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Suppose you have an interaction term of the form

$$\mathcal{L}_{int} = \frac{h g}{3!}\phi^3\partial^2\phi$$ where $h $ and $g$ are both couplings. Now if I draw a diagram of the form given in the figure (please ignore the $\lambda$s and $J$'s.) Then suppose that the propagator is the field with the the derivative on: what would the amplitude be in this case? I know how to compute vertex factors, but this one confuses me. Any suggestions?

Had there be no propagator involved I would simply write $$S_{int} = i\frac{hg}{3!}\int d^4x \phi^3(x)\partial^2\phi(x)\\= i\frac{hg}{3!}\int d^4xDK\widetilde{\phi_{1234}}\exp[i(k_1+\dots +k_4)](ik_4)^2$$ which would then yield the vertex factor $V$ of (taking care of the $3!$ by permuting the three $\phi$'s) $$V = -i hg k_4^2.$$ I've taken all momenta as incoming into the vertex and $DK = \frac{d^4 k_1d^4 k_2d^4 k_3d^4 k_4}{(2\pi...2\pi)^4}$ the tilde is the fourier transform of each separate field.

But this is for a vertex, how about if there's a propagator involved?

enter image description here

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    $\begingroup$ What have you tried ? Best thing is to write down the vertex in Fourier space and everything should follow. $\endgroup$ – Adam Nov 20 '14 at 20:51
  • $\begingroup$ @Adam I've updated $\endgroup$ – Your Majesty Nov 20 '14 at 21:20
  • $\begingroup$ I'm kind of confused by the fact that one field is singled out by the derivative. You should try to rewrite the lagrangian density in a more symmetric way. $\endgroup$ – Adam Nov 20 '14 at 22:06
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Propergator do not have derivative. The interaction term is the vertex in Feynman diagram.

In the following I use the notation convention of Sredniski. Peskin's convention would cause some addtional minus sign.

For example, $\phi ^3 \partial^2 \phi$, the vertex is $\langle 0| \phi ^3 \partial^2 \phi | k_1 k_2 k_3k_4 \rangle = \partial^2_4 \langle0| \phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4)| k_1 k_2 k_3k_4 \rangle = \partial_4^2 (e^{ik_1x_1+ik_2x_2+ik_3x_3+ik_4x_4}+pertumations)|_{x_i =x} = -3! (\sum k_i^2) e^{i(k_1+k_2 + k_3 +k_4) x}$

So, the vertex in momentum space is $-ig/ 3! \cdot -3! (\sum k_i^2) = ig (\sum k_i^2)$.

The propagator will not be affected as it is completely determined by the free field Lagrangian. So $D_f = (\partial^2 + m^2)^{-1}$, in momentum space, $\frac{1}{k^2 + m^2}$

So if you have a derivative coupling you will have a different vertex, but other things are the same with usual $\phi^4$ coupling.

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  • $\begingroup$ Please see my updated question. $\endgroup$ – Your Majesty Nov 20 '14 at 21:14
  • $\begingroup$ Sorry I don't understand what is your question? Why are you confused by the amplitude? Could you please write down how you derive the amplitude? (BTW mabe you make a mistake in deriving the vertex..) $\endgroup$ – Deliang Zhong Nov 20 '14 at 21:31

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