Part II CP Violation in the SM
description
Transcript of Part II CP Violation in the SM
1Chris Parkes
Part IICP Violation in the SM
Chris Parkes
2Chris Parkes
Outline
THEORETICAL CONCEPTS
I. Introductory conceptsMatter and antimatter
Symmetries and conservation laws
Discrete symmetries P, C and T
II. CP Violation in the Standard ModelKaons and discovery of CP violation
Mixing in neutral mesons
Cabibbo theory and GIM mechanism
The CKM matrix and the Unitarity Triangle
Types of CP violation
Kaons anddiscovery of CP violation
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What about the product CP?
+
+
+
+
Intrinsicspin
P C
CP
Initially CP appears to be preservedin weakinteractions …!
Weak interactions experimentally proven to: Violate P : Wu et al. experiment, 1956 Violate C : Lederman et al., 1956 (just think about the pion decay below
and non-existence of right-handed neutrinos)
But is C+P CP symmetryconserved or violated?
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Kaon mesons: in two isospin doubletsPart of pseudo-scalar JP=0- mesons octet with , h
Introducing kaons
K+ = usKo = ds
K- = usKo = ds
S=+1 S=-1
I3=+1/2I3=-1/2
Kaon production: (pion beam hitting a target)Ko : - + p o + Ko
But from baryon number conservation:
Ko : + + p K+ + Ko + pOr
Ko : - + p o + Ko + n +n
Requires higher energy
Much higher
S 0 0 -1 +1
S 0 0 +1 -1 0
S 0 0 +1 -1 0 0
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What precisely is a K0 meson?
Now we know the quark contents: K0 =sd, K0 =sd
First: what is the effect of C and P on the K0 and K0 particles?
(because l=0 q qbar pair)
(because l=0 q qbar pair)
effect of CP :
Bottom line: the flavour eigenstates K0 and K0 are not CP eigenstates
Neutral kaons (1/2)
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Nevertheless it is possible to construct CP eigenstates as linear combinations
Can always be done in quantum mechanics, to construct CP eigenstates
|K1> = 1/2(|K0> + |K0>)
|K2> = 1/2(|K0> - |K0>)
Then:CP |K1> = +1 |K1>
CP |K2> = -1 |K2> Does it make sense to look at these linear combinations?
i.e. do these represent real particles?
Predictions were:The K1 must decay to 2 pions
given CP conservation of the weak interactions
This 2 pion neutral kaon decay was the decay observed and therefore known
The same arguments predict that K2 must decay to 3 pions
History tells us it made sense!
The K2 = KL (“K-long”) was discovered in 1956 after being predicted
(difference between K2 and KL to be discussed later)
Neutral kaons (2/2)
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How do you obtain a pure ‘beam’ of K2 particles?
It turns out that you can do that through clever use of kinematics
Exploit that decay of neutral K (K1) into two pions is much faster than decay of neutral K (K2) into three pions
Mass K0 =498 MeV, Mass π0, π+/- =135 / 140 MeV
Therefore K2 must have a longer lifetime thank K1 since small decay phase space
t1 = ~0.9 x 10-10 sec
t2 = ~5.2 x 10-8 sec (~600 times larger!)
Beam of neutral kaons automatically becomes beam of |K2>as all |K1> decay very early on…
Looking closer at KL decays
Initial K0
beam
K1 decay early (into ) Pure K2 beam after a while!(all decaying into πππ) !
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Incoming K2 beam
Decay of K2 into 3 pions
If you detect two of the three pionsof a K2 decay they will generallynot point along the beam line
Essential idea: Look for (CP violating) K2 decays 20 meters away from K0 production point
The Cronin & Fitch experiment (1/3)
J.H. Christenson, J.W. Cronin,V.L. Fitch, R. Turley PRL 13,138 (1964)
π0
π+
π-
Vector sum of p(π-),p(π+)
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Incoming K2 beam
Decaying pions
If K2 decays into two pions instead ofthree both the reconstructed directionshould be exactly along the beamline(conservation of momentum in K2 decay)
The Cronin & Fitch experiment (2/3)
J.H. Christenson et al.,PRL 13,138 (1964)
Essential idea: Look for (CP violating) K2 decays 20 meters away from K0 production point
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Result: an excess of events at Q=0 degrees!
K2 decays(CP Violation!)
K2 decays
Note scale: 99.99% of K decaysare left of plot boundary
The Cronin & Fitch experiment (3/3)
K2 ++Xp+ = p+ + pq = angle between pK2 and p+If X = 0, p+ = pK2 : cos q = 1If X 0, p+ pK2 : cos q 1
Weak interactions violate CP
Effect is tiny, ~0.05% !
