First interaction

💼 Case: lets consider muon and anti-muon pairs produced from electron positron collision

• Lets first draw the Feynman diagram

  

• Lets look at the conserved quantum numbers

  $$ \begin{matrix} & \text{initial} & \text{final}\\ \text{electric charge} & \sum_i Q_i=0 & \sum F Q_f=0 \\ \text{electron number} & L_e^i=0 & L_e^f=0 \\ \text{muon number} & L\mu^i =0 & L^f_\mu=0 \\

  \end{matrix} $$

  🗒️ Note: electric charge and lepton numbers also conserved at each vertex for EM

• Since at each vertex the coupling strength is $e$, we can write the scattering amplitude

  $$ \mathcal M_{fi}\propto e^2 $$

• We can also write the cross section as

  $$ \sigma_{e^+e^-\to\mu^+\mu^-}=k\alpha ^2_\text{em}=\frac{4\pi\alpha^e_\text{em}}{3s} $$

  where $k=\frac{4\pi}{3s}$ is a constant derived from QFT and $s$ is defined such that $\sqrt{s}$ is the centre of mass energy of the collision (here CoM energy of $e^+e^-$)

• Now we can write down the cross section dependence on the solid angle $\Omega$

  $$ \frac{\text d \sigma}{\text d \Omega} =\frac{\alpha^2_\text{em}}{4s} (1+\cos^2\theta) $$

  where $\theta$ is the polar angle in CoM frame (here of $\mu^\pm$)

  🗒️ Note: this doesn't match experimental data perfectly because we are only considering EM force not electroweak (i.e. $Z$ boson)

Second interaction

💼 Case: consider a electron positron annihilation into a quark antiquark pair

• Again we start with the Feynman diagram

  

  🗒️ Note: included $_r$ for red since each colour $u_r,u_b,u_g$ are treated separately

• Lets look at the conserved quantum numbers

  $$ \begin{matrix} & \text{initial} & \text{final}\\ \text{electric charge} & \sum_i Q_i=0 & \sum _F Q_f=0 \\ \text{electron number} & L_e^i=0 & L_e^f=0 \\ \text{baryon number} & B =0 & B=0 \\

  \end{matrix} $$

  🗒️ Note: quark flavour and baryon number are also conserved at each vertex for EM

• The coupling strength at the first vertex is $e$ and for $u_r\overline u_r$ it is $\frac 23 e$ so scattering amplitude

  $$ \mathcal M\propto \frac 23 e^2 $$

• Again remembering that $\sigma\propto \mathcal M^2$ we get the following cross section

  $$ \sigma_{e^+e^-\to u_r \overline u_r}=k\left ( \frac{4}{9} \right )\alpha^2_\text{em} $$

  ➕ If we now include the 3 quark colours we multiply by 3 $(r,g,b)$

  $$ \sigma_{e^+e^-\to u \overline u}=3k\left ( \frac{4}{9} \right )\alpha^2_\text{em} $$

  ➕ If we include all the flavours of quark we get $(u,d,s,c,t,b)$

  $$ \sigma_{e^+e^-\to u \overline u}=3k\alpha^2_\text{em}\sum Q_f^2 $$

  where $Q$ is the charge of the quark ie $Q_u=\frac 23$ and $Q_d=-\frac 13$

Combining experiments

• We take the ratio $R$ of the 2 cross sections to remove constants and get

  $$ R=\frac{\sigma_{e^+e^-\to u \overline u}}{\sigma_{e^+e^-\to \mu^+ \mu^-}}=3\sum_fQ_f^2 $$