The charmed double bottom

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 . Such a 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 , 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 and color wavefunctions.

I. INTRODUCTION II. THE

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, 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 . 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: {T ,T } = 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 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 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 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 in a 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 (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 ~ ∇b ~ ∇b ∇ 2 dαs 2 3 Hˆ = − 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 |rb1 − rb2 | |rb1 − rc| |rb2 − rc| 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 Hψˆ (q , q , q ) = Eψ(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. Classically speaking, during one αs ' 1 − 2 with t ≡ ln 2 = ln 2 b0t b0 t Λ 4Λ cycle of c, the change in the bb configuration is negligible. (3) This way, if we consider the separation of the b’s at a fixed

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c distance, say R = |rb1 − rb2 |, we can omit the kinetic energy terms of equation (4) to obtain the equation of motion for c:

~r1 ~r ~r2 (Hˆc + Vbb)ψc = Uψc (6) with b b z 2 2 (bc) (bc) ! ˆ ~ ∇c N + 1 αs αs R Hc = − − ~c + (7) ϕ 2mc 2N |rb1 − rc| |rb2 − rc| FIG. 1: Coordinates and

(bb) N + 1 ~cαs Vbb = − . (8) + 2N R looks like the one for the hydrogen molecule H2 , which greatly simplifies our work. Also, we will take coordinates The energy U in equation (6) is the energy of c including as shown in Fig. 1, with the origin on the middle point the attraction between bb, which is constant for a given between the b’s. value of R. Of course, there are infinite possible config- Now we are in the position to use the Variational Prin- urations for bb, and for each of them we can solve the ciple. Consider, in first place, that the distance between Schr¨odinger equation (6) to find a set of wavefunctions both b’s is very large, so c is close to one of them, say b1. and their corresponding energies for c. This way, the Essentially, we have a hydrogen with origin on b1 wavefunctions and energies depend parametrically of the whose fundamental wavefunction is bb configuration: −1/2 −r1 1s1 = π e , (11) ψc = ψc(qc,R) and U = U(R). and we can do the same for b2. These considerations suggest that we can test as variational function As Vbb is constant for a given configuration, omitting this −1/2 −r1 −1/2 −r2 term in the Hamiltonian does not affect the wavefunc- c1π e + c2π e (12) tions and only changes the eigenvalues of the energy by Vbb. We obtain where c1 and c2 are variational parameters. When c is close to b1 the first term predominates, yielding a func- Hˆcψc = Ecψc tion such as (11), as expected. Prior to solving the secular equation, we can improve where the energy Ec (that depends parametrically on the our test function. Consider the limit of the wavefunction + configuration of bb) is related to U by when R goes to 0. Then, our H2 wavefunction becomes the ion He+, whose wavefunction in the ground state is U = Ec + Vbb. (9) (taking Z = 2 in the hydrogen-like function)

3/2 −1/2 −2r Thus, if we get to find Ec for a particular configuration of 2 π e . (13) the bb pair, we will be able to calculate U from equation As we see in Fig. 1, when R goes to 0, ra and rb go to r, (9), where Vbb is easily found from (8). −1/2 −r so the test function goes to (c1 + c2)π e . Clearly, it does not behave well in the limit R = 0, as it tends to e−r instead of e−2r. We can solve this by adding a A. Variational principle variational parameter k = k(R), that will act as a charge factor, such that k(0) = 2 and k(∞) = 1 for the ground From now on, we will work with a generalization of state. So our test function looks like atomic units, wich defines length and energy units as 3/2 −1/2 −kr1 3/2 −1/2 −kr2 follows: c11s1 + c21s2 = c1k π e + c2k π e , 3/2 ~ with the factor k normalizing the function. a0 = = 0.209 fm It can be seen in Chapter 13 of [3] that this test function m cα(bc) N+1 c s 2N leads to: 2 N + 1 2 E = α(bc) m c2 = 0.478 GeV, 1s1 + 1s2 1s1 − 1s2 2N s c ϕ+ = p and ϕ− = p 2(1 + S12) 2(1 − S12)

−kr 2 2 and also ~ = mc = 1. This way, our Hamiltonian, where S12 = e (1+kR+k R /3), with the correspond- ing energies 1 1 1 Hˆ = − ∇2 − − , (10) c c W = t2F (t) + tG (t) (14) 2 |rb1 − rc| |rb2 − rc| ± ± ±

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1 1 ± (1 + t)e−t F = − + , (15) ± 2 1 ± e−t(1 + t + t2/3) t+1 (e−2t − 1) ∓ 2(1 + t)e−t G = t , (16) ± 1 ± e−t(1 + t + t2/3)

and t ≡ kR, which are the ground and first excited states 2−0.04485R+0.5196R2 FIG. 3: k(R) = 1+0.09218R+0.5196R2 (as we will see later, ϕ+ corresponds to the ground state).

