Let's tackle each part step by step: --- ### **Given State (Before Circuit)** The encoded state after a phase flip on the last triplet is: \[ \frac{\alpha}{2^{3/2}} (|000\rangle + |111\rangle)(|000\rangle + |111\rangle)(|000\rangle - |111\rangle) + \frac{\beta}{2^{3/2}} (|000\rangle - |111\rangle)(|000\rangle - |111\rangle)(|000\rangle + |111\rangle) \] Let's denote the 9 qubits as \(q_0, q_1, \ldots, q_8\), grouped as triplets: - First: \(q_0, q_1, q_2\) - Second: \(q_3, q_4, q_5\) - Third: \(q_6, q_7, q_8\) --- ## **(a) State after first two columns (CNOTs and Hadamards)** ### **Step 1: Apply CNOTs** - The first two columns are CNOTs with two targets. - For each triplet, the control is the first qubit in the triplet, targets are the next two qubits. Let's consider \( |000\rangle \) and \( |111\rangle \) for each triplet: - CNOT with control on qubit 0, targets on 1 and 2: - \( |000\rangle \rightarrow |000\rangle \) - \( |111\rangle \rightarrow |111\rangle \) - So the state is unchanged by these CNOTs because in both \( |000\rangle \) and \( |111\rangle \), the two targets are already equal to the control. So after the first column of CNOTs, the state is unchanged. --- ### **Step 2: Apply Hadamard Gates** Hadamard is applied to each qubit in the first and third triplets: \(q_0, q_1, q_2, q_6, q_7, q_8\). Recall: - \( H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \) - \( H|1\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) \) Let's look at what happens to \( |000\rangle + |111\rangle \) when Hadamard is applied to all 3 qubits: \[ H^{\otimes 3}(|000\rangle + |111\rangle) \] - \( H^{\otimes 3}|000\rangle = \frac{1}{2^{3/2}} \sum_{x=0}^{7} |x\rangle \) - \( H^{\otimes 3}|111\rangle = \frac{1}{2^{3/2}} \sum_{x=0}^{7} (-1)^{wt(x)}|x\rangle \) - \( wt(x) \) is the Hamming weight (number of 1's in \( x \)). Thus, \[ H^{\otimes 3}(|000\rangle + |111\rangle) = \frac{1}{2^{3/2}} \sum_{x=0}^{7} (1 + (-1)^{wt(x)})|x\rangle \] - For even Hamming weight: \( 1 + 1 = 2 \) - For odd Hamming weight: \( 1 - 1 = 0 \) So only even-weight basis states survive: - \( |000\rangle, |011\rangle, |101\rangle, |110\rangle \) And similarly, \( |000\rangle - |111\rangle \) would leave only odd-weight states. So, - \( H^{\otimes 3} (|000\rangle + |111\rangle) = \sum_{\text{even } w} |x\rangle \) - \( H^{\otimes 3} (|000\rangle - |111\rangle) = \sum_{\text{odd } w} |x\rangle \) --- ### **Apply to All Triplets** So after the Hadamards (on first and third triplet): - The first and third triplets will become superpositions of even/odd weight states depending on whether you started with \( |000\rangle + |111\rangle \) or \( |000\rangle - |111\rangle \). - The middle triplet remains unchanged. ### **Explicitly:** Let’s expand just the first term for clarity: \[ \frac{\alpha}{2^{3/2}} H^{\otimes 3}|000\rangle_{0,1,2} H^{\otimes 3}|000\rangle_{6,7,8} (|000\rangle + |111\rangle)_{3,4,5} \] - \( H^{\otimes 3}|000\rangle = \frac{1}{2^{3/2}} (|000\rangle + |011\rangle + |101\rangle + |110\rangle) \) - \( H^{\otimes 3}|111\rangle = \frac{1}{2^{3/2}} (|001\rangle + |010\rangle + |100\rangle + |111\rangle) \) So the state after Hadamards is: \[ \frac{\alpha}{2^{3/2}} \sum_{\substack{x_{012} \in \text{even}}} |x_{012}\rangle \otimes (|000\rangle + |111\rangle)_{3,4,5} \otimes \sum_{\substack{x_{678} \in \text{even}}} |x_{678}\rangle \\ + \dots \] (with analogous