Particle Physics Option

Total Page：16

File Type：pdf, Size：1020Kb

Reminder: Spin Algebra for a spin operator ‘J’: J[ i, J j]= ε iijkh k J ‘Isospin operator ‘I’ follows this same algebra Isospin is also additive. Two particles with Isospin Ia and Ib will give a total Isospin I = Ia + Ib By defining I+ = I1 + iI2 and I- = I1 -iI2 we could ‘Raise’ and ‘Lower’ the third component of isospin: 1/2 I-|i,m> = [i(i+1)-m(m-1)] |i,m-1> 1/2 I+|i,m> = [i(i+1)-m(m+1)] |i,m+1> NOTICE: I+|1/2,-1/2> = I+|d> = |u> (or -|d-bar>) All part of what we called SU(2) • Concept Developed Before the Quark Model • Only works because M(up) ? M(down) • Useful concept in strong interactions only • Often encountered in Nuclear physics • From SU(2), there is one key quantum number I3 Up quark ô Isospin = 1/2; I3 = 1/2 Anti-up quark ô I = 1/2; I3 = -1/2 Down quark ô I = 1/2; I3 = -1/2 Anti-down quark ô I = 1/2; I3 = 1/2 Graphical Method of finding all the possible combinations: 1). Take the Number of possible states each I3 particle can have and multiply them. This is the total number you must have in the end. A spin 1/2 particle can have 1 2 states, IF we are combining two particles: 1/2 2 X 2 = 4 total in the end. 0 -1/2 2) Plot the particles as a function of the -1 I3 quantum numbers. Graphical Method of finding all the possible combinations: Triplet Singlet Group A Group B I I3 3 Sum I3 1 1 1 1/2 1/2 1/2 0 0 0 -1/2 -1/2 -1/2 -1 -1 -1 Graphical Method of finding all the possible combinations: We have just combined two fundamental representations of spin 1/2, which is the doublet, into a higher dimensional representation consisting of a group of 3 (triplet) and another object, the singlet. What did we just do as far as the spins are concerned? Quantum states: Triplet I = |I, I3> |1,1> = |1/2,1/2>1 |1/2,1/2>2 |1,0> = 1/÷2 (|1/2,1/2>1 |1/2,-1/2>2 + |1/2,-1/2>1 |1/2,1/2>2 ) |1,-1> = |1/2,-1/2>1 |1/2,-1/2>2 Singlet |0,0> = 1/ ÷ 2 (|1/2,1/2>1 |1/2,-1/2>2 - |1/2,-1/2>1 |1/2,1/2>2) 1 1 d −= , Reminder: u = |1/2,1/2> 2 2 u-bar or d = |1/2,-1/2> Quantum states: Triplet I = 1 |I, I3> |1,1> = |1/2,1/2>1 |1/2,1/2>2 = -|ud> ⎡π + ⎤ ⎡ρ + ⎤ |1,0> = 1/2(|1/2,1/2> |1/2,-1/2> + |1/2,-1/2> |1/2,1/2> ) ⎢ 0 ⎥ ⎢ 0 ⎥ 1 2 1 2 ⎢π ⎥ ⎢ρ ⎥ = 1/2(|uu> - |dd>) ⎢ − ⎥ ⎢ − ⎥ ⎣π ⎦ ⎣ρ ⎦ |1,-1>= |1/2,-1/2>1 |1/2,-1/2>2 = |ud> Singlet |0,0>=1/2(|1/2,1/2>1 |1/2,-1/2>2 - |1/2,-1/2>1 |1/2,1/2>2 =1/2(|uu> + |dd>) Must choose either quark-antiquark states, or q-q states. We look for triplets with similar masses. MESONs fit the bill! π+,π0,π- and ρ+, ρ0, ρ- (q-qbar pairs). w0, f0, and h0 are singlets. WARNING: Ask about |1,0> minus sign or read Burcham & Jobes pgs. 361 and 718 But quarks are also in groups of 3 so we’d like to see that structure I3 too: 3/2 I3 I3 1 a s 1 1 1/2 1/2 1/2 0 0 0 -1/2 -1/2 -1/2 -1 -1 -1 -3/2 Isospins of a few baryon and meson states: + + 3 3 1 1 Δ = , p = , Σ+ 1 = , 1 1 1 2 2 2 2 Ξ0 = , 0 2 2 + 3 1 1 1 Σ1 = , 0 Δ = , n=, − − 1 1 2 2 2 2 − Ξ =, − Σ =1 , − 1 2 2 0 3 1 Δ =, − 1 1 2 2 p= −, − 1 1 2 2 K + = , − 3 3 2 2 Δ =, − 1 1 2 2 n = , 1 1 2 2 K 0=, − 2 2 + π =1 , 1 1 1 K 0= − , π 0 =1 , 0 η0= η′ 0= φ00 = , 0 2 2 1 1 π −=1 , − 1 K −=, − 2 2.

