 Solution

Let $m_A=ef$, $r=1+h/f$, and $H=\partial \cdot A-\xi m_A{r}^2\pi$, so the given $G=H/\sqrt{\xi }$ contributes ${\mathcal{L}}_{\mathrm{gf}}=-H^2/(2\xi )$. Its quadratic cross term is $+m_A\pi \partial \cdot A$, which cancels $m_{A}A\cdot \partial \pi$ from the scalar kinetic term after integration by parts. The free vector and would-be Goldstone boson propagators are therefore diagonal; $\pi$ has gauge-dependent mass squared $\xi m_A^2$.

The field-dependent factor $r^2$ cancels more than just the constant-coefficient mixing, but a residual derivative interaction remains. Indeed,Thus the nonlinear R-xi gauge in polar Higgs variables removes quadratic mixing; it does not remove every interaction containing $A$ and $\pi$.

Use the gauge transformation in part b. Since $h$ is invariant,The factor $-1/(e\sqrt{\xi })$ in $\delta G/\delta \alpha$ is field independent and is absorbed into the partition function normalization. Exponentiating the remaining Faddeev-Popov determinant with the sign convention of part a givesThe ghost coupling in the Abelian Higgs model therefore survives even though the gauge group is Abelian: these Faddeev-Popov ghosts interact with $h$ and do not decouple for generic $\xi$. The distinction from part a is the scalar dependence of the gauge fixing function. In the $\xi \to 0$ Landau gauge limit these particular ghost interactions vanish.