System #0

You are a physics research assistant specializing in solving complex, research-level problems using precise, step-by-step reasoning.

Input Problems will be provided in Markdown format.

Output (Markdown format)

Step-by-Step Derivation - Show every non-trivial step in the solution. Justify steps using relevant physical laws, theorems, or mathematical identities.
Mathematical Typesetting - Use LaTeX for all mathematics: $...$ for inline expressions, $$...$$ for display equations.
Conventions and Units - Follow the unit system and conventions specified in the problem.
Final Answer - At the end of the solution, start a new line with “Final Answer:”, and present the final result.

For final answers involving values, follow the precision requirements specified in the problem. If no precision is specified: - If an exact value is possible, provide it (e.g., \$\sqrt(2)\$, \$\pi/4\$). - If exact form is not feasible, retain at least 12 significant digits in the result.
Formatting Compliance - If the user requests a specific output format (e.g., code, table), provide the final answer accordingly.

User #1

Problem setup:

In quantum error correction, you encode quantum states into logical states made of many qubits in order to improve their resilience to errors. In quantum error detection, you do the same but can only detect the presence of errors and not correct them. In this problem, we will consider a single [[4,2,2]] quantum error detection code, which encodes two logical qubits into four physical qubits, and investigate how robust logical quantum operations in this code are to quantum errors.

Our convention is that the four physical qubits in the [[4,2,2]] code are labelled 0,1,2,3. The two logical qubits are labelled A and B. The stabilizers are $XXXX$ and $ZZZZ$, where $X$ and $Z$ are Pauli matrices. The logical $X$ and $Z$ operators on the two qubits are $X_A = XIXI$, $X_B=XXII$, $Z_A = ZZII$, $Z_B = ZIZI$, up to multiplication by stabilizers.

We will consider different state preparation circuits consisting of controlled not $CNOT_{ij}$ gates, where $CNOT_{ij}$ has control qubit $i$ and target qubit $j$. As a simple model of quantum errors in hardware, we will suppose that each $CNOT_{ij}$ gate in the circuit has a two qubit depolarizing error channel following it that produces one of the 15 non-identity two-qubit Paulis with equal probability $p/15$. The probability $p$ indicates the probability of an error in a single two-qubit gate. We will assess the logical infidelity of certain state preparation protocols as a function of the physical infidelity $p$.

Challenge

Suppose that we prepare a logical two-qubit $|00\rangle_{AB}$ state in the [[4,2,2]] code. To do so, we introduce an ancilla qubit, qubit 4, and use the following state preparation circuit:

\[ M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1) \]

Note that this equation is written in matrix multiplication order, while the quantum operations in the circuit occur in the reverse order (from right-to-left in the above equation). $H$ is a single-qubit Hadamard gate and $M$ is a single-qubit measurement. The ancilla is used to detect errors in the state preparation circuit and makes the circuit fault-tolerant. If the ancilla measurement is $|0\rangle$ ($|1\rangle$), the state preparation succeeds (fails).

What is the logical state fidelity of the final 2-qubit logical state at the end of the circuit as a function of two-qubit gate error rate $p$, assuming the state is post-selected on all detectable errors in the code and on the ancilla qubit measuring $|0\rangle$?

Assistant #2

The problem asks us to calculate the logical state fidelity of a prepared logical $|00\rangle_{AB}$ state in the [[4,2,2]] code, under a specific error model and post-selection conditions.

The [[4,2,2]] code uses 4 physical qubits (0,1,2,3). The stabilizers are $S_1 = X_0 X_1 X_2 X_3$ and $S_2 = Z_0 Z_1 Z_2 Z_3$. The logical operators are $X_A = X_0 I_1 X_2 I_3$, $X_B = X_0 X_1 I_2 I_3$, $Z_A = Z_0 Z_1 I_2 I_3$, $Z_B = Z_0 I_1 Z_2 I_3$. The target state is the logical $|00\rangle_{AB}$, which is an eigenstate of $Z_A$ and $Z_B$ with eigenvalue +1.

