14.5: Determination of Phase-Shifts

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]).

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

  \( \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}\) \(\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}\) \(\newcommand {\btau}{\mbox{\boldmath$\tau$}}\) \(\newcommand {\bmu}{\mbox{\boldmath$\mu$}}\) \(\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}\) \(\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}\) \(\newcommand {\bomega}{\mbox{\boldmath$\omega$}}\) \(\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}\)