What do g 1 (x) and g 2 (x) tell us about the spin and angular momentum content in the nucleon?

Transcription

1 What do g 1 (x) and g (x) tell us about the spin and angular momentum content in the nucleon? Franz Gross -- JLab cake seminar, Dec Introduction DIS hadronic tensor Spin puzzle in DIS Part I - covariant quark model Based on two papers about to be submitted to PRC: Covariant nucleon wave function with S, D, and P-state components handbag diagram wave functions with S, P, and D-state components general formulae for f(x), g 1 (x), and g (x) Spin and angular momentum in the nucleon in collaboration with Teresa Pena Part II - fits to the data predictions for a pure S-state fits f and g 1 with P and D-state components two solutions and predictions for g Conclusions and discussion Gilberto Ramalho Thanks to Christian Weiss and J.P. Chen I welcome your criticism and feedback -- revisions will be made before submission. Franz Gross - JLab/W&M

2 The DIS cross section depends on the hadronic tensor W µν ν µ W µν (P,q) The cross section is proportional to the hadronic tensor, which depends on four structure functions: P W µν (P,q) = π g µν W 1 + µ P ν W M iε µναβ q α S β P q G 1 + G M M + ( S q) iε µναβ q α P β G M 3 unpolarized structure functions; Callen-Gross relation MxW 1 = νw = xf (x) = x q e q f q (x) g µν = g µν qµ q ν q P µ = P µ ( P q)qµ q polarized structure functions νg 1 = g 1 (x) = 1 e q g q 1 (x) ν M G = g (x) = 1 e q g q (x) ν = P q M

3 The DIS cross section depends on the hadronic tensor W µν ν µ W µν (P,q) The cross section is proportional to the hadronic tensor, which depends on four structure functions: P W µν (P,q) = π g µν W 1 + µ P ν W M iε µναβ q α S β P q G 1 + G M M + ( S q) iε µναβ q α P β G M 3 unpolarized structure functions; Callen-Gross relation MxW 1 = νw = xf (x) = x q e q f q (x) g µν = g µν qµ q ν q P µ = P µ ( P q)qµ q polarized structure functions νg 1 = g 1 (x) = 1 e q g q 1 (x) ν M G = g (x) = 1 e q g q (x) ν = P q M

4 The spin puzzle (very short version) Experimental moments of the proton structure function g 1 p are too small (at Q =1 GeV ): For a pure S-state nucleon (i.e. nucleon spin = sum of quark spins), our model predicts (at Q ) Γ 1 p Γ 1 n 1 dx g p 1 (x) = 0.18 ± dx g n 1 (x) = 0.04 ± Γ 1 p Γ 1 n 1 dx g p 1 (x) = 5 18 = dx g n 1 (x) = 0 0 proton neutron This discrepancy seems huge -- what explains it? S-state only Experiment (LO at Q =1)

5 Motivation for this study Light cone dynamics does not handle rotational invariance, and angular momentum, very well. Interpretation is unclear. If angular momentum is the explanation, we need a covariant formalism with manifest rotational invariance and clearly defined angular momentum states. Use the Covariant Spectator Theory (CST ); used previously for NN scattering and the deuteron Three nucleon bound states Deuteron and triton form factors "N scattering and the study of baryon resonances qqbar models of mesons Relativistic quark models of the nucleon form factor and γn N* transitions** **Previous work with T. Pena and G. Ramalho

6 Part I: Covariant quark model motivated by CST Assumptions: NO color gauge invariance; NO Wilson loops or gluon radiation!! (It s NOT QCD) Quarks are dressed hadrons with mass and form factors dressed by gluons Nucleons consist of three valence quarks only; (sea quark contributions could be added later) Matching: take Q limit of the model -- compare to data at Q = Q 0 Assume high Q > Q 0 behavior predicted by QCD evolution NOT included in model Choose Q 0 = 1 GeV Evaluate covariant Feynman diagrams

7 DIS hadronic tensor computed from the handbag diagram Handbag diagram: All intermediate quarks are on-shell Off-shell quarks have dressed masses; their propagator is absorbed into the wave function with no singularities (confinement) Spectator pair of non-interacting quarks (the diquark ) can have a continuous mass distribution starting at m q ; fix at m s -- a parameter. Model the nucleon wave functions adjustible parameters fit to DIS data strength of the P and D-state components determined by fit to data DIS scattering condition constrains the momentum of the intermediate state S(P k) Γ(P, k)

