Determination of phase-shifts

Let us now consider how the phase-shifts $\delta_l$ in Eq. (1285) can be evaluated. Consider a spherically symmetric potential $V(r)$ which vanishes for $r>a$, where $a$ is termed the range of the potential. In the region $r>a$, the wave-function $\psi({\bf r})$ satisfies the free-space Schrödinger equation (1264). The most general solution which is consistent with no incoming spherical-waves is

$$\psi({\bf r}) = \sqrt{n}\, \sum_{l=0}^\infty {\rm i}^l\, (2\,l+1) \, {\cal R}_l(r)\, P_l(\cos\theta),$$ (1288)

where

$${\cal R}_l(r) = \exp(\,{\rm i} \,\delta_l)\, \left[\cos\delta_l \,j_l(k\,r) -\sin\delta_l\, y_l(k\,r)\right].$$ (1289)

Note that $y_l(k\,r)$ 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 wave-function, ${\cal R}_l(r)$, just outside the range of the potential is given by

$$\beta_{l+} = k\,a \left[\frac{ \cos\delta_l\,j_l'(k\,a) - \s... ...}{\cos\delta_l \, j_l(k\,a) - \sin\delta_l\,y_l(k\,a)}\right],$$ (1290)

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

$$\tan \delta_l = \frac{ k\,a\,j_l'(k\,a) - \beta_{l+}\, j_l(k\,a)} {k\,a\,y_l'(k\,a) - \beta_{l+}\, y_l(k\,a)}.$$ (1291)

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 ($r<a$) which does not depend on the azimuthal angle $\phi$ is

$$\psi({\bf r}) = \sqrt{n}\,\sum_{l=0}^\infty {\rm i}^{\,l} \,(2\,l+1)\,{\cal R}_l(r)\,P_l(\cos\theta),$$ (1292)

where

$${\cal R}_l (r) = \frac{u_l(r)}{r},$$ (1293)

and

$$\frac{d^2 u_l}{d r^2} +\left[k^2 -\frac{l\,(l+1)}{r^2} -\frac{2\,m}{\hbar^2} \,V\right] u_l = 0.$$ (1294)

The boundary condition

$$u_l(0) = 0$$ (1295)

ensures that the radial wave-function 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

$$\beta_{l-} = \left.\frac{1}{(u_l/r)} \frac{d(u_l/r)}{dr}\right\vert _{r=a}.$$ (1296)

Since $\psi({\bf r})$ and its first derivatives are necessarily continuous for physically acceptible wave-functions, it follows that

$$\beta_{l+} = \beta_{l-}.$$ (1297)

The phase-shift $\delta_l$ is then obtainable from Eq. (1291).