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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?

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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$$c^2_e+c^2_e\sum_{k}{\frac{|g_k|^2}{(E-\Delta_k)^2}}=1$$

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