Trial wave function: zero energy scattering state

On the BCS side of the resonance the scattering length is negative $a<0$. Here the attractive square well potential well of strength $V_{0}=\hbar ^{2}\varkappa ^{2}/m$. There is no bound state anymore and instead one has a solution with positive energy ${\cal E}=\hbar ^{2}k^{2}/m$. The scattering solution is

$$f(r)=\left\{ \begin{array}{cc} \frac{A}{r}\sin {\cal K}r, & r<R \\ \frac{B}{r}\sin (kr+\delta ), & r>R \end{array}\right.$$ (2.117)

The scattering length $a$ phase is related to the phase of the scatters wave. In the low-energy limit the phase is simply $\delta =-ka$. In this limit ${\cal K}\varkappa$ and we have simple solution

$$f(r)=\left\{ \begin{array}{cc} \frac{A}{r}\sin kr, & r<R \\ B(1+\frac{\vert a\vert}{r}), & r>R \end{array}\right.$$ (2.118)

$$f^{\prime }(r)=\left\{ \begin{array}{cc} \frac{A}{r}\sin kr\left(... ...1}{r} \right), & r<R \\ -\frac{B\vert a\vert}{r^{2}}, & r>R \end{array}\right.$$ (2.119)