e+

16. Francesca Giordano 9
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 X
h = ⇡, K
Collins Fragmentation
h1
h2
Back-to-Back jets
¯q
q
e-
e+

17. 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 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane define
lepton momenta and that of one hadron and the second hadron’s transverse mo
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was
in the previous formulas for the sake of clarity. The kinematic prefactors are defin
A(y) = (1
2
− y − y2
)
CMS
=
1
4
(1 + cos2
θ)
B(y) = y(1 − y)
CMS
=
1
(sin2
θ) ,
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Thrust axis= proxy for the qq axis-

18. Francesca Giordano
Reference frames
11
e+
e ! q ¯q ! h1 h2 X
h = ⇡, K
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Thrust axis= proxy for the qq axis-
h1
h2
Back-to-Back jets
¯q
q
e-
e+

19. Francesca Giordano
Reference frames
11
e+
e ! q ¯q ! h1 h2 X
h = ⇡, K
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Thrust axis= proxy for the qq axis-
h1
h2
Back-to-Back jets
¯q
q
e-
e+

20. Francesca Giordano
Reference frames
11
e+
e ! q ¯q ! h1 h2 X
h = ⇡, K
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Thrust axis= proxy for the qq axis-
h1
h2
Back-to-Back jets
¯q
q
e-
e+

21. Francesca Giordano 12
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
φ1,2 =
ˆn
·
ˆz × ˆn
×
ˆn × Ph1,2
acos
ˆz × ˆn
·
ˆn × Ph1,2
, (4)
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane define
lepton momenta and that of one hadron and the second hadron’s transverse mo
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was
in the previous formulas for the sake of clarity. The kinematic prefactors are defin
A(y) = (1
2
− y − y2
)
CMS
=
1
4
(1 + cos2
θ)
B(y) = y(1 − y)
CMS
=
1
4
(sin2
θ) ,
where y = (1 + cos θ)/2 is a measure of the forwardness of the hard scattering
𝜙0 method:
hadron 1 azimuthal angle with respect
to hadron 2
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Reference frames
F[X] =
X
q¯q
Z
[2ˆh · kT1
ˆh · kT2 kT1 · kT2]
d2
kT1d2
kT2
2
(kT1 + kT2 qT)X
kT i = zi pT i
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i⌘
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 1 + 2)
H
?[1]
1 (z1) ¯H
?[1]
1 (z2)
D
[0]
1 (z1) ¯D
[0]
1 (z2)
⌘
F[n]
(zi) ⌘
Z
d|kT |2
h|kT |
Mi
i[n]
F(zi, |kT |2
)
D. Boer
Nucl.Phys.B806:23,2009

22. Francesca Giordano 12
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
φ1,2 =
ˆn
·
ˆz × ˆn
×
ˆn × Ph1,2
acos
ˆz × ˆn
·
ˆn × Ph1,2
, (4)
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane define
lepton momenta and that of one hadron and the second hadron’s transverse mo
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was
in the previous formulas for the sake of clarity. The kinematic prefactors are defin
A(y) = (1
2
− y − y2
)
CMS
=
1
4
(1 + cos2
θ)
B(y) = y(1 − y)
CMS
=
1
4
(sin2
θ) ,
where y = (1 + cos θ)/2 is a measure of the forwardness of the hard scattering
𝜙0 method:
hadron 1 azimuthal angle with respect
to hadron 2
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Reference frames
F[X] =
X
q¯q
Z
[2ˆh · kT1
ˆh · kT2 kT1 · kT2]
d2
kT1d2
kT2
2
(kT1 + kT2 qT)X
kT i = zi pT i
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i⌘
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 1 + 2)
H
?[1]
1 (z1) ¯H
?[1]
1 (z2)
D
[0]
1 (z1) ¯D
[0]
1 (z2)
⌘
F[n]
(zi) ⌘
Z
d|kT |2
h|kT |
Mi
i[n]
F(zi, |kT |2
)
D. Boer
Nucl.Phys.B806:23,2009

