Scattering on a hard sphere potential

As pointed out in Sec. 1.3.2.1, in the limit of low energy collisions the information about the interaction potential enters in the terms of only one parameter, the $s$-wave scattering length and scattering on all potentials having the same scattering length is the same (the scattering becomes universal). This allows us to choose as simple potential as one can think of. If we consider the scattering on a repulsive potential, then the easiest choice is the hard sphere (HS) potential:

$$V^{HS}(r)= \left\{ \begin{array}{cc} +\infty, & r<a_{3D}\\ 0,& r\ge a_{3D} \end{array}\right.$$ (1.48)

This potential has only one parameter, which we name $a_{3D}$ in the definition (1.48). Obviously it has the meaning of the range of the potential (1.47). At the same time it has meaning of the scattering length, as introduced in (1.44). It will come out naturally from the solution of the scattering problem.

The Schrödinger equation (1.41) becomes ($\mu = m/2$)

$$-\frac{\hbar^2}{m}u''(r)+V^{HS}(r)u(r) = {\cal E}u(r)$$ (1.49)

A particle can not penetrate the hard core of the potential and the solution vanishes for distances smaller than the size of the hard sphere^1.4:

$$\left\{ {\begin{array}{ll} \displaystyle u(r) = 0,& \vert r\vert ... ...isplaystyle u''(r) - k^2 u(r) = 0,&\vert r\vert \ge a_{3D} \end{array}} \right.$$ (1.50)

The solution of the differential equation (1.50) can be easily found. Together with (1.39) we obtain:

$$f(r)= \left\{ {\begin{array}{ll} \displaystyle 0,& \vert r\vert <... ...playstyle A\sin(k(r-a_{3D}))\,/r,&\vert r\vert \ge a_{3D} \end{array}} \right.,$$ (1.51)

where $A$ is an arbitrary constant and $\k$ is given by (1.43). The phase shift is linear in the wave vector of the incident particle $\delta(k) = -ka_{3D}$ and from (1.44) we prove that the range of the potential (1.48) has indeed meaning of the three-dimensional scattering length as stated in the beginning of this section.