Copy Link

Recommended publications

Fundamentals of Particle Physics
Fundamentals of Par0cle Physics Particle Physics Masterclass Emmanuel Olaiya 1 The Universe u The universe is 15 billion years old u Around 150 billion galaxies (150,000,000,000) u Each galaxy has around 300 billion stars (300,000,000,000) u 150 billion x 300 billion stars (that is a lot of stars!) u That is a huge amount of material u That is an unimaginable amount of particles u How do we even begin to understand all of matter? 2 How many elementary particles does it take to describe the matter around us? 3 We can describe the material around us using just 3 particles . 3 Matter Particles +2/3 U Point like elementary particles that protons and neutrons are made from. Quarks Hence we can construct all nuclei using these two particles -1/3 d -1 Electrons orbit the nuclei and are help to e form molecules. These are also point like elementary particles Leptons We can build the world around us with these 3 particles. But how do they interact. To understand their interactions we have to introduce forces! Force carriers g1 g2 g3 g4 g5 g6 g7 g8 The gluon, of which there are 8 is the force carrier for nuclear forces Consider 2 forces: nuclear forces, and electromagnetism The photon, ie light is the force carrier when experiencing forces such and electricity and magnetism γ SOME FAMILAR THE ATOM PARTICLES ≈10-10m electron (-) 0.511 MeV A Fundamental (“pointlike”) Particle THE NUCLEUS proton (+) 938.3 MeV neutron (0) 939.6 MeV E=mc2. Einstein’s equation tells us mass and energy are equivalent Wave/Particle Duality (Quantum Mechanics) Einstein E
[Show full text]
Exploring the Spectrum of QCD Using a Space-Time Lattice
ExploringExploring thethe spectrumspectrum ofof QCDQCD usingusing aa spacespace--timetime latticelattice Colin Morningstar (Carnegie Mellon University) New Theoretical Tools for Nucleon Resonance Analysis Argonne National Laboratory August 31, 2005 August 31, 2005 Exploring spectrum (C. Morningstar) 1 OutlineOutline z spectroscopy is a powerful tool for distilling key degrees of freedom z calculating spectrum of QCD Æ introduction of space-time lattice spectrum determination requires extraction of excited-state energies discuss how to extract excited-state energies from Monte Carlo estimates of correlation functions in Euclidean lattice field theory z applications: Yang-Mills glueballs heavy-quark hybrid mesons baryon and meson spectrum (work in progress) August 31, 2005 Exploring spectrum (C. Morningstar) 2 MonteMonte CarloCarlo methodmethod withwith spacespace--timetime latticelattice z introduction of space-time lattice allows Monte Carlo evaluation of path integrals needed to extract spectrum from QCD Lagrangian LQCD Lagrangian of hadron spectrum, QCD structure, transitions z tool to search for better ways of calculating in gauge theories what dominates the path integrals? (instantons, center vortices,…) construction of effective field theory of glue? (strings,…) August 31, 2005 Exploring spectrum (C. Morningstar) 3 EnergiesEnergies fromfrom correlationcorrelation functionsfunctions z stationary state energies can be extracted from asymptotic decay rate of temporal correlations of the fields (in the imaginary time formalism) Ht −Ht z evolution in Heisenberg picture φ ( t ) = e φ ( 0 ) e ( H = Hamiltonian) z spectral representation of a simple correlation function assume transfer matrix, ignore temporal boundary conditions focus only on one time ordering insert complete set of 0 φφ(te) (0) 0 = ∑ 0 Htφ(0) e−Ht nnφ(0) 0 energy eigenstates n (discrete and continuous) 2 −−()EEnn00t −−()EEt ==∑∑neφ(0) 0 Ane nn z extract A 1 and E 1 − E 0 as t → ∞ (assuming 0 φ ( 0 ) 0 = 0 and 1 φ ( 0 ) 0 ≠ 0) August 31, 2005 Exploring spectrum (C.
