Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 1

Let the bare fields obey the wave-function renormalization definitions ${\psi }_0=Z_2^{1/2}\psi$, $A_{0\mu }=Z_3^{1/2}{A}_{\mu }$, and put $m_0=Z_{m}m$. Define the vertex renormalization factor by $e_0={\mu }^{ϵ/2}e{Z}_1{Z}_2^{-1}{Z}_3^{-1/2}$. Substitution into the bare QED Lagrangian density givesEquivalently, write $Z_i=1+{\delta }_i$ and ${\delta }_m=Z_2{Z}_m-1$. The counterterm Lagrangian isFor the subsequent Feynman gauge calculation include $-(\partial \cdot A{)}^2/2$ in the gauge-fixed action. The Ward identity for QED renormalization constants implies $Z_1=Z_2$ when the regularization preserves gauge invariance; this is a consequence, rather than an extra independent definition of the renormalization constants.

The Clifford algebra of the gamma matrices gives ${\gamma }^{\nu }{\gamma }^{\mu }=2g^{\nu \mu }-{\gamma }^{\mu }{\gamma }^{\nu }$ and ${\gamma }^{\nu }{\gamma }_{\nu }=D1$. ThereforeThe contraction $g^{\nu }{}_{\nu }=D$ must be retained until the dimensional regularization calculation is complete; replacing it prematurely by four can change finite terms multiplying a ultraviolet pole.

There is one one-particle-irreducible Feynman diagram: an internal photon joins two points of the open fermion line, with the external photon attached to that line between them. It is the upper-left panel of the figure. Route internal photon four-momentum $k$ from the incoming side to the outgoing side; the adjacent internal fermion four-momenta are $p-k$ and $p^{\prime }-k$. The symmetry factor is one.

The vertex renormalization counterterm is an additional local contribution when forming the renormalized Feynman vertex. External-leg self-energy insertions belong to the complete three-point correlation function, rather than this amputated one-particle-irreducible Feynman diagram.

With $q=p^{\prime }-p$ and the routing just described, multiplication of the three QED Feynman rules for the Feynman vertices, the two Dirac propagators and the photon propagator gives $(-i{e}_D{)}^3{i}^2(-i)=-e_D^3$. Hence the unrenormalized amputated Feynman vertex through this order isFor the renormalized Feynman vertex, add $-i{e}_D{\delta }_1{\gamma }^{\mu }$ to this expression. Contributions from replacing parameters inside the one-loop term by their bare counterparts enter at higher loop order.

Keep $m>0$ and set $p=p^{\prime }=0$, which is an off shell choice. The numerator isTerms odd in $k$ integrate to zero. Lorentz covariance and dimensional regularization give, inside this integral,Applying the gamma matrices identity from part (ii) once more yieldsConsequentlyThis massive zero-momentum QED vertex integral uses an integrated tensor identity for the replacement of $N^{\mu }$, not an equality for each fixed $k$. The physical photon remains massless. The retained fermion mass supplies a scale and makes this zero-external-four-momentum calculation suitable for extracting the ultraviolet pole without a scaleless integral.

After a Wick rotation, the term in question has radial behaviorThe radial integrand is proportional to $k_E$ at the origin and to $k_E^{-3}$ at infinity. Thus it has neither an infrared divergence nor an ultraviolet divergence. In fact the area of the unit three-sphere givesThis also follows from the Feynman parameter identitywith $A=k^2+i0$, $B=k^2-m^2+i0$. Although the parameterized denominator becomes massless at $x=1$, its numerator $1-x$ cancels the apparent endpoint singularity. Since $\beta \to -2$, multiplication by $\beta$ preserves finiteness. No artificial photon mass is needed.

Only the term proportional to $\alpha$ has an ultraviolet pole. Its $k^2$ cancels the massless denominator, leaving the standard Feynman integralIn the present dimensional regularization convention, $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\alpha =1+O(ϵ)$. ThereforeThe counterterm contribution is $-i{e}_D{\delta }_1{\gamma }^{\mu }$. Cancellation of the ultraviolet pole fixeswhere the displayed expression uses the minimal subtraction scheme; a different renormalization scheme may add finite $O(e^2)$ terms. In the frequently used convention $D=4-2\epsilon$, the same result is $Z_1=1-e^2/(16{\pi }^2\epsilon )+O(e^4)$. The sign agrees with $Z_1=Z_2$ from the Ward identity for QED renormalization constants in Feynman gauge.

The proper quark-gluon vertex colour factors come from two one-particle-irreducible Feynman diagrams, shown in the upper-middle and upper-right panels. One has a virtual gluon joining the open quark line on opposite sides of the external gluon insertion. The other has two virtual gluons meeting the external gluon at a three-gluon Feynman vertex, with their other ends joined by an internal quark. A Faddeev-Popov ghost field has no direct quark coupling, so it supplies no additional one-loop Feynman vertex graph here.

Let $T^a$ be the Hermitian generators of the fundamental representation of $SU(N)$, with $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, $[T^a,T^b]=i{f}^{abc}{T}^c$, ${\sum }_b{T}^b{T}^b=C_{F}1$, and $f^{abc}{f}^{dbc}=C_A{\delta }^{ad}$. Here $C_F=(N^2-1)/(2N)$ and $C_A=N$ are quadratic Casimir operators evaluated in the fundamental representation and the Adjoint representation.

For the first graph, the ordered product is ${\sum }_b{T}^b{T}^a{T}^b$. Expand the double commutator:Thus its colour factor isFor the second graph choose the orientation of the three-gluon Feynman vertex so that its group tensor is $f^{abc}$ and the ordered product along the quark line is $T^b{T}^c$. Antisymmetry removes the anticommutator part:The corresponding real group coefficient is $C_A/2=3/2$ after the conventional factor of $i$ is incorporated into the remaining Feynman rules. If the vertex is written with $i{f}^{abc}$, its raw matrix product is instead $-C_A{T}^a/2$. Reversing the ordered product reverses its sign together with the orientation-dependent three-gluon Feynman rule; these conventions describe the same scattering amplitude. External-leg corrections and local counterterms are separate from these two proper one-loop graphs.