HADRONet p. 1 The non-perturbative N/D method and the 1 S NN partial wave 2nd Spanish Hadron Network Days Madrid, September 216 D.R. Entem, J.A. Oller University of Salamanca
HADRONet p. 2 Summary The framework: Chiral EFT Singular interactions Substractive renormalization The N/D method The non-pertubative (A) Results: LO, NLO and NNLO
HADRONet p. 3 Chiral EFT Effective degrees of freedom nucleons Chiral symmetry breaking pions Simetries: non linear realization of chiral symmetry Lagrangian: L = L ππ +L πn +L NN +... = all terms consistent with the symmetries Non renormalizable theory Power counting ν = 4 N +2(L C)+ i V i i i = d i + n i 2 2 Low energy expansion Contact terms.
HADRONet p. 4 Chiral EFT Order ν = V 1π ( p, p) = g2 A σ 1 q σ 2 q 4fπ 2 τ 1 τ 2 q 2 +m 2 π V () ct ( p, p) = C S +C T σ 1 σ 2
HADRONet p. 5 2π contributions One loop contributions ν = 2L+ i i Q 2 (NLO) Q 3 2 (N LO)
HADRONet p. 6 The perturbative amplitude All amplitudes are evaluated using dimensional regularization and MS. The ampitude is organize as V LO = V () ct +V () 1π V NLO = V LO +V (2) ct +V (2) 1π +V(2) 2π V NNLO = V NLO +V (3) 1π +V(3) 2π V N 3 LO = V NNLO +V (4) ct +V (4) 1π +V(4) 2π +V(4) 3π Contacts exactly absorb the infinities due to loop diagrams
HADRONet p. 7 Weinberg s proposal Evidences of non-perturbative nature: large scattering lengths and a bound state in NN Iterative diagrams breaks the Chiral expansion (non-perturbative) m N d 3 q (2π) 3V(p,q) k 2 q 2 +iǫ V(q,p) If V(p,p) = C d 3 q m N (2π) 3C k 2 q 2 +iǫ C = ic 2 m N k 4π Compute the potential using Chiral EFT and include it in a Lippmann-Schwinger Equation to account for the non-perturbative contribution The infinities are polynomial in the external momenta naive dimensional power counting (needed for renormalization of the irreducible terms)
HADRONet p. 8 Singular potentials Higher orders in the Chiral expansion produces singular potentials More singular than 1/r 2 when r. Divergent when q. The Lippman-Schwinger Eq. has to be regularized V(p,p) f(p,p;λ)v(p,p) In order to obtain regularization independence contacts are needed and we add (at zero order) V c (p,p) = C f(p,p;λ) C is fixed to some low energy data In order to obtain regularization independence there are two points of view Use Λ Λ χ Lepage plots point of view, use Λ between the low energy and the high energy scales so Λ < Λ χ. The tensor part of the LO potential is already singular Pavón-Valderrama and Ruiz-Arriola in coordinate space Nogga, Timmermans and van Kolck in momentum space
HADRONet p. 9 Singular tensor force 1 8 6 4 2 δ -2-4 -6-8 -1 2 4 6 8 1 Λ (MeV) T lab =6 MeV T lab =2 MeV T lab =3 MeV 3 D 2 LO Tensor goes as 1 r 3
HADRONet p. 1 Singular tensor force -.5-1 -1.5-2 δ -2.5-3 -3.5-4 T lab =6 MeV T lab =2 MeV T lab =3 MeV -4.5-5 2 4 6 8 1 Λ (MeV) 3 F 3 LO Tensor goes as 1 r 3
HADRONet p. 11 Substractive Renormalization T. Frederico, V.S. Timoteo, L. Tomio, Nucl. Phys. A 653 C.-J. Yang, Ch. Elster, D.R. Phillips, arxiv:nucl-th/76.1242 Zero Energy T Λ (p,;) = V(p,)+C + 2 Λ ( V(p π M dp p 2,p ) )+C T Λ (p,;) T Λ (,;) = V(,)+C + 2 π M Λ p 2 dp p 2 ( V(,p )+C p 2 ) T Λ (p,;) T Λ (,;) = T(,;) = a M T Λ (p,;) = V(p,)+ a M + 2 π M Λ ( V(p dp p 2,p ) V(,p ) ) p 2 T Λ (p,;)
