Strategies to determine the X(3872) energy from QCD lattice - PowerPoint PPT Presentation

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions Strategies to determine the X(3872) energy from QCD lattice simulations R. Molina 1 , E. J. Garzon 2 , A. Hosaka 3 and E. Oset 2 1 George Washington University, 2 IFIC-Valencia University, 3 RCNP-Osaka University

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions Contents 1. Introduction 2. The X(3872) in the continuum limit 2. Formalism in finite volume 3. The inverse problem 4. Results 5. Conclusions

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions Introduction Search of the X(3872) in lattice QCD: L. Liu, G. Moir, M. Peardon, S. M. Ryan, C. E. Thomas, P . Vilaseca, J. J. Dudek, R. G. Edwards, B. Joó, and D. G. Richards, J. High Energy Phys. 07 (2012) G. Bali, S. Collins, and P . Perez-Rubio, J. Phys. Conf. Ser. 426, 012017 (2013). D. Mohler, S. Prelovsek, and R. M. Woloshyn, Phys. Rev. D87, 034501 (2013). ... S. Prelovsek and L. Leskovec, Phys. Rev. Lett. 111, 192001 (2013): Bound state in dynamical N f = 2 lattice simulation with 11 ± 7 MeV below the D ∗ threshold and quantum numbers 1 ++ . D ¯ We develop a method to determine accurately the binding energy of the X(3872) from lattice data. The analysis of the data requires the use of coupled channels D 0 ¯ D ∗ 0 and D + D ∗− .

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions The X(3872) in the continuum limit Hidden gauge Lagrangian L III = − 1 4 � V µν V µν � + 1 V � [ V µ − i 2 M 2 g Γ µ ] � (1) where V µν = ∂ µ V ν − ∂ ν V µ − ig [ V µ , V ν ] , and g = MV 2 f . → L PPV = − ig � V µ [ P , ∂ µ P ] � − → L 3 V = ig � ( V µ ∂ ν V µ − ∂ ν V µ V µ ) V ν � − Approximation: | k i | 2 / M 2 ∼ 0 for external mesons, D D D D q 2 ∼ 0 ρ, ω, J/ψ ( i V i ) � = ⇒ D ∗ ¯ D ∗ ¯ D ∗ ¯ D ∗ ¯ Fig. 1. Pointlike pseudos.-vector interaction. M. Bando, T. Kugo, S. Uehara, K. Yamawaki and T. Yanagida, Phys. Rev. Lett. 54 , 1215 (1985), M. Bando, T. Kugo and K. Yamawaki, Phys. Rept. 164 , 217 (1988), M. Harada and K. Yamawaki, Phys. Rept. 381 , 1 (2003), U. G. Meissner, Phys. Rept. 161 , 213 (1988).

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions The X(3872) in the continuum limit In the approximation | k i | 2 / M 2 V ∼ 0, this is equivalent to, Lagrangian L PPVV = − 1 4 f 2 Tr ( J µ J µ ) , Currents J µ = ( ∂ µ P ) P − P ∂ µ P , J µ = ( ∂ µ V ν ) V ν − V ν ∂ µ V ν . (L-L: J 88 µ , H-L: J 83 , H-H: J 3 ¯ 3 ) Breaking Parameters m 8 ∗ = m L = 800 MeV, m 3 ∗ = m H = 2050 MeV, m 1 ∗ = m J /ψ = 3097 MeV, f π = 93, f D = 165 MeV. � 2 � 2 = m 2 m 2 � � m 8 ∗ m 8 ∗ γ = L H , ψ = = L m 3 ∗ m 2 m 1 ∗ m 2 J /ψ D. Gamermann and E. Oset, Eur. Phys. J. A 36 , 189 (2008) V ≡ ρ 0   ω ρ + K ∗ + D ∗ 0 ¯ 2 + D. Gamermann and E. Oset, Phys. Rev. D 80 , √ √ 2   − ρ 0 014003 (2009)  K ∗ 0  ρ − ω D ∗− 2 + √ √   2   D. Gamermann, J. Nieves, E. Oset and  K ∗ 0 ¯ D ∗−  K ∗− φ  s  D ∗ 0 D ∗ + D ∗ + E. Ruiz Arriola, Phys. Rev. D 81 , 014029 J /ψ s µ (2010)

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions The X(3872) in the continuum limit V = − ξ ij T = − [ I + VG ] − 1 V Bethe-Salpeter Eq.: 4 f 2 ( s − u ) � ǫ · � ǫ ′ � ǫ · � ǫ ′ 2 ( K ∗− K + − c . c . ) , K ∗ 0 K 0 − c . c . ) , 2 ( D ∗ + D − − c . c . ) , 2 ( ¯ 1 1 1 Channels: √ √ √ 2 ( D ∗ 0 ¯ D 0 − c . c . ) , 1 1 2 ( D ∗ + s D ∗− − c . c . ) . √ √ s − 3 − 3 0 − γ γ   − 3 − 3 − γ 0 γ   ξ ij ≡ 0 − γ − ( 1 + ψ ) − 1 − 1     − γ 0 − 1 − ( 1 + ψ ) − 1   γ γ − 1 − 1 − ( 1 + ψ ) γ = 0 . 14 ψ = 0 . 07 For G , dim. regularization formula or cutoff method can be used, m 2 m 2 2 − m 2 m 2 � √ 1 1 + s G DR ( 1 2 s ) = α ( µ ) + ln µ 2 + ln + 16 π 2 m 2 2 s 1 q √ √ � ln ( s − ( m 2 2 − m 2 s ) + ln ( s + ( m 2 2 − m 2 + 1 ) + 2 q 1 ) + 2 q s ) √ s √ √ �� − ln ( − s + ( m 2 2 − m 2 s ) − ln ( − s − ( m 2 2 − m 2 1 ) + 2 q 1 ) + 2 q s ) , (2) d 3 q ω 1 + ω 2 √ � 1 G co ( P 0 = s ) = (3) ( P 0 ) 2 − ( ω 1 + ω 2 ) 2 + i ǫ ( 2 π ) 3 2 ω 1 ω 2 q < qmax

