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I'm interested in a QFT model featuring a fermion derivative coupling like $XX^* \chi^*\gamma^\mu∂_\mu \psi$ where X is some other field operator. Has anybody seen a paper containing something like this? How would Feynman-Graphs for such an interaction look like? Are derivatives of fermions ruled out by some consideration i am missing?

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Derivative couplings on Fermions are fine. Only issue is, as written, your interaction term violates Lorentz symmetry because it has only one spinor field... – QuantumDot Feb 16 '13 at 16:07 clarify: you need at least some other fermionic field, say $\bar{\chi}$, and $\bar{\chi} \partial \psi$ already has dimension 4, so you're out of luck if you're looking for an interacting perturbative CFT with such a term. – Vibert Feb 16 '13 at 17:28
... and although this isn't completely rigorous and you could use mixed SU(2) operators ($X^{\dot{a b} c}$ etc.), that'd only make things worse. – Vibert Feb 16 '13 at 17:52
thx, I obv. forgot the $\chi$ there on the first try. – NoEscape Feb 16 '13 at 23:27

In four dimensions, the interaction term you have written is expected to render your field theory non-renormalizable. This is OK as far as modeling low energy physics is concerned. But, there is another problem. The interaction is not Hermitian (which would lead not non-unitary time evolution).

Now, the Feynman rule that would be generated from this is a four-point interaction, involving the transformation of a spin-1/2 $\psi$ quanta into a $\chi$ quanta, with an associated emission of a $X$/$X*$ pair. The Feynman rule for such a vertex should be dependent on the momentum: $\text{vertex}=\gamma^\mu p_\mu$.

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hermitian: i have a habit of leaving out unneccesary i's – NoEscape Feb 17 '13 at 13:46
the vertex will add $p$ to a graph. so far so simple. but how does $\psi$ propagate? (e.g. a graph describing the interaction of two X via $\psi$ and $\chi$) The $p$ will cancel the fermionic propagator $S=1/p$, or am I mistaken there? – NoEscape Feb 17 '13 at 13:48
renormalization: don't worry about that :o) – NoEscape Feb 17 '13 at 13:54
Yes, this is true, provided the $\psi$ field is massless. – QuantumDot Feb 18 '13 at 3:18

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