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Almost but not quite!
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with |ε| <<1
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Key Points So Far
• K0, K0 are not CP eigenstates – need to make linear combination
• Short lived and long-lived Kaon states
• CP Violated (a tiny bit) in Kaon decays
• Describe this through Ks, KL as mixture of K0 K0
Mixingin neutral mesons
HEALTH WARNING :We are about to change notationP1,P2 are like Ks, KL (rather than K1,K2)
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Particle can transform into its own anti-particleneutral meson states Po, Po
P could be Ko, Do, Bo, or Bso
Kaon oscillations
So say at t=0, pure Ko, – later a superposition of states
ds
u, c, t W
W+_s
d_u, c, t
ds u, c, t
W W+_s
du, c, t
K0K0
_
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20Chris Parkes Here for general derivation we have labelled states 1,2
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neutral meson states Po, Po
P could be Ko, Do, Bo, or Bso
with internal quantum number F
Such that F=0 strong/EM interactions but F0 for weak interactions
obeys time-dependent Schrödinger equation
M, : hermitian 2x2 matrices, mass matrix and decay matrix
mass/lifetime particle = antiparticle
Solution of form
oo PtbPtat )()()( +
No Mixing – Simplest Case
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neutral meson states Po, Po
P could be Ko, Do, Bo, or Bso
with internal quantum number F
Such that F=0 strong/EM interactions but F0 for weak interactions
obeys time-dependent Schrödinger equation
M, : hermitian 2x2 matrices, mass matrix and decay matrix
H11=H22 from CPT invariance (mass/lifetime particle = antiparticle)
oo PtbPtat )()()( +
bai
ba
ba
dtdi )
2( ΓMH
Time evolution of neutral mesons mixed states (1/4)
*
12
12*12
12
2i
MMMM
H
H is the total hamiltonian:
EM+strong+weak
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Solve Schrödinger for the eigenstates of H :
of the form
with complex parameters p and q satisfying
Time evolution of the eigenstates:
oo
oo
PqPpP
PqPpP
+
2
1
Time evolution of neutral mesons mixed states (2/4)
122 + qp
timi
timi
ePtP
ePtP
)2
(
22
)2
(
11
22
11
)(
)(
Compare with Ks, KL as mixtures of K0, K0
If equal mixtures, like K1 K2
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Some facts and definitions:
Characteristic equation
Eigenvector equation:
0)(
q
pEIH
12*
12*
1212
2
2
iM
iM
qp
Time evolution of neutral mesons mixed states (3/4)
0 EIH )2
)(2
()2
( *1212
*1212
2 iMiMEiM
12
12
mmm
)2
(2
)2
(2
)2
(2
)2
(2
222
111
+
+
imMimE
imMimE
22,1mMm
e.g.
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Evolution of weak/flavour eigenstates:
Time evolution of mixing probabilities:
22
22
P(
P(
(t)fpq(t)PP;t)PP
(t)f(t)PP;t)PP
oooo
oooo
+
Interference term
mx
000
000
)()()(
)()()(
PtfqpPtftP
PtfpqPtftP
+
+
+
+
timitimieetf
)2
()2
( 2211
21)(
Time evolution of neutral mesons mixed states (4/4)
221 +
i.e. if start with P0, what is probability that after time t that have state P0 ?
decay terms
Parameter x determines “speed” of oscillations
compared to the lifetime
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Hints: for proving probabilitiesStarting point
Turn this around, gives
Time evolution
Use these to find
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Δmd = 0.507 ± 0.004 ps−1
xd = 0.770 ± 0.008 Δms = 17.719 ± 0.043 ps−1
xs = 26.63 ± 0.18
x = 0.00419 ± 0.00211
Lifetimes very different (factor 600)
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Key Points So Far
• K0, K0 are not CP eigenstates – need to make linear combination
• Short lived and long-lived Kaon states
• CP Violated (a tiny bit) in Kaon decays
• Describe this through Ks, KL as mixture of K0 K0
• Neutral mesons oscillate from particle to anti-particle• Can describe neutral meson oscillations through mixture of P0 P0
• Mass differences and width determine the rates of oscillations• Very different for different mesons (Bs,B,D,K)
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Cabibbo theoryand
GIM mechanism
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In 1963 N. Cabibbo made the first step to formally incorporate strangeness violation in weak decays1)For the leptons, transitions
only occur within a generation2)For quarks the amount of
strangeness violation can be neatly described in terms of a rotation, where qc=13.1o
Cabibbo rotation and angle (1/3)
, ,e LL L
e ud
cos sinc cL L
uud sd q q
+
Weakforce
transitions
u
d’ = dcosqc + ssinqc
W+
Idea: weak interaction couples to different eigenstates than strong interaction
weak eigenstates can be writtenas rotation of strongeigenstates
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Cabibbo’s theory successfully correlated many decay rates by counting the number of cosqc and sinqc terms in their decay diagram:
Cabibbo rotation and angle (2/3)
4
4 2
0 4 2
cos 0
sin 1
purely leptonic semi-leptonic, semi-leptonic,
e
e C
e C
e g
n pe g S
pe g S
q
q
cos Cg qg sin Cg q
E.g.
