Nonintegrable threshold singularities of two-point functions in perturbation theory


In perturbation theory, the spectral densities of two-point functions develop nonintegrable threshold singularities at higher orders. In QCD, such singularities emerge when calculating the diagrams in terms of the pole quark mass, and they become stronger when one rearranges the perturbative expansion in terms of the running quark mass. In this paper, we discuss the proper way to handle such singularities.

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Fig. 1


  1. 1.

    In addition to the continuum contribution, there is also the contribution of positronium poles located below the threshold (see, e.g., [2]). Expanding the latter in powers of \(\alpha \) and adding to the perturbative expansion of the continuum contribution, one gets the full perturbative spectral density. For our interest in this paper, only the perturbative expansion of the continuum is relevant: The contributions of the positronium poles are localized slightly below the threshold and fall on the threshold being expanded in powers of \(\alpha \). Such contributions are integrable at the threshold and therefore are beyond the scope of our interest. To avoid confusion, we only recall that the positronium poles lead to nonzero small-circle contributions [10] [see Eq. (2.4)].

  2. 2.

    One should not confuse even powers of 1/v in the spectral density of the two-point function considered here and even powers of 1/v in the two-point function itself [10]: the latter emerge in the elastic two-point function in QED and correspond to the \(\delta \)-function and its derivatives in the spectral density and therefore do not represent nonintegrable threshold singularities.


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We are grateful to M. Steinhauser for illuminating comments concerning the application of his program rhad [7]. D. M. was supported by RFBR under project 19-52-15022. S. S. thanks the Italian Ministry of Research (MIUR) for support under the grant PRIN 20172LNEEZ.

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Correspondence to Dmitri Melikhov.


Appendix A: Proof by induction for \(F_{n+\frac{1}{2}}\) [odd powers \(1/v^{2n+1}\)]

We start with the integral over the cut

$$\begin{aligned} F_{n+\frac{1}{2}}(z,\epsilon )= \frac{z}{\pi }\int _{1 + \epsilon }^\infty \mathrm{d}z' \frac{1}{(1- \frac{1}{z'})^{n+\frac{1}{2}}} \frac{1}{z'(z'-z)}. \end{aligned}$$

By virtue of (2.6), we come to a recursive relation

$$\begin{aligned} F_{n+\frac{1}{2}}(z,\epsilon )= \frac{z}{z-1} F_{n-\frac{1}{2}}(z,\epsilon ) + \frac{1}{\pi }\frac{z}{1 - z}\Delta F_{n+\frac{1}{2}}(\epsilon ), \end{aligned}$$


$$\begin{aligned} \Delta F_{n+\frac{1}{2}}(\epsilon )= & {} \int _{1 + \epsilon }^\infty \mathrm{d}z' \frac{1}{(1- \frac{1}{z'})^{n+\frac{1}{2}}} \frac{1}{z'^2}\nonumber \\= & {} \frac{\left( \frac{\epsilon }{1+\epsilon }\right) ^{\frac{1}{2}-n}-1}{n-\frac{1}{2}}. \end{aligned}$$

This function is represented by an infinite sum of powers of \(\sqrt{\epsilon }\), the most singular term being of the order \({\epsilon }^{-(n-\frac{1}{2})}\). Noteworthy, the \(\epsilon \)-independent term in \(\Delta F_{n+\frac{1}{2}}(\epsilon )\) is

$$\begin{aligned} -\frac{1}{(n-\frac{1}{2})}. \end{aligned}$$

We now conjecture that the \(\epsilon \)-independent part of the recursive relation (A.2) yields the recursive relation for \(F_{n+\frac{1}{2}}(z)\):

$$\begin{aligned} F_{n+\frac{1}{2}}(z)-F_{n+\frac{1}{2}}(0)= & {} \frac{z}{z-1}\left\{ F_{n-\frac{1}{2}}(z)-F_{n-\frac{1}{2}}(0)\frac{}{}\right\} \nonumber \\&+\frac{1}{\pi }\frac{z}{z-1} \frac{1}{(n-\frac{1}{2})},\quad n\ge 1, \end{aligned}$$

where the knowledge of the \(\epsilon \)-independent term in \(\Delta F_{n+\frac{1}{2}}(\epsilon )\) is used.

