The renormalization conditions in $\phi^4-$theory are given in Eqn. 10.19 of Peskin and Schroeder, are supposed to define the physical mass and the physical couplings. The second condition is fine; calculating the diagram on the LHS, multiplying that by $i$ and setting $s=4m^2,t=u=0$, one can read off the physical coupling $\lambda$.

However, it is not clear to me how is the first equation useful in defining the physical mass $m$. In Eqn. 10.28, the book says that the renormalization condition is $$\frac{i}{p^2-m^2-M^2(p^2)}=\frac{i}{p^2-m^2}+\text{terms regular at} \hspace{0.2cm}p^2=m^2,\tag{a}$$ which is equivalent to$$M^2(p^2=m^2)=0; \hspace{0.3cm}\frac{d}{dp^2}M^2(p^2)|_{p^2=m^2}=0.\tag{1}$$

  1. How is the first condition of Eqn.(1) obtained fom (a)? My problem is that if I put $p^2=m^2$ in (a), the RHS has a singularity. Moreover, what happens to the regular part?

  2. The renormalization conditions are also expressed as $$\Gamma^{(2)}(0)=m^2; \hspace{0.3cm}\Gamma^{(4)}(0)=-\lambda.\tag{2}$$ Why are these relations not used by Peskin?

  3. I'm also having trouble in deriving the second condition. A Taylor expansion of $M^2(p^2)$ about $p^2=m^2$ goes like $$M^2(p^2)=M^2(m^2)+\frac{d}{dp^2}M^2(p^2)|_{p^2=m^2}(p^2-m^2)+...\tag{3}$$ But how to proceed next?


The logic is that we want the exact propagator

$$\Delta(p^2) = \frac{1}{p^2 - m^2 - M(p^2)} $$

to behave like the free propagator $1/(p^2 - m^2)$ near the pole $p^2 = m^2$. This is because the location of the pole determines the physical mass, and the residue enters into the LSZ formula, see formula 10.14.


  1. We want $\Delta(p^2)$ to have a pole at $m^2$, that way $m$ is the actual, physical mass of our particle. Well, $\Delta(m^2) = 1/M(m^2)$, so we need $M(m^2) = 0$.

  2. We want the residue to be 1. This means that near $p^2 = m^2$, we should have $\Delta(p^2) = 1/(p^2 - m^2) + \text{non-singular terms}$; the residue is the $1$ on top of the fraction. We can calculate it by Taylor expanding $M$ around $m^2$ and using our above result or by using that the residue is $\lim_{p^2 \to m^2} (p^2 - m^2)\Delta(p^2)$; either way, we must have $M'(m^2) = 0$.

As for your second question, I think Peskin just uses a different notation, that is, all that we've just said. If I'm not mistaken it's just a different way of saying the same thing.


The first question has been answered by Javier. For the second question, you may be refer the section 11.5 of Peskin and Schroeder. I guess the $\Gamma$ in your question is effective action and $\Gamma^{(n)}$ is an abbreviation of $$\frac{\delta^n \Gamma[\phi_{\rm cl}]}{\delta \phi_{\rm cl}(x_1) \cdots \delta \phi_{\rm cl}(x_n)}.$$ Equation 11.90 of Peskin and Schroeder says, $$\Gamma^{(2)}(x,y) = iD^{-1}(x,y)$$ where $D(x,y)$ is the exact propagator of the $\phi^4$ theory. In momentum space, we have $$\Gamma^{(2)}(p^2) = p^2-m^2-M^2(p^2)$$ So, $M^2(p^2=m^2) = 0$ is the same as $\Gamma^{(2)}(p^2=m^2) = 0$, not $\Gamma^{(2)}(0) = m^2$. Equation 11.96 of Peskin and Schroeder says, $$\Gamma^{(4)}(x_1,x_2,x_3,x_4) = -i \langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle_{\rm 1PI}$$ In momentum space, $\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle_{\rm 1PI}$ is just amputated Feynman diagram with four external legs. So the renormalization condition that amputated Feynman diagram with four external legs equals $-i\lambda$ when $s=4m^2,t=u=0$ is the same as the condition that $\Gamma^{(4)}(p_1^2,p_2^2,p_3^2,p_4^2 = m^2) = -\lambda$. As for the condition $$\left. \frac{d M^2(p^2)}{d p^2} \right|_{p^2=m^2} = 0.$$ It is the same as $$\left. \frac{d \Gamma^{(2)}(p^2)}{d p^2} \right|_{p^2=m^2} = 1.$$

For your third question, I guess you do not understand Javier's answer totally. If you are not familiar with complex analysis and residue theorem, I can give you a simple but rough explanation. $$\frac{i}{p^2-m^2-M^2(p^2)} = \frac{i}{p^2-m^2+M^2(m^2) + \left.\frac{d M^2(p^2)}{d p^2} \right|_{p^2=m^2}(p^2-m^2)+\cdots}\\ = \frac{i}{\left( 1 + \left.\frac{d M^2(p^2)}{d p^2} \right|_{p^2=m^2}\right)(p^2-m^2)+\cdots}\\ = \frac{i}{p^2-m^2} + \mbox{ regular terms}$$ By comparison, we can get $$\left.\frac{d M^2(p^2)}{d p^2} \right|_{p^2=m^2} = 0$$


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