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There was however one major exception which Cabibbocould not describe: K0 + - (branching ratio ~7.10-9)
Observed rate much lower than expected from Cabibbo’s ratecorrelations (expected rate g8
sin2qc cos2qc)
Cabibbo rotation and angle (3/3)
d
+ -
ucosqc sinqc
WW
s
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The GIM mechanism (1/2)
In 1970 Glashow, Iliopoulos and Maiani publish a model forweak interactions with a lepton-hadron symmetry
The weak interaction couples to a rotated set of down-type quarks:the up-type quarks weakly decay to “rotated” down-type quarks
The Cabibbo-GIM model postulates the existence of a 4th quark :the charm (c) quark !
… discovered experimentally in 1974: J/Y cc state
'
,'
,,sc
due
e
sd
sd
cc
cc
qqqq
cossinsincos
''
Leptonsector
unmixed
Quark section mixed throughrotation of weak w.r.t. strong eigenstates by qc
2D rotation matrix
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The GIM mechanism (2/2) There is also an interesting symmetry between quark generations:
u
d’=cos(qc)d+sin(qc)s
W+
c
s’=-sin(qc)d+cos(qc)s
W+
sd
sd
cc
cc
qqqq
cossinsincos
''
Cabibbo mixing matrix
The d quark as seen by the W, the weak eigenstate d’,
is not the same as the mass eigenstate (the d)
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GIM suppression
d
+ -
ucosqc sinqc
WW
s d
+ -
ccosqc-sinqc
WW
s
expected rate (g4 sinqc cosqc - g4 sinqc cosqc)2
The cancellation is not perfect – these are only the vertex factors – as the masses of c and u are different
See also Bs + - discussion later
The model also explains the smallness of the K0 + - decay
The CKM matrix and theUnitarity Triangle
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How to incorporate CP violation in the SM?
hence “anti-unitary” T (and CP) operation corresponds to complex conjugation !
Simple exercise:
Since H = H(Vij), complex Vij would generate [T,H] 0 CP violation
W +
jDiU
i jU D
ijV
i j i jA U D A U D
W
jDiU
i jU D
ijV = only if:
ij ijV V
How does CP conjugation (or, equivalently, T conjugation)act on the Hamiltonian H ?
CP conservation is: (up to unphysical phase)
Recall:
ˆ ˆˆ ˆ, ˆ ˆˆ ˆ,
Px x Pp pTx x Tp p
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Brilliant idea from Kobayashi and Maskawa(Prog. Theor. Phys. 49, 652(1973) )
Try and extend number of families (based on GIM ideas).E.g. with 3:
… as mass and flavour eigenstates need not be the same (rotated)
In other words this matrix relates the weak states to the physical states
ud’
c s’
t b’
sd
sd
cc
cc
qqqq
cossinsincos
''
bsd
VVVVVVVVV
bsd
tbtstd
cbcscd
ubusud
'''
The CKM matrix (1/2)
Kobayashi Maskawa
Imagine a newdoublet of quarks
bsd
Vbsd
CKM
'''
2D rotation matrix3D rotation matrix
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Standard Model weak charged current
Feynman diagram amplitude proportional to Vij Ui Dj
• U (D) are up (down) type quark vectors
Vij is the quark mixing matrix, the CKM matrix
• for 3 families this is a 3x3 matrix
U =uct
D =dsb
The CKM matrix (2/2)
W +
jDiU
i jU D
ijV
tbtstd
cbcscd
ubusud
CKM
VVVVVVVVV
V
Can estimate relative probabilitiesof transitions fromfactors of |Vij |2
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As the CKM matrix elements are connected to probabilities of transition, the matrix has to be unitary:
CKM matrix – number of parameters (1/2)
Values of elements:a purely experimental matterikjk
jijVV *
In general, for N generations, N2 constraints
Sum of probabilities must add to 1 e.g. t must decay to either b, s, or d so
Freedom to change phase of quark fields
