Photon-Photon-scattering (Feynman diagram) The feynman diagram for the Delbrück-scattering (photon-photon-scattering) in the lowest order looks like as in the picture: 

But why is the following diagram equal to one of the upper three?
There will be six possibillity to order the indices in the loop, but there are only three independent diagrams. Why?

 A: There are three inequivalent permutations of the external legs (with the same orientation of the loop momenta) which give rise to the three independent diagrams that you have drawn. Start off by labelling each of the external photons by $1,2,3$ and $4$ say. Then the possible configurations are given by, for example, sandwiching $1$ between $2$ and $4$, $1$ between $2$ and $3$ and $1$ between $3$ and $4$. This gives the following,

The $4$-tuplets $1432, 1243$ and $1324$ can all be obtained from each other by even permutations, as must be the case for orientation preserving loop momentum.
Another set of three diagrams, belonging to the equivalence class of diagrams similar to that of the above but with the loop momenta reversed may also be considered. Those again are related by even permutations but differ from the ones above by an odd permutation, again to be expected.
The diagram you propose to be different is in fact the same as the diagram far left in the OP. This can be seen by labelling the external momenta, choosing an orientation for the loop momenta and writing out its $4$-tuplet, as defined above. You will see this coincides with either $1432$ or $1234$, corresponding to the choice of clockwise and anticlockwise loop momenta respectively. Intuitively, this amounts to a 'twisting' of the diagram, thereby sending the loop momentum $\ell \rightarrow - \ell.$
At the level of the one loop amplitude for $\gamma \gamma \rightarrow \gamma \gamma$ , all six diagrams should be considered.
A: Its the same as the one on the left.
Recall in an integral how you can integrate $$\int_{t=0}^{t=1}t^2dt=\frac{1}{3}$$ or we can integrate $$\int_{x=0}^{x=1}x^2dx=\frac{1}{3}$$ and it's really the same thing? And not just because they are both 1/3? They are the same because they are instructions to take the interval from zero to one and break it into pieces and pick a value in each piece and square it and multiply that result by the length of the piece and add it up for each piece and then to keep doing that for a series of finer and finer chopped up meshes on zero to one until we find out what number the sequence approaches (which is 1/3) and the name we assign to the dummy index was totally and 100% irrelevant.
Yeah, that. The same thing is going on in a Feynman diagram. You are really doing perturbation theory and the diagram corresponds to an integral and the internal lines are just to tell you the bounds and ranges like the interval does. Just like the $\int_{t=0}^{t=1}$ and the $\int_{x=0}^{x=1}$ were telling you the bounds in 1d. Buts it's more like in 2d or 3d where your cutoff in x depends on y in an iterated integral in 2d. But really you could choose to use a rotated coordinate system or even polar coordinates in 2d and have lots of choices but the integral is the same. Chop in pieces evaluate add up, take the limit.
In this case you are really trying to identify each possible energy and momentum that the internal lines can have that allow them to conserve energy and momentum at each vertex and line up to the fixed external lines.
So the internal line is supposed to correspond to any direction that works. So the last one is just like the first one on the left. They both say you can have an internal line that goes between the bottom left and the upper left and another one that goes between the upper left and the upper right and a third one that goes between the upper right and the lower left and a fourth one that goes between the lower left and the upper right. So they both stand for the family of all such possibilities. So they are as identical as $\int_{t=0}^{t=1}$ and $\int_{x=0}^{x=1}$ and they are identical for the same reason.
