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The Born approximation

Equation (1245) is not particularly useful, as it stands, because the quantity $f({\bf k},{\bf k}')$ depends on the, as yet, unknown wave-function $\psi({\bf r})$ [see Eq. (1240)]. Suppose, however, that the scattering is not particularly strong. In this case, it is reasonable to suppose that the total wave-function, $\psi({\bf r})$, does not differ substantially from the incident wave-function, $\psi_0({\bf r})$. Thus, we can obtain an expression for $f({\bf k},{\bf k}')$ by making the substitution $\psi({\bf r})\rightarrow\psi_0({\bf r}) = \sqrt{n}\,\exp(\,{\rm i}\,
{\bf k}\cdot{\bf r})$ in Eq. (1240). This procedure is called the Born approximation.

The Born approximation yields

\begin{displaymath}
f({\bf k},{\bf k}') \simeq \frac{m}{2\pi\,\hbar^2}
\int {\rm...
...}\,({\bf k}-{\bf k'})\cdot{\bf r}'}\,V({\bf r'})\,d^3{\bf r}'.
\end{displaymath} (1246)

Thus, $f({\bf k},{\bf k}')$ is proportional to the Fourier transform of the scattering potential $V({\bf r})$ with respect to the wave-vector ${\bf q} = {\bf k}-{\bf k}'$.

For a spherically symmetric potential,

\begin{displaymath}
f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi\, \hbar^2} \int\...
...a') \, V(r')\,r'^{\,2}\, dr'\,\sin\theta'
\,d\theta'\,d\phi',
\end{displaymath} (1247)

giving
\begin{displaymath}
f({\bf k}', {\bf k}) \simeq - \frac{2\,m}{\hbar^2\,q}
\int_0^\infty r' \,V(r') \sin(q \,r') \,dr'.
\end{displaymath} (1248)

Note that $f({\bf k}', {\bf k})$ is just a function of $q$ for a spherically symmetric potential. It is easily demonstrated that
\begin{displaymath}
q \equiv \vert{\bf k} - {\bf k}'\vert = 2\, k \,\sin (\theta/2),
\end{displaymath} (1249)

where $\theta $ is the angle subtended between the vectors ${\bf k}$ and ${\bf k}'$. In other words, $\theta $ is the scattering angle. Recall that the vectors ${\bf k}$ and ${\bf k}'$ have the same length, via energy conservation.

Consider scattering by a Yukawa potential

\begin{displaymath}
V(r) = \frac{V_0\,\exp(-\mu \,r)}{\mu \,r},
\end{displaymath} (1250)

where $V_0$ is a constant, and $1/\mu$ measures the ``range'' of the potential. It follows from Eq. (1248) that
\begin{displaymath}
f(\theta) = - \frac{2\,m \,V_0}{\hbar^2\,\mu} \frac{1}{q^2 + \mu^2},
\end{displaymath} (1251)

since
\begin{displaymath}
\int_0^\infty \exp(-\mu \,r') \,\sin(q\,r') \, dr' = \frac{q}{q^2+ \mu^2}.
\end{displaymath} (1252)

Thus, in the Born approximation, the differential cross-section for scattering by a Yukawa potential is
\begin{displaymath}
\frac{d\sigma}{d \Omega} \simeq \left(\frac{2\,m \,V_0}{ \hb...
...u}\right)^2
\frac{1}{[2\,k^2\, (1-\cos\theta) + \mu^2]^{\,2}},
\end{displaymath} (1253)

given that
\begin{displaymath}
q^2 = 4\,k^2\, \sin^2(\theta/2) = 2\,k^2\, (1-\cos\theta).
\end{displaymath} (1254)

The Yukawa potential reduces to the familiar Coulomb potential as $\mu \rightarrow 0$, provided that $V_0/\mu \rightarrow
Z\,Z'\, e^2 / 4\pi\,\epsilon_0$. In this limit, the Born differential cross-section becomes

\begin{displaymath}
\frac{d\sigma}{d\Omega} \simeq \left(\frac{2\,m \,Z\, Z'\, e...
..._0\,\hbar^2}\right)^2
\frac{1}{ 16 \,k^4\, \sin^4( \theta/2)}.
\end{displaymath} (1255)

Recall that $\hbar\, k$ is equivalent to $\vert{\bf p}\vert$, so the above equation can be rewritten
\begin{displaymath}
\frac{d\sigma}{d\Omega} \simeq\left(\frac{Z \,Z'\, e^2}{16\pi\,\epsilon_0\,E}\right)^2
\frac{1}{\sin^4(\theta/2)},
\end{displaymath} (1256)

where $E= p^2/2\,m$ is the kinetic energy of the incident particles. Of course, Eq. (1256) is the famous Rutherford scattering cross-section formula.

The Born approximation is valid provided that $\psi({\bf r})$ is not too different from $\psi_0({\bf r})$ in the scattering region. It follows, from Eq. (1237), that the condition for $\psi({\bf r})
\simeq \psi_0({\bf r})$ in the vicinity of ${\bf r} = {\bf0}$ is

\begin{displaymath}
\left\vert \frac{m}{2\pi\, \hbar^2} \int \frac{ \exp(\,{\rm i}\, k \,r')}{r'}
\,V({\bf r}')\,d^3{\bf r'} \right\vert \ll 1.
\end{displaymath} (1257)

Consider the special case of the Yukawa potential. At low energies, (i.e., $k\ll \mu$) we can replace $\exp(\,{\rm i}\,k\, r')$ by unity, giving
\begin{displaymath}
\frac{2\,m}{\hbar^2} \frac{\vert V_0\vert}{\mu^2} \ll 1
\end{displaymath} (1258)

as the condition for the validity of the Born approximation. The condition for the Yukawa potential to develop a bound state is
\begin{displaymath}
\frac{2\,m}{\hbar^2} \frac{\vert V_0\vert} {\mu^2} \geq 2.7,
\end{displaymath} (1259)

where $V_0$ is negative. Thus, if the potential is strong enough to form a bound state then the Born approximation is likely to break down. In the high-$k$ limit, Eq. (1257) yields
\begin{displaymath}
\frac{2\,m}{\hbar^2} \frac{\vert V_0\vert}{\mu \,k} \ll 1.
\end{displaymath} (1260)

This inequality becomes progressively easier to satisfy as $k$ increases, implying that the Born approximation is more accurate at high incident particle energies.


next up previous contents
Next: Partial waves Up: Scattering theory Previous: Fundamentals   Contents
Richard Fitzpatrick 2006-12-12