2N-1 phases are irrelevant(choose i and j, i≠j)
Rotation matrix has N(N-1)/2 angles
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CKM matrix – number of parameters (2/2)
Example for N = 1 generation: 2 unknowns – modulus and phase:
unitarity determines |V | = 1
the phase is arbitrary (non-physical)
| | iV e
no phase, no CPV
NxN complex element matrix: 2N2 parametersTotal - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases)
Number of phases
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CKM matrix – number of parameters (2/2)
NxN complex element matrix: 2N2 parametersTotal - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases)
Number of phases
Example for N = 2 generations: 8 unknowns – 4 moduli and 4 phases
unitarity gives 4 constraints :
for 4 quarks, we can adjust 3 relative phases
† 1 00 1
VV
only one parameter, a rotation (= Cabibbo angle) left: no phase no CPV
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CKM matrix – number of parameters (2/2)
NxN complex element matrix: 2N2 parametersTotal - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases)
Number of phases
Example for N = 3 generations: 18 unknowns – 9 moduli and 9 phases
unitarity gives 9 constraints
for 6 quarks, we can adjust 5 relative phases
4 unknown parameters left: 3 rotation (Euler) angles and 1 phase CPV !
In requiring CP violation with this structureof weak interactions K&M predicted
a 3rd family of quarks!
tbtstd
cbcscd
ubusud
CKM
VVVVVVVVV
V
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C12 S12 0-S12 C12 0 0 0 1
1 0 00 C23 S230 -S23 C23
3 angles q12, q23, q13 phase
VCKM = R23 x R13 x R12
R12 =R23 =
R13 =C13 0 S13 e-i
0 1 0-S13 e-i 0 C13
CKM matrix – Particle Data Group (PDG) parameterization
Define:
Cij= cos qij
Sij=sin qij
3D rotation matrix form
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A ~ 1, ~ 0.22, ≠ 0 but h ≠ 0 ???
21ˆ ,
21ˆ
22 hh
Introduced in 1983: 3 angles = S12 , A = S23/S2
12 , = S13cos/ S13S23
1 phase h = S13sin/ S12S23
VCKM(3) terms in up to 3
CKM terms in 4,5
CKM matrix - Wolfenstein parameters
Note:smallest couplings are complex ( CP-violation)
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A ~ 1, ~ 0.22, ≠ 0 but h ≠ 0 ???
21ˆ ,
21ˆ
22 hh
Introduced in 1983: 3 angles = S12 , A = S23/S2
12 , = S13cos/ S13S23
1 phase h = S13sin/ S12S23
VCKM(3) terms in up to 3
CKM terms in 4,5
CKM matrix - Wolfenstein parameters
Note:smallest couplings are complex ( CP-violation)
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A ~ 1, ~ 0.22, ≠ 0 but h ≠ 0 ???
21ˆ ,
21ˆ
22 hh
Introduced in 1983: 3 angles = S12 , A = S23/S2
12 , = S13cos/ S13S23
1 phase h = S13sin/ S12S23
VCKM(3) terms in up to 3
CKM terms in 4,5
CKM matrix - Wolfenstein parameters
1ˆˆ12
1
21
423
22
52
32
hh
h
h
iAAiA
AiA
iA
VCKM
Note:smallest couplings are complex ( CP-violation)
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CKM matrix - hierarchy
)1()()()()1()()()()1(
23
2
3
OOOOOOOOO
VVVVVVVVV
V
tbtstd
cbcscd
ubusud
CKM
~ 0.22top
charm
up down
strange
bottom
Charge: +2/3 Charge: 1/3
flavour-changing transitions by weak charged current (boldness indicates transition probability |Vij|)
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CKM – Unitarity Triangle
*cbcdVV
0*** ++ tbtdcbcdubud VVVVVV
*
*
cbcd
ubud
VVVV
• Three complex numbers, which sum to zero• Divide by so that the middle element is 1 (and real)• Plot as vectors on an Argand diagram• If all numbers real – triangle has no area – No CP violation
Real
Imag
inar
y
• Hence, get a triangle
‘Unitarity’ or ‘CKM triangle’• Triangle if SM is correct.