The circuit for state preparation involves 5 physical qubits (0,1,2,3,4) and is given in reverse order of operation (right to left):

\[ U_{circuit} = M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1) \]

We assume the initial state is $|00000\rangle$.

Ideal State Calculation: Let’s trace the initial state $|00000\rangle$ through the circuit unitaries. The Hadamard gate $H_1$ acts on qubit 1. All CNOT gates are of the form $CNOT_{c,t}$. $|\psi_0\rangle = |00000\rangle$ $H_1: |\psi_1\rangle = |0\rangle_0 \frac{|0\rangle_1+|1\rangle_1}{\sqrt{2}} |0\rangle_2 |0\rangle_3 |0\rangle_4$ $CNOT_{12}: |\psi_2\rangle = |0\rangle_0 \frac{|0\rangle_1|0\rangle_2+|1\rangle_1|1\rangle_2}{\sqrt{2}} |0\rangle_3 |0\rangle_4$ $CNOT_{10}: |\psi_3\rangle = \frac{1}{\sqrt{2}} (|00000\rangle + |11100\rangle)$ $CNOT_{23}: |\psi_4\rangle = \frac{1}{\sqrt{2}} (|00000\rangle + |11110\rangle)$ $CNOT_{34}: |\psi_5\rangle = \frac{1}{\sqrt{2}} (|00000\rangle + |11111\rangle)$ $CNOT_{04}: |\psi_6\rangle = \frac{1}{\sqrt{2}} (|00000\rangle + |11110\rangle)$ This is the ideal state before measurement on qubit 4. Let’s call this $|\Psi_{ideal}\rangle = \frac{1}{\sqrt{2}} (|0000\rangle_{0123} + |1111\rangle_{0123}) |0\rangle_4$. After measurement $M_4$ yields $|0\rangle$, the state of qubits 0-3 is $|\Phi_L\rangle = \frac{1}{\sqrt{2}} (|0000\rangle + |1111\rangle)$. Let’s check if $|\Phi_L\rangle$ is the logical $|00\rangle_{AB}$ state: $S_1 |\Phi_L\rangle = XXXX |\Phi_L\rangle = |\Phi_L\rangle$. $S_2 |\Phi_L\rangle = ZZZZ |\Phi_L\rangle = |\Phi_L\rangle$. $Z_A |\Phi_L\rangle = Z_0 Z_1 I_2 I_3 |\Phi_L\rangle = |\Phi_L\rangle$. $Z_B |\Phi_L\rangle = Z_0 I_1 Z_2 I_3 |\Phi_L\rangle = |\Phi_L\rangle$. Indeed, $|\Phi_L\rangle$ is the logical $|00\rangle_{AB}$ state.
Error Model and Fidelity Definition: Each CNOT gate $C_k$ is followed by a two-qubit depolarizing error channel $\mathcal{E}_k$. $\mathcal{E}_k(\rho) = (1-p)\rho + \frac{p}{15} \sum_{P_{ij} \neq I} P_{ij} \rho P_{ij}$. We work to first order in $p$. The state after all operations and error channels is $\rho_f$. The fidelity is $F = \frac{\langle \Phi_L | \text{Tr}_4(P_{code} P_{M_0} \rho_f P_{M_0} P_{code}) | \Phi_L \rangle}{\text{Tr}(\text{Tr}_4(P_{code} P_{M_0} \rho_f P_{M_0} P_{code}))}$, where $P_{M_0} = I_{0123} \otimes |0\rangle\langle 0|_4$ and $P_{code} = \frac{1}{4}(I+S_1)(I+S_2)$.

