Charmed spectroscopy in a model

Tokyo tech, RCNP A,RIKEN B, Tetsuya Yoshida,Makoto Oka , Atsushi HosakaA , Emiko HiyamaB, Ktsunori SadatoA Contents

Ø Motivation Ø Formalism Ø Result ü Spectrum of single charmed baryon ü λ-mode and ρ-mode Ø Summary Motivation

Many unknown states in heavy

ü We know the baryon spectra in light sector but still do not know heavy baryon spectra well.

ü Constituent is successful in describing Many unknown state baryon spectra and we can predict unknown states of heavy baryons by using the model.

Σ Difference from light sector ΛC C

ü λ-mode state and ρ-mode state split in heavy quark sector ü Because of HQS, we expect that there is spin-partner Motivation light quark sector vs heavy quark sector What is the role of ?

How is it in How do spectrum and the heavy quark limit？ change? heavy quark limit m m q Q ∞ λ , ρ mode ü we can see how the spectrum and wave-funcon change ü Is sector near from heavy quark limit (or far) ? Hamiltonian

ij ij ij Confinement H = ∑Ki +∑(Vconf + Hhyp +VLS ) +Cqqq i i< j " % " % 2 π 1 1 brij 2α (m p 2m ) $ ' (r) $ Coul ' Spin-Spin = ∑ i + i i + αcon ∑$ 2 + 2 'δ +∑$ − ' i 3 i< j # mi mj & i< j # 2 3rij & ) # &, 2αcon 8π 3 2αten 1 3Si ⋅ rijS j ⋅ rij + + Si ⋅ S jδ (rij )+ % − Si ⋅ S j (. ∑ 3 % 2 ( Coulomb i< j *+3m i mj 3 3mimj rij $ rij '-. the cause of mass spling Tensor 1 2 2 4 l s C +∑αSO 2 3 (ξi +ξ j + ξiξ j ) ij ⋅ ij + qqq i< j 3mq rij Spin orbit

ü We determined the parameter that the result of the Strange baryon will agree with experimental results . Using the parameter , we performed spectral calculaon of the charm baryon. Heavy Quark Symmetry(HQS)

If we take mQ → ∞ heavy quark spin operator S Q → 0 mQ ü In heavy quark limit, the hamiltonian does not depend on heavy quark spin operator, so the angular momentum of the light component j l become good quantum number. ü Two states that direction of heavy quark spin is different become degenerate. ∗ ΣQ ΣQ doublet

* Σ * ΣC Σ* b 20MeV 194MeV 65MeV ΣC Σb Σ Heavy Quark Symmetry(HQS)

Λ- (positive ) Σ-particle (positive parity) s-wave s-wave 1 1 1 3 jρ = 0 lλ = 0 jl = 0 ( 2 ) jρ = 0 lλ = 0 jl =1 ( 2 , 2 )

1 3 1 1 3 p-wave 1 1 ( 2 ) ,( 2 , 2 ) p-wave 1 1 3 ( 2 ) ,( 2 , 2 ) j = 0,1, 2 lλ =1 j = 0,1, 2,3 2 1 ρ l 3 5 5 7 j 1 l =1 j = 0,1, 2 1 , , , ρ = λ l 3 5 ( 2 2 ) ( 2 2 ) ( 2 , 2 )

d-wave d-wave 2 2 2 1 3 3 5 3 5 ( 2 , 2 ) ,( 2 , 2 ) l = 2, 0 jl = 2 , jρ = 0, 2 λ ( 2 2 ) j =1, 2,3 lλ = 0, 2 jl =1, 2,3 2 ρ 5 7 ( 2 , 2 )

Λ,Σ-particle (negative parity) lλ 1 p-wave 1 2 J = jl ± 1 1 3 ( 2 ) ,( 2 , 2 ) j 2 j 0,1, 2 ρ j =1 lλ = 0,1 l = 1 ρ 3 5 jl = jρ + lλ ( 2 , 2 ) λ-mode and ρ-mode

ϕ q ϕLl l q Llλlρ λ ρ

lρ=0 Q lρ=1 Q l =0 q lλ=1 q λ

In heavy quark sector, 2 modes split ρ-mode

λ-mode

light sector heavy quark sector Gaussian Expansion Method

r jacobi coodinate 1 2 2 2 3 3 R 3 R 2 1 r3 R3 r2 1 1 1

Channel1 Channel2 Channel3 Wave funcon Trial funcon

2 (C=1) (C=2) (C=3) G l −νnr ˆ Ψ JM = ΦJM (r1, R1)+ ΦJM (r2, R2 )+ ΦJM (r3, R3 ) φnlm (r) = Nnlr e Ylm (r)

