Energy calibration
Dong Liu, Liang Yan, Guangshun Huang
University of Science and Technology of China
2014-11-21
1
Introduction
• 104 energy points taken from 3.850 GeV to
4.590 GeV with steps cover from 2 MeV to 20
MeV. There are various energies, like
requested or nominal, BEPCII set, displayed,
BEMS measured, but all of them are not
consistent with each other.
• Using multi-prong events based on the well
calibrated momentum, the total energy can be
reconstructed, e.g. the CLEO-c case as shown.
( PhysRevLeZ.95.062001)
2
Data samples
• Data samples: 104 energy points from 3.85
GeV – 4.59 GeV
• Boss version: 6.6.4 p01
• Channel:
ψ’ ⟶ π- π+ J/ψ
J/ψ ⟶ e- e+
J/ψ ⟶ μ- μ+
ϕ ⟶ K- K+
3
Method
• Maximum Likelihood Method
single event pdf : f(X|θ)
likelihood function for a sample:
(
) (
L X |q = L X1 ,..., X n | q
) = Õ f ( X |q )
n
i
i=1
the purpose is to select a suitable θ to maximize the
function.
• In general, the upper form is hard to deal with, we use
the logarithmic form function. They have the same
maximum value.
4
Method
• PDF
2
f (x| m ,s )=
1
2ps
( x-m )
e 2s 2
2
• Likelihood function
x
m
( )
1
l(x| m ,s )= Õ f (x | m ,s ) = Õ
e 2s 2
2ps
• Logarithmic likelihood function
n
i=1
i
n
i
i=1
n
ll(x| m ,s )= -2lnl(x| m ,s )= -2å ln
i=1
1
e
2ps
2
x -m )
(
i
2s 2
5
Our case
• Calculate invariant mass
E1 = m12 + p12
E2 = m22 + p22
• The mass distribution
f (minv ) = Gaus(minv | m, s )
In experiment, μ is not equal to pdg value.
• Purpose
try to find a correction factor for momentum, so
the mean value μ is equal to pdg value.
p®k×p
6
Our case
• Calculate invariant mass
E1 = m12 + (kp1 )2
E2 = m22 + (kp2 )2
• The mass distribution
f (minv | k) = Gaus(minv | m, s )
if the factor k is suitable, μ will be equal to pdg
value.
• Modification
f (minv | k, w) = w×Gaus(minv | m, s ) + (1- w) / t
Consider background, w: signal weight, t: width of
background
7
Decay channel
• Before correction
J / y ® e+ e-
mean = 3.09570 ± 0.00216
s = 0.018 ± 0.002
signal = 94 ±11
back = 17 ± 7
Events / ( 0.005 GeV )
fit e
18
16
14
12
10
8
6
4
2
mJ /y = 3.096916 ± 0.000011
0
3
3.02
3.04
3.06
3.08
3.1
3.12
3.14
3.16 3.18
3.2
energy (GeV)
8
Decay channel
J / y ® e+ e-
• Likelihood function
-20
-30
400
-40
200
0
-50
-200
-60
-400
-70
-80
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1 0.95
0.98
0.96 0.97
1.01
0.99 1
1.04
1.02 1.03
1.05
X direction: factor
Y direction: signal weight
Z direction: log likelihood function value
Minimum point:
factor=1.00053 , weight=81.8%
-90
-100
0.95
0.96
0.97
0.98
0.99
1
1.01
1.02
1.03
1.04
1.05
X direction: factor
Y direction: log likelihood function value
Minimum point:
factor=1.00053±0.00066
9
Decay channel
• After correction
J / y ® e+ e-
mean = 3.09724 ± 0.00211
s = 0.018 ± 0.002
signal = 93 ±11
back = 19 ± 7
Events / ( 0.005 GeV )
fit e
20
18
16
14
12
10
8
6
4
2
mJ /y = 3.096916 ± 0.000011
0
3
3.02
3.04
3.06
3.08
3.1
3.12
3.14
3.16 3.18
3.2
energy (GeV)
10
Decay channel
• Before correction
J /y ® m + m -
mean = 3.09598 ± 0.00138
s = 0.016 ± 0.001
signal = 143 ±12
back = 11± 5
Events / ( 0.005 GeV )
fit mu
25
20
15
10
5
