Post on 13-Dec-2015
transcript
Different distorted initial matrices
• Distorted matrix sample 6 (dist6)• Distorted matrix sample 7 (dist7)• Distorted matrix sample 8 (dist8)
Standard deviation due to matrices(dist6)
σ = sqrt(σmat+σstat)
σmat = sqrt(σ-σstat)
After matrix optimization
Results for (dist8)(g.s. fitting mean )– Mgen
[keV]A
σmat [keV]
BLambda -4.211.0 21.2Sigma -15.313.0 44.37
ΛHe +44.06.8 23.09
ΛLi -25.435.0 17.510
ΛBe +48.83.3 19.912
ΛB +12.75.8 15.652
ΛCr +65.34.2 5.2
Results for (dist7)(g.s. fitting mean )– Mgen
[keV]A
σmat [keV]
BLambda -2.410.6 6.8Sigma -26.113.1 26.77
ΛHe -11.99.2 16.69
ΛLi -72.37.0 29.110
ΛBe -40.68.1 24.212
ΛB -10.111.1 26.652
ΛCr -61.411.6 27.6
Results for (dist6)(g.s. fitting mean )– Mgen
[keV]A
σmat [keV]
BLambda -11.110.9 32.9Sigma +1.712.5 43.07
ΛHe +20.57.0 32.89
ΛLi -66.15.2 29.210
ΛBe -10.05.2 9.612
ΛB -48.56.2 33.052
ΛCr -19.17.2 37.0
Summary1[Systematic errors]
Fitting mean - Mgen < 80 keV (A: shift of mean value)σmat < 50 keV (B: standard deviation in the interested region)
ΔBmat = sqrt(80*80+50*50) = 94 keVΔEmat = 50 keV
5% of target thickness uncertaintyΛ, Σ0 : < 50 keVHypernucleus : < 20 keV
ΔBmat = 110 keVΔEmat = 50 keV
Test conditions
Common conditions for testsNpoint = 100Step = 0.002 MeVMean1: -11.45 – Step * i (i<Npoint)Mean2: -11.45 + Step * i (i<Npoint)
Parameters• Chunhua’s spectrum
• σ = 0.231 (fixed)
• Han’s spectrum• σ = 0.300 (fixed)
• Toshi’s spectrum • σ = 0.200 (fixed)
|χ2-1.0| distribution (Chunhua2009)
Mean1 [MeV]
Mean2 [MeV]
σ = 231 keV (fixed)
paper
Mean1 – Mean2 = 162 keV
Mean1 [MeV]
Mea
n2 [M
eV]
|χ2 -1
.0|
|χ2-1.0|
|χ2-1.0| distribution(Han2009)
Mean1 [MeV]
Mean2 [MeV]
σ = 300 keV (fixed)
paper
Mean1 – Mean2 = 162 keV
Mean1 [MeV]
Mea
n2 [M
eV]
|χ2 -1
.0|
|χ2-1.0|
|χ2-1.0| distribution (Toshi2009)
Mean1 [MeV]
Mean2 [MeV]
σ = 220 keV (fixed)
Mean1 – Mean2 = 162 keV
Mean1 [MeV]
Mea
n2 [M
eV]
|χ2 -1
.0|
|χ2-1.0|
Test conditions (Simulation)
Npoint = 100Step = 0.002 MeVMean1: -11.32 – Step * i (i<Npoint)Mean2: -11.32 + Step * i (i<Npoint)
Generated dummy data
Assumptions 100 + 600 = 700 events 160 keV separation (-BΛ
g.s. = -11.38 MeV) Gaussian distribution (σ = 0.200 MeV) No background
-BΛ [MeV]
[Cou
nts/
240
keV]
SIMULATION
Fitting results (1)
Mean1 [MeV]
Mean2 [M
eV]
Mean1 [MeV]
Mea
n2 [M
eV]
X
σ = 180 keV σ = 190 keV σ = 200 keV
Assumed width for fitting
Answer
Answer Answer
Mean1 – Mean2 = 160 keV
SIMULATION SIMULATION SIMULATIONAnswer
Fitting results (2)
Mean1 [MeV]Mean2 [MeV]
Mean1 [MeV]
Mea
n2 [M
eV]
X
σ = 210 keV σ = 220 keV σ = 230 keV
Assumed width for fitting
Answer
Answer Answer
Mean1 – Mean2 = 160 keV
SIMULATION SIMULATION SIMULATION
Fitting results (4)
Mean1 [MeV]
Mean2 [MeV]
Mean1 [MeV]
Mea
n2 [M
eV]
X
σ = 240 keV
Answer
Mean1 – Mean2 = 160 keV
SIMULATION
Summary 2[About 12
ΛB(1-,2-) separation]• What we can say from this study
• We could not reproduce paper values….(E01-011 data could be reproduced though)(I would like to see Hampton’s side study)
• What we thought from this study• In the case of simple Gaussian
• Difficult to find the answer with this method…• In the real case (peaks are not simple Gaussian distributions)
• If this method works well after some analysis improvements, we need to confirm whether the method works or not for peaks which affected by energy straggling, production point displacement from matrix origin, detector resolutions, spectrometer acceptance, beam raster and so on. Can be checked by blind analyses.
• Is there any good cut (selection) condition to find the answer ? need further study.
English ver.