flavour oscillation and cp violation flavour oscillation: mixing
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Flavour Oscillation and CP Violation
• Quarks Mixing and the CKM Matrix
• Flavour Oscillation: Mixing of Neutral Mesons
• CP violation
• Neutrino Mixing
Stephanie Hansmann-Menzemer 1
Quark Mixing in SMstrong, elm, weak NC conserve flavour and have identical coupling (constants) for
all up-type and down-type quarks, all leptons and all neutrinos:
• uγµu, cγµc, tγµt
• eγµe, µγµµ, τγµτ
• dγµd, sγµs, bγµb
• νeγµνe, νµγµνµ, ντγ
µντ
these interactions leave abmiguity for definition of quark and lepton eigenstates
e.g. rotational freedom in space of up-type quark:
x = 1√2(u + c) would still have the same form of Lagrangian with same coupling:
xγµx
weak CC couples up and down type quarks: uγµ(1 − γ5)d, ...
Stephanie Hansmann-Menzemer 2
Quark Mixing in SMYukawa term (coupling of Higgs to fermions) introduces:
“
u, c, t”
mu
0
B
B
@
u
c
t
1
C
C
A
+“
d, s, b”
md
0
B
B
@
d
s
b
1
C
C
A
; mu, md (3x3) mass matrices
choice of mass eigenstates for representation of Lagrangien introduceds a (3x3)
rotation matrix for u-type quarks and one for d-type quarks.
U†muU =
0
B
B
@
mu 0 0
0 mc 0
0 0 mt
1
C
C
A
V †mdV =
0
B
B
@
md 0 0
0 ms 0
0 0 mb
1
C
C
A
U, V are unitary matrices, q are mass eigenstates
→
0
B
B
@
u
c
t
1
C
C
A
= U
0
B
B
@
u
c
t
1
C
C
A
0
B
B
@
d
s
b
1
C
C
A
= V
0
B
B
@
d
s
b
1
C
C
A
All terms of the Lagrangian but CC are invariant under this rotation.
LCC =“
u c t”
U†Uγµ(1 − γ5)V †V
0
B
B
@
d
s
b
1
C
C
A
=“
u c t”
Uγµ(1 − γ5)V †
0
B
B
@
d
s
b
1
C
C
A
Stephanie Hansmann-Menzemer 3
Quark Mixing in SM
“
u c t”
Uγµ(1 − γ5)V †
0
B
B
@
d
s
b
1
C
C
A
=“
u c t”
γµ(1 − γ5)UV †
0
B
B
@
d
s
b
1
C
C
A
0
B
B
@
d′
s′
b′
1
C
C
A
= UV †
0
B
B
@
d
s
b
1
C
C
A
=
0
B
B
@
Vud Vus Vub
Vcd Vcs Vcb
Vtd Vts Vtb
1
C
C
A
×
0
B
B
@
d
s
b
1
C
C
A
flavour CKM matrix mass
JCCµ ∝
“
u, c, t”
㵓
1 − γ5”
VCKM
0
B
B
@
d
s
b
1
C
C
A
for quarks: flavour eigenstates 6= mass eigentstates!direct result of coupling of Higgs to quark flavours!
Stephanie Hansmann-Menzemer 4
CKM Matrix I
d′
s′
b′
=
Vud Vus Vub
Vcd Vcs Vcb
Vtd Vts Vtb
d
s
b
flavour CKM matrix mass
18 parameters (9 complex elements)-5 relative quark phases (unobservable)-9 unitarity conditions————————-= 4 independent parameters 3 Euler angles and 1 Phase
4 fundamental Standard Model Parameters (out of ∼18 (28))
Stephanie Hansmann-Menzemer 5
CKM Matrix II
Lagrangian insensitive to phases of left-handed fields:possible redefinition:
uL → eiφ(u)uL cL → eiφ(c)cL tL → eiφ(t)tL
dL → eiφ(d)dL sL → eiφ(s)sL bL → eiφ(b)bL
φ(q): real numbers
V =
e−iφ(u) 0 0
0 e−iφ(c) 0
0 0 e−iφ(t)
Vud Vus Vub
Vcd Vcs Vcb
Vtd Vts Vtb
eiφ(d) 0 0
0 eiφ(s) 0
0 0 eiφ(b)
5 unobservable phase differences:
Vlj → ei(φ(j)−φ(l))VljStephanie Hansmann-Menzemer 6
CKM Matrix III
u c t
d s b
u
d ’
c
s ’
t
b ’
Diagonal elements of CKM matrix are close to one.Only small of diagonal contributions.Mixing between quark families is “CKM suppressed”.
