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  1. We know that in quantum field theory we include infinities at each order of the perturbative expansion of the renormalization $Z$ factors about the coupling constant $\lambda$ to absorb the divergences of the loop diagrams, so it seems $Z$ must be infinite.

  2. On the other hand, if we turn the coupling constant $\lambda$ to zero, the interacting theory then becomes a free theory, so the $Z$ must be $1$ in this case. This means that the $Z$ should be a small variation of $1$ when $\lambda$ is small.

  3. Moreover, according to the Kallen-Lehmann spectral form we must have $Z \in [0, 1]$.

Combining the above arguments, does it mean that although there are infinities in each $\lambda^n$ order term in the expansion of $Z$, their total sum turns out to be a finite number which is a small variation around $1$? That is, when people say the renormalization $Z$ factors are infinite, do they actually mean that the $Z$'s are infinite at each order?

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  • $\begingroup$ related: physics.stackexchange.com/questions/39241, physics.stackexchange.com/questions/262284 $\endgroup$
    – AccidentalFourierTransform
    Commented Oct 21, 2017 at 15:27
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    $\begingroup$ Your second and third points are correct, but the first point is not. It is possible to show to that even though Z has an infinite correction order by order, the value of Z taken for all loop correction amounts to a finite correction. This is known to be true for theory with asymptotic freedom or a theory in which Beta function approaches a fixed point. I don't have time to write a full fledged answer but this discussion can be found in Collins "Renormalization" Chapters 3 and chapter 7. $\endgroup$
    – ved
    Commented Dec 18, 2017 at 5:45

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As you are familiar, the idea is to introduce renormalised parameters and fields in terms of bare quantities, related by various renormalisation factors, $Z_i$.

We then expand $Z_i$ around some classical tree level values; this corresponds to $Z_i = 1$ followed by an infinite series of corrections $\delta_i$, so $Z_i = 1 + \delta_i$.

In applying perturbation theory to calculate amplitudes we find that the $\delta_i$ counterterms contain terms with whatever chosen regulator which means they are infinite in the final limit. For example, the counterterm from computing the photon self-energy gives,

$$\delta_3 = -\frac{e^2_R}{6\pi^2}\frac{1}{\varepsilon} - \frac{e^2_R}{12\pi^2}\ln \frac{\tilde \mu^2}{m^2_R}$$

which is clearly infinite as $\varepsilon \to 0$. So in general, we have,

$$Z_i = \mathrm{finite} + \sum_{n=1}^\infty \frac{c_n}{\varepsilon^n}.$$

The S-matrix itself is generally an asymptotic series, which means the scattering amplitudes themselves may not converge to anything finite. As for whether the sum can converge to a $f(\varepsilon)$ that is finite as $\varepsilon \to 0$ is as far as I know not addressed in most QFT books.

However, if $f(\varepsilon) \to \mathrm{finite}$ as $\varepsilon \to 0$, that makes $Z_i$ finite, implying there were no divergences that needed to be absorbed if we were able to sum the entire perturbation series.

But we know that the reason for renormalisation is not the fact that we cannot compute all the terms of the S-matrix, hence, in my mind at least, a contradiction.

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    $\begingroup$ There are several reasons that make me think about this question. 1. there is a problem in Srednicki's QFT(13.1 on page 95) that asks you to express $Z_{\varphi}$ in terms of the spectral density $\rho(s)$ for a scalar field theory $\mathcal L=-\frac{1}{2}Z_{\varphi}\partial^{\mu}\varphi\partial_{\mu}\varphi-\frac{1}{2}Z_mm^2\varphi^2-\mathcal L_1(\varphi)$, the answer turns out to be $Z_{\varphi}=(1+\int^{\infty}_{4m^2}ds\,\rho(s))^{-1}$, $\rho(s)\geq 0$, so $0\leq Z_{\varphi}\leq 1$ which shows that $Z_{\varphi}$ is finite. $\endgroup$
    – Hsuen Wei
    Commented Oct 21, 2017 at 15:42
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    $\begingroup$ 2.also in Srednicki's QFT in chapter 28, in $\overline{MS}$ scheme, the renormalization $Z$ factors can be written as $Z_{\varphi}=1+\sum^{\infty}_{n=1}\frac{a_n(\alpha)}{\varepsilon^n}$, $Z_{g}=1+\sum^{\infty}_{n=1}\frac{c_n(\alpha)}{\varepsilon^n}$, then he plugs this two expressions into the formula $G(\alpha,\varepsilon)\equiv\mathrm{ln}(Z^2_gZ^{-3}_{\varphi})$ and then he gets $G(\alpha,\varepsilon)=\sum^{\infty}_{n=1}\frac{G_n(\alpha)}{\varepsilon^n}$, here it seems that he treats the $Z^2_gZ^{-3}_{\varphi}$ inside the logrithm as a small variation around $1$ so that he can do the series $\endgroup$
    – Hsuen Wei
    Commented Oct 21, 2017 at 16:05
  • $\begingroup$ expansion, he also has to treat the $\sum^{\infty}_{n=1}$ part in $Z_{\varphi}$ and $Z_g$ as a small variation around $1$ to do the expansion in $Z^2_g$ and $Z^{-3}_{\varphi}$ in order to get the resulting expression for $G(\alpha,\varepsilon)$ as a series of $\varepsilon^{-1}$. If the $\sum^{\infty}_{n=1}$ part in $Z_{\varphi}$ and $Z_g$ is divergent, how could he do the series expansion? $\endgroup$
    – Hsuen Wei
    Commented Oct 21, 2017 at 16:22
  • $\begingroup$ @Shing-Yue note that you don't need parameters to be small in order to perform a Taylor expansion. The precise mathematical term is forma power series. $\endgroup$
    – AccidentalFourierTransform
    Commented Oct 23, 2017 at 17:40
  • $\begingroup$ @AccidentalFourierTransform The formal power series do not seem to have much help. When you formally expand a function of X in terms of Taylor series, you actually define the function of X as the polynomial of the Taylor series as if the X is a number. You need not worry about the convergence problem because here you are dealing with definitions, but when you replace X with numbers, you still can't escape from concerning about the convergence problem. $\endgroup$
    – Hsuen Wei
    Commented Oct 24, 2017 at 15:52

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