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I have two simple questions about Feynman rules for Lagrangian that have a derivative of fields.

  1. For example for Lagrangian in this link part a with derivative couplings. Is the vertex depend on direction of momentum? If we have a vertex and then change the momentum direction does it change? (depend on incoming and outgoing direction.)

  2. And other simple question is that the symmetry factor of diagram is invariant under the change of interaction term and only depends on geometry of the diagram, is it right?for example if we change $1/4!$ of $\phi^4$ interaction to $1$, the same diagram have a same symmetry factor?

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  1. Yes, the Feynman rule (in momentum space) for interaction vertices with derivatives may in general depend on the orientation of the $4$-momentum of its legs. (If there is an even number of derivatives, the orientation doesn't matter.) Be aware that different consistent conventions exist in the literature.

  2. Yes, the symmetry factor of a Feynman diagram does by definition not depend on the normalization of the coupling constant. On the other hand, (the value of) a Feynman diagram will in general obviously depend on the coupling constant and its normalization. Concerning vertex factors, see also this related Phys.SE post.

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  • $\begingroup$ so can I say that for the Lagrangian in the link(suppose we have one derivative of $\phi$ and other derivative is for another scalar field) we can use plus sign for incoming and minus for out going momentum for vertex? $\endgroup$
    – a.p
    Commented May 8, 2020 at 11:12
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    $\begingroup$ In principle yes, if your conventions are otherwise consistent. $\endgroup$
    – Qmechanic
    Commented May 8, 2020 at 11:15
  • $\begingroup$ and for part 2 , the standard coupling constant is good becuase we can compute the symmetry factor from Taylor series and count number of contractions without see the geometry of diagram.if we have combination of scalar fields like $\phi . \eta ^2$ we can put 1/2 behind it and compute symmetry factor in usual way from Taylor series without see the diagram? $\endgroup$
    – a.p
    Commented May 8, 2020 at 11:21
  • $\begingroup$ It sounds like that calculation will depend on the normalization of the coupling constant. $\endgroup$
    – Qmechanic
    Commented May 8, 2020 at 11:36

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