Nuclear states and spectra in holographic QCD Yoshinori Matsuo - PowerPoint PPT Presentation

Nuclear states and spectra in holographic QCD Yoshinori Matsuo Osaka University Based on arXiv:1807.11352 with Koji Hashimoto (Osaka U.), Takeshi Morita (Shizuoka U.) Aug 19, 2019@Strings and Fields 2019

Introduction Baryons in Sakai-Sugimoto model D- branes (D4’) wrapping color D -branes (D4) 𝑉 (radial direction) and 𝑇 4 ( 𝜄 1 ⋯ 𝜄 4 ) 𝑦 5 ⋯ 𝑦 9 D4 (at 𝑉 = 0 ) D4’ on 𝑇 4 𝒚 𝟏 𝒚 𝟐 𝒚 𝟑 𝒚 𝟒 𝒚 𝟓 𝜾 𝟐 𝜾 𝟑 𝜾 𝟒 𝜾 𝟓 𝑽 ✓ ✓ ✓ ✓ ✓ D4 ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓ D8 ✓ ✓ ✓ ✓ ✓ D4’ D4 Background geometry (holography) 𝑇 4 Integrate out Effective theory on D8 Effective theory of mesons D4’ in D8 effective theory Instanton on D8 Skyrmion Effective fields on D4’ ADHM data of instantons

Introduction Baryons in holographic QCD solitonic D4-brane geometry 3 3 𝑒𝑉 2 𝑉 2 𝑆 2 𝑒𝑡 2 = −𝑒𝑢 2 + 𝑒𝑦 2 + 𝑔 𝑉 𝑒𝑦 4 2 + 𝑔 𝑉 + 𝑉 2 𝑒Ω 4 2 𝑆 𝑉 Anti-periodic b.c. for 𝑦 4 D8-brane Baryon D4-brane A similar factor to BH 𝑔(𝑉) Geometry ends at some 𝑉 Baryon is located near the tip of geomtry Nuclear matrix model Matrix model of Baryon vertex (D4-brane) with bosonic field of D4-D8 open string near the tip of solitonic (color) D4-brane background

ሶ Nuclear matrix model Action for 𝐵 baryons [Hashimoto-Iizuka- Yi,’10] 𝑇 = 𝑇 0 + 𝑂 𝑑 න𝑒𝑢 tr𝐵 𝑢 𝑇 0 = න 𝑒𝑢 tr ቈ1 2 𝐸 𝑢 𝑌 𝐽 2 + 1 𝛽𝑗 − 1 2 𝑁 2 ഥ 𝑥 ሶ 𝛽𝑗 𝑥 ሶ 𝛽𝑗 𝑥 ሶ 2 𝐸 𝑢 ഥ 𝐸 𝑢 𝑥 ሶ 𝛽𝑗 + 1 4𝜇 𝐸 𝐽 2 + 𝐸 𝐽 2𝑗𝜗 𝐽𝐾𝐿 𝑌 𝐾 𝑌 𝐿 + ഥ 𝛽𝑗 𝜐 𝐽 ሶ𝛾 𝑥 ሶ 𝑥 ሶ ቉ 𝛾𝑗 𝛽 𝐸 𝑢 = 𝜖 𝑢 − 𝑗𝐵 𝑢 : covariant derivative 𝑌 𝐽 : D4-D4 scalar 𝐵 × 𝐵 matrix Diagonal comp.: position of D-brane ( = baryon) Off-diagonal: interaction between D-branes 𝑥 ( ഥ 𝑥 ): D4-D8 scalar carries charges of quarks (spin, flavor, baryon number)

