Playing with Spin Quantum Numbers at Colliders · Playing with Spin Quantum Numbers at Colliders Ritesh K. Singh Indian Institute of Science Education & Research Kolkata at Collider

Playing with Spin Quantum Numbers at Colliders

Ritesh K. Singh

Indian Institute of Science Education & Research Kolkata

at

Collider Physics: Events, Analysis and QCD

Indian Insitute of Technology Guwahati

March 29, 2017

Ritesh Singh Spin & polarization 1 / 30

The Spin CodeThe spin quantum numberSpin through the polar angleSpin through the azimuthal angle

The angular distributionsThe density matrixSpin-1/2 caseSpin-1 case

Application: Anomalous Triple Gauge CouplingsThe Anomalous LagrangianILC at 500 GeV and 100 fb−1

Z boson at LHC

Extending the observables

Conclusions


The Spin Code

The Spin Code


The Spin CodeThe spin quantum number

Spin quantum number

I Spin is the only internal quantum number of a particle that isrelated to the space-time transformation.

I Spin determines the Lorentz structure of the couplings of theparticles with other particles of known spins.

I i.e. the production and decay mechanisms are almost determined bythe spin of the particle.

Helicity amplitude for the decay |s, λ〉 → |s1, l1〉+ |s2, l2〉 is

Msλl1 l2 (θ, φ) =

√2s + 1

4πDs∗λl (φ, θ,−φ)Ms

l1,l2

=

√2s + 1

4πe i(λ−l)φd s

λl(θ)Msl1,l2 , l = l1 − l2.



Spin quantum number






√2s + 1


l1,l2

=

√2s + 1

4πe i(λ−l)φd s

λl(θ)Msl1,l2 , l = l1 − l2.



Spin quantum number






√2s + 1


l1,l2

=

√2s + 1

4πe i(λ−l)φd s

λl(θ)Msl1,l2 , l = l1 − l2.



Spin quantum number






√2s + 1


l1,l2

=

√2s + 1

4πe i(λ−l)φd s

λl(θ)Msl1,l2 , l = l1 − l2.



Spin quantum number






√2s + 1


l1,l2

=

√2s + 1

4πe i(λ−l)φd s

λl(θ)Msl1,l2 , l = l1 − l2.



Determination of spin

The spin can be determined by

I exploiting the behaviour of the total cross-section at threshold forpair production or the threshold behaviour in the off-shell decay ofthe particle.

I distribution in the production angle relying on a known productionmechanism.

I comparing different spin hypotheses for a given collider signature, forexample, SUSY vs UED

I extracting the (cos θ)2s polar angle dependence or cos 2sφ azimuthalangle dependence of the decay distributions


































The Spin CodeSpin through the polar angle

Polar angle distribution

A

B

E

D

C

MλA,λB

λD ,λE(θBD , φ) = (2s + 1) d s

λi ,λf(θBD) e iφ(λi−λf ) Ms

λi ,λf

λi = λB − λA and λf = λD − λE

d sλi ,λf

(θ) for spin s particle is 2s degree polynomial incos(θ/2) and sin(θ/2), This distribution is frame invariant and can beused in decay chains to determine the spins of all on-shell particles.




A

B

E

D

C

MλA,λB

λD ,λE(θBD , φ) = (2s + 1) d s

λi ,λf(θBD) e iφ(λi−λf ) Ms

λi ,λf

λi = λB − λA and λf = λD − λE

d sλi ,λf

(θ) for spin s particle is 2s degree polynomial incos(θ/2) and sin(θ/2),

This distribution is frame invariant and can beused in decay chains to determine the spins of all on-shell particles.




dΓ(A→ BDE )

d cos θBD= Q0 + Q1 cos θBD + ... + Q2s cos2s θBD .

(pD + pE )2 = p2C = m2

C = constant. C is on-shell.

With m2BD = (pB + pD)2 and dm2

BD = 2EBEDβBβDd cos θBD

dΓ(A→ BDE )

dm2BD

= P0 + P1 m2BD + ... + P2s (m2

BD)2s .

This distribution is frame invariant and can be used in decay chains todetermine the spins of all on-shell particles.




dΓ(A→ BDE )


(pD + pE )2 = p2C = m2




dΓ(A→ BDE )

dm2BD

= P0 + P1 m2BD + ... + P2s (m2

BD)2s .





dΓ(A→ BDE )


(pD + pE )2 = p2C = m2




dΓ(A→ BDE )

dm2BD

= P0 + P1 m2BD + ... + P2s (m2

BD)2s .