8 DIS hadronic tensor computed from the handbag diagram Handbag diagram: All intermediate quarks are on-shell Off-shell quarks have dressed masses; their propagator is absorbed into the wave function with no singularities (confinement) Spectator pair of non-interacting quarks (the diquark ) can have a continuous mass distribution starting at m q ; fix at m s -- a parameter. Model the nucleon wave functions adjustible parameters fit to DIS data strength of the P and D-state components determined by fit to data DIS scattering condition constrains the momentum of the intermediate state m s S(P k) Γ(P, k) Ψ(P,k)

9 DIS hadronic tensor computed from the handbag diagram Handbag diagram: All intermediate quarks are on-shell Off-shell quarks have dressed masses; their propagator is absorbed into the wave function with no singularities (confinement) Spectator pair of non-interacting quarks (the diquark ) can have a continuous mass distribution starting at m q ; fix at m s -- a parameter. Model the nucleon wave functions adjustible parameters fit to DIS data strength of the P and D-state components determined by fit to data DIS scattering condition constrains the momentum of the intermediate state m s S(P k) Γ(P, k) Ψ(P,k) ( p' ) = (P + q k) = m q 1 = M + Q x 1 ( M + ν)e s + q k 1 Q x 1 E s Q Mx + Q + Mx Q ( Mx E s k cosθ M (1 x) ) DIS condition k cosθ

10 Two views of the DIS condition Light cone variables CST variables k = E s k z = M (1 x) k + = E s + k z = m s + k M (1 x) E s k cosθ = M (1 x) cosθ = E s M (1 x) k k fixed; x and k are the variables; range of k integration unconstrained by x cosθ fixed; x and k are variables, with a constraint. Scale k and E k k = Mκ; E k = ME κ = M r +κ ; r m s M then cosθ 1 κ r +κ + (1 x) E κ (1 x) 4( r +κ )(1 x) r + (1 x) κ r (1 x) (1 x) = κ min

11 Two views of the DIS condition k min Light cone variables Boundary in κ as a function of x k = E s k z = M (1 x) k + = E s + k z = M + k M (1 x) k fixed; region x and of r=1.5 k κ are the variables; range integration of k integration for unconstrained r=1 by x r= We chose r = 1 x r=0.75 CST variables E s k cosθ = M (1 x) cosθ = E S M (1 x) k cosθ fixed; x and k are variables, with a constraint. Scale k and E k k = Mκ; E k = ME κ = M r +κ ; r m s M then cosθ 1 κ r +κ + (1 x) E κ (1 x) 4( r +κ )(1 x) r + (1 x) κ r (1 x) (1 x) = κ min

12 CST model of the DIS hadronic tensor Ψ (P, k) Ψ(P, k) k = Mκ; r m s M = 1 The handbag diagram gives the hadronic tensor, an integral over the spectator momentum κ of a trace (with cosθ constrained) W µν (P,q) = 3 Λ κdκ Mx 16πE κ Q tr Ψ Λ (P,κ ) j ν (q)(m q + p ') j µ (q)ψ Λ (P,κ )(M + P )(1 + γ 5 S ) The nucleon wave functions have the general form { } nucleon wave function quark propagator quark current operators nucleon wave function Ψ Λ (P,κ ) = n L O Λ L (P,k)ψ L (P, k) { } L = S,P,D nucleon (spin S) projection operator j ν (q) = Q q γ ν qν q q

13 CST model nucleon wave functions Philosophy: use the principles of the CST to construct the most general wave function k Λ X P k X P λ P introduce parameters to describe the shape and size of the wave functions (NO dynamics at this time) S-state a sum of (Spin,Isospin) diquarks with (0,0) and (1,1) O S,0 (P, k) = 1 φ 0 ψ S (P, k) 1 (0,0) O S,1 (P,k) = 1 6 φ1 ψ S (P,k)γ 5 ε Λ * (1,1) P-state the same with a factor of O P,0 (P, k) = 1 φ 0 ψ P (P, k) k (0,0) O P,1 (P, k) = 1 6 φ1 ψ P (P,k) kγ 5 ε Λ * (1,1) D-state has three contributions k Determine the parameters by fitting data Wave function has the operator form: Ψ Λ L,n (P, k) = O Λ L,n (P, k) u(p,λ P ) O Λ D,0 (P, k) = 3 10 φ 0 k ψ D (P, k) ε Λ * O D,1 Λ (P,k) = 1 30 φ1 k * ψ D (P,k) θ DΛ O Λ D, (P, k) = 3 5 φ1 k ψ D (P,k) ε Λ * ( ) α G αβ ( k,ς ν )γ 5 γ β ( ) α γ 5 γ α ( ) α D αβ (P,k)γ 5 γ β ( k α = k α Pα k P) ; γ α = γ α Pα P M M ; gαβ = g αβ Pα P β M G αβ ( k,ς ν ) = k α ς β α ν + ς ν k β 3 ( gαβ k ςν ); D αβ (P,k) = k α k β 1 3 gαβ k

