Uncovering quirk signal via energy loss inside tracker ... · M =1 0G eV,¤ 2 M =1 0G eV,¤ 2 M =1 0G eV,¤ 6 M=500 GeV, ¤=200 eV M=500 GeV, ¤=2000 eV M=500 GeV, ¤=600 eV Figure:The

Uncovering quirk signal via energy loss inside trackerWenxing Zhang

Wenxing Zhang

Institute of Theoretical Physics,Chinese Academy of Sciences,

Beijing, China

November 27, 2019

Wenxing Zhang ITP, CAS [email protected] November 27, 2019 1 / 17

Quirks are long lived exotic particles that are charged under both the StandardModel (SM) gauge group and a new confining gauge group with the confiningenergy equal to Λ .

Little Hierarchy Problem

Folded Supersymmetry

Quirky Little Higgs

Twin Higgs

` = 1cm × (1 keV

Λ)2 × (

mQ100 GeV

) . (1)

Λ� mQ , Fs ∝ Λ2

Λ & O(10 MeV)→ Intensive oscillations

Λ ∈ [10 keV, 10 MeV]→ Microscopic

Λ . O(10eV)→ Helical trajectory

Λ ∼ 100 eV-10 keV→ Macroscopic andVisible


X/cm

10050

050

100150

Y/cm

5025

025

5075

100

Z/cm

100

50

0

50

100

150

X/cm

020

4060

80100

Y/cm

0

20

40

60

80

100

Z/cm

100

50

0

50

100

X/cm

020

4060

80100

Y/cm

0

2040

6080

100

Z/cm

100

50

0

50

100

Figure: The quirk pair trajectories inside the CMS tracker, with the confinement scale chosen as 1 eV, 1 keV and 2keV (from left to right) for illustration. The quirk system in different panels have common initial momentumand quirk mass is 100 GeV. The cylinder segments indicate the tracking layer.


Quirks at the LHC

Dc = (3, 3, 1, 2/3) , (2)

Ec = (1, 1, 1,−2) , (3)

Dc = (3, 3, 1, 2/3) , (4)

Ec = (1, 1, 1,−2) , (5)

Dc and Ec are spin zero particles, and Dc

and Ec are fermions. The electric chargesof Dc /Dc and Ec /Ec are 1

3 and -1respectively.Due to the color confinement, one can onlyobserve hadrons of quirk-quark boundstate for Dc and Dc . The probability forfinal state hadrons having charge ±1 isaround 30%.

Figure: Production processes of quirks with different rep-resentations at the LHC.

Figure: The production cross sections for different quirksat 13 TeV LHC. We have also required an ISR jetwith pt (jet) > 100 GeV in production processes.

Initial state radiated (ISR) jet satisfypT > 100 GeV.


Quirk equation of motion: synchronism

The equation of motion for quirks is

∂~P∂t

= Λ2√

1− ~v2⊥s − Λ2 v‖~v⊥√

1− ~v2⊥

+ ~Fext

~Fext = q~v × ~B − 〈dEdx〉v (6)

Interaction between two particles is happening in the centre of mass (CM) framesimultaneously.Suppose the two quirks are simultaneous in the CM frame at some point tCM

i , i = 1, 2.

tCM2 = γ(tLab

2 − ~β ·~rLab2 ) (7)

tCM1 = γ(tLab

1 − ~β ·~rLab1 ) (8)

tCM2 − tCM

1 = γ[tLab1 − tLab

2 − ~β · (~rLab1 −~rLab

2 )]

= 0 (9)

tLab1 − tLab

2 = ~β · (~rLab1 −~rLab

2 ) (10)

Simultaneity conditionthus translates intosatisfying the eq. (10).


Suppose at one point two quirks are simultaneous in the CM frame, and map the pointfrom CM frame to Lab frame. All the variables below are defined in the Lab frame.

In the next step, we have:

t ′i = ti + εi

~P′i = ~Pi + εi ~Fi

E ′i =

√m2 + ~P′2i

~r ′i = ~ri +εi

2(~vi +

~P′iE ′i

)

~β′ =~P′1 + ~P′1E ′1 + E ′2

i = 1, 2. (11)

If in the next step the two quirks are alsosimultaneous in CM frame, they should satisfy

t ′1 − t ′2 = t1 + ε1 − (t2 + ε2)

= ~β′

(~r1 − ~r2 +

ε1

2(~v1 +

~p′1E ′1

)−ε2

2(~v2 +

~p′2E ′2

)

)(12)

Taylor-expand Eq(12) to O(εi ), we have

ε1[1− ~v1 · ~β −~r1 −~r2

E1 + E2· (~F1 − ~v1 · ~F1~β)] =

ε2[1− ~v2 · ~β −~r2 −~r1

E1 + E2· (~F2 − ~v2 · ~F2~β)] (13)