HADRONet p. 11 Substractive Renormalization T. Frederico, V.S. Timoteo, L. Tomio, Nucl. Phys. A 653 C.-J. Yang, Ch. Elster, D.R. Phillips, arxiv:nucl-th/76.1242 Zero Energy T Λ (p,p;) = V(p,p)+C + 2 Λ ( V(p π M dp p 2,p ) )+C T Λ (p,p;) T Λ (,p;) = V(,p)+C + 2 π M Λ p 2 dp p 2 ( V(,p )+C p 2 ) T Λ (p,p;) T Λ (,;) = T(,;) = a M T Λ (p,;) = V(p,)+ a M + 2 π M Λ ( V(p dp p 2,p ) V(,p ) ) p 2 T Λ (p,;)
HADRONet p. 11 Substractive Renormalization T. Frederico, V.S. Timoteo, L. Tomio, Nucl. Phys. A 653 C.-J. Yang, Ch. Elster, D.R. Phillips, arxiv:nucl-th/76.1242 Zero Energy T Λ (p,p;) = V(p,p)+C + 2 Λ ( V(p π M dp p 2,p ) )+C T Λ (p,p;) T Λ (,p;) = V(,p)+C + 2 π M Λ p 2 dp p 2 ( V(,p )+C p 2 ) T Λ (p,p;) T Λ (,;) = T(,;) = a M T Λ (p,;) = V(p,)+ a M + 2 π M Λ T Λ (p,;) = T Λ (,p;) ( V(p dp p 2,p ) V(,p ) ) p 2 T Λ (p,;) T Λ (p,p;) T Λ (,p;) = V(p,p) V(,p) + 2 π M Λ dp p 2 ( V(p,p ) V(,p ) p 2 ) T Λ (p,p;)
HADRONet p. 12 Substractive Renormalization T. Frederico, V.S. Timoteo, L. Tomio, Nucl. Phys. A 653 C.-J. Yang, Ch. Elster, D.R. Phillips, arxiv:nucl-th/76.1242 Non-zero energy T Λ () = (V +C)+T Λ ()G Λ ()(V +C) (V +C) = T Λ ()(1+T Λ ()G Λ ()) 1 T Λ (E) = (V +C)+(V +C)G Λ (E)T Λ (E) (V +C) = T Λ (E)(1+G Λ (E)T Λ (E)) 1 T Λ (E) = T Λ ()+T Λ ()(G Λ (E) G Λ ())T Λ (E) Half off-shell T-Matrix at E = Full off-shell T-Matrix at E = Full off-shell T-Matrix at E The value of T(,;) is fixed
HADRONet p. 13 Half-offshell R-matrix 1e-5 5e-6 R(p,p) (MeV -2 ) -5e-6-1e-5-1.5e-5-2e-5-2.5e-5 2 4 6 8 1 p (MeV) N 3 LO NNLO NLO OPE Sharp cutoff (Λ = 4GeV ) T lab = 5MeV
HADRONet p. 13 Half-offshell R-matrix 1e-5 5e-6 R(p,p) (MeV -2 ) -5e-6-1e-5-1.5e-5-2e-5-2.5e-5 2 4 6 8 1 p (MeV) N 3 LO NNLO NLO OPE S N 3 LO S NNLO S NLO S OPE Sharp cutoff (Λ = 4GeV ) T lab = 5MeV
HADRONet p. 13 Half-offshell R-matrix 1e-5 5e-6 R(p,p) (MeV -2 ) -5e-6-1e-5-1.5e-5-2e-5-2.5e-5 5 1 15 2 25 3 35 4 p (MeV) N 3 LO NNLO NLO OPE Sharp cutoff (Λ = 4GeV ) T lab = 5MeV
HADRONet p. 13 Half-offshell R-matrix 1e-5 5e-6 R(p,p) (MeV -2 ) -5e-6-1e-5-1.5e-5-2e-5-2.5e-5 5 1 15 2 25 3 35 4 p (MeV) N 3 LO NNLO NLO OPE S N 3 LO S NNLO S NLO S OPE Sharp cutoff (Λ = 4GeV ) T lab = 5MeV
HADRONet p. 14 Half-offshell R-matrix 8e-6 6e-6 4e-6 R(p,p) (MeV -2 ) 2e-6-2e-6-4e-6-6e-6-8e-6-1e-5 2 4 6 8 1 p (MeV) N 3 LO NNLO NLO Sharp cutoff (Λ = 4GeV ) T lab = 5MeV
HADRONet p. 14 Half-offshell R-matrix 8e-6 6e-6 4e-6 R(p,p) (MeV -2 ) 2e-6-2e-6-4e-6-6e-6-8e-6-1e-5 N 3 LO NNLO NLO S N 3 LO Λ=2 GeV S NNLO Λ=2 GeV S NLO Λ=2 GeV 2 4 6 8 1 p (MeV) Sharp cutoff (Λ = 4GeV ) T lab = 5MeV
HADRONet p. 15 The N/D method We write the T matrix as T(A) = N(A) D(A) with A = k 2 (k the on-shell momentum) and so N(A) has a left hand cut (LHC) (for us A < m2 π 4 L) D(A) has a right hand cut (RHC) (A > ) The general Equations with 2n substractions and a substraction point C given in Z.-H. Guo, J.A. Oller and G. Ríos, Phys. Rev. C 89, 142 (214) D(A) = N(A) = n i=1 n i=1 δ i (A C) n i (A C)n π ν i (A C) n i (A C)n π L dω R ρ(ω R )N(ω R ) (ω R A)(ω R C) n dω L (ω L )D(ω L ) (ω L A)(ω R C) n
