In Peskin & Schroeder Problem 10.3 pg. 345 they renormalize the field in $\phi^4$ theory using the following 2-loop sunset diagram.

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When looking at the correlation function $G^{(2)}_0$ this would seem the obvious thing to do since:

$$G_0^{(2)}=\frac{1}{p^2+m_0^2}-\frac{\lambda_0}{(p^2+m_0^2)^2}\int \frac{d^Dk}{(2\pi)^D}\frac{1}{k^2+m_0^2}$$ $$+\frac{\lambda_0^2}{3!(p^2+m_0^2)^2}\int \frac{d^Dk_1}{(2\pi)^D}\frac{d^Dk_2}{(2\pi)^D}\frac{1}{(k_1^2+m_0^2)(k_2^2+m_0^2)((p-k_1-k_2)^2+m_0^2)}+\cdots $$

where for ease of notation let me write:

$$G_0^{(2)}=\frac{1}{p^2+m_0^2}-\frac{\lambda_0}{(p^2+m_0^2)^2}I_1+\frac{\lambda_0^2}{3!(p^2+m_0^2)^2}I_2+\cdots $$

But as far as I am aware it is not this that we renormalize but the vertex function $\Gamma_0^{(2)}$ which is given by:

$$\Gamma_0^{(2)}=\frac{1}{G_0^{(2)}}$$ $$=(p^2+m_0^2)+\lambda_0I_1-\frac{\lambda_0^2}{3!}I_2+\frac{\lambda_0^2}{p^2+m^2}I_1^2+\cdots$$

thus should the $I_1^2$ term not be included in the field renormalization and why is it often omitted?

  • $\begingroup$ I am not sure I understood your question. Have a look to my answer here physics.stackexchange.com/q/375360 (the second part). It may be useful $\endgroup$ – apt45 Dec 20 '17 at 21:35

I will drop the subscripts $0$ and superscript $(2)$ for ease of notation. If you are going to compute $\frac{1}{G}$ up to second order in $\lambda$, you better make sure that you have at your disposal an expression for $G$ which is complete up to second order. Yours is not. You are missing a $\lambda^2$ term which basically is $I_1^2$ with three free propagators. The $I_1^2$ term in $\Gamma$ is just not there. It is cancelled by the term you missed in $G$. Also note that even if it was there for some weird reason it would not contribute to the wave function/field strength renormalization. The latter is obtained from the $p^2$ coefficient of the Taylor expansion around zero external momentum. Since $I_1$ is an "infinite constant", it has no dependence in the external momentum $p$ and so no $p^2$ term. The first graph which contributes a $p^2$ term is the sunshine graph $I_2$.

  • $\begingroup$ If the diagram you are talking about is the one I am thinking of then it is not 1PI and as such should not be included in $\Gamma$ which is calculated using only 1PI diagrams. $\endgroup$ – Quantum spaghettification Dec 22 '17 at 4:52
  • $\begingroup$ @Quantumspaghettification: If you read my answer carefully you will realize that my point is: the graph $I_1^2$ which is not 1PI but still connected, should be included in $G$ which you forgot to do. If you did, then you would not see it appear in $1/G$ which is $\Gamma$ that should only have 1PIs in it. $\endgroup$ – Abdelmalek Abdesselam Dec 22 '17 at 10:36
  • $\begingroup$ The lecture notes I am using give $\Gamma^{(N)}$ as (in general): $$\Gamma^{(N)}_0=\left. \left[ \left( \prod^N_{j=1}\tilde G_0^{(2)}(p_j)^{-1} \right) \tilde G_0^{(N)}(p_1,\ldots, p_N)\right] \right|_{1PI}$$ It seems like you are saying it is actually given by: $$\Gamma^{(N)}_0=\left( \prod^N_{j=1}\tilde G_0^{(2)}(p_j)^{-1} \right) \left. \left[ \tilde G_0^{(N)}(p_1,\ldots, p_N)\right] \right|_{1PI}$$ or even $$\Gamma^{(N)}_0=\left( \prod^N_{j=1}\tilde G_0^{(2)}(p_j)^{-1} \right) \tilde G_0^{(N)}(p_1,\ldots, p_N)$$ is this correct? $\endgroup$ – Quantum spaghettification Dec 22 '17 at 11:05
  • $\begingroup$ where $\left.\left[ \ldots \right] \right|_{1PI}$ means we only consider the 1PI diagrams for everything in the square brackets (e.g. $\tilde G_0^{(2)}$ in the first case would only being evaluated to include 1PI. $\endgroup$ – Quantum spaghettification Dec 22 '17 at 11:09
  • $\begingroup$ I can write a thoughtful answer about the precise definitions of vertex functions and the Legendre transform vs 1PI. But it would have to be an answer to a new post about that (unless someone already asked) and I can't do that right at the moment. My answer was entirely based on how you phrased your question. I was only talking about $N=2$ and I used the definition that you wrote above, i.e., $\Gamma:=1/G$. Since there is still some communication problem that prevents you from seeing that my answer does indeed settle your question, let me try the Socratic method and ask you questions... $\endgroup$ – Abdelmalek Abdesselam Dec 23 '17 at 9:50

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