Quirk equation of motion: direction of s

Boost the system from the CM frame to thelab frame

α

x

y

2

1

𝛽

y

x 1

2

θ

F ′x =Fx

1 + βvy

√1− β2

F ′y = Fy +βvx Fx

1 + βvy(14)

F ′|| = −√

1− v ′2⊥Λ2 =Fx

1 + βvy

√1− β2

F ′⊥ = −v⊥′v||′√1− v ′2⊥

Λ2 = Fy +βvx Fx

1 + βvy

(15)

F ′|| = F ′x sinα+ F ′y cosα

F ′⊥ = −F ′x cosα+ F ′y sinα (16)

tanα =tan θ − βvx + βvy tan θ√

1− β2(17)

~rs1 = (1− β2)[(~r1 −~r2)− ~β(t1 − t2)] + [(~r1 −~r2)− ~β(t1 − t2)] · ~β(~β − ~v1)

~rs2 = (1− β2)[(~r2 −~r1)− ~β(t2 − t1)] + [(~r2 −~r1)− ~β(t2 − t1)] · ~β(~β − ~v2) (18)


Quirk equation of motion: 〈dEdx 〉

The average ionization energy loss of charged particle can be expressed as function ofvelocity in the Bethe-Bloch (BB) and Lindhard-Scharff (LS) regions.

〈dEdx

(v)LS〉 = A1v

〈dEdx

(v)BB〉 = A2q2

v2ln

(A3v2

1− v2− v2

)

A1 = (3.1× 10−11 GeV2)ρ

g/cm3

q7/6ZA(q2/3 + Z 2/3)3/2

,

A2 = (6.03× 10−18 GeV2)ρ

g/cm3

ZA,

A3 =102200

Z.

with the A, Z and ρ correspond to therelative atomic mass, atomic number anddensity of material, respectively.

The ionization energy loss function inthe region between LS and BB areinterpolated from experimental data.

Si

Cu

PbWO4

Iron

10-4 0.001 0.010 0.100 1 10

5

10

50

100

500

1000

βγ

<dE

/dx>

(MeV

/cm)

Energy loss


0.0 0.2 0.4 0.6 0.8 1.0

vT/c

0.00

0.05

0.10

0.15

0.20

0.25

Pro

babili

ty

MQ=100 GeV

MQ=500 GeV

MQ=700 GeV

0 2 4 6 8 10

Energy Loss [GeV]

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Pro

babili

ty

ECal MQ=100 GeV

ECal MQ=500 GeV

ECal MQ=700 GeV

HCal MQ=100 GeV

HCal MQ=500 GeV

HCal MQ=700 GeV

Figure: The transverse velocity of the quirk-antiquirk system for different quirk masses (left panel). The reconstructedmissing transverse energy for the background process and a few signal processes (right panel).


Numerical solution of the EoM

The equation of motion for quirks is

∂~P∂t

= Λ2√

1− ~v2⊥s − Λ2 v‖~v⊥√

1− ~v2⊥

+ ~Fext

~Fext = q~v × ~B − 〈dEdx〉v (19)

Confinement Scale: Λ = 1keV.

∆t ≤ T ∝ mQ/Λ2,∆t ∼ 10−4ns

Evolution time no longer than 25 ns.

Propagating outside HCal. 0 10 20 30 40 50Hit Number

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

Pro

babili

ty

M=100 GeV, Λ =200 eV

M=100 GeV, Λ =2000 eV

M=100 GeV, Λ =600 eV

Figure: The total number hits for quirk system travelingthrough the tracker of CMS detector.


For quirk-antiquirk system propagates inthe tracker, it leaves N hits located at ~hi(i = 1, 2, . . . ,N) when crosses the detectorlayers. Define

d(~n) =

√√√√ 1N − 1

N∑i=1

(~n · ~hi )2 (20)

dplane = d(~n)min (21)0 200 400 600 800 1000

thickness [µm]

10-4

10-3

10-2

10-1

100

Pro

babili

ty

M=100 GeV, Λ =200 eVM=100 GeV, Λ =2000 eVM=100 GeV, Λ =600 eVM=500 GeV, Λ =200 eVM=500 GeV, Λ =2000 eVM=500 GeV, Λ =600 eV

Figure: The reconstructed thickness of two quirks trajec-tories arises from the recorded hits.

dmin ∼ 1.16(

m100 GeV

)(keV

Λ

)4 ( |q|e

)(Bxy

T

)µm, (22)


Signal and BKG Analysis at the LHC

Simulation for SignalConfinement Scale: Λ = 1keV.