HADRONet p. 16 The N/D method Unitarity and Analitycal properties are used to get the equations and, if necessary, substractions We need as input the LHC discontinuity 2i (A) Usually this is computed perturbatively using χpt The contact terms does not give any contribution to (A) so the infinite contribution of loop diagrams does not contribute Contacts are taken into account through substraction constants In principle any number of substractions can be used, however for singular interactions not always a solution exists. To solve the Equations Including the Eq. for N in D one gets an integral Eq. only for D Once D is known in the LHC, D, N and T are known in all the complex plane
HADRONet p. 17 The N/D 1 We use the N/D equations with one substraction in D Always needed due to the invariance N αn and D αd Results for non-singular interactions should be the same as the result of Lippman-Schwinger Equation The usual prescription is D() = 1 The Equations are D(A) = 1 A π N(A) = 1 π L dω R ρ(ω R )N(ω R ) (ω R A)ω R dω L D(ω L ) (ω L ) (ω L A) where ρ(a) = M N A 4π T -matrix. is the phase space and 2i (A) is the LHC discontinuity of the
HADRONet p. 18 The N/D 11 We perform an additional substraction in N D(A) = 1 A π N(A) = ν 1 + A C π ρ(ω R )N(ω R ) dω R (ω R A)ω R L dω L D(ω L ) (ω L ) (ω L A)(ω L C) We use the Effective range expansion to fix the substraction constant and the substraction point kcotδ = 1 a + 1 2 rk2 + i=2 v i k 2i We get the Equations D(A) = 1+ia A+i M N 4π 2 N(A) = 4πa M N + A π L L dω L D(ω L ) (ω L ) ω L dω L D(ω L ) (ω L ) (ω L A)ω L They are independent of the substraction constant and point A A+ ωl
HADRONet p. 19 The N/D 12 We perform an additional substraction now in D We fix now the Effective Range parameter r We get the Equations D(A) = 1+ia A ar 2 A im NA 4π 2 [ A ( ω L + A) i ω L aω L N(A) = 4πa M N + A π L L ] dω L D(ω L ) (ω L ) (ω L A)ω L They are independent of the substraction constant and point dω L D(ω L ) (ω L ) ω L
HADRONet p. 2 The N/D 22 We perform an additional substraction now in N We fix now the next coefficient v 2 We get the Equations D(A) = (1 2v 2 r A)(1+ia A) ar 2 A L +i M N 4π 2 A D(ω L ) (ω L ) dω L ωl 2 [ A 2 +i A+ ωl ra 2 (1+ia ω L )(1+ia ] A) ω L L N(A) = 4πa +A 8πav 2 M N M N r + A D(ω L ) (ω L ) dω L π ωl 2 [ A (ω L A) + 2 (1+ia ] ω L ) raω L They are independent of the substraction constant and point
HADRONet p. 21 Leading Order - LO The potential is V( q) = g2 A 4f 2 π σ 1 q σ 2 q q 2 +m 2 τ 1 τ 2 +C S +C T σ 1 σ 2 π for the singlet 1 S partial wave The LHC discontinuity is g2 A m 2 π V(p,p) = 4fπ 2 2p p Q (z)+c Q (z) = 1 ( ) z +1 2 log z 1 z = p 2 +p 2 +m 2 π 2p p 1π (A) = θ(l A) g2 A 4f 2 π πm 2 π 4A
Iterative diagrams 2π exchange 2π (A) = θ(4l A) ( g 2 A m 2 π 16f 2 π ) 2 ( M N 2 A A A log m π ) 1 3π exchange 3π (A) = θ(9l A) 4π exchange θ(µ 1 2m π ) 4π (A) = θ(16l A) θ(µ 1 3m π ) θ(µ 2 2m π ) ( g 2 A m 2 π 4fπ 2 µ1 m π ) 3(MN 4π ) 2 π 4A 2 A mπ m π dµ 2 1 µ 2 (2 A µ 2 ) ( g 2 A m 2 π 4fπ 2 µ1 m π ) 4(MN 4π ) 3 π 4A 2m π dµ 1 1 µ 1 (2 A µ 1 ) 2 A mπ 1 dµ 2 2m π µ 2 (2 A µ 2 ) µ2 m π 3m π dµ 1 1 µ 1 (2 A µ 1 ) m π dµ 3 1 µ 3 (2 A µ 3 ). HADRONet p. 22
