Quantum Field Theory – How to Understand Renormalization Conditions in the $\phi^4$-theory

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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?

Best Answer

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.

So:

  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.

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