An introduction to

3WCTMFlavour Physics @ ! Part 1 ! Tom Blake ! Warwick Week 2018

1 An introduction to Flavour Physics

• What’s covered in these lectures: 1. An introduction to flavour in the SM. ➡ A few concepts and a brief history of flavour physics. 2. CP violation (part 1) 3. CP violation (part 2). 4. Flavour changing neutral current processes.

T. Blake 2 Questions for the LHC era?

• What is the mechanism for EW symmetry breaking? ✓ We now know that this is the Higgs mechanism. • What “new physics” lies between the EW and Planck scale?

x No idea! There is no evidence for new particles being directly produced at the TeV scale. • What is dark matter?

x No idea! Dark matter could be heavy or it could be light. There is no evidence for large missing energy in BSM searches at CMS and ATLAS, which constrains EW scale dark matter, or in direct detection experiments.

• Can we explain the matter antimatter asymmetry of the Universe?

x No! The SM cannot explain the baryon asymmetry of the Universe.

T. Blake 3 SM particle content

{ x3 colours

T. Blake 4 Flavour in the SM

• Particle physics can be described to excellent precision by a very simple theory:

! = (A , )+ (,A , ) LSM LGauge a i LHiggs a i with: ➡ Gauge terms that deals with the free fields and their interactions by the strong and electroweak interactions. ➡ Higgs terms that gives mass to the SM particles.

T. Blake 5 Flavour in the SM

• The Gauge part of the Lagrangian is experimentally very well verified, 1 = i ¯ ⇢⇢D F a F µ⌫,a . ! LGauge j j 4g2 µ⌫ a a Xj, X • The fields are arranged a left-handed doublets and right-handed singlets = QL,uR,dR,LL,eR ! u ⌫ Q = L ,L = L ! L d L e ✓ L◆ ✓ L◆

• The Lagrangian is invariant under a specific set of symmetry groups: SU(3) SU(2) U(1) c ⇥ L ⇥ Y • There are three replicas of the basic fermion families, which without the Higgs would be identical (huge degeneracy).

T. Blake 6 Flavour in the SM

• The Higgs part of the Lagrangian, on the other hand, is much more ad-hoc. It is necessary to understand the data but is not stable with respect to quantum corrections (often referred to as the Hierarchy problem). It is also the origin of the flavour structure of the SM.

!

• Masses of the fermions are generated by the Yukawa mechanism:

Q¯i Y ijdj + ... d¯i M ijdj + ... L D R ! L D R Q¯i Y ijdj + ... u¯i M ijuj + ... L U R c ! L U R

T. Blake 7 Flavour in the SM

• Can pick a basis in which one of the Yukawa matrices is diagonal, e.g. yd 00 yu 00 ! mi Y = 0 y 0 ,Y = V † 0 y 0 ,y D 0 s 1 U 0 c 1 i ⇡ 174GeV ! 00yb 00yt @ A @ A !

• The matrix V is complex and unitary (V†V = 1).

• To diagonalise both mass matrices it is necessary to rotate uL and dL separately. Consequently, V also appears in charged current interactions, J µ =¯u µd u¯ V µd W L L ! L L • V is known as the Cabibbo, Kobayashi, Maskawa matrix.

T. Blake 8 Flavour in the SM

mass eigenstates ≠ weak eigenstates

! d, s, b ↔ d’, s’, b’

T. Blake 9 Free parameters of the SM

• 3 gauge couplings

• Higgs mass and vacuum expectation value Flavour parameters • 6 quark masses

• 3 quark mixing angles and one complex phase (in V)

• 3 charged lepton masses

• 3 neutrino masses

• 3 lepton mixing angles and (at least) one complex phase

T. Blake 10 Why is flavour important?

• Most of the free parameters of the SM are related to the flavour sector. ➡ Why are there as so many parameters and why do they have the values they do?

• The flavour sector provides the only source of CP violation in the SM.

• Flavour changing neutral current processes can probe mass scales well beyond those accessible at LHC. ➡ If there are new particles at the TeV-scale, why don’t they manifest themselves in FCNC processes (the so-called flavour problem)?

T. Blake 11 Puzzles?

• Why do we have a flavour structure with three generations? ➡ We know we need 3+ generations to get CP violation. Are there more generations to discover?

• Why do the quarks have a flavour structure that exhibits both smallness and hierarchy? ➡ Why is the neutrino flavour sector so different (neither small nor hierarchical)?

