Pion-Nucleon Light-Cone Distribution Amplitudes Exclusive Reactions 007 May 1-4, 007, Jefferson Laboratory Newport News, Virginia, USA Andreas Peters Regensburg Universtity in collaboration with V.Braun, D.Yu.Ivanov and A.Lenz supported by the Studienstiftung des deutschen Volkes
Outline 1 Motivation/Introduction Technical Issues 3 πn-da Results 4 Outlook
Why πn DAs? 1 Deep Inelastic Pion-Electroproduction: See recent articles: V. M. Braun, D. Y. Ivanov, A. Lenz and A. Peters, Deep inelastic pion electroproduction at threshold, Phys. Rev. D 75 (007) 01401 [arxiv:hep-ph/0611386]. J. P. Lansberg, B. Pire and L. Szymanowski, Hard exclusive electroproduction of a pion in the backward region, Phys. Rev. D 75 (007) 074004 [arxiv:hep-ph/070115]. Handle to learn more about nucleon DAs themselves
Motivation Polyakov, Pobylitsa, Strikman 006 p = φs(x) 6 u d u u u d d u u + φa(x) u u d d u u p π 0 = φs(x) 6f π 6u d u + u u d + d u u φa(x) f π u u d d u u n π + = φs(x) 1fπ u d u 3u u d 3d u u φa(x) f π u u d d u u Rewrite in operator language Extend to higher twists
Leading-Twist Nucleon DAs Three-quark matrix element: with 4 0 ε ijk u i α(a 1 z)u j β (a z)d k γ(a 3 z) p(p, λ) twist 3 = V p 1 (v 1) αβ,γ + A p 1 (a 1) αβ,γ + T p 1 (t 1) αβ,γ (v 1 ) αβ,γ = ( pc) αβ `γ5n + (a 1 ) αβ,γ = ( pγ 5C) αβ N + γ (t 1 ) αβ,γ = (iσ p C) αβ `γ γ 5N + γ γ
Leading-Twist Nucleon DAs Symmetry between the two up quarks: V 1 (1,, 3) = V 1 (, 1, 3), A 1 (1,, 3) = A 1 (, 1, 3) Isospin: T 1 (1,, 3) = [V 1 A 1 ](1, 3, ) + [V 1 A 1 ](, 3, 1) In Brodsky notation V 1 = φ s (x) and A 1 = φ a (x) V 1, A 1 can be expanded in contributions of operators with increasing conformal spin φ N = [V 1 A 1 ](1,, 3) =Nucleon DA T 1 is not independent
Leading-Twist Pion-Nucleon DAs Aim: Find a representation of V πn 1 etc. in terms of nucleon DAs Basic Ansatz: 4 0 ε ijk u i α(a 1 z)u j β (a z)d k γ (a 3 z) π(k)n(p,λ) twist 3 = -i/f π (γ 5 ) δγ {V πn 1 (v 1 ) αβ,δ + A πn 1 (a 1 ) αβ,δ + T πn 1 (t 1 ) αβ,δ }
Soft-Pion theorem: Basic concepts Take the Soft-Pion Theorem (SPT): 0 O π a (k)n i (P, h) = i f π 0 [Q a 5, O] N i(p, h) + Bremsstrahlungs contributions Q a 5 = d 3 x q(x)γ 0 γ 5 τ a q(x) N N π
Leading-Twist Pion-Nucleon DAs For nπ + specify: 0 O π + (k)n(p, h) = i 0 [Q + 5, O] n(p, h) f π Define trilocal operator as: together with O uud αβγ = ɛijk u i α(a 1 z)u j β (a z)d k γ (a 3 z) [Q5, 1 Oαβγ] uud = 1 (γ5) αλ Oλγβ ddu + (γ 5) βλ Oλγα ddu + (γ 5) γλ Oαβλ uuu [Q5, Oαβγ] uud = i (γ5) αλ Oλγβ ddu + (γ 5) βα Oλγα ddu (γ 5) γλ Oαβλ uuu
Results for nπ + The final comparison with the Ansatz leads us to: V nπ+ 1 (1,, 3) = 1 nv n 1 (1, 3, ) + V n 1 (1,, 3) + V n 1 (, 3, 1) o +A n 1(1, 3, ) + A n 1(, 3, 1), A nπ+ 1 (1,, 3) = 1 nv n 1 (3,, 1) V n 1 (1, 3, ) + A n 1(, 1, 3) o +A n 1(, 3, 1) + A n 1(3, 1, ), T1 nπ+ (1,, 3) = 1 n A n 1(, 3, 1) + A n 1(1, 3, ) V1 n (, 3, 1) o -V n 1(1, 3, ).
