3N bound state formalism based on 3N free basis states

A spin–isospin dependent three-dimensional approach has been applied for the formulation of the three-nucleon bound state, and a new expression for Faddeev equation based on three-nucleon free basis states has been obtained. The advantage of this new expression is that the Faddeev integral equation has been simpler for numerical calculation.

In the case of the three-body bound state, the Faddeev equation has been formulated for three identical bosons as a function of vector Jacobi momenta, with the specific stress upon the magnitudes of the momenta and the angles between them [2]. Adding the spin-isospin to the 3D formalism was a major additional task which was carried out in Ref. [5]. In this paper we have attempted to reformulate the three-nucleon (3N) bound state and have obtained a new expression for Faddeev integral equation. To this end, we have used 3N free basis state for representation of 3N wave function.
This manuscript is organized as follows. In Sect. 2, we have derived a new expression for Faddeev equation in a realistic 3D scheme as a function of Jacobi momenta vectors and the spin-isospin quantum numbers. Then, we have chosen suitable coordinate system for describing Faddeev components of total 3N wave function as function of five independent variables for numerical calculations. Finally, in Sect. 3, a summary and an outlook have been presented.

3N bound state in a 3D momentum representation
Faddeev equation for the 3N bound state with considering pairwise interactions is described by [14]: where jw M t i is Faddeev component of the total 3N wave function, M t being the projection of total angular momentum along the quantization axis, P = P 12 P 23 ? P 13 P 23 is the sum of cyclic and anti-cyclic permutations of three nucleons, t denotes the two-body transition operator which is determined by a Lippmann-Schwinger equation and G 0 is the free 3N propagator which is given by: where E is the binding energy of 3N bound state. To solve Eq. (1) in the momentum space, we introduce the 3N free basis state in a 3D formalism as [6]: This basis state involves two standard Jacobi momenta p and q which are the relative momentum vector in the subsystem and the momentum vector of the spectator with respect to the subsystem, respectively [14]. jci jm s 1 m s 2 m s 3 m t 1 m t 2 m t 3 i is the spin-isospin parts of the basis state where the quantities m s_i (m t_i ) are the projections of the spin (isospin) of each three nucleons along the quantization axis. The introduced basis states are completed and normalized as: In the last equality, we have used the antisymmetry of Faddeev component of the 3N wave function as: and also we have considered: The antisymmetrized two-body t-matrix is introduced as where jpm s 2 m s 3 m t 2 m t 3 i a is the antisymmetrized two-body state which is defined as: Hence, the final expression for Faddeev equation is explicitly written: As a simplification, we rewrite this equation as: where we have used indexc instead of m 00 s m s 1 m 0 s m 00 t m t 1 m 0 t for simplicity. The previous expression which has been presented in Ref. [5] is: It is clear that in the new Faddeev integral equation, Clebsch-Gordan coefficients g ac and g c 0 a 0 and summing up on a 0 does not exist. Thus, this new expression is simpler for numerical calculations in comparison with the previous one.
For solving Eq. (15), one needs the matrix elements of the antisymmetrized two-body t-matrix. We connect this quantity to its momentum-helicity representation in ''Appendix 1''. To solve this integral equation numerically, we have to define a suitable coordinate system. It is convenient to choose the spin polarization direction parallel to the z axis and express the momentum vectors in this coordinate system. With this selection, we can write the two-body t-matrix and 3N wave function as (see Appendices 1 and 2): ðp; x p ; cosu pp ; x p ; p; y pp ; Þ; ð17Þ where x 0 ¼q 0 Áẑ; u 0 ¼ u q 0 and the labels ± are related to the signs of sin u pp ; sin u pq and sin u p 0 q 0 : With considering: Our final Faddeev equation from Eq. (15) is rewritten as: where the variables are developed similar to the 3N scattering as [13]: It is clear that the Faddeev component of the wave function w is explicitly calculated as a function of five independent variables. In Appendices 4 and 5, we discuss about the x 0 -and u 0 -integration and also determination of the signs of sine functions without any ambiguity. The Faddeev integral equation (21) represents a set of threedimensional homogenous integral equations, which after discretization turns into a huge matrix eigenvalue equation. The huge matrix eigenvalue equation requires an iterative solution method. We can use a Lanczos-like scheme that is proved to be very efficient for nuclear fewbody problems [15].
