Testing Saturation Physics with pa collisions at NLO Accuracy David Zaslavsky with Anna Staśto and Bo-Wen Xiao Penn State University January 24, 2014 Prepared for University of Jyväskylä HEP group seminar
Outline 1 1 Overview of the Color Glass Condensate Factorization Models Evolution Equations 2 Calculation of the NLO pa Cross Section Theoretical Calculation Numerics 3 Results and Analysis: Relation to Experimental Data Results Resummation 4 Conclusion
Color Glass Condensate Structure of Protons and Nuclei 2 x = 1 ln 1 x small x small Q ln Q2 Q 2 0 large Q
Color Glass Condensate Structure of Protons and Nuclei 2 x = 1 ln 1 x small x saturation small Q ln Q2 Q 2 0 large Q Saturation regime: gluon self-interactions become important
Color Glass Condensate Gluon Distributions 3 Goal: describe gluons in a proton or nucleus Integrated gluon distribution xg(x, Q 2 ) (at right) Unintegrated gluon distributions f(x, k 2 ): Weisäcker-Williams UGD and dipole UGD How to avoid unitarity violations at small x? g(x, Q 2 ) 200 150 100 50 extrapolation Q 2 = 100 GeV 2 data Q 2 = 10 GeV 2 0 10 8 10 4 10 0 x
Color Glass Condensate Why pa? 4 Saturation regime is ( ) λ x0 x = 1 ln 1 x small x saturation small Q ln Q2 Q 2 0 large Q Q 2 Q 2 s = ca 1/3 Q 2 0 x Heavy ions (large A) make saturation more accessible Light projectiles (protons) prevent QGP and medium effects
Color Glass Condensate pa Collisions 5 p + x pp p p h π Momentum fraction of identified particle k z z = p k A x gp A X Momentum fraction of projectile parton Y = rapidity of pion p = transverse momentum of detected pion snn = CM energy of proton and nucleus x p = p z s NN e Y Momentum fraction of target parton x g = p z s NN e Y 0 figure adapted from Dominguez et al. 2011.
Color Glass Condensate» Factorization Models Factorization 6 k T factorization Central rapidity Y 0 x p, x g 0.01 p + Projectile and target treated in same model x pp p p h h Hybrid model a A x gp A k z X Forward rapidity Y 3 to 6 x p x g 10 4 Projectile treated in parton model Target treated as color glass condensate a Dumitru, Hayashigaki, and Jalilian-Marian 2005.
Color Glass Condensate» Factorization Models Factorization 6 k T factorization Central rapidity Y 0 x p, x g 0.01 p + Projectile and target treated in same model x pp p p h h Hybrid model a A x gp A k z X Forward rapidity Y 3 to 6 x p x g 10 4 Projectile treated in parton model Target treated as color glass condensate a Dumitru, Hayashigaki, and Jalilian-Marian 2005.
Color Glass Condensate» Factorization Models Hybrid Model 7 Cross section in the hybrid formalism: d 3 σ dy d 2 = ( dz dx p z 2 x xf i(x, µ)d h/i (z, µ)f x, p ) P(ξ)(...) z i Parton distribution (initial state projectile) Gluon distribution (initial state target) Fragmentation function (final state) Kinematic factors A p + xppp xgp A k z p h h X
Color Glass Condensate» Evolution Equations Evolution Equations 8 Fundamental tool of CGC: Balitsky-JIMWLK equations W Y [ρ] = 1 d 2 [x ]d 2 δ [y ] Y 2 δρ a Y (x ) χ δ ab(x, y ) δρ b Y (y ) W Y (ρ) S (2) Y (x y ) Y = N cα s π N c d 2 b 2π (x y ) 2 (x b ) 2 (y b ) 2 [ S (2) Y (x y ) S (2) Y (x b )S (2) Y (y b ) BK equation (large N c limit) used in practice (here LO) ]