ሶ ሶ ሶ ሶ ሶ 𝐵 𝑢 : gauge field (baryon 𝑇𝑉(𝐵) ) Non-dynamical field EOM gives constraints 0 = 𝜀𝑇 = 𝜀𝑇 0 − 𝑂 𝑑 𝕁 = 𝑅 𝑉 𝐵 − 𝑂 𝑑 𝕁 𝜀𝐵 𝑢 𝜀𝐵 𝑢 Eigenstates of Hamiltonian must be 𝑅 𝑇𝑉 𝐵 = 0 Singlet in baryon 𝑇𝑉(𝐵) symmetry Baryon (quark) number must be 𝑂 𝑑 𝐵 𝑅 𝑉 1 𝐶 = 𝑂 𝑑 𝐵 𝑊 : perturbation Perturbation around harmonic potential 𝐼 = 𝐼 0 + 𝑊 𝐼 0 = 1 2 tr Π I 2 + 1 2 𝑛 2 tr 𝑌 𝐽 2 + 1 𝛽𝑗 + 1 𝑏 𝜌 𝑏 2 𝑁 2 ഥ 𝑏 𝛽𝑗 𝑥 ሶ 2 ത 𝜌 ሶ 𝑥 𝑏 𝛽𝑗 𝛽𝑗 𝑊 = − 1 2 𝑛 2 𝑌 𝐽 2 − 2𝜇 𝑌 𝐽 , 𝑌 𝐾 2 2 ሶ𝛾 𝑢 𝐷 𝑐 + 𝜇 ഥ 𝐵𝐶 ഥ 𝑏 𝑥 ሶ 𝛽𝑗 𝜐 𝐽 𝛽𝑗 𝜐 𝐽 𝐾 𝑌 𝐶 ሶ𝛾 𝑥 ሶ −4𝑗𝜇𝜗 𝐽𝐾𝐿 𝑌 𝐵 𝐿 𝑔 𝑥 ሶ 𝑥 𝑏 𝐷 𝑐 𝛾𝑗 𝛽 𝛽 𝛾𝑗

Ground state and constraint Harmonic oscillators of 𝑌 𝐽 , 𝑥 and ഥ 0-th order Hamiltonian 𝐼 0 𝑥 . Constraint 1: baryon 𝑉(1) charge must be 𝑂 𝑑 𝐵 𝑌 𝐽 : 0 Baryon 𝑉 1 charges 𝑥 : 1 𝑥 : −1 ഥ Constraint for excitation of harmonic oscillators Number of 𝑥 − Number of ഥ 𝑥 = 𝑂 𝑑 𝐵 Lowest energy state Smallest number of excitations Number of 𝑥 = 𝑂 𝑑 𝐵 Number of ഥ 𝑥 = 0 Constraint 2: physical state must be singlet of baryon 𝑇𝑉(𝐵) Physical ground state for 𝐵 ≤ 2𝑂 𝑔 𝑏 1 ⋯ 𝑥 ሶ 𝑏 𝐵 × ⋯ × 𝜗 𝑐 1 ⋯𝑐 𝐵 𝑥 𝑐 1 ⋯ 𝑥 𝑐 𝐵 𝜔 0 = 𝜗 𝑏 1 ⋯𝑏 𝐵 𝑥 ሶ 0 𝛽 1 𝑗 1 𝛽 𝐵 𝑗 𝐵 𝑂 𝑑 of 𝜗 𝑏 1 ⋯𝑏 𝐵 𝑥 𝑏 1 ⋯ 𝑥 𝑏 𝐵 𝑥 𝑏 : raising operator of oscillator 0 : ground state of harmonic oscillators

Physical ground state for 𝐵 ≤ 2𝑂 𝑔 𝑏 1 ⋯ 𝑥 ሶ 𝑏 𝐵 × ⋯ × 𝜗 𝑐 1 ⋯𝑐 𝐵 𝑥 𝑐 1 ⋯ 𝑥 𝑐 𝐵 𝜔 0 = 𝜗 𝑏 1 ⋯𝑏 𝐵 𝑥 ሶ 0 𝛽 1 𝑗 1 𝛽 𝐵 𝑗 𝐵 𝑂 𝑑 of 𝜗 𝑏 1 ⋯𝑏 𝐵 𝑥 𝑏 1 ⋯ 𝑥 𝑏 𝐵 Only 2 × 𝑂 𝑔 of different 𝑥 ሶ 𝛽𝑗 : 2 different spins, 𝑂 𝑔 different flavors 𝑔 ) of 𝑥 𝑏 cannot form antisymmetric combination 𝐵 (> 2𝑂 construct different operator by using 𝑌 𝐽 : 𝑏 𝑥 𝑐 , 𝑌 𝐽 𝑥 𝑏 = 𝑌 𝐽 𝑌 𝐽 𝑌 𝐾 𝑌 𝐿 𝑥 𝑏 , 𝑌 𝐽 𝑌 𝐾 𝑥 𝑏 , ⋯ 𝑐 Physical ground state for 𝐵 > 2𝑂 𝑔 𝑏 2𝑂𝑔 𝑌 𝐽 𝑥 ⋯ 𝑌 𝐾 𝑥 ⋯ 𝑌 𝐿 ⋯ 𝑌 𝑀 𝑥 𝑏 𝐵 𝜔 0 = 𝜗 𝑏 1 ⋯𝑏 𝐵 𝑥 𝑏 1 ⋯ 𝑥 𝑐 2𝑂 𝑔 𝑌𝑥 ⋯ 𝑌 ⋯ 𝑌𝑥 𝑐 𝐵 ] 0 × ⋯ × [𝜗 𝑐 1 ⋯𝑐 𝐵 𝑥 𝑐 1 ⋯ 𝑥 𝑂 𝑑 of 𝜗 𝑏 1 ⋯𝑏 𝐵 𝑥 𝑏 1 ⋯ (𝑌 𝐽 ⋯ 𝑌 𝐾 )𝑥 𝑏 𝐵