In this stage, we discuss about the total number of coupled integral equations. The total number of coupled Faddeev equations for the 3N bound state in a realistic 3D formalism according to the spin-isospin states is given by: where N s and N t are the total number of spin and isospin states, respectively, and N m_s_i is the number of spin states for each nucleon. It is clear that N m_s_i = 2 and N t = 3 for our problem. The factor 2 is related to signs of sine functions of azimuthal angles which is explained in ''Appendix 5''. Consequently, the total number of coupled Faddeev equations for either 3 H or 3 He is N = 48. The total 3N wave function jW M t i is given by [14]: Now, we derive an expression for the matrix elements of the total 3N wave function by inserting the 3N free basis state as follows: By applying the permutation operator P 12 P 23 and P 13 P 23 to the 3N free basis state, Eq. (25) can be written as [6]: with: q; c 2 m s2 m s3 m s1 m t2 m t3 m t1 ; As a simplification, Eq. (26) is rewritten as: Now, we rewrite this equation in the selected coordinate system as: By considering: Eq. (29) can be written as: AE W Mt c ðp; x p ; cosu pq ; x q ; qÞ ¼ AE w Mt c ðp; x p ; cosu pq ; x q ; qÞ þ e iðms 2 þms 3 Þu pq Â e Ài½ðms 3 þms 1 Þu p 2 q þðms 2 ÀMtÞu q 2 q AE w Mt c 2 ðp 2 ; x p2 ; cosu p2q2 ; x q2 ; q 2 Þ n þ e Ài½ðms 1 þms 2 Þu p 3 q þðms 3 ÀMtÞu q 3 q AE w Mt where: cosu p2q2 ¼p 2 Áq 2 À ðp 2 ÁẑÞðq 2 ÁẑÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À ðp 2 ÁẑÞ 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À ðq 2 ÁẑÞ 2 q ¼ cosu p2q ¼p 2 Áq À ðp 2 ÁẑÞðq ÁẑÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À ðp 2 ÁẑÞ 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 1 À ðq ÁẑÞ 2 q ¼ cosu q2q ¼q 2 Áq À ðq 2 ÁẑÞðq ÁẑÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À ðq 2 ÁẑÞ 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 1 À ðq ÁẑÞ 2 q ¼ The labels ± are related to the signs of sin u pq ; sin u p 2 q 2 and sin u p 3 q 3 which are determined in ''Appendix 5''.

Summary and outlook
We extend the recently developed formalism for a new treatment of the Nd scattering in three dimensions for the 3N bound state [13]. We propose a new representation of the 3D Faddeev equation for the 3N bound state including the spin and isospin degrees of freedom in the momentum space. This work provides the necessary formalism for the calculation of the 3N bound state observables which is under preparation.
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Appendix 1: Connection between the antisymmetrized NN t-matrix and its helicity representation
In our formulation, we need the matrix elements of the antisymmetrized two-body t-matrix. We connect these matrix elements to the corresponding ones in the momentum-helicity representation. The antisymmetrized momentumhelicity basis state which is parity eigenstate is given by [4]: jp;pS 23 k; t 23 i pa ¼ 1 ffiffi ffi 2 p ð1 À P 23 Þjp;pS 23 ki p jt 23 i ¼ 1 ffiffi ffi 2 p ð1 À g p ðÀÞ S 23 þt 23 Þjp;pS 23 ki p jt 23 i; Here, S 23 is the total spin, k is the spin projection along relative momentum of two nucleons, t 23 is the total isospin and jt 23 i jt 23 si is the total isospin state of the two nucleons. s is the isospin projection along its quantization axis which reveals the total electric charge of system. For simplicity, s is suppressed since electric charge