Hadrons and Nuclei Single Hadrons Lattice Summer School

Hadrons and Nuclei : Single Hadrons Lattice Summer School Martin Savage Summer 2007 University of Washington

Mass Spectrum of Mesons | - M esons S= C = B = T = 0, qq ¼§ m ¼§ = 139: 57 Me. V ¼ 0 m ¼ 0 = 134: 96 Me. V ´ 0 m ´ = 548: 6 Me. V ´ 00 m ´ 0 = 957: 6 Me. V ½ 0; § m ½ = 770 Me. V ! 0 m ! = 783 Me. V ¿ = 2: 6 £ 10¡ ¿ = 0: 83 £ 10¡ 1 ¿¡ ¿¡ 1 s J ¼ = 0¡ 16 s J ¼ = 0¡ = 0: 9 ke. V J ¼ = 0¡ 1 ¿¡ 8 = 0: 3 Me. V J ¼ = 0¡ = 154 Me. V J ¼ = 1¡ = 9: 9 Me. V J ¼ = 1¡ Á0 m Á = 1020 Me. V A 1 m A 1 = 1275 Me. V ¿¡ 1 » 300 Me. V J ¼ = 1+ m J =à = 3: 1 Ge. V ¿¡ 1 = 88 ke. V J ¼ = 1¡ J=à ¨ m ¨ = 9: 5 Ge. V ¿¡ ¿¡ 1 1 = 4: 2 Me. V J ¼ = 1¡ = 52 ke. V J ¼ = 1¡

Mass Spectrum of Light Baryons | - B ar yons S = C = B = T = 0, qqq p m p = 938: 28 Me. V n m n = 939: 57 Me. V ¢ m ¢ » 1230 Me. V ¿> 1033 yr s J¼ = 1+ 2 ¿ = 898 § 16 s J ¼ = ¿¡ 1 » 120 Me. V J ¼ = 3+ 2 | - B ar yons S = 1 C = B = T = 0, sqq ¤ m ¤ = 1115: 6 Me. V ¿ = 2: 6 £ 10¡ §§ m § § = 1197: 3 Me. V § 0 m § 0 = 1192 Me. V 10 s ¿ = 1: 5 £ 10¡ ¿ = 6 £ 10¡ 20 s = J¼ 10 s 1+ 2 J¼ = 1+ 2

SU(2) Flavor Symmetry : Isospin If mu = md then SU(2) would be an exact symmetry of QC mu - md << Lc so SU(2) is an approximate symmetry of Q u u u d d d SU(2) Global flavor transformations Isospin SU(3)C Local color transformations

SU(3) Flavor Symmetry If mu = md = ms then SU(3) would be an exact symmetry of Q mi - mj << Lc so SU(3) is an approximate symmetry of Q SU(3) Global flavor transformations u u u d d d s s s SU(3)C Local color transformations

Mesons : SU(2) qa = M ab () u d µ Vq Vector symmetry : L = R = V = qb° 5 qa ¡ M M = q 1 2 ±ab qc ° 5 qc VMV p ¶ ¼ 0 = 2 ¼+ p = ¼¡ ¡ ¼ 0 = 2 p 1 ¼a ¿a 2

Mesons : SU(3) qa = M ab () u d s q = qb° 5 qa ¡ M 1 3 Vq ±ab qc ° 5 qc VMV p p 1 + + 0 ¼ = 2+ ´= 6 K p¼ p ¼¡ ¡ ¼ 0 = 2 + ´ = 6 K 0 A M = @ 0 K¡ K ¡ 2=6´ 0 ( Lectures by Claude Bernard )

Meson Correlation Functions and Interpolating Fields Z ^ y)i h. O(x; e. g. ¼+ » ^ y) ei Dq Dq DA ¹ O(x; ¡ ¢y hd(x)° 5 u(x) d(y)° 5 u(y) i = = R d 4 z L ( q; q; A ¹ ) h. Tr [ D (x ! y) ° 5 U(y ! x) ° 5 ]i £ ¤ y h. Tr D (x ! y) U (x ! y) i d-quark propagator u-quark propagato Z d 3 xhd(x)° 5 u(x) ¡ ¢y d(y)° 5 u(y) i ! e¡ m ¼ t E Z¼ + : : 2 m ¼

