an introduction to flavour physics - warwick
TRANSCRIPT
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An introduction to Flavour Physics
!
Part 3 !
Thomas Blake Warwick Week 2015
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An introduction to Flavour Physics
• What’s covered in these lectures:
1.An introduction to flavour in the SM.
2.Neutral meson mixing and sources of CP violation.
3.CP violation in the SM.!
➡Angles and sides of the Unitary triangle, constraints from kaon physics, CPT and low energy observables.!
4.Flavour changing neutral current processes.
2
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Unitarity triangles• From lecture 1:
➡ There are 6 different unitarity triangles, all with equal area.
3
Re
Im
Re
Im
VudV⇤ub + VcdV
⇤cb + VtdV
⇤tb = 0
sides ⇠ �3 sides ⇠ �4,�2,�2
VusV⇤ub + VcsV
⇤cb + VtsV
⇤tb = 0
focus on triangle (one of two) with approximately equal length sides.
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Sides and angles of “the” unitarity triangle
4
Re
Im(⇢, ⌘)
↵
��
(1, 0)(0, 0)
Ru Rt
Rt =
����VtdV ⇤
tb
VcdV ⇤cb
����
� = arg
✓�VudV ⇤
ub
VcdV ⇤cb
◆
� = arg
✓�VcdV ⇤
cb
VtdV ⇤tb
◆
Ru =
����VudV ⇤
ub
VcdV ⇤cb
����sides:
angles: ↵ = arg
✓� VtdV ⇤
tb
VudV ⇤ub
◆
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CKM elements• Magnitude of most CKM elements is measurable using
semileptonic decays
!
!
!
!
!
• Exceptions are the elements Vtd and Vts. These come from mixing measurements in the Bd and Bs system (from ∆md and ∆ms).
5
0
@Vud Vus Vub
Vcd Vcs Vcb
Vtd Vts Vtb
1
A
Nuclear �-decay
K ! ⇡`⌫`b ! u`�⌫`
b ! c`�⌫`
t ! b`+⌫`c ! s`+⌫`
⌫` + d ! c+ `�
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An aside: lifetimes• Smallness of |Vub| and |Vcb|:
➡ B mesons are long lived.
6
⌧(B0) = 1.520± 0.004 ps
⌧(B+) = 1.638± 0.004 ps
⌧(B0s ) = 1.509± 0.004 ps
see http://www.slac.stanford.edu/xorg/hfag/ for details
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CKM
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Vtd and Vts
8
Rt extracted from mixing
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Vtd and Vts from mixing• Extracted from ∆m. Amplitude is given by
9
from the box diagram perturbative CQD corrections
A(B0q ! B0
q) =G2
F
4⇡2m2
W (V ⇤tbVtq)
2S0(m2t/m
2W )⌘B(µ)
⇥ hB0q|(bL�µdL)(bL�µdL)|B0
q i
A(B0q ! B0
q) =G2
F
6⇡2m2
W (V ⇤tbVtq)
2S0(m2t/m
2W )⌘f2
BB
decay constant and bag parameter from lattice
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Vub
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Vub and Vcb set the length of one side of the triangle
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Determining Vub• Three ways to determine Vub
1.Inclusive decays of
➡ No bare quarks, really looking at a sum of exclusive decays.
2.Exclusive decays, e.g.
3.Leptonic decays of e.g.
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b ! u`�⌫`
B0 ! ⇡+`�⌫`
B+ ! `+⌫` B+ ! ⌧+⌫⌧
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Inclusive Vub• Experimentally challenging due to
backgrounds from b → c semileptonic decays. ➡ Reduce backgrounds by cutting
on the mass of the Xu system or the lepton energy (cutting at the end point to reject Xc).
➡ Need a hermetic detector → BaBar and Belle.
• Cuts to reject b → c introduce larger theoretical uncertainties.
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Exclusive Vub• Much simpler experimentally,
but more challenging for theory. ➡ Dependence on form-
factors for the transition.