is conserved. In Eq. (33), P 23 is the permutation operator which exchanges the two nucleons labels in all spaces i.e., momentum, spin and isospin spaces, and jp;pS 23 ki p is parity eigenstate which is given by: where P p is the parity operator, g p = ± 1 are the parity eigenvalues and jp;pS 23 ki is momentum-helicity state. The antisymmetrized two-body t-matrix is given by [6]: hp;pS 23 k; q;qS 1 Kjw M t i Jðp; x p ; cosu pq ; x q ; qÞ ¼ Appendix 5: The u 0 -integration According to Eq. (21), the u 0 -integration for fixed p, q, x p , x q , cosu pq and q 0 can be written as: du 0 e im 1 ðu q Àu 0 Þ e þim 2 u pq 0 e Àim 3 u p 0 q 0 Â A AE ½cosðu q À u 0 Þ; cosðu p À u 0 Þ; cosu pq where the A ± and B ± are known functions determined by t a ± and ± w, respectively. As we know, the exponential functions e þim 2 u pq 0 and e Àim 3 u p 0 q 0 are functions of cosu pq 0 and cosu p 0 q 0 by considering their sine functions as: Also, the cosine functions cosu pq 0 and cosu p 0 q 0 are function of u q À u 0 : Substituting u 00 ¼ u 0 À u q leads to: du 00 e Àim 1 u 00 e þim 2 u pq 0 e Àim 3 u p 0 q 0 A AE ½cosu 00 ; cosðu pq À u 00 Þ; cosu pq B AE ½cosu 00 where the labels of I AE ðcosu pq Þ depend on the sign of sin u pq : It is clear that the angles u pq 0 and u p 0 q 0 belong to the interval [-p, 0] when u 00 varies in the interval [0, p] and they belong to the interval [0, p] when u 00 varies in the interval [p, 2p]. Furthermore, since the labels of B ± depend on the sign of sin u p 0 q 0 ; thus for u 00 2 ½0; p and u 00 2 ½p; 2p, we can choose negative and positive labels, respectively. Consequently, the integral I AE ðcosu pq Þ can be decomposed as: du 00 e Àim 1 u 00 e Àim 2 ju pq 0 j e þim 3 ju p 0 q 0 j A AE Â ½cosu 00 ; cosðu pq À u 00 Þ; cosu pq B À ½cosu 00 þ Z 2p p du 00 e Àim 1 u 00 e þim 2 u pq 0 e Àim 3 u p 0 q 0 A AE Â ½cosu 00 ; cosðu pq À u 00 Þ; cosu pq B þ ½cosu 00 : ð67Þ Now, we discuss about the labels of A ± . As we know, the labels of A ± are related to the sign of sin u pp : We can write u pp ¼ u pq À u pq ; and then we have: sin u pp ¼ sin u pq cosu pq À cosu pq sin u pq ; where: cosu pq ¼p Áq À ðp ÁẑÞðq ÁẑÞ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 1 À ðp ÁẑÞ 2 q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 1 À ðq ÁẑÞ 2 q ¼ y pq À x p x q ffiffiffiffiffiffiffiffiffiffiffiffiffi 1 À x p p ffiffiffiffiffiffiffiffiffiffiffiffi ffi 1 À x 2 q q ; It is clear that the angle u pq belongs to the interval [0, p] when u 00 varies in the interval [0, p] and belong to the interval [p, 2p] when u 00 varies in the interval [p, 2p]. Thus, depending on various intervals of variables u pq and u 00 ; we can choose the positive or negative sign for sin u pq and sin u pq ; and then we can calculate sin u pp from Eq (68). Consequently, for sin u pp 2 ½0; 1 and sin u pp 2 ½À1; 0, we can consider positive and negative signs of A ± , respectively. Substituting u 000 ¼ 2p À u 00 ; in the second integral of Eq. (67) yields: Z p 0 du 000 e þim 1 u 000 e þim 2 u pq 0 e Àim 3 u p 0 q 0 A AE Â ½cosu 000 ; cosðu pq þ u 000 Þ; cosu pq B þ ½cosu 000 ; ð70Þ Therefore, Eq. (67) can be rewritten: du 00 e Àim 1 u 00 e Àim 2 ju pq 0 j e þim 3 jup0 q 0 j A AE Â ½cosu 00 ; cosðu pq À u 00 Þ; cosu pq B À ½cosu 00 þ Z p 0 du 00 e þim 1 u 00 e þim 2 u pq 0 e Àim 3 u p 0 q 0 A AE Â ½cosu 00 ; cosðu pq þ u 00 Þ; cosu pq B þ ½cosu 00 :