Pauli-Villars regularization for Vacuum Polarization

[Korybut]

TL;DR Summary

In most of the textbooks corresponding integral is computed in $k^2 <4m^2$ assumption. How to extend?

Hello!

I am currently reading Itzykson Zuber QFT book and on Chapter 7 where for the first time loops are considered. Particular method of dealing with divergences namely Pauli-Villars regularization is considered in section 7-1-1 considering vacuum polarization diagram. I do understand physics here but lost in math. The integral is computed in the textbook but in the assumption $k^2<4m^2$ and the result is the following

$$\bar{\omega}(k^2,m,\Lambda)=-\frac{\alpha}{3\pi}\Bigg\{-\log \frac{\Lambda^2}{m^2}+\frac{1}{3}+2\Bigg(1+\frac{2m^2}{k^2}\Bigg[\Bigg(\frac{4m^2}{k^2}-1\Bigg)^{\frac{1}{2}} \times\\ \mathrm{arccot}\Bigg(\frac{4m^2}{k^2}-1\Bigg)^{\frac{1}{2}}-1\Bigg]\Bigg\} $$

I am completely fine with this calculation however the following phrase from the textbook is not clear

The calculation has been carried out under the assumptio $k^2<4m^2$. The corresponding function may be continued in the complex $k^2$ plane. The values for $k^2>4m^2$ are obtained by taking a limiting value form above the cut, starting at the point $k^2=4m^2$. The discontinuity across the cut is $$\bar{\omega}(k^2+i\varepsilon)-\bar{\omega}(k^2-i\varepsilon)=2i\,\mathrm{Im}\, \bar{\omega}(k^2+i\varepsilon)$$

Please explain why to obtain values for $k^2>4m^2$ one should take this limit. And why discontinuity across the cut does matter here at all. Completely lost...

Many thanks in advance.

 

[Korybut]

Please, don’t get me wrong I do understand how analytic continuation works. And all functions in the expression are analytic. The question is why “above the cut”? Analytic continuation below the cut is also available but why it is considered to be unphysical?

 

[vanhees71]

This is indeed a subtle point. When calculating the vacuum polarization from the Feynman diagrams involved you already see that the corresponding "self energy" $\Pi(k^2)$ must be a meromorphic function with an exceptional singularity at $k^2=4m^2$, where $m$ is the electron mass (assuming you talk about standard QED with electrons, positrons and photons).

To define your function completely you must define a branch cut and the Riemann sheet you want to define the meromorphic function on. In standard vacuum QFT you want to calculate the time-ordered Green's functions or, as here, building blocks to calculate them, which leads to the conclusion that you need to calculate the truncated one-particle irreducible diagrams (aka the vertex functions).

The time-ordering prescription tells you that for the vacuum polarization you want to define the cut along the real $k^2$ axis, i.e., along $k^2>4 m^2$, and for $k^2<4m^2$ the function should be real. Along the cut the values of the function when taking the limits above and below the cut then yield complex conjugate values. Except along the cut (and maybe some poles on the real axis describing bound states, but these you get only if doing corresponding resummations of "ladder diagrams" to find them) the vacuum polarization is a analytic function and thus you can calculate the function everywhere from knowing the discontinuity along the cut, i.e., from the imaginary part. This leads to the Kallen-Lehmann spectral representation.

You can also renormalize self-energy diagrams in this way: There are Feynman rules to calculate the discontinuity along the cut (the socalled Cutkosky or "cutting" rules), which is also called the spectral function, $\rho=-2 \mathrm{Im} \Pi(k^2+\mathrm{i} \epsilon)$, $k^2 \in \mathbb{R}$, and this is usually finite. Now in standard QED the spektral function goes logarithmically with $k^2$, i.e., the dispersion integral from the Kallen-Lehman representation diverges logarithmically, but you can subtract it according to the physical renormalization description $\Pi(0)=0$, and then the resulting dispersion integral is finite, leading directly to the renormalized vacuum polarization without the need of a regularization.