Color Glass Condensate» Evolution Equations BK Equation 9 S (2) Y (r ) = N cα s d 2 b [ ] Y π 2π K S (2) Y (r ) S (2) Y (s )S (2) Y (t ) y where t K = b r2 s 2 +α s (rc corrections) + t2 }{{} LO r s S (2) Y is the position space CGC gluon distribution x
Color Glass Condensate» Evolution Equations Initial Conditions 10 BK (or JIMWLK) gives evolution in rapidity; initial conditions are unknown and must be determined via fits GBW MV ( ) S (2) Y (r ) = exp r2 Q2 s 4 [ ( S (2) Y (r ) = exp r2 Q2 s ln e + 4 AAMQS S (2) Y (r ) = exp [ (r2 Q2 s) γ 4 etc. ( ln e + 1 r Λ MV 1 r Λ MV )] )]
Color Glass Condensate» Summary Recap: Ingredients of the CGC 11 Hybrid factorization p + xppp p h h k z xgp A X A Evolution equation S (2) Y (r ) = N cα s d 2 b [ ] Y π 2π K S (2) Y (r ) S (2) Y (s )S (2) Y (t ) with free parameters: choice of terms included in kernel choice of form of initial condition choice of parameters in initial condition
NLO pa Cross Section Previous Calculations 12 Dumitru and Jalilian-Marian (2002) Dumitru, Hayashigaki, and Jalilian-Marian (2005) Fujii et al. (2011) Albacete et al. (2013) Rezaeian (2013) Staśto, Xiao, and Zaslavsky (2013) (our project) Gluon dist Cross section GBW MV/AAMQS LO LO BK rc BK inel NLO b-cgc NLO BK other NLO
NLO pa Cross Section Previous Calculations 12 Dumitru and Jalilian-Marian (2002) Dumitru, Hayashigaki, and Jalilian-Marian (2005) Fujii et al. (2011) Albacete et al. (2013) Rezaeian (2013) Staśto, Xiao, and Zaslavsky (2013) (our project) Gluon dist Cross section GBW MV/AAMQS LO LO BK rc BK inel NLO b-cgc NLO BK other NLO
NLO pa Cross Section Previous Calculations 12 Dumitru and Jalilian-Marian (2002) Dumitru, Hayashigaki, and Jalilian-Marian (2005) Fujii et al. (2011) Albacete et al. (2013) Rezaeian (2013) Staśto, Xiao, and Zaslavsky (2013) (our project) Gluon dist Cross section GBW MV/AAMQS LO LO BK rc BK inel NLO b-cgc NLO BK other NLO
NLO pa Cross Section Previous Calculations 12 Dumitru and Jalilian-Marian (2002) Dumitru, Hayashigaki, and Jalilian-Marian (2005) Fujii et al. (2011) Albacete et al. (2013) Rezaeian (2013) Staśto, Xiao, and Zaslavsky (2013) (our project) Gluon dist Cross section GBW MV/AAMQS LO LO BK rc BK inel NLO b-cgc NLO BK other NLO
NLO pa Cross Section Previous Calculations 12 Dumitru and Jalilian-Marian (2002) Dumitru, Hayashigaki, and Jalilian-Marian (2005) Fujii et al. (2011) Albacete et al. (2013) Rezaeian (2013) Staśto, Xiao, and Zaslavsky (2013) (our project) Gluon dist Cross section GBW MV/AAMQS LO LO BK rc BK inel NLO b-cgc NLO BK other NLO
NLO pa Cross Section Previous Calculations 12 Dumitru and Jalilian-Marian (2002) Dumitru, Hayashigaki, and Jalilian-Marian (2005) Fujii et al. (2011) Albacete et al. (2013) Rezaeian (2013) Staśto, Xiao, and Zaslavsky (2013) (our project) Gluon dist Cross section GBW MV/AAMQS LO LO BK rc BK inel NLO b-cgc NLO BK other NLO
NLO pa Cross Section» Theoretical Calculation Full NLO Cross Section 13 Complete NLO corrections to the cross section for pa h + X: 1 d 3 σ dy d 2 p = S jk = + α s 2π + α s 2π dzdξ z 2 [ xqi (x, µ) xg(x, µ) ] d 2 r (2π) 2 S(2) d 2 r Y (r )H (0) (2π) 2 S(2) 2jk Y (r )H (1) 2jk [ Sqq S qg S gq d 2 s d 2 t (2π) 2 S (4) Y (r, s, t )H (1) 4jk + etc. S gg ] [ ] Dh/qi (z, µ) D h/g (z, µ) LO dipole NLO dipole NLO quadrupole Note: we also use S (4) Y (r, s, t ) S (2) Y (s )S (2) Y (t ) 1 Chirilli, Xiao, and Yuan 2012.
