Fundamentals

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

$$H = H_0 + V({\bf r}),$$ (1250)

where

$$H_0 = \frac{p^2}{2 m} \equiv - \frac{\hbar^2}{2 m} \nabla^2$$ (1251)

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

$$\psi_0({\bf r}) = \sqrt{n} {\rm e}^{ {\rm i} {\bf k}\cdot {\bf r}}$$ (1252)

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

$$H_0 \psi_0= E \psi_0,$$ (1253)

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

$$(H_0+V) \psi = E \psi,$$ (1254)

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

The above equation can be rearranged to give

$$(\nabla^2+k^2) \psi = \frac{2 m}{\hbar^2} V \psi.$$ (1255)

Now,

$$(\nabla^2+k^2) u({\bf r}) = \rho({\bf r})$$ (1256)

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

$$u({\bf r}) = u_0({\bf r}) - \int \frac{{\rm e}^{ {\rm i} k... ...\pi \vert{\bf r}-{\bf r}'\vert} \rho({\bf r}') d^3{\bf r}'.$$ (1257)

Here, $u_0({\bf r})$ is any solution of $(\nabla^2+k^2) u_0 = 0$. Hence, Eq. (1255) can be inverted, subject to the boundary condition $\psi\rightarrow\psi_0$ as $V\rightarrow 0$, to give

$$\psi({\bf r}) = \psi_0({\bf r})- \frac{2 m}{\hbar^2} \int\f... ...f r}-{\bf r}'\vert} V({\bf r}') \psi({\bf r}') d^3{\bf r}'.$$ (1258)

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

$$\vert{\bf r}-{\bf r}'\vert \simeq r - \hat{\bf r}\cdot {\bf r}'$$ (1259)

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 wavevector 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 (1258) reduces to

$$\psi({\bf r}) \simeq \sqrt{n}\left[{\rm e}^{ {\rm i} {\bf ... ... + \frac{e^{ {\rm i} k r}}{r} f({\bf k}, {\bf k}')\right],$$ (1260)

where

$$f({\bf k},{\bf k}') = -\frac{m}{2\pi \sqrt{n} \hbar^2}\int... ...f k}'\cdot{\bf r}'} V({\bf r}') \psi({\bf r}') d^3{\bf r}'.$$ (1261)

The first term on the right-hand side of Eq. (1260) 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 wavefunction $\psi$ is

$${\bf j} = \frac{\hbar}{m} {\rm Im}(\psi^\ast \nabla\psi).$$ (1262)

Thus, the particle flux associated with the incident wavefunction $\psi_0$ is

$${\bf j} = n {\bf v},$$ (1263)

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

$${\bf j}' = n \frac{\vert f({\bf k},{\bf k}')\vert^2}{r^2} {\bf v}',$$ (1264)

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

$$\frac{d\sigma}{d\Omega} d\Omega = \frac{r^2 d\Omega \vert{\bf j}'\vert}{\vert{\bf j}\vert},$$ (1265)

which yields

$$\frac{d\sigma}{d\Omega} = \vert f({\bf k},{\bf k}')\vert^2.$$ (1266)

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 $\vert{\bf k}'\vert=\vert{\bf k}\vert$.