[Show full text]
Particle Physics Dr Victoria Martin, Spring Semester 2012 Lecture 12: Hadron Decays
Particle Physics Dr Victoria Martin, Spring Semester 2012 Lecture 12: Hadron Decays !Resonances !Heavy Meson and Baryons !Decays and Quantum numbers !CKM matrix 1 Announcements •No lecture on Friday. •Remaining lectures: •Tuesday 13 March •Friday 16 March •Tuesday 20 March •Friday 23 March •Tuesday 27 March •Friday 30 March •Tuesday 3 April •Remaining Tutorials: •Monday 26 March •Monday 2 April 2 From Friday: Mesons and Baryons Summary • Quarks are confined to colourless bound states, collectively known as hadrons: " mesons: quark and anti-quark. Bosons (s=0, 1) with a symmetric colour wavefunction. " baryons: three quarks. Fermions (s=1/2, 3/2) with antisymmetric colour wavefunction. " anti-baryons: three anti-quarks. • Lightest mesons & baryons described by isospin (I, I3), strangeness (S) and hypercharge Y " isospin I=! for u and d quarks; (isospin combined as for spin) " I3=+! (isospin up) for up quarks; I3="! (isospin down) for down quarks " S=+1 for strange quarks (additive quantum number) " hypercharge Y = S + B • Hadrons display SU(3) flavour symmetry between u d and s quarks. Used to predict the allowed meson and baryon states. • As baryons are fermions, the overall wavefunction must be anti-symmetric. The wavefunction is product of colour, flavour, spin and spatial parts: ! = "c "f "S "L an odd number of these must be anti-symmetric. • consequences: no uuu, ddd or sss baryons with total spin J=# (S=#, L=0) • Residual strong force interactions between colourless hadrons propagated by mesons. 3 Resonances • Hadrons which decay due to the strong force have very short lifetime # ~ 10"24 s • Evidence for the existence of these states are resonances in the experimental data Γ2/4 σ = σ • Shape is Breit-Wigner distribution: max (E M)2 + Γ2/4 14 41.
[Show full text]
Three Lectures on Meson Mixing and CKM Phenomenology
Three Lectures on Meson Mixing and CKM phenomenology Ulrich Nierste Institut für Theoretische Teilchenphysik Universität Karlsruhe Karlsruhe Institute of Technology, D-76128 Karlsruhe, Germany I give an introduction to the theory of meson-antimeson mixing, aiming at students who plan to work at a flavour physics experiment or intend to do associated theoretical studies. I derive the formulae for the time evolution of a neutral meson system and show how the mass and width differences among the neutral meson eigenstates and the CP phase in mixing are calculated in the Standard Model. Special emphasis is laid on CP violation, which is covered in detail for K−K mixing, Bd−Bd mixing and Bs−Bs mixing. I explain the constraints on the apex (ρ, η) of the unitarity triangle implied by ǫK ,∆MBd ,∆MBd /∆MBs and various mixing-induced CP asymmetries such as aCP(Bd → J/ψKshort)(t). The impact of a future measurement of CP violation in flavour-specific Bd decays is also shown. 1 First lecture: A big-brush picture 1.1 Mesons, quarks and box diagrams The neutral K, D, Bd and Bs mesons are the only hadrons which mix with their antiparticles. These meson states are flavour eigenstates and the corresponding antimesons K, D, Bd and Bs have opposite flavour quantum numbers: K sd, D cu, B bd, B bs, ∼ ∼ d ∼ s ∼ K sd, D cu, B bd, B bs, (1) ∼ ∼ d ∼ s ∼ Here for example “Bs bs” means that the Bs meson has the same flavour quantum numbers as the quark pair (b,s), i.e.∼ the beauty and strangeness quantum numbers are B = 1 and S = 1, respectively.
