Uncovering quirk signal via energy loss inside tracker ... · M =1 0G eV,¤ 2 M =1 0G eV,¤ 2 M =1...
Transcript of Uncovering quirk signal via energy loss inside tracker ... · M =1 0G eV,¤ 2 M =1 0G eV,¤ 2 M =1...
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
Wenxing Zhang ITP, CAS [email protected] November 27, 2019 2 / 17
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.
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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.
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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).
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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)
Wenxing Zhang ITP, CAS [email protected] November 27, 2019 6 / 17
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)
Wenxing Zhang ITP, CAS [email protected] November 27, 2019 7 / 17
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
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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).
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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.
Wenxing Zhang ITP, CAS [email protected] November 27, 2019 10 / 17
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)
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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.
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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.
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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)
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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).
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Figure: The 95% C.L. exclusion limit for quirks with different quantum numbers at varying integrated luminosity of 13TeV LHC.
Wenxing Zhang ITP, CAS [email protected] November 27, 2019 16 / 17
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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Wenxing Zhang ITP, CAS [email protected] November 27, 2019 17 / 17