Some comments on calculations of the scalar radius of the pion and the chiral constant $\bar{l}_4$

Physics – High Energy Physics – High Energy Physics - Phenomenology

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The pion scalar radius is given by $=(6/\pi)\int_{4M^2_\pi}^\infty{\rm d}t \delta_S(t)/t^2$, with $\delta_S$ the phase of the scalar form factor. Below $\bar{K}K$ threshold, $\delta_S=\delta_0$, $\delta_0$ being the isoscalar, S-wave $\pi\pi$ phase shift. Between $\bar{K}K$ threshold and $t^{1/2}\sim 1.5 {\rm GeV}$ I argued, in two previous letters, that one can approximate $\delta_S\sim\delta_0$, because inelasticity is small, compared with the errors. This gives $=0.75\pm0.07 {\rm fm}^2$ and the value $\bar{l}_4=5.4\pm0.5$ for the one-loop chiral perturbation theory constant, compared with the values given by Leutwyler and collaborators, $=0.61\pm0.04 {\rm fm}^2$ and $\bar{l}_4=4.4\pm0.3$. At high energy, $t^{1/2}>1.5 {\rm GeV}$, I remarked that the value of $\delta_S$ that follows from perturbative QCD agrees with my interpolation and disagrees with that of Leutwyler and collaborators. In a recent article, Caprini, Colangelo and Leutwyler claim that my estimate of the asymptotic phase $\delta_S$ is incorrect as it neglects higher twist contributions. Here I remark that, when correctly calculated, higher twist contributions are likely negligible. I also show that chiral perturbation theory gives $\bar{l}_4=6.60\pm0.43$, compatible with my estimate but widely off the value $\bar{l}_4=4.4\pm0.3$ of Leutwyler and collaborators.

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