The analytic solution \(F_{\frac{1}{2}}(z)\) is easily found:

$$\begin{aligned} F_{\frac{1}{2}}(z)=F_{\frac{1}{2}}(0) -\frac{2}{\pi }\sqrt{\frac{z}{z-1}}\log \left( \sqrt{-z}+\sqrt{1-z}\right) . \end{aligned}$$

Solving (A.5) iteratively, we find the conjectured solution \(F_{n+\frac{1}{2}}(z)\). We still have to prove that this function indeed satisfies the integral equation (2.4). To this end, we need to calculate the small-circle integral of the function \(F_{n+\frac{1}{2}}(z)\), see Eq. (2.4).

Using (2.6), the small-circle integral satisfies a recursive relation similar to (A.2):

$$\begin{aligned} I_{n+\frac{1}{2}}(z,\epsilon )= & {} \frac{z}{z-1} I_{n-\frac{1}{2}}(z,\epsilon ) + \frac{1}{\pi }\frac{z}{1 - z}\Delta I_{n+\frac{1}{2}}(\epsilon ), \end{aligned}$$


$$\begin{aligned} \Delta I_{n+\frac{1}{2}}(\epsilon )= & {} \frac{1}{2\pi i}\oint _{1 + \epsilon } \mathrm{d}z' F_{n+\frac{1}{2}}(z')\frac{1}{z'^2}. \end{aligned}$$

To calculate this small-circle integral, we expand the integrand near \(z=1\). Making use of the expansion

$$\begin{aligned} F_{n+\frac{1}{2}}(z)= & {} (-1)^n\left( \frac{z}{1-z}\right) ^{n+\frac{1}{2}}+O((z-1)^0) \end{aligned}$$

and expanding also \(1/z^2\) near \(z=1\), we find

$$\begin{aligned} \Delta F_{n+\frac{1}{2}}(\epsilon )+\Delta I_{n+\frac{1}{2}}(\epsilon )=-\frac{1}{(n-\frac{1}{2})}, \end{aligned}$$


$$\begin{aligned}&F_{n+\frac{1}{2}}(z,\epsilon )+I_{n+\frac{1}{2}}(z,\epsilon )\nonumber \\&\quad =\frac{z}{z-1}\left\{ F_{n-\frac{1}{2}}(z,\epsilon )+I_{n-\frac{1}{2}}(z,\epsilon ) \right\} \nonumber \\&\qquad +\frac{z}{\pi (z-1)}\frac{1}{(n-\frac{1}{2})}. \end{aligned}$$

This equation coincides with Eq. (A.5) for

$$\begin{aligned} F_{n+\frac{1}{2}}(z)-F_{n+\frac{1}{2}}(0) =F_{n+\frac{1}{2}}(z,\epsilon )+I_{n+\frac{1}{2}}(z,\epsilon ) \end{aligned}$$

and verifies by induction that the sum \(F_{n+\frac{1}{2}}(z,\epsilon )+I_{n+\frac{1}{2}}(z,\epsilon )\) does not depend on \(\epsilon \) (since \(F_{\frac{1}{2}}(z)-F_{\frac{1}{2}}(0)\) is \(\epsilon \)-independent). We therefore prove our conjecture that the solution \(F_{n+\frac{1}{2}}(z)\) is obtained from the dispersion integral over the cut, \(F_{n+\frac{1}{2}}(z,\epsilon )\), by omitting all \(\epsilon \)-dependent terms. Let us emphasize that the only role of the small-circle integral \(I_{n+\frac{1}{2}}(z,\epsilon )\) is to cancel the \(\epsilon \)-dependent terms in \(F_{n+\frac{1}{2}}(z,\epsilon )\); it leaves no other traces.

Appendix B: Proof by induction for \(F_n\) [even powers \(1/v^{2n}\)]

This appendix gives the analysis for the case \(F_{n}\) which has some subtleties compared to the analysis for \(F_{n+\frac{1}{2}}\) of Appendix A.

