Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 1

There are four topologies for the one-loop gluon self-energy, shown in the upper row of the figure:

The two internal-gluon graphs have symmetry factor $1/2$. The directed quark and ghost loops have symmetry factor one and a minus sign for a closed anticommuting loop. The four-gluon tadpole diagram must be included in the list even though its massless Feynman integral vanishes in dimensional regularization. Local counterterms are excluded here.

Route a directed ghost loop so that its internal four-momenta are $l$ and $l+k$. At the external gluon with indices $(\mu ,a)$, let the incoming ghost carry colour $c$ and the outgoing ghost carry colour $d$; the other Feynman vertex has the reverse order. The stated ghost-gluon Feynman rules supply $g_D^2{f}^{acd}{f}^{bdc}(l+k{)}^{\mu }{l}^{\nu }$. Their colour factor isThe two ghost propagators supply $i^2=-1$, while the closed ghost loop supplies another $-1$. Thus, with the insertion convention stated above,This establishes every prefactor before integration. To isolate its ultraviolet pole, take generic off shell $k$, initially $k^2<0$, so the massless integral has no overlapping infrared divergence. Introduce a Feynman parameter, shift $r=l+xk$, and set $\Delta =-x(1-x)k^2>0$. Odd powers of $r$ vanish, givingThe needed dimensional regularization integrals follow from the supplied gamma function formula. In $D=4-ϵ$,Here $\Gamma (ϵ/2)=2/ϵ+O(1)$, rather than $1/ϵ$. Using $1/D=1/4+O(ϵ)$ and ${\int }_0^{1}x(1-x)dx=1/6$, obtainTherefore the ghost-loop pole of the gluon self-energy isThe factor ${\mu }^{ϵ}$ in $g_D^2$ affects only finite terms at this order. The isolated ghost loop is not transverse: contracting this result gives $g^2{C}_A{\delta }^{ab}{k}^2{k}^{\nu }/(32{\pi }^2ϵ)$. Transversality is a condition on the complete gluon self-energy, including the other loop graphs.

There is one four-gluon tadpole diagram. Temporarily replace its internal gluon propagator by $-i{g}_{\rho \sigma }{\delta }^{cd}/(l^2-m_g^2+i0)$. Contract the supplied four-gluon Feynman vertex with $g_{\rho \sigma }{\delta }^{cd}$. Its first colour factor vanishes because $f^{cce}=0$. Each of the other two terms contributes $C_A{\delta }^{ab}(D-1)g^{\mu \nu }$, soInclude the symmetry factor $1/2$ and the factor $-i$ from the internal gluon propagator:The exact regulated tadpole integral isIn particular,The entire expression is proportional to $(m_g^2{)}^{1-ϵ/2}$ and vanishes as the auxiliary mass tends to zero near four dimensions. Equivalently, its expanded terms are proportional to $m_g^2/ϵ$, $m_g^2\log m_g^2$ and $m_g^2$, all of which disappear when the regulator is removed. HenceThis is the massless scaleless integral result obtained directly in dimensional regularization. The artificial mass is only an intermediate device for this graph; retaining it in the full theory would modify the gauge symmetry and cannot be used when applying the massless transversality identity in part (iv).

The Slavnov-Taylor identity implies transversality of the full perturbative gluon self-energy:The quark loop is separately transverse. This follows without evaluating its Feynman integral: contract its vector Feynman vertex with $k_{\mu }$ and use $/k=(/l+/k-m)-(/l-m)$ between its two Dirac propagators. One obtains a difference of two loop integrals that cancel after a shift of $l$ in dimensional regularization. The colour factor is proportional to $\mathrm{tr}(T^a{T}^b)$ and does not obstruct this argument.

The four-gluon tadpole diagram vanishes after removal of $m_g$. Thus the nontransverse parts of the two-three-gluon-vertex loop and the ghost loop must cancel. Contracting the preceding ghost-loop pole of the gluon self-energy givesFor QCD, set $C_A=3$. No evaluation of the remaining three-gluon Feynman diagram is required. Its uncontracted transverse part is not fixed by this identity.

For the one-loop renormalization-group beta function, write $t=\log \mu$. Separating variables givesThis expression resums the leading logarithms of the one-loop running coupling, while omitting higher-order terms in the renormalization-group beta function. Assume a positive initial coupling.

For $b_0<0$, increasing the renormalization scale decreases the running coupling; the zero-coupling renormalization-group fixed point is attractive in the ultraviolet. This is asymptotic freedom. Toward lower scales the one-loop expression has a formal pole at ${\mu }_i\exp [1/(b_0{\alpha }_i)]<{\mu }_i$. The strong-coupling neighborhood of that pole lies outside perturbative control. QCD provides the main example; the minimal Standard Model $SU(2)$ coupling is also asymptotically free at one loop.

For $b_0>0$, the running coupling grows with the renormalization scale and formally diverges at the ultraviolet Landau pole ${\mu }_i\exp [1/(b_0{\alpha }_i)]>{\mu }_i$. Toward the infrared it decreases to zero within this approximation. The Standard Model $U(1)$ weak hypercharge coupling has this sign, as does QED in a fixed charged-particle regime. A perturbative Landau pole marks failure of the approximation, rather than proving that the exact theory has that divergence.

For $b_0=0$, the running coupling is constant at this order; higher-loop terms decide whether it runs. Vanishing of the leading coefficient alone does not establish exact scale invariance.

For a concrete common convention, define ${\alpha }_i=g_i^2/(4\pi )$ with $g_Y$ in its usual, non-unified weak hypercharge normalization. In the minimal Standard Model with three generations and one Higgs doublet, above its particle thresholds,Thus the question's $b_0$ is $b_i/(2\pi )$, with the sign regimes just described. Crossing a mass threshold changes the active field content and requires the appropriate piecewise renormalization-group beta function.