Exercise 10.5#

Consider the 9-qubits Shor’s code with codeword as

(1)#

\begin{array}{r} \begin{matrix} | 0_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \\ | 1_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \end{matrix} \end{array}

We can encode an arbitrary state in the form $| ψ ⟩ = a | 0_{L} ⟩ + b | 1_{L} ⟩$ .

No error occur#

If there is no phase flip error occur, we have an arbitrary state as $| ψ ⟩ = a | 0_{L} ⟩ + b | 1_{L} ⟩$ . Since $X | 0 ⟩ = | 1 ⟩$ and $X | 1 ⟩ = | 0 ⟩$ we have

(2)#

\begin{array}{r} \begin{aligned} \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ + | 000 ⟩) (| 111 ⟩ + | 000 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} = | 0_{L} ⟩ \\ \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ - | 1111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ - | 000 ⟩) (| 111 ⟩ - | 000 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} = | 1_{L} ⟩ \end{aligned} \end{array}

Then measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ for $| ψ ⟩ = a | 0_{L} ⟩ + b | 1_{L} ⟩$ , the outcome is $+ 1$ , and the probability of getting $_{1}$ is

(3)#

\begin{array}{r} \begin{aligned} p (m_{123456} = - 1) & = ⟨ ψ | (X_{1} X_{2} X_{3} X_{4} X_{5} X_{6})^{†} X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩ \\ = (a^{*} ⟨ 0_{L} | + b^{*} ⟨ 1_{L} |) (a | 0_{L} ⟩ + b | 1_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(4)#

\frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩}{p (m_{123456} = - 1)} = a | 0_{L} ⟩ + b | 1_{L} ⟩

After measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ we perform the measurement $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ , and we have

(5)#

\begin{array}{r} \begin{matrix} \begin{matrix} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | 0_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 111 ⟩ + | 000 ⟩) (| 111 ⟩ + | 000 ⟩)}{2 \sqrt{2}} = | 0_{L} ⟩ \end{matrix} \\ X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | 1_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 111 ⟩ - | 000 ⟩) (| 111 ⟩ - | 000 ⟩)}{2 \sqrt{2}} = | 1_{L} ⟩ \end{matrix} \end{array}

Then measuring $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ for $| ψ ⟩ = a | 0_{L} ⟩ + b | 1_{L} ⟩$ , the outcome is $+ 1$ , and the probability of getting $+ 1$ is

(6)#

\begin{array}{r} \begin{aligned} p (m_{456789} = + 1) & = ⟨ ψ | (X_{4} X_{5} X_{6} X_{7} X_{8} X_{9})^{†} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩ \\ = (a^{*} ⟨ 0_{L} | + b^{*} ⟨ 1_{L} |) (a | 0_{L} ⟩ + b | 1_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(7)#

\frac{X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩}{p (m_{456789} = + 1)} = a | 0_{L} ⟩ + b | 1_{L} ⟩

Phase flip at first three qubits set#

If there is a phase flip error in the first three qubit, the state $| 0_{L} ⟩$ and $| 1_{L} ⟩$ becomes

(8)#

\begin{array}{r} \begin{matrix} | 0_{L} ⟩ \to | {\tilde{0}}_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \\ | 1_{L} ⟩ \to | {\tilde{1}}_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \end{matrix} \end{array}

and the arbitrary state becomes $| ψ ⟩ = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩$ . Since $X | 0 ⟩ = | 1 ⟩$ and $X | 1 ⟩ = | 0 ⟩$ we have

(9)#

\begin{array}{r} \begin{aligned} \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ - | 000 ⟩) (| 111 ⟩ + | 000 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} = - | {\tilde{0}}_{L} ⟩ \\ \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 1111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ + | 000 ⟩) (| 111 ⟩ - | 000 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} = - | {\tilde{1}}_{L} ⟩ \end{aligned} \end{array}

Then measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ for $| ψ ⟩ = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩$ , the outcome is $- 1$ , and the probability of getting $- 1$ is

(10)#

\begin{array}{r} \begin{aligned} p (m_{123456} = - 1) & = ⟨ ψ | (X_{1} X_{2} X_{3} X_{4} X_{5} X_{6})^{†} X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩ \\ = (a^{*} ⟨ {\tilde{0}}_{L} | + b^{*} ⟨ {\tilde{1}}_{L} |) (a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(11)#

\frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩}{p (m_{123456} = - 1)} = - a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩

After measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ we perform the measurement $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ , and we have

(12)#

\begin{array}{r} \begin{matrix} \begin{matrix} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | {\tilde{0}}_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 111 ⟩ + | 000 ⟩) (| 111 ⟩ + | 000 ⟩)}{2 \sqrt{2}} = | {\tilde{0}}_{L} ⟩ \end{matrix} \\ X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | {\tilde{1}}_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 111 ⟩ - | 000 ⟩) (| 111 ⟩ - | 000 ⟩)}{2 \sqrt{2}} = | {\tilde{1}}_{L} ⟩ \end{matrix} \end{array}

Then measuring $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ for $| ψ ⟩ = - a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩$ , the outcome is $+ 1$ , and the probability of getting $+ 1$ is

(13)#

\begin{array}{r} \begin{aligned} p (m_{456789} = + 1) & = ⟨ ψ | (X_{4} X_{5} X_{6} X_{7} X_{8} X_{9})^{†} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩ \\ = (- a^{*} ⟨ {\tilde{0}}_{L} | - b^{*} ⟨ {\tilde{1}}_{L} |) (- a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(14)#

\frac{X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩}{p (m_{456789} = + 1)} = - a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩

Phase flip at second three qubits set#

If there is a phase flip error in the second three qubit, the state $| 0_{L} ⟩$ and $| 1_{L} ⟩$ becomes

(15)#

\begin{array}{r} \begin{matrix} | 0_{L} ⟩ \to | {\tilde{0}}_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \\ | 1_{L} ⟩ \to | {\tilde{1}}_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \end{matrix} \end{array}

and the arbitrary state becomes $| ψ ⟩ = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩$ . Since $X | 0 ⟩ = | 1 ⟩$ and $X | 1 ⟩ = | 0 ⟩$ we have

(16)#

\begin{array}{r} \begin{aligned} \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ + | 000 ⟩) (| 111 ⟩ - | 000 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} = - | {\tilde{0}}_{L} ⟩ \\ \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ - | 000 ⟩) (| 111 ⟩ + | 000 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} = - | {\tilde{1}}_{L} ⟩ \end{aligned} \end{array}

Then measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ for $| ψ ⟩ = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩$ , the outcome is $- 1$ , and the probability of getting $- 1$ is

(17)#

\begin{array}{r} \begin{aligned} p (m_{123456} = - 1) & = ⟨ ψ | (X_{1} X_{2} X_{3} X_{4} X_{5} X_{6})^{†} X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩ \\ = (a^{*} ⟨ {\tilde{0}}_{L} | + b^{*} ⟨ {\tilde{1}}_{L} |) (a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(18)#

\frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩}{p (m_{123456} = - 1)} = - a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩

After measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ we perform the measurement $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ , and we have

\begin{array}{r} \begin{matrix} \begin{matrix} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | {\tilde{0}}_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 111 ⟩ - | 000 ⟩) (| 111 ⟩ + | 000 ⟩)}{2 \sqrt{2}} = - | {\tilde{0}}_{L} ⟩ \end{matrix} \\ X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | {\tilde{1}}_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 111 ⟩ + | 000 ⟩) (| 111 ⟩ - | 000 ⟩)}{2 \sqrt{2}} = - | {\tilde{1}}_{L} ⟩ \end{matrix} \end{array}

Then measuring $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ for $| ψ ⟩ = - a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩$ , the outcome is $- 1$ , and the probability of getting $- 1$ is

(19)#

\begin{array}{r} \begin{aligned} p (m_{456789} = - 1) & = ⟨ ψ | (X_{4} X_{5} X_{6} X_{7} X_{8} X_{9})^{†} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩ \\ = (- a^{*} ⟨ {\tilde{0}}_{L} | - b^{*} ⟨ {\tilde{1}}_{L} |) (- a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(20)#

\frac{X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩}{p (m_{456789} = - 1)} = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩

Phase flip at third three qubit set#

If there is a phase flip error in the second three qubit, the state $| 0_{L} ⟩$ and $| 1_{L} ⟩$ becomes

