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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?

knzhou
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Hsuen Wei
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1 Answers1

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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.

JamalS
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