The Spin CodeSpin through the azimuthal angle

Azimuthal angle distribution

Owing to the quantum interference of the different helicity amplitudes,i.e., the presence of transverse polarization, the azimuthal distribution ofdecay particles are given by

dΓ

dφ= a0 +

2s∑j=1

aj cos(jφ) +2s∑j=1

bj sin(jφ),

in the rest frame of decaying particle.

I aj are the CP-even contributions

I bj are the CP-odd contributions

I a2s/a0 6= 0 or b2s/a0 6= 0 implies spin of the decaying particle is s.

Next we will show how the above form of azimuthal distribution arises.




Owing to the quantum interference of the different helicity amplitudes,

i.e., the presence of transverse polarization, the azimuthal distribution ofdecay particles are given by

dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),









Owing to the quantum interference of the different helicity amplitudes,i.e., the presence of transverse polarization,

the azimuthal distribution ofdecay particles are given by

dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),










dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),










dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),










dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),










dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),










dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),










dΓ

dφ= a0 +

2s∑j=1


bj sin(jφ),







The angular distributions

The angular distributions


The angular distributionsThe density matrix

The process

We look at the production process B1B2 → A A1 ... An−1 followed by thedecay of A as A→ BC . The differential cross-section is given by

dσ =∑λ,λ′

[(2π)4

2Iρ(λ, λ′)δ4

(kB1 + kB2 − pA −

( n−1∑i

pi

))

× d3pA2EA(2π)3

n−1∏i

d3pi2Ei (2π)3

]

×[

1

ΓA

(2π)4

2mAΓ′(λ, λ′)δ4(pA − pB − pC )

d3pB2EB(2π)3

d3pC2EC (2π)3

]

First bracket = σ(λ, λ′) = σA PA(λ, λ′)

Second bracket =BBC (2s + 1)

4πΓA(λ, λ′)dΩB



The process


dσ =∑λ,λ′

[(2π)4

2Iρ(λ, λ′)δ4

(kB1 + kB2 − pA −

( n−1∑i

pi

))

× d3pA2EA(2π)3

n−1∏i

d3pi2Ei (2π)3

]

×[

1

ΓA

(2π)4


d3pB2EB(2π)3

d3pC2EC (2π)3

]






The process


dσ =∑λ,λ′

[(2π)4

2Iρ(λ, λ′)δ4

(kB1 + kB2 − pA −

( n−1∑i

pi

))

× d3pA2EA(2π)3

n−1∏i

d3pi2Ei (2π)3

]

×[

1

ΓA

(2π)4


d3pB2EB(2π)3

d3pC2EC (2π)3

]






The angular distribution

1

σ

dσ

dΩB=

2s + 1

4π

∑λ,λ′

PA(λ, λ′) ΓA(λ, λ′),

I σ = BBC σA is the cross-section of production of A and its decayinto BC .

I BBC is the branching ration of A into BC .

I PA(λ, λ′) = σ(λ, λ′)/σA is the polarization density matrix.

I ΓA(λ, λ′) is the normalized decay density matrix in the rest frame ofA.




1

σ

dσ

dΩB=

2s + 1

4π

∑λ,λ′









1

σ

dσ

dΩB=

2s + 1

4π

∑λ,λ′









1

σ

dσ

dΩB=

2s + 1

4π

∑λ,λ′









1

σ

dσ

dΩB=

2s + 1

4π

∑λ,λ′








The decay density matrix

The decay density matrix for the decay proess A→ BC is given by

Γ′s(λ, λ′) =∑l1,l2

Msλl1l2M

∗sλ′l1l2

=

(2s + 1

4π

)e i(λ−λ

′)φ∑l1,l2

d sλl(θ)d s

λ′l(θ) |Msl1,l2 |

2

= e i(λ−λ′)φ∑l

d sλl(θ)d s

λ′l(θ)

[∑l1

(2s + 1

4π

)|Ms

l1,l1−l |2

]= e i(λ−λ

′)φ∑l

d sλl(θ)d s

λ′l(θ) asl

|l1| ≤ s1, |l1 − l | ≤ s2, |l | ≤ s and Tr(Γ′s(λ, λ′)) =∑

l asl .





