Rare decays with the KLOE detector at DA NE

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Transcript Rare decays with the KLOE detector at DA NE

Rare  decays with the KLOE
detector at DANE
Biagio Di Micco
Università degli Studi Roma Tre
I.N.F.N sezione di Roma III
22 May 2003
LNF Spring School "Bruno
Touschek"
C violating decay 
  
  
22 May 2003
Largely:
Br() = .39
C() = C(2) = +1
C violating decay channel C(3) = -1
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Theoretical estimation
Dicus [2] did an estimation for 0 3 using a quark loop model assuming a parity
violating, CP conserving weak interaction interaction, through the following
diagrams:
For the  case, using mq > 1/5 mN Herczeg [3] found Br(3) < 3  10-12
A general flavor conserving
C and CP violating interaction,
taking in account the limit on the Br(3) < 10-19
electric dipole moment of the
neutron (P.Herczeg)
[2] D.A. Dicus, Phys. Rev. D 12, 2133 (1975)
[3] P.Herczeg, in Production and decay of light mesons,
P.Fleury (World Scientific, Singapore, 1988)
22 May 2003
QFT on not commutative
spaces can give relevant
contribution [4]
Br(03) < 10-19
[5]
[4] Hinchliffe, I. and Kersting, N., Phys. Rev. D 64, (2001)
[5] Grosse, H. and Liao, Y., Phys. Lett. B520, 63(2001)
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Experimental situation
510-495%CL
PDG 2002
(Serpukhov-140
experiment)
Br(
4.5×10-5 95% CL (2000)
1.8×10-5 90% CL (2002)
Crystal Ball
preliminary [1]
 production method
- + p   + n
[1] Prakhov, S., Ucla Crystall Ball Report CB-01-008 (2001) in
B.M.K. Nefkens and J.W. Price – Phys.Scripta T99:114-122,2002
22 May 2003
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The KLOE calorimeter and used data
sample
EMC parameters
2001 :  150 pb-1 of data
 Hermetical coverage
2002 :  260 pb-1 of data
 98 % coverage of solid angle
 High efficiency for low energy photons
E> 20 MeV
 Time resolution
t = 54ps / E (GeV) 50 ps  125 ps
Intrinsic calorimeter
costant term due to
miscalibration
 production chain
e  e -    
E = 363 MeV

N. produced = 1.5109
vertex position
uncertainty due to
the bunch length
Br() = 1.3 %
N.  produced  2107
22 May 2003
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4
Background channels
5
(pb)
bkg channel
4
e e   (    )
 -
0
1100
cut with
0 veto
  0 (   )
90
  0 (   0 )
440
 f 0 ( f 0   0 0 )
230
 K s K L  5 0
3
> 5
   0
4000
 (   )
17000
ee-  
22 May 2003
340000
4 if
70000
 K s K L  4 0
300
  (  3 0 )
42000
accidental cluster or cluster splitting,
cut with
Emin > 50 MeV
|cos| < 0.91 ( and bhabha)
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4 if
loose or merging of
some photons, cut
with Etot > 800
MeV and 0 veto
Analysis strategy
N*prompt-clusters = 4
Emin > 50 MeV
|cos()| < 0.91

Kinematical fit
constraint: 1) energy momentum
conservation
2) t = r/c (photons)
 
2
( yi - i ) 2
i
 2y
i
   j f j (k )
j
yi: (54) measured variables (x,y,z,t,E) of the clusters
i:output variables, improved by the fit
fj: 8 constraints
22 May 2003
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*Prompt cluster definition:
|t-r/c| < min(5t,2 ns)
2min cut
The 2min is used to reject background for > 4 channels
(energy momentum conservation)
2min distribution
Total Energy 2001 DATA
missing momentum
events
2001 DATA
accepted = 0.6
before
2min cut
2min
MeV
MC 3
after
2min cut
pure phase space
sig = 0.93
2min
22 May 2003
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MeV
0 veto
To reject background channels with 0 we cut on m of any pair of photons .
the main background channels with 0
m combinatorial
e+e-
 