23. Francesca Giordano 12
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
φ1,2 =
ˆn
·
ˆz × ˆn
×
ˆn × Ph1,2
acos
ˆz × ˆn
·
ˆn × Ph1,2
, (4)
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane define
lepton momenta and that of one hadron and the second hadron’s transverse mo
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was
in the previous formulas for the sake of clarity. The kinematic prefactors are defin
A(y) = (1
2
− y − y2
)
CMS
=
1
4
(1 + cos2
θ)
B(y) = y(1 − y)
CMS
=
1
4
(sin2
θ) ,
where y = (1 + cos θ)/2 is a measure of the forwardness of the hard scattering
𝜙0 method:
hadron 1 azimuthal angle with respect
to hadron 2
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Reference frames
F[X] =
X
q¯q
Z
[2ˆh · kT1
ˆh · kT2 kT1 · kT2]
d2
kT1d2
kT2
2
(kT1 + kT2 qT)X
kT i = zi pT i
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i⌘
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 1 + 2)
H
?[1]
1 (z1) ¯H
?[1]
1 (z2)
D
[0]
1 (z1) ¯D
[0]
1 (z2)
⌘
F[n]
(zi) ⌘
Z
d|kT |2
h|kT |
Mi
i[n]
F(zi, |kT |2
)
D. Boer
Nucl.Phys.B806:23,2009

24. Francesca Giordano
Product of 2 Collins FFs
13
Like-sign couples

25. Francesca Giordano
Product of 2 Collins FFs
13
Like-sign couples
Fav Ł UnFav Fav Ł UnFav
}

26. Francesca Giordano
Product of 2 Collins FFs
14
Like-sign couples
Fav Ł UnFav Fav Ł UnFav
}Unlike-sign couples
All charges couples
Fav Ł Fav UnFav Ł UnFav
Fav + unFav Ł UnFav + Fav
}

27. Francesca Giordano 15
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
are formed between the scattering plane and their transverse momenta Phi⊥
thrust axis ˆn.
effect can only be visible in the combination of 2 functions being able to create
n asymmetry each. Accordingly the combination of a quark and an antiquark
tion in opposing hemispheres gives a product of two sin(φ) modulations for
muthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
azimuthal angles are defined as
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane define
lepton momenta and that of one hadron and the second hadron’s transverse mo
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was
in the previous formulas for the sake of clarity. The kinematic prefactors are defin
A(y) = (1
2
− y − y2
)
CMS
=
1
4
(1 + cos2
θ)
B(y) = y(1 − y)
CMS
=
1
(sin2
θ) ,
e+
e ! q ¯q ! h1 h2 X
h = ⇡, K
𝜙0 method:
hadron 1 azimuthal angle with respect
to hadron 2
𝜙1+𝜙2 method:
hadron azimuthal angles with respect
to the qq axis proxy
-
Reference frames
R0 =
N0( 0)
hN0i
R12 =
N12( 1 + 2)
hN12i

28. Francesca Giordano 16
But! Acceptance and radiation
effects also contribute to azimuthal
asymmetries!
Double-ratios

29. Francesca Giordano 16
But! Acceptance and radiation
effects also contribute to azimuthal
asymmetries!
Double-ratios
Simulation: unlike-sign, like-sign

30. Francesca Giordano 16
But! Acceptance and radiation
effects also contribute to azimuthal
asymmetries!
Double-ratios
Simulation: unlike-sign, like-signData: unlike-sign, like-sign

31. Francesca Giordano 16
But! Acceptance and radiation
effects also contribute to azimuthal
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
Dh1h2
ul = RU
/RL
Dh1h2
uc = RU
/RC