13
B0 ! ⇡+`�⌫`
B ! ⇡
Simultaneous fit of BaBar, Belle data & lattice data, using Boyd-Grinstein-Lebed parameterisation.
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Leptonic Vub • Branching fraction of leptonic decays
!
!
!
!
• Experimentally challenging, only has a large branching fraction. ➡ To reduce backgrounds in e+e- collisions can fully reconstruct
the other B meson in the event.
14
B(B� ! `�⌫`) =G2
FmBm2`
8⇡
✓1� m2
`
m2B
◆2
f2B |Vub|2⌧B
helicity suppression decay constant
from latticeB+ ! ⌧+⌫⌧
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Vub “tension”• Long-standing tension between
Vub measured by exclusive and inclusive measurements.
15
New “exclusive” result from LHCb at Moriond EW 2015 using ⇤0
b ! pµ�⌫µ
3.1𝜎
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Vub interpretation• Can attempt to explain the Vub
tension by introducing a RH current
!
!
• Can explain tension between old inclusive and exclusive measurements, but it’s difficult to reconcile with Vub from Λb decays.
16
[arXiv:1408.2516]
Le↵ / V Lub(u�µPLb+
"Ru�µPRb)(⌫�µPL`) + h.c
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Determining Vcb • Can determine Vcb in a similar manner to Vub, using inclusive or
exclusive decays.
• PDG gives, ➡ Inclusive Vcb
!
➡ Exclusive Vcb, using and decays
17
|Vcb| = (42.2± 0.7)⇥ 10�3
|Vcb| = (39.5± 0.8)⇥ 10�3
B ! D`⌫ B ! D⇤`⌫
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D(*)𝜏ν• There is also an interesting
“tension” between experiment and theory in D(*)𝜏ν decays.
18
BaBa
r Phy
s. R
ev. L
ett.
109,
101
802
(201
2)
RD(⇤) =�[B ! D(⇤)⌧�⌫⌧ ]
�[B ! D(⇤)`�⌫`]
RD,SM = 0.297± 0.017
RD⇤,SM = 0.252± 0.003
RD,exp = 0.462± 0.067
RD⇤,exp = 0.341± 0.029
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D(*)𝜏ν• Charged current interaction
couples universally to the different lepton flavours
due to phasepsace.
• Third generation 𝜏 lepton can be sensitive to extensions of the SM, e.g. to models with a second Higgs doublet that give rise to charged Higgs contributions.
19
RD(⇤) 6= 1
type II 2HDMBaBar analysis
model prediction
BaBa
r Phy
s. R
ev. L
ett.
109,
101
802
(201
2)
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CKM angle gamma
20
a
a
_
_
dm6K¡
K¡
sm6 & dm6
ubV
`sin 2
(excl. at CL > 0.95) < 0`sol. w/ cos 2
excluded at CL > 0.95
_
`a
l-1.0 -0.5 0.0 0.5 1.0 1.5 2.0
d
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5excluded area has CL > 0.95
Summer 14
CKMf i t t e r
From interference between and
� = arg
✓�VudV ⇤
ub
VcdV ⇤cb
◆
b ! cusb ! ucs
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Gamma
• Relative weak between the diagrams is -𝛾.
• To determine 𝛾, need to decay to a common final state.
• Neutral D mixing phase is known to be small.
21
B�
K�
D0
D0
b
u
c
c
u
s
s
VcbV⇤us VubV
⇤cs
B�K�
b
u
u
Colour favoured Colour suppressed
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Gamma• Need decay of and meson to a common final state.
• Two options:
22
D ! fCP D0 ! K+⇡� D0 ! K+⇡�or
Cabibbo favoured
Doubly Cabibbo suppressed
Atwood, Dunietz and Soni (ADS) method
Gronau, London and Wyler (GLW) method using or !Giri, Grossman, Soffer and Zupan (GGSZ) method using
D0 ! ⇡+⇡�
D0 ! K+K�
D0 ! K0S⇡
+⇡�
D0 D0
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Gamma
• Also need to account for relative suppression of the colour suppressed diagram and the relative strong phase difference, rB and 𝛿B.
