hadronic structure of the nucleon

Transcription

1 hadronic structure of the nucleon Kim Somfleth PhD Student, CSSM, University of Adelaide August 6, 2018 Collaborators: Jacob Bickerton, Alex Chambers, Roger Horsley, Yoshifumi Nakamura, Holger Perlt, Paul Rakow, Gerrit Schierholz, Arwed Schiller, Hinnurk Stüben, Ross Young & James Zanotti (QCDSF Collaboration)

2 proton structure spin sum rule 1 2 momentum sum rule 1 2

3 proton structure spin sum rule 1 2 = 1 2 q q momentum sum rule 1 = q x q 2

4 proton structure spin sum rule (Ji Decomposition) 1 2 = 1 2 q + L q + J g q momentum sum rule 1 = q x q + x g 2

5 form factors Contain information about charge radius (low Q 2 ) and distribution (high Q 2 ) 3

6 parton distribution functions 1 NNPDF3.0 (NNLO) xf(x,µ 2 =10 GeV ) g/10 u v Contain information about longitudional momentum of partons e e s 2 10 c x u d v d N X 4

7 generalised parton distribution functions GPDs unify form factors and parton momentum fraction 5

8 generalised parton distribution functions GPDs unify form factors and parton momentum fraction Want theoretical input for experiments to more completely understand strongly bound systems 5

9 generalised parton distribution functions GPDs unify form factors and parton momentum fraction Want theoretical input for experiments to more completely understand strongly bound systems Full image of GPDs on lattice still a challenge 5

10 lattice

11 feynman diagram calculation Hard problem Infinite series of problems Value = O(1) + O(α) + O(α 2 )

12 feynman diagram calculation Hard problem Infinite series of problems Value = O(1) + O(α) + O(α 2 )

13 path integrals O = 1 Dψ Z f Dψ f DA µ O [ ] A µ, ψ f, ψ f e S[A µ,ψ f,ψ f] f 8

14 path integrals O = 1 Dψ Z f Dψ f DA µ O [ ] A µ, ψ f, ψ f e S[A µ,ψ f,ψ f] f All possible field configurations 8

15 path integrals O = 1 Dψ Z f Dψ f DA µ O [ ] A µ, ψ f, ψ f e S[A µ,ψ f,ψ f] f All possible field configurations Field Configuration Importance 8

16 the lattice O = 1 Dψ Z f Dψ f DA µ O [ ] A µ, ψ f, ψ f e S[A µ,ψ f,ψ f] f ν Uµ(x) ψ(x) ψ(x + aˆµ) µ 9

17 the lattice O = 1 Dψ Z f Dψ f DA µ O [ ] A µ, ψ f, ψ f e S[A µ,ψ f,ψ f] f After discretisation and weighted Monte Carlo: ν Uµ(x) O = 1 N N i O [ ] U (i) µ ψ(x) µ ψ(x + aˆµ) Weighted by det [D f (U µ )] e Sg[Uµ] 9 f

18 lattice calculations 0 χ(0) Ω 10

19 lattice calculations 0 t G (2) (t) = Ω χ(t)χ(0) Ω 10

20 lattice calculations 0 t G (2) (t) = Ω χ(t)χ(0) Ω = X A X e E Xt 10

21 lattice calculations 0 t G (2) (t) = Ω χ(t)χ(0) Ω = X A X e E Xt A N e E Nt 10

22 gluons

23 sum rules spin sum rule (Ji Decomposition) 1 2 = 1 2 q + L q + J g q momentum sum rule 1 = q x q + x g 12

24 form factors of energy momentum tensor T g µν = Tr c G µα G α ν p T µν p =Su(p ) [γ µp νa 20(Q 2 )+ iσ µαq α P 2m νb N 20(Q 2 )+ q µqν C m N 20(Q 2 ) ] u(p) 13

25 form factors of energy momentum tensor T g µν = Tr c G µα G α ν p T µν p =Su(p ) [γ µp νa 20(Q 2 )+ iσ µαq α P 2m νb N 20(Q 2 )+ q µqν C m N 20(Q 2 ) ] u(p) A 20 (0) = x (A 20 + B 20 ) (0) = J 13

26 form factors of energy momentum tensor T g µν = Tr c G µα G α ν p T µν p =Su(p ) [γ µp νa 20(Q 2 )+ iσ µαq α 2m N =Su(p ) [P µ P ν M 2(Q 2 )+ iσ µαq α 2m N P νb 20(Q 2 )+ q µqν C m N 20(Q 2 ) ] u(p) P ν J(Q 2 )+ q µqν d m N 1(Q 2 ) ] u(p) A 20 (0) = x (A 20 + B 20 ) (0) = J A 20 = M 2 A 20 + B 20 = J C 20 = d 1 13