Baryons : SU(2) spin Bi°j k » h qi®; a qj¯ ; bqk° ; c ¡ qi®; a qj° ; c qk¯ ; b i ² abc (C° 5 ) ®¯ flavor Babc = 1 p ( ² ab N c + ² ac N b ) 6 µ N= p n ¶ B V®a V¯b² ab ! ! VVVB ² ®¯

Baryons : SU(3) spin Bi°j k » h flavor Babc = 0 B = @ qi®; a qj¯ ; bqk° ; c ¡ p 1 6 p § 0= ¡ qi®; a qj° ; c qk¯ ; b i ² abc (C° 5 ) ®¯ ² abd B dc + ² acd B db ¢ p 1 + 2+ ¤= 6 p p§ p A §¡ ¡ § 0= 2 + ¤ = 6 n ¥¡ ¥ 0 ¡ 2=6¤ B V®a V¯b² abc ! ! VVVB ² ®¯ ° Vcy°

Light Baryons : SU(3) S=0, I=3/2 J¼ = 3+ 2 S=-1, I=1 S=-2, I=1/2 S=-3, I=0 S=0, I=1/2 J¼ = 1+ 2 S=-1, I=0, 1 S=-2, I=1/2

Charmed (or Bottom) Baryons : SU(3) Sl=1, I=1/2 Sl=1, I=0 Mixing / ms¡ m m. Q Sl=0, I=0 Sl=0, I=1/2

Light Quark Masses and Spurions ¹ = q i°¹ @q ¡ L SU(3) invariant 0 0 mu mq = @ 0 md 0 0 BUT : let m q ! qm q q breaks SU(3) 1 0 0 A ms V m q V y and t hen bot h t erms are SU(3) invariant T hen we can simply use t he W igner-Eckhart T heorem t o const ruct invariant mat rix element s.

p, K, h Masses : Gell-Mann—Okubo Mass Relation L = = ® T r [ M M mq ] + ¯ T r [ M M ] Tr [ mq ] + : : ¡ m 2¼ ¼+ ¼¡ + : : Construct all possible group invariants that can contribute More insert ions of M and m q m 2¼ = 2® m + 2¯(2 m + m s ) m 2 K = ® (m + m s ) + 2¯(2 m + ms ) m 2´ = 2® 3 2 4 m K (m + 2 m s ) + 2¯(2 m + m s ) ¡ 2 m¼ = 3 m 2´

p, n, L, S, X Masses : Gell-Mann—Okubo Mass Relation L = MN £ ¤ ¡ M 0 Tr B B ¡ ® T r B m q B £ ¤ ¡ ¯ T r B B mq ¡ ° T r B B T r [ m q ] + : : = M 0 + (2 m + m s )° + m® + m s ¯ 2 M N + 2 M ¥ = M § + 3 M ¤

Quark Masses from Lattice ( Claudes lectures ) m u =m d M s =m = = 0: 43 § 0: 01 § 0: 08 27: 4 § 0: 1 § 0: 4 § 0: 0 § 0: 1 MILC collaborat ion , ¹ = 2 Ge. V

Homework 1 : ® Check the validity of GMO mass relation amongst the pseudo-Goldstone bosons using Particle Data group compliations. ® What is the violation as a percent of the pion mass? the masses of the S, L , X at oneinsertion of the light quark mass matrix ® Derive

Latest Lattice results : LHPC : DW on Staggered mp=350 Me. V LHPC, Negele et al

Lattice Result for GMO

Electromagnetism ¹ j em = = 0 Q = 2 ¹ u° u ¡ 3 q Q °¹ q 2 3 + @ 0 0 1 ¹ d° d ¡ 3 1 0 ¡ 1 3 0 Singlet plus Triplet of SU(2) Q = spurion field 1 ¹ s° s 3 0 0 A ¡ 1 3 Octet of SU(3) Q ! V QV y

