Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 303 / 2

A momentum-shell renormalization group step has three parts:

1. Split the field into slow modes ${\varphi }^{-}$ with $|k|<\Lambda /\zeta$ and fast modes ${\varphi }^{+}$ with $\Lambda /\zeta <|k|<\Lambda$, then integrate out ${\varphi }^{+}$ to obtain a Wilsonian effective action for ${\varphi }^{-}$.
2. Rescale momenta by $k^{\prime }=\zeta k$, equivalently coordinates by $x^{\prime }=x/\zeta$, so the reduced cutoff returns from $\Lambda /\zeta$ to $\Lambda$.
3. Rescale the field, at the Gaussian fixed point by ${\varphi }^{\prime }(x^{\prime })={\zeta }^{(d-2)/2}{\varphi }^{-}(x)$, so the coefficient of $(\nabla \varphi {)}^2/2$ retains its chosen normalization.

The resulting functional has the original cutoff but changed coefficients. Iterating the operation gives a renormalization-group flow on masses and interaction couplings.

At the Gaussian fixed point, the engineering dimension of the field isThe operator ${\varphi }^n({\nabla }^2\varphi {)}^m$ contains $n+m$ fields and $2m$ derivatives. Since its integral must be dimensionless,It is marginal when this vanishes, namelywhen $n+m\ne 2$. Since $n,m$ are positive, the exceptional case is $n=m=1$: $\varphi {\nabla }^2\varphi$ has a dimensionless coupling in every $d$ and differs from the kinetic term by integration by parts. If the displayed formula gives no positive $d$, there is no positive spatial dimension in which that operator is naively marginal.

At an interacting fixed point, field and composite operators acquire anomalous dimensions, and operators with the same symmetries can mix under renormalization. Their full scaling dimensions can therefore differ from these naive engineering dimensions.

Use two external slow-field legs and internal fast-mode propagators. Through the requested orders, the connected mass-correction topologies are:

A nominal one-line bridge at order ${\lambda }_0{g}_0$ vanishes in a sharp momentum-shell scheme at small external momentum because that line would have to carry momentum outside the fast shell. These descriptions specify the same diagrams without depending on a particular drawing convention for vertices and external legs.

Writeand, for the three-line topology,and define the four-line integralwith every propagator momentum restricted to the fast shell.

The first cumulant containsSince the quadratic free energy is $\frac{1}{2}{\mu }^2({\varphi }^{-}{)}^2$, this gives

The mixed term in the second cumulant isWick contraction with the printed normalization $V_6={\lambda }_0\int {\varphi }^6$ and $V_4=g_0\int {\varphi }^4$ gives, at zero external momentum,The coefficients respectively combine the two placements of both external legs in the two-line topology, the three-line topology with one external leg on each vertex, and the four-line topology. Couplings normalized as $g_0{\varphi }^4/4!$ and ${\lambda }_0{\varphi }^6/6!$ absorb the corresponding factorials, which is why formulas in that convention have much smaller numerical coefficients.