14 Some details about the structure of the D-state Three contributions, all of the form (J=1)+1/ 1/ (no J=0) O D,1 Λ (P, k) = 1 30 φ1 k * ψ D (P, k) ( θ DΛ ) α γ 5 γ α Average over the internal structure of the diquark (j=1)+1/ 1/ diquark ( =)+(S=1) (j=1): θ DΛ O D, Λ (P, k) = 3 5 φ1 k * ψ D (P, k) ( ε Λ ) α D αβ (P, k)γ 5 γ β D αβ (P,k) = k α k β 1 3 gαβ k (L=)+(S=1)+1/ 1/ diquark ( = 0,S=1): ε Λ O Λ D,0 (P, k) = 3 10 φ 0 k ψ D (P,k) ε Λ * G αβ ( k,ς ν ) = k α ς ν β + ς ν α k β 3 gαβ ( ) α G αβ ( k,ς ν )γ 5 γ β ( k ςν ) (L=1)+( =1)+(S=1) +1/ 1/ diquark ( =1,S=1): ζ ν ε Λ ( k α = k α Pα k P) ; γ α = γ α Pα P M M ; gαβ = g αβ Pα P β M

15 Some details about the structure of the D-state Three contributions, all of the form (J=1)+1/ 1/ (no J=0) O D,1 Λ (P, k) = 1 30 φ1 k * ψ D (P, k) ( θ DΛ ) α γ 5 γ α Average over the internal structure of the diquark (j=1)+1/ 1/ diquark ( =)+(S=1) (j=1): θ DΛ O D, Λ (P, k) = 3 5 φ1 k * ψ D (P, k) ( ε Λ ) α D αβ (P, k)γ 5 γ β D αβ (P,k) = k α k β 1 3 gαβ k (L=)+(S=1)+1/ 1/ diquark ( = 0,S=1): ε Λ O Λ D,0 (P, k) = 3 10 φ 0 k ψ D (P,k) ε Λ * G αβ ( k,ς ν ) = k α ς ν β + ς ν α k β 3 gαβ ( ) α G αβ ( k,ς ν )γ 5 γ β ( k ςν ) (L=1)+( =1)+(S=1) +1/ 1/ diquark ( =1,S=1): ζ ν ε Λ ( k α = k α Pα k P) ; γ α = γ α Pα P M M ; gαβ = g αβ Pα P β M

16 Some details about the structure of the D-state Three contributions, all of the form (J=1)+1/ 1/ (no J=0) O D,1 Λ (P, k) = 1 30 φ1 k * ψ D (P, k) ( θ DΛ ) α γ 5 γ α Average over the internal structure of the diquark (j=1)+1/ 1/ diquark ( =, S=1) (j=1): θ DΛ O D, Λ (P, k) = 3 5 φ1 k * ψ D (P, k) ( ε Λ ) α D αβ (P, k)γ 5 γ β D αβ (P,k) = k α k β 1 3 gαβ k (L=)+(S=1)+1/ 1/ diquark ( = 0,S=1): ε Λ O Λ D,0 (P, k) = 3 10 φ 0 k ψ D (P,k) ε Λ * G αβ ( k,ς ν ) = k α ς ν β + ς ν α k β 3 gαβ ( ) α G αβ ( k,ς ν )γ 5 γ β ( k ςν ) (L=1)+( =1)+(S=1) +1/ 1/ diquark ( =1,S=1): ζ ν ε Λ ( k α = k α Pα k P) ; γ α = γ α Pα P M M ; gαβ = g αβ Pα P β M

17 Structure of the scalar parts of the wave functions There are three scalar wave functions: P k ψ L (P, k) L = { S, P, D} k X X P Because k = m s and P = M, these can depend on only one variable, the square of the off-mass-shell quark, (P-k) Write this variable in terms of χ ( ) ( P k) χ = M m s Mm s = ( r +κ 1) 1 +κ 1 r ( ) Then the integrals that determine W µν become κ min = χ min = ξ = x( x) (1 x) x (1 x) κdκ ψ L (P, k)ψ L ' (P, k) = 16πE κ κ min ξ dχ ψ L (χ)ψ L ' (χ) 3π