27 generation G (3) (J, t, τ, p, q) = d 3 ye iq y G (2) (t, p ) O (y, τ) 14

28 generation G (3) (J, t, τ, p, q) = d 3 ye iq y G (2) (t, p ) O (y, τ) Current multiplied in configuration by configuration 14

29 generation G (3) (J, t, τ, p, q) = d 3 ye iq y G (2) (t, p ) O (y, τ) Current multiplied in configuration by configuration Done for all possible unique combinations of p and q 14

30 generation G (3) (J, t, τ, p, q) = d 3 ye iq y G (2) (t, p ) O (y, τ) Current multiplied in configuration by configuration Done for all possible unique combinations of p and q No extra (expensive) quark propagator calculations 14

31 generation G (3) (J, t, τ, p, q) = d 3 ye iq y G (2) (t, p ) O (y, τ) Current multiplied in configuration by configuration Done for all possible unique combinations of p and q No extra (expensive) quark propagator calculations Take ratio R to remove time dependence 14

32 operators T µν = Tr c G µα G α ν 15

33 operators T µν = Tr c G µα G α ν Rotational Symmetry Hypercubic Symmetry H 4 = To avoid mixing use traceless combinations that transform irreducibly under H 4 15

34 operators T µν = Tr c G µα G α ν Rotational Symmetry Hypercubic Symmetry H 4 = To avoid mixing use traceless combinations that transform irreducibly under H 4 Operator T 4i Interpretation (ExB) i 15

35 operators T µν = Tr c G µα G α ν Rotational Symmetry Hypercubic Symmetry H 4 = To avoid mixing use traceless combinations that transform irreducibly under H 4 Operator T 4i T (T 33 + T 22 + T 11 ) Interpretation (ExB) ( i B 2 E 2) 15

36 form factor rewritten Interested to extract M 2 ( Q 2 = 0 ) and J ( Q 2 = 0 ) = (A 20 + B 20 ) ( Q 2 = 0 ) 16

37 form factor rewritten Interested to extract M 2 ( Q 2 = 0 ) and J ( Q 2 = 0 ) = (A 20 + B 20 ) ( Q 2 = 0 ) p T µν p =Su(p ) [P µp νm 2(Q 2 )+ iσµαqα P 2m νj(q 2 )+ qµqν d N m N 1(Q 2 ) ] u(p) 16

38 form factor rewritten Interested to extract M 2 ( Q 2 = 0 ) and J ( Q 2 = 0 ) = (A 20 + B 20 ) ( Q 2 = 0 ) p T µν p =Su(p ) [P µp νm 2(Q 2 )+ iσµαqα P 2m νj(q 2 )+ N ] u(p) Focus on operator T 4i and, q i = 0 16

39 Results 17

40 lattice details L 3 T β κ m π (MeV) N cfg N src/cfg (3)

41 lattice details L 3 T β κ m π (MeV) N cfg N src/cfg (3) Quenched study to compare with previous UKQCD/QCDSF results 18

42 lattice details L 3 T β κ m π (MeV) N cfg N src/cfg (3) Quenched study to compare with previous UKQCD/QCDSF results Current from Clover Plaquette on Wilson flowed Gauge Fields (Lüscher /JHEP08(2010)071) 18

43 three point function fit τ 0 t 0 τ t 0 << τ << t 19

44 three point function fit 19

45 3888 sets of three point functions for 49 data points 20

46 21

47 22

48 23

49 sum rules ( 1 2 = 1 A q 20 2 (0) + q q 1 = q B q 20 (0) + Ag 20 (0) + Bg 20 (0) ) x q + x g 24

50 sum rules ( 1 2 = 1 A q 20 2 (0) + q q 1 = q B q 20 (0) + Ag 20 (0) + Bg 20 (0) ) x q + x g µ = 2 GeV B 20 (0) J 24

51 sum rules ( 1 2 = 1 A q 20 2 (0) + q q 1 = q B q 20 (0) + Ag 20 (0) + Bg 20 (0) ) x q + x g µ = 2 GeV B 20 (0) J (u + d) con (32) 24