HADRONet p. 23 Iterative diagrams nπ exchange (with µ 2 A) ( ) g nπ (A) = θ(n 2 2 L A) A m 2 n(mn π 4π 4f 2 π ) n 1 π 4A n 1 j=1 θ(µ j 1 (n+1 j)m π ) µj 1 m π (n j)m π dµ j 1 µ j (2 A µ j ) This is the formal solution of the integral equation (with (A) = (A,2 A)) ( ) MN A (A, µ) = 1π (A)+ θ( µ 2m π ) π 2 µ mπ m π dµ 1π(A) (A,µ) µ(2 A µ) When A m π (A) = λπ 2 ( M N A e 2λ arctanh 1 m ) π A A λ = g2 A m2 π 4f 2 π M N 4π
HADRONet p. 24 General Case The general Equation is T(ν,k) = V(ν,k) θ(k m π )θ(k ν 2m π ) M N π 2 V(ν 1,ν) T(ν 1,k) k mπ ν+m π dν 1 ν 2 1 k 2 ν 2 1 with 2 T(ν, k) lím ǫ lím δ ImT(iν +ǫ δ,ik +ǫ) ImT(iν +ǫ+δ,ik +ǫ) 2 V(ν 1,ν) lím ǫ lím δ {ImV(iν +ǫ,iν 1 +ǫ+δ) ImV(iν +ǫ,iν 1 +ǫ δ)} = lím ǫ lím δ {ImV(iν 1 +ǫ δ,iν +ǫ) ImV(iν 1 +ǫ+δ,iν +ǫ)} and for OPE in the 1 S partial wave V(ν 1,ν) = π g2 A m2 π 16f 2 π 1 ν 1 ν sign(ν 1 ν)θ[ ν 1 ν m π ])
HADRONet p. 25 General Case Or if we define we get ˆT(ν 1,ν 2 ) = ν 1+l 1 ν 1+l 2 T(ν 1,ν 2 ) ˆV(ν 1,ν 2 ) = ν 1+l 1 ν 1+l 2 V(ν 1,ν 2 ) ˆT(ν,k) = ˆV(ν,k) M Nθ(k m π )θ(k ν 2m π ) π 2 k mπ We get the previous Equation with ν+m π dν 1 1 k 2 ν 2 1 1 ν 2l 1 For OPE in the 1 S partial wave V(ν 1,ν) = 1π (A) Making the change µ = k ν Identifying (A,µ) = k 2 ˆT(ν = k µ,k) ˆV(ν,ν 1 ) ˆT(ν 1,k)
HADRONet p. 26 Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 1.1 (A) ( L -1 ).1.1.1 1e-5 1e-6 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 26 Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 1.1 (A) ( L -1 ).1.1.1 1e-5 1e-6 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 26 Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 1.1 (A) ( L -1 ).1.1.1 1e-5 1e-6 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 26 Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 1.1 (A) ( L -1 ).1.1.1 1e-5 1e-6 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 26 Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 1.1 (A) ( L -1 ).1.1.1 1e-5 1e-6 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 26 Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 1.1 (A) ( L -1 ).1.1.1 1e-5 1e-6 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 27
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 27
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 27
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 27
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 27
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 27
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 28
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 28
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 28
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 28
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 28
Regular case At LO only OPE is present which is a non-singular interaction Results for the physical g A 14 12 1 δ1 S 8 6 4 2 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 28
