# Asymptotic LSZ reduction formula (Peskin & Schroeder)

Peskin & Schroeder, An Introduction to Quantum Field Theory, write at page 224

$$\int d^{4} x e^{i p \cdot x}\left\langle\Omega\left|T\left\{\phi(x) \phi\left(z_{1}\right) \cdots\right\}\right| \Omega\right\rangle$$ $$\underset{p^0\to+E_{\mathbf{p}}}{\sim}\frac{i}{p^{2}-m^{2}+i \epsilon} \sqrt{Z}\left\langle\mathbf{p}\left|T\left\{\phi\left(z_{1}\right) \cdots\right\}\right| \Omega\right\rangle\tag{7.37},$$

but I don't understand the meaning of this notation ($$\sim$$). At the beginning of section 7.2. Peskin & Schroeder also write

Here and throughout this section we use the symbol $$\sim$$ to mean that the poles of both sides are identical [...],

but I don't understand the meaning of this sentence.

• I think it simply means "asymptotically equal to." Namely $f(x) \underset{x\to y}{\sim} g(x)\,\Leftrightarrow \lim_{x\to y} f(x)/g(x) = 1$. – MannyC Apr 16 at 14:39
• @MannyC, I'm not sure about this. At the beginning of this section Peskin also writes Here and throughout this section we use the symbol $\sim$ to mean that the poles of both sides are identical, but I don't understand the meaning of this sentence. – Alessandro Greco Apr 16 at 15:01
• I think he means both the poles and the residues, but I'm not sure. I don't have the book here. If the poles and the residues are the same, if you get arbitrarily close to where the poles blow up the functions are approximately equal. – MannyC Apr 16 at 15:04

Here is a simple example of the meaning of this notation. Consider $$f(x)=\frac{x-3}{x^2-3x+2}=\frac{x-3}{(x-1)(x-2)}.$$ Then $$f(x)\underset{x\rightarrow 1}{\sim} \frac{2}{x-1},$$ that is, the pole of $$f(x)$$ at $$x=1$$ is the same as the pole of $$2/(x-1)$$ at $$x=1$$. As pointed out by @MannyC, this means nothing other than that $$f(x)$$ and $$2/(x-1)$$ are asymptotically equal as $$x\rightarrow 1$$, $$\lim_{x\rightarrow 1} f(x)/(2/(x-1)) = 1$$.
The additional finite term Peskin refers to is given by $$f(x) \underset{x\rightarrow 1}{\sim} \frac{2}{x-1} + 1.$$ Higher order terms can be found straightforwardly. Note that $$f(x) = \frac{2}{x-1} - \frac{1}{x-2},$$ so $$f(x)\underset{x\rightarrow 2}{\sim} -\frac{1}{x-2}.$$
Figure 1. Plot of $$f(x)$$ and $$2/(x-1)$$ in black and red, respectively.