52 sum rules ( 1 2 = 1 A q 20 2 (0) + q q 1 = q B q 20 (0) + Ag 20 (0) + Bg 20 (0) ) x q + x g µ = 2 GeV B 20 (0) J (u + d) con (32) gluon (68) 0.209(29) 24

53 sum rules ( 1 2 = 1 A q 20 2 (0) + q q 1 = q B q 20 (0) + Ag 20 (0) + Bg 20 (0) ) x q + x g µ = 2 GeV B 20 (0) J (u + d) con (32) gluon (68) 0.209(29) Total (29) 24

54 25

55 quarks

56 deep inelastic scattering e e N X ω = 2p Q Q 2 = m2 X m2 N Q

57 deep inelastic scattering Hadron Tensor 2 Compton Amplitude Hadron Tensor has structure functions F 1, F 2 Compton Amplitude has Lorenzt-scalar functions T 1, T 2 F i = 1 2π ImT i 28

58 compton amplitude in experiment T µν=ρ ss d 4 ξe iq ξ p,s TJ µ(ξ)j ν(0) p,s 29

59 compton amplitude in experiment T µν=ρ ss d 4 ξe iq ξ p,s TJ µ(ξ)j ν(0) p,s Two photon exchange part of Hydrogen spectroscopic transition s contribution to proton charge radius uncertainty 29

60 compton amplitude in experiment T µν=ρ ss d 4 ξe iq ξ p,s TJ µ(ξ)j ν(0) p,s Two photon exchange part of Hydrogen spectroscopic transition s contribution to proton charge radius uncertainty Unpolarised Compton amplitude proton neutron difference contribution to P-N mass splitting 29

61 compton amplitude in experiment T µν=ρ ss d 4 ξe iq ξ p,s TJ µ(ξ)j ν(0) p,s Two photon exchange part of Hydrogen spectroscopic transition s contribution to proton charge radius uncertainty Unpolarised Compton amplitude proton neutron difference contribution to P-N mass splitting Subtraction term T(ω = 0, Q 2 ) 29

62 compton amplitude in experiment T µν=ρ ss d 4 ξe iq ξ p,s TJ µ(ξ)j ν(0) p,s Two photon exchange part of Hydrogen spectroscopic transition s contribution to proton charge radius uncertainty Unpolarised Compton amplitude proton neutron difference contribution to P-N mass splitting Subtraction term T(ω = 0, Q 2 ) Muonic hydrogen lamb shift uncertainty 29

63 compton amplitude in experiment T µν=ρ ss d 4 ξe iq ξ p,s TJ µ(ξ)j ν(0) p,s Two photon exchange part of Hydrogen spectroscopic transition s contribution to proton charge radius uncertainty Unpolarised Compton amplitude proton neutron difference contribution to P-N mass splitting Subtraction term T(ω = 0, Q 2 ) Muonic hydrogen lamb shift uncertainty P-N self energy uncertainty 29

64 compton amplitude in experiment T µν=ρ ss d 4 ξe iq ξ p,s TJ µ(ξ)j ν(0) p,s Two photon exchange part of Hydrogen spectroscopic transition s contribution to proton charge radius uncertainty Unpolarised Compton amplitude proton neutron difference contribution to P-N mass splitting Subtraction term T(ω = 0, Q 2 ) Muonic hydrogen lamb shift uncertainty P-N self energy uncertainty Reggeon dominance hypothesis 29

65 compton amplitude T µν = ρ ss d 4 ξe iq ξ p, s TJ µ (ξ) J ν (0) p, s ( = g µν + q ) ( µq ν T q p µ 1 ) ( 2 ωq µ p ν 1 ) 2 ωq T2 ν ν [ ] s + ϵ µναβ q α β ν G 1 + νmsβ s qp β G ν 2 2 where ν = p Q ω = 2p Q Q 2 30

66 compton amplitude T µν = ρ ss d 4 ξe iq ξ p, s TJ µ (ξ) J ν (0) p, s ( = g µν + q ) ( µq ν T q p µ 1 ) ( 2 ωq µ p ν 1 ) 2 ωq T2 ν ν [ ] s + ϵ µναβ q α β ν G 1 + νmsβ s qp β G ν 2 2 where restrict to a subset ν = p Q ω = 2p Q Q 2 p 3 = q 3 = 0 µ = ν = 3 ρ = 1 2 I 30

67 compton amplitude T 33 = d 4 ξe iq ξ p, s TJ 3 (ξ) J 3 (0) p, s ( = T ) 1 ω, Q 2 where restrict to a subset ν = p Q ω = 2p Q Q 2 p 3 = q 3 = 0 µ = ν = 3 ρ = 1 2 I 30