e+
0
e-
 ISR


*

0
2001 DATA
2001 DATA
 peak
before veto
MeV/c2
90 MeV
22 May 2003
after veto
180 MeV
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m 0
MeV/c2
E max distribution
To evaluate the upper limit we study the distribution of the most energetic photon energy
Energy range used to evaluate
the upper limit
2001 DATA
2002 DATA
events
with E > Erad
0
MeV
E max
MeV
E max
MC signal
22 May 2003
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Background and signal estimate
The background shape and normalization is estimated from the data,
using the energy range: (280,340)  (380,480)
2001+2002
DATA
polynomial fit
3 degree
 2 / ndf  92 / 87
MeV
MeV
MeV
E max
E max
fitted function: f(xi) = pol3(xi) + Nsigfi
22 May 2003
fraction of events in the “i” bin of the
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Touschek"
Preliminary result
We normalize our sample to the number of 30 events
selected in the same data sample.
 3  0.42
0
(N = 17106)
N 3 0 (2001)  887046
 3 ( MC )  0.207  0.001( stat.)
Nsig=1.830
N 3 0 (2002)  1561655
NU(95 % C.L)=1.8 + 1.64×30 = 51
NUp- sig    0 0 0
Br (   )
-5


4
.
3

10
Br (   0 0 0 ) N  0 0 0   
Br (  3 )  1.4 10
22 May 2003
-5
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95 % C.L.
(systematic to check)
0
Interesting decay for ChPT:
ChPT prediction at O(p4)
from L2 loops


+,K+
-loops = 0.8410-3 eV
k-loops = 2.4510-3 eV
0

-,K-
Experimental value:
Br(0) = (7.21.4)x10-4
0.84 0.18eV
PDG-2002 (GAMS 1984)
ChPT all order estimation using
resonance saturation and a0(980) e
a2(1232) contribution:
ChPT = 0.4  0.2 eV
Preliminary result from E927 (Slac)
Crystal Ball detector (2003)
Br(0) = ( 3.2  0.9)  10-4
 0.2 0.10 0.11eV
see Ameteller; Physica Scripta T99 and references therein
22 May 2003
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->0at KLOE
N.  = 17106
Lint  400 pb-1
Br(0) = 3.2  10-4
Br(0)
=7.2
   0
expected
10-4
efficiency from a preliminary
analysis:  = 0.17
main baclground:
22 May 2003
30 background
due to cluster merging
and lost photons
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5400
12000
The END
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You hoped it was true, isn’t it?
22 May 2003
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Some DATA-MC comparisons
2 sample
kinematic fit equal to that of 3
E distribution (all photons)
MC
DATA run 22000-22152
=362.1 MeV
=4.88 MeV
22 May 2003
=362.7 MeV
=4.60 MeV
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 sample selection
m o
after m cut
DATA
 sample
m 0
MC
3
sample
127 < m < 141
(best 2 combination
under  hypothesis)
3%
systematic
22 May 2003
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Touschek"
MC time correction DATAREC V. 14
ALL
E > 200 MeV
ns
ns
ns
Barrel t+8 ps (before -)
Time correction:
ns
Endcap t+27 ps
22 May 2003
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Residuals distribution 1/2
a bit high, and
different to endcap y
22 May 2003
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Residuals distribution 2/2
significative asimmetry,
residual correction
to t-r/c?
22 May 2003
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Residuals distribution of data
wait the general meeting
(perhaps)
22 May 2003
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Systematic effects
Variation with energy window:
Discrepancy in the
Efficiency curve
(Bini’s routine)
[340,380]
NU/() = 245
[330,390]
NU/() = 267
+ 9%
weigth = 24341/120000
[350,370]
Nu/() = 208
-15%
mc = 24552/120000
What is the correct range for variation 1, 5, 10 MeV ?
22 May 2003
LNF Spring School "Bruno
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0.9 %
Number of signal events
in the sample at 95% C.L
Ntot = 1935
fitted function
x  bw / 2
f  pol 3( x)  N sig
380
E max
Fit result:
f
sig
( x)dx  1
340
Nsig = -4  29
22 May 2003
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g
sig
x -bw / 2
MeV
( x' )dx'