2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 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. $\endgroup$ – Alessandro Greco Apr 16 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 at 15:04
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.