B. Positive region C. Bound state wave function and match at x = 0 17 With the help of the continuation formula of the Hypergeometric function [26] we can extract the behavior of the solution in the vicinity of x = 0 (recall that for x → 0, y, z → 1 and 1 − y ≈ e − a ( x − L ) Recall the parametric transformation for φ$_{L,R}$ : φ$_{R}$ = z − $^{ν}$(1 − z ) − $^{ϵ}$g ( z ) and φ$_{L}$ = y $^{ν}$(1 − y ) $^{ϵ}$h ( y ) and that in the limit of x → ±∞ the variable y → 0 as well as z → 0. Therefore imposing the boundary condition of a bound state (vanishing wave function at infinity) we obtain B ′ = C ′ = 0 and we are left with: We note that the wave function in the x > 0 region can be obtained from that of the x < 0 region simply letting ν → − ν and ϵ → − ϵ . The general solutions to Eqs. (48,50) are: Following a line of thought similar to that outlined in the previous subsection ( x < 0) we obtain τ 2 = − ( E 2 − m 2 $_{0}$) /a 2 = σ 2 = ν 2 and η = − i ( E + W ) /a = ϵ so that Eq. 49 reduces to: We consider the same substitution of scattering states for the variable x in the positive region. So we use the variable 1 /z = 1 + e a ( x − L $^{)}$, and the transformation φ = z − $^{τ}$(1 − z ) − $^{η}$g ( z ) z (1 − z ) d $^{2}$g ( z ) dz 2 + [1 − 2 τ − z (1 − 2 η − 2 τ )] d g ( z ) dz + + 1 a $^{2}$z (1 − z ) [( E + W z ) 2 − m 2 $_{0}$(1 − z ) 2 + iaW z (1 − z ) + iaz ( E + W z )] g ( z ) + 1 z (1 − z ) [ τ ( τ + 1)(1 − z ) 2 + η ( η + 1) z 2 − τ (1 − z ) + ηz − 2 ηz 2 − τ z (1 − z ) + ηz $^{2}$] g ( z ) +(2 τ − 2 τ η ) g ( z ) = 0 (49) z (1 − z ) d $^{2}$f ( z ) dz 2 + [1 − 2 ν − (1 − 2 ϵ − 2 ν ) z ] d f ( z ) dz 2 − ( − ϵ − ν + λ )( − ϵ − ν − λ ) f ( z ) = 0 . (50) h ( y ) = A ′ $_{2}$F$_{1}$ ( ϵ + ν + λ, ϵ + ν − λ ; 1 + 2 ν ; y ) + B ′ y − 2 ν $_{2}$F$_{1}$ ( ϵ − ν + λ, ϵ − ν − λ ; 1 − 2 ν ; y ) , g ( z ) = C ′ $_{2}$F$_{1}$ ( − ϵ − ν + λ, − ϵ − ν − λ ; 1 − 2 ν ; z ) + D ′ z 2 ν $_{2}$F$_{1}$ ( − ϵ − ν + λ, − ϵ − ν − λ ; 1 + 2 ν ; z ) . φ$_{L}$ ( y ) = A ′ y $^{ν}$(1 − y ) ϵ $_{2}$F$_{1}$ ( ϵ + ν + λ, ϵ + ν − λ, 1 + 2 ν ; y ) φ$_{R}$ ( z ) = D ′ z $^{ν}$(1 − z ) − ϵ $_{2}$F$_{1}$ ( − ϵ + ν + λ, − ϵ + ν − λ, 1 + 2 ν ; z )