B. Finding k

The relation k(R) (for each W ) is found by minimizing W as a function of k, and the condition ∂W/∂k = 0 leads to

2 G(t) + tG0(t) FIG. 4: k(R) = 2+0.9391R k = − . (17) 1+0.9391R2 2F (t) + tF 0(t)

This allows us to find k for a given value of t, and then C. The Schr¨odinger equation we use R = t/k to find the value of R corresponding to each value of k. Using Mathematica, we generate a large Now that we have solved the motion of c with the bb pair number of values of t and then we plot k vs R. Here we fixed at a distance R, we are able to use the code on [6] can see both k(R) curves, corresponding to W+ and W−. to solve the Schr¨odinger equation for the motion of the b’s, whose Hamiltonian is, explicitly 2.0 1.0 2 2 1.8 ∇ ∇ Hˆ = − b1 − b2 + U(R) 1.6 0.8 bb 2mb/mc 2mb/mc 1.4 0.6 1.2 with the potential U(R) given by (9), using 0.4 1.0 0 5 10 15 20 0 5 10 15 20 α(bb)/α(bc) 0.687 E = W (R) and V = − s s = − c bb R R FIG. 2: Numerical data for k+(R) and k−(R)

and the mass mb/mc = 2.713. Note that everything As we see, the first one (k+(R)) is almost monotonically is given in units of mc for consistency. The resulting decreasing from k(0) = 2 to k(∞) = 1, as expected, and energy for the ground state (n = 1, l = 0) of bb is the other one increases from k(0) = 0.4 to k(∞) = 1. As Eg = −1.71 = −0.816 GeV and the normalized radial neither of them is greater than 2, it is easy to see using wavefunction R10(R) is shown in Fig. 5. (14), (15) and (16) that W+ < W−, so the ground state 2.5 energy corresponds to W+ ≡ W . From now on we will stick to it, since our approximations are not valid for the 2.0 excited states. Now, we try adjusting our plot with functions of the type 1.5 2+aR+bR2 2+aR2 k(R) = 2 and k(R) = 2 (which behave as 1+cR+bR 1+aR 1.0 wanted at R = 0 and R → ∞) using the built-in Non-

LinearModel function (Mathematica). The adjusted 0.5 functions are shown in red and the plotted points in blue, in Figures 3 and 4. 0.0 0.5 1.0 1.5 2.0 2.5 3.0 We can see in Fig. 3 that the first function is a bet- ter adjustment for k(R). Now we can express W = FIG. 5: Normalized radial wavefunction as a function of R. (kR)2F (kR) + (kR) G(kR) as a function of just R. The angular component is the l = m = 0 spherical har- q 1 monic Y00(θ, φ) = 4π .

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IV. CONCLUSIONS as the intrinsic of each quark is 1. In conclu- + + sion, our baryon is expected to have J P = 1 , 3 . • The c quark wavefunction is 2 2

• The final mass M 0 of the baryon corresponds to 1s1 + 1s2 Ωcbb ψc = ϕ+(r1, r2) = p the sum of the masses of each quark and the bind- 2(1 + S12) ing energy of the whole system, Eg. In addition, + we need to estimate the error made in (1), where which corresponds to σg1s in the notation of H2 additional terms depending on powers of αs, which wavefunctions. It has Lc = 0 angular momentum and is symmetric under the interchange of the b’s. correspond to the exchange of more than one gluon, should be accounted for. A simple (and gross) way The wavefunction of the bb is to do this is to multiply the binding energy by αs. In our case, we will use α(bc) since it is an upper ψ = R (R)Y (θ, φ), s bb 10 00 bound for all the couplings.

and has Lbb = 0 so it is also symmetric. Thus, (bc) MΩ0 =2mb + mc + Eg(1 ± α ) = the total spatial wavefunction ψspatial = ψcψbb is cbb s symmetric under the interchange b1 ↔ b2. =11.10 ± 0.62 GeV. Now, the total wavefunction of our baryon is made up of three components: ψspatial, ξcolor and χspin. It might be possible to reduce the level of error by We have already seen that ψspatial is symmetric, analyzing the contribution of each αs in the final and the color wavefunction is antisymmetric be- result, and also by noting that αs always appears 2 cause it must correspond to the color singlet state multiplied by the color factor #qq = 3 . 1 ξcolor = √ (rgb − rbg + gbr − grb + brg − bgr) 6 V. APPENDIX due to confinement, being r, g, b the three color • Hydrogen-like wavefunctions: states of the quarks. Since the Pauli Exclusion Principle must hold for ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ). our particle, the total wavefunction must be anti- symmetric, which implies that the spin component – Radial wavefunctions:  3/2 χspin is symmetric in the interchange of the b’s. Z −Zr R10(r) = 2 exp( ). This implies that the bb pair must have spin 1, cor- a0 a0 responding to parallel quark spins and symmetric – Spherical harmonics: wavefunction. Adding the spin of the third quark p 1 3 Y00(θ, φ) = 1/4π. gives J = S = 2 , 2 , so we should expect our par- ticle to be in a degenerate energy state with both values of J. Observing the light baryons, though, 1 Acknowledgments we see that those with J = 2 tend to be lighter, mainly due to the spin-spin interaction [7], but for heavy quarks this is not clear since this interaction I would like to express my gratitude to my advisor Dr. −1 is proportional to m . Joan Soto i Riera for his valuable and enlightening sug- The parity of the baryon is gestions during the development of this research project. I would also wish to thank my family and friends for their Lc Lbb P = PcPbPb(−1) (−1) = 1 constant support and encouragement.

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