Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2025 / 2

The first panel is the one-particle-irreducible Feynman diagram for photon vacuum polarization. The two internal Dirac propagators carry $k$ and $k+q$, and the external photons carry $q$. The other panels fix the routings used in parts c and d.

The closed fermion loop sign contributes a minus sign; there is no extra symmetry factor. Applying the given Feynman rules, with the external propagators amputated,Equivalently, the required amplitude isThe Feynman slash notation is $/k={\gamma }^{\rho }{k}_{\rho }$ and $\mathrm{tr}1=4$ in the chosen dimensional regularization convention.

The Ward identity implies $q^{\mu }{\Pi }_{\mu \nu }=0$. Lorentz invariance then gives ${\Pi }_{\mu \nu }=(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })\Pi$, so $g^{\mu \nu }{\Pi }_{\mu \nu }=(d-1)q^2\Pi$. The gamma matrices satisfy ${\gamma }_{\mu }/k{\gamma }^{\mu }=(2-d)/k$. Odd traces vanish, and thereforeSubstituting gives the requested scalar coefficient:To extract the ultraviolet divergence without introducing a Feynman parameter, set $A=(k+q{)}^2-m^2$, $B=k^2-m^2$ and useTranslation invariance of dimensional regularization gives ${\int }_k{A}^{-1}={\int }_k{B}^{-1}={\mathcal{I}}_1$. ConsequentlyIn the pole coefficient one may set $d=4$: the bracket becomesThus the ultraviolet pole of QED vacuum polarization isThe cancellation of the apparent $m^2/q^2$ divergence is essential: gauge invariance forbids a photon mass counterterm. The scalar result at $q^2=0$ is understood by continuation after this cancellation.

The middle panel in part a has an internal photon of momentum $k$ joining two points on a fermion line; the internal fermion has momentum $p-k$. It is a one-particle-irreducible Feynman diagram with no closed fermion loop sign. In Feynman gauge,orThe decomposition in the source gives $\mathrm{tr}(/p\Sigma )=4p^{2}S(p^2)$. In particular,For a direct ultraviolet divergence extraction, use a Feynman parameter $x$ multiplying the massive denominator, shift $r=k-xp$, and define $\Delta =x{m}^2-x(1-x)p^2$. Then $/p-/k=(1-x)/p-/r$ andOdd $r$ terms integrate to zero. Taking the trace explicitly supplies the weight $(2-d)(1-x)$, whose integral tends to $-1$. HenceAs a check on the same fermion self-energy calculation,This is the ultraviolet pole of QED fermion self-energy. It is safe to extract it at generic external momentum; an on shell or massless limit must distinguish any infrared divergence from this ultraviolet divergence.

In the right panel of part a, $p$ enters the fermion line, $p^{\prime }=p+q$ leaves it, and a virtual photon of momentum $k$ bridges the external photon vertex. Write ${\Gamma }^{\mu }={\gamma }^{\mu }+{\Lambda }^{\mu }$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop amplitude isThis Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive mass $M$ while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }$. Symmetric integration replaces $k_{\rho }{k}_{\sigma }$ by $g_{\rho \sigma }{k}^2/d$. Two contractions of gamma matrices then giveFor the auxiliary-mass extraction of ultraviolet poles,The second term is ultraviolet convergent. ThusThe ultraviolet pole of the QED vertex is therefore ${\Lambda }^{\mu }{|}_{\mathrm{UV}}=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. The Pauli structure has zero trace against ${\gamma }_{\mu }$ and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local ${\gamma }^{\mu }$ coefficient, which is all that is required here.

In $d=4-2ϵ$, the modified minimal subtraction scheme subtracts ${\Delta }_{\overline{\mathrm{MS}}}=1/ϵ-{\gamma }_E+\log (4\pi )$ for each one-loop pole, with no further finite, kinematics-dependent subtraction. Equivalently, use a rescaled renormalization scale ${\overline{\mu }}^2=4\pi e^{-{\gamma }_E}{\mu }_R^2$ and subtract only $1/ϵ$ in that convention. This differs from the minimal subtraction scheme by a fixed scale convention.

Let $C=e^2{\Delta }_{\overline{\mathrm{MS}}}/(16{\pi }^2)$. With the source's counterterm insertion signs, cancellation proceeds as follows. The vertex has $-ie{\gamma }^{\mu }C$ from the loop and $-ie{\gamma }^{\mu }{\delta }_1$ from its counterterm, giving ${\delta }_1=-C$. The kinetic part of the fermion self-energy insertion is $+i/pC$ and its counterterm is $+i/p{\delta }_2$, giving ${\delta }_2=-C$. Finally the photon vacuum polarization insertion is $-i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })(4C/3)$ and its counterterm is $-i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu }){\delta }_3$, givingWriting $Z_i=1+{\delta }_i$ therefore verifies the Ward identity for QED renormalization constants, $Z_1=Z_2$, at one loop. The proper-vertex Ward-Takahashi identity in QED independently gives ${\Lambda }^{\mu }(p,p)=-\partial \Sigma (p)/\partial p_{\mu }$, whose divergent part reproduces the opposite signs of $F_1^{(1)}$ and $S$ found above. The equality holds in a common gauge fixing and subtraction prescription; the individual constants $Z_1,Z_2$ are gauge dependent.