Γ′s(λ, λ′) =∑l1,l2

Msλl1l2M

∗sλ′l1l2

=

(2s + 1

4π

)e i(λ−λ

′)φ∑l1,l2

d sλl(θ)d s


2


d sλl(θ)d s

λ′l(θ)

[∑l1

(2s + 1

4π

)|Ms

l1,l1−l |2

]= e i(λ−λ

′)φ∑l

d sλl(θ)d s

λ′l(θ) asl


l asl .





Γ′s(λ, λ′) =∑l1,l2

Msλl1l2M

∗sλ′l1l2

=

(2s + 1

4π

)e i(λ−λ

′)φ∑l1,l2

d sλl(θ)d s


2


d sλl(θ)d s

λ′l(θ)

[∑l1

(2s + 1

4π

)|Ms

l1,l1−l |2

]= e i(λ−λ

′)φ∑l

d sλl(θ)d s

λ′l(θ) asl


l asl .





Γ′s(λ, λ′) =∑l1,l2

Msλl1l2M

∗sλ′l1l2

=

(2s + 1

4π

)e i(λ−λ

′)φ∑l1,l2

d sλl(θ)d s


2


d sλl(θ)d s

λ′l(θ)

[∑l1

(2s + 1

4π

)|Ms

l1,l1−l |2

]


d sλl(θ)d s

λ′l(θ) asl


l asl .





Γ′s(λ, λ′) =∑l1,l2

Msλl1l2M

∗sλ′l1l2

=

(2s + 1

4π

)e i(λ−λ

′)φ∑l1,l2

d sλl(θ)d s


2


d sλl(θ)d s

λ′l(θ)

[∑l1

(2s + 1

4π

)|Ms

l1,l1−l |2

]= e i(λ−λ

′)φ∑l

d sλl(θ)d s

λ′l(θ) asl


l asl .





Γ′s(λ, λ′) =∑l1,l2

Msλl1l2M

∗sλ′l1l2

=

(2s + 1

4π

)e i(λ−λ

′)φ∑l1,l2

d sλl(θ)d s


2


d sλl(θ)d s

λ′l(θ)

[∑l1

(2s + 1

4π

)|Ms

l1,l1−l |2

]= e i(λ−λ

′)φ∑l

d sλl(θ)d s

λ′l(θ) asl


l asl .



The final distribution

The normalized decay density matrix is given by

ΓA(λ, λ′) = e i(λ−λ′)φ

∑l d

sλl(θ)d s

λ′l(θ)asl∑l a

sl

= e i(λ−λ′)φ γA(λ, λ′; θ),

and the final distribution is given by

1

σ

dσ

dΩB=

2s + 1

4π

[ ∑λ

PA(λ, λ) γA(λ, λ)

+∑λ 6=λ′

<[PA(λ, λ′)] γA(λ, λ′) cos((λ− λ′)φ)

−∑λ6=λ′

=[PA(λ, λ′)] γA(λ, λ′) sin((λ− λ′)φ)

.






∑l d

sλl(θ)d s

λ′l(θ)asl∑l a

sl

= e i(λ−λ′)φ γA(λ, λ′; θ),


1

σ

dσ

dΩB=

2s + 1

4π

[ ∑λ


+∑λ 6=λ′

<[PA(λ, λ′)] γA(λ, λ′) cos((λ− λ′)φ)

−∑λ6=λ′

=[PA(λ, λ′)] γA(λ, λ′) sin((λ− λ′)φ)

.






∑l d

sλl(θ)d s

λ′l(θ)asl∑l a

sl

= e i(λ−λ′)φ γA(λ, λ′; θ),


1

σ

dσ

dΩB=

2s + 1

4π

[ ∑λ


+∑λ 6=λ′

<[PA(λ, λ′)] γA(λ, λ′) cos((λ− λ′)φ)

−∑λ6=λ′

=[PA(λ, λ′)] γA(λ, λ′) sin((λ− λ′)φ)

.


The angular distributionsSpin-1/2 case

The density matrix: s = 1/2

For | 12 , l〉 → |s1, l1〉+ |s2, l2〉

Γ 12(λ, λ′) =

1+α cos θ2

α sin θ2 e iφ

α sin θ2 e−iφ 1−α cos θ

2

,Here α = (a

1/21/2 − a

1/2−1/2)/(a

1/21/2 + a

1/2−1/2) and

a1/21/2 =

(1

2π

)∑l1

|M1/2l1,l1−1/2|

2 |l1| ≤ s1, |l1 − 1/2| ≤ s2

a1/2−1/2 =

(1

2π

)∑l1

|M1/2l1,l1+1/2|

2 |l1| ≤ s1, |l1 + 1/2| ≤ s2.



