Neutral kaon decay and CP violation  TUNLmep/phy305/kaondecay_cpviolation.pdf · 2008. 4. 6. ·...
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Haiyan Gao Physics 305, Spring2008
Weak interactions
Neutral kaon decay and CP violation

Haiyan Gao Physics 305, Spring2008
Quark mixing and CKM matrix
• Observation: ΔS=1 weak decay suppressedthan ΔS=0 weak process
!
n" p + e#
+ $ e (1)
%#" n + e
#+ $ e (2)
Process (2) is suppressed by a factor of 20 compared with (1)
Look at the diagrams at the quark level for these two processes(diagram on board)

Haiyan Gao Physics 305, Spring2008
!
u
d
"
# $ %
& ' weak
=u
d
"
# $ %
& ' strong
?
In analogous to doublet,
Introduce quark doublet:
!
u
dcos"c
+ ssin"c
#
$ %
&
' (
Then the effective coupling for
!
n" p + e#
+ $ e
%#" n + e
#+ $ e
!
Gcos"c
Gsin"c
!
e
"e
#
$ %
&
' (
!
µ
" µ
#
$ %
&
' (

Haiyan Gao Physics 305, Spring2008
!
20 = (cos"
c
sin"c
)2#"
c$12.6
0
Cabibbo angle
In 1970, Glashow, Iiiopoulos and Maiani (GIM)
!
u
d'
"
# $ %
& ' =
u
dcos(c
+ ssin(c
"
# $
%
& '
!
c
s'
"
# $ %
& ' =
c
scos(c) d sin(
c
"
# $
%
& '
Which can be expressed in matrix form (charm discovered in 1974)
!
d'
s'
"
# $ %
& ' =
cos(c
sin(c
)sin(ccos(
c
"
# $
%
& ' d
s
"
# $ %
& '
Quark state in the weak interaction

Haiyan Gao Physics 305, Spring2008
Neutral current selection rule Neutral current selection rule ΔS=0Only consider u. d, s quarks, the neutral current interaction (diagram on board):
!
(uu + dd cos2"
c+ ss sin
2"c) + (sd + ds )sin"
ccos"
c
ΔS=0 ΔS≠0
To fix this, introduce a second quark doublet by having a new quark c
!
(uu + cc ) + (dd + ss )cos2"
c+ (dd + ss )sin
2"c
+
(sd + ds # s d # sd )sin"ccos"
c

Haiyan Gao Physics 305, Spring2008
With six quark flavors, the weak currents will be described by unitary transformations among three quark doublets
!
d'
s'
b'
"
#
$ $ $
%
&
' ' '
=V
d
s
b
"
#
$ $ $
%
&
' ' '
Where the transformation matrix V is 3 by 3 unitary

Haiyan Gao Physics 305, Spring2008
Cabbibo, Kobayashi, Maskawa Matrix• Standard model:
– The CKM matrix: the complex phase of CKMmatrix leads to CP violation
!
d'
s'
b'
"
#
$ $ $
%
&
' ' '
=
Vud
Vus
Vub
Vcd
Vcs
Vcb
Vtd
Vts
Vtb
"
#
$ $ $
%
&
' ' '
d
s
b
"
#
$ $ $
%
&
' ' '
= ˆ V CKM
d
s
b
"
#
$ $ $
%
&
' ' '
For n generations:n(n1)/2 angles 3(n1)(n2)/2 phases 1
⇒
Weak eigenstates
⇐
Mass eigenstates

Haiyan Gao Physics 305, Spring2008
Charged hadronic weak current
!
J"+
= (u ,c ,t )#"(1$ # 5) %
c1 s1c3 s1s3
$s1c2 c1c2c3 $ s2s3ei&
c1c2c3 + s2c3ei&
$s1s2 c1s2c3 + c2s3ei&
c1s2s3 $ c2c3ei&
'
(
) ) )
*
+
, , ,
d
s
b
'
(
) ) )
*
+
, , , !
ci= cos("
i)
si= sin("
i)
i =1,2,3!
d'
s'
b'
"
#
$ $ $
%
&
' ' '
=
c1
s1c3
s1s3
(s1c2
c1c2c3( s
2s3ei)
c1c2c3
+ s2c3ei)
(s1s2
c1s2c3
+ c2s3ei)
c1s2s3( c
2c3ei)
"
#
$ $ $
%
&
' ' '
d
s
b
"
#
$ $ $
%
&
' ' '

Haiyan Gao Physics 305, Spring2008
The phase enters the wavefunction as not invariant under t→t. This phase introduces the importantPossibility of Tviolation, i.e., CP violation
!
exp[i("t + #)]
The various elements of the matrix have been determined in a range of experiments
!
Vcd2
+ Vcs2
+ Vcb2=1
!
VCKM
=
Vud
= 0.975 Vus
= 0.221 Vub
= 0.005
Vcd
= 0.221 Vcs
= 0.974 Vcb
= 0.04
Vtd
= 0.01 Vts
= 0.041 Vtb
= 0.999
"
#
$ $ $
%
&
' ' '

