Why does gauge invariance imply zero one-loop photon and Z corrections at q²=0?

[Worldsheep]

Does anyone know a simple explanation for the following statement:

Gauge invariance ⇒ $Π^μν_ϒϒ(0) = Π_μνϒZ(0) = 0$

Where Π_VV' is the V to V' one loop correction, ϒ is the photon field and Z is the Z-boson field. The argument of Π is the incoming momentum q² = 0

 

[vanhees71]

Your photon propagator is
$$D_{\mu \nu}(q)=-\frac{1}{q^2-q^2 \Pi(q^2)+\mathrm{i} 0^+}[g_{\mu \nu}-q_{\mu} q_{\nu}] + A(q^2) q^{\mu} q^{\nu},$$
where $A(q^2)$ is a gauge-dependent non-interacting piece, which doesn't enter any physical result.

Now the photon has strictly 0 mass. Together with the Ward-Takahashi identity of the photon polarization tensor, which makes it purely 4-transverse, this implies that
$$\Pi_{\mu \nu}=q^2 \Pi(q^2) (g_{\mu \nu}-q_{\mu} q_{\nu}).$$
$\Pi$ is a logarithmically divergent scalar. Now to make the residuum of the photon propgator 1 at $q^2=0$, you impose the renormalization condition
$$\Pi(q^2=0)=0.$$
The same argument holds for the $\gamma$-Z mixing piece too.

Note that the above renormalization condition is dangerous with regard to infrared divergences, which must be resummed. For this purpose it's better to choose the renormalization point in the space-like.