32. Francesca Giordano 17
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Figure 3: Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
the angles are formed between the scattering plane and their transverse momenta Phi⊥
around the thrust axis ˆn.
the Collins effect can only be visible in the combination of 2 functions being able to create
a single spin asymmetry each. Accordingly the combination of a quark and an antiquark
Collins function in opposing hemispheres gives a product of two sin(φ) modulations for
the two azimuthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
e+
e−
these azimuthal angles are defined as
φ1,2 =
ˆn
|ˆn|
·
ˆz × ˆn
|ˆz||ˆn|
×
ˆn × Ph1,2
|ˆn||Ph1,2|
acos
ˆz × ˆn
|ˆz||ˆn|
·
ˆn × Ph1,2
|ˆn||Ph1,2|
, (4)
where ˆz is just a unit vector in the z-axis defined by the e+
e−
axis and ˆn 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 polarization
of both quarks. Since the e+
e−
process does not exhibit a well defined polarization axis
only an average transverse polarization can yield this property. In fact the virtual photon
emitted has to be spin 1 which in the helicity basis of the incoming leptons can be created
by the combinations +− and −+. In the case of creating a quark-antiquark pair under
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane defined by the
lepton momenta and that of one hadron and the second hadron’s transverse momentum
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was omitted
in the previous formulas for the sake of clarity. The kinematic prefactors are defined as:
A(y) = (1
2
− 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 the
intrinsic 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
Q2
z2
1z2
2 e2
q/4(1 + cos2
θ)D
q,[0]
1 (z1)D
q,[0]
1 (z2)
+e2
q/4 sin2
θ cos(φ1 + φ2)H
⊥,[1],q
1 H
⊥,[1],q
1 , (13)
𝜙0 method𝜙1+𝜙2 method
Double-ratios
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)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(2 0))
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 1 + 2)
H
?[1]
1 (z1) ¯H
?[1]
1 (z2)
D
[0]
1 (z1) ¯D
[0]
1 (z2)
⌘
Dh1h2
uc = RU
/RC
Dh1h2
ul = RU
/RL

33. Francesca Giordano 17
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Figure 3: Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
the angles are formed between the scattering plane and their transverse momenta Phi⊥
around the thrust axis ˆn.
the Collins effect can only be visible in the combination of 2 functions being able to create
a single spin asymmetry each. Accordingly the combination of a quark and an antiquark
Collins function in opposing hemispheres gives a product of two sin(φ) modulations for
the two azimuthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
e+
e−
these azimuthal angles are defined as
φ1,2 =
ˆn
|ˆn|
·
ˆz × ˆn
|ˆz||ˆn|
×
ˆn × Ph1,2
|ˆn||Ph1,2|
acos
ˆz × ˆn
|ˆz||ˆn|
·
ˆn × Ph1,2
|ˆn||Ph1,2|
, (4)
where ˆz is just a unit vector in the z-axis defined by the e+
e−
axis and ˆn 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 polarization
of both quarks. Since the e+
e−
process does not exhibit a well defined polarization axis
only an average transverse polarization can yield this property. In fact the virtual photon
emitted has to be spin 1 which in the helicity basis of the incoming leptons can be created
by the combinations +− and −+. In the case of creating a quark-antiquark pair under
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane defined by the
lepton momenta and that of one hadron and the second hadron’s transverse momentum
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was omitted
in the previous formulas for the sake of clarity. The kinematic prefactors are defined as:
A(y) = (1
2
− 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 the
intrinsic 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
Q2
z2
1z2
2 e2
q/4(1 + cos2
θ)D
q,[0]
1 (z1)D
q,[0]
1 (z2)
+e2
q/4 sin2
θ cos(φ1 + φ2)H
⊥,[1],q
1 H
⊥,[1],q
1 , (13)
𝜙0 method𝜙1+𝜙2 method
Double-ratios
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)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(2 0))
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 1 + 2)
H
?[1]
1 (z1) ¯H
?[1]
1 (z2)
D
[0]
1 (z1) ¯D
[0]
1 (z2)
⌘
A0 =
sin2
✓
1 + cos2 ✓
F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i
A12 =
sin2
✓
1 + cos2 ✓
H
?[1]
1 (z1) ¯H
?[1]
1 (z2)
D
[0]
1 (z1) ¯D
[0]
1 (z2)

34. Francesca Giordano
Published results:
18
PRD 78, 032011 (2008)PRD 78, 032011 (2008)
𝜙0 method𝜙1+𝜙2 method
𝜋𝜋
Belle publications
PRL 96,232002, (2006)
PRD 78, 032011 (2008)