Let $U_k$ be the $k$-th CNOT gate in execution order (from right to left in the problem statement). $U_1 = H_1$ (no error channel) $U_2 = CNOT_{12}$ $U_3 = CNOT_{10}$ $U_4 = CNOT_{23}$ $U_5 = CNOT_{34}$ $U_6 = CNOT_{04}$ An error $P_{ij}$ occurring after $U_k$ is effectively equivalent to an error $E_{prop} = U_{post,k} P_{ij} U_{post,k}^\dagger$ acting on the ideal state $|\Psi_{ideal}\rangle$, where $U_{post,k} = U_6 \dots U_{k+1}$. $E_{prop}$ is a 5-qubit Pauli operator, $E_{prop} = E_{0123} \otimes E_4$.

We are interested in errors that are not detected by post-selection: 1. The ancilla measurement $M_4$ yields $|0\rangle$. This means $E_4 \in \{I_4, Z_4\}$. 2. The state must be in the code space. This means $E_{0123}$ must commute with both stabilizers $S_1$ and $S_2$. If $E_{0123}$ anti-commutes with $S_1$ or $S_2$, it’s detected. Among these undetected errors: - If $E_{0123}|\Phi_L\rangle = |\Phi_L\rangle$ (e.g., $E_{0123}$ is a product of stabilizers and logical Z operators), it contributes to the numerator ($N_0$). - If $E_{0123}|\Phi_L\rangle \neq |\Phi_L\rangle$ but is still a valid logical state (e.g., $E_{0123}$ is a logical X operator), it contributes only to the denominator ($N_L$).

The infidelity is $p \frac{N_L}{15}$, where $N_L$ is the total count of errors from all 5 CNOT gates that satisfy conditions 1 and 2, but result in a logical state different from $|\Phi_L\rangle$.

Let’s use the $(\vec{x}, \vec{z})$ formalism for Pauli propagation. A Pauli $P$ on 5 qubits is represented by two binary vectors $(\vec{x}, \vec{z})$, where $P = \prod_{j=0}^4 X_j^{x_j} Z_j^{z_j}$. The transformation for $CNOT_{c,t}$ on an error $E=(\vec{x},\vec{z})$ is: $x_c \to x_c \oplus x_t$ $z_c \to z_c$ $x_t \to x_t$ $z_t \to z_t \oplus z_c$

Let’s analyze each CNOT gate:

1. Errors after $CNOT_{04}$ (qubits 0,4, $U_{post,6}=I$) An original error $P_{04}$ (on qubits 0 and 4) is directly $E_{prop} = P_{04}$. - $E_4 \in \{X_4, Y_4\}$ (8 errors): These flip qubit 4, so they are detected by $M_4$. - $E_4 \in \{I_4, Z_4\}$ (7 errors, e.g. $X_0 I_4, Y_0 I_4, Z_0 I_4, I_0 Z_4, X_0 Z_4, Y_0 Z_4, Z_0 Z_4$). For these, $E_{0123}$ is $P_0$. - $P_0=X_0$: anti-commutes with $S_1=XXXX$. Detected. - $P_0=Y_0$: anti-commutes with $S_1=XXXX$ and $S_2=ZZZZ$. Detected. - $P_0=Z_0$: anti-commutes with $S_2=ZZZZ$. Detected. Therefore, for $CNOT_{04}$, $N_0=0$ and $N_L=0$.