Stephanie Hansmann-Menzemer 7
Definition of CKM Angles
VCKM =
Vud Vus Vub
Vcd Vcs Vcb
Vtd Vts Vtb
=
1 − λ2/2 λ Aλ3e−iγ
−λ 1 − λ2/2 Aλ2
Aλ3e−iβd −Aλ2 1
+ O(λ4)
cos θC = λ ∼ 0.22;
one unitarity relation: VudV∗ub + VcdV
∗cb + VtdV
∗tb = 0
Bd triangle:α = arg
(
− VtdV ∗tb
VudV ∗ub
)
;
β = arg(
−VcdV ∗cb
VtdV ∗tb
)
;
γ = arg(
−VudV ∗ub
VcdV ∗cb
)
;
area of triangle 6= 0 → complex contributions of CKM elementsStephanie Hansmann-Menzemer 8
Unitarity Triangle
γ
α
α
dm∆
Kε
Kεsm∆ & dm∆
ubV
βsin 2(excl. at CL > 0.95)
< 0βsol. w/ cos 2
α
βγ
ρ-0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0
η
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
excl
uded
are
a ha
s C
L >
0.95
Winter 12
CKMf i t t e r
Current status of knowledge on “the” (Bd) CKM triangle.Sofar all measurements consistent with each other.
Stephanie Hansmann-Menzemer 9
Complex Elements and CP Violation
Stephanie Hansmann-Menzemer 10
Neutral Meson Mixing
Stephanie Hansmann-Menzemer 11
Phenomenology of Mixing ISchrödinger equation for unstable mesons (at rest):
i ddt|Ψ >= H|Ψ >= (m − i
2Γ)|ψ >
→ |Ψ(t) >= |Ψ0 > e−imte−1
2Γt
→ ||Ψ(t) > |2 = ||Ψ0 > |2e−Γt
For neutral mesons, consider 2 components (formulas equivalent valid for K0, D0, B0s mesons):
id
dt
0
@
B0
B0
1
A = H
0
@
B0
B0
1
A =
0
@
H11 H21
H12 H22
1
A
0
@
B0
B0
1
A
=“
M − i2Γ
”
0
@
B0
B0
1
A =
0
@
m11 − i2Γ11 m21 − i
2Γ21
m12 − i2Γ12 m22 − i
2Γ22
1
A
0
@
B0
B0
1
A
CPT theorem:
m11 = m22 = m(B0) = m(B0)
Γ11 = Γ22 = Γ
= 1τ(B0)
= 1
τ(B0)
off-diagonal elements ⇒ mixing
M , Γ hermetic:
m12 = m∗21, Γ12 = Γ∗
21
Stephanie Hansmann-Menzemer 12
Mass Eigenstates
diagonlizing matrix → mass eigenstates x with eigenvalues mx + iΓx2
BL = p|B0 > +q|B0 > with mL, ΓL
BH = p|B0 > −q|B0 > with mH , ΓH
|p2| + |q2| = 1, complexe coefficients
Flavour eigenstates:
B0 = 12p(|BL > +|BH >)
B0 = 12q (|BL > −|BH >)
Parameters of mass states:
mH,L = m ± Re√
H12H21
ΓH,L = Γ ∓ 2Im√
H12H21
∆m = mH − MK = 2Re√
H12H21
∆Γ = ΓH − ΓL = −4Im√
H12H21
x ≡ ∆mΓ ; y ≡ ∆Γ
2Γ
Stephanie Hansmann-Menzemer 13
Phenomenology of Mixing II
H =
m − i2Γ ∆m
2 − i2∆Γ
∆m2 − i
2∆Γ m − i2Γ
Two mixing mechanisms:
• mixing through decay: y = ∆ΓΓ ∼ O(1)
• mixing through oscillation: x = ∆mΓ ∼ O(1)
(K0K0, B0B0, B0s B0
s and D0D0 show different behavior)
long distant, on shell states
important for K not for B mesons
→ ∆Γ
short distant, virtual states
→ ∆m Stephanie Hansmann-Menzemer 14
Time Evolution I|BH/L(t) >= bH/L(t)|BH,L > with bH/L(t) = e−imH/Lte−ΓH/Lt/2
|B0(t) > =|BL(t) > +|BH(t) >
2p
=1
2p(bL(t)(p|B0 > +q|B0 >) + bH(t)(p|B0 > −q|B0 >))
= f+(t)|B0 > − q
pf−(t)|B0 >
|B0(t) > = f+(t)|B0 > +p
qf−(t)|B0 >
f±(t) = 12
“
eimH te−ΓH t/2 ± e−imLte−ΓLt/2”
CP-violation in mixing: P (B0 → B0) 6= P (B0 → B0) ⇔ | qp| 6= 1
Stephanie Hansmann-Menzemer 15
Time Evolution II
f±(t) = 12
(
eimH te−ΓH t/2 ± e−imLte−ΓLt/2)
