3
$\begingroup$

I'm working on the Eq.(7.57) in Peskin(page 236).

enter image description hereSo I try to verify it with LSZ formula.

According to Eq (7.42)

enter image description here

So $\mathcal{M}(p \rightarrow p)=-Z M^{2}\left(p^{2}\right)$

In this I have two question:

① Consider S=1+iT, why did the "1" vanish?

② Is Eq.(2)->Eq.(3) correct?

Consider Eq (7.22)in Peskin, it seems to lack a free propagator.

Eq.(2)= 1 + (1PI) + (1PI-1PI) + (1PI-1PI-1PI) + ...

Eq.(3)= (1PI) + (1PI-1PI) + (1PI-1PI-1PI) + ...

enter image description here

$\endgroup$
0

1 Answer 1

1
$\begingroup$
  1. I think that your equality (2)=(3) is not correct. Moreover I am not sure what is the right interpretation of (3) in case of diagrams with exactly 2 exterior legs. Furthermore (4) seems to contradict the Kallen-Lehmann formula (see (7.9) in Peskin-Schroeder book). The latter implies that $$\int d^4x_1d^4x_2 e^{ip_1x_1}e^{-ip_2x_2}\langle\Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega\rangle\approx (2\pi)^4\delta^{(4)}(p_1-p_2)\frac{iZ}{p_1^2-m^2+i\epsilon}.$$ Namely the exact propagator has a pole of first order as $p^2\to m^2$.

  2. Another remark is that as far as I understand, in the Peskin-Schroeder book one particles states are assumed to be stable. That means in particular that $$\langle p_1|S|p_2\rangle=\delta^{(4)}(p_1-p_2).$$ That means that in the decomposition $S=1+iT$ the $1$ term is present, but the $iT$ term vanishes.

$\endgroup$
4
  • 1
    $\begingroup$ (4) comes from Eq (7.42) in peskin with n=1 and m=1, $$\left(\prod_{i=1}^{n} \frac{\sqrt{Z} i}{p_{i}^{2}-m^{2}+i \epsilon}\right)\left(\prod_{j=1}^{m} \frac{\sqrt{Z} i}{k_{j}^{2}-m^{2}+i \epsilon}\right)\left\langle\mathbf{p}_{1} \cdots \mathbf{p}_{n}|S| \mathbf{k}_{1} \cdots \mathbf{k}_{m}\right\rangle$$, why is this wrong? $\endgroup$
    – sky
    Feb 5, 2020 at 11:52
  • 1
    $\begingroup$ @sky Formally speaking you are right: this is what is written there. However I think that the book has a sloppy writing at this point. The $S$-matrix $S$ should be replaced by its connected part. This is actually is discussed few lines below, but again not in the most precise form. An equivalent, but precise form of the LSZ for scalar particles can be fount in the book "QFT" by Itzykson-Zuber, see the formula (5-28) there, where the presence of disconneced terms is explicitly indicated. $\endgroup$
    – MKO
    Feb 5, 2020 at 14:20
  • 1
    $\begingroup$ @sky In the case when $n=m=1$ the disconnected part is actually connected and represents 1 in the decomposition $S=1+iT$. The connected part vanishes. That means that the expression $$\int d^4x_1d^4x_2 e^{ip_1x_1}e^{-ip_2x_2}<\Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega>$$ has vanishing coefficient near $$(p_1^2-m^2)^{-1}(p_2^2-m^2)^{-1}.$$ $\endgroup$
    – MKO
    Feb 5, 2020 at 14:22
  • 1
    $\begingroup$ @sky The last claim is certainly consistent with (7.9) (Kallen-Lehmann) in Peskin-Schroeder. $\endgroup$
    – MKO
    Feb 5, 2020 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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