HADRONet p. 29 Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 1 1 1 (A) ( L -1 ) 1 1.1.1.1.1 1e-5 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 29 Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 1 1 1 (A) ( L -1 ) 1 1.1.1.1.1 1e-5 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 29 Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 1 1 1 (A) ( L -1 ) 1 1.1.1.1.1 1e-5 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 29 Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 1 1 1 (A) ( L -1 ) 1 1.1.1.1.1 1e-5 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 29 Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 1 1 1 (A) ( L -1 ) 1 1.1.1.1.1 1e-5 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
HADRONet p. 29 Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 1 1 1 (A) ( L -1 ) 1 1.1.1.1.1 1e-5 1 1 1 1 1 1 1e+6 A ( L ) 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 45 4 35 3 δ1 S 25 2 15 1 5 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 3
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 45 4 35 3 δ1 S 25 2 15 1 5 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 3
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 45 4 35 3 δ1 S 25 2 15 1 5 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 3
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 45 4 35 3 δ1 S 25 2 15 1 5 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 3
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 45 4 35 3 δ1 S 25 2 15 1 5 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 3
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative 45 4 35 3 δ1 S 25 2 15 1 5 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 1 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 3
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative but it looks perturbative 6 4 2-2 δ1 S -4-6 -8-1 -12-14 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 31
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative but it looks perturbative 6 4 2-2 δ1 S -4-6 -8-1 -12-14 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 31
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative but it looks perturbative 6 4 2-2 δ1 S -4-6 -8-1 -12-14 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 31
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative but it looks perturbative 6 4 2-2 δ1 S -4-6 -8-1 -12-14 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 31
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative but it looks perturbative 6 4 2-2 δ1 S -4-6 -8-1 -12-14 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 31
Regular case Results for the unphysical g A = 6,8 Now the problem is non-perturbative but it looks perturbative 6 4 2-2 δ1 S -4-6 -8-1 -12-14 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11 1π (A) 2π (A) 3π (A) 4π (A) (A) (A) (for A m π HADRONet p. 31