Magnetic Moments in SU(3): Coleman-Glashow Relations L = £ ¹ F Tr B ¾¹ º F¹ º ¤ £ [Q; B ] + ¹ D Tr B ¾¹ º F¹ º ¤ f Q; B g Limit of exact SU(3) symmetry…. mu=md=ms ¹N = ¹F + 1 3 ¹D ¹ p ; ¹ n ; ¹ ¤ ; ¹ § + ; ¹ § ¡ ; ¹ ¥ 0 ; ¹ ¥¡ ; ¹ § ¤

Magnetic Moments : Coleman-Glashow Relations L ¹ §+ = ¹ p 2¹ ¤ = ¹ n ¹ ¥ 0 = ¹ n ¹ §¡ + ¹ n = ¡ ¹ p ¹ ¥¡ = ¹ §¡ p 2¹ ¤ § 0 = 3¹ n : : : = e ¹ i B ¾¹ º F ¹ º B 4 M N 2: 42 § 0: 05 NM = 2: 7928 NM ¡ 1: 226 § 0: 008 NM = ¡ 1: 9130 NM ¡ 1: 25 § 0: 01 NM = ¡ 1: 9130 NM ¡ 3: 07 § 0: 03 NM = ¡ 2: 7928 NM ¡ 0: 6507 § 0: 0025 NM = ¡ 1: 16 § 0: 03 NM 3: 22 § 0: 16 NM = 3: 31 NM Works as well as can be expected for SU(3) symmetry

Homework 2 : ® Explore the validity of the Coleman-Glashow relations between the magnetic moments of the baryon octet. ® Find analogous relations between the baryon decuplet, and find relations between the EM transition rates between the decuplet and octet assuming M 1 transition.

Matrix Elements in Nucleon (1) · = hpjq. Q° ¹ qjpi Up F 1( p) ° ¹ + F 2( p) i ¾¹ º qº + F 3( p) q¹ 2 M N Fi( p) @¹ j ¹ ( p) F 1 (0) ( p) F 2 (0) CONSTRAINTS ( p) 2 F 3 (q ) = 0 ! = +1 = · p = ( 2: 79 ¡ ´ GE = F 1 ¡ F 2 Up Fi( p) (q 2 ) = 0 1 ) NM Similarly for the neutron j. Q 2 j 4 M N 2 ¸ GM = F 1 + F 2

Perdisat et al Proton : 4 Q (p) G M / mp Pert urbat ive QCD : ( p) GM / Q 4

Dipole Form Factors for Nucleon !! GD i p ol e(Q 2 ) = 1 ( 1+ Q 2 =0: 71) 2 Perdisat et al

Recent Comprehensive Lattice Study : S. Boinepalli et al. , hep-lat/0604022 ® Clover on Quenched ® Not QCD (unfortunately), likely close to nature from all previous experiences. ® Clover gives O(a 2) errors in the quarks ® Good step toward fully-dynamical ® Disconnected diagrams evaluated phenomenologically…. . computationally expensive

Lattice Contractions

Baryon Charge Radii hr E 2 i = ¡ 6 d. Qd 2 Zanotti et al ¯ GE (Q 2 ) ¯ Q 2 = 0 Larger m ¼ t he smaller lat t ice can be !!

Zanotti et al Baryon Magnetic Moments

Zanotti et al Baryon Magnetic Radii

Isovector-Vector Form Factors : Lattice ¹ GE =GM Wilson on Quenched Domain-Wall on Staggered GM (0) Alexandru et al hep-lat/0611008

Isovector-Vector Form Factor ( George Fleming, LHPC ) Domain-Wall on Staggered

Just how Strange is the Proton ?