NLO pa Cross Section» Theoretical Calculation Sample Fenyman Diagrams 14 LO real virtual p + p + p + A A A p + p + Squared amplitudes for qq channel A p + A p + A A p + p + A A
NLO pa Cross Section» Theoretical Calculation Factorization Scale Dependence 15 All parts depend on factorization scale µ F or renormalization scale µ R (where we set µ F = µ R = µ) Theory Sum of perturbative series independent of µ In practice: hope NLO corrections reduce µ dependence More on this later...
NLO pa Cross Section» Numerics Master Formula 16 Numerical integration does not deal with delta functions and plus prescriptions. These have to be eliminated manually. Each term can be split up into a singular portion (regulated by a plus prescription) F s, a portion proportional to a delta function F d, and a normal portion F n. 1 τ = dz 1 τ z 1 τ [ ] Fs (z, ξ) dξ + F n (z, ξ) + F d (z, ξ)δ(1 ξ) (1 ξ) + dz 1 τ dy z τ [ Fs (z, ξ) F s (z, 1) z(1 τ) 1 ξ 1 [ ( + dz F s (z, 1) ln 1 τ z τ ] + F n (z, ξ) ) ] + F d (z, 1)
NLO pa Cross Section» Numerics Regularization 17 Some terms of quadrupole hard factors take a form like H (n) 4ji = Ae ik t δ(1 ξ) 1 [ e dξ P (ξ iχ(ξ )k r ) 0 r 2 δ (2) (r ) d 2 r e ik ] r r 2 Position space: regularization is necessary F (n) d2 ij (z, 1, r ) = A (2π) 3 ( ) n αs S 2π z 2 1 S (4) Y (0, r, r )e ik r i x p f i (x p, µ)d h/i (z, µ) 0 dξ P (ξ ) ln 1 χ(ξ )
NLO pa Cross Section» Numerics Regularization 17 Some terms of quadrupole hard factors take a form like H (n) 4ji = Ae ik t δ(1 ξ) 1 [ e dξ P (ξ iχ(ξ )k r ) 0 r 2 δ (2) (r ) d 2 r e ik ] r r 2 Position space: regularization is necessary F (n) d4ij (z, 1, s, t ) = A (2π) 4 ( ) n αs S 2π z 2 [ S (4) Y (r, s, t ) S (4) ] Y (0, t, t ) 1 0 i x p f i (x p, µ)d h/i (z, µ) e ik t dξ P (ξ ) e iχ(ξ )k r r 2
NLO pa Cross Section» Numerics Regularization 17 Some terms of quadrupole hard factors take a form like H (n) 4ji = Ae ik t δ(1 ξ) 1 [ e dξ P (ξ iχ(ξ )k r ) 0 r 2 Momentum space: use identity δ (2) (r ) d 2 r e ik ] r r 2 d δ 2 2 r (r ) r 2 e ik r 1 r 2 e iξ k r = 1 d 2 k r 4π e ik ln (k ξ k ) 2 k 2
NLO pa Cross Section» Numerics Transformation to Momentum Space 18 Multidimensional Fourier integrals suffer from large numerical instabilities d 2 r S (2) Y (r )e ik r (...) d 2 s S (4) Y (r, s, t )e ik r (...) Easiest solution: transform to momentum space to remove phase factors F (k ) = 1 (2π) 2 d 2 r S (2) Y (r )e ik r = 1 2π 0 dr S (2) Y (r )J 0 (k r ) This offloads the numerical instabilities into the Fourier transform of the gluon distribution
NLO pa Cross Section» Numerics Transformation to Momentum Space 18 Multidimensional Fourier integrals suffer from large numerical instabilities d 2 r S (2) Y (r )e ik r (...) d 2 s S (4) Y (r, s, t )e ik r (...) Alternate solution: algorithms for direct evaluation of multidimensional Fourier integrals exist, but this requires a lot of coding We have not explored this
