Feynman rule involving derivative of a field.

1. Jul 21, 2008

arroy_0205

Can anybody help me by explaining how to write down the Feynman rule corresponding to a term like
$$ieA^{\mu}\phi^{\ast}\partial_{\mu}\phi$$
I am confused about the presence of the derivative term. Terms have usual meaning.

2. Jul 21, 2008

CompuChip

Usually, when you have a term like
$$A^\mu \phi^* \phi$$
you would take the Fourier transform
$$A^\mu(x) = \int dk A^\mu(k) e^{-i k \cdot x}, \qquad \phi(x) = \int dk \phi(k) e^{-i k \cdot x}, \qquad \phi^*(x) \int dk \phi^*(k) e^{i k \cdot x}$$
Then you plug this in and you get the Feynman rule for the interaction. I assume you have seen this calculation and you know how to do it (if not, I suggest you review it - it's much more handy to know how to do the calculation than to remember what all the Feynman rules look like).

In this case, you can do exactly the same. You will end up having the derivative act on the phi field, which will produce an extra factor $-i k_\mu$. Usually, this is indicated in the graphical rule by drawing a double line for the field which the derivative acts on (in this case, the phi field).

3. Jul 21, 2008

arroy_0205

Thanks for the reply. However I wondered is there any significance of the derivative itself? Normally I have seen product of fields to represent interaction vertex and inverse of quadratic part of field related to propagator; that way is the first derivative of any special significance?

4. Jul 21, 2008

CompuChip

As I said, it introduces a term with $k_\mu$. So in a sense, its special significance is that it makes the coupling dependent on the momentum of the field. Or, in the non-interaction term $\partial^\mu \phi^* \partial_\mu \phi$ you get a $k_\mu k^\mu = k^2$ dependence, which indicates that it's a kinetic term.