7.5 Reconstruction-Induced E ff ects in the Υ -Shape
7.5.4 From correctly reconstructed 1p1n to all reconstructed 1p1n Taus
The final step is to move from the set of correctly reconstructed 1p1n taus to all reconstructed 1p1n taus, as they are available in data. The reconstructed 1p1n taus produce theΥdistributions shown in Figure 7.11. In that figure, the blue triangles are the same distribution as the black dots from Figure7.10. The red triangles correspond toΥdistributions obtained from the (larger) set of all true taus, reconstructed as 1p1n.
The change for left-handed taus is such that the distribution is broader than the one for correctly classified 1p1n taus. For right-handed taus, the left peak is less pronounced when considering all true modes that are reconstructed as 1p1n. The contributions of the other true modes, mainly 1p0n and 1pXn, are shown in Figure7.12. The black dashes (red circles) in this figure correspond to the red (blue) triangles in Figure7.11. For both helicities, the misclassification 1pXn→1p1n (1p0n→1p1n) can be found at negative values and values around zero (at large positive values).
9AlthoughΥuses the energies of theπ±andπ0, it also connectsET(π±) andET(π0), because theηvalues of theπ±andπ0 are very similar. The distribution of∆R(π±, π0) is shown in Figure5.12c.
) / GeV π0 T( ) - E π± T( E
0 5 10 15 20 25 30 35 40 45 50
arbitrary units
0.02 0.04 0.06 0.08 0.1
(MC) τ τ
→ Sample: Z Cut: No cuts
τ Matched true True 1p1n, Reco 1p0n Truth helicity
L T, Match
τ
R T, Match
τ
(a)Difference in transverse energy of theπ±andπ0. The large tail in the distribution forτRight makes a correct reconstruction of theπ0harder.
) / GeV π0 T( E
0 1 2 3 4 5 6 7 8 9 10
arbitrary units
0.02 0.04 0.06 0.08 0.1
0.12 Sample: Z →ττ (MC) Cut: No cuts
τ Matched true True 1p1n, Reco 1p0n Truth helicity
L T, Match
τ
R T, Match
τ
(b)Transverse energy of theπ0. The majority of true 1p1n taus withΥ >0.7fromZ → ττdecays have aπ0 that falls below theET threshold inπ0reconstruction.
) / GeV π0 T( E
0 1 2 3 4 5 6 7 8 9 10
) / GeV0π( T) - E±π(TE
0 20 40 60 80 100 120
140 Sample: Z →ττ (MC) Cut: No cuts
τ Matched true True 1p1n, Reco 1p0n Truth helicity
L T, Match
τ
R T, Match
τ
(c)Energy difference as a function ofET(π0). The sharp edge is due to the requirementΥ>0.7.
Figure7.9: Transverse energy difference ofπ± and π0 (7.9a), transverse energy of π0 (7.9b) and the former as a function of the latter (7.9c) for true 1p1n taus, reconstructed as 1p0n withΥ>0.7to investig-ate the dent in Figure7.7.
In summary, the dent arises from low energetic π0 that fail the preselection cut, orπ0 that are badly re-constructed because of a much higher energyπ±that is hard to subtract correctly. The bad reconstruction most likely leads to a reconstructed energy below the threshold, so that theπ0 is not reconstructed at all.
More details are given in the text.
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
arbitrary units
0.01 0.02 0.03 0.04 0.05
0.06 Sample: Z →ττ (MC)
had
τtrue
matched to τreco
Cut:
τ Matched PanTau True & Reco 1p1n Truth helicity
true) π0 true, π± L (
R, Match
τ
reco) π0 true, π± L (
R, Match
τ
true) π0 reco, π± L (
R, Match
τ
reco) π0 reco, π± L (
R, Match
τ
(a)Left-handed, correctly reconstructed 1p1n taus.
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
arbitrary units
0.02 0.04 0.06 0.08 0.1
(MC) τ τ
→ Sample: Z
had
τtrue
matched to τreco
Cut:
τ Matched PanTau True & Reco 1p1n Truth helicity
true) π0 true, π± R (
R, Match
τ
reco) π0 true, π± R (
R, Match
τ
true) π0 reco, π± R (
R, Match
τ
reco) π0 reco, π± R (
R, Match
τ
(b)Right-handed, correctly reconstructed 1p1n taus.
Figure7.10:Υdistributions for left- and right-handed, correctly reconstructed 1p1n taus, using different variants of theπ±andπ0. The magenta stars calculateΥusing the trueπ±andπ0. The red (blue) triangles exchange the trueπ0(π±) with the reconstructed one. Finally, the black dots represent theΥdistribution calculated when using both reconstructedπ±andπ0.
