content_image/336002.jpg content_image/336003.jpg content_image/336004.jpg content_image/336005.jpg content_image/336006.jpg content_image/336007.jpg
## B. Positive region 

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 ) 

<img src='content_image/336002.jpg'>

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: 

<img src='content_image/336003.jpg'>

## C. Bound state wave function and match at x = 0 

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: 

<img src='content_image/336004.jpg'>

<img src='content_image/336005.jpg'>

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: 

<img src='content_image/336006.jpg'>

<img src='content_image/336007.jpg'>

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 ) 

