# How can I solve this secular (eigen) equation to get non-zero solution?

I have the Hamiltonian of the studied system which contains a two-level atom (with excited state $$\left|e \right>$$ and ground state $$\left|g \right>$$) interacting with a continuum waveguide:$$H=\sum_{k}{\Delta_ka^{\dagger}_ka_k}+\sum_{k}{g_ka_k\sigma_++g^*_ka^{\dagger}_k\sigma_-}$$where $$\Delta_k$$ is detune, $$g_k$$ and $$g^*_k$$ is coupling strength. In the single-excitation regime, the bound state $$\left|E \right>$$ is$$\left|E \right>=c_e\left|e \right>\left|vac \right>+\sum_{k}{c_k\left|g \right>\left|1_k \right>}$$and satisfies secular equation$$H\left|E \right>=E\left|E \right>$$pluging $$H$$ and $$\left|E \right>$$ into the secular equation, I got the equation set that reads\left\{\begin{aligned} \sum_{k}{g^*_kc_k}=Ec_e \\ \Delta_kc_k+g_kc_e=Ec_k \\ \end{aligned} \right.Firstly I solved the second equa and got $$c_k=\frac{g_kc_e}{E-\Delta_k}$$ but once I pluged $$c_k=\frac{g_kc_e}{E-\Delta_k}$$ into the first equation would result in$$c_e\sum_{k}{\frac{|g_k|^2}{E-\Delta_k}=Ec_e}$$ whcih will cancel $$c_e$$ on both sides. In paper the author got $$c_k=\frac{g_kc_e}{E-\Delta_k}$$ then he made an approximation that $$c_e\simeq 1$$ and solved $$c_k$$. Is it possible to solve that secular equation without any approximation?

All I need to do to solve this equation is using the probability conservation condition$$|c_e|^2+\sum_{k}{|c_k|^2}=1$$assuming $$c_e$$ is real, which means$$c^2_e+c^2_e\sum_{k}{\frac{|g_k|^2}{(E-\Delta_k)^2}}=1$$