35. Francesca Giordano
Belle vs. Babar
19

36. Francesca Giordano
SIDIS
20
DF
σ
FF
}chiral even
Chiral odd!
⇥
chiral oddchiral odd
X H1
⊥

37. Francesca Giordano
SIDIS
20
DF
σ
FF
H?
1h1 ⇥⊥
H?
1h1 ⇥
}chiral even
Chiral odd!
⇥
chiral oddchiral odd
X H1
⊥

38. Francesca Giordano
[Airapetian et al., Phys. Lett. B 693 (2010) 11-16]
Collins amplitudes in SIDIS
H?
1h1 ⇥AUT /
π+/−
Κ+/−
Deuterium
21
𝜋 Collins FF

39. Francesca Giordano
[Airapetian et al., Phys. Lett. B 693 (2010) 11-16]
Collins amplitudes in SIDIS
H?
1h1 ⇥AUT /
π+/−
Κ+/−
Deuterium
21
𝜋 Collins FF
Anselmino et al.
Phys.Rev. D75 (2007)
transversity

40. Francesca Giordano
quark pol.
U L T
nucleonpol.
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
Transversity dist
(Collins fragme
+
u
[A. Airapetian et al, Phys. Rev. Le
2005: First evidence from
SIDIS on proton
Non-zero transvers
Non-zero Collins fu
-0.1
0
0.1
0.2
-0.2
-0.1
0
0.1 0.2 0.3
2!sin("+"S)#
UT
$
$
+
x
$-
0.3
H?,unfav
1 ⇡ H?,fav
1
[Airapetian et al., Phys. Lett. B 693 (2010) 11-16]
Collins amplitudes in SIDIS
H?
1h1 ⇥AUT /
π+/−
Κ+/−
Deuterium
21
𝜋 Collins FF
Anselmino et al.
Phys.Rev. D75 (2007)
transversity

41. Francesca Giordano
Collins amplitudes in SIDIS
22
H?
1h1 ⇥AUT /
AUU / h?
1 ⇥ H?
1

42. Francesca Giordano
Collins amplitudes in SIDIS
22
H?
1h1 ⇥AUT /
K+ amplitudes larger
than 𝜋+?
AUU / h?
1 ⇥ H?
1

43. Francesca Giordano
More recently...
23
z qT sin2 𝜭/(1+cos2 𝜭) pT
New!
New!
New!
New!New!New!
New! New! New!
z qT sin2 𝜭/(1+cos2 𝜭) pT
New!
New!
New!
New!New!New!
New! New! New!
D⇡⇡
uc = RU⇡⇡
/RC⇡⇡
D⇡k
uc = RU⇡k
/RC⇡k
Dkk
uc = RUkk
/RCkk
Dkk
ul = RUkk
/RLkk
D⇡k
ul = RU⇡k
/RL⇡k
D⇡⇡
ul = RU⇡⇡
/RL⇡⇡

44. Francesca Giordano
More recently...
23
z qT sin2 𝜭/(1+cos2 𝜭) pT
New!
New!
New!
New!New!New!
New! New! New!
z qT sin2 𝜭/(1+cos2 𝜭) pT
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
D⇡⇡
uc = RU⇡⇡
/RC⇡⇡
D⇡k
uc = RU⇡k
/RC⇡k
Dkk
uc = RUkk
/RCkk
Dkk
ul = RUkk
/RLkk
D⇡k
ul = RU⇡k
/RL⇡k
D⇡⇡
ul = RU⇡⇡
/RL⇡⇡

