Exercise 10.3#

For \(|\psi\rangle = a|000\rangle + b|111\rangle\)#

Consider three qubit bit flip code \(|0_{L}\rangle = |000\rangle\) and \(|1_L\rangle = |111\rangle\), if we measure \(Z_1Z_2\) of an arbitrary noiseless state \(|\psi\rangle = a|0_L\rangle + b|1_L\rangle = a|000\rangle + b|111\rangle\), the outcome should be eigenvalue of \(Z_1Z_2\). Since \(Z|0\rangle = |0\rangle\) and \(Z|1\rangle = -|1\rangle\), we have

(1)#\[ Z_1Z_2 |000\rangle = |000\rangle, Z_1Z_2 |111\rangle = |111\rangle \]

and

Therefore, \(|\psi\rangle\) is the eigenstate of \(Z_1Z_2\) and the outcome of measuring \(Z_1Z_2\) for \(|\psi\rangle\) is \(+1\). The probability of getting \(+1\) is then obtained by

(3)#\[\begin{split} \begin{align} {\rm Pr}(m_{12}=+1) =& \langle \psi|(Z_1Z_2)^{\dagger}Z_1Z_2|\psi\rangle \\ =& (a^*\langle 000| + b^*\langle 111| )(a| 000\rangle + b| 111\rangle ) = 1 \end{align} \end{split}\]

After measuring \(Z_1Z_2\) we obtain post-measurement state \(Z_1Z_2|\psi\rangle/P(m_{12}=1) = |\psi\rangle\). Note that

it means that the outcome of measuring \(Z_2Z_3\) for \(|\psi\rangle\) is \(+1\), and similarly, the probability of getting \(+1\) is

(5)#\[\begin{split} \begin{align} {\rm Pr}(m_{23}=+1) =& \langle \psi|(Z_2Z_3)^{\dagger}Z_2Z_3|\psi\rangle \\ =& (a^*\langle 000| + b^*\langle 111| )(a| 000\rangle + b| 111\rangle ) = 1 \end{align} \end{split}\]

and we obtain post-measurement state \(Z_2Z_3|\psi\rangle/P(m_{23}=1) = |\psi\rangle\).

(6)#\[\begin{split} \begin{align} P_0 = |000\rangle\langle 000| + |111\rangle\langle 111|\\ P_1 = |100\rangle\langle 100| + |011\rangle\langle 011|\\ P_2 = |010\rangle\langle 010| + |101\rangle\langle 101|\\ P_3 = |001\rangle\langle 001| + |110\rangle\langle 110|\\ \end{align} \end{split}\]

Note that \(P_0, P_1, P_2, P_3\) are all Hermitian operators, so the measurement outcome of them are their eigenvalues. Notice that \(\langle \psi|P_{0}|\psi\rangle = 1\) and \(\langle \psi|P_{1}|\psi\rangle = \langle \psi|P_{2}|\psi\rangle=\langle \psi|P_{3}|\psi\rangle=0\), so the probability of measuring \(P\) is

(7)#\[ \begin{align} p =& \langle \psi|(P)^{\dagger}P|\psi\rangle = \langle \psi|P_0|\psi\rangle = 1 \end{align} \]

and the post measurement state is

(8)#\[ \frac{P_0|\psi\rangle}{\langle \psi|P_{0}|\psi\rangle} = (|000\rangle\langle 000| + |111\rangle\langle 111|)(a|000\rangle + b|111\rangle) = a|000\rangle + b|111\rangle = |\psi\rangle \]

Thus, the probability of getting some measurement outcome is all \(1\) for both measuring projectors and measuring \(Z_1Z_2\) and \(Z_2Z_3\), and the post-measurement state is the same as the post-measurement state after measuring \(Z_1Z_2\) and \(Z_2Z_3\).

