Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 2

At the stated order the renormalization-group beta function is separable:ConsequentlyFor a positive initial running coupling, this expression is used on the connected interval containing ${\mu }_i$ where its denominator is positive and perturbation theory remains valid.

If $b_0<0$, the running coupling decreases at high renormalization scale and tends to zero: this is asymptotic freedom. Decreasing the renormalization scale increases the running coupling, with a formal infrared pole at $\Lambda ={\mu }_i\exp [1/(b_0{\alpha }_i)]$. Near $\Lambda$, perturbation theory fails; the pole alone is not a proof of confinement. Both QCD and the SU(2) weak gauge coupling have this sign for the Standard Model particle content above the relevant thresholds.

If $b_0>0$, the running coupling decreases towards low renormalization scale and grows towards high renormalization scale. The formal ultraviolet Landau pole is ${\mu }_L={\mu }_i\exp [1/(b_0{\alpha }_i)]$. The pole signals failure of this perturbative evolution, rather than a controlled physical infinity. The hypercharge gauge coupling and ordinary QED with charged matter have this sign.

If $b_0=0$, the running coupling is constant at this order; higher-order terms may still make it run. No Standard Model gauge factor has a vanishing one-loop order coefficient for the usual particle content.

For precise normalization, take ${\alpha }_i=g_i^2/(4\pi )$ and $Q=T_3+Y$. The Standard Model one-loop gauge coefficients then giveThese coefficients use $d/d\ln \mu$, not $d/d\ln {\mu }^2$. Threshold matching changes the active particle content, so the single constant-$b_0$ expression is applied piecewise across thresholds.

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

The two Bhabha scattering tree-level Feynman diagrams are the two panels of Solution with a photon mediator: $s$-channel annihilation with internal momentum $p_1+p_2$, and $t$-channel exchange with internal momentum $p_1-p_3$. Each interaction vertex preserves the electron species. There is no independent $u$-channel Feynman diagram, since the outgoing electron and Positron are distinguishable.

Removing the color charge factor from Solution and replacing $g$ by $e$ givesCrossing symmetry interchanges $s$ and $t$ for the exchange Feynman diagram, givingSquaring the sum of the two scattering amplitudes also produces the interference term $2\mathrm{Re}\overline{{\mathcal{M}}_s{\mathcal{M}}_t^{\ast }}$. The given third term must be this cross term: it has both propagator denominators $st$, whereas a squared individual Feynman diagram has $s^2$ or $t^2$. With the correct relative fermionic sign, its value is $4e^4{u}^2/(st)$. ThusThe coefficient of the last displayed fraction is positive, but $s>0$ and $t<0$ in the physical scattering region, so this interference is destructive. The relative sign originates in the two possible external fermion contractions; changing an external-state phase changes both scattering amplitudes together.^[ref]

Both $s$-channel annihilation and $t$-channel gluon exchange are now allowed. Their spin sum factors are the Bhabha scattering factors with $e\mapsto g$. Their color charge tensors, with the external indices of Solution, areEach squared term has $K_{ss}=K_{tt}=(N_c^2-1)/(4N_c^2)$. The cross term instead has the color averageTo evaluate it, use the SU(N) fundamental completeness relationThis follows by completing the orthogonal basis of traceless Hermitian matrices with the identity. Since $\mathrm{tr}{T}^b=0$, it gives ${\sum }_a{T}^a{T}^b{T}^a=-T^b/(2N_c)$, and henceMultiplying the two squared Bhabha scattering terms by $K_{ss}$ and its cross term by $K_{st}$ givesFor the SU(3) gauge group, this isThe negative color charge overlap reverses the sign of the QED interference. Since $st<0$, this cross contribution is positive in the physical region. No additional spinor calculation is needed; all the new information is in the SU(N) fundamental completeness relation.