Past exam of the quantum field theory course of the University of Oxford / 2018 / 3

A phi-fourth theory interaction has mass dimension $4-D$, so its bare coefficient must contain a renormalization scale factor ${\mu }^{4-D}=({\mu }^2{)}^{2-D/2}$ when expressed through a dimensionless renormalized coupling. The tree-level four-leg interaction vertex is linear in that coupling. At the next order, two quartic vertices form the three one-loop bubble diagram graphs shown in c, with external pairings $12|34$, $13|24$ and $14|23$.

For each pairing there are two identical internal lines and Feynman-diagram symmetry factor 2. The loop integral has the large-momentum behaviourIt is logarithmically divergent in four dimensions. Dimensional regularization represents that local divergence by a simple pole proportional to $1/(D-4)$. Its residue is independent of external momenta, so a local quartic counterterm cancels it. The three graphs sum to one constant coefficient; this argument establishes its structure without requiring its value.

The relation consequently has the formThe sign of $C$ incorporates the chosen pole denominator and vertex convention. In particular, a negative $C$ here corresponds to a positive coefficient of $1/(4-D)$ in the bare coupling. Finite pieces depend on the renormalization scheme; a pure-pole relation is natural in minimal subtraction scheme. One-loop two-point tadpole diagrams contribute mass renormalization, but not the momentum-dependent field renormalization that could change this coupling relation at order $g^2$. The four-point object relevant here is the amputated one-particle-irreducible vertex; external-leg self-energy insertions are dealt with separately.

Let $\delta =4-D$ and $F(g)=g-C{g}^2/\delta +O(g^3)$. Since changing the arbitrary renormalization scale must leave the bare theory fixed,Write $\beta (g)=-\delta g+b{g}^2+O(g^3)$. Substitution gives, through order $g^2$,Therefore $b=-C$, and the specified residue gives the renormalization-group beta functionThe derivative is at fixed bare parameters. It does not say the physical theory changes when $\mu$ changes; the running coupling compensates the scale used to describe it. The coefficient also depends on defining the derivative with respect to $\log \mu$, rather than $\log {\mu }^2$; using the latter would divide both terms by two.

In four dimensions set $b=3/(16{\pi }^2)$ and retain the one-loop beta function $dg/d\log \mu =b{g}^2$. On an interval where $g\ne 0$, differentiate its inverse:Integrating between ${\mu }_s$ and $\mu$ yieldsThe $g_s=0$ solution is identically zero and is also the continuous limit of this expression. Expanding the denominator gives $g(\mu )=g_s+b{g}_s^2\log (\mu /{\mu }_s)+\cdots$, directly checking the perturbative sign.

For positive $g_s$, the running coupling decreases towards the infrared and increases towards the ultraviolet. The one-loop expression has a Landau pole atThis is the scale where the truncated solution diverges, not a controlled perturbative prediction of a physical singularity at arbitrarily strong coupling. The formula is quantitatively reliable while the running coupling remains small enough that omitted higher-loop terms are negligible. It resums the leading logarithms of that approximate renormalization group equation; it is not the exact all-orders solution of the theory.

For a fixed point $g^{\ast }$ with $\beta (g^{\ast })=0$, put $g=g^{\ast }+h$. To first order in the displacement,Hence ${\beta }^{\prime }(g^{\ast })<0$ means ultraviolet attraction, since $h\to 0$ as $\mu \to \infty$; ${\beta }^{\prime }(g^{\ast })>0$ means infrared attraction, since $h\to 0$ as $\mu \to 0$. If the derivative vanishes, the linear test is inconclusive and higher terms determine the direction of flow.

For $\delta =4-D>0$ and $b=3/(16{\pi }^2)$, the retained beta function is $-\delta g+b{g}^2$. Its two zeros and slopes areThus the positive interacting Wilson-Fisher fixed point isand is infrared attractive in the coupling direction. Couplings between zero and $g^{\ast }$ increase towards $g^{\ast }$ when the scale decreases; those just above $g^{\ast }$ decrease towards it. The Gaussian fixed point $g=0$ is ultraviolet attractive for $D<4$.

The conclusion is local and perturbative. For $4-D\ll 1$, $b{g}^{\ast }=4-D$ is small, making the expansion controlled; extrapolation far below four dimensions requires more information. It is also a statement about the displayed coupling flow, not stability against every possible deformation of the full theory. In particular, the mass term is a relevant perturbation of the critical theory and must be tuned to reach the massless critical surface. Calling this fixed point infrared stable must not erase that additional relevant direction.