(C) (C) G G 2 Φ = A "φ (r )ψ (R )$ G L −λNr ˆ JM ∑ nClC .NCLC # nClC C NCLC C % ψNLM (R) = N NL R e YLM (R) nCl,C NCLC Eigen value problem

Hc = ENc ü We describe baryon wave ! H H  H $! c $ ! N N  N $! c $ # 11 12 1N &# 1 & # 11 12 1N &# 1 & funcon as sum of channels (i) ( j) # H21 H22  H2 N &# c2 & # N21 N22  N2 N &# c2 & N = φ φ # &# & = E# &# & ij JM JM ü We use Gaussian basis  (i) ( j) #     &# & #     &#  & H = φ H φ # H H  H &# c & # N N N &# c & ij JM JM funcon " nN nN NN %" N % " nN nN  NN %" N %

E. Hiyama, Y. Kino and M. Kamimura, Prog. Part. Nucl. Phys. 51 (2003) 223 Result

1. Mass of charmed lambda, charmed sigma ü we predict unkown states

2. Mass and wave function of single heavy baryon in the heavy quark limit ü We will see which states are doublet (or singlet) ü What is different from light sector

3. Wave function and baryon mass in different quark mass ü We will see how baryon mass and wave function change Baryon spectroscopy of single charmed baryon

Λc

2940MeV 2880MeV

2595MeV 2625MeV

2285MeV Baryon spectroscopy of single charmed baryon

Σc

2800MeV

2518MeV 2455MeV Lambda baryon (negative parity)

(heavy quark limit) Λ 1 − , 3 − C C Φ ( 2 2 ) = λψλ + ρψρ

jl = 2 jl =1 Λ 5 − Φ ( 2 ) = ψρ jl = 0 λ-wave ρ-wave l = 0 lλ = 0 λ

l =1 lρ = 0 ρ

Λ,Σ-particle (negative parity)

1 2 p-wave 1 1 3 ( 2 ) ,( 2 , 2 ) j =1 lλ = 0,1 jl = 0,1, 2 j 1 ρ 1 l = 3 5 ( 2 , 2 )

ΛGR = 0 one singlet and three doublet Lambda baryon (negative parity)

(heavy quark limit) Λ 1 − , 3 − C C Φ ( 2 2 ) = λψλ + ρψρ

jl = 2 jl =1 Λ 5 − Φ 2 = ψρ j = 0 ( ) l 2 2 Cλ = 0, Cρ =1 λ-wave ρ-wave

l = 0 lλ = 0 pure ρ-mode λ l =1 lρ = 0 ρ

Λ,Σ-particle (negative parity)

1 2 p-wave 1 1 3 2 2 ( 2 ) ,( 2 , 2 ) Cλ =1, Cρ = 0 j =1 lλ = 0,1 jl = 0,1, 2 j 1 ρ 1 l = 3 5 pure λ-mode ( 2 , 2 )

ΛGR = 0 one singlet and three doublet Lambda baryon (negative parity)

2 C 2 λ C Λ 1 − 3 − ρ P − Φ , = C ψ +C ψ P 1 − ρ 1 2 2 λ λ ρ ρ λ J = J = 2 Λ ( ) 2 m E m ms c 1 ms c E2

2 C C 2 ρ ρ λ λ

2 C 2 ρ Cλ P − P − ρ 1 λ 3 J = 2 J = 2

m E3 m ms c ms c E1

2 2 Cλ C λ ρ ρ

2 C ρ P − P 3 − 3 ρ ρ J = 2 J = 2

m mc E m E s 2 ms c 3

C 2 λ λ λ Sigma baryon (negative parity)

Σ 1 − , 3 − C C Φ ( 2 2 ) = λψλ + ρψρ

Σ 5 − Φ ( 2 ) = ψλ λ-wave ρ-wave l = 0 lλ =1 λ

l = 0 lρ = 0 ρ

Λ,Σ-particle (negative parity)

1 2 p-wave 1 1 3 ( 2 ) ,( 2 , 2 ) j =1 l = 0,1 jl = 0,1, 2 ρ λ 1 3 5 ( 2 , 2 ) one singlet and three doublet

ΛGR = 0

(heavy quark limit) Sigma baryon (negative parity)

Σ 1 − , 3 − C C Φ ( 2 2 ) = λψλ + ρψρ

j 1 l = Σ 5 − Φ 2 = ψλ 2 2 ( ) Cλ = 0, Cρ =1 ρ-wave pure ρ-mode λ-wave l = 0 lλ = 0 λ

l =1 lρ = 0 ρ

Λ,Σ-particle (negative parity)

1 2 p-wave 1 1 3 ( 2 ) ,( 2 , 2 ) j 0,1, 2 jρ =1 lλ = 0,1 l = jl =1 1 j = 2 3 5 l ( 2 , 2 ) jl = 0 2 2 C =1, C = 0 λ ρ one singlet and three doublet pure λ-mode ΛGR = 0 (heavy quark limit) Sigma baryon (negative parity)

