Scattering on a square-well potential

Let us consider an attractive version of the soft sphere potential (1.52):

$$V^{SW}(r) = \left\{ {\begin{array}{ll} -V_0,& r < R\\ 0, & r \ge R \end{array}} \right.$$ (1.89)

Interaction (1.89) is called a square-well potential, with $V_0$ (positive) being its depth and $R$ being its range. The Schrödinger equation (1.41) for a pair of particles in the center of mass system is given by

$$\left\{ {\begin{array}{ll} u''(r)+(k^2+\varkappa ^2)\,u(r) = 0,& r<R\\ u''(r)+k^2u(r) = 0, &r\ge R \end{array}} \right.,$$ (1.90)

where, as usual, $k^2 = m{\cal E}/\hbar^2$ and

$$\varkappa ^2 = -mV_0/\hbar^2>0$$ (1.91)

We are interested in finding solutions with positive energies, as that are the solutions corresponding to a scattered state, instead solutions with negative energy are localized. On the opposite to the situation described in Sec. 1.3.5.1, the interaction potential is always lower than the value of the scattering energy $V_{int}(r)<{\cal E}$. For convenience we introduce ${\cal K}^2=\varkappa ^2+k^2>0$. In both regions the solution is a free-wave like:

$$u(r) = \left\{ {\begin{array}{ll} A\sin({\cal K}r+\delta_1),& r<R\\ B\sin(kr+\delta), &r\ge R \end{array}} \right.$$ (1.92)

The condition (1.40) immediately fixes the phase $\delta_1 = 0$. The matching equations for the function and its derivative read as

$$\left\{ {\begin{array}{lll} A\sin({\cal K}R) &=& B\sin(kR+\delta)\\ A{\cal K}\cos({\cal K}R) &=& Bk\cos(kR+\delta)\\ \end{array}} \right.$$ (1.93)

Condition of the continuity of the logarithmic derivative ${\cal K}\mathop{\rm ctg}\nolimits ({\cal K}R)= k\mathop{\rm ctg}\nolimits (kR+\delta)$ fixes the phase $\delta(k)$ of the solution

$$\delta(k) = \mathop{\rm arctg}\nolimits \left(\frac{k}{{\cal K}}\mathop{\rm tg}\nolimits {\cal K}R\right)-kR$$ (1.94)

This builds the relation between constants $A$ and $B$:

$$A^2 = \frac{B^2}{\sin^2 kR+\left(\frac{{\cal K}}{k}\cos kR\right)^2}$$ (1.95)

By taking limit of low energy in (1.94) and using the definition (1.44) one obtains the expression for the $s$-wave scattering length:

$$a_{3D} = R\left[1-\frac{\mathop{\rm tg}\nolimits \varkappa R}{\varkappa R}\right]$$ (1.96)

The dependence of the scattering length of the scattering on a soft sphere potential (Eq. 1.59) looks similar to (1.96) with the only difference that the trigonometric tangent is substituted with the hyperbolic one. The difference is crucial. Indeed, as $0<\th(x)/x\le 1$, the scattering length on the SS potential is always smaller than the range of the potential. Instead, the term $\mathop{\rm tg}\nolimits (x)/x$ is unbound. When the scattering happens at resonant momentum $\varkappa R = \pi/2+\Delta(\varkappa )$ with small detuning $\vert\Delta(\varkappa )\vert\ll 1$, the scattering length becomes extremely large and changes its sign.

The square well potential is attractive and in principle can have the bound state solution with energy $E_b = -\hbar^2k_b^2/m<0$. In outer region $r>R$ the solution (1.92) gets modified and decays exponentially fast. The condition of the continuity of the logarithmic derivative in the limit $k\to 0$ is $\varkappa \mathop{\rm tg}\nolimits \varkappa R = k_b\th k_bR$. This condition can not be satisfied before crossing the resonance, as inequality $\mathop{\rm tg}\nolimits x > \th x$ holds for arguments $x<0<\pi/2$. Instead immediately after the resonance position $\Delta(\varkappa )>0$ a shallow bound state appears in the system.