For \(|\psi\rangle = a|100\rangle + b|011\rangle\)#

Consider a bit flip happen on the first qubit, if we measure \(Z_1Z_2\) the outcome should be eigenvalue of \(Z_1Z_2\). Since \(Z|0\rangle = |0\rangle\) and \(Z|1\rangle = -|1\rangle\), we have

(9)#\[ Z_1Z_2 |100\rangle = -|100\rangle, Z_1Z_2 |011\rangle = -|011\rangle \]

and

Therefore, \(|\psi\rangle\) is the eigenstate of \(Z_1Z_2\) and the outcome of measuring \(Z_1Z_2\) for \(|\psi\rangle\) is \(-1\). The probability of getting \(-1\) is then obtained by

(11)#\[\begin{split} \begin{align} {\rm Pr}(m_{12}=-1) =& \langle \psi|(Z_1Z_2)^{\dagger}Z_1Z_2|\psi\rangle \\ =& (a^*\langle 100| + b^*\langle 011| )(a| 100\rangle + b| 011\rangle ) = 1 \end{align} \end{split}\]

After measuring \(Z_1Z_2\) we obtain post-measurement state \(Z_1Z_2|\psi\rangle/P(m_{12}=1) = -|\psi\rangle\). Note that

it means that the outcome of measuring \(Z_2Z_3\) for \(|\psi\rangle\) is \(+1\), and the probability of getting \(+1\) is

(13)#\[\begin{split} \begin{align} {\rm Pr}(m_{23}=+1) =& \langle \psi|(Z_2Z_3)^{\dagger}Z_2Z_3|\psi\rangle \\ =& (-a^*\langle 000| - b^*\langle 111| )(-a| 000\rangle - b| 111\rangle ) = 1 \end{align} \end{split}\]

and we obtain post-measurement state \(Z_2Z_3|\psi\rangle/P(m_{23}=1) = -|\psi\rangle\).

Now we measure state \(|\psi\rangle = a|100\rangle + b|011\rangle\) with projector \(P = P_0 + P_1 + P_2 + P_3\). Notice that \(\langle \psi|P_{1}|\psi\rangle = 1\) and \(\langle \psi|P_{0}|\psi\rangle = \langle \psi|P_{2}|\psi\rangle=\langle \psi|P_{3}|\psi\rangle=0\), so the probability of measuring \(P\) is

(14)#\[ \begin{align} p =& \langle \psi|(P)^{\dagger}P|\psi\rangle = \langle \psi|P_1|\psi\rangle = 1 \end{align} \]

and the post measurement state is

(15)#\[ \frac{P_1|\psi\rangle}{\langle \psi|P_{1}|\psi\rangle} = (|100\rangle\langle 100| + |011\rangle\langle 011|)(a|100\rangle + b|011\rangle) = a|100\rangle + b|011\rangle = |\psi\rangle \]

For \(|\psi\rangle = a|010\rangle + b|101\rangle\)#

Consider a bit flip happen on the first qubit, if we measure \(Z_1Z_2\) the outcome should be eigenvalue of \(Z_1Z_2\). Since \(Z|0\rangle = |0\rangle\) and \(Z|1\rangle = -|1\rangle\), we have

(16)#\[ Z_1Z_2 |010\rangle = -|010\rangle, Z_1Z_2 |101\rangle = -|101\rangle \]

and

Therefore, \(|\psi\rangle\) is the eigenstate of \(Z_1Z_2\) and the outcome of measuring \(Z_1Z_2\) for \(|\psi\rangle\) is \(-1\). The probability of getting \(-1\) is then obtained by

(18)#\[\begin{split} \begin{align} {\rm Pr}(m_{12}=-1) =& \langle \psi|(Z_1Z_2)^{\dagger}Z_1Z_2|\psi\rangle \\ =& (a^*\langle 010| + b^*\langle 101| )(a| 010\rangle + b| 101\rangle ) = 1 \end{align} \end{split}\]

After measuring \(Z_1Z_2\) we obtain post-measurement state \(Z_1Z_2|\psi\rangle/P(m_{12}=1) = -|\psi\rangle\). Note that

it means that the outcome of measuring \(Z_2Z_3\) for \(|\psi\rangle\) is \(-1\), and the probability of getting \(-1\) is

(20)#\[\begin{split} \begin{align} {\rm Pr}(m_{23}=-1) =& \langle \psi|(Z_2Z_3)^{\dagger}Z_2Z_3|\psi\rangle \\ =& (-a^*\langle 000| - b^*\langle 111| )(-a| 000\rangle - b| 111\rangle ) = 1 \end{align} \end{split}\]

and we obtain post-measurement state \(-Z_2Z_3|\psi\rangle/P(m_{23}=1) = |\psi\rangle\).

