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# 14.1: Fundamentals of Scattering Theory

Consider time-independent, energy conserving scattering in which the Hamiltonian of the system is written $H = H_0 + V({\bf r}),$ where $H_0 = \frac{p^{\,2}}{2\,m} \equiv - \frac{\hbar^{\,2}}{2\,m}\,\nabla^{\,2}$ is the Hamiltonian of a free particle of mass $$m$$, and $$V({\bf r})$$ the scattering potential. This potential is assumed to only be non-zero in a fairly localized region close to the origin. Let $\psi_0({\bf r}) = \sqrt{n}\,{\rm e}^{\,{\rm i}\,{\bf k}\cdot {\bf r}}$ represent an incident beam of particles, of number density $$n$$, and velocity $${\bf v} = \hbar\,{\bf k}/m$$. [See Equation ([e14.14gg]).] Of course, $H_0\,\psi_0= E\,\psi_0,$ where $$E = \hbar^{\,2}\,k^{\,2}/(2\,m)$$ is the particle energy. Schrödinger’s equation for the scattering problem is $(H_0+V)\,\psi = E\,\psi,$ subject to the boundary condition $$\psi\rightarrow\psi_0$$ as $$V\rightarrow 0$$.

The previous equation can be rearranged to give $\label{e15.6} (\nabla^{\,2}+k^{\,2})\,\psi = \frac{2\,m}{\hbar^{\,2}}\,V\,\psi.$ Now, $(\nabla^{\,2}+k^{\,2})\,u({\bf r}) = \rho({\bf r})$ is known as the Helmholtz equation. The solution to this equation is well known $u({\bf r}) = u_0({\bf r}) - \int \frac{{\rm e}^{\,{\rm i}\,k\,|{\bf r}-{\bf r}'|}} {4\pi\,|{\bf r}-{\bf r}'|}\,\rho({\bf r}')\,d^{\,3}{\bf r}'.$ Here, $$u_0({\bf r})$$ is any solution of $$(\nabla^{\,2}+k^{\,2})\,u_0 = 0$$. Hence, Equation ([e15.6]) can be inverted, subject to the boundary condition $$\psi\rightarrow\psi_0$$ as $$V\rightarrow 0$$, to give

[\label{e15.9} \psi({\bf r}) = \psi_0({\bf r})- \frac{2\,m}{\hbar^{\,2}} \int\frac{{\rm e}^{\,{\rm i}\,k\,|{\bf r}-{\bf r}'|}} {4\pi\,|{\bf r}-{\bf r}'|}\,V({\bf r}')\,\psi({\bf r}')\,d^{\,3}{\bf r}'.\]

Let us calculate the value of the wavefunction $$\psi({\bf r})$$ well outside the scattering region. Now, if $$r\gg r'$$ then $|{\bf r}-{\bf r}'| \simeq r - \hat{\bf r}\cdot {\bf r}'$ to first-order in $$r'/r$$, where $$\hat{\bf r}/r$$ is a unit vector that points from the scattering region to the observation point. It is helpful to define $${\bf k}'=k\,\hat{\bf r}$$. This is the wavevector for particles with the same energy as the incoming particles (i.e., $$k'=k$$) that propagate from the scattering region to the observation point. Equation ([e15.9]) reduces to

$\label{e15.11} \psi({\bf r}) \simeq \sqrt{n}\left[{\rm e}^{\,{\rm i}\,{\bf k}\cdot{\bf r}} + \frac{e^{\,{\rm i}\,k\,r}}{r}\,f({\bf k}, {\bf k}')\right],$ where

$\label{e5.12} f({\bf k},{\bf k}') = -\frac{m}{2\pi \sqrt{n}\,\hbar^{\,2}}\int {\rm e}^{-{\rm i}\,{\bf k}'\cdot{\bf r}'}\,V({\bf r}')\,\psi({\bf r}')\,d^{\,3}{\bf r}'.$ The first term on the right-hand side of Equation ([e15.11]) represents the incident particle beam, whereas the second term represents an outgoing spherical wave of scattered particles.

The differential scattering cross-section, $$d\sigma/d{\mit\Omega}$$, is defined as the number of particles per unit time scattered into an element of solid angle $$d{\mit\Omega}$$, divided by the incident particle flux. From Section [s7.2], the probability flux (i.e., the particle flux) associated with a wavefunction $$\psi$$ is ${\bf j} = \frac{\hbar}{m}\,{\rm Im}(\psi^\ast\,\nabla\psi).$ Thus, the particle flux associated with the incident wavefunction $$\psi_0$$ is

$\label{e14.14gg} {\bf j} = n\,{\bf v},$ where $${\bf v}=\hbar\,{\bf k}/m$$ is the velocity of the incident particles. Likewise, the particle flux associated with the scattered wavefunction $$\psi-\psi_0$$ is ${\bf j}' = n\,\frac{|f({\bf k},{\bf k}')|^{\,2}}{r^{\,2}}\,{\bf v}',$ where $${\bf v}' = \hbar\,{\bf k}'/m$$ is the velocity of the scattered particles. Now, $\frac{d\sigma}{d{\mit\Omega}}\,d{\mit\Omega} = \frac{r^{\,2}\,d{\mit\Omega}\,|{\bf j}'|}{|{\bf j}|},$ which yields

$\label{e15.17} \frac{d\sigma}{d{\mit\Omega}} = |f({\bf k},{\bf k}')|^{\,2}.$ Thus, $$|f({\bf k},{\bf k}')|^{\,2}$$ gives the differential cross-section for particles with incident velocity $${\bf v}=\hbar\,{\bf k}/m$$ to be scattered such that their final velocities are directed into a range of solid angles $$d{\mit\Omega}$$ about $${\bf v}'=\hbar\,{\bf k}'/m$$. Note that the scattering conserves energy, so that $$|{\bf v}'|=|{\bf v}|$$ and $$|{\bf k}'|=|{\bf k}|$$.

# Contributors

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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