Undergrad What Insights Can the Feynman Diagram of γ+γ→γ+γ Provide?

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The Feynman diagram for the process γ+γ→γ+γ consists of box diagrams with four photons at each vertex. There are six distinct diagrams that represent all possible topologies of the external lines and two orientations of the loop. Despite being superficially divergent, the four-photon vertex is actually convergent due to the Ward-Takahashi identities related to electromagnetic gauge invariance. It is crucial to combine all six diagrams to observe the cancellation of divergences. Understanding these aspects is essential for accurate analysis in quantum electrodynamics.
ghbeefcake
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Photon photon scattering diagram
If you have γ+γ→γ+γ what would the Feynman diagram look like (time-ordering implied).

I think it will be a square with four photons on each vertex but is this all there is to it or am I missing something?
 
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ghbeefcake said:
am I missing something?
You're missing quite a lot. :oldsmile:
 
In leading order (fourth order perturbation theory) you have indeed box diagrams. There are 6 diagrams of this kind with all topologies of the external lines and two orientations of the loop. The remarkable property of the four-photon vertex is that it is superficially divergent but in fact convergent due to the Ward-Takahashi identities of em. gauge invariance. It's important to add all the 6 diagrams together to see the cancellation of the divergence.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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