Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2017 / 1

Gauge invariance and power counting permit the same local operators as in the original renormalizable action. With renormalized fields, a convenient counterterm parametrization isHere $D_{\mu }={\partial }_{\mu }-i{\mu }^{ϵ/2}e{A}_{\mu }$. Expanding the first term supplies kinetic, scalar-scalar-photon, and seagull counterterms with the relative coefficients required by the Ward identity. Equivalently introduce separate vertex constants first and then impose that identity. For ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$, $A_0=Z_A^{1/2}A$, and $e_0={\mu }^{ϵ/2}{Z}_{e}e$, it gives $Z_e{Z}_A^{1/2}=1$. Thus charge renormalization can be represented through the photon field factor, rather than adding an independent gauge-noninvariant interaction.

The mass coefficient defined here is $\delta M^2=Z_{\varphi }{m}_0^2-m^2$; it must be distinguished from $m_0^2-m^2$. Gauge fixing may require the corresponding gauge-parameter redefinition, and vacuum diagrams can require an additive vacuum-energy counterterm. No photon mass counterterm is allowed by the unbroken gauge symmetry. The quartic scalar counterterm remains necessary for the full theory even at a scale where the renormalized $\lambda$ is zero, because gauge interactions generate scalar four-point divergences.

The one-particle-irreducible scalar two-point function has three one-loop topologies: a scalar-photon exchange loop built from two cubic vertices; a photon tadpole at one seagull vertex; and a scalar tadpole at one quartic vertex. The last is absent when $\lambda =0$. Reducible chains of self-energy insertions belong to the Dyson expansion and are not additional one-loop self-energy graphs. The scalar-QED photon-tadpole symmetry factor is $1/2$ because its two identical photon legs are joined; the oriented complex-scalar tadpole has no such identical-real-line factor.

Route the photon momentum as $l$ and the internal scalar momentum as $l+k$. The transverse projector obeys $P_{\alpha \beta }{l}^{\beta }=0$, so the two cubic vertices giveMultiplying the two vertices and propagators therefore givesFor the seagull, $g^{\alpha \beta }{P}_{\alpha \beta }=d-1$. Including the photon-loop symmetry factor gives insteadThe printed seagull coefficient is twice this result. The vertex's factor two differentiates the two photon fields; closing them into one line requires the compensating symmetry factor. Direct expansion of the action gives the same check: the interaction density $-e^2{A}^2{\varphi }^{†}\varphi$ has one photon contraction, yielding coefficient $-e^2$, not $-2e^2$. The identical-photon origin of the vertex factor is also explicit in the standard scalar-electrodynamics rules.^[ref] This discrepancy does not affect the massless-photon limit in part d or the ultraviolet pole in part f.

Apply the dimensionally regulated tadpole integral before removing the infrared regulator:At fixed regulator $0<\mathrm{Re}ϵ<2$ away from gamma-function poles, the power of $m_{\gamma }^2$ has positive real part. ConsequentlyBoth the correctly weighted tadpole and the source's doubled version vanish. This is the near-four-dimensional limit appropriate to the question; the order of limits is important. Equivalently the exactly massless tadpole is a scaleless integral and is zero in dimensional regularization. The statement is not valid for every arbitrary fixed dimension: for example, $ϵ=5/2$ gives a nonzero prefactor and growth as $m_{\gamma }^{-1/2}$. One must keep the regulator in its stated near-four-dimensional domain.

First remove the projector denominator explicitly:Let $A=(l+k{)}^2+m^2$ and $B=l^2$. The Feynman parameter for a simple and a squared propagator follows by differentiating the elementary two-denominator Feynman formula with respect to $B$ givesShift $q=l+xk$. Then $xA+(1-x)B=q^2+D$, with $D=x(1-x)k^2+x{m}^2$. The numerator is unchanged by this parallel shift:Therefore the correctly combined integral isThe printed weight $x/2$ is not compatible with this denominator and prefactor. Exchanging $x$ with $1-x$ changes the mass term in $D$ as well and cannot supply the missing normalization. A concrete counterexample is $m=0$ with nonzero spacelike external momentum and $2<d<4$: after Wick rotation, the remaining integral is nonzero and symmetric under $x\leftrightarrow 1-x$. The source's $4e^2\int (x/2)dx$ expression is then exactly one quarter of the correct expression. Thus the claimed equality cannot be imposed when computing the pole.

Lorentz-covariant tensor integration replaces $q_{\mu }{q}_{\nu }$ by $g_{\mu \nu }{q}^2/d$. The numerator consequently contributes $(d-1)k^2{q}^2/d$. The supplied master integral with powers $a=1,b=3$ givesInserting the corrected parameter integral givesThe pole uses $\Gamma (ϵ/2)=2/ϵ+O(1)$, $D^{-ϵ/2}=1+O(ϵ)$ and ${\int }_0^1(1-x)dx=1/2$. ThusThere is no momentum-independent pole at $\lambda =0$ in this transverse gauge after removing the photon tadpole. The two-point counterterm insertion is $-i(\delta Z_{\varphi }{k}^2+\delta M^2)$, so minimal subtraction requiresThese coefficients cancel the actual one-loop insertion derived from the action. Applying the misprinted weight in part e would give a wave-function pole four times too small. In a $d=4-2\epsilon$ convention the same answer is $\delta Z_{\varphi }=3e^2/(16{\pi }^2\epsilon )$. Finally, $\delta M^2=0$ does not mean that the bare mass parameter is unchanged: $Z_{\varphi }{m}_0^2=m^2$ gives $m_0^2-m^2=-m^2\delta Z_{\varphi }+O(e^4)$. Photon, charge and quartic counterterms require their respective Green functions and cannot be determined from this scalar two-point function alone.