$$\require{cancel}$$
Let us now consider how the phase-shifts, $$\delta_l$$, in Equation ([e17.73]) can be evaluated. Consider a spherically symmetric potential, $$V(r)$$, that vanishes for $$r>a$$, where $$a$$ is termed the range of the potential. In the region $$r>a$$, the wavefunction $$\psi({\bf r})$$ satisfies the free-space Schrödinger equation ([e17.54]). The most general solution that is consistent with no incoming spherical-waves is $\psi({\bf r}) = \sqrt{n}\, \sum_{l=0,\infty} {\rm i}^{\,l}\, (2\,l+1) \, {\cal R}_l(r)\, P_l(\cos\theta),$ where $\label{e17.80} {\cal R}_l(r) = \exp(\,{\rm i} \,\delta_l)\, \left[\cos\delta_l \,j_l(k\,r) -\sin\delta_l\, y_l(k\,r)\right].$ Note that $$y_l(k\,r)$$ functions are allowed to appear in the previous expression because its region of validity does not include the origin (where $$V\neq 0$$). The logarithmic derivative of the $$l$$th radial wavefunction, $${\cal R}_l(r)$$, just outside the range of the potential is given by $\beta_{l+} = k\,a \left[\frac{ \cos\delta_l\,j_l'(k\,a) - \sin\delta_l\, y_l'(k\,a)}{\cos\delta_l \, j_l(k\,a) - \sin\delta_l\,y_l(k\,a)}\right],$ where $$j_l'(x)$$ denotes $$dj_l(x)/dx$$, et cetera. The previous equation can be inverted to give $\label{e17.82} \tan \delta_l = \frac{ k\,a\,j_l'(k\,a) - \beta_{l+}\, j_l(k\,a)} {k\,a\,y_l'(k\,a) - \beta_{l+}\, y_l(k\,a)}.$ Thus, the problem of determining the phase-shift, $$\delta_l$$, is equivalent to that of obtaining $$\beta_{l+}$$.
The most general solution to Schrödinger’s equation inside the range of the potential ($$r<a$$) that does not depend on the azimuthal angle $$\phi$$ is $\psi({\bf r}) = \sqrt{n}\,\sum_{l=0,\infty} {\rm i}^{\,l} \,(2\,l+1)\,{\cal R}_l(r)\,P_l(\cos\theta),$ where ${\cal R}_l (r) = \frac{u_l(r)}{r},$ and $\label{e17.85} \frac{d^{\,2} u_l}{d r^{\,2}} +\left[k^{\,2} -\frac{l\,(l+1)}{r^{\,2}} -\frac{2\,m}{\hbar^{\,2}} \,V\right] u_l = 0.$ The boundary condition $\label{e17.86} u_l(0) = 0$ ensures that the radial wavefunction is well behaved at the origin. We can launch a well-behaved solution of the previous equation from $$r=0$$, integrate out to $$r=a$$, and form the logarithmic derivative $\beta_{l-} = \left.\frac{1}{(u_l/r)} \frac{d(u_l/r)}{dr}\right|_{r=a}.$ Because $$\psi({\bf r})$$ and its first derivatives are necessarily continuous for physically acceptable wavefunctions, it follows that $\beta_{l+} = \beta_{l-}.$ The phase-shift, $$\delta_l$$, is then obtainable from Equation ([e17.82]).
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