18 Formulae of the structure functions (1) f q (x) After the traces are done, and Q, the three structure functions, g q 1 (x), and g q (x) become sums of the following 10 elementary structure functions f q L (x) = Mm s 16π g q L (x) = Mm s 16π ξ ξ dχ k L ψ L q (χ) dχ P (z 0 )k L ψ L q (χ) L=0,1, (or S, P, D) L=1, d q (x) = Mm s 16π dχ P (z 0 )k ψ S q (χ)ψ D q (χ) SD interference ξ h L q (x) = Mm s dχ z 16π 0 k L +1 ψ L q (χ)ψ P q (χ) ξ L h +1 q (x) = Mm s dχ 1 z 16π 0 ξ ( ) k L + ψ L q (χ)ψ P q (χ) 4Mx with z 0 = ξ = x (1 x) χ + x χ(χ + 4) L=0,; SP or DP interference Interpretation: Note that S-state wave function can be extracted from f q S (x) ψ q S (ξ) 16π = Mm s dx df S q (x) dξ dx = 16π (1 x) df S q (x) Mm s x( x) dx

19 Formulae for the structure functions () To get a picture of the content of a structure function, write as an array z = S a SS a SP a SD P a PS a PP a PD = D a DS a DP a DD LL ' a LL ' n S = 1 n P n D S P D n S f S q (x) n S n P h 0 q (x) 0 f q (x) = n S n P h 0 q (x) n P f P q (x) n D f D q (x) = n S f q S (x) + n P f q P (x) + n D f q D (x) n P n S h q 0 (x) 0 0 n n d g u 1 = 3 5 S D u f + u 0 8 n 9 P( f P u g P u ) n n h n n d 3 5 P D u ; g d 1 = 1 5 S D d 3 f + 4 d 0 n 9 P( f P d g P d ) 4 n n h 3 5 P D d ; n n d 3 5 S D u n n h 3 5 P D u n D ( f D u 9 g D 60 u ) n n d 5 S D d 4 n n h P D d n g D D 15 d g u = 0 n n (h 1 3 S P u h 0 u ) n n d 5 S D u n n (h 1 3 S P u h 0 u ) 4 n P 3 Pg u n n (h P D u h u ) n n d 5 S D u n n (h P D u h 9 u ) n g D D 40 u ; g d = 1 0 n n (h 1 3 S P d h 0 d ) n n d 5 S D d 1 n n (h 1 3 S P d h 0 d ) n P 3 Pg d 4 n n (h P D d h d ) n n d 5 S D d 4 n n (h P D d h 4 d ) n g D D 5 d ;

20 Summary of Part I Using a covariant quark model, formulae for the structure functions have been obtained from the handbag diagram Valence quarks ONLY; matching to QCD at Q = 1 GeV The strength of the S, P and D-state components, parameterized by n S, n P, and n D are to be determined from data The shape of the wave functions are also fit to data, we have chosen the effective diquark mass, m s = M (r = 1) Normalization: 1 1 = dx 0 f q (x) d 3 k 1 = e q0 k L ψ L (π ) 3 q (k) e q0 dx E k valence quark distributions (multiply by for the u quark distribution in the proton) 1 f q L (x) charge normalization, where e qo < 1 is the dressed quark charge at Q = 0 (in units of the bare quark charge Q q )

21 Part II - three steps to fitting the data Step 1: look at the prediction for a pure S-state: n S = 1; (n P = n D = 0) Choose shape of wave function to fit f q (x) using the valence fits of Martin, Roberts, Stirling, and Thorne (MRST0)* See the fits to g 1 (x); use the analysis of Leader, Sidorov, and Stamenov (LSS10)** Look at the qualitative effect of adding P and D-state components Step : choose the same parameters for the S P and D-states, find n P and n D which give the correct Γ 1 p and Γ 1 n Two solutions Look at shapes without adjusting parameters Step 3: adjust P and D-state parameters top get a good fit to g 1 (x) Predict g (x) and compare to data *Phys. Lett. B 531, 16 (005) ** Phys. Rev. D8, (010)

22 Fits to unpolarized data with S-state only Formulae (review) f q (x) = f q S (x) = M Fits ξ = 16π ξ x (1 x) dχ ψ S q (χ) ψ q S (ξ) 16π = M (1 x) x( x) df q S (x) dx u quark d quark ψ (χ) = N β cosθ + χ sinθ β θ n 0 n 1 e 0 u " χ n 0 (β + χ)n 1 n " d