mJ /y = 3.096916 ± 0.000011
0
3
3.02
3.04
3.06
3.08
3.1
3.12
3.14
3.2
3.16 3.18
energy (GeV)
11
Decay channel
• Likelihood function
J /y ® m + m -
200
600
0
400
200
-200
0
-200
-400
-400
-600
-800
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
-600
0.1
1.01 1.02
0.99 1
0.98
0.97
0.95 0.96
1.03 1.04
1.05
X direction: factor
Y direction: signal weight
Z direction: log likelihood function value
Minimum point:
factor=1.00034 , weight=91.4%
-800
0.95
0.96
0.97
0.98
0.99
1
1.01
1.02
1.03
1.04
1.05
X direction: factor
Y direction: log likelihood function value
Minimum point:
factor=1.00034+0.00764
-0.00752
12
Decay channel
• After correction
J /y ® m + m -
mean = 3.09696 ± 0.00142
s = 0.016 ± 0.001
signal = 144 ±12
back = 10 ± 5
Events / ( 0.005 GeV )
fit mu
25
20
15
10
5
mJ /y = 3.096916 ± 0.000011
0
3
3.02
3.04
3.06
3.08
3.1
3.12
3.14
3.16 3.18
3.2
energy (GeV)
13
Decay channel
• Before correction
J / y ® p +p -l +l -
mean = 3.68571± 0.00022
s = 0.00284 ± 0.00002
signal = 219 ±15
back = 22 ± 6
Events / ( 0.002 GeV )
fit pi
90
80
70
60
50
40
30
20
my ' = 3.686109 +0.000012
-0.000014
10
0
3.65
3.66
3.67
3.68
3.69
3.7
3.71
3.72
3.73
energy (GeV)
14
Decay channel
-1200
-2000
0.5
0.4
0.3
0.2
0.1 0.95 0.96
0.97
1.02
1 1.01
0.98 0.99
1.03 1.04
X direction: factor
Y direction: signal weight
Z direction: log likelihood function value
Minimum point:
factor=1.00134 , weight=76.4%
1.05
1.01 1.02 1.03 1.04 1.05
-1500
0.9
0.8
0.7
0.6
-1400
1
-1000
-1600
-500
-1800
0.95 0.96 0.97 0.98 0.99
0
-2000
• Likelihood function
J / y ® p +p -l +l -
X direction: factor
Y direction: log likelihood function value
Minimum point:
factor=1.00134+0.00838
-0.00839
15
Decay channel
• After correction
J / y ® p +p -l +l -
mean = 3.68623 ± 0.00219
s = 0.00283± 0.00003
signal = 218 ±15
back = 23 ± 6
Events / ( 0.002 GeV )
fit pi
70
60
50
40
30
20
10
my ' = 3.686109 +0.000012
-0.000014
0
3.65
3.66
3.67
3.68
3.69
3.7
3.71
3.72
3.73
energy (GeV)
16
Decay channel
f ® K +K -
• Before correction
mean = 1.01921± 0.00021
s = 0.00250 ± 0.00021
signal = 345 ± 28
back = 942 ± 37
Events / ( 0.0005 GeV )
fit kaon
50
40
30
20
10
mf = 1.019455 ± 0.000020
0
1
1.005
1.01
1.015
1.02
1.025
1.03
1.035
1.04 1.045 1.05
energy (GeV)
17
Decay channel
f ® K +K -
• Likelihood function
-6180
-6200
-4800
-6220
-5000
-5200
-6240
-5400
-5600
-6260
-5800
-6000
-6280
-6200
-6400
-6300
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
-6320
0.1
0.95
0.98
0.96 0.97
1.01
0.99 1
1.04
1.02 1.03
1.05
X direction: factor
Y direction: signal weight
Z direction: log likelihood function value
Minimum point:
factor=0.997482 , weight=40.5%
-6340
-6360
0.95
0.96
0.97
0.98
0.99
1
1.01
1.02
1.03
1.04
1.05
X direction: factor
Y direction: log likelihood function value
Minimum point:
factor= 0.99748+0.00368
-0.00367
18
Decay channel
• After correction
f ® K +K -
mean =1.01909 ± 0.00021
s = 0.00250 ± 0.00021
Events / ( 0.0005 GeV )
fit kaon
50
40
signal = 343 ± 28
30
back = 945 ± 37
20
10
mf = 1.019455 ± 0.000020
0
1
1.005
1.01
1.015
1.02
1.025
1.03
1.035
1.04 1.045 1.05
energy (GeV)
| mean - mf |> d mean ?
