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?


1 Answer 1


The gauge invariance argument is dealt with in section 4 (§4) of the book, it basically is just saying the polarisation is perpendicular to the momentum of the photon (I think).

As for the rest of that, by the gauge invariance,

$$ (e^{mu} + c k^{mu}) M_{mu} = e^{mu} M_{mu} $$ 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.

  • $\begingroup$ Thanks, the bit about gauge invariance wasn't my problem. What I wanted to know is if this result makes the computation of the cross section easier, as it seems rather involved (I can't get an exact result after pages of calculations, and I was wondering if it could lead to discarding terms of the amplitude). As for the other result, I thought I was missing something since, as you mention it, it is rather obvious. Anyways, thank you very much! $\endgroup$ Commented Sep 29, 2016 at 4:33
  • 1
    $\begingroup$ @gradStudent I would probably look at the B. De Tollis paper mentioned in the book. $\endgroup$ Commented Sep 29, 2016 at 4:35

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