Results for nπ + The amplitudes have the natural symmetry property V nπ+ 1 (1,, 3) = V nπ+ 1 (, 1, 3), A nπ+ 1 (1,, 3) = A nπ+ 1 (, 1, 3) They are always built nucleon DAs of the same twist But they do not fulfill the Isospin relation anymore as they contain both I = 1/ and I = 3/ contributions
Results for pπ 0 The calculation for pπ 0 DAs is much easier as we do not need any Fierz transform [Q5, 3 Oαβγ] uud = 1 (γ5) αλ Oλβγ uud + (γ 5) βλ Oαλγ uud (γ 5) γλ Oαβλ uud V pπ0 1 (1,, 3) = 1 V p 1 (1,, 3), A pπ0 1 (1,, 3) = 1 Ap 1 (1,, 3), T pπ0 1 (1,, 3) = 3 T p 1 (1,, 3).
Results for pπ + pπ + DAs are built of nπ + DAs and nucleon DAs O uuu αγβ = ɛ ijk u i α(a 1 z)u j β (a z)u k γ(a 3 z) [Q5, 1 Oαβγ(z)] uuu = 1 n o (γ 5) αλ O duu λβγ(z) + (γ 5) βλ Oαλγ(z) udu + (γ 5) γλ Oαβλ(z) uud [Q5, Oαβγ(z)] uuu = i n o (γ 5) αλ O duu λβγ(z) + (γ 5) βλ Oαλγ(z) udu + (γ 5) γλ Oαβλ(z) uud V pπ+ i = ±V nπ+ i + V p i, A 1 and T 1 similar
Higher Twists General Definition of Higher Twists 4 0 ε ijk u i α(a 1 z)u j β (a z)d k γ(a 3 z) N(P, λ)π(k) = (γ 5) γδ i f π h S πn 1 (s 1 ) αβ,δ + S πn (s ) αβ,δ + P πn 1 (p 1 ) αβ,δ + P πn (p ) αβ,δ +V πn 1 (v 1 ) αβ,δ + V πn (v ) αβ,δ + 1 V πn 3 (v 3 ) αβ,δ + 1 V πn 4 (v 4 ) αβ,δ +V πn 5 (v 5) αβ,δ + V πn 6 (v 6 ) αβ,δ + A πn 1 (a 1 ) αβ,δ + A πn (a ) αβ,δ +A πn 3 (a 3 ) αβ,δ + 1 AπN 4 (a 4 ) αβ,δ + A πn 5 (a 5) αβ,δ + A πn 6 (a 6 ) αβ,δ +T πn 1 (t 1 ) αβ,δ + T πn (t ) αβ,δ + T πn 3 (t 3 ) αβ,δ + T πn 4 (t 4 ) αβ,δ +T πn 5 (t 5) αβ,δ + T πn 6 (t 6 ) αβ,δ + 1 T πn 7 (t 7) αβ,δ + 1 T πn i 8 (t 8 ) αβ,δ. Higher twist calculation follows the same procedure but much more extensive x -corrections can also be calculated
Summary We defined a trilocal Operator O and made use of the SPT to calculate πn DAs We obtained πn DAs which have the natural symmetry property. They are expressed in terms of nucleon DAs having the same twist
Outlook Apply to deep inelastic Pion Electroproduction (see talk of V.Braun) etc.... References: V. M. Braun, D. Y. Ivanov, A. Lenz and A. Peters, Phys. Rev. D 75 (007) 01401 [arxiv:hep-ph/0611386].