1
$\begingroup$

I know how to treat Feynman interactions without derivatives by Wick contraction. But now, take for example $$\mathcal{L}_{int}=\lambda \phi (\partial_{\mu}\phi)(\partial^{\mu}\phi).$$

Now many books write that in momentum space the derivatives turn into momenta. While I can imagine this happening, I don't really know how to write this down explicitly. At what point do I consider the Fourier transform of the field? Am I still using Wick contractions, but now with the field depending on the momenta? I have not found a source doing this explicitly.

$\endgroup$
1
$\begingroup$

For your case, starting with this interaction term, let us substitute the expansion of $\phi$ in Fourier modes: $$ \phi = \sum_k \phi_k e^{i kx} $$ The action of derivative produces a factor of $ik$. Then, in the action you sum(integrate) over all $x$ : $$ \sum_x \sum_{k_1, k_2, k_3} (ik_{2 \mu}) (ik^{3 \mu}) \lambda \phi_{k_1} \phi_{k_2} \phi_{k_3} e^{i (k_1 + k_2 + k_3) x} = \sum_{k_1, k_2, k_3} (ik_{2 \mu}) (ik^{3 \mu}) \lambda \phi_{k_1} \phi_{k_2} \phi_{k_3} \delta (k_1 + k_2 + k_3) $$ Where in the last expression we have employed the well-known integral for exponent. The change from derivatives to momenta is simply results of change from positional basis, to momentum basis and has nothing to do with Wick theorem.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ And how do I get the Feynman rules from there? $\endgroup$ – korni1990 Aug 12 at 14:16
  • $\begingroup$ @korni1990 in momentum space the vertex with momenta $k_1, k_2, k_3$, which we will take to be outcoming, to each vertex one has to assign $-\lambda (k_2 \cdot k_3) + \text{permutations}$., where the conservation of momenta has to be imposed $\endgroup$ – spiridon_the_sun_rotator Aug 12 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.