这是一份umass麻省大学 MATH 397C作业代写的成功案例
A quantum circuit which models the phase-flip channel is shown in Fig. 6. Let $\rho_{S}$ be the first qubit input state while $(1-p)|0\rangle\langle 0|+p| 1\rangle\langle 1|$ be the second qubit input state. The circuit is the inverted controlled $-\sigma_{z}$ gate
$$
V=I \otimes|0\rangle\left\langle 0\left|+\sigma_{z} \otimes\right| 1\right\rangle\langle 1| .
$$
The output of this circuit is
$$
\begin{aligned}
&V\left(\rho_{S} \otimes[(1-p)|0\rangle\langle 0|+p| 1\rangle\langle 1|]\right) V^{\dagger} \
&=(1-p) \rho_{S} \otimes|0\rangle\left\langle 0\left|+p \sigma_{z} \rho_{S} \sigma_{z} \otimes\right| 1\right\rangle\langle 1|
\end{aligned}
$$
from which we obtain
$$
\mathcal{E}\left(\rho_{S}\right)=(1-p) \rho_{S}+p \sigma_{z} \rho_{S} \sigma_{z}
$$
The second qubit input state may be a pure state
$$
\left|\psi_{E}\right\rangle=\sqrt{1-p}|0\rangle+\sqrt{p}|1\rangle,
$$
for example. Then we find
$$
\mathcal{E}\left(\rho_{S}\right)=\operatorname{tr}{E}\left[V \rho{S} \otimes\left|\psi_{E}\right\rangle\left\langle\psi_{E}\right| V^{\dagger}\right]=E_{0} \rho_{S} E_{0}^{\dagger}+E_{1} \rho_{S} E_{1}^{\dagger}
$$
where the Kraus operators are
$$
E_{0}=\left\langle 0|V| \psi_{E}\right\rangle=\sqrt{1-p} I, E_{1}=\left\langle 1|V| \psi_{E}\right\rangle=\sqrt{p} \sigma_{z}
$$
MMATH 397C COURSE NOTES ：
(1) For any $a, b, c \in G$ we have
$$
\mu(a, \mu(b, c))=\mu(\mu(a, b), c) .
$$
(2) For any $a \in G$ there exists an element $e \in G$ such that
$$
\mu(a, e)=\mu(e, a)=a,
$$
where $e$ is called a unit element of $G$.
(3) For any $a \in G$ there exists an element $a^{\prime} \in G$ such that
$$
\mu\left(a, a^{\prime}\right)=\mu\left(a^{\prime}, a\right)=e,
$$
where $a^{\prime}$ is called an inverse element of $a$ and we write $a^{-1} \in G$.