P (B0 → B0) = P (B0 → B0) = |f+(t)|2 =14
(
e−ΓLt + e−ΓH t + 2e−(ΓL+ΓH)t/2 cos ∆mt)
P (B0 → B0) = | qp |2|f−(t)|2 =
14 |
qp |2
(
e−ΓLt + e−ΓH t − 2e−(ΓL+ΓH)t/2 cos ∆mt)
P (B0 → B0) = |pq |2|f−(t)|2 =
14 |
pq |2
(
e−ΓLt + e−ΓH t − 2e−(ΓL+ΓH)t/2 cos ∆mt)
Mixing asymmetry :P (B0→B0)−P (B0→B0)
P (B0→B0)+P (B0→B0)= 2e−(ΓH+ΓL)t/2 cos ∆mt
e−ΓHt+e−ΓLt
assume | qp | = 1Stephanie Hansmann-Menzemer 16
Phenomenology of Mixing IIITwo mixing mechanisms:
• mixing through decay: y = ∆ΓΓ ∼ O(1)
• mixing through oscillation: x = ∆mΓ ∼ O(1)
K0/K0 D0/D0 B0/B0 Bs/Bs
τ [ps]∗ 89 0.4 1.6 1.5
51700
Γ [ps−1] 5.6× 10−3 2.4 0.64 0.62
y= ∆Γ2Γ
-0.997 0.01 |y|<0.01 0.03±0.03
∆m [ps−1] 5.3 × 10−3 0.02 0.5 17.8
x= ∆mΓ
0.95 0.01 0.8 26
xxxxxxxx∗) at LHCb energies lifetime in ps ∼ decay length in cm
K0: mixing in decay and mixing in oscillation (medium) -
D0: very slow mixing - 2008
Bd: dominantly mixing in oscillation (medium) - 1987
Bs: dominantly mixing in oscillation (very fast) - 2006Stephanie Hansmann-Menzemer 17
How to Measure Mixing?
• identify initial flavour of meson
• identify final flavour of meson→ identify Meson as mixed or unmixed
• if mixing is fast ... need to measured decay time:
decay length L+ momentum p → t = mLp
• if mixing is slow ... can deduce information from time integratedasymmetry
Stephanie Hansmann-Menzemer 18
Neutral Kaonsassume no CP violation, Ks, KL correspond to mass eigenstates:
K short
• τ(K0L) = 51.7 ± 0.44 ns
• typical decay length: 5-20 m
• KL → 3π BR > 99%
• CP = -1
|KL >= 1√
2|K0 > −|K0 >
K long
• τ(K0S) = 0.089 ± 0.001 ns
• typical decay length: few cm
• KS → 2π BR > 99%
• CP = +1
|KS >= 1√
2|K0 > +|K0 >
Stephanie Hansmann-Menzemer 19
CPLEAR Experiment
Stephanie Hansmann-Menzemer 20
CPLEAR Experiment
Stephanie Hansmann-Menzemer 21
CPLEAR Experiment
A = 2e−(ΓS+ΓL)t/2 cos ∆mte−ΓSt+e−ΓLt
Stephanie Hansmann-Menzemer 22
B-Mixing
B mesons are produced in high energetic collitions:
• B factories:e+e− → Y (4S) → B0B0
e+e− → Y (4S) → B+B−
50% : 50% for charged and neutral B meson productionm(Y(4S)) = 10.58 GeV; not sufficient energy to produce Bse−
e+
b
b
Y(4S)γ
b
b
uu
B−
B+
e.g. ARGUS, BELLE, BABAR
• pp, pp collidere.g. TEVATRON, LHC
Stephanie Hansmann-Menzemer 23
ARGUS 1987
e+e− → Y (4S) → B0B0
√s = 10.58 GeV
unmixed:
B0B0 → ℓ+ℓ−
mixed:
B0B0 → ℓ+ℓ+
B0B0 → ℓ−ℓ−
time integrated measurement:
∼ 18% of B0 mix before they decay
decay of other B meson is used to tagged production
tt
d
d
b
b
Bd Bd
V V
VV
tb td
td tb*
*
(u,c) (u,c)
Stephanie Hansmann-Menzemer 24
B0s
−
¯B0
sMixing Analysis
xxx t = L Bm
p
L−D
−π
+K−
K
+π
b hadronPV
Bs0
s
signal sideopposite side
1) Bs selection & reconstruction2) Measurement of proper decay time ct
3) Flavor tagging (main challenge at hadron colliders)
Time dependent asymmetry measurement:
A(t) ≡ N(t)mixed−N(t)unmixed
N(t)mixed+N(t)unmixed∝ cos(∆mst)
