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

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-301/3/a/solution