Angular distribution: s = 1/2

Polarization density matrix:

P 12(λ, λ′) =

1

2

1 + η3 η1 − iη2

η1 + iη2 1− η3

,Thus the angular distribution becomes:

1

σ1

dσ1

dΩB=

1

4π[1 + αη3 cos θ + αη1 sin θ cosφ+ αη2 sin θ sinφ] .

The cos θ averaged azimuthal distribution is given by

1

σ1

dσ1

dφ=

1

2π

[1 +

αη1π

4cosφ+

αη2π

4sinφ

].



Angular distribution: e+e− → tt√s = 400GeV, η1 = −0.75, η2 ≈ 0, η3 = −0.19

CT_l0Entries 1000000Mean -0.06222RMS 0.5739

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

5000

10000

15000

20000

25000

CT_l0Entries 1000000Mean -0.06222RMS 0.5739

Cos(th_l0)

PH_l0Entries 1000000

Mean -0.0008276

RMS 0.6728

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

5000

10000

15000

20000

25000

30000

35000

PH_l0Entries 1000000

Mean -0.0008276

RMS 0.6728

Phi_l0

CT_b0Entries 1000000Mean 0.02348RMS 0.5769

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

5000

10000

15000

20000

25000

CT_b0Entries 1000000Mean 0.02348RMS 0.5769

Cos(th_b0)

PH_b0Entries 1000000Mean 0.001216RMS 0.5348

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 10

5000

10000

15000

20000

25000

30000

35000

PH_b0Entries 1000000Mean 0.001216RMS 0.5348

Phi_b0



Angular distribution: e+e− → tt√s = 400GeV, η1 = −0.75, η2 ≈ 0, η3 = −0.19

PH_bEntries 1000000Mean 0.4225RMS 0.2838

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5000

10000

15000

20000

25000

30000

35000

PH_bEntries 1000000Mean 0.4225RMS 0.2838

Phi_b

PH_lEntries 1000000Mean 0.4467RMS 0.2861

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5000

10000

15000

20000

25000

30000

35000PH_l

Entries 1000000Mean 0.4467RMS 0.2861

Phi_l


The angular distributionsSpin-1 case

The density matrix: s = 1

For |1, l〉 → |s1, l1〉+ |s2, l2〉,

Γ1(l, l′) =

1+δ+(1−3δ) cos2 θ+2α cos θ

4sin θ(α+(1−3δ) cos θ)

2√

2eiφ (1 − 3δ)

(1−cos2 θ)4

ei2φ

sin θ(α+(1−3δ) cos θ)

2√

2e−iφ δ + (1 − 3δ) sin2 θ

2sin θ(α−(1−3δ) cos θ)

2√

2eiφ

(1 − 3δ)(1−cos2 θ)

4e−i2φ sin θ(α−(1−3δ) cos θ)

2√

2e−iφ 1+δ+(1−3δ) cos2 θ−2α cos θ

4

,

where,

α =a1

1 − a1−1

a11

+ a10

+ a1−1

, δ =a1

0

a11

+ a10

+ a1−1

and

a11 =

( 3

4π

)∑l1

|M1l1,l1−1|

2 |l1| ≤ s1, |l1 − 1| ≤ s2

a10 =

( 3

4π

)∑l1

|M1l1,l1|2 |l1| ≤ min(s1, s2)

a1−1 =

( 3

4π

)∑l1

|M1l1,l1+1|

2 |l1| ≤ s1, |l1 + 1| ≤ s2



Angular distribution: s = 1

Polarization density matrix:

P1(λ, λ′) =

13

+ pz2

+ Tzz√6

px−ipy

2√

2+

Txz−iTyz√3

Txx−Tyy−2iTxy√6

px+ipy

2√

2+

Txz+iTyz√3

13− 2Tzz√

6

px−ipy

2√

2− Txz−iTyz√

3

Txx−Tyy +2iTxy√6

px+ipy

2√

2− Txz+iTyz√

3

13− pz

2+ Tzz√

6

,The angular distribution is:

1

σ

dσ

dΩ=

3

8π

[(2

3− (1− 3δ)