Stephanie Hansmann-Menzemer 25
Time Dependent Asymmetries
decay time, ps0.0 0.5 1.0 1.5 2.0
prob
abili
ty d
ensi
ty
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
totalunmixedmixed
decay time, ps
asym
met
ry
������������������������������������������������������������������
������������������������������������������������������������������
A = #unmixed − #mixed#unmixed + #mixed
0 2.5 5 7.5 10proper decay time, t [ps]
asym
met
ry Bd mixing ∆md = 0.5 ps−1
Bs mixing ∆ms= 18 ps−1
A = 2e−(ΓH+ΓL)t/2 cos ∆mte−ΓHt+e−ΓLt ∼ cos ∆m
ΓH ∼ ΓL
Stephanie Hansmann-Menzemer 26
Time dependent Asymmetry
???Stephanie Hansmann-Menzemer 27
Tagging Dilution
Measured Asymmetrie:
Ameasured(t) =N(B0)
′
(t) − N(B0)′
(t)
N(B0)′
(t) + N(B0)′
(t)
=N(B0)(t)(1 − Pmt) + N(B0)(t)Pmt − N(B0)(t)(1 − Pmt) − N(B0)(t)Pmt
NRS(t) + NWS(t)
= (1 − 2Pmt)NB0
(t) − NB0(t)
NB0(t) + NB0
(t)= (1 − 2Pmt)A(t) = DA(t)
N′
B0/N
′
B0: as B0/B0 tagged decays
Pmt : Mistag-Probability
D : Tagging Dilution
Stephanie Hansmann-Menzemer 28
Auswirkung der Tagging Dilution
Mis-tag dilutes the observed Oszillation!
decay time, ps0.0 0.5 1.0 1.5 2.0
prob
abili
ty d
ensi
ty
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
perfect tagging
totalunmixedmixed
decay time, ps0.0 0.5 1.0 1.5 2.0
prob
abili
ty d
ensi
ty
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
P(mistag) = 0.4
totalunmixedmixed
Amplitude: 1-2P(mistag)
Stephanie Hansmann-Menzemer 29
Asymmetry Measurement
Tagging dilution D = 1 − 2Pmt
Always true ↔ 100% dilution; radom Tag (Pmt = 50%) ↔ 0% Dilution;
Tagging efficiency ǫ = Ntagged
Nall
effective statist. size of sample
→ Neff = Nall × ǫD2.
[%] ǫD2 Reduktion des Datensatzes
D0/CDF 2.5 - 5.0 × 20-50
BABAR/BELLE ≈ 30 × 3-4
LHCb 3.0 × 30
e+e− experiments factor of 10 better!
Stephanie Hansmann-Menzemer 30
Flavour Tagging Methoden
K+
sBbl−
Opposite Side Tagging
bb Paarproduktion→ korrelierte ProduktionsflavourInklusive Reko. des OS B
xxxxxx Same Side Tagging
xb
u
u
Bb
s
K +
s
s
Nicht bei Y (4S) möglich!(keine Fragmentation)Stephanie Hansmann-Menzemer 31
Mixing @ Babar & CDF
B0B0
x
B0s B
0s (2006)
[ps]sm∆/πDecay Time Modulo 20 0.05 0.1 0.15 0.2 0.25 0.3 0.35
Fitt
ed A
mpl
itude
-2
-1
0
1
2
data
cosine with A=1.28
CDF Run II Preliminary -1L = 1.0 fb
Stephanie Hansmann-Menzemer 32
Bs − Bs Mixing at LHCb
∼ 9.250 Bs candidates in 3 channels
Proper time resolution: σt = 45 fs
∆ms = 17.725 ± 0.041 ± 0.026 ps−1
world best measurement!
]2 mass [MeV/csB5400 5600 5800
2#
even
ts /
15 M
eV/c
0
2000
datafitsignalmisid. bkg.
Ks D→sB
comb. bkg.
LHCb preliminary
= 7 TeVs
-1341 pb
[ ps ]sm∆ / πt modulo 20 0.05 0.1 0.15 0.2 0.25 0.3 0.35
mix
A
-0.4
-0.2
0
0.2
0.4
LHCb preliminary
= 7 TeVs
-1341 pb
Aufl
ösun
gre
d.A
mpl
itude
LHC
b-C
ON
F-2
011-
050
Stephanie Hansmann-Menzemer 33
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