terms for the phase-flipped triplet, using odd-weight states). --- ### **Summary (a):** **After the first two columns (CNOTs and Hadamards), the state of the 9 qubits is:** - The first and third triplets are in superpositions of even or odd weight basis states (depending on original signs). - The middle triplet is still in \((|000\rangle + |111\rangle)\) or \((|000\rangle - |111\rangle)\) (unchanged). --- ## **(b) Parity Extraction** The next part of the circuit (middle CNOTs) uses ancilla qubits to extract the phase parity of adjacent triplets. - Each ancilla qubit is connected via CNOTs to qubits in the triplets. - After these CNOTs, the ancillas will be in states that reflect the parity (even/odd) of the corresponding triplets. The measurement of the ancilla qubits will thus reveal which triplet (if any) had a phase flip, because the parity will be odd for a triplet with a phase error. --- ## **(c) State at the End** After the full circuit (including final Hadamards and CNOTs), the original logical qubit will be re-encoded, and the error will be identified by the ancillas. --- ## **(d) Correction** From the measured ancilla parities, you can determine which triplet experienced the phase flip. You would then apply a \( Z \) gate (phase flip correction) to any one qubit in the affected triplet. --- ## **Final Answers (in clear steps):** ### **(a) State after first two columns:** The state is: \[ \frac{\alpha}{2^{3/2}} \sum_{\substack{x_{012} \in \text{even}}} |x_{012}\rangle \otimes (|000\rangle + |111\rangle) \otimes \sum_{\substack{x_{678} \in \text{odd}}} |x_{678}\rangle + \frac{\beta}{2^{3/2}} \sum_{\substack{x_{012} \in \text{odd}}} |x_{012}\rangle \otimes (|000\rangle - |111\rangle) \otimes \sum_{\substack{x_{678} \in \text{even}}} |x_{678}\rangle \] where \(x_{012}\), \(x_{678}\) are 3-bit strings with even/odd Hamming weight. --- ### **(b) Parities after middle CNOTs:** The ancilla qubits will indicate which triplet had the phase flip: - If a triplet had a phase flip, its parity will be odd. - The ancillas "store" these parities. --- ### **(c) State at the end:** After the final gates, the logical qubit is re-encoded, and the ancillas indicate the error syndrome. --- ### **(d) Correction:** - From the ancilla readings, identify the triplet with odd parity. - Apply a \( Z \) gate to any one qubit in that triplet to correct the phase flip. --- If you want the full, explicit state in Dirac notation for (a), let me know!

Let's carefully work through each part of the problem with detailed calculations, step-by-step, to understand how the circuit operates on the given encoded state. --- ## **Part (a): State after the first two columns (CNOTs and Hadamards)** ### **Initial State (after phase flip on last triplet):** Given the original logical state: \[ |\psi\rangle = \frac{\alpha}{2^{3/2}} (|000\rangle + |111\rangle)_{abc} \; |\text{rest}\rangle + \frac{\beta}{2^{3/2}} (|000\rangle - |111\rangle)_{abc} \; |\text{rest}\rangle \] where the triplets are labeled: - Triplet 1: qubits 0,1,2 - Triplet 2: qubits 3,4,5 - Triplet 3: qubits 6,7,8 The key is that the last triplet experienced a phase flip, changing the overall state to a superposition involving \(|000\rangle \pm |111\rangle\) in the respective triplets. --- ### **Step 1: Apply the first set of CNOTs** - **CNOTs** with control on the first qubit of each triplet and targets