68 analytic structure of compton amplitude Imω ω 2 < 1 ω T ( ω, Q 2) 1 1 Reω 31

69 analytic structure of compton amplitude Imω ω 2 < 1 ω T ( ω, Q 2) 1 1 Reω 31

70 analytic structure of compton amplitude Imω ω 2 < 1 ω T ( ω, Q 2) T ( 0, Q 2) 1 1 Reω 31

71 analytic structure of compton amplitude Imω ω 2 < 1 ω 1 1 T ( ω, Q 2) T ( 0, Q 2) = 4ω2 dω ImT ( 1 ω, Q 2) 2π 1 ω (ω 2 ω 2 ) Reω 31

72 analytic structure of compton amplitude Imω ω 2 < 1 ω 1 1 T ( ω, Q 2) T ( 0, Q 2) = 4ω2 dω ImT ( 1 ω, Q 2) 2π 1 ω (ω 2 ω 2 ) Reω 1 = 4ω 2 dxx F ( ) 1 x, Q 2 1 (ωx) 2 0 ω = 1 x 31

73 moments Have T 1 ( ω, Q 2 ) T 1 ( 0, Q 2 ) = 4ω dx xf ( ) 1 x, Q 2 1 (ωx) 2 32

74 moments Have T 1 ( ω, Q 2 ) T 1 ( 0, Q 2 ) = 4ω 2 1 use geometric sum = n 1 ( 4ω 2n dxx 2n 1 F ) 1 x, Q dx xf ( ) 1 x, Q 2 1 (ωx) 2 32

75 moments Have T 1 ( ω, Q 2 ) T 1 ( 0, Q 2 ) = 4ω 2 1 use geometric sum but we know = n F 2 ( x, Q 2 ) = 2xF 1 ( x, Q 2 ) 1 ( 4ω 2n dxx 2n 1 F ) 1 x, Q 2 F 2 ( x, Q 2 ) = e 2 qx (q (x) + q (x)) 0 0 dx xf ( ) 1 x, Q 2 1 (ωx) 2 (Callan-Gross) (Parton Model) 32

76 moments Have T 1 ( ω, Q 2 ) T 1 ( 0, Q 2 ) = 4ω 2 1 use geometric sum but we know so = n F 2 ( x, Q 2 ) = 2xF 1 ( x, Q 2 ) 1 ( 4ω 2n dxx 2n 1 F ) 1 x, Q 2 F 2 ( x, Q 2 ) = e 2 qx (q (x) + q (x)) = n 0 0 dx xf ( ) 1 x, Q 2 1 (ωx) 2 (Callan-Gross) (Parton Model) 2ω 2n x 2n 1 32

77 Let s take it to lattice QCD 33

78 lattice situation»just«calculate the four point function d 3 xd 4 yd 4 ze ip x e iq y e iq z χ (x) T {J µ (y) J ν (z)} χ (0) 34

79 lattice situation»just«calculate the four point function d 3 xd 4 yd 4 ze ip x e iq y e iq z χ (x) T {J µ (y) J ν (z)} χ (0) Just remember three point functions from our gluons and do it in an extra dimension 34

80 lattice situation»just«calculate the four point function d 3 xd 4 yd 4 ze ip x e iq y e iq z χ (x) T {J µ (y) J ν (z)} χ (0) Just remember three point functions from our gluons and do it in an extra dimension Other techniques to get at PDFs Heavy quark currents Current-current correlators Quasi-PDFs Pseudo-PDFs Good Lattice Cross Sections 34

81 lattice situation»just«calculate the four point function d 3 xd 4 yd 4 ze ip x e iq y e iq z χ (x) T {J µ (y) J ν (z)} χ (0) Just remember three point functions from our gluons and do it in an extra dimension Other techniques to get at PDFs Heavy quark currents Current-current correlators Quasi-PDFs Pseudo-PDFs Good Lattice Cross Sections Complementary new method using Feynman-Hellmann theorem 34

82 feynman-hellmann theorem Calculate matrix elements using two point methods, ie. energy eigenstates 35

83 feynman-hellmann theorem Calculate matrix elements using two point methods, ie. energy eigenstates Modify Action S S + λ d 4 x cos (q x) J (x) 35

84 feynman-hellmann theorem Calculate matrix elements using two point methods, ie. energy eigenstates Modify Action S S + λ d 4 x cos (q x) J (x) Feynman-Hellmann Theorem de X,p dλ = 1 X, p J (0) X, p ± q λ=0 2E X,p 35

