13.4: Collision between two unpolarised electron beams

Last updated
Save as PDF

Page ID: 28693

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

In this case, half the collisions will be between like-polarised electrons, so will involve interference, and half will be between unlike electrons: so there would be no interference. In both cases \(|f(\theta )|^2\) represents Rutherford scattering. The differential cross section of finding an electron scattered through an angle \(\theta\) is thus:

\[I = \frac{1}{2} (I_{dis} + I_{ind}) = \frac{1}{2} (|f(\theta )|^2 + |f(\pi − \theta )|^2 ) + \frac{1}{2} |f(\theta ) − f(\pi − \theta )|^2 \nonumber\]

Consider \(\theta = \pi /2\). The like polarised beams give zero probability, so unpolarised beams give only half what we would expect from Coulomb scattering of distinguishable particles. Furthermore, the spins of pairs of electrons scattered through \(\theta = \pi /2\) are always observed to be opposite.

An alternate philosophy is that we should treat the spins as a symmetric triplet and an antisymmetric singlet, with probabilities \(\frac{3}{4}\) and \(\frac{1}{4}\). Then the spatial scattering process must be antisymmetric in the first case and symmetric in the second. This gives the same answer!