Flavor Structure of the Nucleon : Tree-Level e q V¹ and A ¹ 0 °; Z

Relevant parts of the Standard Model D ¹ = @¹ + i g 2 W¹a T a + i g 1 12 Y B ¹ ¡ ¢ 2 Z D ¹ = @¹ + i e. QA ¹ + i s ec T 3 ¡ Qsw ¹ w t an µw B¹ = p 1 g 12 + g 22 W¹ 3 = p YÁ = + 1 1 g 12 + g 22 = w g 1 g 2 (g 1 Z ¹ + g 2 A ¹ ) (g 1 A ¹ ¡ g 2 Z ¹ ) µ hÁi = 0 p v= 2 ¶

0 Z -couplings L int : = ¡ e 4 cw sw u u £¡ 1¡ 8 s 2 3 w ¢ 0 Z ¤ °¹ ¡ °¹ ° 5 u Z¹ u

Flavor Structure of the Nucleon : EM ¹ j em Transforms as an octet under SU(3) 2 ¹ 1 ¹ = d° d ¡ u° u ¡ s° s 3 3 3 = q Q °¹ q 0 1 1 0 0 1 @ = 0 ¡ 1 0 A °¹ q q I=1 2 0 0 1 1 0 0 1 @ 0 1 0 A °¹ q I = 0 q + 6 0 0 ¡ 2

Flavor Structure of the Nucleon : Z 0 µ j Z¹ 0 = = ¶ µ 1 8 1 1 ¡ s 2 w u° ¹ u ¡ 2 3 2 1 1 ¹ d° ° 5 d + u° ¹ ° 5 u ¡ 2 2 0 1 0 ¢ 1 ¡ 2 q @ 0 ¡ 1 1 ¡ 2 sw 2 0 0 1¡ ¡ µ d° ¹ d ¡ 1 ¹ s° ° 5 s 2 1 0 0 A °¹ q ¡ 0 Vector Current 0 4 2 s 3 w ¶ 1 4 1 ¡ s 2 w 2 3 0 1 1 2 @ 0 s q 3 w 0 1 ¹ s° ¡ 2 ¡ 0 0 1 0 A °¹ q 0 ¡ 2 s s° ¹ ° 5 s Axial-Vector Current I=1 s° ¹ s 1 1 1 0 0 1 @ 0 ¡ 1 0 A ° ¹ ° 5 q + q 2 0 0 0 ¶ I=0

Matrix Elements in Nucleon (2) 0 1 1 0 0 hpjq @ 0 ¡ 1 0 A ° ¹ qjpi 0 0 0 = Up · ³ ´ F 1(p) ¡ F 1(n ) ° ¹ + ³ F 2(p) ¡ F 2(n ) ´ q i ¾¹ º º 2 M N ¸ Up Isovector 0 1 1 0 0 hpjq @ 0 1 0 A ° ¹ qjpi 0 0 ¡ 2 = · ³ 3 Up F 1( p) + F 1( n ) ° ¹ + · hpjs° ¹ sjpi = Up ´ F 1( s) (q 2 )° ¹ ³ F 2( p) + F 2( n ) ´ q i ¾¹ º º 2 M N ¸ Up Isoscalar ¸ qº ( s) 2 ¹ º + F 2 (q )i ¾ Up 2 M N strange CONSTRAINTS F 1( s) (0) = 0

Tree-Level ( e) g. V e q 2 , » 1 ¡ 4 sw ( e) g. A 0 °; Z » 1

Radiative Corrections q e. g. e 0 °; Z

Hadronic Corrections e Parity-violating verte °; Z 0 q q Z 0

¹ h. N js° sj. N i is small !! Q 2 = 0. 1 Ge. V 2 GE(s) = F 1(s) ¡ j. Q 2 j 4 M N 2 F 2(s) G(Ms) = F 1( s) + F 2( s) Liu, Mc. Keown and Ramsey-Musolf, ar. Xiv: 0706. 0226 v 2 Jlab and Bates