85 lattice caveat Connected Modify quark propagator High correlation for different λ 36

86 lattice caveat Connected Disconnected Modify quark propagator High correlation for different λ Modify weighting No correlation for different λ 36

87 things feynman-hellmann can do Large momentum EM form factors Disconnected spin contribution to nucleon spin 37

88 FHT of 2nd Order 38

89 fht setup Two point function now dependent on λ. Simplify calculation no excited states, ω 1 = Ep = 0 λ e ip x χ (x) χ (0) A p (λ) e E p(λ)x 4 39

90 fht setup Two point function now dependent on λ. Simplify calculation no excited states, ω 1 = Ep = 0 λ e ip x χ (x) χ (0) A p (λ) e E p(λ)x 4 Differentiate RHS [ ] Ap λ x E p 4A p e E p(λ)x 4 λ 39

91 fht setup Two point function now dependent on λ. Simplify calculation no excited states, ω 1 = Ep = 0 λ e ip x χ (x) χ (0) A p (λ) e E p(λ)x 4 Differentiate RHS [ ] Ap λ x E p 4A p e E p(λ)x 4 λ Twice [ 2 A p λ 2 A p E p 2x 4 λ λ x 2 E p 4A p λ 2 ( ) ] 2 + x 2 Ep 4A p e Ep(λ)x 4 λ 39

92 fht setup Two point function now dependent on λ. Simplify calculation no excited states, ω 1 = Ep = 0 λ e ip x χ (x) χ (0) A p (λ) e E p(λ)x 4 Differentiate RHS [ ] Ap λ x E p 4A p e E p(λ)x 4 λ Twice [ 2 A p λ 2 A p E p 2x 4 λ λ x 2 E p 4A p λ 2 ( ) ] 2 + x 2 Ep 4A p e Ep(λ)x 4 λ 39

93 lhs Differentiate LHS e ip x χ (x) χ (0) twice, ignoring disconnected terms ( ) 2 S e χ ip x (x) χ (0) λ 40

94 lhs Differentiate LHS e ip x χ (x) χ (0) twice, ignoring disconnected terms ( ) 2 S e χ ip x (x) χ (0) λ Pick S (λ) = d 4 y 2 cos (q y) J (y) 40

95 lhs Differentiate LHS e ip x χ (x) χ (0) twice, ignoring disconnected terms ( ) 2 S e χ ip x (x) χ (0) λ Pick S (λ) = d 4 y 2 cos (q y) J (y) Current outside operator has e Ex 4 like time dependence = corresponds to 2 A p λ 2 40

96 lhs Differentiate LHS e ip x χ (x) χ (0) twice, ignoring disconnected terms ( ) 2 S e χ ip x (x) χ (0) λ Pick S (λ) = d 4 y 2 cos (q y) J (y) Current outside operator has e Ex 4 like time dependence = corresponds to 2 A p λ 2 Current inbetween operators has x 4 e Ex 4 like time dependence = corresponds to 2 E p λ 2 40

97 lhs (cont.) LHS now e ip x x4 0 x4 y 4 z 4 0 d 3 yd 3 z 4 cos (q y) cos (q z) χ (x) T {J (y) J (z)} χ (0) 41

98 lhs (cont.) LHS now e ip x x4 0 x4 y 4 z 4 0 d 3 yd 3 z 4 cos (q y) cos (q z) χ (x) T {J (y) J (z)} χ (0) Which becomes A p x 4 e E px 4 2E p d 4 ξ 2 cos (q ξ) p T {J (ξ) J (0)} p 41

99 fht Second Order FHT d 2 E p d 4 ξ 2 cos (q ξ) TJ (ξ) J (0) p dλ 2 = λ=0 2E 42

100 renormalisation Need to renormalise T 33 43

101 renormalisation Need to renormalise T 33 T phys 33 = Z 2 V Tlatt 33 43

102 renormalisation Need to renormalise T 33 T phys 33 = Z 2 V Tlatt 33 Same renormalisation as for electro-magnetic form factor 43

103 discretised momentum 44

104 the structure function recipe S S + λ d 4 x ( e iq x + e iq x) qγ 3 q 45

105 the structure function recipe S S + λ d 4 x ( e iq x + e iq x) qγ 3 q Do this for λ = 0.0, ±0.0125, ±0.025, ± and calculate corresponding two point functions C (λ, t) at m π 470MeV 45

