# How to obtain the tensor-product of two quantum operations (superoperators) explicitly?

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

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.

• 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 $1/2$? – Jacey Li Dec 5 '19 at 0:59
• Yes, given that $\sum_iE_i^\dagger E_i=I$. No. – DaftWullie Dec 5 '19 at 10:18