How can I get an unitary operator acting on the composed system from an unitary operator that acts on a subspace of the composed system? In $S[42]$ in the supplemental material of this paper Vacancy-like dressed states in topological waveguide QED, the author wanted to construct an unitary operator $\widetilde{U}_S$ from
another unitary operator $U_s$, namely$$\widetilde{U}_S=\alpha\left|e \right>\left<e \right|+\sum_{i}{\beta_i\left|e \right>\left<i \right|+\gamma_i\left|i \right>\left<e \right|+U_s}$$where $i=1, 2, ..., N$ denotes the state that there's a photon on the i-th lattice in the subspace spaned by $\{\left|vac \right>,\left|1 \right>, \left|2 \right>, ..., \left|N \right>\}$ and $U_s$ is an unitary operator corresponding to this subspace, $\left|e \right>$ denotes the excited state in another two-level subsystem spaned by $\{\left|e \right>,\left|g \right>\}$, $\alpha, \beta_i$ and $\gamma_i$ are unknown constants, the Hamiltonian of this composed system was commute with the total excitations $\hat{N}=\sigma_+\sigma_-+\sum_{i}{a^{\dagger}_ia_i}$. So the key is to determine these constants to make $\widetilde{U_s}$ become unitary. The author said that "the unitary condition implies that $|\alpha|=1, \beta_i=\gamma_i=0\ \forall i$".Question 1: how can I get that result by the condition$$\widetilde{U}^{\dagger}_s\widetilde{U}_s=1,U^{\dagger}_sU_s=1$$I've tried to write the complete expression $\widetilde{U}_s\widetilde{U}^{\dagger}_s$ $$(|\alpha|^2+\alpha^* U_s+\sum_{i}{|\beta_i|^2})\left|e \right>\left<e \right|+\alpha \left|e \right>\left<e \right|U^{\dagger}_s+(\sum_{i}{\alpha \gamma^*_i+\gamma^*_iU_s})\left|e \right>\left<i \right|+\sum_{i}{\beta_i}\left|e \right>\left<i \right|U^{\dagger}_s+(\sum_{i}{\alpha^*\gamma_i+\beta^*_iU_s})\left|i \right>\left<e \right|+\sum_{i}{\gamma_i\left|i \right>\left<e \right|U^{\dagger}_s}+\sum_{i}{|\gamma_i|^2}\left|i \right>\left<i \right|+I=I$$but it's not obvious for me that $|\alpha|=1, \beta_i=\gamma_i=0\ \forall i$.Question 2: Under the condition $|\alpha|=1, \beta_i=\gamma_i=0\ \forall i$, $\widetilde{U}_s$ is written as$$\widetilde{U}_s=e^{i\phi}\left|e \right>\left<e \right|+U_s$$How can I get $\widetilde{U}^{\dagger}_s\widetilde{U}_s=1$ from this expression for $\widetilde{U}_s$?Much appriciation for help!
 A: I'm pretty sure I can answer my question. To do so, we notice that the Hilbert space of this composed system was spanned by the set of basis$$\{\left|e \right>,\left|g \right>\}\otimes\{\left|vac \right>,\left|1 \right>,\left|2 \right>,... ,\left|N \right>\}$$Then we consider the single-excitation regime, it's clear that we just need the basis like $\{\left|e,vac \right>,\left|g,i \right>\}$, whose the number of excitation equals to 1. Hence that we can wirte the matrix form of this unitary operator $\widetilde{U}_s$ under the set of basis$$\begin{pmatrix}\alpha & \beta_1 & \beta_2 & \cdots & \beta_N \\ \gamma_1 & U_{11} & U_{12} & \cdots & U_{1N} \\ \vdots & & \ddots & & \vdots \\ \gamma_N & U_{N1} & U_{N2} & \cdots & U_{NN}\end{pmatrix}=\begin{pmatrix}\alpha & \vec{\beta} \\ \vec{\gamma}^T & U_s\end{pmatrix}$$then apply $\widetilde{U}_s\widetilde{U}^{\dagger}_s=1$, we have$$\begin{pmatrix}\alpha & \vec{\beta} \\ \vec{\gamma}^T & U_s\end{pmatrix}\begin{pmatrix}\alpha^* & \vec{\gamma}^* \\ {\vec{\beta}^T}^* & U^{\dagger}_s\end{pmatrix}=\begin{pmatrix}|\alpha|^2+\vec{\beta}\cdot{\vec{\beta}^T}^* & \alpha\vec{\gamma}^*+\vec{\beta}U^{\dagger}_s \\ \alpha^*{\vec{\gamma}^*}^T+U_s{\vec{\beta}^*}^T & \vec{\gamma}^T\cdot \vec{\gamma}^*+U_sU^{\dagger}_s\end{pmatrix}$$Now look at $\vec{\gamma}^T\cdot \vec{\gamma}^*+U_sU^{\dagger}_s$ the factor $\vec{\gamma}^T\cdot \vec{\gamma}^*$ equals to $0$ due to the unitary condition for $\widetilde{U}_s$ and $U_s$. Then look at the $\alpha^*{\vec{\gamma}^*}^T+U_s{\vec{\beta}^*}^T=U_s{\vec{\beta}^*}^T$, this term will also be $0$ due to the unitary condition, so we have$$U_s{\vec{\beta}^*}^T=0$$Now you can see $U_s$ is a unitary matrix, it's determinant $\det (U_s)\ne 0$ and this equation for $\vec{\beta}^*$ has the only solution $0$.