HADRONet p. 32 Regular case Results for the unphysical g A = 6,8 N/D 1 as (fm) r (fm) v 2 (fm 3 ) v 3 (fm 5 ) v 4 (fm 7 ) v 5 (fm 9 ) v 6 (fm 11 ) 1π 1.66.714 -.168.847-5.35 35.5-241 2π 3.53 2.3-5.7 1 2 3.38-27.4 234-1.99 1 3 3π 1.8 1.15-8.71 1 2.924-5.69 36.9-247 4π -6.89 13.7 47.5 356 2.63 1 3 1.97 1 4 1.46 1 5 Non Perturbative -23.75 8.9 18.7 89.8 411 2. 1 3 8.98 1 3 N/D 11 1π -23.75 8.56 15.3 6.5 221 96 3.8 1 3 2π -23.75 8.8 17.7 8.4 346 1.6 1 3 6.71 1 3 3π -23.75 8.88 18.4 87.5 395 1.9 1 3 8.4 1 3 4π -23.75 8.9 18.6 89.3 48 1.98 1 3 8.87 1 3 Non-perturbative -23.75 8.9 18.7 89.8 411 2. 1 3 8.98 1 3
HADRONet p. 33 Regular case We encrease g A = 7,45 to have a near threshold bound state N/D 1 N/D 11 Schrödinger 1π 2.2 2π 2.18 3π 2.21 4π.89 2.22 Non-perturbative 2.22 2.22 2.22
HADRONet p. 34 Regular case Results for the physical g A LO N/D 11 compare with other renormalization procedures 8 7 6 5 δ1 S 4 3 2 1 5 1 15 2 25 3 35 4 k (MeV)
HADRONet p. 35 Regular case LO N/D 11 comparison with other renormalization procedures Renormalization with a counter term with different regulators Monopole form factor (same analytical properties as LO) V 1 (p,p) = C 1 Λ 2 2p p Q (z Λ ) Q (z Λ ) = 1 2 log ( zλ +1 z Λ 1 z Λ = p 2 +p 2 +Λ 2 2p p Gaussian type form factor (different analytical properties of LO) ) ( V 2 (p,p) = Ce 2 p ) 4 Λ ( p Λ )4 V 3 (p,p) = C 3 e ( p Λ Substractive renormalization with a sharp cutoff. ) 6 ( p Λ )6
HADRONet p. 36 Regular case LO N/D 11 comparison with other renormalization procedures 78.5 78. δ1 S (k=4 MeV) 77.5 77. 76.5 76. 75.5 75. 1 2 1 4 1 6 1 8 1 1 1 1 1/Λ (MeV -1 )
HADRONet p. 36 Regular case LO N/D 11 comparison with other renormalization procedures 78.5 78. δ1 S (k=4 MeV) 77.5 77. 76.5 76. 75.5 75. 1 2 1-5 4 1-5 6 1-5 8 1-5 1 1-4 1/Λ (MeV -1 ) V 1 (monopole) V 2 (G4) V 3 (G6)
HADRONet p. 36 Regular case LO N/D 11 comparison with other renormalization procedures 78.5 78. δ1 S (k=4 MeV) 77.5 77. 76.5 76. 75.5 75. 1 2 1-5 4 1-5 6 1-5 8 1-5 1 1-4 1/Λ (MeV -1 ) V 1 (monopole) V 2 (G4) V 3 (G6)
HADRONet p. 36 Regular case LO N/D 11 comparison with other renormalization procedures 78.5 78. δ1 S (k=4 MeV) 77.5 77. 76.5 76. 75.5 75. 1 2 1-5 4 1-5 6 1-5 8 1-5 1 1-4 1/Λ (MeV -1 ) V 1 (monopole) V 2 (G4) V 3 (G6)
HADRONet p. 37 Regular case - more substractions LO N/D 11 comparison with other renormalization procedures 8 7 6 5 δ1 S 4 3 2 1-1 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 12 N/D 22
HADRONet p. 37 Regular case - more substractions LO N/D 11 comparison with other renormalization procedures 8 7 6 5 δ1 S 4 3 2 1-1 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 12 N/D 22
HADRONet p. 37 Regular case - more substractions LO N/D 11 comparison with other renormalization procedures 8 7 6 5 δ1 S 4 3 2 1-1 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 12 N/D 22
HADRONet p. 38 Singular atractive case - NLO Now 1 1 V(ν 1,ν 2 ) = θ( ν 1 ν 2 µ )sign(ν 1 ν 2 ) ν 1 ν 2 1536πfπ 4 { ν 12 ˆω(ν 12 ) (1+52gA 2 158g4 A )m4 π + 1 2 (197g4 A 46g2 A 7)m2 π ν2 12 1 } 4 (59g4 A 1g2 A 1)ν4 12 +6(5gA 4 +2g2 A +1)m4 πˆl(ν 12 ) with ν 12 = ν 1 ν 2 and ˆω(ν 12 ) ν12 2 4m2 π ˆL(ν 12 ) ˆω(ν ( ) 12) ν12 + ˆω(ν 12 ) log ν 12 2m π
