Physics Asked on June 29, 2021
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$$widetilde{U}_S=alphaleft|e right>left<e right|+sum_{i}{beta_ileft|e right>left<i right|+gamma_ileft|i right>left<e right|+U_s}$$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$$widetilde{U}^{dagger}_swidetilde{U}_s=1,U^{dagger}_sU_s=1$$I’ve tried to write the complete expression $widetilde{U}_swidetilde{U}^{dagger}_s$ $$(|alpha|^2+alpha^* U_s+sum_{i}{|beta_i|^2})left|e right>left<e right|+alpha left|e right>left<e right|U^{dagger}_s+(sum_{i}{alpha gamma^*_i+gamma^*_iU_s})left|e right>left<i right|+sum_{i}{beta_i}left|e right>left<i right|U^{dagger}_s+(sum_{i}{alpha^*gamma_i+beta^*_iU_s})left|i right>left<e right|+sum_{i}{gamma_ileft|i right>left<e right|U^{dagger}_s}+sum_{i}{|gamma_i|^2}left|i right>left<i right|+I=I$$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$$widetilde{U}_s=e^{iphi}left|e right>left<e right|+U_s$$How can I get $widetilde{U}^{dagger}_swidetilde{U}_s=1$ from this expression for $widetilde{U}_s$?
Much appriciation for help!
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$$begin{pmatrix}alpha & beta_1 & beta_2 & cdots & beta_N gamma_1 & U_{11} & U_{12} & cdots & U_{1N} vdots & & ddots & & vdots gamma_N & U_{N1} & U_{N2} & cdots & U_{NN}end{pmatrix}=begin{pmatrix}alpha & vec{beta} vec{gamma}^T & U_send{pmatrix}$$then apply $widetilde{U}_swidetilde{U}^{dagger}_s=1$, we have$$begin{pmatrix}alpha & vec{beta} vec{gamma}^T & U_send{pmatrix}begin{pmatrix}alpha^* & vec{gamma}^* {vec{beta}^T}^* & U^{dagger}_send{pmatrix}=begin{pmatrix}|alpha|^2+vec{beta}cdot{vec{beta}^T}^* & alphavec{gamma}^*+vec{beta}U^{dagger}_s alpha^*{vec{gamma}^*}^T+U_s{vec{beta}^*}^T & vec{gamma}^Tcdot vec{gamma}^*+U_sU^{dagger}_send{pmatrix}$$Now look at $vec{gamma}^Tcdot vec{gamma}^*+U_sU^{dagger}_s$ the factor $vec{gamma}^Tcdot vec{gamma}^*$ 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 $0$.
Correct answer by Guoqing on June 29, 2021
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