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions The X(3872) in the continuum limit g i g j Near to a pole: T ∼ ( s − s p ) � ǫ · � ǫ ′ D. Gamermann, J. Nieves, E. Oset and E. R. Arriola, √ s 0 = ( 3871 . 6 − i 0 . 001 ) MeV PRD 81, 014029 Generalized compositeness condition: Channel | g i | [MeV] ∂ G i g 2 − � ∂ s = 1 2 ( K ∗− K + − c . c ) 1 i 53 √ K ∗ 0 K 0 − c . c ) 2 ( ¯ 1 49 √ Probability of finding the i ch. in 2 ( D ∗ + D − − c . c ) 1 3638 √ the wave func., 2 ( D ∗ 0 ¯ D 0 − c . c ) 1 3663 √ 0 . 86 for D ∗ 0 ¯ D 0 − c . c , 1 2 ( D ∗ + s D − s − c . c ) 3395 √ 0 . 124 for D ∗ + D − − c . c and 0 . 016 for D ∗ + s D − s − c . c . Table 1 . Couplings of the pole to the channel i with α H = − 1 . 265. However ( 2 π ) 3 / 2 ψ ( 0 ) i = g i G i (wave function at the origin) are nearly equal, and this usually enters the evaluation of observables.

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions Formalism in finite volume ω 1 ( � q i )+ ω 2 ( � G → ˜ ˜ q i ) G ( P 0 ) = 1 1 � G : � L 3 q i 2 ω 1 ( � q i ) ω 2 ( � ( P 0 ) 2 − ( ω 1 ( � q i )+ ω 2 ( � q i )) 2 q i ) 1 q i I ( P 0 ,� = � q ) L 3 � � q i | 2 and the momentum � m 2 i + | � where ω i = q is √ m i , q i = 2 π q i | = 2 π quantized as � L � n i , | � L z , i = m i and n max = q max L n 2 x , i + n 2 y , i + n 2 (symmetric box). 2 π A. M. Torres, L. R. Dai, C. Koren, D. Jido, and E. Oset, PRD 85, 014027   d 3 q  1 � G DR + ˜ � I ( P 0 ,� ( 2 π ) 3 I ( P 0 ,� G = lim q ) − q )  qmax →∞ L 3 q < qmax q < qmax G DR + ≡ qmax →∞ δ G lim (4) T → ˜ T = ( I − V ˜ ˜ G ) − 1 V T : Energy levels in the box: det ( I − V ˜ G ) = 0

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions Formalism in finite volume δ G converges as q max → ∞ . We take and average for different values between q max = 1500 − 2500 MeV. 0.0000 � 0.0001 � 0.0002 � � G G � 0.0003 � 0.0004 � 0.0005 2000 3000 4000 5000 6000 7000 q max G − G for D + D ∗− in function of q max for √ s = 3850 MeV. The thick Fig. 2 . Representation of δ G = ˜ line represents the average.

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions Formalism in finite volume One channel case : T = (˜ G ( E i ) − G ( E i )) − 1 0.005 0.000 � 0.005 � � � � 0.010 3800 3850 3900 3950 4000 4050 E � MeV � Fig. 3 . ˜ G (solid) and V − 1 (dashed) energy dependence of D + D ∗− for Lm π = 2 . 0. Dotted lines are the free energies. Energy levels in a cubic box for one channel : 4000 4000 3980 3980 3960 3960 3940 3940 E � MeV � E � MeV � 3920 3920 3900 3900 3880 3880 3860 3860 1.5 2.0 2.5 3.0 3.5 1.5 2.0 2.5 3.0 3.5 Lm Π Lm Π Fig. 4 . L dependence of the energies for a single channel, D 0 ¯ D ∗ 0 and D + D ∗− respectively.

Introduction The X(3872) in the continuum limit Formalism in finite volume The inverse problem Results Conclusions Formalism in finite volume At Lm π = 1 . 4 ( L = 2 fm), ∆ E = E 2 ( L ) − E 1 ( L ) = 137 MeV (if V = 0 ∆ E 0 = 194 MeV). While in the approach of S. Prelovsek and L. Leskovec, it is 161 MeV: Both approaches have an attractive interaction with similar strength, E 1 = 3860 vs. (3853 ± 8) MeV, E 2 = 3997 vs. (4014 ± 11) MeV. S. Prelovsek and L. Leskovec, Phys. Rev. Lett. 111 , 192001 (2013) Two channel case : For simplicity, we redone the calculation of Table 1 with α = − 1 . 153 in order to get the same position of the pole of the X(3872) with two channels instead of five. 4050 4000 E � MeV � 3950 3900 3850 1.5 2.0 2.5 3.0 3.5 Lm Π Fig. 5 . L dependence of the energies for the two first levels of D 0 ¯ D ∗ 0 and D + D ∗− . Dotted lines correspond to the free energies.