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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.

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    $\begingroup$ I think it simply means "asymptotically equal to." Namely $f(x) \underset{x\to y}{\sim} g(x)\,\Leftrightarrow \lim_{x\to y} f(x)/g(x) = 1$. $\endgroup$
    – MannyC
    Apr 16, 2019 at 14:39
  • $\begingroup$ @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. $\endgroup$ Apr 16, 2019 at 15:01
  • $\begingroup$ 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. $\endgroup$
    – MannyC
    Apr 16, 2019 at 15:04

1 Answer 1

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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 $$f(x)\underset{x\rightarrow 2}{\sim} -\frac{1}{x-2}.$$

enter image description here

Figure 1. Plot of $f(x)$ and $2/(x-1)$ in black and red, respectively.

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