on the next two qubits in that triplet. For each triplet: \[ \text{CNOT control} = qubit 0,3,6 \] Target qubits: +1, +2 in each triplet. - For triplet 1: control \(q_0\), targets \(q_1, q_2\) - For triplet 2: control \(q_3\), targets \(q_4, q_5\) - For triplet 3: control \(q_6\), targets \(q_7, q_8\) --- **Effect of CNOTs:** - Since the initial states are either \(|000\rangle\) or \(|111\rangle\): 1. \( |000\rangle \): \[ q_0=0, q_1=0, q_2=0 \Rightarrow \text{no change} \] 2. \( |111\rangle \): \[ q_0=1, q_1=1, q_2=1 \] - CNOT with control 1, targets 1,2: \[ q_1 \rightarrow q_1 \oplus q_0 = 1 \oplus 1=0 \] \[ q_2 \rightarrow q_2 \oplus q_0=1 \oplus 1=0 \] - Similarly for the second and third triplets. **Result:** - The CNOTs map: - \( |000\rangle \rightarrow |000\rangle \) - \( |111\rangle \rightarrow |111\rangle \) so **the overall state remains unchanged** after the CNOTs at this stage. --- ### **Step 2: Apply Hadamard gates \(H^{\otimes 3}\) to triplets 1 and 3** - Hadamard transforms: \[ H|0\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, \quad H|1\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}} \] - The states \(|000\rangle\) and \(|111\rangle\) transform as: \[ H^{\otimes 3}|000\rangle = \frac{1}{2^{3/2}} \sum_{x \in \{0,1\}^3} |x\rangle \] (all basis states equally superposed with amplitude \(1/2^{3/2}\)). Similarly, \[ H^{\otimes 3}|111\rangle = \frac{1}{2^{3/2}} \sum_{x} (-1)^{w(x)} |x\rangle \] where \(w(x)\) is the Hamming weight (number of 1's) in \(x\). - Sum of these: \[ H^{\otimes 3} (|000\rangle + |111\rangle) = \frac{1}{2^{3/2}} \left[\sum_{x} |x\rangle + \sum_{x} (-1)^{w(x)} |x\rangle \right] \] - For each basis state \(|x\rangle\): \[ \text{Amplitude} = \frac{1}{2^{3/2}} \left[ 1 + (-1)^{w(x)} \right] \] - The sum: \[ \Rightarrow \text{Only states with even } w(x) \text{ survive with amplitude } \frac{2}{2^{3/2}} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}} \] and states with odd \(w(x)\) cancel out. --- ### **Result after Hadamards:** - **For states originating from \(|000\rangle + |111\rangle\):** \[ H^{\otimes 3} |000\rangle + |111\rangle \rightarrow \sum_{x: w(x)\text{ even}} |x\rangle \] - **For states from \(|000\rangle - |111\rangle\):** \[ H^{\otimes 3} |000\rangle - |111\rangle \rightarrow \sum_{x: w(x)\text{ odd}} |x\rangle \] --- ### **Summary for part (a):** - The **first** and **third** triplets become superpositions of basis states with either **even** or **odd** weights, depending on initial signs. - The **middle** triplet remains in the \((|000\rangle + |111\rangle)\) or \((|000\rangle - |111\rangle)\) states, unaffected by these operations. --- ## **Part (b): Show that ancilla qubits store phase parities** - The next layers involve CNOTs from qubits in the triplets to ancilla qubits. - These CNOTs "extract" the parity (sum mod 2) of the triplet's basis states. - Because the superpositions are over states with definite parity (even or odd), the ancilla qubits' states encode the phase flip information. **In particular:** - If a triplet is in a superposition of even weight states, the ancilla connected via CNOTs will end up in \(|0\rangle\), indicating **no phase flip**. - If in odd weight superpositions, the ancilla will be \(|1\rangle\), indicating **phase flip**. Thus, **after the middle CNOT layers**, the ancilla qubits **store the parity information** of each triplet, revealing