NLO pa Cross Section» Numerics Evaluation Errors 19 Many sources of error Inaccuracy of Fourier integrals Monte Carlo statistical error Always the possibility for bugs in the code! Two parallel implementations of selected parts (esp. qq channel): Mathematica, for rapid prototyping C++, for execution speed
Results and Analysis Experiments 20 RHIC: BRAHMS PHENIX STAR 2000-2006 2000-now 2000-now Small x Charged particles Hadrons LHC: ATLAS CMS ALICE 2011-now 2011-now 2011-now General General + Muons Heavy Ions
Results and Analysis» Results RHIC Comparison to Data 21 BRAHMS η = 2.2, 3.2 STAR η = 4 [ GeV 2 ] d 3 N dηd 2 p 10 1 10 1 10 3 10 5 η = 2.2 η = 3.2 ( 0.1) 10 2 10 3 10 4 10 5 LO NLO data 10 7 0 1 2 3 10 6 1 1.2 1.4 1.6 1.8 2 p [GeV] p [GeV]
Results and Analysis» Results RHIC Comparison to Data 21 BRAHMS η = 2.2, 3.2 STAR η = 4 [ GeV 2 ] d 3 N dηd 2 p 10 1 10 1 10 3 10 5 low p : decent match η = 2.2 η = 3.2 ( 0.1) 10 2 10 3 10 4 10 5 LO NLO data 10 7 0 1 2 3 p [GeV] 10 6 1 1.2 1.4 1.6 1.8 2 p [GeV]
Results and Analysis» Results RHIC Comparison to Data 21 BRAHMS η = 2.2, 3.2 STAR η = 4 [ GeV 2 ] d 3 N dηd 2 p 10 1 10 1 10 3 10 5 10 7 high p : negative! 0 1 2 3 η = 2.2 η = 3.2 ( 0.1) 10 2 10 3 10 4 10 5 10 6 LO NLO data 1 1.2 1.4 1.6 1.8 2 p [GeV] p [GeV]
Results and Analysis» Results Comparison of Gluon Distributions 22 [ GeV 2 ] [ GeV 2 ] d 3 N dηd 2 p d 3 N dηd 2 p 10 1 GBW 10 3 10 7 S (2) xg = exp [ r2 4 Q2 0 ( x0 xg ) λ ] 10 1 BK 10 3 10 7 0 1 2 3 p [GeV] [ S (2) xg = exp r2 4 Q2 0( x0 xg MV ) ( )] λ ln e + 1 Λr rcbk 0 1 2 3 p [GeV]
Results and Analysis» Results µ Dependence 23 LO NLO (fixed) NLO (running) 10 2 [ GeV 2 ] d 3 N dηd 2 p 10 3 10 4 10 5 LHC at η = 6.375 RHIC at η = 4.0 10 6 0 2 4 6 8 10 (µ/p ) 2
Results and Analysis» Resummation Sources of Negativity 24 [ GeV 2 ] d 3 N dηd 2 p 10 0 LO qq and gg 10 1 10 2 10 3 10 4 10 5 10 6 0.5 1 1.5 2 2.5 3 3.5 p [GeV] Plot shows magnitude of channel contribution Coloring indicates where value is Negative Positive Negativity comes from NLO diagonal channels: qq and gg
Results and Analysis» Resummation Sources of Negativity 24 [ GeV 2 ] d 3 N dηd 2 p 10 0 NLO qq 10 1 10 2 10 3 10 4 10 5 10 6 0.5 1 1.5 2 2.5 3 3.5 p [GeV] Plot shows magnitude of channel contribution Coloring indicates where value is Negative Positive Negativity comes from NLO diagonal channels: qq and gg
Results and Analysis» Resummation Sources of Negativity 24 [ GeV 2 ] d 3 N dηd 2 p 10 0 NLO gg 10 1 10 2 10 3 10 4 10 5 10 6 0.5 1 1.5 2 2.5 3 3.5 p [GeV] Plot shows magnitude of channel contribution Coloring indicates where value is Negative Positive Negativity comes from NLO diagonal channels: qq and gg
