Past exam of the quantum field theory course of the University of Oxford / 2017 / 3

A free contraction must connect one $\varphi$ and one ${\varphi }^{\ast }$ field. In momentum space,Represent the nonzero propagator by a directed line so the two conjugate ends remain distinguishable; an undirected line alone would incorrectly allow same-charge contractions. Choose one consistent arrow convention, for example pointing from the starred end to the unstarred end.

The interaction contains two identical $\varphi$ fields and two identical ${\varphi }^{\ast }$ fields. Their $2!2!$ assignments cancel the coefficient $1/4$, givingThe vertex has two incoming and two outgoing arrows. Include $(2\pi {)}^d{\delta }^d(\sum p_i)$ at every vertex, integrate $d^{d}k/(2\pi {)}^d$ for each independent loop, and divide by the oriented graph's symmetry factor. There is no fermion-loop sign.

The two real fields have canonical normalization becauseThereforeThe interaction is invariant under $O(2)$ rotations and reflections of $({\varphi }_1,{\varphi }_2)$. The continuous rotations are the original complex-field $U(1)$ symmetry. The factor $1/16$ follows from both the original $1/4$ and the square of the $1/2$ in ${\varphi }^{\ast }\varphi$.

The propagator carries a real-species label rather than a charge-flow arrow:Differentiating the interaction with respect to its four external real fields gives the compact tensor ruleThus four ${\varphi }_1$ legs or four ${\varphi }_2$ legs give $-3i{\lambda }_0/2$, while two of each give $-i{\lambda }_0/2$. Vertices with an odd number of either species vanish. For example, $4!({\lambda }_0/16)=3{\lambda }_0/2$, whereas $2!2!(2{\lambda }_0/16)={\lambda }_0/2$. These factors already account for assignments at one vertex; graph symmetry factors must still be included separately. Momentum conservation and loop measures are the same as in part a.

Use renormalized propagators of mass $m$ and tree vertex $-ig{\mu }^{ϵ}$. The interaction counterterms have Feynman factors $-i{\delta }_{m^2}$ for a two-leg mass insertion and $-i{\mu }^{ϵ}g{\delta }_g$ for a four-leg insertion. At one loop the two-point one-particle-irreducible vertex isThe directed tadpole has $S=1$: interchanging its two ends would reverse the field labels, so the real-scalar tadpole's factor $1/2$ does not apply. It is independent of external momentum, so no wave-function counterterm is required at this order.

For the four-point vertex, label the two unstarred incoming fields by $1,2$ and the starred ones by $3,4$, with all momenta incoming. The tree contact, the three oriented bubbles and the contact counterterm giveThe first bubble has two equivalent arrows along the same direction between its vertices, giving $S=2$. In the other two channels the arrows oppose one another, and interchanging the lines does not preserve their oriented attachments, giving $S=1$. This is the oriented complex-scalar quartic graph factors rule.

The mass tadpole is order $g{m}^2$, while the four-point loop is order $g^2$. In the multiplicative coupling convention of the source,The distinction between ${\delta }_g$ and the whole additive four-leg coefficient $g{\delta }_g$ matters for the requested order counting.

Source correction: equation 11 on the original page has $\Gamma (\alpha -1/2)$. The dimensionally regulated integral instead contains $\Gamma (\alpha -d/2)$. This is not an interchangeable convention: for $\alpha =2$ and $d=4-ϵ$, the correct factor $\Gamma (ϵ/2)$ has a logarithmic ultraviolet pole, whereas the printed $\Gamma (3/2)$ is finite.

To obtain the correct expression, shift the loop momentum and perform a Wick rotation. For positive Euclidean ${\mathcal{M}}^2$, introduce a Schwinger parameterization:The derivation first holds in the convergent range and defines the result elsewhere by analytic continuation. For integer $\alpha$, rotating back giveswith the branch inherited from $i0$. This is the Lorentzian continuation of the Euclidean massive loop integral.

Set $p=0$, $\alpha =1$, $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,Consequently the directed tadpole in part d givesThe counterterm is $-i{\delta }_{m^2}$, so cancelling only the pole in the minimal subtraction scheme requiresThis is the quartic complex-scalar mass renormalization result. The scale factors and logarithms contribute finite terms at this order and do not change the displayed pole coefficient. A real-field check gives a tadpole coefficient $\frac{1}{2}(3g/2+g/2)=g$, in agreement with the oriented complex-field computation.

To first order, invert $m_0^2=m^2[1+g/(8{\pi }^2ϵ)]$ and use $g={\lambda }_0{\mu }^{-ϵ}+O({\lambda }_0^2)$:The remainder is a perturbative statement at fixed regulator. When differentiating, hold the bare mass $m_0^2$ fixed as well as ${\lambda }_0$; otherwise the mass running would not be defined by a fixed bare theory. Since $\mu {\partial }_{\mu }{\mu }^{-ϵ}=-ϵ{\mu }^{-ϵ}$,Thus, with the source's extra factor of one half,The factor $-ϵ$ from the running dimensionless bare coupling cancels the counterterm pole. Differentiating as though $g$ were independent of $\mu$ would miss this finite result. Some conventions call $\mu d\log m^2/d\mu$ itself the mass anomalous dimension; that convention is twice the one defined here.