フェルミの4点相互作用を用いたパイオン相互作用の考察
Lorentz不変の相互作用として \begin{align} \mathcal{H}_{int} = \sum_{i} (\bar{\psi}_{u}\Gamma_{i} \psi_{d}) [\bar{\psi}_{e} \Gamma_{i}(C_{i}+C_{i}^{\prime}\gamma_{5})\psi_{\nu}] + H.C \end{align} ただし、 \[ \Gamma_{i} = \begin{cases} 1 & Schalar\\ \gamma_{\lambda} & Vector \\ \sigma_{\lambda\sigma} & Tensor \\ i\gamma_{5}\gamma_{\lambda} & Axial-Vector \\ \gamma_{5} & Pseudo-Scalar \end{cases} \]
散乱振幅は \begin{align} \langle f| \mathcal{H}_{int} |i\rangle &= \frac{1}{V^{2}}\sum_{i} (\bar{v}_{\bar{u}}(p_{\bar{u}}\Gamma_{i} u_{d}(p_{d})) [\bar{u}_{e}(p_{e}) \Gamma_{i}(C_{i}+C_{i}^{\prime}\gamma_{5})v_{\bar{\nu}}(p_{\bar{\nu}})] \\ & \times (2\pi)^{4}\delta^{(4)}(p_{e}+p_{\bar{\nu}} - p_{\bar{u}} - p_{d}) \end{align}
各項をまとめると \begin{align} &(p_{\bar{u}}\Gamma_{i} u_{d}(p_{d})) [\bar{u}_{e}(p_{e}) \Gamma_{i}(C_{i}+C_{i}^{\prime}\gamma_{5})v_{\bar{\nu}}(p_{\bar{\nu}})] \\ &= -(v^{\dagger}_{\bar{u}} u_{d}) [\bar{u}_{e}(p_{e}) \gamma_{5}(C_{P}+C_{P}^{\prime}\gamma_{5})v_{\bar{\nu}}(p_{\bar{\nu}}) - \bar{u}_{e}(p_{e}) \gamma_{5}\gamma_{0}(C_{A}+C_{A}^{\prime}\gamma_{5})v_{\bar{\nu}}(p_{\bar{\nu}})] \\ & -(v^{\dagger}_{\bar{u}} \sigma^{i} u_{d}) [\bar{u}_{e}(p_{e}) \gamma^{i}(C_{V}+C_{V}^{\prime}\gamma_{5})v_{\bar{\nu}}(p_{\bar{\nu}}) - 2\bar{u}_{e}(p_{e}) \sigma^{0i}(C_{T}+C_{T}^{\prime}\gamma_{5})v_{\bar{\nu}}(p_{\bar{\nu}})] \end{align}
$(v^{\dagger}_{\bar{u}} u_{d})$の部分はパイオンと見ることができるが、 \[ \mathcal{H}_{int} \propto \partial_{\lambda}\phi_{\pi}[\bar{\psi}_{e}\gamma_{5}\gamma^{\lambda}(1+\gamma_{5})\psi_{\nu}] \] の形とは異なっている。
おそらくはじめにパイオンの静止系を仮定したことによる。