An introduction to Flavour Physics

!

Part 3 !

Thomas Blake Warwick Week 2015

1

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

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.

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

◆

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+ `�

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

CKM

7

Vtd and Vts

8

Rt extracted from mixing

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

Vub

10

Vub and Vcb set the length of one side of the triangle

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+ ! ⌧+⌫⌧

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.

12

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.

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+ ! ⌧+⌫⌧

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

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

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⇤`⌫

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

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)

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

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

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

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

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.

CP in GLW

25

CP violation shows � 6= 0

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]

Gamma• Combining measurements

for GLW+ADS and GGSZ (for many modes)

!

!

• Least well known of the angles.

27

� = (73.2+6.3�7.0)

�

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

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↵

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.

30

B+ ! ⇡+⇡0

Alpha (result)

31

↵ = (87.7+3.5�3.3)

�

ruled out by direct CP

violation in ππ

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)

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 = ⌘+�

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

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

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/

36

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

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†

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)

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.

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

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)

Low energy flavour conserving observables

43

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]

44

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.

45

de < 8.7⇥ 10�29

dµ < 1.9⇥ 10�19

dn < 3.0⇥ 10�26

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,

46

~µ = � e

2m~� = �g

µB

~~S

g = 2 +↵

⇡

✓1

2m(~p+ e ~A)2 +

e

2m� · ~B � eA0

◆ = E

�E = �~µ · ~B

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).

47

(g � 2)e = (1159.652186± 0.000004)⇥ 10�6

(g � 2)µ = (11659208± 6)⇥ 10�6

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.

48

Fin

49