1 / 28

¡(1 S ) Production at D Ø

¡(1 S ) Production at D Ø. Daniela Bauer, Jundong Huang, Andrzej Zieminski. Indiana University. Why measure ¡ (1S) production at D Ø ?. Because we can: The ¡ (1S) cross-section had been measured at the Tevatron (Run I measurement by CDF) up to a rapidity of 0.4.

isla
Télécharger la présentation

¡(1 S ) Production at D Ø

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ¡(1S) Production at DØ Daniela Bauer, Jundong Huang, Andrzej Zieminski Indiana University

  2. Why measure ¡(1S) production at DØ ? • Because we can: The ¡(1S) cross-section had been measured at the Tevatron (Run I measurement by CDF) up to a rapidity of 0.4. DØ has now measured this cross-section up to a rapidity of 1.8 at √s = 1.96 TeV • Measuring the ¡(1S) production cross-section provides an ideal testing ground for our understanding of the production mechanisms of heavy quarks. There is considerable interest from theorists in these kinds of measurements: E.L. Berger, J.Qiu, Y.Wang, Phys Rev D 71 034007 (2005) and hep-ph/0411026; V.A. Khoze , A.D. Martin, M.G. Ryskin, W.J. Stirling, hep-ph/0410020

  3. The Analysis Goal: Measuring the ¡(1S) cross-section in the channel ¡(1S) → μ+μ-as a function of pt in three rapidity ranges: 0 < | y¡| < 0.6, 0.6 < | y¡| < 1.2 and 1.2 < | y¡| < 1.8 Sample selection: • Opposite sign muons • Muon have hits in all three layers of the muon system • Muons are matched to a track in the central tracking system • pt (μ) > 3 GeV and |η (μ)| < 2.2 • At least one isolated muon • Track from central tracking system must have at least one hit in the Silicon Tracker

  4. The Analysis d2σ(¡(1S)) N(¡) = L × Δpt × Δy × εacc× εtrig× kdimu× ktrk× kqual dpt× dy L Luminosity kdimu local muon reconstruction y rapidity ktrk tracking εacc Acceptance kqual track quality cuts εtrig Trigger 0.0 < y < 0.6 0.6 < y < 1.2 1.2 < y < 1.8 εacc 0.15 - 0.26 0.19 – 0.28 0.20 - 0.27 εtrig 0.70 0.73 0.82 kdimu 0.85 0.88 0.95 ktrk 0.99 0.99 0.95 kqual 0.85 0.85 0.93

  5. Data Monte carlo comparison To determine our efficiencies, we only need an agreement between Monte Carlo and data within a give pTϒ and yϒbin and not an agreement over the whole pTϒ and yϒ range at once. MC Data* η(μ) φ(μ) pt(μ) in GeV 0 5 10 15 20 -2 -1 0 1 2 0 3 6 0.6 < | y¡| < 1.2 *9.0 GeV < m(μμ) < 9.8 GeV

  6. Fitting the signal Signal: 3 Gaussians: ¡(1S), ¡(2S), ¡(3S) Fitting a single Gaussian recovers ~95 % of the signal. Background: 3rd order polynomial m(¡(2/3S)) = m(¡(1S)) + ∆mPDG(¡(2/3S)-¡(1S)) σ(¡(2/3S)) = m(¡(2/3S)/m(¡(1S)) * σ(¡(1S)) →5 free parameters in signal fit: m(¡(1S)), σ(¡(1S)), c(¡(1S)), c(¡(2S)), c(¡(3S)) All plots: 4 GeV < pt(¡) < 6 GeV PDG: m(¡(1S)) = 9.460 GeV m(¡) = 9.419 ± 0.007 GeV m(¡) = 9.412 ± 0.009 GeV m(¡) = 9.437 ± 0.010 GeV 1.2 < |y¡ | < 1.8 0 < |y¡ | < 0.6 0.6 < |y¡ | < 1.2

  7. Questions: a) Why do we parametrize the signal as two Gaussians ? b) How do we determine the shape of these two Gaussians ? J/ψ → µµ double Gauss single Gauss

  8. Question: Why do we use different fitting ranges for different bins ? • The fit assumes a smooth background. We choose the fitting range accordingly. ptϒ 1-2 GeV yϒ 1.2 - 1.8 • With increasing ptϒ the mass peaks widens, but background is better behaved →increased fitting range • We derive a systematic error from varying our fitting range.

