Past exam of the quantum field theory course of the University of Oxford / 2023 / 1 / e

The additional six-leg interaction vertex has Feynman rule $-i{g}_6$. Since both interactions lack derivatives, the graph identity still gives $\omega =2(1-V)$, with $V=V_4+V_6$. The difference is that a one-vertex contraction of the phi-six theory can leave four external legs: choosing the contracted pair gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$. Its quartic interaction correction isConsequently a one-loop quartic interaction counterterm is now necessary:The one-loop mass counterterm from c is unchanged. Contracting four legs of a single six-leg interaction vertex gives $\left(\genfrac{}{}{0ex}{}{6}{2}\right)3=45$ double contractions, hence a two-loop raw self-energy $g_6{T}_{\Lambda }^2/8$. Its subtraction must include the quartic interaction counterterm insertion, rather than subtracting this raw expression in isolation.

All divergent field-dependent coefficients are most simply organized by the normal-ordering subtraction scheme in two dimensions. With $T=T_{\Lambda }$, express a potential normal ordered at mass $m$ in ordinary fields:Thus its divergent subtraction coefficients, in this scheme, areIndeed the two-loop self-energy terms are $g_6{T}^2/8+\delta g_{4}T/2=-g_6{T}^2/8$, which the displayed mass counterterm cancels. The constant terms only affect vacuum energy. After removing all one-vertex self-contractions, every remaining subgraph is ultraviolet finite. Additional finite on-shell renormalization scheme adjustments are subject to the same qualification as d.