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Determination of Phase-Shifts

Let us now consider how the phase-shifts $\delta_l$ in Eq. (1306) 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 wavefunction $\psi({\bf r})$ satisfies the free-space Schrödinger equation (1285). The most general solution which is consistent with no incoming spherical-waves is
\begin{displaymath}
\psi({\bf r}) = \sqrt{n}  \sum_{l=0}^\infty
{\rm i}^l  (2 l+1)   {\cal R}_l(r)  P_l(\cos\theta),
\end{displaymath} (1309)

where
\begin{displaymath}
{\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].
\end{displaymath} (1310)

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 wavefunction, ${\cal R}_l(r)$, just outside the range of the potential is given by
\begin{displaymath}
\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],
\end{displaymath} (1311)

where $j_l'(x)$ denotes $dj_l(x)/dx$, etc. The above equation can be inverted to give
\begin{displaymath}
\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)}.
\end{displaymath} (1312)

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

\begin{displaymath}
\psi({\bf r}) = \sqrt{n} \sum_{l=0}^\infty
{\rm i}^{ l}  (2 l+1) {\cal R}_l(r) P_l(\cos\theta),
\end{displaymath} (1313)

where
\begin{displaymath}
{\cal R}_l (r) = \frac{u_l(r)}{r},
\end{displaymath} (1314)

and
\begin{displaymath}
\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.
\end{displaymath} (1315)

The boundary condition
\begin{displaymath}
u_l(0) = 0
\end{displaymath} (1316)

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
\begin{displaymath}
\beta_{l-} = \left.\frac{1}{(u_l/r)} \frac{d(u_l/r)}{dr}\right\vert _{r=a}.
\end{displaymath} (1317)

Since $\psi({\bf r})$ and its first derivatives are necessarily continuous for physically acceptible wavefunctions, it follows that
\begin{displaymath}
\beta_{l+} = \beta_{l-}.
\end{displaymath} (1318)

The phase-shift $\delta_l$ is then obtainable from Eq. (1312).


next up previous
Next: Hard Sphere Scattering Up: Scattering Theory Previous: Partial Waves
Richard Fitzpatrick 2010-07-20