{wiki}

A kinetic term contains derivatives of a field and determines its free propagation. Its quadratic differential operator becomes the inverse <quantum field theory propagator>.

Kinetic term

Varying the displayed gauge-fixed <Lagrangian density> gives the kinetic operator
$$
D^{\mu\rho}=\eta^{\mu\rho}\Box-\left(1-\frac1\alpha\right)\partial^\mu\partial^\rho.
$$
Its <quantum field theory propagator> should satisfy
$$
D_x^{\mu\rho}\Delta_{\rho\nu}(x-y)=i\delta^\mu_\nu\delta^{(4)}(x-y),
$$
and inversion into transverse and longitudinal projectors gives the numerator
$$
\Pi_{\mu\nu}^{\mathrm{correct}}
=-\eta_{\mu\nu}+(1-\alpha)\frac{k_\mu k_\nu}{k^2}.
$$

The paper instead prints $-\eta_{\mu\nu}+(\alpha-1)k_\mu k_\nu/k^2$. Except at $\alpha=1$, that is not the inverse of the displayed Lagrangian's kinetic operator. Taken literally, for $\alpha\ne2$ it is the Green function of
$$
\left[\eta^{\mu\rho}\Box+\frac{\alpha-1}{2-\alpha}
\partial^\mu\partial^\rho\right]\Delta_{\rho\nu}
=i\delta^\mu_\nu\delta^{(4)},
$$
and at $\alpha=2$ its longitudinal part is noninvertible. Thus the longitudinal sign in the printed propagator is a typographical error; the two forms coincide in <Feynman gauge>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 / 3 / a / Solution

The first two terms are the <kinetic term> and <mass term> of a <real scalar field>; together they determine the free <quantum field theory propagator>. The term $-\lambda\phi^4/4!$ is its quartic <field interaction term>. The remaining three terms are <counterterms>: $\delta_Z$ renormalizes the field normalization, $\delta_m$ renormalizes the mass, and $\delta_\lambda$ renormalizes the quartic coupling. Their regulator dependence cancels the ultraviolet divergences of loop diagrams, while their finite parts implement the chosen <renormalization conditions>.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 1 / a / Solution

Let $K_M=-\partial^2+M^2$ and let $\Delta_M=K_M^{-1}$ be its Euclidean <quantum field theory propagator>. With the source-sign convention suited to the expression in the question, define
$$
Z_0[J]=N_0\int\mathcal D\chi\,
\exp\left[-S_2[\chi]-\int d^dx\,J(x)\chi(x)\right]
=\exp\left[\frac12\int d^dx\,d^dy\,
J(x)\Delta_M(x-y)J(y)\right],
$$
where $N_0$ makes $Z_0[0]=1$. Since inserting $\chi(x)$ is equivalent to acting with $-\delta/\delta J(x)$, the <integrating out a field>[integral over $\chi$] gives
$$
W[\phi]=S_1[\phi]-\log\left.
\left\{\exp\left[-S_3\left(\phi,-\frac{\delta}{\delta J}\right)\right]Z_0[J]\right\}\right|_{J=0}.
$$

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 3 / b / i / Solution

{wiki=Propagator}

A propagator is a time-ordered two-point function and a Green function for the quadratic kinetic operator with an $i\epsilon$ boundary prescription.


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 Incoming links: Quantum field theory propagator