45. Francesca Giordano 24
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Figure 3: Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
the angles are formed between the scattering plane and their transverse momenta Phi⊥
around the thrust axis ˆn.
the Collins effect can only be visible in the combination of 2 functions being able to create
a single spin asymmetry each. Accordingly the combination of a quark and an antiquark
Collins function in opposing hemispheres gives a product of two sin(φ) modulations for
the two azimuthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
e+
e−
these azimuthal angles are defined as
φ1,2 =
ˆn
|ˆn|
·
ˆz × ˆn
|ˆz||ˆn|
×
ˆn × Ph1,2
|ˆn||Ph1,2|
acos
ˆz × ˆn
|ˆz||ˆn|
·
ˆn × Ph1,2
|ˆn||Ph1,2|
, (4)
where ˆz is just a unit vector in the z-axis defined by the e+
e−
axis and ˆn 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 polarization
of both quarks. Since the e+
e−
process does not exhibit a well defined polarization axis
only an average transverse polarization can yield this property. In fact the virtual photon
emitted has to be spin 1 which in the helicity basis of the incoming leptons can be created
by the combinations +− and −+. In the case of creating a quark-antiquark pair under
the CMS angle of θ = π/2 (see Fig. 3) both lepton helicity combinations would be equally
contributing and transverse polarization of the quarks has to average out. Hence the quark-
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane defined by the
lepton momenta and that of one hadron and the second hadron’s transverse momentum
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was omitted
in the previous formulas for the sake of clarity. The kinematic prefactors are defined as:
A(y) = (1
2
− 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 the
intrinsic 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
Q2
z2
1z2
2 e2
q/4(1 + cos2
θ)D
q,[0]
1 (z1)D
q,[0]
1 (z2)
+e2
q/4 sin2
θ cos(φ1 + φ2)H
⊥,[1],q
1 H
⊥,[1],q
1 , (13)
where the fragmentation functions appear as the zeroth[0] or first[1] moments in the
absolute value of their corresponding transverse momenta:
n
𝜙0 method𝜙1+𝜙2 method
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i⌘
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 1 + 2)
H
?[1]
1 (z1) ¯H
?[1]
1 (z2)
D
[0]
1 (z1) ¯D
[0]
1 (z2)
⌘
More recently...

46. Francesca Giordano 24
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Figure 3: Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
the angles are formed between the scattering plane and their transverse momenta Phi⊥
around the thrust axis ˆn.
the Collins effect can only be visible in the combination of 2 functions being able to create
a single spin asymmetry each. Accordingly the combination of a quark and an antiquark
Collins function in opposing hemispheres gives a product of two sin(φ) modulations for
the two azimuthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
e+
e−
these azimuthal angles are defined as
φ1,2 =
ˆn
|ˆn|
·
ˆz × ˆn
|ˆz||ˆn|
×
ˆn × Ph1,2
|ˆn||Ph1,2|
acos
ˆz × ˆn
|ˆz||ˆn|
·
ˆn × Ph1,2
|ˆn||Ph1,2|
, (4)
where ˆz is just a unit vector in the z-axis defined by the e+
e−
axis and ˆn 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 polarization
of both quarks. Since the e+
e−
process does not exhibit a well defined polarization axis
only an average transverse polarization can yield this property. In fact the virtual photon
emitted has to be spin 1 which in the helicity basis of the incoming leptons can be created
by the combinations +− and −+. In the case of creating a quark-antiquark pair under
the CMS angle of θ = π/2 (see Fig. 3) both lepton helicity combinations would be equally
contributing and transverse polarization of the quarks has to average out. Hence the quark-
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane defined by the
lepton momenta and that of one hadron and the second hadron’s transverse momentum
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was omitted
in the previous formulas for the sake of clarity. The kinematic prefactors are defined as:
A(y) = (1
2
− 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 the
intrinsic 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
Q2
z2
1z2
2 e2
q/4(1 + cos2
θ)D
q,[0]
1 (z1)D
q,[0]
1 (z2)
+e2
q/4 sin2
θ cos(φ1 + φ2)H
⊥,[1],q
1 H
⊥,[1],q
1 , (13)
where the fragmentation functions appear as the zeroth[0] or first[1] moments in the
absolute value of their corresponding transverse momenta:
n
𝜙0 method𝜙1+𝜙2 method
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i⌘
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 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...