ሶ Magic numbers Nuclei are stable (small energy) for some specific numbers of proton (neutron) Magic number for 𝑣 (or 𝑒 ) quarks ( 𝑂 𝑑 = 1 case for simplicity) 𝑥 = 𝑣 ሶ 𝛽 𝛽 = 1, 2 ( 𝑗 = 1 ) 2 of 𝑣 (spin ↑ and ↓ ) Additional energy of 𝑌 𝐽 𝐵 = 1 𝜔 0 = 𝜗𝑣 ↑ |0〉 Magic 𝜔 0 = 𝜗𝑣 ↑ 𝑣 ↓ |0〉 𝐵 = 2 number 𝜔 0 = 𝜗𝑣 ↑ 𝑣 ↓ 𝑣 ↑ 0 = 0 𝜔 0 = 𝜗𝑣𝑣 𝑌𝑣 |0〉 𝐵 = 3 𝐵 = 4 𝜔 0 = 𝜗𝑣𝑣(𝑣𝑌)(𝑣𝑌)|0〉 𝑌 𝐽 𝐽 = 1, 2, 3 ⋮ ⋮ Magic 𝜔 0 = 𝜗𝑣𝑣 𝑌𝑣 ⋯ 𝑌𝑣 |0〉 𝐵 = 8 number 2 × 3 = 6 of 𝑌𝑣 6 of 𝑌𝑣 𝐵 = 9 𝜔 0 = 𝜗𝑣𝑣 𝑌𝑣 ⋯ 𝑌𝑣 (𝑌𝑣) 0 = 0 6 of 𝑌𝑣 𝜔 0 = 𝜗𝑣𝑣 𝑌𝑣 ⋯ 𝑌𝑣 (𝑌𝑌𝑣) 0 𝐵 = 9 6 of 𝑌𝑣

Magic numbers Magic number for either proton or neutron ( 𝑂 𝑑 = 3 ) 𝑂 𝑔 = 2 Both quarks and nucleons have isospin 𝐽 = 1/2 Example: ( 𝑂 𝑞 = 8 , 𝑂 𝑜 = 2 ) 𝜔 0 = 𝑒𝑒 𝑣𝑣(𝑌𝑣)(𝑌𝑣)(𝑌𝑣)(𝑌𝑣)(𝑌𝑣)(𝑌𝑣) 3 sets of anti-sym. × 𝑒𝑒 𝑣𝑣(𝑌𝑣)(𝑌𝑣)(𝑌𝑣)(𝑌𝑣)(𝑌𝑣)(𝑌𝑣) combinations 𝜗𝑥 ⋯ 𝑌𝑥 × 𝑣𝑣 𝑒𝑒 𝑌𝑒 𝑌𝑒 𝑌𝑒 𝑌𝑒 𝑌𝑒 𝑌𝑒 0 2 neutrons 8 protons Proton and neutron Quark configurations ? configurations Example: 𝑂 𝑣 = 6 , 𝑂 𝑒 = 3 ; same to 3 protons (3 𝑞 ’s cannot be in ground state) 𝜔 0 = 𝑣𝑣 𝑒 𝑣𝑣 𝑒 𝑣𝑣 𝑒 0 State without 𝑌 excitation is possible corresponds to nucleus with Δ will be heavier if 1 st order perturbation 𝑊 is taken into account

