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9.6: Determination of Phase-Shifts

Let us now consider how the phase-shifts $ \delta_l$ 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 x})$ satisfies the free-space Schrödinger equation (958). The most general solution that is consistent with no incoming spherical waves is


$\displaystyle \psi({\bf x}) = \frac{1}{(2\pi)^{3/2}} \sum_{l=0,\infty} {\rm i}^l\, (2\,l+1) \, A_l(r)\, P_l(\cos\theta),$ (985)





$\displaystyle A_l(r) = \exp(\,{\rm i} \,\delta_l)\, \left[ \cos\delta_l \,j_l(k\,r) -\sin\delta_l\, \eta_l(k\,r)\right].$ (986)



Note that Neumann functions are allowed to appear in the above expression, because its region of validity does not include the origin (where $ V\neq 0$ ). The logarithmic derivative of the $ l$ th radial wavefunction $ A_l(r)$ just outside the range of the potential is given by


$\displaystyle \beta_{l+} = k\,a \left[\frac{ \cos\delta_l\,j_l'(k\,a) - \sin\de...
... \eta_l'(k\,a)}{\cos\delta_l \, j_l(k\,a) - \sin\delta_l\,\eta_l(k\,a)}\right],$ (987)



where $ j_l'(x)$ denotes $ dj_l(x)/dx$ , etc. The above equation can be inverted to give


$\displaystyle \tan \delta_l = \frac{ k\,a\,j_l'(k\,a) - \beta_{l+}\, j_l(k\,a)} {k\,a\,\eta_l'(k\,a) - \beta_{l+}\, \eta_l(k\,a)}.$ (988)



Thus, the problem of determining the phase-shift $ \delta_l$ is equivalent to that of determining $ \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 $ \varphi$ is


$\displaystyle \psi({\bf x}) = \frac{1}{(2\pi)^{3/2}}\sum_{l=0,\infty} {\rm i}^l \,(2\,l+1)\,R_l(r)\,P_l(\cos\theta),$ (989)





$\displaystyle R_l (r) = \frac{u_l(r)}{r},$ (990)





$\displaystyle \frac{d^2 u_l}{d r^2} +\left[k^2 - \frac{2m}{\hbar^2} \,V - \frac{l\,(l+1)} {r^2}\right] u_l = 0.$ (991)



The boundary condition


$\displaystyle u_l(0) = 0$ (992)



ensures that the radial wavefunction is well-behaved at the origin. We can launch a well-behaved solution of the above equation from $ r=0$ , integrate out to $ r=a$ , and form the logarithmic derivative


$\displaystyle \beta_{l-} = \left.\frac{1}{(u_l/r)} \frac{d(u_l/r)}{dr}\right\vert _{r=a}.$ (993)



Because $ \psi({\bf x})$ and its first derivatives are necessarily continuous for physically acceptible wavefunctions, it follows that


$\displaystyle \beta_{l+} = \beta_{l-}.$ (994)


The phase-shift $ \delta_l$ is obtainable from Equation (988).