9.6: Determination of Phase-Shifts

Let us now consider how the phase-shifts $V(r)$ that vanishes for $a$ is termed the range of the potential. In the region $\psi({\bf x})$ satisfies the free-space Schrödinger equation $\ref{958}$. The most general solution that is consistent with no incoming spherical waves is

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

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

$\ref{987}$

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

$\delta_l$ is equivalent to that of determining $r<a$ ) that does not depend on the azimuthal angle $\psi({\bf x}) = \frac{1}{(2\pi)^{3/2}}\sum_{l=0,\infty} {\rm i}^l \,(2\,l+1)\,R_l(r)\,P_l(\cos\theta),$ $\ref{989}$

where

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

The boundary condition

$r=0$, integrate out to $\beta_{l-} = \left.\frac{1}{(u_l/r)} \frac{d(u_l/r)}{dr}\right\vert _{r=a}.$ $\ref{993}$

Because $\beta_{l+} = \beta_{l-}.$ $\ref{994}$

The phase-shift $\delta_l$ is obtainable from Equation $\ref{988}$.

Contributors

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

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