• To maximise sensitivity to 𝛾 need large interference
➡ Interference is large for ADS, because we compare (favoured x suppressed) with (suppressed x favoured), i.e. similar magnitude. 23
B�
K�
D0
D0
b
u
c
c
u
s
s
VcbV⇤us VubV
⇤cs
B�K�
b
u
u
Colour favoured Colour suppressed
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GLW/ADS observables
24
RCP+ =
�[B± ! D[⇡+⇡�,K+K�]K±
]
�[B± ! Dfav.K±]
= 1 + r2B + 2rB cos �B cos �
RADS =
�[B± ! Dsupp.K±]
�[B± ! Dfav.K±]
=
r2B + r2D + 2rBrD cos(�B + �D) cos �
1 + (rBrD)
2+ 2rBrD cos(�B � �D) cos �
ACP+ =
�[B� ! DCPK�]� �[B+ ! DCPK+
]
�[B� ! DCPK�] + �[B+ ! DCPK+
]
=
2rB sin �B sin �
1 + r2B + 2rB cos �B cos �
AADS =
�[B� ! DADSK�]� �[B+ ! DCPK+
]
�[B� ! DADSK�] + �[B+ ! DADSK+
]
=
2rBrD sin(�B + �D) sin �
r2B + r2D + 2rBrD cos(�B + �D) cos �
• Large number of observables sensitive to 𝛾.
!
!
!
!
!
!
• rB ~ 0.1, rD from measurements at CLEO-c.
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CP in GLW
25
CP violation shows � 6= 0
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GGSZ• Need to perform an amplitude (Dalitz) analysis or bin in regions of
the Dalitz plot to extract 𝛾 when using
26
Resonant structure provides important phase information that can be used to remove ambiguities in determination of 𝛾
K0S⇡
+⇡�
e.g [PRL 105 (2010) 121801]
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Gamma• Combining measurements
for GLW+ADS and GGSZ (for many modes)
!
!
• Least well known of the angles.
27
� = (73.2+6.3�7.0)
�
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CKM angle alpha
28
a
a
_
_
dm6K¡
K¡
sm6 & dm6
ubV
`sin 2
(excl. at CL > 0.95) < 0`sol. w/ cos 2
excluded at CL > 0.95
_
`a
l-1.0 -0.5 0.0 0.5 1.0 1.5 2.0
d
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5excluded area has CL > 0.95
Summer 14
CKMf i t t e r
↵ = arg
✓� VtdV ⇤
tb
VudV ⇤ub
◆
Time dependent CP violation in b ! uud
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Measuring alpha• Can measure alpha from time
dependent CP violation in tree level decays.
!
• Unfortunately can also receive contributions from penguin decays to the same final state !!i.e. direct CP violation is possible.
29
⇢+
⇡�
⇢+
⇡�
B0
B0tb
b u
V ⇤ub
V ⇤tb Vtd
Vud
b ! uud
C = 0 , S = ⌘CP sin 2↵
b ! duu
C 6= 0 , S 6= ⌘CP sin 2↵
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Measuring alpha in ππ• To resolve penguin pollution can either find modes where pollution
is small or exploit the approximate isospin symmetry. ➡ Reminder, Isospin (u ↔ d) symmetry is an approximate
symmetry of the strong interaction (only broken by the difference between mu and md)
• The B is an I = 1/2 state. ➡ Tree decay can be ∆I = 1/2 or 3/2. ➡ Penguin decay is ∆I = 1/2. ➡ Bose statistics imply pions in I = 0 or I = 2.
Therefore, only allowed at tree level.
• Can relate the 2π final states using isospin and solve for alpha.