ሶ First order perturbation Energy at first order = expectation value of 𝐼 for 𝜔 0 𝐹 = 𝜔 0 𝐼 𝜔 0 = 𝐹 0 + 𝜔 0 𝑊|𝜔 0 〉 𝐹 0 = 𝑂 𝑑 𝐵𝑁 Energy at 0-th order 𝑊 for 𝑂 𝑑 𝐵 excitations of 𝑥 Linear order correction 2 𝛽𝑗 𝜐 𝐽 No 𝑌 𝐽 excitation for 𝐵 ≤ 2𝑂 ሶ𝛾 𝑥 ሶ 𝑥 ሶ 𝑊 = 𝜇 ഥ 𝑔 𝛾𝑗 𝛽 No excitations of 𝑌 𝐽 or ഥ 𝑥 (only 𝑥 excitations) 2𝐵 − 𝑂 𝑊 = 4𝜇 𝑔 + 𝜇 𝑔 2 𝑁 2 𝐷 𝑂 𝑥 𝑁 2 𝑂 𝑔 𝐵 𝐷 𝑔 : quadratic Casimir of flavor 𝑇𝑉(𝑂 𝑔 ) 𝑂 𝑥 : Number of excitations of 𝑥 Smaller flavor charge more stable

Allowed states ( 𝑂 𝑔 = 2 ) 𝑥 ሶ 𝛽𝑗 : raising operator of 𝑥 ሶ 𝛽𝑗 𝐵 = 1 and 𝑂 𝑑 = 3 𝜔 0 = 𝑥 ሶ 𝛽 1 𝑗 1 𝑥 ሶ 𝛽 2 𝑗 2 𝑥 ሶ 𝛽 3 𝑗 3 0 (No baryon index) 𝑥 are bosonic ( = symmetric) Spin 𝐾 and isospin 𝐽 are in same rep. 1 2 , 1 2 , 3 3 𝐾, 𝐽 = 𝐾, 𝐽 = or 2 2 proton or neutron Δ Mass 𝐹 = 3𝑁 + 4𝜇 𝑁 2 𝐽(𝐽 + 1) proton and neutron have smaller mass than Δ

Allowed states ( 𝑂 𝑔 = 2 ) 𝐵 = 2 and 𝑂 𝑑 = 1 𝑏 1 𝑥 ሶ 𝑏 2 𝜔 0 = 𝜗 𝑏 1 𝑏 2 𝑥 ሶ 0 𝛽 1 𝑗 1 𝛽 2 𝑗 2 spin antisymmetric or antisymmetric symmetric 𝑥 are antisymmetric isospin is symmetric 𝐾, 𝐽 = 1,0 𝐾, 𝐽 = 0,1 stable 𝐵 = 2 and 𝑂 𝑑 = 3 Symmetric combination of 3 sets of 𝐵 = 2 and 𝑂 𝑑 = 1 𝐾, 𝐽 = 1,0 𝐾, 𝐽 = 3,0 𝐾, 𝐽 = 1,2 𝐾, 𝐽 = 0,1 𝐾, 𝐽 = 0,3 𝐾, 𝐽 = 2,1 Most stable states 𝐾, 𝐽 = 1,0 𝐾, 𝐽 = 3,0 Dibaryon 𝐸 03 Deuteron

ሶ ሶ ሶ ሶ ሶ Hyperon 𝑣 , 𝑒 , 𝑡 quarks 𝑂 𝑔 = 3 𝑡 quark has larger mass Put larger mass for 𝑥 (𝑡) = 𝑥 𝑗=3 by hand 3 2 𝑁 2 ഥ 𝑏 + 𝑁 𝑇 𝑁 2 ഥ 2 (ഥ 𝑏 𝑏 𝛽𝑗 𝑥 ሶ 𝛽 (𝑥 (𝑡) ) ሶ 𝛽𝑗 𝑥 ሶ 𝑥 𝑏 𝑥 (𝑡) ) 𝑏 𝑥 𝑏 ෍ ෍ 𝛽𝑗 𝛽 𝛽𝑗 𝑗=1 𝑗=1 Number of 𝑥 𝑡 = 𝑥 𝑗=3 in |𝜔 0 〉 Number of 𝑡 quarks in nucleus 2 𝛽𝑗 𝜐 𝐽 ሶ𝛾 𝑥 ሶ 𝑏 Assumption: no 𝑇𝑉(3) breaking effect in 𝑊 = 𝜇 ഥ 𝑥 𝑏 𝛽 𝛾𝑗 Effect of mass in raising and lowering operators 𝑏 † : raising (creation) operator of 𝑥 𝑥 = 1 𝑏 † + ത 𝑏 𝑁 𝑏 : lowering (annihilation) operator of ഥ ത 𝑥