Past exam of the quantum field theory course of the University of Oxford / 2024 / 3

In $2+1$ spacetime dimensions, the dimensionless action gives $[\mathcal{L}]=3$ and $[{\partial }_{\mu }]=1$. The scalar kinetic term requires $2[\varphi ]+2=3$, whereas the fermion kinetic term requires $2[\psi ]+1=3$. Thus the engineering dimensions of relativistic fields areMatching each interaction to mass dimension three givesHere the brackets mean exponents of mass in natural units. The coefficients $g_1,g_3$ are marginal couplings by classical power counting, and $g_2$ is a relevant coupling.

Let $I_b,I_f$ count internal scalar and fermion lines and $L$ count independent loops in a connected amputated Feynman diagram. At large loop momentum, the scalar propagator falls as $k^{-2}$ and the Dirac propagator falls as $k^{-1}$, including its numerator. Each loop integration is $d^{3}k$. The nonderivative interaction vertices add no powers. ThereforeCount endpoints of each kind of line and use the connected-graph loop order identity:Eliminating $L,I_b,I_f$ givesThe cancellation of $n_1,n_3$ reflects their dimensionless couplings; every insertion of the positive-dimensional $g_2$ improves the superficial degree of divergence by one.

The printed exact homogeneity is shorthand for ultraviolet power counting, not an identity for a massive integrated correlation function. For example, even the free amputated scalar two-point vertex ${\Gamma }^{(2)}(p)=p^2-M^2$ obeys ${\Gamma }^{(2)}(\lambda p)={\lambda }^2{p}^2-M^2$, which differs from ${\lambda }^2{\Gamma }^{(2)}(p)$ when $M\ne 0$. Logarithmic divergences, actual numerator cancellations and ultraviolet subdivergences also require separate treatment.

Use one-particle-irreducible correlation functions to identify independent local counterterms. The exact internal symmetry $\varphi \mapsto -\varphi$ excludes odd $b$, and Dirac fermion number conservation requires equal numbers of $\psi$ and $\bar{\psi }$, so $f$ is even. From b, the complete list of potentially overall divergent vertices with external legs is:

The zero-leg vacuum diagrams, with $\omega =3-n_2$, may also require a vacuum energy density subtraction. Four-fermion vertices have $\omega \le -1$ and do not require a new four-fermion counterterm. These are superficial bounds: a particular allowed diagram can vanish or be convergent.

Locality and Lorentz invariance restrict the divergent parts to a finite basis. The scalar two-point vertex can contain a constant and $p^2$; there is no Lorentz scalar linear in a single momentum. The fermion two-point vertex can contain $I$ and $/p$. The remaining listed vertices have no allowed derivative divergence. A complete counterterm Lagrangian, up to total derivatives, is thereforeThe coefficients supply wavefunction renormalization, mass counterterms and renormalizations of the three interactions already present. Recursive subtraction of ultraviolet subdivergences then leaves only these local overall divergences. This finite counterterm closure is why the complete theory is expected to have perturbative renormalizability. It does not mean every listed coefficient first becomes nonzero at one loop, or that setting an allowed coupling to zero is stable under renormalization.

Set the renormalized $g_3$ to zero. For a six-scalar vertex, b gives $\omega =-n_2$, so a new overall logarithmic divergence can only occur in graphs with $n_2=0$. With $b=6,f=0$, the line counts become $I_f=n_1$ and $I_b=n_1-3$, hence $L=n_1-2$.

At one loop this forces $n_1=3$ and $I_b=0$: the only candidate is a closed triangle Feynman diagram with two external scalar legs at each vertex. Its numerator at large loop momentum is proportional toThis vanishes for the specified two-component gamma matrices, each of which is traceless. Replacing one leading $/k$ by the fermion mass or an external momentum lowers the superficial degree of divergence from zero to at most $-1$. Thus the apparent logarithm cancels. Graphs with a $g_2$ vertex already have negative overall degree; the pure-scalar one-loop triangle, for instance, has $\omega =-3$. There is no one-loop ultraviolet subdivergence inside these proper one-loop vertices. ThereforeThe six-point correlation function need not vanish: finite terms are allowed. Nor is it correct to say that all odd gamma-matrix traces vanish in $2+1$ dimensions. In this representation,The cancellation above follows from repeating the same leading loop momentum, not from a false four-dimensional trace rule.

At two loops, $n_1=4$, $I_f=4$ and $I_b=1$ are possible. A closed fermion box with a scalar line joining two of its vertices leaves six external scalar legs. It has $L=2$, $\omega =0$, and four leading gamma matrices; the one-loop odd-numerator argument no longer applies.

A background-field check demonstrates that its logarithm is actually nonzero. Put $\varphi =v+\eta$ with constant $v$. Define the background fermion mass $\mu =m+g_1{v}^2/2$, the background scalar squared mass ${\mathcal{B}}^2=M^2+g_2{v}^2/2$, and the induced three-leg Yukawa coupling $y=g_{1}v$. The two-loop quantum effective potential contains the sunset diagram with two fermion lines and one scalar line. After a Wick rotation, its relevant integral, apart from a nonzero overall combinatorial factor, isHere the dots and gamma-matrix traces are Euclidean. With $A=k^2+{\mu }^2$, $B=l^2+{\mu }^2$, $C=(k-l{)}^2+{\mathcal{B}}^2$, its numerator is exactlyThe first three terms reduce to products of one-loop tadpole integrals, which have no logarithmic pole near three dimensions in dimensional regularization. The remaining term isFor positive regulator masses, the large-momentum region of $H$ has a positive, nonzero logarithmic divergence. One can see this directly by restricting the relative angles and the ratio $|k|/|l|$ to compact intervals away from zero: the remaining common radial integral is proportional to ${\int }^{\Lambda }dr/r$. Its proper one-loop subgraphs are ultraviolet convergent.

Finally, the coefficient multiplying $H$ containsThe two-loop sextic counterterm in quartic Yukawa theory follows because its $v^6$ coefficient is $g_1^4\ne 0$ for a nonzero coupling. Other two-loop vacuum topologies reduce to products of finite-in-dimensional-regularization tadpoles and do not cancel this logarithmic pole. ThusA sextic counterterm is required even if the renormalized sextic coupling is set to zero at a chosen renormalization scale. A theory constrained to $g_3=0$ at every scale lacks counterterm closure when $g_1\ne 0$. The decoupled exception $g_1=0$ is the three-dimensional scalar quartic interaction, which needs no sextic counterterm by power counting.