Because of the good momentum resolution, exchanging the trueπ±with the reconstructed ones does not make a large difference. Analogously, due to the worse energy resolution ofπ0, moving from trueπ0to reconstructedπ0 changes the distribution.
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
arbitrary units
0.01 0.02 0.03 0.04
τ Matched PanTau
Truth helicity τLR, Match 1p1n → 1p1n
(a)Left-handed taus.
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
arbitrary units
0.01 0.02 0.03 0.04 0.05 0.06
0.07 Matched PanTau τ
Truth helicity τRR, Match 1p1n → 1p1n
(b)Right-handed taus.
Figure 7.11:Υ distributions for correctly reconstructed 1p1n true taus (blue) and all true taus that are recon-structed as 1p1n (red). TheΥfor reconstructed 1p1n left-handed taus is a bit broadened compared to the one for correctly reconstructed 1p1n taus. In the set of right-handed taus, the negativeΥvalues are shifted towards zero.
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
Number of PanTau taus
2000 4000 6000 8000 10000 12000 14000 16000 18000
20000 Sample: Z →ττ (MC) Cut: No cuts
τ Matched PanTau Truth helicity
1p1n
→ 1p0n
L R, Match
τ
1p1n
→ 1p1n
L R, Match
τ
1p1n
→ 1pXn
L R, Match
τ
1p1n
→ 3p0n
L R, Match
τ
1p1n
→ 3pXn
L R, Match
τ
(a)Left-handed taus.
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
Number of PanTau taus
2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 22000 24000
(MC) τ τ
→ Sample: Z Cut: No cuts
τ Matched PanTau Truth helicity
1p1n
→ 1p0n
R R, Match
τ
1p1n
→ 1p1n
R R, Match
τ
1p1n
→ 1pXn
R R, Match
τ
1p1n
→ 3p0n
R R, Match
τ
1p1n
→ 3pXn
R R, Match
τ
(b)Right-handed taus.
Figure7.12:Breakdown of theΥshape for true taus reconstructed as 1p1n into the different true modes that con-tribute. The black dashed line is the sum of all the other curves. The true 1pXn and true 1p0n that are reconstructed as 1p1n are the main sources for the changes observed here. Contributions fromτ3-prongare negligible.
This is somewhat expected, as negative Υ values indicate a high energeticπ0. Such a π0 can be produced if the twoπ0 from a 1pXn decay are close by and are not resolved10, so that they appear as one neutral pion. This pion then has the energy of the two pions from 1pXn, lowering the resultingΥ value.
For 1p0n→1p1n it is the other way around. Theπ0in the 1p1n is not a realπ0, but stems from either noise, pile-up or imperfect subtraction, making it very low energetic in comparison to the π± which really originates from the tau decay.
10There are such taus which are classified as 1pXn - although they only have one reconstructedπ0- due to the number of photons found in the shots in EM1, c.f. Section5.3.2. However, not all 1pXn taus are correctly classified because of this, and many 1pXn taus are migrated to 1p1n byPanTau(c.f. Table5.12).
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
arbitrary units
0.01 0.02 0.03 0.04 0.05 0.06
0.07 Sample: Z →ττ (MC) Polarisation variables τL
Matched PanTau True Any, Reco 1p1n Lab. rest frame
had
τtrue
matched to τreco
) > 20 GeV τreco T( p
τ from
µtrue
matched to µreco
) > 26 GeV τreco T( p
≤ 2 N(SCT Holes)
(a) Υ for reconstructed 1p1n taus, matched to a left-handed true tau.
0)]
π ) + E(
π±
)] / [E(
π0
) - E(
π±
[E(
-1.5 -1 -0.5 0 0.5 1
arbitrary units
0.02 0.04 0.06 0.08 0.1
0.12 Sample: Z →ττ (MC) Polarisation variables τR
Matched PanTau True Any, Reco 1p1n Lab. rest frame
had
τtrue
matched to τreco
) > 20 GeV τreco T( p
τ from
µtrue
matched to µreco
) > 26 GeV τreco T( p
≤ 2 N(SCT Holes)
(b)Υ for reconstructed 1p1n taus, matched to a right-handed true tau.
Figure7.13:Υshapes of reconstructed 1p1n taus that have a truth match at selected stages of the event selection.
The main differences can be observed for right-handed taus when applying thepT(τ)cut, as discussed in the text in Section7.3. Theno cutstage is not shown because the set of taus shown here implicitly has a cut applied to it, namely the match to a true tau.