$$

\mathcal{\alpha}=\frac{\mathcal{B}(K_1\to K^*\pi)}{\mathcal{B}(K_1\to K\rho)} $$

$$ \delta_\alpha=\frac{\mathcal{B}(K_1\to KX^0)}{\mathcal{B}(K_1\to K\rho)} $$

$$ \mathcal{B}_{3body}=\mathcal{B}(K_1\to K\rho)+\mathcal{B}(K_1\to K^*\pi)+\mathcal{B}(K_1\to KX^0) $$

where $\mathcal{B}(K_1\to KX^0)= \mathcal{B}(K_1\to K\omega(\to \pi\pi))+\mathcal{B}(K_1\to Kf_0(1370))$.

推导:

$$ \frac{\mathcal{B}{3body}}{\mathcal{B}(K_1\to K\rho)}=1+\alpha+\delta\alpha\Rightarrow\mathcal{B}(K_1\to K\rho)=\frac{\mathcal{B}{3body}}{1+\alpha+\delta\alpha}\\ \frac{\mathcal{B}{3body}}{\mathcal{B}(K_1\to K^*\pi)}=\frac{1}{\alpha}(1+\alpha+\delta\alpha)\Rightarrow\mathcal{B}(K_1\to K^\pi)=\frac{\mathcal{B}_{3body}\alpha}{1+\alpha+\delta_\alpha} $$

$$ \delta_\alpha=\frac{\mathcal{B}(K_1\to KX^0)}{\mathcal{B}(K_1\to K\rho)}\Rightarrow \mathcal{B}(K_1\to KX^0)=\frac{\delta_\alpha*\mathcal{B}{3body}}{1+\alpha+\delta\alpha} $$

$$ \mathcal{B}(K_1^+\to K^+\pi^-\pi^+)=\frac{1}{3}\times \mathcal{B}(K_1\to K\rho)+\frac{4}{9}\times \mathcal{B}(K_1\to K^(892)\pi)+\frac{4}{9}\times \mathcal{B}(K_1\to K^0(1430)\pi)+\mathcal{B}(K_1\to KX^0)=\frac{3\mathcal{B}{3body}+4\mathcal{B}{3body}*\alpha+9\delta\alpha*\mathcal{B}{3body}}{9(1+\alpha+\delta\alpha)} $$

$$ \mathcal{B}(K_1^0\to K^+\pi^-\pi^0)=\frac{2}{3}\times \mathcal{B}(K_1\to K\rho)+\frac{4}{9}\times \mathcal{B}(K_1\to K^(892)\pi)+\frac{4}{9}\times \mathcal{B}(K_1\to K^0(1430)\pi)=\frac{6\mathcal{B}{3body}+4\mathcal{B}{3body}*\alpha}{9(1+\alpha+\delta\alpha)} $$

$$

\beta^{-1} = 1 - \frac{\mathcal{B}(K_1^+\to K^+\pi^-\pi^+)}{\mathcal{B}(K^0_1\to K^+\pi^-\pi^0)} = 1-\frac{3+4\alpha+9\delta_\alpha}{6+4\alpha}=\frac{3-9\delta_\alpha}{6+4\alpha} $$

$$ \alpha=\frac{3}{4}[\beta(1-3\delta_\alpha)-2] $$