(21)#

\begin{array}{r} \begin{matrix} | 0_{L} ⟩ \to | {\tilde{0}}_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \\ | 1_{L} ⟩ \to | {\tilde{1}}_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \end{matrix} \end{array}

and the arbitrary state becomes $| ψ ⟩ = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩$ . Since $X | 0 ⟩ = | 1 ⟩$ and $X | 1 ⟩ = | 0 ⟩$ we have

(22)#

\begin{array}{r} \begin{aligned} \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ + | 111 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ + | 000 ⟩) (| 111 ⟩ + | 000 ⟩) (| 000 ⟩ - | 111 ⟩)}{2 \sqrt{2}} = | {\tilde{0}}_{L} ⟩ \\ \frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ - | 111 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} \\ = & \frac{(| 111 ⟩ - | 000 ⟩) (| 111 ⟩ - | 000 ⟩) (| 000 ⟩ + | 111 ⟩)}{2 \sqrt{2}} = | {\tilde{1}}_{L} ⟩ \end{aligned} \end{array}

Then measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ for $| ψ ⟩ = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩$ , the outcome is $+ 1$ , and the probability of getting $+ 1$ is

(23)#

\begin{array}{r} \begin{aligned} p (m_{123456} = + 1) & = ⟨ ψ | (X_{1} X_{2} X_{3} X_{4} X_{5} X_{6})^{†} X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩ \\ = (a^{*} ⟨ {\tilde{0}}_{L} | + b^{*} ⟨ {\tilde{1}}_{L} |) (a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(24)#

\frac{X_{1} X_{2} X_{3} X_{4} X_{5} X_{6} | ψ ⟩}{p (m_{123456} = + 1)} = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩

After measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ we perform the measurement $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ , and we have

(25)#

\begin{array}{r} \begin{matrix} \begin{matrix} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | {\tilde{0}}_{L} ⟩ = \frac{(| 000 ⟩ + | 111 ⟩) (| 111 ⟩ + | 000 ⟩) (| 111 ⟩ - | 000 ⟩)}{2 \sqrt{2}} = - | {\tilde{0}}_{L} ⟩ \end{matrix} \\ X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | {\tilde{1}}_{L} ⟩ = \frac{(| 000 ⟩ - | 111 ⟩) (| 111 ⟩ - | 000 ⟩) (| 111 ⟩ + | 000 ⟩)}{2 \sqrt{2}} = - | {\tilde{1}}_{L} ⟩ \end{matrix} \end{array}

Then measuring $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$ for $| ψ ⟩ = a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩$ , the outcome is $- 1$ , and the probability of getting $- 1$ is

(26)#

\begin{array}{r} \begin{aligned} p (m_{456789} = - 1) & = ⟨ ψ | (X_{4} X_{5} X_{6} X_{7} X_{8} X_{9})^{†} X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩ \\ = (a^{*} ⟨ {\tilde{0}}_{L} | + b^{*} ⟨ {\tilde{1}}_{L} |) (a | {\tilde{0}}_{L} ⟩ + b | {\tilde{1}}_{L} ⟩) = 1 \end{aligned} \end{array}

The post-measurement state is then

(27)#

\frac{X_{4} X_{5} X_{6} X_{7} X_{8} X_{9} | ψ ⟩}{p (m_{456789} = - 1)} = - a | {\tilde{0}}_{L} ⟩ - b | {\tilde{1}}_{L} ⟩

From above calculation we know that,

If there is no error, we measure $+ 1$ for $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ and $+ 1$ for $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$
If there is a phase flip error at the first three qubit set, we measure $- 1$ for $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ and $+ 1$ for $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$
If there is a phase flip error at the second three qubit set, we measure $- 1$ for $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ and $- 1$ for $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$
If there is a phase flip error at the third three qubit set, we measure $+ 1$ for $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ and $- 1$ for $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$

From above summary, we can easily see that, the syndrome measurement for detecting phase flip errors in the Shor code corresponds to measuring $X_{1} X_{2} X_{3} X_{4} X_{5} X_{6}$ and $X_{4} X_{5} X_{6} X_{7} X_{8} X_{9}$