2. Errors after $CNOT_{34}$ (qubits 3,4, $U_{post,5}=CNOT_{04}$) Original error $P_{34}$ on qubits 3,4. Initial $(\vec{x},\vec{z})=(0,0,0,x_3,x_4), (0,0,0,z_3,z_4)$. Applying $CNOT_{04}$ (control 0, target 4): $x_0^{new} = x_0 \oplus x_4 = 0 \oplus x_4 = x_4$. $z_0^{new} = z_0 = 0$. $x_4^{new} = x_4$. $z_4^{new} = z_4 \oplus z_0 = z_4 \oplus 0 = z_4$. Propagated error $(\vec{x}',\vec{z}')=(x_4,0,0,x_3,x_4), (0,0,0,z_3,z_4)$. So $E_{0123} = X_0^{x_4} X_3^{x_3} Z_3^{z_3}$. And $E_4 = X_4^{x_4} Z_4^{z_4}$. - Ancilla detection ($x_4=1$): If $x_4=1$, $E_4$ contains $X_4$ or $Y_4$, detected by $M_4$. (8 errors) - No ancilla detection ($x_4=0$): $P_4 \in \{I_4, Z_4\}$. $E_4 = Z_4^{z_4}$. $E_{0123} = X_3^{x_3} Z_3^{z_3} = P_3$. (7 errors: $X_3 I_4, Y_3 I_4, Z_3 I_4, I_3 Z_4, X_3 Z_4, Y_3 Z_4, Z_3 Z_4$). We check $P_3$ for detection by code ($S_1, S_2$) and logical state change ($Z_A, Z_B$). - $P_3 = I_3$: (error $I_3 Z_4$, $x_3=0, z_3=0$) $E_{0123}=I$. Commutes with $S_1, S_2, Z_A, Z_B$. $N_0 \leftarrow N_0+1$. - $P_3 = X_3$: Anti-commutes $S_1$. Detected. - $P_3 = Y_3$: Anti-commutes $S_1, S_2$. Detected. - $P_3 = Z_3$: Anti-commutes $S_2$. Detected. For $CNOT_{34}$, $N_0=1$, $N_L=0$.

3. Errors after $CNOT_{23}$ (qubits 2,3, $U_{post,4}=CNOT_{34} CNOT_{04}$) Original error $P_{23}$. Initial $(\vec{x},\vec{z})=(0,0,x_2,x_3,0), (0,0,z_2,z_3,0)$. Applying $CNOT_{34}$ (control 3, target 4): $x_3^{new} = x_3$. $z_3^{new} = z_3$. $x_4^{new} = x_4=0$. $z_4^{new} = z_4 \oplus z_3 = 0 \oplus z_3 = z_3$. Intermediate $(\vec{x}_a,\vec{z}_a)=(0,0,x_2,x_3,0), (0,0,z_2,z_3,z_3)$. Applying $CNOT_{04}$ (control 0, target 4): $x_0^{new} = x_0 \oplus x_4 = 0 \oplus 0 = 0$. $z_0^{new} = z_0 = 0$. $x_4^{new} = x_4 = 0$. $z_4^{new} = z_4 \oplus z_0 = z_3 \oplus 0 = z_3$. Propagated error $(\vec{x}',\vec{z}')=(0,0,x_2,x_3,0), (0,0,z_2,z_3,z_3)$. So $E_{0123} = X_2^{x_2} Z_2^{z_2} X_3^{x_3} Z_3^{z_3} = P_{23}$. And $E_4 = Z_4^{z_3}$. - Ancilla detection: $E_4=Z_4^{z_3}$. No $X_4$ component, so no ancilla detection. All 15 errors pass. - Code detection & logical change: We check $P_{23}$. Commutes with $S_1=XXXX$ (even $X/Y$ count on $Q_2, Q_3$) and $S_2=ZZZZ$ (even $Z/Y$ count on $Q_2, Q_3$). - $I_2 I_3$: ($x_2=0,z_2=0,x_3=0,z_3=0$) $E_{0123}=I$. $E_4=I_4$. Contributes to $N_0$. $N_0 \leftarrow N_0+1$. - $X_2 X_3$: ($x_2=1,z_2=0,x_3=1,z_3=0$) $E_{0123}=X_2 X_3$. $E_4=I_4$. Commutes $S_1, S_2$. Anti-commutes $Z_B=Z_0 I_1 Z_2 I_3$ ($X_2, Z_2$). Contributes to $N_L$. $N_L \leftarrow N_L+1$. - $Y_2 Y_3$: ($x_2=1,z_2=1,x_3=1,z_3=1$) $E_{0123}=Y_2 Y_3$. $E_4=Z_4$. Commutes $S_1, S_2$. Anti-commutes $Z_B=Z_0 I_1 Z_2 I_3$ ($Y_2, Z_2$). Contributes to $N_L$. $N_L \leftarrow N_L+1$. - $Z_2 Z_3$: ($x_2=0,z_2=1,x_3=0,z_3=1$) $E_{0123}=Z_2 Z_3$. $E_4=Z_4$. Commutes $S_1, S_2$. Commutes $Z_A=Z_0 Z_1$, $Z_B=Z_0 Z_2$. Contributes to $N_0$. $N_0 \leftarrow N_0+1$. All other 11 errors on $P_{23}$ anti-commute with $S_1$ or $S_2$. Detected by code. For $CNOT_{23}$, $N_0=2$, $N_L=2$. Cumulative: $N_0=1+2=3$. $N_L=0+2=2$.