T. Blake 12 Flavour hierarchy?

CKM matrix for quark sector PMNS matrix for neutrino sector

T. Blake 13 Mass hierarchy?

• 12 Large hierarchy in 10 scale between the 9 masses of the fermions. 10

mass [eV] 6 • Equivalent to having a 10 large hierarchy in the 3 10 Yukawa couplings. 1 • Why is this hierarchy so -3 large? Why is yt ~ 1? 10

- - - ν ν ν e µ τ u d s c b t e µ τ

T. Blake 14 Symmetries and flavour *KUVQTKECNRGTURGEVKXGQPHNCXQWTKPVJG5/

15 Isospin

• Proton and neutron have: ➡ Different charges but almost identical masses. ➡ An identical coupling to the strong force.

• In 1932 Heisenberg proposed that they are both part of the an Isospin doublet and can be treated as the same particle with different isospin projections.

p(I, IZ) = (½,½), n(I,IZ) = (½,-½)

• Pions can be arranged as an Isospin triplet,

π+(I, IZ) = (1,+1), π0(I, IZ) = (1, 0) , π-(I, IZ) = (1,-1)

T. Blake 16 Isospin

• Heisenberg proposed that strong interactions are invariant under rotations in Isospin space (an SU(2) invariance). ➡ Isospin was conserved in strong interactions. ➡ Isospin is violated in weak interactions.

! (p + p d + ⇡+):(p + n d + ⇡0)=2:1 ! ! !

• We now know that this is not the correct model but it is still a very useful concept.

• It works because mu ~ md < ΛQCD and can be used to predict interaction rates.

T. Blake 17 Clebsch-Gordon coefficients

T. Blake 18 Kaon observation

• In 1947, Rochester and Butler observed two new particles with masses around 500 MeV and relatively long lifetimes.

0 + + + K ⇡ ⇡ K µ ⌫µ S ! !

T. Blake 19 Quark model

Iz • Concept of quarks introduced by Gell- Mann, Nishijima and Ne’eman to explain the “zoo” of particles.

• Now understood to be real “fundamental” particles.

• Strangeness conserved by the strong interaction but violated in weak decays.

• Organised by

Y = B + SQ= e(Iz + Y/2)

baryon strangeness Isospin number

T. Blake 20 Quark model

meson } • Can only make colour neutral objects: q q¯ ➡ Quark anti-quark or three quark combinations (mesons and baryons). q q q }

baryon

T. Blake 21 SU(2) flavour mixing

• Four possible combinations from u and d quarks ! uu, dd, ud, du • Under SU(2) symmetry,

➡ π0 is a member of an isospin triplet, � is a isosinglet

1 1 ⇡0 = (uu dd), ⌘ = (uu + dd) p2 p2

T. Blake 22 SU(3) flavour mixing

• Introducing the strange quark, under SU(3) symmetry we now have an octuplet and a singlet.

! 0 1 1 1 ⇡ = (uu dd), ⌘1 = (uu + dd + ss), ⌘8 = (uu + dd 2ss) ! p2 p3 p6

• Physical states involve a further mixing

⌘ = ⌘ cos ✓ + ⌘ sin ✓⌘0 = ⌘ sin ✓ + ⌘ cos ✓ 1 8 1 8

T. Blake 23 Quark model

• Can only make colour neutral objects: q q¯ ➡ quark anti-quark or three quark combinations (mesons and baryons). Nearly all known q q q particles fall into one of these two categories. ➡ Can also build colour neutral states containing more quarks (e.g. 4 or 5 quark states).

pentaquark hybrid

tetraquark q q q q q¯ + … q¯ q q q q¯ q¯ q q q¯ q¯ weakly bound molecule glueball

T. Blake 24 Exotic charmonium states

• Several exotic states have recently been discovered: ➡ X(3872) by CDF, Z(4430)+ and Y(4140) by Belle etc

• These states decay to charmonia and have masses below the b- quark mass. ➡ Z(4430)+ is charged, therefore has minimal quark content ccud

• States could be weakly bound molecular D D states or genuine four quark states or an admixture of the two.

T. Blake 25 Dalitz plot formalism

• Often analyse three body decays in terms off the Dalitz plot formalism.

• n-body decay rate: (2⇡)4 ! d = 2d(p ...p ) 2M |M| 1 n !

• For a 3-body decay:

! 1 1 2 2 2 A,B parallel B d = 3 3 dm12dm23 (2⇡) 32M |M| A at rest ! C C ie 3-body phasespace is flat in the Dalitz plot.

T. Blake 26 Dalitz plot formalism

• Often analyse three body decays in terms off the Dalitz plot formalism. spin-1 • Resonances appear as bands resonance in the Dalitz plot. ➡ Number of lobes is related to spin of the resonance.

