# Photon-photon scattering

I was reading the section on photon-photon scattering in Lifshitz-Berestetski-Pitaevskii's book on QED (Section 127). Here, they write the total scattering amplitude as $$M_{fi} = e^{\lambda}_{1} e^{\mu}_{2} e^{\nu}_{3} e^{\rho}_{4} M_{\lambda \mu \nu \rho},$$ where $e^{\mu}_{i}$ is a polarization vector, and $M_{\lambda \mu \nu \rho} = M_{\lambda \mu \nu \rho}(k_{1},k_{2},k_{3},k_{4})$ is the photon-photon scattering tensor. Later on, they state the following:

... Because of the gauge invariance, the amplitude $M_{fi}$ is unchanged when $e$ is replaced by $e + \text{constant} \cdot k$. Thus, we have $$k_{1}^{\lambda} M_{\lambda \mu \nu \rho} = k_{2}^{\mu} M_{\lambda \mu \nu \rho}= \cdots = 0.$$ It is easily deduced from this that, in particular, the expansion of the scattering tensor in powers of the $4$-momenta $k_{1}$, $k_{2}$, $\cdots$ must begin with terms containing quaternary products of the components, and certainly $$M_{\lambda \mu \nu \rho}(0,0,0,0) = 0.$$

What does this last paragraph actually mean? Also, I had the impression that the condition $k_{1}^{\lambda} M_{\lambda \mu \nu \rho} = k_{2}^{\mu} M_{\lambda \mu \nu \rho}= \cdots = 0$ could be useful when calculating the total cross-section, but I can't see why, can anybody help me understand this?

$$(e^{mu} + c k^{mu}) M_{mu} = e^{mu} M_{mu}$$ so $$k^{mu} M_{mu} = 0.$$
The other bit, $$M(0,0,0,0) = 0$$ seems obvious - photons with no momenta have no scattering cross section.