  9. Fitting - Results ● The fitted width is pt independent ● The resolution is ~30% worse than in MC ● The (relative) widening of the peak in the forward region is reproduced by MC Width ϒ(1S) 0 0.25 0.5

  10. Fitting - results Ratio of n((2S+3S))/n((1S)

  11. Normalized differential cross section There is very little difference in shape of the distribution as a function of the ϒϒ rapidity.

  12. Results: dσ(ϒ(1S))/dy × B(ϒ(1S)) → µ+µ- 0.0 < yϒϒ < 0.6 732 ± 19 (stat) ± 73 (syst)± 48 (lum) pb 0.6 < yϒϒ < 1.2 762 ± 20 (stat) ± 76 (syst) ± 50 (lum) pb 1.2 < yϒϒ < 1.8 600 ± 19 (stat) ± 56 (syst) ± 39 (lum) pb 0.0 < yϒϒ < 1.8 695 ± 14 (stat) ± 68 (syst) ± 45 (lum) pb CDF Run I: 680 ± 15 (stat) ± 18 (syst) ± 26 (lum) pb Largest contribution to the systematic error is the uncertainty on determing kdimu, followed CDF can separate the three Upsilon states.

  13. Comparison with previous results Pythia σ(1.2 < yϒ < 1.8)/σ(0.0 < yϒ < 0.6) actually we aren't that bad

  14. Questions: Why does the mass peak widen in the forward region ? How is the increased resulting smearing corrected in the cross section ? • The tracking resolution is worse in the forward region, as the tracks have a higher momentum and intersect with the tracking detectors at an angle. • This trend is reproduced in our Monte Carlo. • We increase the Monte carlo resolution by 30% over the default to match what we see in data. • The momentum scale is corrected for using the difference between measured and PDG ϒ(1S) mass. Question: Does the mass resolution really scale with the mass of the particle ? Yes. Both in data (compared to J/ψ→µµ) and in Monte-Carlo.

  15. Questions: Do we understand the rapidity dependence of our correction factors k_qual and k_trk ? Track quality cuts: At least 1 SMT hit on each track. At least one of the tracks forming an ϒhas to be isolated. k_trk (Tracking efficiency): The Monte-Carlo overestimates the tracking efficiency in the forward region as it does not take the faulty SMT disks into account. k_qual: SMT hit requirement: If we do find a track in the forward region in data, it will have an SMT hit, as there is nothing else to make a track. Therefore the SMT hit requirement affects the forward region less than the central region. Isolation: In MC the isolation requirement is 100% efficient, i.e. we do not loose any signal when applying it. In data we loose up to 6% of the signal, but considerably less in the forward region.

  16. Question: What is the L1 trigger efficiency per muon ? • We only determine the trigger efficiency for dimuons. • Approximately half the muons fromϒ(1S) have a pt < 5 GeV. • Trigger turn-on curve from Rob McCroskey µ from ϒ(1S) Plot by Rob McCroskey log scale 0 3 5 9 GeV

  17. Question: CDF measured ϒ(1S) polarization for |yϒ| < 0.4. How can we be sure that our forward ϒ(1S) are not significantly polarized ? • So far there is no indication for ϒ(1S) polarization. CDF measured α = -0.12 ± 0.22 for pTϒ > 8 GeV α = 1 (-1) ⇔ 100% transverse (longitudinal) polarization The vast majority of our ϒ(1S) has pTϒ < 8 GeV Theory predicts that if there is polarization it will be at large pT. • We checked for polarization in our signal (|y| < 1.8) and found none. There is not enough data for a fit in the forward region alone. • We estimated the effect of ϒ(1S) polarization on our cross-section. Even at α = ± 0.3 the cross-section changes by 15% or less in all pT bins. The effect is the same in all rapidity regions.

