# Connected graphs and phi^4 interacting theory

1. Feb 4, 2016

### CAF123

I am considering the $\psi^4$ interacting theory and a lagrangian of the form $$\mathcal L = \frac{1}{2}Z_{\psi} \partial^{\mu} \psi \partial_{\mu} \psi - \frac{1}{2}Z_m m^2 \psi^2 - \frac{1}{4!}Z_g g \psi^4$$ with generating functional for the interacting theory $$Z(J) = \int \mathcal D [\psi] \exp(i \int d^4 x(\mathcal L_o + \mathcal L_I + J\psi))$$ which is then written like $$\exp\left(i [\int d^4 y \frac{\delta}{i\delta J(y)} (-\frac{1}{2}(Z_{\psi}-1)\partial^2 - \frac{1}{2}(Z_m -1) m^2) \frac{\delta}{i\delta J(y)} - \frac{Z_g g}{4!}\left(\frac{\delta}{i\delta J(y)}\right)^4]\right) Z_o(J)$$

My notes then go onto say that the feynman rules can be easily deduced from this. I am wondering how so? I realise that the part in brackets between the two derivatives is like a propagator term sandwiched between two source terms and the last term is like a vertex rule, but I am not sure how to formally identify or extract the rules.

Thanks!

2. Feb 4, 2016

3. Feb 4, 2016

### CAF123

I see, thanks. But I am wondering, in my case, I have these renormalised scales. Do they give rise to further feynman rules? In $$Z_o(J) = \exp \left[-\frac{1}{2} \int d^4 x \int d^4 x' J(x) D_F(x-x') J(x')\right],$$ this would give rise to a free theory propagator represented pictorially with a line connecting two source vertices at positions x and x' which are subsequently integrated over. What makes it clear what, in the interacting functional, the term $$(-\frac{1}{2} (Z_{\psi} - 1)\partial^2 - \frac{1}{2}(Z_m - 1) m^2)$$ sandwiched between two derivatives wrt the same source is? I think I remember my lecturer saying that it is obvious by looking at the term what it corresponds to.

Thanks!