[Show full text]
Electro-Weak Interactions
Electro-weak interactions Marcello Fanti Physics Dept. | University of Milan M. Fanti (Physics Dep., UniMi) Fundamental Interactions 1 / 36 The ElectroWeak model M. Fanti (Physics Dep., UniMi) Fundamental Interactions 2 / 36 Electromagnetic vs weak interaction Electromagnetic interactions mediated by a photon, treat left/right fermions in the same way g M = [¯u (eγµ)u ] − µν [¯u (eγν)u ] 3 1 q2 4 2 1 − γ5 Weak charged interactions only apply to left-handed component: = L 2 Fermi theory (effective low-energy theory): GF µ 5 ν 5 M = p u¯3γ (1 − γ )u1 gµν u¯4γ (1 − γ )u2 2 Complete theory with a vector boson W mediator: g 1 − γ5 g g 1 − γ5 p µ µν p ν M = u¯3 γ u1 − 2 2 u¯4 γ u2 2 2 q − MW 2 2 2 g µ 5 ν 5 −−−! u¯3γ (1 − γ )u1 gµν u¯4γ (1 − γ )u2 2 2 low q 8 MW p 2 2 g −5 −2 ) GF = | and from weak decays GF = (1:1663787 ± 0:0000006) · 10 GeV 8 MW M. Fanti (Physics Dep., UniMi) Fundamental Interactions 3 / 36 Experimental facts e e Electromagnetic interactions γ Conserves charge along fermion lines ¡ Perfectly left/right symmetric e e Long-range interaction electromagnetic µ ) neutral mass-less mediator field A (the photon, γ) currents eL νL Weak charged current interactions Produces charge variation in the fermions, ∆Q = ±1 W ± Acts only on left-handed component, !! ¡ L u Short-range interaction L dL ) charged massive mediator field (W ±)µ weak charged − − − currents E.g.
[Show full text]
Introduction to Flavour Physics
Introduction to flavour physics Y. Grossman Cornell University, Ithaca, NY 14853, USA Abstract In this set of lectures we cover the very basics of flavour physics. The lec- tures are aimed to be an entry point to the subject of flavour physics. A lot of problems are provided in the hope of making the manuscript a self-study guide. 1 Welcome statement My plan for these lectures is to introduce you to the very basics of flavour physics. After the lectures I hope you will have enough knowledge and, more importantly, enough curiosity, and you will go on and learn more about the subject. These are lecture notes and are not meant to be a review. In the lectures, I try to talk about the basic ideas, hoping to give a clear picture of the physics. Thus many details are omitted, implicit assumptions are made, and no references are given. Yet details are important: after you go over the current lecture notes once or twice, I hope you will feel the need for more. Then it will be the time to turn to the many reviews [1–10] and books [11, 12] on the subject. I try to include many homework problems for the reader to solve, much more than what I gave in the actual lectures. If you would like to learn the material, I think that the problems provided are the way to start. They force you to fully understand the issues and apply your knowledge to new situations. The problems are given at the end of each section.
[Show full text]
Lecture 5: Quarks & Leptons, Mesons & Baryons
Physics 3: Particle Physics Lecture 5: Quarks & Leptons, Mesons & Baryons February 25th 2008 Leptons • Quantum Numbers Quarks • Quantum Numbers • Isospin • Quark Model and a little History • Baryons, Mesons and colour charge 1 Leptons − − − • Six leptons: e µ τ νe νµ ντ + + + • Six anti-leptons: e µ τ νe̅ νµ̅ ντ̅ • Four quantum numbers used to characterise leptons: • Electron number, Le, muon number, Lµ, tau number Lτ • Total Lepton number: L= Le + Lµ + Lτ • Le, Lµ, Lτ & L are conserved in all interactions Lepton Le Lµ Lτ Q(e) electron e− +1 0 0 1 Think of Le, Lµ and Lτ like − muon µ− 0 +1 0 1 electric charge: − tau τ − 0 0 +1 1 They have to be conserved − • electron neutrino νe +1 0 0 0 at every vertex. muon neutrino νµ 0 +1 0 0 • They are conserved in every tau neutrino ντ 0 0 +1 0 decay and scattering anti-electron e+ 1 0 0 +1 anti-muon µ+ −0 1 0 +1 anti-tau τ + 0 −0 1 +1 Parity: intrinsic quantum number. − electron anti-neutrino ν¯e 1 0 0 0 π=+1 for lepton − muon anti-neutrino ν¯µ 0 1 0 0 π=−1 for anti-leptons tau anti-neutrino ν¯ 0 −0 1 0 τ − 2 Introduction to Quarks • Six quarks: d u s c t b Parity: intrinsic quantum number • Six anti-quarks: d ̅ u ̅ s ̅ c ̅ t ̅ b̅ π=+1 for quarks π=−1 for anti-quarks • Lots of quantum numbers used to describe quarks: • Baryon Number, B - (total number of quarks)/3 • B=+1/3 for quarks, B=−1/3 for anti-quarks • Strangness: S, Charm: C, Bottomness: B, Topness: T - number of s, c, b, t • S=N(s)̅ −N(s) C=N(c)−N(c)̅ B=N(b)̅ −N(b) T=N( t )−N( t )̅ • Isospin: I, IZ - describe up and down quarks B conserved in all Quark I I S C B T Q(e) • Z interactions down d 1/2 1/2 0 0 0 0 1/3 up u 1/2 −1/2 0 0 0 0 +2− /3 • S, C, B, T conserved in strange s 0 0 1 0 0 0 1/3 strong and charm c 0 0 −0 +1 0 0 +2− /3 electromagnetic bottom b 0 0 0 0 1 0 1/3 • I, IZ conserved in strong top t 0 0 0 0 −0 +1 +2− /3 interactions only 3 Much Ado about Isospin • Isospin was introduced as a quantum number before it was known that hadrons are composed of quarks.
