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 $\rho=1/2|00\rangle\langle 00|+1/2|11\rangle\langle 11|$?


1 Answer 1


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.

  • $\begingroup$ 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, 2019 at 0:59
  • 1
    $\begingroup$ Yes, given that $\sum_iE_i^\dagger E_i=I$. No. $\endgroup$
    – DaftWullie
    Dec 5, 2019 at 10:18
  • $\begingroup$ Hello @DaftWullie , how do you derive those 4 operator elements? Which properties of Tensor product you had to use? Could you please elaborate? $\endgroup$ Jun 28, 2020 at 13:50
  • $\begingroup$ @DaftWullie I have same the question. Are these elements derived by seeing how the operation acts on separable states $\rho_1 \otimes \rho_2$, or even on elements of the same form where $\rho_1$ or $\rho_2$ have zero trace (aren't density ops)? Since every density operator of the product of two systems is a linear combination of such elements. When you do this, you get the operator elements you posted. $\endgroup$
    – dylan7
    Oct 21, 2020 at 0:48
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    $\begingroup$ If you just act $\mathcal{E}$ on the first qubit, the operators would be $E_1\otimes I$ and $E_2\otimes I$. Similarly, if you act it on the second qubit, the operators would be $I\otimes E_1$ and $I\otimes E_2$. Now, $\mathcal{E}\otimes \mathcal{E}$ is entirely equivalent to doing the noise on one qubit and then on the other. The net effect? All possible pairs of operators occur, the net effect of each being the product of the individual terms. $\endgroup$
    – DaftWullie
    Oct 21, 2020 at 6:44

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