which triplet experienced the phase flip. --- ## **Part (c): State after the entire circuit** - The final layers **re-encode** the logical qubit. - The **error syndrome** (parity bits in the ancilla) indicate if and where a phase flip occurred. - Applying the correction (a \(Z\) gate) to the affected triplet restores the original logical state. --- ## **Part (d): Gate to fix the phase flip** - **From the ancilla parities**, identify which triplet has an odd parity. - Apply a **phase flip \(Z\)** to **any** qubit in that triplet. - This corrects the phase flip in the code. --- ## **Summary of final answers:** | Part | Explanation | Final Result | |--------|--------------|--------------| | (a) | State after first two columns (CNOTs + Hadamards) | Superpositions where first and third triplets are in states of definite parity (even/odd), middle triplet unchanged. Specifically, from initial \(|000\rangle \pm |111\rangle\), the superpositions become sums over basis states with definite weight parity. | \[ \boxed{ \begin{aligned} & \frac{\alpha}{\sqrt{2}} \left( \sum_{w(x)\text{ even}} |x\rangle_{1} \right) \otimes (|000\rangle + |111\rangle)_{2} \otimes \left(\sum_{w(x)\text{ even}} |x\rangle_{3}\right) \\ + & \frac{\beta}{\sqrt{2}} \left( \sum_{w(x)\text{ odd}} |x\rangle_{1} \right) \otimes (|000\rangle - |111\rangle)_{2} \otimes \left(\sum_{w(x)\text{ odd}} |x\rangle_{3}\right) \end{aligned} } \] where subscript 1 = triplet 1, 2 = triplet 2, 3 = triplet 3. | (b) | Parity qubits encode phase error information | Ancillas obtain parity bits: \(|0\rangle\) for even parity, \(|1\rangle\) for odd parity, indicating phase flip location. | | (c) | State after full circuit | The original logical qubit is restored, and ancilla measurement reveals the error syndrome. | | (d) | Correction | Apply a \(Z\) gate to any qubit in the triplet with odd parity (detected from ancillas) to fix the phase flip. | --- If further explicit basis states or detailed amplitude calculations are needed, I can expand accordingly!

Certainly! Here are the concise final answers for each part: **(a)** After the first two columns (CNOTs and Hadamards), the 9-qubit state transforms such that the first and third triplets are in superpositions of basis states with definite parity (even or odd), depending on their initial signs, while the middle triplet remains in a superposition of \(|000\rangle\) and \(|111\rangle\). Specifically, the superpositions become sums over basis states with even or odd Hamming weight, reflecting the initial phase flip. **(b)** The middle layers of CNOTs transfer the parity information of each triplet into the ancilla qubits. As a result, these ancillas store the phase parity (whether a phase flip has occurred) of each adjacent triplet, with ancilla \(|0\rangle\) indicating even parity (no phase flip) and \(|1\rangle\) indicating odd parity (phase flip). **(c)** At the end of the circuit, the original logical state is re-encoded, and the ancillas' measurement outcomes provide the error syndrome, pinpointing which triplet experienced a phase flip. The overall state is corrected accordingly, restoring the initial encoded state. **(d)** From the measured parities, apply a phase flip (\(Z\) gate) to any one qubit within the triplet identified as having odd parity. This operation rectifies the phase error introduced by the original phase flip.

