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Next: Determination of Phase-Shifts Up: Scattering Theory Previous: Born Approximation

Partial Waves

We can assume, without loss of generality, that the incident wavefunction is characterized by a wavevector ${\bf k}$ which is aligned parallel to the $z$-axis. The scattered wavefunction is characterized by a wavevector ${\bf k}'$ which has the same magnitude as ${\bf k}$, but, in general, points in a different direction. The direction of ${\bf k}'$ is specified by the polar angle $\theta $ (i.e., the angle subtended between the two wavevectors), and an azimuthal angle $\phi$ about the $z$-axis. Equations (1269) and (1270) strongly suggest that for a spherically symmetric scattering potential [i.e., $V({\bf r}) = V(r)$] the scattering amplitude is a function of $\theta $ only: i.e.,
f(\theta, \phi) = f(\theta).
\end{displaymath} (1282)

It follows that neither the incident wavefunction,
\psi_0({\bf r}) = \sqrt{n} \exp( {\rm i} k z)= \sqrt{n} \exp( {\rm i} k r\cos\theta),
\end{displaymath} (1283)

nor the large-$r$ form of the total wavefunction,
\psi({\bf r}) = \sqrt{n}
\left[ \exp( {\rm i} k r\cos\theta) + \frac{\exp( {\rm i} k r)  f(\theta)}
{r} \right],
\end{displaymath} (1284)

depend on the azimuthal angle $\phi$.

Outside the range of the scattering potential, both $\psi_0({\bf r})$ and $\psi({\bf r})$ satisfy the free space Schrödinger equation

(\nabla^2 + k^2) \psi = 0.
\end{displaymath} (1285)

What is the most general solution to this equation in spherical polar coordinates which does not depend on the azimuthal angle $\phi$? Separation of variables yields
\psi(r,\theta) = \sum_l R_l(r)  P_l(\cos\theta),
\end{displaymath} (1286)

since the Legendre functions $P_l(\cos\theta)$ form a complete set in $\theta $-space. The Legendre functions are related to the spherical harmonics, introduced in Cha. 8, via
P_l(\cos\theta) = \sqrt{\frac{4\pi}{2 l+1}}  Y_{l,0}(\theta,\varphi).
\end{displaymath} (1287)

Equations (1285) and (1286) can be combined to give
r^2 \frac{d^2 R_l}{dr^2} + 2 r  \frac{dR_l}{dr} + [k^2  r^2 -
l (l+1)]R_l = 0.
\end{displaymath} (1288)

The two independent solutions to this equation are the spherical Bessel functions, $j_l(k r)$ and $y_l(k r)$, introduced in Sect. 9.3. Recall that
$\displaystyle j_l(z)$ $\textstyle =$ $\displaystyle z^l\left(-\frac{1}{z}\frac{d}{dz}\right)^l\left(\frac{\sin z}{z}\right),$ (1289)
$\displaystyle y_l(z)$ $\textstyle =$ $\displaystyle -z^l\left(-\frac{1}{z}\frac{d}{dz}\right)^l\left(\frac{\cos z}{z}\right).$ (1290)

Note that the $j_l(z)$ are well-behaved in the limit $z\rightarrow 0$ , whereas the $y_l(z)$ become singular. The asymptotic behaviour of these functions in the limit $z\rightarrow\infty$ is
$\displaystyle j_l(z)$ $\textstyle \rightarrow$ $\displaystyle \frac{\sin(z - l \pi/2)}{z},$ (1291)
$\displaystyle y_l(z)$ $\textstyle \rightarrow$ $\displaystyle - \frac{\cos(z-l \pi/2)}{z}.$ (1292)

We can write

\exp( {\rm i} k r \cos\theta) = \sum_l a_l  j_l(k r)  P_l(\cos\theta),
\end{displaymath} (1293)

where the $a_l$ are constants. Note there are no $y_l(k r)$ functions in this expression, because they are not well-behaved as $r\rightarrow 0$. The Legendre functions are orthonormal,
\int_{-1}^1 P_n(\mu)  P_m(\mu) d\mu = \frac{\delta_{nm}}{n+1/2},
\end{displaymath} (1294)

so we can invert the above expansion to give
a_l  j_l(k r) = (l+1/2)\int_{-1}^1 \exp( {\rm i} k r  \mu)  P_l(\mu)  d\mu.
\end{displaymath} (1295)

It is well-known that
j_l(y) = \frac{(-{\rm i})^l}{2} \int_{-1}^1 \exp( {\rm i}  y \mu)
 P_l(\mu) d\mu,
\end{displaymath} (1296)

where $l=0, 1, 2, \cdots$ [see M. Abramowitz and I.A. Stegun, Handbook of mathematical functions, (Dover, New York NY, 1965), Eq. 10.1.14]. Thus,
a_l = {\rm i}^{ l}  (2 l+1),
\end{displaymath} (1297)

\psi_0({\bf r}) = \sqrt{n} \exp( {\rm i} k r \cos\theta)...
... \sum_l {\rm i}^{ l} (2 l+1)  j_l(k r)  P_l(\cos\theta).
\end{displaymath} (1298)

The above expression tells us how to decompose the incident plane-wave into a series of spherical waves. These waves are usually termed ``partial waves''.

