The charmed double bottom baryon Author: Marcel Roman´ı Rod´es. Facultat de F´ısica, Universitat de Barcelona, Diagonal 645, 08028 Barcelona, Spain. Advisor: Dr. Joan Soto i Riera (Dated: January 15, 2018) Abstract: The aim of this project is to calculate the wavefunction and energy of the ground 0 state of the Ωcbb baryon, which is made up of 2 bottom and 1 charm quarks. Such a particle has ++ not been found yet, but recent observation of the doubly charmed baryon Ξcc (ucc) indicates that a baryon with three heavy quarks may be found in the near future. In this work, we will use the fundamental representation of the SU(3) group to compute the interaction between the quarks, then we will follow the Born-Oppenheimer approximation to find the effective potential generated by the motion of the c quark, which will allow us to solve the Schr¨odinger equation for the bb system. The total spatial wavefunction we are looking for results from the product of the wavefunctions of the two components (c and bb). Finally, we will discuss the possible states taking into account the spin and color wavefunctions. I. INTRODUCTION II. THE STRONG INTERACTION The interaction between quarks is explained by Quan- In July 2017, the LHCb experiment at CERN reported tum Chromodynamics (QCD), which is a gauge theory the observation of the Ξ++ baryon [1], indicating that, cc based on the SU(3) symmetry group. At short distances, sooner rather than later, baryons made up of three heavy where the confinement term is negligible, the interaction quarks will be found. potential between two heavy quarks is Coulomb-like be- In this project we are going to study the specific case of cause it approximately corresponds to the exchange of a 0 the charmed double bottom Omega baryon Ωcbb, whose single gluon. It is given, in natural units, by mass and properties have not been established yet. The #α fact that we are dealing with three heavy quarks im- V (r) = s ; (1) plies that, at first approximation, the QCD potential r is Coulomb-like, which allows us to use the well-known techniques of atomic and molecular physics, accessible at where αs is the strong interaction coupling constant and the Bachelor's degree level. # is the color factor, which depends on the representa- tion of the two quarks. The value of # is found using We will study the strong interactions using the funda- the properties of the fundamental representation of the mental representation of the SU(3) group, and we will SU(N) Lie algebra. This properties are: apply the Born-Oppenheimer approximation to calculate the possible wavefunctions of the 3-body system. • Antisymmetric structure constants: [T a;T b] = if abcT c In order to do so, we will split the problem into two parts, which is a good path to follow thanks to the mass • Traceless: T r(T a) = 0 difference between the bottom and charm quarks, being the former much heavier than the latter. First, we will a b 1 • Normalization: T r(T T ) = 2 δab suppose that the b's are fixed and at a relative distance R, and we will use the variational principle to find the a b 1 • Symmetric structure constants: fT ;T g = N δab + ground state energy of the c quark for each possible value dabcT c of R, which will act as a potential energy to the bb system. To do this more accurately, we must find a charge factor a a 1 1 • Projection operator: TijTkl = 2 δilδjk − N δijδkl function k(R) using numerical methods, as it will appear to be the solution to a transcendental equation. The Feynman diagram for the quark-quark and the quark-antiquark interactions are the following Then, we will solve the Schr¨odinger equation for the mo- tion of the bb system numerically, running a code on a a T 0 0 T 0 0 α α α α 0 α α α α 0 Mathematica, letting the total potential be the sum of q1 q1 q q the energy of the fundamental state of the c quark and the QCD potential between the bb pair. g g Finally, we will discuss the possible states of the total β β0 β β0 0 0 wavefunction taking into account the color and spin of the q2 q2 q q T a T a constituent particles and the Pauli Exclusion Principle. β0β