C 2 λ λ λ P − Σ 1 − 3 − J = 1 P 1 − , C C 2 J = 2 Φ ( 2 2 ) = λψλ + ρψρ E1 E ms m m 2 c s mc

2 C ρ ρ ρ

ρ λ P 1 − P 3 − J = 2 J = 2

E E1 m 3 m s mc s mc λ ρ

λ P 3 − ρ P 3 − J = 2 J = 2

E2 E3 ms m ms m Σ c c ρ λ Lambda baryon (positive parity)

Λ 1 + C C Φ ( 2 ) = sψs + pψ p jl = 2 Λ 3 + 5 + Φ 2 , 2 = Csψs +Cpψ p +Cd ψd +C ψ ( ) ρ ρ dλ dλ Λ 7 + Φ ( 2 ) = ψ p s-wave p-wave dρ-wave dλ-wave l = 0 l =1 l = 0 λ λ λ lλ = 2

l = 0 l =1 l = 2 ρ ρ ρ lρ = 0

jl = 2

jl = 0 jl = 3 jl =1

jl = 2

jl = 0

jl =1

jl = 0 jl = 2

jl =1 Lambda baryon (positive parity)

Λ 1 + C C Φ ( 2 ) = sψs + pψ p jl = 2 Λ 3 + 5 + 2 2 2 2 C = C = C = 0, C =1 Φ 2 , 2 = Csψs +Cpψ p +Cd ψd +C ψ s p d d dρ-wave state d d λ ρ ( ) ρ ρ λ λ Λ 7 + Φ ( 2 ) = ψ p s-wave p-wave dρ-wave dλ-wave l = 0 l =1 l = 0 λ λ λ lλ = 2

l = 0 l =1 l = 2 ρ ρ ρ lρ = 0 p-wave state 2 2 2 2 C = C = C = 0, C =1 s dλ dρ p jl = 2

jl = 0 jl = 3 jl =1

jl = 2

jl = 0 s-wave(1s) 2 2 2 2 Cs = Cp = Cd = 0, Cd =1 state and dλ ρ λ jl =1 state jl = 0 jl = 2 2 2 2 2 C = C = C = 0, C =1 p dλ dρ s

jl =1 Lambda baryon (positive parity)

In charm sector (M=1841MeV)

Λ 1 + C C Φ ( 2 ) = sψs + pψ p Λ 3 + 5 + Φ 2 , 2 = Cpψ p +Cd ψd +C ψ ( ) ρ ρ dλ dλ

1 + 2 E2 2 2 Cs = 0.998, Cp = 0.002

3 + 2 2 E1 Cp = 0.0006 2 2 C = 0.0004, C = 0.9990 dρ dλ

5 + E1 2 2 Cp = 0.0007

2 2 C = 0.0001, C = 0.9998 dρ dλ Sigma baryon (positive parity)

Λ 1 + C C Φ ( 2 ) = sψs + pψ p Λ 3 + 5 + Φ 2 , 2 = Csψs +Cpψ p +Cd ψd +C ψ ( ) ρ ρ dλ dλ Λ 1 + Φ ( 2 ) = ψ p s-wave p-wave dρ-wave dλ-wave l = 0 l =1 l = 0 λ λ λ lλ = 2

l = 0 l =1 l = 2 ρ ρ ρ lρ = 0

jl = 3

jl = 2

jl =1

jl =1 jl = 2 jl =1

j 2 l = jl = 3 jl =1

jl =1 Sigma baryon (positive parity)

Λ 1 + C C Φ ( 2 ) = sψs + pψ p Λ 3 + 5 + Φ 2 , 2 = Csψs +Cpψ p +Cd ψd +C ψ ( ) ρ ρ dλ dλ Λ 1 + Φ ( 2 ) = ψ p s-wave p-wave dρ-wave dλ-wave l = 0 l =1 l = 0 λ λ λ lλ = 2

l = 0 l =1 l = 2 ρ ρ ρ lρ = 0

jl = 3

jl = 2

jl =1 2 2 2 2 C = C = C = 0, C =1 s p dλ dρ dρ-wave state

2 2 2 2 C = C = C = 0, C =1 p dλ dρ s 2 2 2 2 Cs = Cd = Cd = 0, Cp =1 λ ρ p-wave state

jl =1 j = 2 jl =1 l 2 2 2 2 C = C = C = 0, C =1 p dλ dρ s 2 2 2 2 s-wave state C = C = C = 0, C =1 s-wave(1s) state and dλ state s p dρ dλ

j 2 l = jl = 3 jl =1

jl =1 Sigma baryon (positive parity)

In charm sector (M=1841MeV)

