Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 1 / v / Solution

Use the Feynman parameters and $J_3$ defined in Solution. With $A=4C$, the finite term just found is $-C\Delta J_3$. Combining it with the remaining $g_{\mu \nu }$ coefficient givesThe other coefficient is $(B-4Cxy)J_3$. Thus, writing ${\mathcal{I}}_{\mu \nu }$ for the Feynman integral without the overall prefactor,For $T_{\mu \nu }=\frac{s}{2}{g}_{\mu \nu }-p_{2,\mu }{p}_{1,\nu }$, the two coefficients must have the corresponding ratio. The mismatch is simply $(B-C)J_3$, independent of $xy$. Therefore the required relation iswhich is satisfied. In the primitive normalization the result isand the full top quark scattering amplitude, in our stated $S$ convention, isIndeed $p_1^{\mu }{T}_{\mu \nu }=p_2^{\nu }{T}_{\mu \nu }=0$: replacing either polarization vector by its photon momentum makes the scattering amplitude vanish. This is the Ward identity, expressing electromagnetic gauge invariance. The result also is symmetric under exchange of the two photons. The finite term in Solution is essential for this Ward identity.

As an independent low-energy check, for $s/m^2\to 0$ the Feynman parameter integral isThe coefficient is consequently $-N_{c}y{e}^2{Q}^2/(6{\pi }^{2}m)$ at leading order. It has the inverse-mass dimension required for an effective $h{F}_{\mu \nu }{F}^{\mu \nu }$ coupling. If $y=m/v$, this coefficient stays finite in the large-$m$ limit at fixed $v$; that limit increases the Yukawa coupling, so it is not a fixed-coupling heavy-field decoupling limit.