Let us consider a two-qubit system (A and R) initially entangled as given by: $$ \left| {{\Psi _{AR}}} \right\rangle = \alpha \left| {{0_A}} \right\rangle \otimes \left| {{1_R}} \right\rangle + \beta \left| {{1_A}} \right\rangle \otimes \left| {{0_R}} \right\rangle $$ where ${\left| \alpha \right|^2} + {\left| \beta \right|^2} = 1$. By using the fact that Rob’s detector is the only one which interacts with the field, the initial state of the total system is: $$\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle = \left| {{\Psi _{AR}}} \right\rangle \otimes \left| {{0_M}} \right\rangle $$ where $\left| {{0_M}} \right\rangle $ the Minkowski vacuum. The final state of the qubit-field system can be described as: $$\left| {\Psi _\infty ^{AB\phi }} \right\rangle = \left( {I + a_{RI}^\dagger(\lambda )R - {a_{RI}}(\bar \lambda ){R^\dagger}} \right)\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle $$ where $I$ is the identity matrix with the same dimension of $\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle $. ${D^\dagger}\left| 1 \right\rangle = D\left| 0 \right\rangle = 0$, $D\left| 1 \right\rangle = \left| 0 \right\rangle $, and ${D^\dagger}\left| 0 \right\rangle = \left| 1 \right\rangle $, ($D \in A,R$) are the raising and lowering operators for the qubit.
Using the first two equations in the last one, the authors got: $$\left| {\Psi _\infty ^{AB\phi }} \right\rangle = \left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle + \alpha \left| {{0_A}} \right\rangle \otimes \left| {{0_R}} \right\rangle \otimes \left( {a_{RI}^\dagger(\lambda )\left| {{0_M}} \right\rangle } \right)\\ + \beta \left| {{1_A}} \right\rangle \otimes \left| {{1_R}} \right\rangle \otimes \left( {{a_{RI}}(\bar \lambda )\left| {{0_M}} \right\rangle } \right)$$ Please, could someone explain to me how the authors got this equation?
The intended equation is Eq. (3.6) in 10.1103/PhysRevA.80.032315
First, $$ \left| {\Psi _\infty ^{AB\phi }} \right\rangle = \left( {I + a_{RI}^\dagger(\lambda )R - {a_{RI}}(\bar \lambda ){R^\dagger}} \right)\left| {\Psi _{ - \infty }^{AB\phi }} \right\rangle = \left( {I + a_{RI}^\dagger(\lambda )R - {a_{RI}}(\bar \lambda ){R^\dagger}} \right) \left( \left| {{\Psi _{AR}}} \right\rangle \otimes \left| {{0_M}} \right\rangle \right) \\ = \left| {{\Psi _{AR}}} \right\rangle \otimes \left| {{0_M}} \right\rangle + a_{RI}^\dagger(\lambda )R \left( \left| {{\Psi _{AR}}} \right\rangle \otimes \left| {{0_M}} \right\rangle \right) - {a_{RI}}(\bar \lambda ){R^\dagger}\left( \left| {{\Psi _{AR}}} \right\rangle \otimes \left| {{0_M}} \right\rangle \right) $$ Then, for the second term, $ a_{RI}^\dagger(\lambda )R \left( \left| {{\Psi _{AR}}} \right\rangle \otimes \left| {{0_M}} \right\rangle \right) = R\left| {{\Psi _{AR}}} \right\rangle \otimes a_{RI}^\dagger(\lambda )\left| {{0_M}} \right\rangle, $ where $$ R\left| {{\Psi _{AR}}} \right\rangle = R \left( \alpha \left| {{0_A}} \right\rangle \otimes \left| {{1_R}} \right\rangle + \beta \left| {{1_A}} \right\rangle \otimes \left| {{0_R}} \right\rangle \right) = \alpha \left| {{0_A}} \right\rangle \otimes R\left| {{1_R}} \right\rangle + \beta \left| {{1_A}} \right\rangle \otimes R \left| {{0_R}} \right\rangle \\ = \alpha \left| {{0_A}} \right\rangle \otimes \left| {{0_R}} \right\rangle + \beta \left| {{1_A}} \right\rangle \otimes R 0 = \alpha \left| {{0_A}} \right\rangle \otimes \left| {{0_R}} \right\rangle . $$ Likewise for the third term, $- {a_{RI}}(\bar \lambda ){R^\dagger}\left( \left| {{\Psi _{AR}}} \right\rangle \otimes \left| {{0_M}} \right\rangle \right)$.