B. Determination of the propagator

To determine the propagator in the covariant $\alpha$-gauge, ${{\mathcal{D}} ^{-1}} ^{\mu \nu} (k ; \alpha)$, with the four dimensional quantities that are at our disposal we consider the ansatz:

$$A \left( g ^{\mu \nu} - \frac{k ^{\mu} k ^{\nu}}{k ^{2}} \ri... ...bar{k} ^{\nu} + D \frac{k ^{\mu} k ^{\nu}}{(k ^{2}) ^{2}} \,.$$

To determine the coefficients $A$, $B$, $C$, $D$ we can now compute ${{\mathcal{D}} ^{-1}} ^{\mu \nu} ( k ; \alpha ) {{\mathcal{D}} ^{-1}} _{\nu \alpha} ( k ; \alpha )$:

$${\mathcal{D}} ^{-1 \, \mu \nu} {\mathcal{D}} _{\nu \alpha} = \left[ \left( k ^{2} g ^{\mu \nu} - k ^{\mu} k ^{\nu} \right) - \... ...\mu} \bar{k} ^{\nu} \right) + \frac{1}{\alpha} k ^{\mu} k ^{\nu} \right] \times$$

$$\times \left[ A \left( g _{\nu \alpha} - \frac{k _{\nu} k _{\alph... ..._{\nu} \bar{k} _{\alpha} + D \frac{k _{\nu} k _{\alpha}}{(k ^{2}) ^{2}} \right]$$

$$= A k ^{2} \delta ^{\mu} _{\alpha} - A k ^{\mu} k _{\alpha} - \frac... ...r{k} ^{\mu} \bar{k} _{\alpha} \right) + \frac{A}{\alpha} k ^{\mu} k _{\alpha} -$$

$$- \, A k ^{\mu} k _{\alpha} + A k ^{\mu} k _{\alpha} + \frac{\kap... ...}} \bar{k} ^{\mu} k _{\alpha} \right) - \frac{A}{\alpha} k ^{\mu} k _{\alpha} +$$

$$+ \, B k ^{2} \bar{\delta} ^{\mu} _{\alpha} - B k ^{\mu} \bar{k} ... ...{\mu} \bar{k} _{\alpha} \right) + \frac{B}{\alpha} k ^{\mu} \bar{k} _{\alpha} +$$

$$+ \, C k ^{2} \bar{k} ^{\mu} \bar{k} _{\alpha} - C \bar{k} ^{2} k... ... _{\alpha} \right) + \frac{C}{\alpha} \bar{k} ^{2} k ^{\mu} \bar{k} _{\alpha} +$$

$$+ \, D \frac{k ^{\mu} k _{\alpha}}{k ^{2}} - D \frac{k ^{\mu} k _... ...u} k _{\alpha} \right) + \frac{D}{\alpha} \frac{k ^{\mu} k _{\alpha}}{k ^{2}} .$$

Now comparing terms with the same tensorial character

$$\delta ^{\mu} _{\alpha} : A k ^{2} = 1$$

$$k ^{\mu} k _{\alpha} : \frac{D}{\alpha} \frac{1}{k ^{2}} = A$$

$$\bar{\delta} ^{\mu} _{\alpha} : B k ^{2} - \frac{\kappa (A + B) \bar{k} ^{2}}{k ^{2} - m _{\mathrm{A}} ^{2}} = 0$$

$$k ^{\mu} \bar{k} _{\alpha} : - B - C \bar{k} ^{2} = 0$$

$$\bar{k} ^{\mu} \bar{k} _{\alpha} : C k ^{2} - \frac{\kappa ( B + A )}{k ^{2} - m _{\mathrm{A}} ^{2}} = 0 \,.$$

An independent subset of (four of) these gives (consistently with the remaining equation) the final result for the coefficients:

$$A = \frac{1}{k ^{2}}$$

$$B = \frac{\kappa \bar{k} ^{2} A} { k ^{2} \left( k ^{2} - m _{\mathrm{A}} ^{2} \right) - \kappa \bar{k} ^{2} }$$

$$= \frac{\kappa \bar{k} ^{2}} { k ^{2} \left( k ^{2} \left( k ^{2} - m _{\mathrm{A}} ^{2} \right) - \kappa \bar{k} ^{2} \right) }$$

$$C = - \frac{\kappa} { k ^{2} \left( k ^{2} \left( k ^{2} - m _{\mathrm{A}} ^{2} \right) - \kappa \bar{k} ^{2} \right) }$$

$$D = \alpha \,,$$

so that we can finally write

$${\mathcal{D}} ^{\mu \nu} ( k ; \alpha ) = \frac{1}{k ^{2}} \left( g ^{\mu \nu} - \frac{k ^{\mu} k ^{\nu}}{k... ... \left( k ^{2} - m _{\mathrm{A}} ^{2} \right) - \kappa \bar{k} ^{2} \right) } -$$

$$- \, \frac{\kappa \bar{k} ^{\mu} \bar{k} ^{\nu}} { k ^{2} \left( ... ...kappa \bar{k} ^{2} \right) } + \alpha \frac{k ^{\mu} k ^{\nu}}{(k ^{2}) ^{2}} .$$