ββ0 The charmed double bottom baryon Marcel Roman´ıRod´es (cc¯) being α and β the color of the ingoing quarks (antiquarks) to find suitable values for αs and mc. We took 0 0 and α and β the color of the outgoing quarks (anti- Λ ≡ ΛMS = 332 ± 17MeV (a constant of integration quarks). The Feynman rule for a vertex involving two whose value is indicative of the energy range where non- µ a quarks and a gluon is −igsγijTα0α. We are interested in perturbative dynamics dominates) from eq. (24.d) in a a the product Tα0αTβ0β, which will yield the suitable color [4], corresponding to nf = 3, and Mexp(cc¯) from [5]. factor when we multiply it by , connecting the in- (cc¯) αβγ The solution is αs = 0:624 and mc = 1:86 GeV. Re- dexes of the ingoing quarks so the resulting particle has peating this calculation for the b's using experimental zero net color charge. We have (b¯b) values of the bottomium (b¯b) we get αs = 0:373 and mb = 5:03 GeV. a a 1 1 T 0 T 0 = δ 0 δ 0 − δ 0 δ 0 = Now that we have the masses m and m , we are able to α α β β αβγ 2 α β β α N α α β β αβγ b c calculate the coupling at the scales we are interested in, 1 1 N + 1 namely the C.o.M of bc and that of bb. We will do it by = 0 0 − 0 0 = − 0 0 2 β α γ N α β γ 2N α β γ solving (3) with the right expressions for t: " # N+1 N=3 2 m m 2 #2 α2 showing that #qq = − 2N = − 3 . b c qq s (bc) t = ln 2 −! αs = 0:762 For the quark-antiquark interaction we are interested in mb + mc Λ a a β the product Tα0αTββ0 multiplied by δα, so the net charge " # m 2 #2 α2 of the resulting meson is zero. t = ln b qq s −! α(bb) = 0:524: 2 Λ2 s a a 1 1 T 0 T 0 δ = δ δ 0 0 − δ 0 δ 0 δ = α α ββ αβ 2 αβ β α N β β α α αβ 1 1 III. THE BORN-OPPENHEIMER = δααδβ0α0 − δβ0αδαα0 = APPROXIMATION 2 N 1 1 N 2 − 1 = Nδβ0α0 − δβ0α0 = δβ0α0 The Born-Oppenheimer approximation in molecular 2 N 2N physics consists in the assumption that the motion of atomic nuclei and electrons in a molecule can be sepa- N 2−1 N=3 4 thus, the color factor is CF = #qq = 2N = − 3 . rated due to the difference in the velocity of the protons (slow) and the electrons (fast). In mathematical terms, it allows the wavefunction of a molecule to be broken into A. Running coupling its electronic and nuclear components. A similar assumption can be made regarding the motion of our baryon, breaking down the system in two compo- The αs coupling is a function of a renormalization scale µ. When µ is taken close to the scale of momentum trans- nents: the bb pair, which will act as the nucleus, and the 2 2 c quark. It is possible to do so thanks to the mass differ- fer Q in a given process, then αs(µ ' Q ) is indicative of the effective strength of the strong interaction in that ence between the quarks, 2mb >> mc, which allows us process. The coupling satisfies the following renormal- to treat each component separately. ization equation Let's take the Hamiltonian of our system: 2 2 2 2 2 2 ~ rb ~ rb r 2 dαs 2 3 H^ = − 1 − 2 − ~ c − (4) µ = −(b0αs + b1αs + ··· ) (2) dµ2 2mb 2mb 2mc ! N + 1 α(bb) α(bc) α(bc) 2 s s s with b0 = (33 − 2nf )=12π, b1 = (153 − 19nf )=(24π ) − ~c + + 2N jrb1 − rb2 j jrb1 − rcj jrb2 − rcj as found in [4], and nf = 3 the number of active quark flavors at our energy scale. where b , b and c subscripts refer to the two b and c We need to calculate this value at two different scales: the 1 2 quarks respectively. The wavefunctions and energies of center-of-mass (C.o.M) of the bc system (for the motion this system are obtained from the Schr¨odinger equation of c) and the C.o.M energy of the bb system (for the interaction of the b's). We use the experimental values H^ (q ; q ; q ) = E (q ; q ; q ) (5) of the charmonium (cc¯) mass and a solution of equation b1 b2 c b1 b2 c (2) at second order to solve the self-consistent system where the q's denote the coordinates of the quarks. The key remains on the fact that c is much lighter than the bb 2 m (cc¯) c pair. Thus, it will move much faster than the \nucleus", Mexp(cc¯) = 2mc − CF αs 2 and, in good approximation, we can consider the bb fixed 2 2 2 2 (cc¯) 1 b1 ln t µ mc cF αs while c moves around.