Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2021 / 2 / a

Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The external on shell relations are $p_1^2=p_2^2=m_e^2$ and $p_3^2=p_4^2=m_t^2$. Expanding the three Mandelstam variables and using four-momentum conservation, $p_3+p_4=p_1+p_2$, gives

Let $\Sigma =m_e^2+m_t^2$, $Q_e=-1$, $Q_t=2/3$ and $N_c=3$. The single tree-level Feynman diagram is annihilation through a virtual photon. Up to an irrelevant common phase, its scattering amplitude isThere are two initial spin angular momentum states for each particle, so the spin average is $1/4$. The final color charge sum gives ${\sum }_{c,d}|{\delta }_{cd}{|}^2=N_c$. There is no additional photon polarization average for a virtual photon and no separate polarization label beyond the Dirac spinor spin sums.

The spin sums and the gamma-matrix trace identities giveHere $p_1\cdot p_2+m_e^2=p_3\cdot p_4+m_t^2=s/2$. Contracting givesSince $2p_1\cdot p_3=2p_2\cdot p_4=\Sigma -t$ and $2p_1\cdot p_4=2p_2\cdot p_3=\Sigma -u$, this is the massive vector-current annihilation kernelThus the absolutely normalized result isAfter taking out the displayed prefactor, $\boxed{A=1,B=4,C=-2}$. Because the source uses a proportionality sign, only their ratio is fixed until that prefactor is specified. Equivalently, after taking out just $e^4/s^2$, their values are $8/3,32/3,-16/3$. No mass has been neglected. At $m_e=0$ and $s=4m_t^2$, the bracket yields the finite isotropic threshold value of the scattering amplitude; the vanishing threshold scattering cross-section comes from relativistic two-body phase space.

The exchanged particle is now a gluon, and the incoming up quark mass replaces $m_e$. DefineThe Dirac trace and spin average are unchanged. Normalize the fundamental gauge generators by $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. Averaging over the $N_c^2$ initial color charge choices and summing the final color charge choices givesReplace the factor $N_c{e}^4{Q}_e^2{Q}_t^2$ from part (ii) by $g_s^4{C}_g$. ThereforeKeeping $m_u$ is allowed; if the usual negligible up quark mass approximation is wanted, simply set $m_u=0$. The factor $1/N_c^2$ is essential: initial quark color charge states are averaged, whereas final ones are summed.

There are two tree-level Feynman diagrams, with intermediate photon and gluon respectively. Their color charge tensors are ${\delta }_{ji}{\delta }_{kl}$ and ${\sum }_A{T}_{ji}^A{T}_{kl}^A$. The pure photon factor isTheir interference vanishes after the color charge sum because of photon-gluon colour orthogonality:Thus there is no term proportional to $e^2{g}_s^2$, and with the $F_u$ of part (iii),Here $Q_u=Q_t=2/3$. The absence of interference concerns the specified averaged and summed result; a fixed choice of external color charge need not have zero interference.