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Suppose we have a Lagrangian with a fermion of mass $m$ and a 4-point interaction of the form

$$\mathcal L=\bar{\psi}(i\partial\!\!\!/-m)\psi-g\bar{\psi}\partial\!\!\!/\psi\bar{\psi}\partial\!\!\!/\psi$$

where $g$ has mass dimension $-4$. I'd like to determine the Feynman rule for this vertex, and the usual procedure is differentiating the action

$$i\frac{\delta^4 S}{\delta\psi\delta\bar\psi\delta\psi\delta\bar\psi}$$

and then transforming into momentum space, which gives for a vertex like

4-point vertex

a rule of

$$ig(p\!\!\!/_{1\,ac}p\!\!\!/_{2\,bd}+p\!\!\!/_{1\,bd}p\!\!\!/_{2\,ac}-p\!\!\!/_{1\,ad}p\!\!\!/_{2\,bc}-p\!\!\!/_{1\,bc}p\!\!\!/_{2\,ad}).$$

Great, this follows the Pauli principle and all that jazz. But the problem is, if I had differentiated the action in some other order, maybe

$$i\frac{\delta^4 S}{\delta\psi\delta\bar\psi\delta\bar\psi\delta\psi},\,\mathrm{or}\,i\frac{\delta^4 S}{\delta\bar\psi\delta\psi\delta\bar\psi\delta\psi},$$

the Feynman rule would have a different sign.

If I had different interactions in the Lagrangian (maybe I added that interaction to QED), which order of differentiation should I use to make things consistent?

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1 Answer 1

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It will sort out when you calculate observables or any quantity. If you take the functional derivative with respect to a specific order, this is the order of the operators in vertex that you calculate. Then, when you do any expansion or calculating S-matrix or whatever, and you come upon a vertex, then the order of the operators in the vertex should match the order of the operators that you evaluated for the diagram. If not - reorder them. That's why it is convenient to keep the order as it appears in the action, as it will usually be the order in which it will appear in the different expansions.

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