Otherwise triangle will not close,
Angles won’t add to 180o
*
*
1cbcd
cbcd
VVVV
*
*
cbcd
tbtd
VVVV
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ikjkj
ijVV *
Plot on Argand diagram: 6 triangles in complex plane
3,2,1,123
1
jVi
ij : no phase info.
kjkjVVi
ikij
,3,2,1,,03
1
*
0
0
0
0
0
0
***
***
***
***
***
***
++
++
++
++
++
++
cbubcsuscdud
tbcbtscstdcd
tbubtsustdud
tstdcscdusud
tbtscbcsubus
tbtdcbcdubud
VVVVVV
VVVVVV
VVVVVV
VVVVVV
VVVVVV
VVVVVVdb:sb:ds:ut:ct:uc:
Unitarity conditions and triangles
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The Unitarity Triangle(s) & the a, b, g angles Area of all the triangles is the same (6A2h)Jarlskog invariant J, related to how much CP violation
Two triangles (db) and (ut) have sides of similar size
• Easier to measure, (db) is often called THE unitarity triangle
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CKM Triangle - Experiment
Find particle decays that are sensitive to measuring the angles (phase difference) and sides (probabilities) of the triangles
• Measurements constrain the apex of the triangle
• Measurements are consistent
We will discuss how to experimentally measure the sides / angles
• CKM model works, 2008 Nobel prize
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Key Points So Far• K0, K0 are not CP eigenstates – need to make linear combination
• Short lived and long-lived Kaon states
• CP Violated (a tiny bit) in Kaon decays
• Describe this through Ks, KL as mixture of K0 K0
• Neutral mesons oscillate from particle to anti-particle
• Can describe neutral meson oscillations through mixture of P0 P0
• Mass differences and width determine the rates of oscillations
• Very different for different mesons (Bs,B,D,K)
• Weak and mass eigenstates of quarks are not the same
• Describe through rotation matrix – Cabibbo (2 generations), CKM (3 generations)
• CP Violation included by making CKM matrix elements complex
• Depict matrix elements and their relationships graphically with CKM triangle
Types of CP violationWe discussed earlier how CP violation
can occur in Kaon (or any P0) mixing if p≠q.
We didn’t consider the decay of the particle –
this leads to two more ways to violate CP
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CP in decay
CP in mixing
CP in interference between mixing and decay
Pff
P
ff P
Pff P
P P P
ff PP P P
+ +
Types of CP violation
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Occurs when a decay and its CP-conjugate decayhave a different probability
Decay amplitudes can be written as:
Two types of phase: Strong phase: CP conserving, contribution from intermediate states Weak phase : complex phase due to weak interactions
fP
fP
PHfA
PHfA
f
f
1
i
iii
i
iii
f
f
ii
ii
eeA
eeA
AA
1) CP violation in decay (also called direct CP violation)
Valid for both charged and neutral particles P
(other types are neutral only since involve oscillations)
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Mass eigenstates being different from CP eigenstates Mixing rate for P0 P0 can be different from P0 P0
If CP conserved :
If CP violated :
oo
oo
PqPpP
PqPpP
+
2
1
21
qpwith
1
2
2
1212
*1212
*2
iM
iM
pq
such asymmetries usually small need to calculate M,,
involve hadronic uncertainties hence tricky to relate to CKM
parameters
2) CP violation in mixing (also called indirect CP violation)
22
11
1
1
PPCP
PPCP
+
(This is the case if Ks=K1, KL=K2)
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Say we have a particle* such thatP0 f and P0 f are both possible
There are then 2 possible decay chains, with or without mixing!
Interference term depends on
Can put and get but
* Not necessary to be CP eigenstate
1pq 1
f
f
AA
1
3) CP violation in the interference of mixing and decay
CP can be conserved in mixing and in decay, and still be violated overall !
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Key Points So Far• K0, K0 are not CP eigenstates – need to make linear combination
• Short lived and long-lived Kaon states
• CP Violated (a tiny bit) in Kaon decays
• Describe this through Ks, KL as mixture of K0 K0
• Neutral mesons oscillate from particle to anti-particle
• Can describe neutral meson oscillations through mixture of P0 P0
• Mass differences and width determine the rates of oscillations
• Very different for different mesons (Bs,B,D,K)
• Weak and mass eigenstates of quarks are not the same
• Describe through rotation matrix – Cabibbo (2 generations), CKM (3 generations)
• CP Violation included by making CKM matrix elements complex
• Depict matrix elements and their relationships graphically with CKM triangle
• Three ways for CP violation to occur
• Decay
• Mixing
• Interference between decay and mixing