Strange Vector Form Factors

Axial-Current Matrix Elements in Nucleon (1) · hpju° ¹ ° 5 djni = Up g 1 (q 2 )° ¹ ° 5 q + g 2 (q 2 )i ¾¹ º ° 5 º + g 3 (q 2 )° 5 q¹ 2 M N CONSTRAINTS T and I g 1 (0) ! g 2 (q 2 ) = 0 = g. A ¯ ¡ decay ¸ Un

Neutron b-decay n p W- O » u° ¹ (1 ¡ ° 5 )d g. A = 1: 26 e u g 3 comes wit h a fact or of m e ¹ ¡ + p ! n + º ¹ sensit ve t o g 3

PCAC p Am a A ¹ (x) = q° ¹ ° 5 a T q(x) h 0 j. A ¹a (x)j¼b(q)i = ¡ i f ¼ q¹ e¡ h 0 j@¹ A ¹a (x)j¼b(q)i = ¡ f ¼ m 2¼ e¡ @¹ A ¹a (x) = ¡ f ¼ m 2¼ ¼a (x) i q: x ±ab T herefore @¹ A ¹ is a good int erpolat ing ¯eld for t he pion.

PCAC : Goldberger-Treiman (1) (1958) h. N j. A a¹ (x)j. N i = h. N j@¹ A a¹ (x)j. N i = U [ g 1 ° ¹ ° 5 + g 3 q¹ ° 5 ] T a U e¡ i q: x £ ¤ a ¹ 2 ¡ i U g 1 q ° ¹ ° 5 + g 3 q ° 5 T U e¡ i q: x In chiral limit, @¹ A a¹ = 0 hence 2 M N g 1 (q 2 ) + q 2 g 3 (q 2 ) = 0

PCAC : Goldberger-Treiman (2) L = i g¼N N N ° 5 T b. N ¼b In chiral limit, · h. N j. A a¹ (x)j. N i g 3 = (q 2 ) ¡ U = g¼N N f ¼ q¹ q 2 ¸ ° 5 g¼N N f ¼ ¡ q 2 T a U e¡ i q: x + non-pole

PCAC : Goldberger-Treiman (3) In chiral limit, g¼N N f ¼ 2 g 3 (q ) = ¡ q 2 2 M N g 1 (q 2 ) + q 2 g 3 (q 2 ) = g 1 (q 2 ) = 0 g¼N N f ¼ 2 M N Away from the chiral limit, g 3 (q 2 ) = g¼N N f ¼ ¡ 2 q ¡ m 2¼ g 1 (q 2 ) = g¼N N f ¼ m 2¼ + O( 2 ) 2 M N ¤

PCAC : Goldberger-Treiman (4) At the physical point, g 3 (q 2 ) g 1 (q 2 ) = = g¼N N f ¼ ¡ 2 q ¡ m 2¼ g¼N N f ¼ + O( 2 ) 2 M N ¤ g. A g¼N N 2 M N g. A 1¡ f ¼g¼N N Sid Coon, Nucl-th/9906011 = = 1: 2654 § 0: 0042 13: 12 ; 13: 02 = 0: 023 ; 0: 015

g. A from Lattice QCD

Axial Charges : D. I. S. Large Q 2 (Deep Inelastic Scattering 2 N ~ N N Operator-Product Expansion

Axial Charges (2) Measure 1 2 ³ 2 ®s ( Q ) ¼ 1 ¡ R 1 0 0 Q 2 = ´ hpjq Q 2 ° ¹ ° 5 qjpi in DIS dx g 1 (x; Q 2 ) 1 0 1 1 0 0 1@ 1 2 @0 1 0 A + @0 1 0 A 0 ¡ 1 0 A + 6 18 9 0 0 0 ¡ 2 0 0 1 Relat ed by Isospin t o g. A Relat ed by SU(3) t o Hyperon Decays

