From P&S consider the $\phi^4$ bare Lagrangian: $$\mathcal{L} = \frac12 (\partial_\mu \phi)^2 - \frac12 m_0^2\phi^2 - \frac{\lambda_0}{4!} \phi^4.\tag{p.323}$$

When using renormalized perturbation theory we rewrite this Lagrangian using physical quantities and counterterms: $$\mathcal{L} = \frac12 (\partial_\mu \phi_r)^2 - \frac12 m^2\phi_r^2 - \frac{\lambda}{4!} \phi_r^4 + \frac12 \delta_Z(\partial_\mu \phi_r)^2 -\frac12 \delta_m \phi_m^2 - \frac{\delta_\lambda}{4!} \phi_r^4.\tag{10.18}$$

The first three terms follow the usual momentum space Feynman rules: each line with momentum $p$ corresponds to the propagator $$\frac{i}{p^2 - m^2 + i\epsilon}\tag{p.325}$$ and each interaction vertex corresponds to a factor of $$-i\lambda.\tag{p.325}$$ Similarly I can convince myself that the interaction vertex of the last term $\frac{\delta_\lambda}{4!} \phi_r^4$ corresponds to a factor of $$-i\delta_\lambda.\tag{p.325}$$

However what is less obvious to me is why the fourth and fifth terms, $\frac12 \delta_Z(\partial_\mu \phi_r)^2 -\frac12 \delta_m \phi_m^2$, get treated together and whose vertex corresponds to a factor of $$i(p^2\delta_Z - \delta_m)\tag{p.325}.$$

Why are these two terms treated as one and how would one see that the corresponding contribution to a Feynman is a line with a single vertex whose value is $i(p^2\delta_Z - \delta_m)$? I think the derivative term is what is confusing me, why does the derivative pull down a factor of the momentum?

  • 2
    $\begingroup$ These are just the standard Feynman rules. Which textbook? Which page? $\endgroup$
    – Qmechanic
    Commented Mar 22 at 4:52
  • $\begingroup$ @Qmechanic This is from Peskin & Schroeder, pages 324-325. I understand the usual Feynman rules, but I'm confused about the term $\frac12 \delta_Z(\partial_\mu \phi_r)^2 -\frac12 \delta_m \phi_m^2$ and why this is treated as a single contribution to the Feynman diagram. Also, how does the derivative term pull down a factor of the momentum, i.e. why does the term $(\partial_\mu \phi_r)^2$ give a contribution $p^2$? $\endgroup$
    – CBBAM
    Commented Mar 22 at 19:18

1 Answer 1


Why are these two terms treated as one

You don't have to treat them as a single contribution. It is perfectly acceptable to consider two different vertices with factors of $ip^2\delta_Z$ and $-i\delta_m$. However, because those two vertices have the same number of external lines and the same order in perturbation theory, each time there is a graph with one present, there will be another graph with it replaced by the other vertex. Taking this into account, you see that the results are the same whether you consider those two vertices or a single one whose Feynman weight is $i(p^2\delta_Z-\delta_m)$

Why does the derivative pull down a factor of the momentum?

In short, this is because we are working in momentum space, having performed a Fourier transform from position space. To see this, let us go over the derivation of Feynman rules. In position space, the Feyman rules organize the perturbative expansion of the $n$-point time-ordered correlation functions as a sum over diagrams. One derivation of position space Feynman rules starts with the Dyson series : $$\langle \phi(x_1)\ldots\phi(x_n)\rangle = \sum_{m=0}^{+\infty} \frac{(-i)^m}{m!} \int \text dy_1\ldots \text dy_m \Big\langle \phi(x_1)\ldots \phi(x_n) \mathcal L_{\rm{int}}[\phi](y_1)\ldots \mathcal L_{\rm{int}}[\phi](y_m)\Big\rangle_0$$ Then, you expand $\mathcal L_{\rm{int}}$ as a sum of monomials and apply Wick's theorem to obtain an expression for the correlation function as a sum over Feynmann diagrams.

The S-matrix elements are defined in momentum space, so you want to compute the Fourier transform of the correlation functions above. To do this, you do a Fourier transform on the variables $y_1,\ldots, y_m$. When you do this, each vertex will be associated with the corresponding term in the Fourier transform of the interaction Lagrangian.

Back to OP's question : the interaction Lagrangian contains a term $\mathcal L_{\rm{int}} \ni \frac12 \delta_Z(\partial_\mu \phi_r(x))^2 $. Its Fourier transform is $-\frac12 \delta_Zp^2 \phi(p)^2$

  • $\begingroup$ Thank you. Unfortunately it seems hft has deleted their answer before I could see it. $\endgroup$
    – CBBAM
    Commented Mar 24 at 3:10
  • $\begingroup$ I added a paragraph to quickly go over the derivations of Feynman rules in position space and the Fourier transform to get the momentum space rules $\endgroup$ Commented Mar 24 at 17:52
  • $\begingroup$ Thank you very much! This was very helpful! $\endgroup$
    – CBBAM
    Commented Mar 24 at 18:17

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