Tzz√6

)+ α pz cos θ +

√3

2(1− 3δ) Tzz cos2 θ

+

(α px + 2

√2

3(1− 3δ) Txz cos θ

)sin θ cosφ

+

(α py + 2

√2

3(1− 3δ) Tyz cos θ

)sin θ sinφ

+ (1− 3δ)

(Txx − Tyy√

6

)sin2 θ cos(2φ) +

√2

3(1− 3δ) Txy sin2 θ sin(2φ)

]Ritesh Singh Spin & polarization 17 / 30


Polarization asymmetries: e+e− → ZZ

Define I (θ, φ) = (1/σ)(dσ/dΩ), then we have

Alr =

[ ∫ π

θ=0

∫ π2

φ=−π2I (θ, φ) sin(θ)dθdφ−

∫ π

θ=0

∫ 3π2

φ=π2

I (θ, φ) sin(θ)dθdφ

]=

3αpx4

=σ(sx .p < 0)− σ(sx .p > 0)

σ(sx .p < 0) + σ(sx .p > 0)

Here sx = (0, 1, 0, 0) in the rest frame of the decaying particle.

Similarly one can define asymmetries for other polarization parameters aswell.




0 500 1000 1500 2000

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

Beam Energy

Px

0 500 1000 1500 20000.0

0.1

0.2

0.3

0.4

Beam EnergyT

zz




0 500 1000 1500 2000-0.24

-0.22

-0.20

-0.18

-0.16

-0.14

-0.12

Beam Energy

Txx

0 500 1000 1500 2000

-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

Beam Energy

Tyy

0 500 1000 1500 2000-0.25

-0.20

-0.15

-0.10

-0.05

0.00

0.05

Beam Energy

Txx-T

yy

0 500 1000 1500 2000

-0.006

-0.004

-0.002

0.000

0.002

0.004

0.006

Beam Energy

Pz



Polarization asymmetries: e+e− → Zγ

0 500 1000 1500 2000-0.05

0.00

0.05

0.10

0.15

Beam Energy

Px

0 500 1000 1500 2000

0.1

0.2

0.3

0.4

Beam EnergyT

zz



Polarization asymmetries: e+e− → Zγ

0 500 1000 1500 2000-0.25

-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

Beam Energy

Txx

0 500 1000 1500 2000-0.24

-0.23

-0.22

-0.21

-0.20

-0.19

-0.18

-0.17

-0.16

Beam Energy

Tyy

0 500 1000 1500 2000-0.05

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

Beam Energy

Txx-T

yy

0 500 1000 1500 2000-0.03

-0.02

-0.01

0.00

0.01

0.02

Beam Energy

Pz


Application: Anomalous Triple Gauge Couplings

Application: Anomalous Triple Gauge Couplings


Application: Anomalous Triple Gauge CouplingsThe Anomalous Lagrangian

The Lagrangian and Vertex

LaTGC =ge

M2Z

[− [f γ4 (∂µF

µβ) + f Z4 (∂µZµβ)]Zα(∂αZβ) + [f γ5 (∂σFσµ) + f Z5 (∂σZσµ)]ZµβZβ

− [hγ1 (∂σFσµ) + hZ1 (∂σZσµ)]ZβFµβ − [hγ3 (∂σF

σρ) + hZ3 (∂σZσρ)]ZαFρα

].

For e+e− → ZZ

ΓµνσV?ZZ

(q, k1, k2) = −ge

M2Z

(q2 −M2

V

)[f V4(qσgµν + qνgµσ

)− f V5 ε

µνσα(k1 − k2)α

].

For e+e− → Zγ

ΓµνσV?γZ

(q, k1, k2) =ge

M2Z

(q2 −M2

V

)[hV1(kµ1 gνσ − kσ1 gµν

)− hV3 ε

µνσαk1α

].