106 the structure function recipe S S + λ d 4 x ( e iq x + e iq x) qγ 3 q Do this for λ = 0.0, ±0.0125, ±0.025, ± and calculate corresponding two point functions C (λ, t) at m π 470MeV Take advantage of correlation C(λ,t)C( λ,t) C(0,t) 2 e 2 E even(λ)t 45

107 the structure function recipe S S + λ d 4 x ( e iq x + e iq x) qγ 3 q Do this for λ = 0.0, ±0.0125, ±0.025, ± and calculate corresponding two point functions C (λ, t) at m π 470MeV Take advantage of correlation C(λ,t)C( λ,t) C(0,t) 2 e 2 E even(λ)t E even = 1 2! λ2 d2 E dλ ! λ4 d4 E dλ

108 the structure function recipe S S + λ d 4 x ( e iq x + e iq x) qγ 3 q Do this for λ = 0.0, ±0.0125, ±0.025, ± and calculate corresponding two point functions C (λ, t) at m π 470MeV Take advantage of correlation C(λ,t)C( λ,t) C(0,t) 2 e 2 E even(λ)t E even = 1 2! λ2 d2 E dλ ! λ4 d4 E d2 E dλ 2 T 33 ( ω = 2p q Q 2, Q 2) dλ

109 the structure function recipe S S + λ d 4 x ( e iq x + e iq x) qγ 3 q Do this for λ = 0.0, ±0.0125, ±0.025, ± and calculate corresponding two point functions C (λ, t) at m π 470MeV Take advantage of correlation C(λ,t)C( λ,t) C(0,t) 2 e 2 E even(λ)t E even = 1 2! λ2 d2 E dλ ! λ4 d4 E d2 E dλ 2 T 33 ( ω = 2p q Q 2, Q 2) take T uu 33 Tdd 33 dλ 4 + and fit vs ω to get moments of x u d 45

110 46

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118 79 total such fits 50

119 51

120 = n 4ω2n 1 0 dxx 2n 1 F 1 (x,q 2 ) 52

121 53

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123 54

124 54

125 < x > u-d Alekhin et al., phenomenology ETMC 10 (TMF, N f =2) ETMC 13 (TMF, N f =2+1+1) ETMC 14 (TMF&clover, N f =2) LHPC 10 (DWF/asqtad, N f=2+1) LHPC 11 (DWF/asqtad, N f=2+1) m 2 π (GeV 2 ) RBC/UKQCD 10 (DWF, N f =2+1) QCDSF/UKQCD 11 (Clover, N f =2) QCDSF 11 (Clover, N f =2) LHPC 11 (DWF N f=2+1) LHPC 12 (Clover, N f=2+1) RQCD 13 (Clover, N f=2) Constantinou [ar Xi v: ] 55

126 56

127 deep inelastic scattering 57

128 twist and ope Non local operator in terms of a series of local operators N = c 1 O 1 + c 2 O Q c 3 O Q Handbag Diagram Cat s Ears Diagram Twist 2 or Leading Twist Higher Twist 58

129 deep inelastic scattering Experiment limited in quark content of targets (Ignoring heavier quark terms) F 2 (x,q 2 ) * 2 i x H1+ZEUS BCDMS E665 NMC SLAC T proton = 4 9 Tuu Tdd 2 9 Tud+du T neutron = 1 9 Tuu Tdd 2 9 Tud+du Q 2 (GeV 2 ) 59

130 deep inelastic scattering Experiment limited in quark content of targets (Ignoring heavier quark terms) F 2 (x,q 2 ) * 2 i x H1+ZEUS BCDMS E665 NMC SLAC T proton = 4 9 Tuu Tdd 2 9 Tud+du T neutron = 1 9 Tuu Tdd 2 9 Tud+du Q 2 (GeV 2 ) Lattice extraction of more combinations T uu, T dd, T uu + T dd + T ud+du, T uu + T dd T ud+du 59

131 60

132 61

133 conclusion

134 gluon structure (A 20 /M 2 ), J and B 20 for a large range of Q 2 using Wilson flow 63

135 gluon structure (A 20 /M 2 ), J and B 20 for a large range of Q 2 using Wilson flow Want to extract (C 20 /d 1 ) with the other operator 63

136 quark structure Looked at higher twist contributions, and quark flavour decomposition Want to determine F 2, g 1 and g 2 64

137 Thank You 65

138 backup

139 What about subtraction terms? 67

140 68

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145 m q qq Q 2 + x q Q 2 73