47. Francesca Giordano 24
Ph1
φ1
Ph2
φ2 − π
θ
Thrust axis
e−
e+
Ph1⊥
Ph2⊥
Figure 3: Definition of the azimuthal angles φ1 and φ2 − π of the two hadrons, where
the angles are formed between the scattering plane and their transverse momenta Phi⊥
around the thrust axis ˆn.
the Collins effect can only be visible in the combination of 2 functions being able to create
a single spin asymmetry each. Accordingly the combination of a quark and an antiquark
Collins function in opposing hemispheres gives a product of two sin(φ) modulations for
the two azimuthal angles φ1 and φ2, resulting in a cos(φ1 + φ2) modulation (see Fig. 3). In
e+
e−
these azimuthal angles are defined as
φ1,2 =
ˆn
|ˆn|
·
ˆz × ˆn
|ˆz||ˆn|
×
ˆn × Ph1,2
|ˆn||Ph1,2|
acos
ˆz × ˆn
|ˆz||ˆn|
·
ˆn × Ph1,2
|ˆn||Ph1,2|
, (4)
where ˆz is just a unit vector in the z-axis defined by the e+
e−
axis and ˆn 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 polarization
of both quarks. Since the e+
e−
process does not exhibit a well defined polarization axis
only an average transverse polarization can yield this property. In fact the virtual photon
emitted has to be spin 1 which in the helicity basis of the incoming leptons can be created
by the combinations +− and −+. In the case of creating a quark-antiquark pair under
the CMS angle of θ = π/2 (see Fig. 3) both lepton helicity combinations would be equally
contributing and transverse polarization of the quarks has to average out. Hence the quark-
Ph1
φ0
Ph2 θ2
e−
e+
Ph1⊥
Figure 4: Definition of the azimuthal angle φ0 formed between the plane defined by the
lepton momenta and that of one hadron and the second hadron’s transverse momentum
Ph1⊥ relative to the first hadron.
The dependence on the transverse momentum and on the fractional energy was omitted
in the previous formulas for the sake of clarity. The kinematic prefactors are defined as:
A(y) = (1
2
− 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 the
intrinsic 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
Q2
z2
1z2
2 e2
q/4(1 + cos2
θ)D
q,[0]
1 (z1)D
q,[0]
1 (z2)
+e2
q/4 sin2
θ cos(φ1 + φ2)H
⊥,[1],q
1 H
⊥,[1],q
1 , (13)
where the fragmentation functions appear as the zeroth[0] or first[1] moments in the
absolute value of their corresponding transverse momenta:
n
𝜙0 method𝜙1+𝜙2 method
⇠ M0
⇣
1 +
sin2
✓2
1 + cos2 ✓2
cos(2 0)F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i⌘
⇠ M12
⇣
1 +
sin2
✓T
1 + cos2 ✓T
cos( 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...

48. Francesca Giordano
Nj,raw
= PijNi
i = ⇡, K
25
PID correction

49. Francesca Giordano
Perfect PID ➯ j = i
Nj,raw
= PijNi
j = e, µ, ⇡, K, p
i = ⇡, K
25
PID correction

50. Francesca Giordano
Perfect PID ➯ j = i
ự(π) ≳ 90% ự(K) ≳ 85%
Nj,raw
= PijNi
BUT!!
j = e, µ, ⇡, K, p
i = ⇡, K
25
PID correction

51. Francesca Giordano
Perfect PID ➯ j = i
ự(π) ≳ 90% ự(K) ≳ 85%
⇡
⇡
90%
Nj,raw
= PijNi
BUT!!
j = e, µ, ⇡, K, p
i = ⇡, K
25
PID correction

52. Francesca Giordano
Perfect PID ➯ j = i
ự(π) ≳ 90% ự(K) ≳ 85%
⇡
e, µ, K, p
⇡
90%
10%
Pij =
0
B
B
B
B
@
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
C
C
C
C
A
Nj,raw
= PijNi
BUT!!
j = e, µ, ⇡, K, p
i = ⇡, K
25
PID correction

53. Francesca Giordano
Perfect PID ➯ j = i
ự(π) ≳ 90% ự(K) ≳ 85%
⇡
e, µ, K, p
⇡
90%
10%
Pij =
0
B
B
B
B
@
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
C
C
C
C
A
Nj,raw
= PijNi
BUT!!
j = e, µ, ⇡, K, p
i = ⇡, K
Ni
= P 1
ij Nj,raw
25
PID correction