23 Fits to unpolarized data with S-state only Formulae (review) f q (x) = f q S (x) = M Fits at Q = 1 16π ξ dχ ψ S q (χ) ξ = x (1 x) ψ q S (ξ) 16π = M (1 x) x( x) df q S (x) dx u quark d quark ψ S (χ) = φ(χ) = N β cosθ + χ sinθ β θ n 0 n 1 e 0 u " χ n 0 (β + χ)n 1 n " d

24 Qualitative effect of the D-state Keeping the D-state parameters equal to the S-state, f(x) is independent of n D and results for xg 1q (x) xδq(x) are n D > 0 ψ D (χ) = 1 κ φ(χ) = 4 χ(χ + 4) φ(χ) f D q (x) = M 16π dχ κ 4 ψ D (χ) q = M 16π dχ φ(χ) ξ f S q (x) (if parameters are the same) n D ξ 0 0. n D > 0 n D < LSS(10) Conclusion: n D = 0.6 solves the proton spin puzzle, but neutron??

25 Qualitative effect of the P-state Keeping the P-state parameters equal to the S-state, f(x) is NO longer independent of n P ψ P (χ) = 1 κ φ(χ) = χ(χ + 4) φ(χ) f P q (x) = M 16π dχ κ ψ P (χ) q = M 16π dχ φ(χ) ξ f q S (x) (if parameters are the same) ξ n P > 0 n P > 0 n P < 0 n P n P > 0 LSS(10) or MRST(0) n P > 0 n P < 0 Conclusion: n P = 0.4 solves the proton spin puzzle, but neutron??

26 Fixing n P and n D Keeping the other parameters the same, fix n P and n D so that Γ 1 p and Γ 1 n fit the experimental data. Two solutions: solution 1: n P = 0.43; n D = 0.18 solution : d quark with opposite phase for BOTH P and D-state n P = 0.08; n D = 0.59 d quark with opposite phase for D-state Keep n P and n D fixed, and adjust the other P and D-state parameters to fit the shapes of f and g 1

27 Solution 1 for u quark (n P = 0.43; n D = 0.18) same parameters adjusted Total S +P +D SP MRST(0) Total xf q all others LSS(10)

28 Solution 1 for d quark (n P = -0.43; n D = -0.18) same parameters adjusted Total S +P +D SP MRST(0) Total xf q all others LSS(10)

29 Solution for u quark (n P = 0.08; n D = 0.59) same parameters, no adjustment Total S +P +D SP MRST(0) Total xf q all others LSS(10)

30 Solution for d quark (n P = 0.08; n D = -0.59) same parameters adjusted Total S +P +D SP MRST(0) Total xf q all others LSS(10)

31 Predictions for g g is a special interest of JLab very small No quark model interpretation(??) Our predictions: total D-state only total D-state only solution solution 1 proton neutron

32 Where is the glue? Light cone formalism has long taught us that about 50% of the proton momentum comes from glue: 1 q 0 1 q 0 dx{ f q (x) f q (x)} = 3 baryon conservation (agrees with expt) xdx{ f q (x) f q (x)} momentum conservation (difference attributed to gluons) In this model glue is contained in the dressing of the quark the second conservation rule is not momentum (always conserved) but charge (at Q = 0): q e q0 Q q 1 dx{ f q (x) f q (x)} = 1 Both can be satisfied if e q0 < 1. Example: if all 1 distributions are identical, then dx f V (x) = 1 0 8(1 x) 3 1 f q (x) f q (x) f V (x) = 3 π xdx 1 f V (x) = 0.17 (x + 1) dx f V (x) = 11. e 0q = 0.089

33 Conclusions and discussion Solution (0.6% P-state and 35% D-state) fits all of the structure functions and explains the spin-puzzle. Solution 1 with 18% P-state and only 3% D-state also explains the spin puzzle, has smaller angular momentum components, but does not fit g! BUT, large errors and lack of data for g make conclusions tentative. Issues: Is 35% D-state believable ( 3 H wave function has a 7-9% D-state)? What would a simultaneous fit to the D-state strength AND the shape yield? Is it meaningful to match at Q = 1? (Some match at Q 0.16 GeV ) Where is the glue? (See comments on previous slide) EM gauge invariance is imposed in an ad-hoc manner. Are the predictions sensitive to this? What does this teach us (if anything) about using the LC formalism to interpret angular-momentum content of the proton?