Exercise 10.2

Consider two eigenstates of \(X\) operator,

(1)\[ |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}} \]

We can write the corresponding projector as

(2)\[\begin{split} \begin{align} P_{+} &= |+\rangle\langle +| =\frac{1}{2}(|0\rangle\langle 0| + |0\rangle\langle 1|+|1\rangle\langle 0|+|1\rangle\langle 1| ) \\ P_{-} &= |-\rangle\langle -| =\frac{1}{2}(|0\rangle\langle 0| - |0\rangle\langle 1|-|1\rangle\langle 0|+|1\rangle\langle 1| ) \\ \end{align} \end{split}\]

Notice that \(P_{+} = P^{\dagger}_{+}\) and \(P_{-} = P^{\dagger}_{-}\). We can compute \( P_+ \rho P_+\)​ as

(3)\[\begin{split} \begin{align} pP_+ \rho P_+ =& \frac{1}{4}p(|0\rangle\langle 0| + |0\rangle\langle 1|+|1\rangle\langle 0|+|1\rangle\langle 1| )\rho \\ &\cdot(|0\rangle\langle 0| + |0\rangle\langle 1|+|1\rangle\langle 0|+|1\rangle\langle 1| ) \\ =& \frac{1}{4}p(|0\rangle\langle 0|\rho|0\rangle\langle 0| + |0\rangle\langle 0|\rho|0\rangle\langle 1|+|0\rangle\langle 0|\rho|1\rangle\langle 0|+|0\rangle\langle 0|\rho|1\rangle\langle 1| ) \\ &+\frac{1}{4}p(|0\rangle\langle 1|\rho|0\rangle\langle 0| + |0\rangle\langle 1|\rho|0\rangle\langle 1|+|0\rangle\langle 1|\rho|1\rangle\langle 0|+|0\rangle\langle 1|\rho|1\rangle\langle 1| ) \\ &+\frac{1}{4}p(|1\rangle\langle 0|\rho|0\rangle\langle 0| + |1\rangle\langle 0|\rho|0\rangle\langle 1|+|1\rangle\langle 0|\rho|1\rangle\langle 0|+|1\rangle\langle 0|\rho|1\rangle\langle 1| ) \\ &+\frac{1}{4}p(|1\rangle\langle 1|\rho|0\rangle\langle 0| + |1\rangle\langle 1|\rho|0\rangle\langle 1|+|1\rangle\langle 1|\rho|1\rangle\langle 0|+|1\rangle\langle 1|\rho|1\rangle\langle 1| ) \\ \end{align} \end{split}\]

Also, we can compute \( P_- \rho P_-\)​ as

(4)\[\begin{split} \begin{align} pP_- \rho P_- =& \frac{1}{4}p(|0\rangle\langle 0| - |0\rangle\langle 1|-|1\rangle\langle 0|+|1\rangle\langle 1| )\rho \\ &\cdot(|0\rangle\langle 0| - |0\rangle\langle 1|-|1\rangle\langle 0|+|1\rangle\langle 1| ) \\ =& \frac{1}{4}p(|0\rangle\langle 0|\rho|0\rangle\langle 0| - |0\rangle\langle 0|\rho|0\rangle\langle 1|-|0\rangle\langle 0|\rho|1\rangle\langle 0|+|0\rangle\langle 0|\rho|1\rangle\langle 1| ) \\ &+\frac{1}{4}p(-|0\rangle\langle 1|\rho|0\rangle\langle 0| + |0\rangle\langle 1|\rho|0\rangle\langle 1|+|0\rangle\langle 1|\rho|1\rangle\langle 0|-|0\rangle\langle 1|\rho|1\rangle\langle 1| ) \\ &+\frac{1}{4}p(-|1\rangle\langle 0|\rho|0\rangle\langle 0| + |1\rangle\langle 0|\rho|0\rangle\langle 1|+|1\rangle\langle 0|\rho|1\rangle\langle 0|-|1\rangle\langle 0|\rho|1\rangle\langle 1| ) \\ &+\frac{1}{4}p(|1\rangle\langle 1|\rho|0\rangle\langle 0| - |1\rangle\langle 1|\rho|0\rangle\langle 1|-|1\rangle\langle 1|\rho|1\rangle\langle 0|+|1\rangle\langle 1|\rho|1\rangle\langle 1| ) \\ \end{align} \end{split}\]

Combine eq. (3) and eq. (4), we have

(5)\[\begin{split} \begin{align} pP_+ \rho P_+ + pP_- \rho P_- =& \frac{1}{2}p(|0\rangle\langle 0|\rho|0\rangle\langle 0| +|0\rangle\langle 0|\rho|1\rangle\langle 1| ) +\frac{1}{2}p( |0\rangle\langle 1|\rho|0\rangle\langle 1|+|0\rangle\langle 1|\rho|1\rangle\langle 0|) \\ &+\frac{1}{2}p(|1\rangle\langle 0|\rho|0\rangle\langle 1|+|1\rangle\langle 0|\rho|1\rangle\langle 0|) +\frac{1}{2}p(|1\rangle\langle 1|\rho|0\rangle\langle 0|+|1\rangle\langle 1|\rho|1\rangle\langle 1| ) \\ =& \frac{1}{2}p(|0\rangle\langle 0|+|1\rangle\langle 1| )\rho(|0\rangle\langle 0| +|1\rangle\langle 1|) +\frac{1}{2}p(|0\rangle\langle 1|+|1\rangle\langle 0|)\rho(|0\rangle\langle 1|+|1\rangle\langle 0|) \\ =& \frac{1}{2}p I\rho I + \frac{1}{2}p X\rho X = \frac{1}{2}p \rho + \frac{1}{2}p X\rho X \end{align} \end{split}\]

From eq. (5), we could find that

(6)\[ 2pP_+ \rho P_+ + 2pP_- \rho P_- = p\rho + pX\rho X \iff pX\rho X = 2pP_+ \rho P_+ + 2pP_- \rho P_- - p\rho \]

With above equation, we could re-write the bit flip channel as

(7)\[\begin{split} \begin{align} \mathcal{E}(\rho) &= (1-p)\rho + pX\rho X \\ &= (1-p)\rho+2pP_+ \rho P_+ + 2pP_- \rho P_- - p\rho \\ &= (1-2p)\rho+2pP_+ \rho P_+ + 2pP_- \rho P_- \end{align} \end{split}\]