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

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

where

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

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

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

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

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

where

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

and

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

The boundary condition

$$u_l(0) = 0$$ (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

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

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

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

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