19
problem
• For f ® K + K • The correction factor will cause that the
invariant mass of 2 kaon a little small than the
pdg value of f , it is not in the range of error.
• It is caused by background? Toy MC shows it is
not the case.
20
Toy Monte Carlo
• Generate sample
distribution: gaus(m,sigma), no background
events: 3000
mean = 1.01946 ± 0.00005
s = 0.00249 ± 0.00003
signal = 3000 ± 55
back = 0 +1
Events / ( 0.0005 GeV )
fit kaon
250
200
150
100
50
0
1
1.005
1.01
1.015
1.02
1.025
1.03
1.035
1.04 1.045 1.05
energy (GeV)
21
Toy Monte Carlo
• Likelihood function
f ® K +K -
-22000
X direction: factor
Y direction: log likelihood function value
Minimum point:
factor=0.99692±0.00072
-23000
-24000
-25000
-26000
-27000
0.95
0.96
0.97
0.98
0.99
1
1.01
1.02
1.03
1.04
1.05
22
Toy Monte Carlo
• After correction
mean = 1.01926 ± 0.00005
s = 0.00247 ± 0.00003
signal = 3000 ± 55
+2
back = 0
Events / ( 0.0005 GeV )
fit kaon
250
200
150
100
50
0
1
1.005
1.01
1.015
1.02
1.025
1.03
1.035
1.04 1.045 1.05
energy (GeV)
23
backup
• Single event in mc
fit kaon
Events / ( 0.0005 GeV )
Events / ( 0.0005 GeV )
fit kaon
3
2.5
2
3
2.5
2
1.5
1.5
1
1
0.5
0.5
0
1
1.005
1.01
1.015
1.02
1.025
1.03
Before correction
mass=1.01733
factor=1.0354+/-0.0016
1.035
1.04 1.045 1.05
energy (GeV)
0
1
1.005
1.01
1.015
1.02
1.025
1.03
1.035
1.04 1.045 1.05
energy (GeV)
After correction
Mass=1.01945
24
backup
• 2 events in mc, 10 times
Mean before correct
factor
Mean after correct
1.02123
0.972597
1.01943
1.02045
0.984074
1.01942
1.01898
1.00758
1.01945
1.01882
1.00864
1.01936
1.02189
0.962710
1.01940
1.02096
0.975424
1.01936
1.02202
0.961676
1.01945
1.02135
0.970480
1.01941
1.07151
1.02608
1.01908
1.01817
1.02026
1.01942
25
backup
• Invariant mass change with factor
1.1
1.05
1
0.95
0.9
0.9
0.95
Graph
1
1.05
1.1
Black line refer to factor multiply momentum
Red line refer to factor multiply invariant mass directly
26
Event Cuts
• Global cut:
Good Charged track number: 4
Total Net Charge: 0
• Pion cut: p < 0.5 GeV/c
• Electron cut: p > 1.0 GeV/c & E/p > 0.7
• Muon cut: p > 1.0 GeV/c
& emc shower energy < 0.35 GeV
27