Haiyan Gao Physics 305, Spring2008
Wolfenstein parametrization 1983• Expansion in terms λ=sinθc :
• From Blifetime:• CP violation effects smaller third order:• Keep and real;
Use unitarity to calculate!
Vus
= 0.22 = " #Vud
= (1$ "2 /2)
!
Vcb
= 0.04 ~ 0.06 = A"2
!
A"3(# $ i%)
!
Vud,V
us,V
ts
!
Vtb
!
"4
!
1"#2
2# A#3($ " i% + i%
#2
2)
"# 1"#2
2" i%A#4 A#2(1+ i%#2)
A#3(1" $ " i%) "A#2 1
&
'
( ( ( ( ( (
)
*
+ + + + + +

Haiyan Gao Physics 305, Spring2008
Neutral K mesons• Two isospin doublets
!
s = +1
K+(us )
K0(ds )
"
# $
%
& '
!
s = "1
K 0(d s)
K"(su )
#
$ %
&
' (
They are particle and antiparticle and by the CPT theorem have the same mass. Experimentally we find:
!
 (mK
o "mK
o )  /m < 9 #10"19
!
CP K0
>= " K 0
>
CP K 0
>= " K0
>
!
K1
0 =1
2 K
0 > +  K 0 >[ ], CP = 1
K2
0 =1
2 K
0 > "  K 0 >[ ], CP = +1

Haiyan Gao Physics 305, Spring2008
Weak decay ΔS=±1 to nonstrange mesons, leptons
!
Ko"# +#$
!
K+"# +# 0
!
KS
0 mean lifetime (0.8923± 0.0022) "10
#10 (sec)
KS
0$% +%#, % 0% 0, CP = +1
KL
0 mean lifetime (5.183± 0.040) "10
#8 (sec)
KL
0$% 0% 0% 0, % +%#% 0 , CP = #1

Haiyan Gao Physics 305, Spring2008
Strangeness Productions
!
" + p#K +$0
!
" + p#K +K$p
!
" + p#K 0$+
!
" + p#K 0K 0p!
"# + p$% + K 0
" + + p$K + + K o + p
Eigen states of production
!
K0,K
0
Strong and EM interactions conserve strangeness

Haiyan Gao Physics 305, Spring2008
Experimental observation
Starting off with a beam of pure K0 beam one could end up aftera few meters, with a beam of mixed strangeness K0 and
!
K 0
!
K(t) >="(t) K 0 > +#(t) K 0 >
``Object decay by weak interactions are eigenstates of CP, not of Strangeness’’
Strangeness eignstatesCP eigenstatesMass eigenstates
production
decay

Haiyan Gao Physics 305, Spring2008
Strangeness oscillationThe weak interaction kaon eigenstates
are those concerned when kaonspropagate through space. Because KLand KS have different life times and
decay modes, they can be expected tohave a difference in mass, which willalter the relative phase of these two
eignstates

Haiyan Gao Physics 305, Spring2008
The amplitude of the state Ks as a function of time, measured inThe particle rest frame, with E=m is
!
As(t) = A
S(0)e
"(#S/ 2+ im
S)t
#S
=h
$S
Is the width of the KS state and ms is its rest mass.
Similarly for KL
!
AL(t) = A
L(0)e
"(#L/ 2+ im
L)t
Assume at t=0, we start with pure K0 beamAt time t the K0 intensity
!
AL(0) = A
S(0) =
1
2

Haiyan Gao Physics 305, Spring2008!
I(K0) =1
2[A
S(t) + A
L(t)][A
S
*(t) + A
L
*(t)]
=1
4[e
"#St+ e
"#Lt+ 2e
"(#S
+#L)t
2 cos$mt
$m = mL"m
S
!
I(K 0) =1
4[e
"#S
t+ e
"#L
t" 2e
"(#S
+#L)
t
2 cos$mt]
$m = (3.491± 0.009) %10"12
MeV
!
K0 >=
1
2K
L
0
> + KS
0
>[ ]
K 0 >=
1
2K
L
0> " K
S
0>[ ]

Haiyan Gao Physics 305, Spring2008
Strangeness Oscillationsthe probability that a beam initially consisting of K0’s contains K0’s at a later time:
!
K o
Ko(t)
2
=1
4e"#
St + e"#L t " 2e
"t
2(#
S+#
L)
cos $mt( )%
& '
(
) *
We can measure the strangeness content ofa beam as a function of time (distance) byputting some material in the beam andcountingthe number of strong interactions that haveS=+1in the final state vs. the number with S=1.
Kop! K
+n
K o
p ! "p
S= +1
S= 1
Strangeness oscillations for aninitially pure K0 beam. A valueof (m2m1)τS=0.5 is used.