HADRONet p. 39 Singular atractive case - NLO N/D 12 does not converge 7 6 5 4 δ1 S 3 2 1-1 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 22
HADRONet p. 39 Singular atractive case - NLO N/D 12 does not converge 7 6 5 4 δ1 S 3 2 1-1 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 22
HADRONet p. 39 Singular atractive case - NLO N/D 12 does not converge 7 6 5 4 δ1 S 3 2 1-1 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 22
HADRONet p. 4 Singular atractive case - NNLO Now 1 V(ν 1,ν 2 ) = θ( ν 1 ν 2 µ )sign(ν 1 ν 2 ) ν 1 ν 2 g 2 A 384fπ 4 ( {( 48c 1 336c 3 128c 4 4 25g2 A M N 4M N 72c 1 +36c 3 3 M N + 3g2 A M N ( 12c 1 12c 3 +4c 4 + 2 ( 18c 3 6c 4 3 M N + 45g2 A 4M N ) m 4 πν 12 95g2 A M N 2M N ) } ν12 5 ) m 2 πν 3 12 ) m 5 π Dependent on the c i LECS
HADRONet p. 41 Singular atractive case - NNLO N/D 12 does not converge 7 6 5 4 3 δ1 S 2 1-1 -2-3 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 22
HADRONet p. 41 Singular atractive case - NNLO N/D 12 does not converge 7 6 5 4 3 δ1 S 2 1-1 -2-3 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 22
HADRONet p. 41 Singular atractive case - NNLO N/D 12 does not converge 7 6 5 4 3 δ1 S 2 1-1 -2-3 5 1 15 2 25 3 35 4 k (MeV) N/D 11 N/D 22
HADRONet p. 42 NNLO Different values for the c i LECS Analysis c 1 c 2 c 3 RS -.74-3.61 2.44 GW-HBChPT -1.13-5.51 3.71 KH-HBChPT -.75-4.77 3.34 GW-EOMS -1.5-6.63 3.68 KH-EOMS -1.26-6.74 3.74 GW-UChPT -1.11-4.78 3.4 KH-UChPT -1.4-4.48 3.
HADRONet p. 43 Singular atractive case - NNLO 8 6 4 δ1 S 2-2 -4 RS KH-EOMS Granada 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 11
HADRONet p. 44 Singular atractive case - NNLO 7 6 5 4 δ1 S 3 2 1-1 -2 RS KH-EOMS Granada 5 1 15 2 25 3 35 4 k (MeV) Solution of N/D 22 The same a, r and v 2 has been used This is probably an upper limit of the discrepancies
HADRONet p. 45 The antibound state Once you have (A) yo have the T matrix in the whole complex plane But you also have the second Riemmann sheet For Im[ A] > T 1 II D II N II = T 1 I +2iρ(A) = D I +2iρ(A)N I N I The 1 S partial wave has an untibound state (pole on the Imaginary axis of the second Riemmann sheet and so zero of D II ). We obtain for N/D 11 LO E =,7 MeV NLO E =,67 MeV NNLO E =,66 MeV With N/D 22 we get E =,66 MeV in all cases
HADRONet p. 46 Summary The N/D method with a non-perturbative (A) has been developed and study for the 1 S NN partial wave χpt is used to organize the finite range contributions of the potential The diagrams generates a perturbative (A) which is finite For LO the interaction is regular and the N/D method agrees with the solution of the LS Equation In the regular case any number of substractions can be taken For the NLO and NNLO the interactions are singular atractive and the N/D 11 is needed The solution of N/D 11 agrees with the renormalization with one counter term, substractive renormalization and so with renormalization with boundary conditions The N/D method can take additional substractions Taking more substractions in the N/D method allows to make a non-perturbative case to look perturbative No regularization is needed even for singular interactions