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B+ ! ⇡+⇡0
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Alpha (result)
31
↵ = (87.7+3.5�3.3)
�
ruled out by direct CP
violation in ππ
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kaon physics constraints
32
a
a
_
_
dm6K¡
K¡
sm6 & dm6
ubV
`sin 2
(excl. at CL > 0.95) < 0`sol. w/ cos 2
excluded at CL > 0.95
_
`a
l-1.0 -0.5 0.0 0.5 1.0 1.5 2.0
d
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5excluded area has CL > 0.95
Summer 14
CKMf i t t e r
From CP violation in the kaon system (q/p)
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CP in the kaon sector• CP violation first observed in 2π decays of KL mesons.
➡ Is it just mixing induced or do we also see direct CP violation?
• If CP violation is mixing induced expect
!
!
• Also see evidence for CP violation in semileptonic decays
!
!
• Relationship to ϵ/ϵ’
33
⌘00 =A(K0
L ! ⇡0⇡0)
A(K0S ! ⇡0⇡0)
, ⌘+� =A(K0
L ! ⇡+⇡�)
A(K0S ! ⇡+⇡�)
� = ACP(K0L ! `+⌫`⇡
�) =�[K0
L ! `+⌫`⇡�]� �[K0L ! `�⌫`⇡+]
�[K0L ! `+⌫`⇡�] + �[K0
L ! `�⌫`⇡+]
⌘00 = ✏� 2✏0 ⌘+� = ✏+ ✏0 � =2Re(✏)
1 + |✏|2
⌘00 = ⌘+�
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NA 48 experiment• Long running saga to establish Re(ϵ/ϵ’) ≠ 0
➡ Confirmed by NA48 at CERN and KTEV experiment in Japan.
34
NA48 beams
Simultaneous beams of KS and KL from separate
targets
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NA 48 experiment• Fixed target experiment in the CERN north area
35
R =|⌘00|2
|⌘+�|2⇡ 1� 6Re
✓✏0
✏
◆
= (13.7± 2.5± 1.8)⇥ 10�4
Double ratio of π0π0 and π+π- decays from KL and KS mesons
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Resources• CKMFitter
➡ http://ckmfitter.in2p3.fr/
• UTFit ➡ http://www.utfit.org/UTfit/
• Heavy Flavour Averaging Group (HFAG) ➡ http://www.slac.stanford.edu/xorg/hfag/
• Particle Data Group (PDG) ➡ http://pdg.lbl.gov/
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Time reversal• Time reversal symmetry maps
!
• Obviously can’t test this experimentally, because we can’t run our experiments backwards in time.
• We observe C violation and P violation, but the product CPT is known to be conserved, therefore CP violation = T violation.
37
t ! �t
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CPT theorem• Cannot write a quantum field theory that is Lorentz invariant
➡ measurements are invariant of the position or Lorentz boost of the system.
with a Hermitian Hamiltonian that violates the product of CPT.
• Several important consequences, CPT implies:
1.Mass and lifetime of particles and antiparticles are identical.
2.Quantum numbers of antiparticles are opposite those of particles.
3.Integer spin particles obey Bose-Einstein statistics and half- integer spin particles obey Fermi-Dirac statistics.
38
H = H†
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T violation in B system• Generalisation of the sin 2β analysis.
• Identify the flavour of the B by tagging the other B in the event. Also separate the decays by CP-odd ( ) or CP-even final state ( )
39
B0 ! B�
B0 ! B+
J/ K0S
J/ K0L
B0(t1) ! B�(t2)
B�(t1) ! B0(t2)
• Time reversal violation would appear as a difference in rates between
and B� ! B0
B+ ! B0
BaBar Phys. Rev. Lett. 109, 211801 (2012)
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T violation in B system• Generalisation of the sin 2β analysis.
• Identify the flavour of the B by tagging the other B in the event. Also separate the decays by CP-odd ( ) or CP-even final state ( )
40
J/ K0S
J/ K0L
B0(t1) ! B�(t2)
B�(t1) ! B0(t2)
• Time reversal violation would appear as a difference in rates between
and
i.e. tagged at time t1 and decaying at later time t2, compared to decaying at time t1, tagged at t2.