Axial Charges (3) Including SU(3)-breaking ¡ 0: 35 < ¢ s < 0 ¡ 0: 1 < ¢ u + ¢ d + ¢ s < + 0: 3 2 ¢ q U p s¹ Up = hpjq° ¹ ° 5 qjpi

Nucleon s-Term (1) H^ (m q ) j. N (m q )i h. N (m q )j H^ (m q ) j. N (m q )i @ E (m q ) mq @mq L Q C D (m q ) = = Feynman-Hellman Thm E (m q )j. N (mq )i = E (m q ) = @ ^ H (m q ) j. N (m q )i h. N (m q )jm q @m q L Q C D (0) ¡ X m i qi qi i @ ^ mi H (m q ) @m i = H^ (m q ) = m i qi qi H^ (0) + X i @ ^ H (m q ) mi @m i

Nucleon s-Term (2) : SU(2) ¾N = X i @M N 2 = m¼ + : : mi 2 @m i @m ¼ = h. N (m q )j m u uu + m d dd j. N (m q )i = m h. N (m q )j uu + dd j. N (m q )i ¾N » 45 Me. V from scat t ering See later Not e t hat : h. N (m q )j. N (m q )i = 1 : convent ional t o use = 2 M N

Nucleon s-Term (3) : SU(3) ¾N = = h. N (m q )j m u uu + m d dd + m s ss j. N (m q )i 1 (2 m + m s ) h. N (m q )j uu + dd + ss j. N (m q )i 3 1 + (m ¡ m s ) h. N (m q )j uu + dd ¡ 2 ss j. N (m q )i 3 SU(3) Singlet SU(3) Octet

Nucleon s-Term (4) : strangeness m h. N (m q )juu + dd ¡ 2 ssj. N (m q )i » 35 § 5 Me. V · ¸ 2 m¼ 3 1 = (M ¥ ¡ M N ) ¡ (M § ¡ M ¤ ) 2 2 4 m. K ¡ m¼ 2 Using ¾N » 45 Me. V gives 2 h. N (m q )jssj. N (m q )i h. N (m q )juu + ddj. N (m q )i » 0: 2 ! 0: 4

Nucleon s-Term (5) h. N (m q )j m s ss j. N (m q )i » 130 Me. V h. N (m q )j H^ (0)j. N (m q )i » 764 Me. V Strange quarks (non-valence) play a nontrivial role on the structure of the Nucleon

s-Term from the Lattice Two methods used presently : 1. Compute MN and take numerical derivatives … poor precision…many configs (QCD) 2. Compute 3 -pt function

Lattice Results: MN vs mp D not included in fit Physical value used in f Staggered Clover

Lattice Results: MN vs mp Overlap fermions MN (Ge. V) Galletly et al, hep-lat/0607024 mp ~ 235 Me. V Physical point Mp 2 Ge. V 2

Lattice Results (2): MN vs mp s. N is derivative of curv --- Much larger uncertainties

Dilatations T¹ º (y) = ± p 2 ¡ g ±g ¹ º ( y ) R Energy-Momentum Tensor d 4 x p ¡ g. L

Improved Energy-Momentum Tensor and Scale (Dilatation) Current ¹ º O @¹ O ¹ º = = ¹ º T + surface t erms 0 Scale-Current ¹ S = ¹ º O ¹ @¹ S = ® O® xº

Scale Transformation x ! = e® x 0 x Z S = Z d 4 x j@¹ Á(x)j 2 ! d 4 x j@¹ ¡ e®dÁ Á(e® x) Z = e 2®( dÁ ¡ 1) S 0 = e 2®( dÁ ¡ 1) dÁ d 4 x 0 j@0¹ Á(x 0)j 2 S = 1 ; dà = Require scale-invariant when massless 3 2 ¢ j 2

Masses Break Scale-Invariance Z Z d 4 xm 2¼jÁ(x)j 2 ¡ @¹ S¹ ! ¡ e®( 2 d Á = 2 m 2¼ jÁj 2 ¡ 4) Z ¡ d 4 x 0 m 2¼jÁ(x 0)j 2 Z d 4 x m N N N @¹ S¹ ! ¡ e®( 2 d Á = m. N N N ¡ 4) d 4 x 0 m N N N