[Show full text]
Chapter 18 Electromagnetic Interactions
Chapter 18 Electromagnetic Interactions 18.1 Electromagnetic Decays There are a few cases of particles which could decay via the strong interactions without violating flavour conservation, but where the masses of the initial and final state particles are such that this decay is not energetically allowed. For example the Σ ∗0 (mass 1385 MeV /c2) can decay into a Λ (mass 1115 MeV /c2) and a π0 (mass 135 MeV /c2). The quark content of the Σ ∗0 and Λ are the same and the π0 consists of a superposition of quark-antiquark pairs of the same flavour. As required in strong interaction processes the isospin is conserved - the Σ ∗0 has isospin I=1, the Λ has isospin I = 0 and the pion has isospin I = 1. On the other hand, the Σ 0 whose mass is 1189 MeV /c2 does not have enough energy to decay into a Λ and a pion. In such cases the decay can proceed via the electromagnetic interactions producing one or more photons in the final state. The dominant decay mode of the Σ 0 is Σ0 Λ + γ → The quark content of the Σ 0 and the Λ are the same, but one of the charged quarks emits a photon in the process. Note that in this decay the isospin is not conserved - the initial state has isospin I=1, whereas the final state has isospin I = 0. Electromagnetic interactions do not conserve isospin. Because the electromagnetic coupling constant, e, is much smaller than the strong cou- pling constant the rates for such decays are usually much smaller than the rates for decays which can proceed via the strong interactions.The lifetime of the Σ 0 is 10 −10 seconds, whereas the Σ ∗0 has a width of 36 MeV, corresponding to a lifetime of about 10 −23 seconds.
[Show full text]
Lecture 7 Isospin
Isospin H. A. Tanaka Announcements • Problem Set 1 due today at 5 PM • Box #7 in basement of McLennan • Problem set 2 will be posted today PRESS RELEASE 6 October 2015 The Nobel Prize in Physics 2015 The Royal Swedish Academy of Sciences has decided to award the Nobel Prize in Physics for 2015 to Takaaki Kajita Arthur B. McDonald Super-Kamiokande Collaboration Sudbury Neutrino Observatory Collaboration University of Tokyo, Kashiwa, Japan Queen’s University, Kingston, Canada “for the discovery of neutrino oscillations, which shows that neutrinos have mass” Metamorphosis in the particle world The Nobel Prize in Physics 2015 recognises Takaaki lenges for more than twenty years. However, as it requires Kajita in Japan and Arthur B. McDonald in Canada, neutrinos to be massless, the new observations had clearly for their key contributions to the experiments which showed that the Standard Model cannot be the complete demonstrated that neutrinos change identities. This theory of the fundamental constituents of the universe. metamorphosis requires that neutrinos have mass. The discovery rewarded with this year’s Nobel Prize The discovery has changed our understanding of the in Physics have yielded crucial insights into the all but innermost workings of matter and can prove crucial hidden world of neutrinos. After photons, the particles to our view of the universe. of light, neutrinos are the most numerous in the entire Around the turn of the millennium, Takaaki Kajita presented cosmos. The Earth is constantly bombarded by them. the discovery that neutrinos from the atmosphere switch Many neutrinos are created in reactions between cosmic between two identities on their way to the Super-Kamiokande radiation and the Earth’s atmosphere.
