4
$\begingroup$

In the following I limit my considerations to 4-point diagrams.

After the introduction of renormalized field operator (in renormalized perturbation theory) $$\phi_r= (\sqrt{Z})^{-1} \phi\tag{10.15}$$ in eq. (10.15) P&S states that "in computing the S-matrix elements, we no longer need the factors of $Z$ in Eq. (7.45);" Eq. (7.45) looks like (I regret not to know well to draw bubbles) $$\langle p_1 p_2|S|k_1k_2\rangle = \tag{7.45}$$ $$ (\sqrt{Z})^4(\mbox{sum of all amputated conn. diagrams with $p_1$, $p_2$ incoming, $k_1$, $k_2$ outgoing}).$$

That means, the factor $(\sqrt{Z})^4$ would no longer appear in eq. (7.45). Actually I don't understand this conclusion.

What I do understand is that when in the LSZ-formula (7.42) the renormalized field operators are used the $Z$-factors disappear:

$$\prod_1^2 \int d^4x_i e^{i p_i x_i} \prod_1^2 \int d^4y_i e^{-i k_i y_i}\langle\Omega|T\{\phi_r(x_1)\phi_r(x_2)\phi_r(y_1)\phi_r(y_2)\}|\Omega\rangle$$ $$\sim \prod_1^2\frac{i}{p_i^2-m^2} \prod_1^2\frac{i}{k_i^2-m^2} \langle p_1 p_2|S|k_1k_2\rangle $$

where for simplicity I neglected the $+i\epsilon$ terms and the limits $p^0_i\rightarrow +E_{p_i}$ and $k^0_i\rightarrow +E_{k_i}$ on the right side of the equation. In the following P&S states that

$$\prod_1^2 \int d^4x_i e^{i p_i x_i} \prod_1^2 \int d^4y_i e^{-i k_i y_i}\langle\Omega|T\{\phi(x_1)\phi(x_2)\phi(y_1)\phi(y_2)\}|\Omega\rangle $$

corresponds to Figure 7.4, i.e. a most general 4-point diagram whose 4 legs contain self-energy bubbles (represented by dark-shaded circles) and the sum of all connected amputated 4-point diagrams in the center. And each self-energy bubble would correspond to a factor like

$$ \frac{i Z}{p^2-m^2}$$

Naively I thought first that this way the $Z$-factors would come in again, but later I thought that in renormalized perturbation theory the self-energy bubbles come also contain the counter terms so that they better correspond to

$$ \frac{i}{p^2-m^2}\tag{10.19}$$

but finally I realized that in renormalized perturbation theory the expression

$$\prod_1^2 \int d^4x_i e^{i p_i x_i} \prod_1^2 \int d^4y_i e^{-i k_i y_i}\langle\Omega|T\{\phi_r(x_1)\phi_r(x_2)\phi_r(y_1)\phi_r(y_2)\}|\Omega\rangle$$

should be considered instead. Looking at this expression I am no longer sure whether it would be still correspond to figure (7.4), i.e.

$$\frac{iZ}{p_1^2-m^2}\frac{iZ}{p_2^2-m^2}\frac{iZ}{k_1^2-m^2}\frac{iZ}{k_2^2-m^2}\cdot (\mbox{sum of all amputated conn. 4-point diagrams})$$

which only seems to hold for non-normalized field operators. For renormalized field operators the $Z$'s in the precedent expression certainly disappear, but does the $$(\mbox{sum of all amputated conn. 4-point diagram})$$ the same? It is not clear for me. This is, however, the prequisite for the validity of eq. (7.45) without $Z$'s, the equation I cited at the beginning of my question.

I would be grateful if somebody with more insight could explain it to me.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

The main points (which hopefully resolves OP's questions) seem to be the following.

  1. The LSZ formalism in section 7.2 is using bare fields (which we will call $\phi_0$ for clarity). If Ref. 1 had instead used renormalized fields $\phi_r$ there would not be any explicit $Z$-factors in the LSZ reduction formulas (7.42) & (7.45).

  2. The generator of 1PI vertices, the effective/proper action $$\Gamma[\phi_{\rm cl}]~=~\sum_{n;k_1,\ldots,k_n}\frac{1}{n!} \Gamma_{n;k_1,\ldots,k_n} \phi^{k_1}_{\rm cl}\ldots \phi^{k_n}_{\rm cl}$$ scales under renormalization $$\phi^k_0~=~\sqrt{Z}\phi^k_r\tag{10.15}$$ as $$\Gamma^r_{n;k_1,\ldots,k_n}~=~Z^{n/2}\Gamma^0_{n;k_1,\ldots,k_n}.$$

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; eqs. (7.42-45) + eq. (10.15).
$\endgroup$

Your Answer

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