54. Francesca Giordano
ijHow to determine the Pij?
26

55. Francesca Giordano
ijHow to determine the Pij?
From data!
26

56. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
⇡+
slowFrom data!
26

57. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fastFrom data!
26

58. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fast
m⇤
D m0
D
Negative hadron = .
(no PID likelihood used)
K
From data!
26

59. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fast
m⇤
D m0
D
Negative hadron = .
(no PID likelihood used)
K
Negative hadron
identified as .⇡
m⇤
D m0
D
From data!
26

60. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fast
m⇤
D m0
D
Negative hadron = .
(no PID likelihood used)
K
Negative hadron
identified as .⇡
m⇤
D m0
D PK !⇡
From data!
26

61. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fast
m⇤
D m0
D
Negative hadron = .
(no PID likelihood used)
K
Negative hadron
identified as .⇡
m⇤
D m0
D PK !⇡
K
PK !K
From data!
26

62. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fast
m⇤
D m0
D
Negative hadron = .
(no PID likelihood used)
K
Negative hadron
identified as .⇡
m⇤
D m0
D PK !⇡
K
PK !K
PK !¯p
¯p
From data!
26

63. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fast
m⇤
D m0
D
Negative hadron = .
(no PID likelihood used)
K
Negative hadron
identified as .⇡
m⇤
D m0
D PK !⇡
µ
PK !µ
K
PK !K
PK !¯p
¯p
From data!
26

64. Francesca Giordano
ijHow to determine the Pij?
D⇤ D0
K
⇡+
slow ⇡+
fast
m⇤
D m0
D
Negative hadron = .
(no PID likelihood used)
K
Negative hadron
identified as .⇡
m⇤
D m0
D PK !⇡
eµ
PK !e
PK !µ
K
PK !K
PK !¯p
¯p
From data!
26

65. Francesca Giordano
2D correction
Detector performance depends on momentum
and scattering angle!
27

66. Francesca Giordano
2D correction
Detector performance depends on momentum
and scattering angle!
Pij Pij(p, ✓)➬
27

67. Francesca Giordano
2D correction
Detector performance depends on momentum
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 ranges
0 [-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

68. 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
B
B
B
B
@
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
C
C
C
C
A
28

69. 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
B
B
B
B
@
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
C
C
C
C
A
28

70. 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
B
B
B
B
@
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
C
C
C
C
A
pπ, K -> j from D* decay
pπ, p -> j from ờ decay
pe, µ -> j from J/ψ decay
28

71. Francesca Giordano
2D correction0
B
B
B
B
@
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
C
C
C
C
A
pπ, K -> j from D* decay
pπ, p -> j from ờ decay
pe, µ -> j from J/ψ decay
29

72. Francesca Giordano
2D correction0
B
B
B
B
@
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
C
C
C
C
A
pπ, K -> j from D* decay
pπ, p -> j from ờ decay
pe, µ -> j from J/ψ decay
29

73. Francesca Giordano
PID correction
⇡
K
< 15%
30% >
30

74. 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

75. Francesca Giordano 32
𝜋K couples
uds-charm-bottom-tau contributions

76. Francesca Giordano 33
KK couples
uds-charm-bottom-tau contributions

77. Francesca Giordano 33
KK couples
uds-charm-bottom-tau contributions
For the moment charm contribution
is not being corrected out
in any of the samples (𝜋𝜋, 𝜋K, KK)

78. Francesca Giordano
yCollins asymmetries
34

79. Francesca Giordano 35
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

80. Francesca Giordano 36
yasymmetries𝜙0
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
But! charm have different contributions,
we need to account for it!