[Show full text]
Quark Masses 1 66
66. Quark masses 1 66. Quark Masses Updated August 2019 by A.V. Manohar (UC, San Diego), L.P. Lellouch (CNRS & Aix-Marseille U.), and R.M. Barnett (LBNL). 66.1. Introduction This note discusses some of the theoretical issues relevant for the determination of quark masses, which are fundamental parameters of the Standard Model of particle physics. Unlike the leptons, quarks are confined inside hadrons and are not observed as physical particles. Quark masses therefore cannot be measured directly, but must be determined indirectly through their influence on hadronic properties. Although one often speaks loosely of quark masses as one would of the mass of the electron or muon, any quantitative statement about the value of a quark mass must make careful reference to the particular theoretical framework that is used to define it. It is important to keep this scheme dependence in mind when using the quark mass values tabulated in the data listings. Historically, the first determinations of quark masses were performed using quark models. These are usually called constituent quark masses and are of order 350MeV for the u and d quarks. Constituent quark masses model the effects of dynamical chiral symmetry breaking discussed below, and are not directly related to the quark mass parameters mq of the QCD Lagrangian of Eq. (66.1). The resulting masses only make sense in the limited context of a particular quark model, and cannot be related to the quark mass parameters, mq, of the Standard Model. In order to discuss quark masses at a fundamental level, definitions based on quantum field theory must be used, and the purpose of this note is to discuss these definitions and the corresponding determinations of the values of the masses.
[Show full text]
Chapter 12 Charge Independence and Isospin
Chapter 12 Charge Independence and Isospin If we look at mirror nuclei (two nuclides related by interchanging the number of protons and the number of neutrons) we ﬁnd that their binding energies are almost the same. In fact, the only term in the Semi-Empirical Mass formula that is not invariant under Z (A-Z) is the Coulomb term (as expected). ↔ Z2 (Z N)2 ( 1) Z + ( 1) N a B(A, Z ) = a A a A2/3 a a − + − − P V − S − C A1/3 − A A 2 A1/2 Inside a nucleus these electromagnetic forces are much smaller than the strong inter-nucleon forces (strong interactions) and so the masses are very nearly equal despite the extra Coulomb energy for nuclei with more protons. Not only are the binding energies similar - and therefore the ground state energies are similar but the excited states are also similar. 7 7 As an example let us look at the mirror nuclei (Fig. 12.2) 3Li and 4Be, where we see that 7 for all the states the energies are very close, with the 4Be states being slightly higher because 7 it has one more proton than 3Li. All this suggests that whereas the electromagnetic interactions clearly distinguish between protons and neutrons the strong interactions, responsible for nuclear binding, are ‘charge independent’. Let us now look at a pair of mirror nuclei whose proton number and neutron number 6 6 diﬀer by two, and also the nuclide between them. The example we take is 2He and 4Be, which are mirror nuclei. Each of these has a closed shell of two protons and a closed shell of 6 6 two neutrons.
[Show full text]
Higgs Bosons with Large Couplings to Light Quarks
YITP-SB-19-25 Higgs bosons with large couplings to light quarks Daniel Egana-Ugrinovic1;2, Samuel Homiller1;3 and Patrick Meade1 1C. N. Yang Institute for Theoretical Physics, Stony Brook University, Stony Brook, New York 11794, USA 2Perimeter Institute for Theoretical Physics, Waterloo, Ontario N2L 2Y5, Canada 3Department of Physics, Brookhaven National Laboratory, Upton, New York 11973, USA Abstract A common lore has arisen that beyond the Standard Model (BSM) particles, which can be searched for at current and proposed experiments, should have flavorless or mostly third-generation interactions with Standard Model quarks. This theoretical bias severely limits the exploration of BSM phenomenology, and is especially con- straining for extended Higgs sectors. Such limitations can be avoided in the context of Spontaneous Flavor Violation (SFV), a robust and UV complete framework that allows for significant couplings to any up or down-type quark, while suppressing flavor- changing neutral currents via flavor alignment. In this work we study the theory and phenomenology of extended SFV Higgs sectors with large couplings to any quark generation. We perform a comprehensive analysis of flavor and collider constraints of extended SFV Higgs sectors, and demonstrate that new Higgs bosons with large arXiv:1908.11376v2 [hep-ph] 31 Dec 2019 couplings to the light quarks may be found at the electroweak scale. In particular, we find that new Higgses as light as 100 GeV with order ∼ 0.1 couplings to first- or second-generation quarks, which are copiously produced at the LHC via quark fusion, are allowed by current constraints. Furthermore, the additional SFV Higgses can mix with the SM Higgs, providing strong theory motivation for an experimental program looking for deviations in the light quark{Higgs couplings.
[Show full text]