We start with the integral over the cut

$$\begin{aligned} F_n(z,\epsilon )= \frac{z}{\pi }\int _{1 + \epsilon }^\infty \mathrm{d}z' \frac{1}{(1- \frac{1}{z'})^n} \frac{1}{z'(z'-z)}. \end{aligned}$$

By virtue of (2.6), one obtains a recursive relation

$$\begin{aligned} F_n(z,\epsilon )= \frac{z}{z-1} F_{n-1}(z,\epsilon )+ \frac{1}{\pi }\frac{z}{1 - z}\Delta F_n(\epsilon ), \end{aligned}$$


$$\begin{aligned} \Delta F_n(\epsilon )= & {} \int _{1 + \epsilon }^\infty \mathrm{d}z' \frac{1}{(1- \frac{1}{z'})^n} \frac{1}{z'^2}\nonumber \\= & {} \frac{ \left[ (1 + \frac{1}{\epsilon })^n - 1\right] \epsilon - 1}{(n-1)(1 + \epsilon )}. \end{aligned}$$

For integer n, this function is represented by a finite sum of poles in \(1/\epsilon \) of the increasing order; it contains no regular structures in \(\epsilon \).

Solving (B.2) for the \(\epsilon \)-independent part and making use of the known expression for \(F_1(z)\), we obtained \(F_n(z)\) in the form [see (2.21)]:

$$\begin{aligned} F_n(z)=F_n(0) - \frac{1}{\pi } \left( \frac{z}{z - 1} \right) ^n \log (1-z). \end{aligned}$$

This is still our conjecture and we have to check that this function satisfies the integral equation (2.4).

Let us calculate the small-circle integral for \(F_n(z)\) given by (B.4). Using again (2.6), we obtain also for the small-circle integral a recursive relation

$$\begin{aligned} I_n(z,\epsilon )= & {} \frac{z}{z-1} I_{n-1}(z,\epsilon )+ \frac{1}{\pi } \frac{z}{1 - z}\Delta I_n(\epsilon ), \end{aligned}$$


$$\begin{aligned} \Delta I_n(\epsilon )= & {} \frac{1}{2\pi i}\oint _{1 + \epsilon } \mathrm{d}z'\log (1-z') \frac{1}{(1- \frac{1}{z'})^n} \frac{1}{z'^2}\nonumber \\= & {} -\frac{ \left[ (1 + \frac{1}{\epsilon })^n - 1\right] \epsilon - 1}{(n-1)(1 + \epsilon )}=-\Delta F_n(\epsilon ), \end{aligned}$$

Thus, \(F_n(z,\epsilon )+I_n(z,\epsilon )\) satisfies the recursive relation

$$\begin{aligned} F_n(z,\epsilon )+I_n(z,\epsilon )=\frac{z}{z-1}\left\{ F_{n-1}(z,\epsilon ) +I_{n-1}(z,\epsilon ) \frac{}{}\right\} . \end{aligned}$$

For \(n=1\) we find

$$\begin{aligned} F_1(z,\epsilon )= & {} -\frac{1}{\pi } \frac{z}{z-1}\log (1-z) +\frac{1}{\pi } \frac{z}{z-1} \log (\epsilon )\nonumber \\&+\frac{z}{(z-1)^2}\epsilon +\frac{z}{2(z-1)^3}\epsilon ^2+O(\epsilon ^3) \end{aligned}$$


$$\begin{aligned} I_1(z,\epsilon )= & {} -\frac{1}{\pi } \frac{z}{z-1} \log (\epsilon )\nonumber \\&-\frac{z}{(z-1)^2}\epsilon -\frac{z}{2(z-1)^3}\epsilon ^2+O(\epsilon ^3), \end{aligned}$$

leading to

$$\begin{aligned} F_1(z,\epsilon )+I_1(z,\epsilon )=-\frac{1}{\pi }\frac{z}{z-1}\log (1-z). \end{aligned}$$

Let us emphasize again that not only singular in \(\epsilon \) terms, but also powers of \(\epsilon \) cancel in the sum, so \(F_1(z,\epsilon )+I_1(z,\epsilon )\) does not depend on \(\epsilon \) at all. As follows from (B.7), for any n the sum \(F_n(z,\epsilon )+I_n(z,\epsilon )\) also does not depend on \(\epsilon \) and has the form

$$\begin{aligned} F_n(z,\epsilon )+I_n(z,\epsilon )=-\frac{1}{\pi }\left( \frac{z}{z-1}\right) ^n\log (1-z). \end{aligned}$$

So our conjectured solution (B.4) indeed solves the integral equation (2.4).

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Melikhov, D., Simula, S. Nonintegrable threshold singularities of two-point functions in perturbation theory. Eur. Phys. J. Plus 136, 106 (2021).

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