Application: Anomalous Triple Gauge CouplingsILC at 500 GeV and 100 fb−1

Process e+e− → ZZ

Observables Linear terms Quadratic terms

σ f Z5 , fγ

5 (f γ4 )2, (f γ5 )2, (f Z4 )2, (f Z5 )2, f γ4 f Z4 , fγ

5 f Z5

σ × Ax f γ5 , fZ

5 −

σ × Ay f γ4 , fZ

4 −

σ × Axy f Z4 , fγ

4 f Z4 f γ5 , fγ

4 f Z5 , fγ

4 f γ5 , fZ

4 f Z5

σ × Ax2−y2 f Z5 , fγ

5 (f γ4 )2, (f γ5 )2, (f Z4 )2, (f Z5 )2, f γ4 f Z4 , fγ

5 f Z5

σ × Azz f Z5 , fγ

5 (f γ4 )2, (f γ5 )2, (f Z4 )2, (f Z5 )2, f γ4 f Z4 , fγ

5 f Z5



Process e+e− → ZZ

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

f4γ(10

-3)

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

f5γ(10

-3)

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

f4z(10

-3)

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

f5z(10

-3)

S( σ)S(Ax)S(Ay)

S(Axy)S(Ax

2-y

2)S(Azz)



Process e+e− → Zγ

Observables Linear terms Quadratic terms

σ hZ3 , hγ3 (hγ1 )2, (hγ3 )2, (hZ1 )2, (hZ3 )2, hγ1 h

Z1 , h

γ3 h

Z3

σ × Ax hZ3 , hγ3 (hγ1 )2, (hγ3 )2, (hZ1 )2, (hZ3 )2, hγ1 h

Z1 , h

γ3 h

Z3

σ × Ay hγ1 , hZ1 −

σ × Axy hγ1 , hZ1 −

σ × Ax2−y2 hγ3 , hZ3 −

σ × Azz hZ3 , hγ3 (hγ1 )2, (hγ3 )2, (hZ1 )2, (hZ3 )2, hγ1 h

Z1 , h

γ3 h

Z3



Process e+e− → Zγ

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

h1γ(10

-3)

S( σ)

S(Ax)

S(Ay)

S(Axy)

S(Ax2-y

2)

S(Azz)

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

h3γ(10

-3)

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

h1z(10

-3)

0

0.5

1

1.5

2

2.5

3

3.5

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

Sensitiv

ity

h3z(10

-3)



Best limits at ILC

ZZ process γZ proces

Coupling Limits Comes from Coupling Limits Comes from

f Z4 0.0± 4.2× 10−3 σ hZ1 0.0± 2.9× 10−3 Ay

f γ4 0.0± 2.4× 10−3 Ay hγ1 0.0± 3.6× 10−3 Axy , σ

f Z5 0.0+8.8×10−3

−2.3×10−3 σ hZ3 0.0± 2.8× 10−3 Ax

f γ5 0.0+2.7×10−3

−2.3×10−3 Ax , σ hγ3 0.0+1.3×10−3

−2.1×10−3 σ

0.0−6.5×10−3

−9.9×10−3


Z boson at LHC

The tri-lepton background at LHC

pp → ZW+ → (µ+µ−)(e+νe)

PH_NmEntries 100000

Mean 0.2997

RMS 0.2596

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1000

2000

3000

4000

5000

6000PH_Nm

Entries 100000

Mean 0.2997

RMS 0.2596

Delta phi nePH_mu

Entries 100000

Mean 0.3423

RMS 0.2729

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

500

1000

1500

2000

2500

3000

3500

4000

4500

5000PH_mu

Entries 100000

Mean 0.3423

RMS 0.2729

Delta phi mu-


Extending the observables

Beyond spin interpretation

Consider tri-lepton signal : µ+µ− e+ 6 E . We can define pll = p+ + p−and let’s choose pll = E (1, β sin θ, 0, β cos θ) and define:

sx = (0, − cos θ, 0, sin θ)

sy = (0, 0, 1, 0)

sz = (β, sin θ, 0, cos θ)/√p2ll

Various polarization asymmetries can be defined with respect tomomentum correlators like : sx .p+, sy .p+, sz .p+, (sx .p+)2 − (sy .p+)2,(sx .p+)(sy .p+) etc.

A distribution of these correlators can also be used as kinematicdiscriminator for enhancing signals (under progress).


Conclusions

to conclude ....

I Spin/polarization sensitive observables can be accessed at colliders.

I The polarization asymmetries can provide strong limits on some ofthe anomalous couplings.

I The distribution of polarization inspired correlator can be used tocut/select events of prefered kind for new physics searches.

Thank you !


Conclusions

to conclude ....




Thank you !


Conclusions

to conclude ....




Thank you !


Conclusions

to conclude ....




Thank you !


Conclusions

to conclude ....




Thank you !


Documents

Playing with Spin Quantum Numbers at Colliders · Playing with Spin Quantum Numbers at Colliders Ritesh K. Singh Indian Institute of Science Education & Research Kolkata at Collider