4. Errors after $CNOT_{10}$ (qubits 1,0, $U_{post,3}=CNOT_{23} CNOT_{34} CNOT_{04}$) Original error $P_{10}$. Initial $(\vec{x},\vec{z})=(x_0,x_1,0,0,0), (z_0,z_1,0,0,0)$. The three gates $CNOT_{23}, CNOT_{34}, CNOT_{04}$ are applied. $CNOT_{23}$ (c=2, t=3): No change as it acts on qubits 2,3. $CNOT_{34}$ (c=3, t=4): No change. $CNOT_{04}$ (c=0, t=4): $x_0^{new} = x_0 \oplus x_4 = x_0 \oplus 0 = x_0$. $z_0^{new} = z_0$. $x_4^{new} = x_4 = 0$. $z_4^{new} = z_4 \oplus z_0 = 0 \oplus z_0 = z_0$. Propagated error $(\vec{x}',\vec{z}')=(x_0,x_1,0,0,0), (z_0,z_1,0,0,z_0)$. So $E_{0123} = X_0^{x_0} Z_0^{z_0} X_1^{x_1} Z_1^{z_1} = P_{10}$. And $E_4 = Z_4^{z_0}$. - Ancilla detection: $E_4=Z_4^{z_0}$. No $X_4$ component, so no ancilla detection. All 15 errors pass. - Code detection & logical change: We check $P_{10}$. Commutes with $S_1=XXXX$ (even $X/Y$ count on $Q_0, Q_1$) and $S_2=ZZZZ$ (even $Z/Y$ count on $Q_0, Q_1$). - $I_0 I_1$: ($x_0=0,z_0=0,x_1=0,z_1=0$) $E_{0123}=I$. $E_4=I_4$. Contributes to $N_0$. $N_0 \leftarrow N_0+1$. - $X_0 X_1$: ($x_0=1,z_0=0,x_1=1,z_1=0$) $E_{0123}=X_0 X_1$. $E_4=I_4$. Commutes $S_1, S_2$. Anti-commutes $Z_A=Z_0 Z_1$ ($X_0, Z_0$). Contributes to $N_L$. $N_L \leftarrow N_L+1$. - $Y_0 Y_1$: ($x_0=1,z_0=1,x_1=1,z_1=1$) $E_{0123}=Y_0 Y_1$. $E_4=Z_4$. Commutes $S_1, S_2$. Anti-commutes $Z_A=Z_0 Z_1$ ($Y_0, Z_0$). Contributes to $N_L$. $N_L \leftarrow N_L+1$. - $Z_0 Z_1$: ($x_0=0,z_0=1,x_1=0,z_1=1$) $E_{0123}=Z_0 Z_1$. $E_4=Z_4$. Commutes $S_1, S_2$. Commutes $Z_A, Z_B$. Contributes to $N_0$. $N_0 \leftarrow N_0+1$. All other 11 errors on $P_{10}$ anti-commute with $S_1$ or $S_2$. Detected by code. For $CNOT_{10}$, $N_0=2$, $N_L=2$. Cumulative: $N_0=3+2=5$. $N_L=2+2=4$.