T. Blake 27 Pentaquark discovery ] 2 • In ⇤ J/ pK decays, the 26 b ! LHCb

LHCb experiment sees a [GeV

p 24 ψ

resonant contribution to the 2 J/ J/ p mass. m 22 J/ѱ p resonance? 20 • This contribution would have minimal quark content ccuud¯ 18 16 • To understand what the 2 3 4 5 6 2 2 contribution is, need to mKp [GeV ] perform an amplitude [LHCb, Phys. Rev. Lett. 115 (2015) 072001] analysis of the J/ pK system.

T. Blake 28 2200 data ] total fit 2000 (a) LHCb background Pentaquark P (4450) 1800 c Pc(4380) 1600 Λ(1405) Λ(1520) 1400 (1600) Events/(15 MeV) Λ 1200 Λ(1670) Λ(1690) 1000 Λ(1800) Λ(1810) • Perform an amplitude analysis of the 800 Λ(1820) 600 Λ(1830) Λ(1890) J/ pK system allowing for 400 Λ(2100) Λ(2110) contributions from all known Λ 200 0 1.4 1.6 1.8 2 2.2 2.4 2.6 resonances. mK p [GeV] • Data can be described by 800 introducing two new J/ p states. 700 (b) LHCb

➡ States have opposite parity, one is 600 (2015) 072001 115 Lett. Rev. Phys. [LHCb, wide with J = 3/2 and the other Events/(15 MeV) 500 400

narrow with J = 5/2. 300

200

100

0 4 4.2 4.4 4.6 4.8 5

mJ/ψp [GeV]

T. Blake 29 Pentaquark

• How can we be sure that these really are genuine new states?

• Resonances should have a pole:

0.15 c ➡ Exploit the phase P 0.1 (a) (b) Im A rotation of a BW. 0.05

0

-0.05 Pc(4450) -0.1 Pc(4380)

-0.15

-0.2

-0.25 -0.3 LHCb -0.35 -0.35 -0.3 -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.115 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 P Pc Re A c Re A [LHCb, Phys. Rev. Lett. 115 (2015) 072001] T. Blake 30 Cabibbo angle

• The quark content of the K+ and K0 are (¯ su ) and (¯ sd)

• The main decays of the K+ are + + + 0 + ! K µ ⌫ and K ⇡ e ⌫ ! µ ! e i.e. it decays via the charged current interaction.

• The charged current interaction couples to left-handed doublets, therefore need to construct a doublet that allows s→u and d→u.

• Cabibbo proposed a solution in terms of quark mixing

u u = d0 d cos ✓ + s sin ✓ ✓ ◆ ✓ C C ◆

T. Blake 31 Cabibbo angle

• The quark mixing angle, �C, is determined experimentally to be

! sin ✓ 0.22 C ⇡ • Cabibbo’s proposed solution also explained a discrepancy between the weak coupling constant between muon decays and nuclear decay.

• However, this opened up a new problem, why is + 0 + ! [K µ⌫] [K µ µ] ? ! L ! • If the doublet of the weak interaction is the one Cabibbo suggested, can have neural currents 0 J = d¯0 (1 )d0 ! µ µ 5 which introduces tree level FCNCs.

T. Blake 32 GIM mechanism

0 2 2 • Expanding, J =¯s (1 )s sin ✓ + d¯ (1 )d cos ✓ µ µ 5 C µ 5 C +¯s (1 )d sin ✓ cos ✓ ! µ 5 C C + d¯ (1 )s sin ✓ cos ✓ ! µ 5 C C • So was Cabibbo wrong? Glashow, Iliopoulos and Maiani provided a solution in 1970 by adding a second doublet

u ! u c c = , = d0 d cos ✓ + s sin ✓ s0 d sin ✓ + s cos ✓ ✓ !◆ ✓ C C ◆ ✓ ◆ ✓ C C ◆ • The second doublet exactly cancels the FCNC term. Quark mixing led to the prediction of the charm quark.

T. Blake 33 GIM at loop order?

• For “strange” decays still have an W effective GIM suppression. s d ! u, c

• Amplitude is: Z0 2 2 (V ⇤ V )f(const + m /m ) ! A ⇠ us ud u W 2 2 +(V ⇤ V )f(const + m /m ) ! cs cd c W

• 2 x 2 unitarity implies

! V ⇤ Vud + V ⇤ Vcd =0 0 us cs A ⇡ • FCNC decays are very rare, 0 + 9 (K µ µ)=(6.8 0.1) 10 B L ! ± ⇥

T. Blake 34 Observation of J/�

• Experimental evidence for charm quark came in 1974.

• Discovery of charmonium (J) at Brookhaven in p Be → e+e−X.

