# Density matrix derivation of an entangled state in amplitude dampling channel

The density matrix for an entangled state when both the qubits decohere with probability D1 and D2 in amplitude damping channel is given as $$\rho_d$$ in this paper on page 2. I am reproducing it here, $$\begin{bmatrix} |\alpha^2|+|\beta^2|D_1 D_2 & 0 & 0 & \sqrt{\bar{D_1}\bar{D_2}}\alpha^*\beta \\ 0 & D_1 \bar{D_2}|\beta^2| & 0 & 0 \\ 0 & 0 & D_2 \bar{D_1}|\beta^2| & 0 \\ \sqrt{\bar{D_1}\bar{D_2}}\alpha\beta^* & 0 & 0 & \bar{D_2} \bar{D_1}|\beta^2| \end{bmatrix}$$ I am deriving the same expression using amplitude damping modeling, but I am not getting these non-diagonal terms $$\sqrt{\bar{D_1}\bar{D_2}}\alpha \beta$$ at matrix element $$a_{34}$$ and $$a_{43}$$ positions. Solving this using amplitude damping modeling we get, $$\alpha \vert 0_S\rangle \vert 0_S\rangle \otimes \vert 0_E\rangle$$ = $$\alpha \vert 0_S\rangle\vert 0_S\rangle \otimes \vert 0_E\rangle$$ similarly, $$$$\begin{split} \beta \vert 1_S\rangle\vert 1_S\rangle \otimes \vert 0_E\rangle & = \beta \bar{D_1}\bar{D_2}\vert 1_S\rangle \vert 1_S\rangle \otimes \vert 0_E\rangle\\ &= \beta \bar{D_1}D_2\vert 1_S\rangle\vert 0_S\rangle \otimes \vert 0_E\rangle\\ &=\beta \bar{D_2}D_1\vert 0_S\rangle\vert 1_S\rangle \otimes \vert 0_E\rangle\\ &=\beta D_1 D_2\vert 0_S\rangle\vert 0_S\rangle \otimes \vert 0_E\rangle \end{split}$$$$ Considering the density matrix of the above terms we get all the diagonal terms of the matrix. But I am not getting these non-diagonal terms. Please explain.

Thank you.