Now we measure state \(|\psi\rangle = a|010\rangle + b|101\rangle\) with projector \(P = P_0 + P_1 + P_2 + P_3\). Notice that \(\langle \psi|P_{2}|\psi\rangle = 1\) and \(\langle \psi|P_{0}|\psi\rangle = \langle \psi|P_{1}|\psi\rangle=\langle \psi|P_{3}|\psi\rangle=0\), so the probability of measuring \(P\) is

(21)#\[ \begin{align} p =& \langle \psi|(P)^{\dagger}P|\psi\rangle = \langle \psi|P_2|\psi\rangle = 1 \end{align} \]

and the post measurement state is

(22)#\[ \frac{P_2|\psi\rangle}{\langle \psi|P_{2}|\psi\rangle} = (|010\rangle\langle 010| + |101\rangle\langle 101|)(a|010\rangle + b|101\rangle) = a|010\rangle + b|101\rangle = |\psi\rangle \]

For \(|\psi\rangle = a|001\rangle + b|110\rangle\)#

Consider a bit flip happen on the first qubit, if we measure \(Z_1Z_2\) the outcome should be eigenvalue of \(Z_1Z_2\). Since \(Z|0\rangle = |0\rangle\) and \(Z|1\rangle = -|1\rangle\), we have

(23)#\[ Z_1Z_2 |001\rangle = |001\rangle, Z_1Z_2 |110\rangle = |110\rangle \]

and

Therefore, \(|\psi\rangle\) is the eigenstate of \(Z_1Z_2\) and the outcome of measuring \(Z_1Z_2\) for \(|\psi\rangle\) is \(+1\). The probability of getting \(+1\) is then obtained by

(25)#\[\begin{split} \begin{align} {\rm Pr}(m_{12}=+1) =& \langle \psi|(Z_1Z_2)^{\dagger}Z_1Z_2|\psi\rangle \\ =& (a^*\langle 001| + b^*\langle 110| )(a| 001\rangle + b| 110\rangle ) = 1 \end{align} \end{split}\]

After measuring \(Z_1Z_2\) we obtain post-measurement state \(Z_1Z_2|\psi\rangle/P(m_{12}=1) =|\psi\rangle\). Note that

it means that the outcome of measuring \(Z_2Z_3\) for \(|\psi\rangle\) is \(-1\), and the probability of getting \(-1\) is

(27)#\[\begin{split} \begin{align} {\rm Pr}(m_{23}=-1) =& \langle \psi|(Z_2Z_3)^{\dagger}Z_2Z_3|\psi\rangle \\ =& (a^*\langle 000| + b^*\langle 111| )(a| 000\rangle +b| 111\rangle ) = 1 \end{align} \end{split}\]

and we obtain post-measurement state \(Z_2Z_3|\psi\rangle/P(m_{23}=1) = -|\psi\rangle\).

Now we measure state \(|\psi\rangle = a|001\rangle + b|110\rangle\) with projector \(P = P_0 + P_1 + P_2 + P_3\). Notice that \(\langle \psi|P_{3}|\psi\rangle = 1\) and \(\langle \psi|P_{0}|\psi\rangle = \langle \psi|P_{1}|\psi\rangle=\langle \psi|P_{2}|\psi\rangle=0\), so the probability of measuring \(P\) is

(28)#\[ \begin{align} p =& \langle \psi|(P)^{\dagger}P|\psi\rangle = \langle \psi|P_3|\psi\rangle = 1 \end{align} \]

and the post measurement state is

(29)#\[ \frac{P_3|\psi\rangle}{\langle \psi|P_{3}|\psi\rangle} = (|001\rangle\langle 001| + |110\rangle\langle 110|)(a|001\rangle + b|110\rangle) = a|001\rangle + b|110\rangle = |\psi\rangle \]

From above calculation we know that, measuring \(Z_1Z_2\) followed by \(Z_2Z_3\) result in the same measurement statistics and post-measurement states as measuring the four projectors.