Certainly! Here's the detailed calculation for **Part (a)**, showing step-by-step how the state evolves after the first two columns of the circuit (CNOTs and Hadamards). --- ### **Initial State Before the Circuit** The logical state after a phase flip on the last triplet (qubits 6,7,8) is: \[ |\psi_0\rangle = \frac{\alpha}{2^{3/2}}\left( |000\rangle_{012} + |111\rangle_{012} \right) \left( |000\rangle_{345} + |111\rangle_{345} \right) \left( |000\rangle_{678} - |111\rangle_{678} \right) \\ + \frac{\beta}{2^{3/2}}\left( |000\rangle_{012} - |111\rangle_{012} \right) \left( |000\rangle_{345} - |111\rangle_{345} \right) \left( |000\rangle_{678} + |111\rangle_{678} \right) \] This state encodes the logical qubit with a phase flip on the last triplet. --- ### **Step 1: Applying CNOTs (First layer)** - The **CNOTs** have control on the first qubit of each triplet and targets on the next two qubits in that triplet: \[ \text{Triplet 1: control } q_0,\; \text{targets } q_1, q_2 \] \[ \text{Triplet 2: control } q_3,\; \text{targets } q_4, q_5 \] \[ \text{Triplet 3: control } q_6,\; \text{targets } q_7, q_8 \] - **Effect**: Since the initial states are either \(|000\rangle\) or \(|111\rangle\): - For \(|000\rangle\): \[ q_0=0 \Rightarrow \text{targets unaffected} \Rightarrow |000\rangle \] - For \(|111\rangle\): \[ q_0=1 \Rightarrow \text{targets } q_1, q_2 \rightarrow q_1 \oplus 1, q_2 \oplus 1 \Rightarrow |111\rangle \rightarrow |1,0,0\rangle \] But in the superpositions, applying CNOTs maps: \[ |000\rangle \rightarrow |000\rangle \] \[ |111\rangle \rightarrow |111\rangle \] because the control bits are identical to the targets in these basis states, and the CNOTs preserve the states \(|000\rangle\) and \(|111\rangle\). **Therefore, the state remains unchanged after these CNOTs**. --- ### **Step 2: Applying Hadamard Gates to the First and Third Triplets** **Apply \(H^{\otimes 3}\) to triplets 1 (qubits 0,1,2) and 3 (qubits 6,7,8).** Recall: \[ H^{\otimes 3} |000\rangle = \frac{1}{2^{3/2}} \sum_{x \in \{0,1\}^3} |x\rangle \] \[ H^{\otimes 3} |111\rangle = \frac{1}{2^{3/2}} \sum_{x \in \{0,1\}^3} (-1)^{w(x)} |x\rangle \] where \(w(x)\) is the Hamming weight (number of 1's in \(x\)). **Transformations:** - For \(|000\rangle + |111\rangle\): \[ H^{\otimes 3} (|000\rangle + |111\rangle) = \frac{1}{2^{3/2}} \left( \sum_{x} |x\rangle + \sum_{x} (-1)^{w(x)} |x\rangle \right) \] - For each basis state \(|x\rangle\), the combined amplitude is: \[ \frac{1}{2^{3/2}} [1 + (-1)^{w(x)}] \] which is: - **2** if \(w(x)\) is even (since \(1 + 1=2\)) - **0** if \(w(x)\) is odd (since \(1 - 1=0\)) **Result:** - The superposition collapses to basis states with **even** Hamming weight: \[ |x\rangle \text{ with } w(x) \text{ even} \] Similarly, the states originating from \(|000\rangle - |111\rangle\) will involve the sum: \[ \frac{1}{2^{3/2}} \left( \sum_x |x\rangle - \sum_x (-1)^{w(x)} |x\rangle \right) \] which selects **odd** weight basis states. --- ### **Explicitly:** - **Superpositions after Hadamards in the first triplet:** \[ \boxed{ H^{\otimes 3} (|000\rangle + |111\rangle) = \sum_{w(x)\text{ even}} |x\rangle } \] - **Similarly for the third triplet.** --- ### **Final combined state after these steps:** \[ |\psi_1\rangle = \frac{\alpha}{2^{3/2}} \left( \sum_{w(x)\text{ even}} |x\rangle_{012} \right) \left( |000\rangle_{345} + |111\rangle_{345} \right) \left( \sum_{w(x)\text{ even}} |x\rangle_{678} \right) \\ + \frac{\beta}{2^{3/2}} \left( \sum_{w(x)\text{ odd}} |x\rangle_{012} \right) \left( |000\rangle_{345} - |111\rangle_{345} \right) \left( \sum_{w(x)\text{ odd}} |x\rangle_{678} \right) \] --- ### **Summary** - The first and third triplets are now in superpositions over basis states with **definite parity (even or odd weights)**, depending on the initial phase. - The middle triplet remains in the superpositions \(|000\rangle \pm |111\rangle\). --- ### **In essence:** \[ \boxed{ \textbf{State after first two columns:} \\ \quad |\text{superposition over even/odd weight basis states in triplets 1 and 3, with triplet 2 unchanged.}} } \] This completes the detailed calculation for **Part (a)**.