• Discovery of charmonium (�) at SLAC in e+e− → hadrons, e+e−, µ+µ−

T. Blake 35 �-� puzzle

• Two decays were found for charged strange mesons

! ✓+ ⇡+⇡0 + ! + + ⌧ ⇡ ⇡⇡ ! ! • The � and � had the same mass and lifetime, but the parity of 2π and 3π is different.

➡ Resolution is that the � and � are the same particle and parity is violated in the decay.

T. Blake 36 C and P

• Prior to 1956, it was thought that the laws of physics were invariant under parity, i.e. mirror image of a process is also a valid physical process.

➡ Shown to be violated in β-decays of Co-60 by C. S. Wu (following an idea by T. D. Lee an C. N. Yang).

• Now know that Parity is maximally violated in weak decays.

➡ No right-handed neutrinos.

• C is also maximally violated in weak decays.

➡ No left-handed anti-neutrino.

• The product CP, distinguishes between matter and anti-matter.

T. Blake 37 Neutral kaon system

• Ignoring CP violation, the two physical states in the neutral kaon system are 0 0 0 0 ! K K K + K K1 = | i| i and K2 = | i | i | i p2 | i p2 ! under Parity and Charge Conjugation

! K0 = K0 , K0 = K0 and K0 = K0 P| i | i C| i | i CP| i | i

• For the physical states

! K1,2 = K1,2 , K1,2 = K1,2 and K1,2 = K1,2 P| i | i C| i ⌥| i CP| i ±| i i.e. they are P, C and CP eigenstates as well.

T. Blake 38 Neutral kaon system

• What does this tell us about their decays?

+ ! ⇡ ⇡ shorter lived K1 ➡ P = +1, C = +1 and CP = +1 }

+! 0 ⇡ ⇡⇡ longer lived K2 ➡ P = -1, C = + 1 and CP = -1 }

• K2 decays to 3π but the 2π decay would be forbidden if CP is conserved.

T. Blake 39 CP violation in the kaon system

• In 1964, Christensen, Cronin, Fitch and Turlay observed 2π decays of K2 mesons (KL). Observation of CP violation in the kaon system.

Fixed target p+Be experiment at Brookhaven AGS

T. Blake 40 The CKM matrix CPF%2XKQNCVKQPKPVJG5/

41 V†V = 1

• The CKM matrix is a complex 3x3 unitary matrix ➡ 9 magnitudes and 9 phases

• Unitary condition gives 9 constraints, e.g.

! VudVub⇤ + VcdVcb⇤ + VtdVtb⇤ =0 V 2 + V 2 + V 2 =1 ! | ud| | us| | ub| • Can remove a single phase (global phase) and absorb phases into external quark fields. ➡ 4 parameters, 3 Euler angles and a single complex phase.

NB If there were only two generations, V would be a real rotation matrix with no complex phase.

T. Blake 42 CKM matrix

• Standard form is to express the CKM matrix in terms of three rotation matrices and one CP violating phase,

! i13 10 0 c13 0 s13e c12 s12 0 VCKM != 0 c23 s23 010 s12 c12 0 0 1 0 +i13 1 0 1 0 s23 c23 s13e 0 c13 001 ! @ A @ A @ A where

cij = cos ✓ij and sij =sin✓ij

T. Blake 43 Im Unitarity triangles

VudVub⇤ Re • Unitarity conditions can be represented by triangles in the complex plane. Im ➡ Six triangles with the same area.

VtdVtb⇤

Re

Im

Re VcdVcb⇤ T. Blake 44 Jarlskog invariant

• All of the unitarity triangles have the same area, called the Jarlskog invariant.

! J =Im(V V V ⇤ V ⇤) for i = k andj = l | | ij kl kj il 6 6 • This is a phase convention independent measure of CP violation in the quark sector. 2 • In the standard notation J = c12c13c23s12s13s23 sin • Small size of the Euler angles means J (and CP violation) is small in the SM.

T. Blake 45 Matter antimatter asymmetry CPFVJG5CMJCTQXEQPFKVKQPU

46 Matter dominated Universe

• From CMB measurements by WMAP + Planck

! nB nB¯ 10 6 10 ! n ⇡ ⇥

• In early (hot) universe expect annihilation to give

! n n ¯ n B ⇡ B ⇡ The matter anti-matter imbalance is small but far too large to be explained by Electroweak Baryogenesis. CP violation in the SM (quark sector) is too small due to the small size of the mixing angles and large hierarchy of quark masses.