  18. Question: Why is CDF's systematic error so much smaller than ours ? • Better tracking resolution --- CDF can separate the three ϒ resonances: → Variations in the fit contribute considerable both to our statistical and systematic error. → We believe we have achieved the best resolution currently feasible without killing the signal. • Poor understanding of our Monte-Carlo and the resulting large number of correction factors. • Signal is right on the trigger turn-on curve.

  19. Conclusions • ¡(1S) cross-section measurement extended to yϒ = 1.8. • First measurement of ¡(1S) cross-section at √s = 1.96 TeV. • Normalized cross-sections show very little dependence on rapidity. • Normalized cross-section is in good agreement with published results.

  20. Backup slides

  21. Where do the ¡(1S) come from ? • ~ 50 % of ¡(1S) are produced directly. • The rest are the result of higher mass states decaying. Bottomonium States

  22. The signal Fitting a single Gaussian recovers ~95 % of the signal. Signal: 3 Gaussians: ¡(1S), ¡(2S), ¡(3S) Background: 3rd order polynomial m(¡(2/3S)) = m(¡(1S)) + mPDG(¡(2/3S)-¡(1S)) σ (¡(2/3S)) = m(¡(2/3S)/m(¡(1S)) * σ σ(¡(1S)) →5 free parameters in signal fit: m(¡(1S)), σ(¡(1S)), c(¡(1S)), c(¡(2S)), c(¡(3S)) PDG: m(¡(1S)) = 9.46 GeV All plots: 3 GeV < pt(¡) < 4 GeV m(¡) = 9.423 ± 0.008 GeV m(¡) = 9.415± 0.009 GeV m(¡) = 9.403 ± 0.013 GeV 7 8 9 10 11 12 13 7 8 9 10 11 12 13 7 8 9 10 11 12 13 GeV GeV GeV 0 < |y¡ | < 0.6 0.6 < |y¡ | < 1.2 1.2 < |y¡ | < 1.8

  23. Width from fit for ¡(1S) with |y¡| < 0.6 Data MC Width (GeV) 0.25 0.15 0.1 pt (GeV) ~ 43000 ¡(1S) candidates

  24. Trigger Level 1: di-muon trigger, scintillator only Level 2: one medium muon (early runs) two muons, at least one medium, separated in eta and phi (later runs) Both triggers at Level 2 are ~ 97 % efficient wrt Level 1 condition. Trigger efficiency for fully reconstructed di-muon events: central region: 65 % forward region: 80 % Trigger efficiency |y(¡)| < 0.6 Data wrt single μ 0.75 MC 0.65 0 4 8 12 16 20 GeV [pt(¡)]

  25. Corrections: Local muon reconstruction efficiency • reconstruct J/ψ: muon & muon and muon & track • ε = muon & muon / muon & track εData εMC |η| central muon detector forward muon detector

  26. Corrections: Tracking efficiency Method: Reconstruct J/ψ using global (i.e. muons matched to a track in the central tracking system) and local (i.e. muons that are only reconstructed in the muon system and not matched to a central track) muons. global-local* (local mu only) global-local* (signal events only) global-local* (all events) 4000 5000 350 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 GeV GeV GeV * i.e. the local momentum of the test muon is used, whether is was matched or not.

  27. Corrections: Tracking Efficiency NJ/ψ(global & global) NJ/ψ(global & local) + NJ/ψ(global & global) Efficiency = 1.0 0.8 |η| 0.0 0.4 0.8 1.2 1.6 2 2.2

  28. Corrections: Isolation and Silicon Hit Requirement From data – Monte Carlo predicts isolation requirement to be 100% efficient. Isolation efficiency for signal | y(¡)| < 1.8 pt in GeV

More Related