 Solution

The quadratic part of the covariant-derivative term contains $g^2{v}^2{A}_{\mu }{A}^{\mu }/2-gv{A}_{\mu }{\partial }^{\mu }b$. Its mixing term cancels the $+gv{A}_{\mu }{\partial }^{\mu }b$ in the stated gauge fixing term. Thus the vector mass and diagonal quadratic action areFor nonzero $p^2$ introduce the transverse and longitudinal transverse and longitudinal momentum projectors $P_T^{\mu \nu }=g^{\mu \nu }-p^{\mu }{p}^{\nu }/p^2$ and $P_L^{\mu \nu }=p^{\mu }{p}^{\nu }/p^2$. After integration by parts, the inverse quadratic kernel isTheir orthogonality makes the inversion immediate. The gauge-boson propagator isEquivalently,The projector formula explains the physical massive pole and the gauge-dependent longitudinal pole. The apparent $1/p^2$ in the projectors cancels in the equivalent expression; it is not an extra massless physical state. At $\xi =1$ the numerator reduces to $g^{\mu \nu }$. At fixed momentum the $\xi \to \infty$ limit gives the unitary gauge propagator $-i(g^{\mu \nu }-p^{\mu }{p}^{\nu }/M_A^2)/(p^2-M_A^2+i0)$. The relation between this limit and ultraviolet power counting is not uniform in momentum.^[ref]