Σ 1 + 3 + Φ 2 , 2 = Csψs +Cpψ p +Cd ψd +C ψ ( ) ρ ρ dλ dλ

Σ 5 + Φ 2 = Cpψ p +Cd ψd +C ψ ( ) ρ ρ dλ dλ 1 + 2 E E2 3 2 2 2 2 Cs = 0.9984, Cp = 0.001 Cs = 0.0002, Cp = 0

2 2 2 2 C = 0.0004, C = 0.0002 C = 0.0004, C = 0.9994 dρ dλ dρ dλ 3 + E3 2 E2 2 2 2 2 C = 0.944, C = 0.0005 Cs = 0.05, Cp = 0.01 s p 2 2 2 2 C = 0.0007, C = 0.05 C = 0.004, C = 0.94 dρ dλ dρ dλ

E 2 2 4 Cs = 0.008, Cp = 0.0

2 2 C = 0.0006, C = 0.991 dρ dλ 5 + E E2 2 1 2 C = 0.011 2 p Cp = 0.0002 2 2 C = 0.005, C = 0.984 2 2 dρ dλ C = 0.001, C = 0.998 dρ dλ Summary

ü We calculate charmed baryon spectra and our result reproduce experimental data.

ü In heavy quark limit, states separate into groups and each group do not mix each other.

ü In charm sector, the wave function is almost same in the heavy quark limit. Change of state

Ξ and Ω parcle

2 ρ ρ 2 ϕ ϕ ϕ ρ ϕ ρ P − ρ J 1 P 1 − = 2 ρ J = 2 Ξ Ξ cc Ω Ωc

2 λ λ λ 2 ϕ ϕ λ λ λ ϕ ϕ

mQ mQ Figur9.Probability of two states(Ξ parcle) Figur10.Probability of two states(Ω parcle) q q q Q q Q s s q Λ and Σ spectrum

(without LS,TENSOR force)

S Λ(Pρ, χ )

(P , ρ ) λ Σ ρ χ Λ(Pρ, χ ) S Σ(Pλ, χ )

λ Σ(Pλ, χ )

ρ Λ(Pλ, χ )

Σ(S, χ S )

Σ(S, χ λ )

mQ[MeV] Harmonic oscillator Baryon spectroscopy of double charmed baryon Angular momentum

+ 1 + Λ 3 Λ 2 2 ( ) N lρ lλ L s_qq S ( ) N lρ lλ L s_qq S 1 0 0 0 0 1/2

2 1 1 0 1 1/2 1 1 1 0 1 3/2 2 1 1 1 1 1/2 3 1 1 1 1 1/2 3 1 1 1 1 3/2 4 1 1 1 1 3/2 4 1 1 2 1 1/2 5 1 1 2 1 3/2 5 1 1 2 1 3/2

+ 6 2 0 2 0 1/2 5 N lρ lλ L s_qq S Λ( 2 ) 7 0 2 2 0 1/2 1 1 1 1 1 3/2

2 1 1 2 1 1/2 lλ 3 1 1 2 1 3/2

4 2 0 2 0 1/2 lρ, sqq

5 0 2 2 0 1/2 Angular momentum

+ 1 + Σ 3 Σ 2 2 ( ) N lρ lλ L s_qq S ( ) N lρ lλ L s_qq S 1 0 0 0 1 1/2

2 1 1 0 0 1/2 1 0 0 0 1 3/2 2 1 1 1 0 1/2 3 1 1 1 0 1/2 3 1 1 2 0 3/2 4 2 0 2 1 3/2 4 2 0 2 1 1/2 5 0 2 2 1 3/2 5 2 0 2 1 3/2

+ 6 0 2 2 1 1/2 5 N lρ lλ L s_qq S Σ( 2 ) 7 0 2 2 1 3/2 1 1 1 2 0 1/2

2 2 0 2 1 1/2 lλ 3 2 0 2 1 3/2

4 0 2 2 1 1/2 lρ, sqq

5 0 2 2 1 3/2 Angular momentum

− − Λ 1 , 3 5 − ( 2 2 ) Λ( 2 )

N lρ lλ L s_qq S N lρ lλ L s_qq S 1 0 1 1 0 1/2 1 1 0 1 1 3/2 2 1 0 1 1 1/2 l 3 1 0 1 1 3/2 λ

lρ, sqq Σ 1 − , 3 − 5 − ( 2 2 ) Σ( 2 ) N lρ lλ L s_qq S N lρ lλ L s_qq S 1 1 0 1 0 1/2 1 0 1 1 1 3/2 2 0 1 1 1 1/2 3 0 1 1 1 3/2 Example of a table

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λ-mode and ρ-mode

What can we find from the analysis in two excitation modes?

We can know the decay paern of ΛQ ΣQ

ρ-mode λ-mode

Which is decay pattern?