# Gauge-invariant vertex structure for $h\to\gamma\gamma$ via fermion loop

I am struggling (a bit) with the following diagram for scalar Higgs to two photons.

$h\to\gamma\gamma$" />

If I put $$q_\mu$$ on-shell (or at the very least if I put both $$q_\mu$$ and $$q'_\nu$$ on-shell), the vertex function should have the following form: $$i\Gamma^{\mu\nu} \sim [\eta^{\mu\nu}qq' - q'^\mu q^\nu]$$ However, from the explicit calculation of the amplitude I have: $$i\Gamma^{\mu\nu} \sim \int \frac{d^4k}{(2\pi)^4} \frac{Tr\{(\not k - \not q' + m_f)\gamma^\nu (\not k + m_f)\gamma^\mu(\not k - \not q + m_f)\}}{[(k-q')^2-m_f^2][k^2-m_f^2][(k-q)^2-m_f^2]} \\ \sim \int_0^1 dx \int_0^{1-x} dy \frac{\eta^{\mu\nu}[m_f^2-x^2 q'^2 + (1-2xy)qq'] +q'^\mu[2x(2x-1)q'^\nu + (4xy-1)q^\nu]}{m_f^2-q'^2x(1-x)+2xyqq'}$$ From this result I can get the gauge invariant part, which agrees for instance with this article (New Barr-Zee contributions to $$(g−2)_μ$$ in two-Higgs-doublet models), but I'm left with additional terms that shouldn't be there. Even if I put $$q'^\nu$$ on-shell, I still get an additional term: $$i\Gamma^{\mu\nu} \sim [\eta^{\mu\nu}qq' - q'^\mu q^\nu] + \eta^{\mu\nu} m_f^2$$ Initially, I was hoping these additional terms might cancel with those from the diagram, with opposite fermion direction, but adding this diagram just results in an overall factor of 2.

The calculation was done just using standard Feynman parameters and then setting $$q^2=0$$ and ignoring terms $$\sim q^\mu$$ (and $$q'^\nu$$ for $$q'$$ on-shell). After shifting the integration momentum, the trace was evaluated using FeynCalc.

Please let me know if you need more details on the calculation.

Ok i made a pretty instructive mistake. Since the diagram is power-counting finite, i assumed it would be fine to take the momentum integral to be 4-dimensional from the start. Then at some point i used $$\int\frac{d^4k}{(2\pi)^4} \frac{4k^\mu k^\nu}{[k^2-\Delta]^3} = \int\frac{d^4k}{(2\pi)^4} \frac{k^2 \eta^{\mu\nu}}{[k^2-\Delta]^3}$$ However, this is wrong, since the resulting integral is divergent. Instead i need to stay in D-dimensions which means $$\int\frac{d^4k}{(2\pi)^4} \frac{Dk^\mu k^\nu}{[k^2-\Delta]^3} = \int\frac{d^4k}{(2\pi)^4} \frac{k^2 \eta^{\mu\nu}}{[k^2-\Delta]^3}$$ Then all the gauge-variant terms magically drop out from the calculation