Connection between the Dipole Polarizabilities of Charged and Neutral $\pi$-Mesons

Abstract

In light of the contribution of the states with isospin $I=0$ in the difference of the amplitudes of the processes $\gamma\gamma\to\pi^{+}\pi^{-}$ and $(\gamma+\gamma)\to(\pi^{0}+\pi^{0})$ being very small, the dispersion sum rules for the difference between the dipole polarizabilities of the charged and neutral pions are analyzed as a function of the $\sigma$ meson’s parameters. Using the value of the chiral theory of perturbations for $(\alpha_{1}-\beta_{1})_{\pi^{0}}=-1.9$, $(\alpha_{1}-\beta_{1})_{\pi^{\pm}}=9.4{-}8.2$ is found for the $\sigma$ meson parameter within the region $m_{\sigma}=400{-}550$ MeV, $\Gamma_{\sigma}=400{-}600$ MeV, $\Gamma_{\sigma\to\gamma\gamma}=0{-}3$ keV. Estimated optimum value of decay width $\sigma\to\gamma\gamma$ yields $\Gamma_{\sigma\to\gamma\gamma}\lesssim 0.7$ keV.

Appendices

Appendix A

The contributions from vector and axial-vector mesons $(\rho,\omega,\phi,a_{1}$, and $b_{1})$ to $ImM_{++}(s,u=\mu^{2})$ were calculated using the expression

$$ImM_{++}^{(V)}(s,u=\mu^{2})$$

$${}=\mp 4g_{V}^{2}s\frac{\Gamma_{0}}{(m_{V}^{2}-s)^{2}+\Gamma_{0}^{2}},$$

(A1)

where $m_{V}$ is the meson mass, sign ‘‘${+}$’’ corresponds to the contribution from $a_{1}$ and $b_{1}$ mesons and

$$g_{V}^{2}=6\pi\sqrt{\frac{m_{V}^{2}}{s}}\left(\frac{m_{V}}{m_{V}^{2}-\mu^{2}}\right)^{3}$$

$${}\times\Gamma_{V\to\gamma\pi}D_{1}(m_{V}^{2})/D_{1}(s),$$

(A2)

$$\Gamma_{0}=\left(\frac{q_{i}^{2}(s)}{q_{i}^{2}(m_{V}^{2})}\right)^{\frac{3}{2}}\frac{m_{V}^{2}}{\sqrt{s}}D_{1}(m_{V}^{2})/D_{1}(s)\Gamma_{V}.$$

(A3)

Here, $D_{1}$ is associated with the centrifugal potential. It is equal to $D_{1}=1+(q_{i}r)^{2}$ [48]; $r=1$fm is the effective radius of interaction; $\Gamma_{V}$ and $\Gamma_{V\to\gamma\pi}$ are the total decay width and the decay width for $\gamma\pi$ of these mesons. Momenta $q_{i}^{2}$ for $(\rho,\,\omega,\,\phi,\,a_{1}$, and $b_{1})$ mesons are equal to $(s-4\mu^{2})/4$, $(s-9\mu^{2})/4$, $(s-4m_{k}^{2})/4$, $(s-(m_{\rho}+\mu)^{2}/)4$, and $(s-16\mu^{2})/4$, respectively.

Appendix B

The amplitude of the contribution from a scalar meson to the process $\gamma\gamma\to\pi\pi$ can be written as

$$T=\frac{g_{s}}{\sqrt{t}-M_{s}-i\frac{1}{2}\Gamma_{s}}.$$

(B1)

It is then easy to show that the imaginary part of amplitude $ImM_{++}^{\sigma}(t)$ of the $\sigma$ meson contributions to the considered process can be presented as

$$ImM_{++}^{\sigma}(t)=\dfrac{g_{\sigma}(\sqrt{t}+M_{s})\Gamma_{0}^{\sigma}(t)}{(t-M_{\sigma}^{2})^{2}+(\Gamma_{0}^{\sigma}(t))^{2}},$$

(B2)

where

$$g_{\sigma}=\frac{8\pi}{t}\left[\frac{2}{3}\dfrac{M_{\sigma}\Gamma_{\gamma\gamma}\Gamma_{\sigma}}{\sqrt{M_{\sigma}^{2}-4\mu^{2}}}\right]^{1/2},$$

(B3)

$$\Gamma_{0}^{\sigma}=\dfrac{M_{\sigma}(\sqrt{t}+M_{\sigma})}{2\sqrt{t}}\left(\dfrac{t-4\mu^{2}}{M_{\sigma}^{2}-4\mu^{2}}\right)^{1/2}\Gamma_{\sigma}.$$

(B4)

Expressions (B2)–(B4) can be very useful in describing scaler mesons with large decay widths.

Since the two $K$ mesons make a large contribution to the decay width of the $f_{0}(980)$ meson and the threshold of the reaction $\gamma\gamma\to K\overline{K}$ is very close to the mass of the $f_{0}(980)$ meson, we consider Flatté’s expression [49] for the $f_{0}(980)$ meson’s contribution to the process $\gamma\gamma\to\pi\pi$.

For $t>4m_{k}^{2}$:

$$ImM_{++}^{f_{0}}=g_{f_{0}}\frac{\Gamma_{0_{f_{0}}}}{(m_{f_{0}}^{2}-t)^{2}+\Gamma_{0_{f_{0}}}^{2}},$$

(B5)

where

$$\Gamma_{0_{f_{0}}}=\Bigg{[}\Gamma_{f_{0}\to\pi\pi}\left(\frac{t-4\mu^{2}}{m_{f_{0}}^{2}-4\mu^{2}}\right)^{1/2}$$

$${}+\Gamma_{f_{0}\to kk}\left(\frac{t-4m_{k}^{2}}{4m_{k}^{2}-m_{f_{0}}^{2}}\right)^{1/2}\Bigg{]}m_{f_{0}}.$$

(B6)

For $t<4m_{k}^{2}$:

$$ImM_{++}=g_{f_{0}}\Gamma_{0_{f_{0}}}\Bigg{(}\Bigg{[}m_{f_{0}}^{2}-t$$

$${}-\left(\dfrac{4m_{k}^{2}-t}{4m_{k}^{2}-m_{f_{0}}^{2}}\right)^{1/2}m_{f_{0}}\Gamma_{f_{0}\to kk}\Bigg{]}^{2}$$

$${}+\Gamma_{0_{f_{0}}}^{2}\Bigg{)}^{-1},$$

(B7)

$$\Gamma_{0_{f_{0}}}=\Gamma_{f_{0}\to\pi\pi}m_{f_{0}}\left(\frac{t-4\mu^{2}}{m_{f_{0}}^{2}-4\mu^{2}}\right)^{1/2}.$$

(B8)