81. Francesca Giordano 37
versus sin2 𝜭/(1+cos2 𝜭)
A0 =
sin2
✓
1 + cos2 ✓
F
hH?
1 (z1) ¯H?
1 (z2)
D?
1 (z1) ¯D?
1 (z2)
i
linear in sin2 𝜭/(1+cos2 𝜭),
go to 0 for sin2 𝜭/(1+cos2 𝜭) ➛0
fit form: p0 + p1 sin2 𝜭/(1+cos2 𝜭)
𝜋𝜋
𝜋K
KK
QCD test?

82. Francesca Giordano 38
Collins fragmentation: u-d-s contributions
(vs 0.1, 27th Sept, 2012)
1 Definition and assumptions
u, d → π (u ¯d, ¯ud)
Dfav = Dπ+
u = Dπ−
d = Dπ−
¯u = Dπ+
¯d
(1)
Ddis = Dπ−
u = Dπ+
d = Dπ+
¯u = Dπ−
¯d
(2)
s → π (u ¯d, ¯ud)
Ddis
s→π = Dπ+
s = Dπ−
s = Dπ+
¯s = Dπ−
¯s (3)
u, d → K (u¯s, ¯us)
Dfav
u→K = DK+
u = DK−
¯u (4)
Ddis
u,d→K = DK−
u = DK+
¯u = DK+
d = DK−
¯d
= DK−
d = DK+
¯d
(5)
s → K (u¯s, ¯us)
Dfav
s→K = DK−
s = DK+
¯s (6)
Ddis
s→K = DK+
s = DK−
¯s (7)
In the end we are left with 7 possible fragmentation functions:
Dfav, Ddis, Ddis
s→π, Dfav
u→K, Ddis
u,d→K, Dfav
s→K, Ddis
s→K (8)
2 Pion-Pion
Assuming charm contribute
only as a dilution
Fragmentation contributions

83. Francesca Giordano 39
For pion-pion couples:
For pion-Kaon couples:
For Kaon-Kaon couples:
Fragmentation contributions

84. Francesca Giordano 39
For pion-pion couples:
For pion-Kaon couples:
For Kaon-Kaon couples:
Not so easy! A full phenomenological study needed!
Fragmentation contributions

85. Francesca Giordano
ySummary & outlook
40
𝜙0 asymmetries
present similar features for 𝜋𝜋 and KK couples
very small/compatible with zero for 𝜋K couples
for 𝜋𝜋 and 𝜋K the sin2 𝜭/(1+cos2 𝜭) dependence of asymmetries are
not inconsistent with a linear dependence going to zero
KK show a more convoluted sin2 𝜭/(1+cos2 𝜭) dependence

86. Francesca Giordano
ySummary & outlook
40
𝜙0 asymmetries
present similar features for 𝜋𝜋 and KK couples
very small/compatible with zero for 𝜋K couples
for 𝜋𝜋 and 𝜋K the sin2 𝜭/(1+cos2 𝜭) dependence of asymmetries are
not inconsistent with a linear dependence going to zero
KK show a more convoluted sin2 𝜭/(1+cos2 𝜭) dependence
𝜙12 asymmetries with Thrust axis in progress
study using jet algorithm instead of Thrust in progress

87. Francesca Giordano
ySummary & outlook
40
𝜙0 asymmetries
present similar features for 𝜋𝜋 and KK couples
very small/compatible with zero for 𝜋K couples
for 𝜋𝜋 and 𝜋K the sin2 𝜭/(1+cos2 𝜭) dependence of asymmetries are
not inconsistent with a linear dependence going to zero
KK show a more convoluted sin2 𝜭/(1+cos2 𝜭) dependence
𝜙12 asymmetries with Thrust axis in progress
study using jet algorithm instead of Thrust in progress
Stay tuned!

88. Francesca Giordano
http://abstrusegoose.com/342
41

89. Francesca Giordano 42
Backups

90. 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
sin2 𝜭/(1+cos2 𝜭) 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 direction
1. 𝜙1+𝜙2 method: the thrust axis
2. 𝜙0 method: hadron 2
pT1, pT2
pT0

91. Francesca Giordano 44
𝜙0 yasymmetries

92. Francesca Giordano 45
𝜙0 yasymmetries

93. Francesca Giordano
yMore asymmetries
46
𝜙0