ISR Jet: pT ≥ 100GeV.

Pile Up: 〈µ〉 = 50.

Error of 〈 dEdx 〉: δ〈

dEdx 〉 = 0.05〈 dE

dx 〉

Analyzation of BKG processesZ (→ νν) + jets: pT (νν) > 200 GeV.

σZ (→νν)jj =3.6pb

Z (νν)e+e−+ jet: pT (νν) > 100GeV,pT (e±) > 1 GeV.

σZ (→νν)e+e− j =2.5fb

Pile Up: 〈µ〉 = 50.

Error of 〈 dEdx 〉: δ〈

dEdx 〉 = 0.05〈 dE

dx 〉

They will give O(104) hits inside the tracker.


Quirk Signal Selection

We have about O(104) hits inside the tracker for both signal and BKG.

a1. Remove all the hits with dEdx ≤ 3.0

MeV/cm.

a2. For those remaining hits, we divide allof them into different classes based ontheir dE/dx .(

dEdx

)a

aver=

1Na

Σi=Nai=1

(dEdx

)i,

(23)

If the next hit satisfies(dEdx

)next−(

dEdx

)c

aver< 1.0 MeV/cm

(24), the hit is assigned to the cth class.

a3. For classes containing less than 80hits are selected out and consideredas quirk hits candidate .

a4. For classes that have more than 80hits, we remove the sets of hits whichcan be reconstructed as helix.The remaining hits are hits of theseclasses are considered as quirk hitscandidate .This step is found to be very useful forsuppressing pileup events.


a5. For classes that have more than 80hits, The location of the possiblecenter of circles (COC) in transverseplane for any two hits in the class isgiven by

x0 =1

2 (k1 − k2)

[k1x2

(1 + k2

2

)− k2x1

(1 + k2

1

)],

(25)

y0 = −x0

k1+

x1

2k1+

y1

2, (26)

k1 =y1

x1, k2 =

y2

x2. (27)

a6. Find the center of circle.

a7. Remove the hits located on the circlewith δR ≤ 0.1cm.

mQ/GeV 200 500 800 BKG

ε1500 0.903 0.899 0.871 0.153

Table: The signal and background reduction efficiencies(ε1500 = Nsig

left/Nsigtot ) after iteration until the total hit

number (Nsig + Npile ) less than 1500.

b1. The plane normal vector can bedetermined for any two of theremaining hits:

~n = 〈~r1 × ~r2〉 (28)

b2. Account the total number Nleft of hitswithin the distance of 30 µm to thegiven plane.

di = |~ri × ~n| (29)


c1. Based on the reconstructed quirkplane, define

Sp =N∑

i=0

e−di/d0 ×(

dEdx

)i, (30)

∆p =|Sp − Sp′|

Sp(31)

The angle between Sp and S′p is π12 .

1. EmissT ≥ 500 GeV,

2. ∆p ≥ 0.8,

3. Sp ≥ 50 MeV/cm,

4. Nleft ≥ 10.

10 20 30 40 50

Nleft

10-4

10-3

10-2

10-1

100

Pro

babili

ty

MQ=200 GeV

MQ=500 GeV

MQ=800 GeV

Backgrounds

0 50 100 150 200 250 300

Sp [MeV/cm]

10-3

10-2

10-1

100

Pro

babili

ty

MQ=200 GeV

MQ=500 GeV

MQ=800 GeV

Backgrounds

Figure: Distributions for number of hits within the d <30 µm of quirk plane (upper) and the distance-weighted ionization energy loss Sp (lower).


Figure: The 95% C.L. exclusion limit for quirks with different quantum numbers at varying integrated luminosity of 13TeV LHC.


Summary

For the parameter space of interest (i.e. λ ∼ O(100− 1000) eV, mQ ∼ O(100)GeV, pT (QQ) & 100 GeV), most of the quirk pair can leave total number of ∼ 26hits inside the tracker. Those hits are lie on the plane with thickness . O(100) µm.

The main background for the quirk signal will be abundant pileup events (which wetake 〈µ〉 = 50) as well as the SM Z (→ νν)+jets process.Three discriminative variables can be defined:

Nleft within the d < 30 µmdistance-weighted ionization energy loss: Spderivative of Sp : ∆p

D/D is excluded up to 2.1/1.1 TeV.E /E is excluded up to 450/150 GeV.

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Documents

Uncovering quirk signal via energy loss inside tracker ... · M =1 0G eV,¤ 2 M =1 0G eV,¤ 2 M =1 0G eV,¤ 6 M=500 GeV, ¤=200 eV M=500 GeV, ¤=2000 eV M=500 GeV, ¤=600 eV Figure:The