T violation has also been seen in Kaon decays.
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Triple product asymmetries• In pseudoscalar decays to a pair of vector mesons, can form
asymmetries of kinematical quantities that are products of 3 independent vectors, e.g.
!
!
• Triple products are T- and P-odd and C-even (hence CP-odd). A non-vanishing asymmetry would be T- and CP-violating.
• Unlike direct CP asymmetries, T-odd asymmetries are not suppressed by small strong phase differences
41
AT =�[~v1 · (~v2 ⇥ ~v3) > 0]� �[~v1 · (~v2 ⇥ ~v3) < 0]
�[~v1 · (~v2 ⇥ ~v3) > 0] + �[~v1 · (~v2 ⇥ ~v3) < 0]
AT / cos(�1 � �2) sin(�1 � �2)
strong phase difference weak phase
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Strong CP problem• In the SM, we only have evidence for CP violation in weak
interactions.
• However, can add natural terms to the QCD lagrangian that are P and T violating, i.e. would violate CP,
!
!
• This term would generate a neutron EDM, if 𝜃 were non-zero. Massive fine tuning problem, 𝜃 < 10-9.
• Most well known solution from Pecci-Quinn, is to introduce a new scalar particle (the axion) — a Nambu Goldstone boson of a U(1) symmetry.
• If one of the quarks were massless 𝜃 would be unobservable. 42
L = �1
4Fµ⌫F
µ⌫ � nfg2✓
32⇡2✏µ⌫��Fµ⌫F�� + (i�µDµ �mei✓
0�5)
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Low energy flavour conserving observables
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Electric dipole moments• Classically, EDMs are a measure of the spatial separation of
positive and negative charges.
• Can also be measured for fundamental particles (electron, muon, neutron etc).
• Tested using the Zeeman effect, i.e. looking for shift in energy levels under an external electric field.
• EDMs are both parity and time reversal violating.
!
[they can also be viewed as a measure of how spherical the shape of the particle is]
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Electric dipole moments• Electron EDM:
Science 343 (2014) 6168
• Muon EDM:
Phys. Rev. D 80 (2009) 052008
• Neutron EDM:
Phys. Rev. Lett. 97 (2006) 131801
• Probing amazingly small charge separation distances
e.g. if the Neutron were the size of the earth, the EDM corresponds to a charge separation of 10 µm.
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de < 8.7⇥ 10�29
dµ < 1.9⇥ 10�19
dn < 3.0⇥ 10�26
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Magnetic dipole moments• “Spinning” charge acts as a magnetic dipole with moment µ
giving an energy shift in external magnetic field,
!
• Prediction of g = 2 (classically g = 1 ) was a big success of the Dirac equation, e.g. in external field Aµ
!
!
• Receives corrections from higher order processes, e.g. at order 𝛼2,
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~µ = � e
2m~� = �g
µB
~~S
g = 2 +↵
⇡
✓1
2m(~p+ e ~A)2 +
e
2m� · ~B � eA0
◆ = E
�E = �~µ · ~B
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Anomalous magnetic moments
• (g-2)e is a powerful precision test of QED
!
• (g-2)µ receives important Weak and QCD contributions. The latest experimental value from the Brookhaven E821 experiment
!
from Phys. Rev. Lett. 92 (2004) 1618102 is ~3𝜎 from the SM expectation.
• Could this be a hint of a NP contribution to (g-2)µ? For a review see Phys. Rept. 477 (2009) 1-110 (arXiv:0902.3360).
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(g � 2)e = (1159.652186± 0.000004)⇥ 10�6
(g � 2)µ = (11659208± 6)⇥ 10�6
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Recap• In this lecture we discussed:
➡ The sides of the unitarity triangle and the tension in Vub.
➡ The CKM angles 𝛼 and 𝛾.
➡ CP violation in the kaon system. ➡ T violation and CPT. ➡ Low energy flavour conserving observables.
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Fin
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