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liberulo
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e+e- --> gamma f0 --> gamma pi0 pi0 cross section with VMD
Find the cross-section of ##e^+e^- \to \gamma f_0(980) \to \gamma \pi^0 \pi^0## using the vector meson dominance model.
Some Feynman's rules:
The photon propagator is [itex] -i \frac{g_{\mu\nu}}{q^2} [/itex].
The propagator of ##\varphi##-meson is ##-i \frac{g_{\mu\nu} - \frac{q_\mu q_\nu}{m_\varphi^2}}{q^2 - m_\varphi^2 + i m_\phi \Gamma_\varphi}##, ##\Gamma_\varphi## - the particle width .
The ##\gamma \varphi##-vertex is ##-i e \frac{m_\varphi^2}{g_\varphi}##.
##g_{\varphi \omega f_0}## is the ##\varphi \omega f_0##-vertex constant.
The effective Lagrangian is
[tex]
\mathcal{L} = \mathcal{L}_{QED} + \mathcal{L}_{em} + \mathcal{L}_{str},
[/tex]
where
[tex]
\mathcal{L}_{str} = g_{\varphi \omega f_0} {F_\varphi}^{\alpha \beta} {F_\omega}^{\mu \nu} \varepsilon_{\alpha \beta \mu \nu} f_0 + g_{f_0 \pi^0 \pi^0} f_0 \pi \pi,
[/tex]
[tex]
\mathcal{L}_{em} = -e \frac{{m_\varphi}^2}{g_\varphi} \Phi^\mu A_\mu -e \frac{{m_\omega}^2}{g_\omega} \Omega^\mu A_\mu.
[/tex]
## \Phi^\mu, \Omega^\mu, A_\mu, f_0, \pi ## - ##\varphi##, ##\omega##, photon, ##f_0##, ##\pi^0## fields.
After that I try to write the matrix element for the ##e^+e^- \to \gamma f_0(980) \to \gamma \pi^0 \pi^0## diagram. There is my trouble.
[tex]
i M = \bar{v} (-i e \gamma_\mu ) u \cdot
\left(-i \frac{g^{\mu \nu}}{q^2} \right)
\left( -ie \frac{m_\varphi^2}{g_\varphi}\right)
\left( -i \right) \frac{g_{\nu\alpha} - \frac{q_\nu q_\alpha}{m_\varphi^2}}{q^2 - m_\varphi^2 + i m_\varphi \Gamma_\varphi} g_{\varphi \omega f_0}
\left( -i \right) \frac{g^{\alpha \beta} - \frac{k^\alpha k^\beta}{m_\omega^2}}{k^2 - m_\omega^2 + i m_\omega \Gamma_\omega}
\left( -ie \frac{m_\omega^2}{g_\omega}\right)
\cdot \\ \cdot
( k^\tau {\epsilon_{\gamma}}^\sigma - k^\sigma {\epsilon_{\gamma}}^\tau )
\varepsilon_{\tau \sigma ? ?}
\cdot
\frac{-i}{r^2 - m_{f_0}^2 + i m_{f_0} \Gamma_{f_0}} g_{f_0 \pi^0 \pi^0}
.
[/tex]
k - the radiative photon four-momentum, ##\epsilon_{\gamma}## - the photon polarization, r - ##f_0## four-momentum.
Homework Statement
Find the cross-section of ##e^+e^- \to \gamma f_0(980) \to \gamma \pi^0 \pi^0## using the vector meson dominance model.
Homework Equations
Some Feynman's rules:
The photon propagator is [itex] -i \frac{g_{\mu\nu}}{q^2} [/itex].
The propagator of ##\varphi##-meson is ##-i \frac{g_{\mu\nu} - \frac{q_\mu q_\nu}{m_\varphi^2}}{q^2 - m_\varphi^2 + i m_\phi \Gamma_\varphi}##, ##\Gamma_\varphi## - the particle width .
The ##\gamma \varphi##-vertex is ##-i e \frac{m_\varphi^2}{g_\varphi}##.
##g_{\varphi \omega f_0}## is the ##\varphi \omega f_0##-vertex constant.
The Attempt at a Solution
The effective Lagrangian is
[tex]
\mathcal{L} = \mathcal{L}_{QED} + \mathcal{L}_{em} + \mathcal{L}_{str},
[/tex]
where
[tex]
\mathcal{L}_{str} = g_{\varphi \omega f_0} {F_\varphi}^{\alpha \beta} {F_\omega}^{\mu \nu} \varepsilon_{\alpha \beta \mu \nu} f_0 + g_{f_0 \pi^0 \pi^0} f_0 \pi \pi,
[/tex]
[tex]
\mathcal{L}_{em} = -e \frac{{m_\varphi}^2}{g_\varphi} \Phi^\mu A_\mu -e \frac{{m_\omega}^2}{g_\omega} \Omega^\mu A_\mu.
[/tex]
## \Phi^\mu, \Omega^\mu, A_\mu, f_0, \pi ## - ##\varphi##, ##\omega##, photon, ##f_0##, ##\pi^0## fields.
After that I try to write the matrix element for the ##e^+e^- \to \gamma f_0(980) \to \gamma \pi^0 \pi^0## diagram. There is my trouble.
[tex]
i M = \bar{v} (-i e \gamma_\mu ) u \cdot
\left(-i \frac{g^{\mu \nu}}{q^2} \right)
\left( -ie \frac{m_\varphi^2}{g_\varphi}\right)
\left( -i \right) \frac{g_{\nu\alpha} - \frac{q_\nu q_\alpha}{m_\varphi^2}}{q^2 - m_\varphi^2 + i m_\varphi \Gamma_\varphi} g_{\varphi \omega f_0}
\left( -i \right) \frac{g^{\alpha \beta} - \frac{k^\alpha k^\beta}{m_\omega^2}}{k^2 - m_\omega^2 + i m_\omega \Gamma_\omega}
\left( -ie \frac{m_\omega^2}{g_\omega}\right)
\cdot \\ \cdot
( k^\tau {\epsilon_{\gamma}}^\sigma - k^\sigma {\epsilon_{\gamma}}^\tau )
\varepsilon_{\tau \sigma ? ?}
\cdot
\frac{-i}{r^2 - m_{f_0}^2 + i m_{f_0} \Gamma_{f_0}} g_{f_0 \pi^0 \pi^0}
.
[/tex]
k - the radiative photon four-momentum, ##\epsilon_{\gamma}## - the photon polarization, r - ##f_0## four-momentum.
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