T. Blake 47 Sakharov conditions

• Three conditions needed to generate a matter dominated universe from a symmetric initial state were proposed by A. Sakharov in 1967:

1. C and CP violation.

2. Baryon number violation.

3. A system out of thermal equilibrium.

T. Blake 48 Sakharov conditions

• If we start with an equal amount of matter (M) and anti-matter (M̅) and M A ! ! M A ! !

where A and A̅ have baryon numbers NA and -NA.

• If C and CP are violated [M A] = [M¯ A¯] ! ! 6 ! but even with different decay rates, after sufficient time there will be equal amounts of matter/antimatter.

T. Blake 49 Sakharov conditions

• Can get round this problem by having two (or more) competing process

! M A with probability p ! M B with probability 1 p ! ! M¯ A¯ with probabilityp ¯ ! ! M¯ B¯ with probability 1 p¯ • Then !

! N = N p + N (1 p) N ¯ p¯ N ¯ (1 p¯) A · B · A · B · ! =(p p¯)(N N ) A B • Need NA different from NB to generate an asymmetry (i.e. baryon number violation as well as CP violation).

T. Blake 50 Sakharov conditions

• Even then, the system needs to be out of thermal equilibrium or

! [A B + C]=[B + C A] ! ! ! and the asymmetry will be destroyed as fast as it is created.

T. Blake 51 Fin

52 Low energy flavour observables 'NGEVTKECPFOCIPGVKEFKRQNGOQOGPVU

53 Magnetic dipole moments

• “Spinning” charge acts as a magnetic dipole with moment µ giving an energy shift in external magnetic field, ! E = µ~ B~ · • Prediction of g = 2 (classically g = 1) was a big success of the Dirac equation, e.g. in external field Aµ

! 1 e e µB (p~ + eA~)2 + B~ eA0 = E µ~ = ~ = g S~ 2m 2m · 2m ~ ✓ ! ◆

• Receives corrections from higher order processes, e.g. at order �2, ↵ g =2+ ⇡

T. Blake 54 Anomalous magnetic moments

• (g-2)e is a powerful precision test of QED 6 (g 2) = (1159.652186 0.000004) 10 ! e ± ⇥ • (g-2)µ receives important Weak and QCD contributions. The latest experimental value from the Brookhaven E821 experiment 6 (g 2) = (11659208 6) 10 ! µ ± ⇥ from [Phys. Rev. Lett. 92 (2004) 1618102] is ~3� 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).

T. Blake 55 Electric dipole moments

• Classically, EDMs are a measure of the spatial separation of positive and negative charges in a particle. ➡ A finite EDM can only exist if the charge centres do not coincide.

• Can also be measured for fundamental particles (electron, muon, neutron etc). ➡ Interpreted as a measure of the sphericity of particle.

• Tested using the Zeeman effect, i.e. looking for shift in energy levels under an external electric field E = d ~ E ~ . ·

T. Blake 56 Electric dipole moments

• A non zero EDM would violate T and P symmetries. ➡ Under time reversal, the magnetic moment would change direction but the EDM would remain unchanged. ➡ Under parity, the EDM would changes direction but the magnetic dipole moment remains unchanged.

• Violation of P and T implies CP violation.

T. Blake 57 Electric dipole moments

• Electron EDM: 29 d < 8.7 10 [Science 343 (2014) 6168] e ⇥ • Muon EDM: 19 d < 1.9 10 [Phys. Rev. D 80 (2009) 052008] µ ⇥ • Neutron EDM:

26 [Phys. Rev. Lett. 97 (2006) 131801] d < 3.0 10 n ⇥ !

• Probing amazingly small charge separation distances!

T. Blake 58 Strong CP problem

• The complicated nature of the QCD vacuum should give rise to a term: ↵s µ⌫ ˜ ✓ = ✓ Fa Fa,µ⌫ ! L 8⇡

• This is both P and T violating but C conserving (and hence CP violating).

• This term will also contribute to the neutron dipole moment but experimentally we know this is small. 2 9 ! dn e ✓ mq/MN → ✓ 10 ' · ·  • What mechanism forces � to be small?

T. Blake 59 Strong CP problem

• The small size of the � parameter is a (another) massive fine tuning problem.

• Peccei-Quin solution is to introduce a U(1) symmetry that removes the strong CP problem by dynamically making � small. ➡ Spontaneous breaking of this symmetry is associated with a pseudo Nambu-Goldstone boson (c.f. Higgs mechanism), the axion. ➡ The axion can be light particle that couples very weakly to known SM particles.

T. Blake 60 Axion searches

• There are a large number of searches for axions produced in particle collisions.

• Could also be detected by converting axions to photons in the presence of a strong magnetic field, e.g. CAST experiment at CERN.

T. Blake 61