5. Errors after $CNOT_{12}$ (qubits 1,2, $U_{post,2}=CNOT_{10} CNOT_{23} CNOT_{34} CNOT_{04}$) Original error $P_{12}$. Initial $(\vec{x},\vec{z})=(0,x_1,x_2,0,0), (0,z_1,z_2,0,0)$. Applying $CNOT_{10}$ (c=1, t=0): $x_1^{new} = x_1 \oplus x_0 = x_1$. $z_1^{new} = z_1$. $x_0^{new} = x_0=0$. $z_0^{new} = z_0 \oplus z_1 = z_1$. $(\vec{x}_a,\vec{z}_a)=(0,x_1,x_2,0,0), (z_1,z_1,z_2,0,0)$. Applying $CNOT_{23}$ (c=2, t=3): $x_2^{new} = x_2 \oplus x_3 = x_2$. $z_2^{new} = z_2$. $x_3^{new} = x_3=0$. $z_3^{new} = z_3 \oplus z_2 = z_2$. $(\vec{x}_b,\vec{z}_b)=(0,x_1,x_2,0,0), (z_1,z_1,z_2,z_2,0)$. Applying $CNOT_{34}$ (c=3, t=4): $x_3^{new} = x_3 \oplus x_4 = 0$. $z_3^{new} = z_3 = z_2$. $x_4^{new} = x_4=0$. $z_4^{new} = z_4 \oplus z_3 = 0 \oplus z_2 = z_2$. $(\vec{x}_c,\vec{z}_c)=(0,x_1,x_2,0,0), (z_1,z_1,z_2,z_2,z_2)$. Applying $CNOT_{04}$ (c=0, t=4): $x_0^{new} = x_0 \oplus x_4 = 0 \oplus 0 = 0$. $z_0^{new} = z_0 = z_1$. $x_4^{new} = x_4 = 0$. $z_4^{new} = z_4 \oplus z_0 = z_2 \oplus z_1$. Propagated error $(\vec{x}',\vec{z}')=(0,x_1,x_2,0,0), (z_1,z_1,z_2,z_2,z_1 \oplus z_2)$. So $E_{0123} = Z_0^{z_1} X_1^{x_1} Z_1^{z_1} X_2^{x_2} Z_2^{z_2} Z_3^{z_2}$. And $E_4 = Z_4^{z_1 \oplus z_2}$. - Ancilla detection: $E_4=Z_4^{z_1 \oplus z_2}$. No $X_4$ component, so no ancilla detection. All 15 errors pass. - Code detection & logical change: We check $E_{0123}$. $E_{0123}$ must commute with $S_1=XXXX$ (even $X/Y$ count). $x_0 \oplus x_1 \oplus x_2 \oplus x_3 = 0 \oplus x_1 \oplus x_2 \oplus 0 = 0 \implies x_1=x_2$. $E_{0123}$ must commute with $S_2=ZZZZ$ (even $Z/Y$ count). $z_0 \oplus z_1 \oplus z_2 \oplus z_3 = z_1 \oplus z_1 \oplus z_2 \oplus z_2 = 0$. This is always true. So, only errors with $x_1=x_2$ are not detected by $S_1$. These are $I_1 I_2, X_1 X_2, Y_1 Y_2, Z_1 Z_2$. (4 errors). - $I_1 I_2$: ($x_1=0,z_1=0,x_2=0,z_2=0$) $E_{0123}=I$. $E_4=I_4$. Contributes to $N_0$. $N_0 \leftarrow N_0+1$. - $X_1 X_2$: ($x_1=1,z_1=0,x_2=1,z_2=0$) $E_{0123}=X_1 X_2$. $E_4=I_4$. Commutes $S_1, S_2$. Anti-commutes $Z_A=Z_0 Z_1$ ($X_1, Z_1$). Contributes to $N_L$. $N_L \leftarrow N_L+1$. - $Y_1 Y_2$: ($x_1=1,z_1=1,x_2=1,z_2=1$) $E_{0123}=Z_0 X_1 Z_1 X_2 Z_2 Z_3 = Y_0 Y_1 Y_2 Z_3$. $E_4=I_4$. Number of $X/Y$ terms is 3 ($Y_0, Y_1, Y_2$). Anti-commutes $S_1$. Detected by code. - $Z_1 Z_2$: ($x_1=0,z_1=1,x_2=0,z_2=1$) $E_{0123}=Z_0 Z_1 Z_2 Z_3 = S_2$. $E_4=I_4$. $S_2$ is a stabilizer, so it belongs to $N_0$. $N_0 \leftarrow N_0+1$. All other 11 errors in $P_{12}$ (where $x_1 \ne x_2$) anti-commute with $S_1$. Detected by code. For $CNOT_{12}$, $N_0=2$, $N_L=1$. Cumulative: $N_0=5+2=7$. $N_L=4+1=5$.

