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The Born approximation

Equation (923) is not particularly useful, as it stands, because the quantity $f({\bf k}', {\bf k})$ depends on the unknown ket $\vert\psi\rangle$. Recall that $\psi({\bf r})=\langle {\bf r}\vert\psi\rangle$ is the solution of the integral equation
\psi({\bf r}) = \phi({\bf r})-\frac{m}{2\pi \hbar^2}
...cdot \!{\bf r}') 
V({\bf r}')  \psi ({\bf r}') d^3{\bf r}',
\end{displaymath} (924)

where $\phi({\bf r})$ is the wave-function of the incident state. According to the above equation the total wave-function is a superposition of the incident wave-function and lots of spherical-waves emitted from the scattering region. The strength of the spherical-wave emitted at a given point is proportional to the local value of the scattering potential, $V$, as well as the local value of the wave-function, $\psi$.

Suppose 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, $\phi({\bf r})$. Thus, we can obtain an expression for $f({\bf k}', {\bf k})$ by making the substitution

\psi({\bf r}) \rightarrow \phi({\bf r}) =
\frac{\exp( {\rm i} {\bf k}\!\cdot \! {\bf r} ) }{(2\pi)^{3/2}}.
\end{displaymath} (925)

This is called the Born approximation.

The Born approximation yields

f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi  \hbar^2} \int ...
... {\bf k}')\!\cdot \!{\bf r}'\right]
V({\bf r}') d^3{\bf r}'.
\end{displaymath} (926)

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 ${\bf q} \equiv {\bf k} - {\bf k}'$.

For a spherically symmetric potential,

f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi  \hbar^2} \int\...
...s\theta')   V(r') r'^2 dr' \sin\theta'
 d\theta' d\phi',
\end{displaymath} (927)

f({\bf k}', {\bf k}) \simeq - \frac{2 m}{\hbar^2 q}
\int_0^\infty r'  V(r') \sin(q  r')  dr'.
\end{displaymath} (928)

Note that $f({\bf k}', {\bf k})$ is just a function of $q$ for a spherically symmetric potential. It is easily demonstrated that
q \equiv \vert{\bf k} - {\bf k}'\vert = 2  k  \sin (\theta/2),
\end{displaymath} (929)

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

Consider scattering by a Yukawa potential

V(r) = \frac{V_0 \exp(-\mu  r)}{\mu  r},
\end{displaymath} (930)

where $V_0$ is a constant and $1/\mu$ measures the ``range'' of the potential. It follows from Eq. (928) that
f(\theta) = - \frac{2 m  V_0}{\hbar^2 \mu} \frac{1}{q^2 + \mu^2},
\end{displaymath} (931)

\int_0^\infty \exp(-\mu  r')  \sin(q r')   dr' = \frac{q}{\mu^2 + q^2}.
\end{displaymath} (932)

Thus, in the Born approximation, the differential cross-section for scattering by a Yukawa potential is
\frac{d\sigma}{d \Omega} \simeq \left(\frac{2 m  V_0}{ \hb...
... \mu}\right)^2
\frac{1}{[2 k^2  (1-\cos\theta) + \mu^2]^2},
\end{displaymath} (933)

given that
q^2 = 4 k^2  \sin^2(\theta/2) = 2 k^2  (1-\cos\theta).
\end{displaymath} (934)

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

\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} (935)

Recall that $\hbar  k$ is equivalent to $\vert{\bf p}\vert$, so the above equation can be rewritten
\frac{d\sigma}{d\Omega} \simeq\left(\frac{Z  Z'  e^2}{16\pi\epsilon_0 E}\right)^2
\end{displaymath} (936)

where $E= p^2/2 m$ is the kinetic energy of the incident particles. Equation (936) is the classical Rutherford scattering cross-section formula.

The Born approximation is valid provided that $\psi({\bf r})$ is not too different from $\phi({\bf r})$ in the scattering region. It follows, from Eq. (907), that the condition for $\psi({\bf r})
\simeq \phi({\bf r})$ in the vicinity of ${\bf r} = 0$ is

\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} (937)

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
\frac{2 m}{\hbar^2} \frac{\vert V_0\vert}{\mu^2} \ll 1
\end{displaymath} (938)

as the condition for the validity of the Born approximation. The condition for the Yukawa potential to develop a bound state is
\frac{2 m}{\hbar^2} \frac{\vert V_0\vert} {\mu^2} \geq 2.7,
\end{displaymath} (939)

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. (937) yields
\frac{2 m}{\hbar^2} \frac{\vert V_0\vert}{\mu  k} \ll 1.
\end{displaymath} (940)

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
Next: Partial waves Up: Scattering theory Previous: The Lipmann-Schwinger equation
Richard Fitzpatrick 2006-02-16