Results and Analysis» Resummation Sources of Negativity 24 [ GeV 2 ] d 3 N dηd 2 p 10 0 NLO gq 10 1 10 2 10 3 10 4 10 5 10 6 0.5 1 1.5 2 2.5 3 3.5 p [GeV] Plot shows magnitude of channel contribution Coloring indicates where value is Negative Positive Negativity comes from NLO diagonal channels: qq and gg
Results and Analysis» Resummation Sources of Negativity 24 [ GeV 2 ] d 3 N dηd 2 p 10 0 NLO qg 10 1 10 2 10 3 10 4 10 5 10 6 0.5 1 1.5 2 2.5 3 3.5 p [GeV] Plot shows magnitude of channel contribution Coloring indicates where value is Negative Positive Negativity comes from NLO diagonal channels: qq and gg
Results and Analysis» Resummation Candidate Resummation Procedure 25 Cross section in large p limit: d 3 σ qq dy d 2 α sn c S 1 p 4π 2 d 3 σ gg dy d 2 α sn c S p 2π 2 τ 1 τ dz z 2 D q(z) Q2 s k 4 dz z 2 D q(z) Q2 s k 4 Dominant contribution comes from τ 1 ξ 1 because ξ τ 1 1 τ z 1 set ξ = 1: x = x 0 = τ z, xq(x) = x 0q(x 0 ) 1 τ z dξ xq(x)j q (ξ) x 0 q(x 0 ) and likewise for gluon channel 1 τ z dξ xq(x)j q (ξ) dξ xq(x)j g (ξ) x 0 dξ J q (ξ) = x 0 q(x 0 )I q (x 0 )
Results and Analysis» Resummation Candidate Resummation Procedure 25 Cross section in large p limit: d 3 σ qq dy d 2 α sn c S 1 p 4π 2 d 3 σ gg dy d 2 α sn c S p 2π 2 τ 1 τ dz z 2 D q(z) Q2 s x 0 q(x 0 )I q (x 0 ) k 4 dz z 2 D q(z) Q2 s x 0 g(x 0 )I g (x 0 ) 10 k 4 I q (x 0 ) and I g (x 0 ) grow logarithmically as x 0 1 leading to large negative contributions at high p 0 0.2 0.4 0.6 0.8 1 0 10 4 ln(1 x 0 ) x 0
Results and Analysis» Resummation Candidate Resummation Procedure 25 Cross section in large p limit: d 3 σ qq dy d 2 α sn c S 1 p 4π 2 d 3 σ gg dy d 2 α sn c S p 2π 2 τ 1 τ dz z 2 D q(z) Q2 s x 0 q(x 0 )I q (x 0 ) k 4 dz z 2 D q(z) Q2 s x 0 g(x 0 )I g (x 0 ) Exponential resummation: add correction term 1 [ ( dz S z 2 D q(z) Q2 s x 0 q(x 0 ) exp c α ) sn c 4π 2 I q(x 0 ) c α ] sn c 4π 2 I q(x 0 ) 1 τ k 4 c is an empirical constant k 4
Results and Analysis» Resummation Resummation Results 26 BRAHMS dau η = 2.2 µ 2 = 10 GeV 2 to 50 GeV 2 BRAHMS dau η = 3.2 µ 2 = 10 GeV 2 to 50 GeV 2 [ GeV 2 ] d 3 N dηd 2 p 10 4 10 1 10 2 10 5 10 4 10 1 10 2 10 5 LO NLO c = 3 c = 10 data 10 8 0.5 1 1.5 2 2.5 3 3.5 p [GeV] 10 8 0 1 2 3 p [GeV] (note: slightly wrong)
Conclusion Summary 27 We have the first numerical implementation of the complete NLO pa h + X cross section Good: Bad: Matches data at low p Reduces µ dependence Negative at high p Resummation to the rescue? We ll see...
Acknowledgments Thanks to: Tuomas Lappi, my host Anna Staśto, my Ph.D. advisor Bowen Xiao, our collaborator Feng Yuan, for many useful discussions Work supported by US DOE OJI grant No. DE-SC0002145, Polish NCN grant DEC-2011/01/B/ST2/03915, and the Sloan Foundation.
