\(\varepsilon_{\mathbf{n}+\mathbf{1}}^{\varepsilon}=\varepsilon_{\mathbf{n}}^{\varepsilon}+\lambda\sqrt{\frac{3}{2}}n_{n+1}\ \ ,\ \ \ \ {\text{\ \ }\alpha}_{n+1}=\alpha_{n}+\lambda\ ,\)