The most general expression for the total wavefunction outside the scattering region is

\psi({\bf r}) = \sqrt{n}\sum_l\left[
A_l j_l(k r) + B_l y_l(k r)\right] P_l(\cos\theta),
\end{displaymath} (1299)

where the $A_l$ and $B_l$ are constants. Note that the $y_l(k r)$ functions are allowed to appear in this expansion, because its region of validity does not include the origin. In the large-$r$ limit, the total wavefunction reduces to
\psi ({\bf r} ) \simeq \sqrt{n} \sum_l\left[A_l 
..._l \frac{\cos(k r -l \pi/2)}{k r}
\right] P_l(\cos\theta),
\end{displaymath} (1300)

where use has been made of Eqs. (1291) and (1292). The above expression can also be written
\psi ({\bf r} ) \simeq \sqrt{n} \sum_l C_l 
\frac{\sin(k r - l \pi/2+ \delta_l)}{k r}  P_l(\cos\theta),
\end{displaymath} (1301)

where the sine and cosine functions have been combined to give a sine function which is phase-shifted by $\delta_l$. Note that $A_l=C_l \cos\delta_l$ and $B_l=-C_l \sin\delta_l$.

Equation (1301) yields

\psi({\bf r}) \simeq \sqrt{n} \sum_l C_l\left[
\frac{{\rm e}...
...\pi/2+ \delta_l)} }{2 {\rm i} k r} \right] P_l(\cos\theta),
\end{displaymath} (1302)

which contains both incoming and outgoing spherical waves. What is the source of the incoming waves? Obviously, they must be part of the large-$r$ asymptotic expansion of the incident wavefunction. In fact, it is easily seen from Eqs. (1291) and (1298) that
\psi_0({\bf r}) \simeq \sqrt{n} \sum_l {\rm i}^{ l} 
... (k r - l \pi/2)}}{2 {\rm i} k r} \right]P_l(\cos\theta)
\end{displaymath} (1303)

in the large-$r$ limit. Now, Eqs. (1283) and (1284) give
\frac{\psi({\bf r} )- \psi_0({\bf r}) }{ \sqrt{n}} =
\frac{\exp( {\rm i} k r)}{r} 
\end{displaymath} (1304)

Note that the right-hand side consists of an outgoing spherical wave only. This implies that the coefficients of the incoming spherical waves in the large-$r$ expansions of $\psi({\bf r})$ and $\psi_0({\bf r})$ must be the same. It follows from Eqs. (1302) and (1303) that
C_l = (2 l+1) \exp[ {\rm i} (\delta_l + l \pi/2)].
\end{displaymath} (1305)

Thus, Eqs. (1302)-(1304) yield
f(\theta) = \sum_{l=0}^\infty (2 l+1) \frac{\exp( {\rm i} \delta_l)}
{k}  \sin\delta_l P_l(\cos\theta).
\end{displaymath} (1306)

Clearly, determining the scattering amplitude $f(\theta)$ via a decomposition into partial waves (i.e., spherical waves) is equivalent to determining the phase-shifts $\delta_l$.

Now, the differential scattering cross-section $d\sigma/d\Omega$ is simply the modulus squared of the scattering amplitude $f(\theta)$ [see Eq. (1266)]. The total cross-section is thus given by

$\displaystyle \sigma_{\rm total}$ $\textstyle =$ $\displaystyle \int \vert f(\theta)\vert^2 d\Omega$  
  $\textstyle =$ $\displaystyle \frac{1}{k^2} \oint d\phi \int_{-1}^{1} d\mu
\sum_l \sum_{l'} (2 l+1) (2 l'+1)
\exp[ {\rm i} (\delta_l-\delta_{l'})]$  
    $\displaystyle \mbox{\hspace{1cm}}\times \sin\delta_l  \sin\delta_{l'} 
P_l(\mu)  P_{l'}(\mu),$ (1307)

where $\mu = \cos\theta$. It follows that
\sigma_{\rm total} = \frac{4\pi}{k^2} \sum_l (2 l+1) \sin^2\delta_l,
\end{displaymath} (1308)

where use has been made of Eq. (1294).

next up previous
Next: Determination of Phase-Shifts Up: Scattering Theory Previous: Born Approximation
Richard Fitzpatrick 2010-07-20