Total counts: $N_0 = 7$ (errors that preserve the desired logical state and are not detected). $N_L = 5$ (errors that map to a different logical state and are not detected). Total undetected errors = $N_0 + N_L = 7+5=12$. Total errors in the depolarizing channels $= 5 \times 15 = 75$.

Logical Fidelity: The fidelity $F$ is given by:

\[ F = \frac{(1-p)^5 + \frac{p}{15}(1-p)^4 N_0}{(1-p)^5 + \frac{p}{15}(1-p)^4 (N_0+N_L)} \]

To first order in $p$:

\[ F \approx \frac{1 - 5p + \frac{p}{15} N_0}{1 - 5p + \frac{p}{15} (N_0+N_L)} \]

\[ F \approx \left(1 - 5p + \frac{N_0}{15}p\right) \left(1 + 5p - \frac{N_0+N_L}{15}p\right) \]

\[ F \approx 1 + \left( \frac{N_0}{15} - \frac{N_0+N_L}{15} \right)p = 1 - \frac{N_L}{15}p \]

Substituting $N_L=5$:

\[ F = 1 - \frac{5}{15}p = 1 - \frac{1}{3}p \]

The final answer is $\boxed{1 - \frac{p}{3}}$

User #3

Populate your final answer into the code template provided below. This step is purely for formatting/display purposes. No additional reasoning or derivation should be performed. Do not import any modules or packages beyond what is provided in the template.

import sympy as sp

p = sp.symbols('p')

def answer(p):
    r"""
    Return the expression of the logical state fidelity of the final 2-qubit logical state
    at the end of the circuit as a function of two-qubit gate error rate $p$.

    Inputs
    ----------
    p: sympy.Symbol, two-qubit gate error rate, $p$

    Outputs
    ----------
    F_logical: sympy.Expr, logical state fidelity of the final 2-qubit logical state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_logical = ...  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical

Assistant #4

Reasoning (model)

Final Answer Implementation

Okay, so I need to take the final answer I calculated, which was 1 - p/3, and integrate it into the provided Python code. The goal here is to define the F_logical variable as a symbolic sympy expression that represents that exact result. Essentially, I’m just translating my earlier, human-readable solution into a format the code can understand and manipulate. I know exactly how to do this within sympy; it’s straightforward substitution of the math into the library. This should be a quick and easy modification.

import sympy as sp

p = sp.symbols('p')

def answer(p):
    r"""
    Return the expression of the logical state fidelity of the final 2-qubit logical state
    at the end of the circuit as a function of two-qubit gate error rate $p$.

    Inputs
    ----------
    p: sympy.Symbol, two-qubit gate error rate, $p$

    Outputs
    ----------
    F_logical: sympy.Expr, logical state fidelity of the final 2-qubit logical state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_logical = 1 - p/3
    # ---------------------------------------------------------------

    return F_logical