Supplemental slides
Supplemental slides Quark-quark terms 28 S qq = + αs 2π d 2 r (2π) 2 S(2) Y (r )H (0) 2qq + αs 2π d 2 s d 2 t (2π) 2 S (4) Y (r, s, t )H (1) 4qq d 2 r (2π) 2 S(2) Y (r )H (1) 2qq H (0) 2qq = e ik r δ(1 ξ) ( H (1) 2qq = C F P qq(ξ) e ik r + 1 ) ξ 2 e ik r /ξ ln c2 0 r 2 3C µ2 F e ik r c 2 0 δ(1 ξ) ln r 2 k2 H (1) 4qq = 4πNce ik r 1 ξ i {e ξ k (x b ) 1 + ξ2 1 x b (1 ξ) + ξ (x b ) 2 y b (y b ) 2 1 δ(1 ξ) dξ 1 + ξ 2 [ e i(1 ξ)k (y b ) 0 (1 ξ ) + (b y ) 2 δ (2) (b y ) d 2 r e ik r r 2 ]}
Supplemental slides Gluon-gluon terms d 2 r S gg = (2π) 2 S(2) Y (r )S (2) Y (r )H (0) 2gg + αs 2π + αs d 2 s d 2 t 2π (2π) 2 S (2) Y (r )S (2) Y (s )S (2) Y (t )H (1) 6gg d 2 r (2π) 2 S(2) Y (r )H (1) 2gg + αs 2π d 2 r (2π) 2 S(2) Y (r )H (1) 2q q 29 H (0) 2gg = e ik r δ(1 ξ) [ H (1) 2gg = 2ξ 2(1 ξ) Nc + (1 ξ) + ξ ( 11 + 2ξ(1 ξ) + 6 2N ) ] f T R δ(1 ξ) 3N c ln c2 ( 0 µ 2 r 2 e ik r + 1 k ξ ) ( ξ 2 e i r 11 3 4N ) f T R N cδ(1 ξ)e ik r ln 3N c H (1) 2q q = 8πN f T R e ik (y b ) δ(1 ξ) 1 [ dξ [ξ 2 + (1 ξ ) 2 e iξ k (x y ) ] 0 (x y ) 2 δ (2) (x y ) d 2 r e ik r ] r 2 { H (1) 6gg = 16πNce ik r e i k (y ξ b ) [1 ξ(1 ξ)] 2 1 x y (1 ξ) + ξ 2 (x y ) 2 b y (b y ) 2 1 [ δ(1 ξ) dξ ξ 0 (1 ξ + 1 ] [ ) + 2 ξ (1 ξ e iξ k (y b ) ) (b y ) 2 δ (2) (b y ) d 2 r e ik r ]} r 2 c 2 0 r 2 k2
Supplemental slides Quark-gluon terms 30 S gq = αs 2π + αs 2π d 2 r (2π) 2 S(2) Y (r )[H (1,1) 2gq + S(2) Y (r )H (1,2) 2gq ] d 2 s d 2 t (2π) 2 S (4) Y (r, s, t )H (1) 4gq H (1,1) 2gq H (1,2) 2gq = Nc 2 1 ξ 2 e ik r /ξ 1 ξ [ 1 + (1 ξ) 2 ] ln c2 0 r 2 µ2 = Nc 2 e ik r 1 [ 2 ] 1 + (1 ξ) ln c2 0 ξ r 2 µ2 H (1) 4gq = 4πNce ik r /ξ ik t 1 [ 2 ] r 1 + (1 ξ) ξ r 2 t t 2
Supplemental slides Gluon-quark terms 31 S qg = αs 2π + αs 2π d 2 r (2π) 2 S(2) Y (r )[H (1,1) 2qg + S(2) Y (r )H (1,2) 2qg ] d 2 s d 2 t (2π) 2 S (4) Y (r, s, t )H (1) 4qg H (1,1) 2qg = 1 2 e ik r [ (1 ξ) 2 + ξ 2]( ln c2 ) 0 r 2 1 µ2 H (1,2) 2qg = 1 2ξ 2 e ik r /ξ[ (1 ξ) 2 + ξ 2] ( ln c2 ) 0 r 2 1 µ2 H (1) 4qg = 4πe ik r ik t /ξ (1 ξ)2 + ξ 2 r ξ r 2 t t 2