Fundamentals

Consider time-independent, energy conserving scattering in which the Hamiltonian of the system is written

$$H = H_0 + V({\bf r}),$$ (1229)

where

$$H_0 = \frac{p^2}{2\,m} \equiv - \frac{\hbar^2}{2\,m}\,\nabla^2$$ (1230)

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

$$\psi_0({\bf r}) = \sqrt{n}\,{\rm e}^{\,{\rm i}\,{\bf k}\cdot {\bf r}}$$ (1231)

represent an incident beam of particles, of number density $n$, and velocity ${\bf v} = \hbar\,{\bf k}/m$. Of course,

$$H_0\,\psi_0= E\,\psi_0,$$ (1232)

where $E = \hbar^2\,k^2/2\,m$ is the particle energy. Schrödinger's equation for the scattering problem is

$$(H_0+V)\,\psi = E\,\psi,$$ (1233)

subject to the boundary condition $\psi\rightarrow\psi_0$ as $V\rightarrow 0$.

The above equation can be rearranged to give

$$(\nabla^2+k^2)\,\psi = \frac{2\,m}{\hbar^2}\,V\,\psi.$$ (1234)

Now,

$$(\nabla^2+k^2)\,u({\bf r}) = \rho({\bf r})$$ (1235)

is known as the Helmholtz equation. The solution to this equation is well-known:⁴

$$u({\bf r}) = u_0({\bf r}) - \int \frac{{\rm e}^{\,{\rm i}\,k... ...\pi\,\vert{\bf r}-{\bf r}'\vert}\,\rho({\bf r}')\,d^3{\bf r}'.$$ (1236)

Here, $u_0({\bf r})$ is any solution of $(\nabla^2+k^2)\,u_0 = 0$. Hence, Eq. (1234) can be inverted, subject to the boundary condition $\psi\rightarrow\psi_0$ as $V\rightarrow 0$, to give

$$\psi({\bf r}) = \psi_0({\bf r})- \frac{2\,m}{\hbar^2} \int\f... ...f r}-{\bf r}'\vert}\,V({\bf r}')\,\psi({\bf r}')\,d^3{\bf r}'.$$ (1237)

Let us calculate the value of the wave-function $\psi({\bf r})$ well outside the scattering region. Now, if $r\gg r'$ then

$$\vert{\bf r}-{\bf r}'\vert \simeq r - \hat{\bf r}\cdot {\bf r}'$$ (1238)

to first-order in $r'/r$, where $\hat{\bf r}/r$ is a unit vector which points from the scattering region to the observation point. It is helpful to define ${\bf k}'=k\,\hat{\bf r}$. This is the wave-vector for particles with the same energy as the incoming particles (i.e., $k'=k$) which propagate from the scattering region to the observation point. Equation (1237) reduces to

$$\psi({\bf r}) \simeq \sqrt{n}\left[{\rm e}^{\,{\rm i}\,{\bf ... ... + \frac{e^{\,{\rm i}\,k\,r}}{r}\,f({\bf k}, {\bf k}')\right],$$ (1239)

where

$$f({\bf k},{\bf k}') = -\frac{m}{2\pi\,\sqrt{n}\,\hbar^2}\int... ...f k}'\cdot{\bf r}'}\,V({\bf r}')\,\psi({\bf r}')\,d^3{\bf r}'.$$ (1240)

The first term on the right-hand side of Eq. (1239) represents the incident particle beam, whereas the second term represents an outgoing spherical wave of scattered particles.

The differential scattering cross-section $d\sigma/d\Omega$ is defined as the number of particles per unit time scattered into an element of solid angle $d\Omega$, divided by the incident particle flux. From Sect. 7.2, the probability flux (i.e., the particle flux) associated with a wave-function $\psi$ is

$${\bf j} = \frac{\hbar}{m}\,{\rm Im}(\psi^\ast\,\nabla\psi).$$ (1241)

Thus, the particle flux associated with the incident wave-function $\psi_0$ is

$${\bf j} = n\,{\bf v},$$ (1242)

where ${\bf v} = \hbar\,{\bf k}/m$ is the velocity of the incident particles. Likewise, the particle flux associated with the scattered wave-function $\psi-\psi_0$ is

$${\bf j}' = n\,\frac{\vert f({\bf k},{\bf k}')\vert^2}{r^2}\,{\bf v}',$$ (1243)

where ${\bf v}' = \hbar\,{\bf k}'/m$ is the velocity of the scattered particles. Now,

$$\frac{d\sigma}{d\Omega}\,d\Omega = \frac{r^2\,d\Omega\,\vert{\bf j}'\vert}{\vert{\bf j}\vert},$$ (1244)

which yields

$$\frac{d\sigma}{d\Omega} = \vert f({\bf k},{\bf k}')\vert^2.$$ (1245)

Thus, $\vert f({\bf k},{\bf k}')\vert^2$ 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 $d\Omega$ about ${\bf v}' = \hbar\,{\bf k}'/m$. Note that the scattering conserves energy, so that $\vert{\bf v}'\vert=\vert{\bf v}\vert$ and $\vert{\bf k}'\vert=\vert{\bf k}\vert$.