Past exam of the quantum field theory course of the University of Oxford / 2017 / 4

At fixed external momenta, fixed regulator and fixed bare coupling, define the total scale derivativeThe bare vertex has no independent dependence on the arbitrary renormalization scale. Differentiating ${\Gamma }^{(n)}=Z_{\varphi }^{n/2}{\Gamma }_0^{(n)}$ therefore givesDefining the two scale-response functions as in the source yieldsThis is the Callan-Symanzik equation. Its minus sign and factor $n/2$ follow from the specified $Z_{\varphi }^{n/2}$ multiplying the bare vertex. In particular, the source's $\gamma =\mu d\log Z_{\varphi }/d\mu$ is twice the field anomalous dimension defined using $\log Z_{\varphi }^{1/2}$. A convention change must be made consistently in both definitions and equation.

The dimensionless bare parameter is $g_0={\lambda }_0{\mu }^{-ϵ}$, so direct differentiation at fixed ${\lambda }_0$ givesThe dimensionful bare vertex itself is scale independent. Reexpressing it as ${\hat{\Gamma }}^{(n)}(p,g_0,\mu )={\Gamma }_0^{(n)}(p,g_0{\mu }^{ϵ})$ introduces only the combination $g_0{\mu }^{ϵ}$. HenceThere is no multiplicative field rescaling in this bare parametrization, soThis does not assert that the interacting renormalized field has zero anomalous dimension. It says that the unrescaled bare field has no additional normalization factor under this change of the arbitrary scale.

Write ${\mathcal{D}}_0=\mu {\partial }_{\mu }+{\beta }_0(g_0){\partial }_{g_0}$. Allow a general bare coefficient ${\gamma }_0$ temporarily, so the starting equation is $({\mathcal{D}}_0-n{\gamma }_0/2){\hat{\Gamma }}^{(n)}=0$. Substitute ${\hat{\Gamma }}^{(n)}=Z_{\varphi }(g_0{)}^{-n/2}{\Gamma }^{(n)}(p,g(g_0),\mu )$ and apply the product and chain rules:Multiplying by $Z_{\varphi }^{n/2}$ identifiesThus the renormalization-group beta function transforms by the chain rule under a coupling reparametrization, while the field coefficient also records the scale dependence of the finite or divergent field normalization. For the actual bare parametrization here, substitute ${\beta }_0=-ϵg_0$ and ${\gamma }_0=0$.

At fixed regulator abbreviateThe coupling expansion is $g=g_0+A{g}_0^2+O(g_0^3)$, whose inverse is $g_0=g-A{g}^2+O(g^3)$. Using part c before removing the regulator,Therefore the regulator-free coefficients at the orders determined by the supplied expansions areThe canonical term $-ϵg$ vanishes, and the factors of $ϵ$ cancel the simple poles. The finite coefficients $a_2,z_2$ and higher positive powers of $ϵ$ do not enter these leading regulator-free terms. Inverting the coupling is essential: substituting $g_0=g$ too early would incorrectly give $-2a_1{g}^2$.

The calculation establishes finiteness only at the displayed perturbative orders. Higher-order pole cancellations require the higher-order coefficients, as the next part makes explicit.

Retain the specified leading poles and write $A=a_1/ϵ$, $B=b_1/{ϵ}^2$. Solving $g=g_0+A{g}_0^2+B{g}_0^3$ recursively givesThe scale derivative must again be reexpressed in the renormalized coupling:The leading-pole consistency under coupling reparametrization shows that a finite renormalization-group beta function requires the two-loop double-pole consistency conditionThe leading two-loop pole is fixed by the one-loop simple pole; it is not an independent parameter. This is an instance of the simple-pole formulas for minimal-subtraction renormalization group functions and their higher-pole consistency relations.

The source's leading-pole truncation does not fix the finite two-loop coefficient of the beta function. For example, adding the omitted term $c_1{g}_0^3/ϵ$ leaves the double-pole condition unchanged but contributes $-2c_1{g}^3$ to the finite beta function. Thus cancelling the displayed $1/ϵ$ term is not a claim that the physical two-loop running vanishes.