I have an amplitude damping channel, denoted as a superoperator $\mathcal{E}$ with operator elements

\begin{matrix} E_1=\begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-r} \end{pmatrix},\quad E_2=\begin{pmatrix} 0 & \sqrt{r} \\ 0 & 0 \end{pmatrix} \end{matrix} I am confused that how to explicitly obtain the $\mathcal{E}^{\otimes 2}$ in matrix form?

Also, I am trying to understand what is $\mathcal{E}^{\otimes 2}(\rho)$, where $\rho=1/2|00\rangle\langle 00|+1/2|11\rangle\langle 11|$?


I guess that what you're after is $ \mathcal{E}^{\otimes 2} $ is defined by the 4 operator elements $$ E_1\otimes E_1,E_1\otimes E_2,E_2\otimes E_1,E_2\otimes E_2. $$

If you apply this to $\rho$, you get $$ \frac{1+r^2}{2}|00\rangle\langle 00|+\frac{(1-r)^2}{2}|11\rangle\langle 11|+\frac{r(1-r)}{2}(|01\rangle\langle|01|+|10\rangle\langle 10|). $$ An important check is that this still has trace 1.

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  • $\begingroup$ Thank you. Should these 4 operator elements satisfy the completeness relationship $\sum_{i,j} (E_i\otimes E_j)^\dagger (E_i\otimes E_j)=I$? If so, would each operator element have a coefficient $1/2$? $\endgroup$ – Jacey Li Dec 5 '19 at 0:59
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    $\begingroup$ Yes, given that $\sum_iE_i^\dagger E_i=I$. No. $\endgroup$ – DaftWullie Dec 5 '19 at 10:18

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