Born Approximation

Equation (10.28) 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 $\psi({\bf x})\equiv \langle {\bf x}\vert\psi\rangle$ is the solution of the integral equation

$$\psi({\bf x}) = \phi({\bf x})-\frac{m}{2\pi\,\hbar^{\,2}} \int d^... ...- {\bf x}'\vert)}{\vert{\bf x} - {\bf x}'\vert}\, V({\bf x}')\, \psi({\bf x}'),$$ (10.29)

where $\phi({\bf x})$ is the wavefunction of the incident state. [See Equation (10.13).] According to the previous equation, the total wavefunction is a superposition of the incident wavefunction and a great many spherical waves emitted from the scattering region. The strength of the spherical wave emitted at a given point in the scattering region is proportional to the local value of the scattering potential, $V({\bf x})$ , as well as the local value of the wavefunction, $\psi({\bf x})$ .

Suppose, however, that the scattering is not particularly intense. In this case, it is reasonable to suppose that the total wavefunction, $\psi({\bf x})$ , does not differ substantially from the incident wavefunction, $\phi({\bf x})$ . Thus, we can obtain an expression for $\psi({\bf x})$ by making the substitution

$$\psi({\bf x}) \rightarrow \phi({\bf x}) = \frac{\exp(\,{\rm i}\,{\bf k} \cdot {\bf x} ) }{(2\pi)^{3/2}}$$ (10.30)

on the right-hand side of Equation (10.29). [See Equation (10.14).] This simplification is known as the Born approximation [11].

The Born approximation yields

$$f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi\, \hbar^{\,2}} \int d... ...\,\exp\left[\, {\rm i}\, ({\bf k} - {\bf k}')\cdot {\bf x}'\right] V({\bf x}').$$ (10.31)

[See Equation (10.21).] Thus, $f({\bf k}', {\bf k})$ is proportional to the Fourier transform of the scattering potential, $V({\bf x})$ , with respect to the relative wavevector, ${\bf q} \equiv {\bf k} - {\bf k}'$ .

For a spherically symmetric scattering potential,

$$f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi\, \hbar^{\,2}} \int_0... ...hi'\,r'^{\,2}\,\sin\theta' \,{\rm e}^{\,{\rm i} \, q \,r'\cos\theta'} \, V(r'),$$ (10.32)

giving

$$f({\bf k}', {\bf k}) \simeq - \frac{2\,m}{\hbar^{\,2}\,q} \int_0^\infty dr\,r \,V(r) \sin(q \,r).$$ (10.33)

Hence, it is clear that, for the special case of a spherically symmetric potential, $f({\bf k}', {\bf k})$ depends only on the magnitude of the relative wavevector, ${\bf q}={\bf k}-{\bf k}'$ , and is independent of its direction. Now, it is easily demonstrated that

$$q \equiv \vert{\bf k} - {\bf k}'\vert = 2\, k \,\sin (\theta/2),$$ (10.34)

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, as a consequence of energy conservation. It follows that, according to the Born approximation, $f({\bf k}',{\bf k})=f(\theta)$ for a spherically symmetric scattering potential, $V(r)$ . Moreover, $f(\theta)$ is real. Finally, the differential scattering cross-section, $d\sigma/d{\mit\Omega}=\vert f(\theta)\vert^{\,2}$ , is invariant under the transformation $V\rightarrow -V$ . In other words, the pattern of scattering is identical for attractive and repulsive scattering potentials of the same strength.

Consider scattering by a Yukawa potential [118],

$$V(r) = \frac{V_0\,\exp(-\mu \,r)}{\mu \,r},$$ (10.35)

where $V_0$ is a constant, and $1/\mu$ measures the ``range'' of the potential. It follows from Equation (10.33) that

$$f(\theta) = - \frac{2\,m \,V_0}{\hbar^{\,2}\,\mu} \frac{1}{q^{\,2} + \mu^{\,2}},$$ (10.36)

because

$$\int_0^\infty dr\, \exp(-\mu \,r) \,\sin(q\,r) = \frac{q}{q^{\,2}+\mu^{\,2}}.$$ (10.37)

(See Exercise 6.) Thus, the Born approximation yields a differential cross-section for scattering by a Yukawa potential of the form

$$\frac{d\sigma}{d {\mit\Omega}} \simeq \left(\frac{2\,m \,V_0}{ \h... ...right)^2 \frac{1}{\left[4\,k^{\,2}\,\sin^2(\theta/2) + \mu^{\,2}\right]^{\,2}}.$$ (10.38)

[See Equations (10.28) and (10.34).]

The Yukawa potential reduces to the familiar Coulomb potential in the limit $\mu \rightarrow 0$ , provided that $V_0/\mu \rightarrow Z\,Z'\, e^{\,2} / 4\pi\,\epsilon_0$ . Here, $Z\,e$ and $Z'\,e$ are the electric charges of the two interacting particles. In the Coulomb limit, the previous Born differential cross-section transforms into

$$\frac{d\sigma}{d{\mit\Omega}} \simeq \left(\frac{2\,m \,Z\, Z'\, ... ...,\epsilon_0\,\hbar^{\,2}}\right)^2 \frac{1}{ 16 \,k^{\,4}\, \sin^4( \theta/2)}.$$ (10.39)

Recalling that $\hbar\, k$ is equivalent to $\vert{\bf p}\vert$ , where ${\bf p}$ is the momentum of the incident particles, the preceding equation can be rewritten

$$\frac{d\sigma}{d{\mit\Omega}} \simeq\left(\frac{Z \,Z'\, e^{\,2}}{16\pi\,\epsilon_0\,E}\right)^2 \frac{1}{\sin^4(\theta/2)},$$ (10.40)

where $E= p^{\,2}/(2\,m)$ is the kinetic energy of the incident particles. Equation (10.40) is identical to the well-known Rutherford scattering cross-section formula of classical physics [67].

The Born approximation is valid provided $\psi({\bf x})$ is not significantly different from $\phi({\bf x})$ in the scattering region. It follows, from Equation (10.29), that the condition that must be satisfied in order that $\psi({\bf x}) \simeq \phi({\bf x})$ in the vicinity of ${\bf x} = {\bf0}$ is

$$\left\vert \frac{m}{2\pi\, \hbar^{\,2}} \int d^{\,3} {\bf x}'\,\frac{ \exp(\,{\rm i}\, k \,r')}{r'} \,V({\bf x}') \right\vert \ll 1.$$ (10.41)

Consider the special case of the Yukawa potential, (10.35). At low energies (i.e., $k\ll \mu$ ), we can replace $\exp(\,{\rm i}\,k\, r')$ by unity, giving

$$\frac{2\,m}{\hbar^{\,2}} \frac{\vert V_0\vert}{\mu^{\,2}} \ll 1$$ (10.42)

as the condition for the validity of the Born approximation. Now, the criterion for the Yukawa potential to develop a bound state turns out to be

$$\frac{2\,m}{\hbar^{\,2}} \frac{\vert V_0\vert} {\mu^{\,2}} \geq 2.7,$$ (10.43)

provided $V_0$ is negative [95]. 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 (i.e., $k\gg \mu$ ), Equation (10.41) yields

$$\frac{2\,m}{\hbar^{\,2}} \frac{\vert V_0\vert}{\mu \,k} \ll 1.$$ (10.44)

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