Haiyan Gao Physics 305, Spring2008
CP Violation In 1964, Christenson, Cronin, Fitch and Turlay
discovered at BNL that the longlived neutral Kmeson with CP=1 could decay occasionally towith CP=+1 about once every 500 decays
!
" +"#
!
Ko
L "#+#$# o
!
Ko
L "#+#$
•CP violations in nuclear systems•More CP violations in experiments: B meson decays, SLAC and KEK

Haiyan Gao Physics 305, Spring2008
CPviolation in neutral kaon system• π0 π0, π+ π
– Bose statistics 1⇔2 the total wavefunction issymmetric
– Spin 0 1⇔2 = operation of C followed by P ⇒
CP=+1•π0 π+ π
CP = +1 for π+ π •Neutral kaon decay has very small Qvalue, 3 pions in relative Sstate•π0 : P=1, C=+1 ⇒ π0 π+ π = 1

Neutral Kaons and CP violationSince CP is not conserved in neutral kaon decay it makes more sense to usemass (or lifetime) eigenstates rather than k1> and k2>:
!
KS
>=1
1+ " 2(K1 >+ " K2 >) "K # short"
KL
>=1
1+ " 2(K2 >+ " K1 >) "K # long"
There can be two types of CP violation in KL decay: indirect (“mixing”): KL →ππ because of its K1 component direct: KL→ππ because the amplitude for K2 allows K2→ππIt turns out that both types of CP violation are present and indirect >> direct!
Long lifetime state with τL ≈ 5x108 sec.
Short lifetime state with τS ≈ 9x1011 sec.
ε is a (small) complex number that allows for CP violation through mixing.
Slides 18, 2124 adapted from notes by Richard Kass at OSU

Neutral Kaons and CP violationThe KL and KS states are not CP (or S) eigenstates since:
!
CP Ks
>=1
1+ " 2CP K
1>+ "CP K
2>( ) =
1
1+ " 2K
1>#" K
2>( ) $Ks >
!
CP KL
>=1
1+ " 2CP K
2>+ "CP K
1>( ) =
1
1+ " 2# K
2>+ " K
1>( ) $KL >
In fact these states are not orthogonal!
!
< KSK
L>=2Re"
1+ " 2
If CP violation is due to mixing (“indirect”only) then the amplitude for kL→ππ is:
!
< K1K
L>=
1
1+ " 2< K
1K
2>+ " < K
1K
1>( ) =
"
1+ " 2
While the amplitude for kL→πππ is:
!
< K2K
L>=
1
1+ " 2< K
2K
2>+ " < K
2K
1>( ) =
1
1+ " 2
From experiments we find that ε = 2.3x103.However, the standard model predicts a small amount of direct CP violation too!

Neutral Kaons and CP violationThe standard model predicts that the quantities η+ and η00 should differ very slightly as a result of direct CP violation (CP violation in the amplitude).
!
"+# $Amp(KL %&
+)
Amp(Ks %&+)
"oo $Amp(KL %&
o& o)
Amp(Ks %&o& o)
BR(KL ! "+"# )
BR(Ks ! "+"# )
BR(KL ! "o"o)
BR(Ks ! "o"o)
=$+#$oo
2
=% + & % 2
% # 2 & % 2 '1 + 6Re
& %
%
( )
* +
CP violation is now described by two complex parameters, ε and ε′, with ε′related to direct CP violation. The standard model estimates Re(ε′/ε) to be (430)x104!
Experimentally what is measured is the ratio of branching ratios:
After many years of trying (starting in 1970’s) and some controversial experiments,a nonzero value of Re(ε′/ε) has been recently been measured (2 different experiments):
Re(ε′/ε)=17.2±1.8x104At this point, the measurement is more precise than the theoretical calculation! Calculating Re(ε′/ε) is presently one of the most challenging HEP theory projects.

Neutral Kaons, Flavor Oscillations, & CP ViolationFrom an experimentalist's point of view a good quantity to measure is the yield(as a function of proper time, time measured in the rest frame of the K) of π+πdecays from a beam that is initially K0. Since we are measuring the sum of thesquare of the amplitude KL→ π+π> +KS→π+π> there will be an interference termin the number of π+π decays/time (≡I(t)). The yield of π+π decays is given by:
I!+!"
(t) = I!+!"
(0) e
"t
#s + $+"2e
"t
#L + 2$+" e"t
2(1
#s+1
#L)
cost
h(m2 "m1) + %+"
& '
( )
&
'
* *
(
)
+ +
Thus by measuring this yield we gain information on the mass difference as well as the CPviolation parameters η+ and φ+.
τ sec x 10105 15
Event rate for π+πdecays as a functionof proper time. The bestfit requires interferencebetween the KL and KSamplitudes.
m2m1=3.491±0.009x106 eV η+  =(2.29±0.01)x103φ+ =(43.7±0.6)0τS=0.893x1010 secτL=0.517x107 sec