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I have the Lagrangian density $$ \mathcal{L} = \frac{1}{2}\partial_\mu \phi_i \partial^\mu \phi_i - \phi_i A_{ij}^\mu \partial_\mu \phi_j, $$ where $\phi_i$ are real scalar fields and $A_{ij}^\mu$ is an external field antisymmetric under the exchange of $i$ and $j$. (The external field is given but does not ensures momentum conservation at each vertex --hence I used crossed dots in the diagram, see below; we still have overall momentum conservation, even though I do not think that this matters)

I want to compute the symmetry factor for the diagram

enter image description here

but I am not sure of how to proceed.

I have tried to consider naively the possible contractions of the amplitude, schematically written $$ \sim \int \phi \phi\int \phi \phi\int \phi \phi\int \phi \phi $$ that can give rise to such a diagram, but I am not confident at all with my argument. I pick the first contraction, let's say with the first field; I have 6 other fields to which I can contract it (the one in the same integral is not allowed), for instance (index = contraction, I don't know how to TeX them) $$ \sim \int \phi_a \phi \int \phi_a \phi\int \phi \phi\int \phi \phi. $$ Now I want to contract the other field of the first integral; I have 4 possibilities (can't use the 3rd field because would give a 2-point loop); one of those allowed is $$ \sim \int \phi_a \phi_b \int \phi_a \phi\int \phi_b \phi\int \phi \phi. $$ Ultimately there are two contractions left, that can be arranged in 2 different ways. So: $$6 \times 4 \times 2$$

But I have no idea if this kind of reasoning is correct. (Moreover I have a clue that the result should be 18 to match with further computations, but I am not 100% sure of this result)

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  • $\begingroup$ Could you include a drawing of the diagram you're trying to compute? I would say if you're following Wick's theorem, it should be fine. $\endgroup$ – MSha Dec 30 '16 at 22:09
  • $\begingroup$ @MSha right, diagram added. I am not sure on how to proceed, in particular because of the additional symmetry given by the antisymmetry of $A_{[ij]}$. $\endgroup$ – yoric Dec 31 '16 at 8:15
  • $\begingroup$ I don't think you have to worry because about the $A_{ij}$, since you can kind of think about it as $U(N)$ or $SU(N)$ gauge theory with bosons. However, instead of $\gamma_\mu$ you have $\partial_\mu$, so this breaks the symmetry between the left and the right bosonic fields when you contract (you have an oriented interaction). Hence, I would imagine you would have different types of box diagrams, though maybe only one class of them contributes. $\endgroup$ – Aaron Dec 31 '16 at 23:26
  • $\begingroup$ @Aaron Thanks for your comment. Yes I understand that there is an asymmetry between the two legs of the vertices, however I think (but maybe I am wrong) that indeed being the $A$ antisymmetric all diagrams with four vertices could be related up to a redefinition of the integrated momentum $\ell$; I am still unable to perform the computation, though. $\endgroup$ – yoric Jan 1 '17 at 10:52

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