Gauge Fields Renormalizat ion Scale ¹ relat ed t o coordinat es via ¹ » 1=x x ! 0 x g = g(Q 2 =¹ 2 ) ! g(¹ ) ! L = ±L ±® = = e® x g(e¡ 2® Q 2 =¹ 2 )) g(e® ¹ ) £ 2¤ 1 Tr G ¡ 4 g 2 QCD b-function £ 2¤ ¯ = @¹ S¹ Tr G 2 g 3

Nucleon Mass h. N j. O®® j. N i = M N £ 2¤ ¯ = h. N j Tr G j. N i + 3 2 g h. N j(1 ¡ ° u )m u uu + (1 ¡ ° d )m d dd + (1 ¡ ° s )m s ssj. N i Anomalous dimension = quantum corrections

Ademollo-Gatto Theorem (1964) ® Corrections to the matrix elements of a charge operator between states in the same irreducible representation first occur as the square of the symmetry breaking parameter ® True if matrix element is analytic function of breaking parameter ® ® NOT valid for vector current matrix elements in light hadrons due to IR behavior of QCD True for heavy quark symmetry. . Luke’s Theorem Vector Current Matrix elements between members of SU(3), SU(2) irreps are protected from symmetry breaking effects, since they are the charge operators

Ademollo-Gatto Theorem (1964) Z ^ ab Q Z = d 3 x qa (x)° 0 qb(x) = h h. K 0 j h ^us Q i ^us ; Q ^ su Q i ^ su j. K 0 i ; Q d 3 x qay ° 0 qb = ^uu ¡ Q ^ ss Q = ^uu ¡ Q ^ ss j. K 0 i h. K 0 j Q ³ P n ´ ^ u s jni hnj Q ^ su j. K 0 i ¡ h. K 0 j Q ^ su jni hnj Q ^ u s j. K 0 i = 0 ¡ (¡ 1) h. K 0 j Q ³ ´ ^ su j. K 0 i j 2 ¡ jhnj Q ^ u s j. K 0 i j 2 = 1 AND t ransit ions out side jhnj Q oct et are O(¸ ) =0 1¡ ¡ h¼ su 0 ^ j Q j. K i = O(¸ l = SU(3) breaking parameter 2)

Baryon Resonance Spectrum ® So far just discussed extracting the ground states from lattice calculations. ® What about excitations ® ® If stable, the correlation function has simple exponential form If unstable, volume dependence required… l (see later)

Flavour, Orbital and Radial Structure • States classified according to SU(2) Flavor • Spatial and radial structure explored using displaced-source (sink) quark propagators Classified wrt transformation under hypercubic group … the symmetry group of the lattice

Methodology: Luscher-Wolff Compute correlation matrix from the r sources and r sinks C®¯ (t; t 0 ) = h 0 j. O® (t) O¯ (t 0 )j 0 i The eigenvalues of A ¸i ! = p 1 C(t 0 ) are³ e¡ E i ( t ¡ t 0 ) C(t) p 1 C(t 0 ) 1 + e¡ ¢ E (t ¡ t 0 ) ´ min ( En – Ei) • Eigenvalues ! Energies = masses of stable particles, (or energy of scattering state for unstable particles) • Eigenvectors ! “wave functions”

Glimpsing nucleon spectrum Adam Lichtl, Ph. D 2006 Spectroscopy Group. . . JLab

Summary ® Huge amount of phenomenology … traditionally the domain of nuclear physics (t > ~ 1970…QCD) ® ® Flavor structure Interactions Excitations Far fewer lattice calculations than for mesons ® ® Correlator falls much faster Signal degrades exponentially faster Requires significantly more effort … people-power and computers Relatively straighforward procedure to follow Go forth and compute the properties of the b blocks of nuclei from QCD !