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 $p$.

Checkpoints

Checkpoint 1

Suppose that we wish to prepare a logical two-qubit GHZ state $(|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}$ in the [[4,2,2]] code. To do so, we use the following state preparation circuit:

\[ (CNOT_{03}) (H_0) (CNOT_{21}) (H_2). \]

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.

What is the physical state fidelity of the final physical 4-qubit state at the end of the circuit as a function of the two-qubit gate error rate $p$?

Assistant #2

Reasoning (model)

The Problem of Quantum Error Correction

My task is to delve into the realm of quantum error correction using the [4,2,2] code. This involves understanding how to protect two logical qubits, A and B, encoded into four physical qubits labeled 0, 1, 2, and 3. The code’s distance of two implies the ability to detect single-qubit errors.

The code’s structure is defined by the following: * Stabilizers: * $S_1 = X_0X_1X_2X_3$ (XXXX) * $S_2 = Z_0Z_1Z_2Z_3$ (ZZZZ) * Logical Operators: * $X_A = X_0I_1X_2I_3$ (XIXI) * $X_B = X_0X_1I_2I_3$ (XXII) * $Z_A = Z_0Z_1I_2I_3$ (ZZII) * $Z_B = Z_0I_1Z_2I_3$ (ZIZI)

The objective is to prepare a two-qubit logical GHZ state: $|\psi_L\rangle = \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB})$.

State Preparation

The circuit to prepare this state is defined by $U = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$. Starting from the initial state $|0000\rangle$, I traced the state through each gate operation. I checked the ideal state and verified it to be a logical GHZ state.

The GHZ state is a simultaneous +1 eigenstate of $Z_A Z_B$ and $X_A X_B$ and belongs to the codespace of the code.

Error Model and Fidelity

The error model states that each CNOT gate is followed by a two-qubit depolarizing error channel $\mathcal{E}_{ij}$. This depolarizing channel acts as a sum of Pauli error operators: $\mathcal{E}_{ij}(\rho) = (1-p) \rho + p \sum_{P \in \{I,X,Y,Z\}^{\otimes 2} \setminus \{II\}} \frac{1}{15} P \rho P^\dagger$. My main goal is to calculate the physical state fidelity, which is $F = \langle \psi_{ideal} | \rho_{final} | \psi_{ideal} \rangle$, as a function of the error probability $p$.

The ideal final state, $|\psi_{ideal}\rangle$, can be obtained by applying the state-preparation circuit U to the initial state, $|0000\rangle$.

Deriving the Final Fidelity

By tracking the effects of the circuit and the error channels, I derived the following expression for the final fidelity:

$F = (1-p)^2 + \frac{6 p(1-p)}{15} + \frac{9 p^2}{225}$

Using some Pauli algebra with the stabilizer elements of the ideal state, I simplified this result:

$F = (1 - \frac{4}{5}p)^2$

This is the central result that describes the fidelity, with the error model, and how the quality of the state is degraded, in this particular error correction problem.

Step-by-Step Derivation

1. Determine the ideal final state

The circuit consists of the sequence of operations $U = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$ applied to the initial state $|\psi_{init}\rangle = |0000\rangle$. The operations are applied from right to left.

Initial state: $|\psi_0\rangle = |0000\rangle$.
Apply $H_2$ (Hadamard on qubit 2):

\[ |\psi_1\rangle = H_2 |\psi_0\rangle = |00\rangle \otimes \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)_2 \otimes |0\rangle_3 = \frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle) \]

Apply $CNOT_{21}$ (control on 2, target on 1): This gate flips qubit 1 if qubit 2 is $|1\rangle$.

\[ |\psi_2\rangle = CNOT_{21} |\psi_1\rangle = \frac{1}{\sqrt{2}}(CNOT_{21}|0000\rangle + CNOT_{21}|0010\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle) \]

Apply $H_0$ (Hadamard on qubit 0):

\[ |\psi_3\rangle = H_0 |\psi_2\rangle = \frac{1}{\sqrt{2}} \left( H_0|0000\rangle + H_0|0110\rangle \right) = \frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)_0|000\rangle + \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)_0|110\rangle \right) \]

\[ |\psi_3\rangle = \frac{1}{2} (|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle) \]

Apply $CNOT_{03}$ (control on 0, target on 3): This gate flips qubit 3 if qubit 0 is $|1\rangle$.

\[ |\psi_{ideal}\rangle = CNOT_{03} |\psi_3\rangle = \frac{1}{2} (CNOT_{03}|0000\rangle + CNOT_{03}|1000\rangle + CNOT_{03}|0110\rangle + CNOT_{03}|1110\rangle) \]

\[ |\psi_{ideal}\rangle = \frac{1}{2} (|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle) \]

This is the pure quantum state produced by the circuit in the absence of errors.

2. Model the circuit with errors

The circuit contains two $CNOT$ gates, each followed by a two-qubit depolarizing error channel. Let $C_1 = CNOT_{21}$ and $C_2 = CNOT_{03}$. The ideal unitaries between the CNOTs are $U_1 = H_2$ and $U_2 = H_0$. The error channel after a CNOT gate on qubits $i,j$ is given by:

\[ \mathcal{E}_{ij}(\rho) = (1-p) \rho + p \sum_{P \in \mathcal{P}_{ij}^*} \frac{1}{15} P \rho P \]

where $\mathcal{P}_{ij}^*$ is the set of 15 non-identity two-qubit Pauli operators on qubits $i$ and $j$.

The evolution of the density matrix $\rho$ from the initial state $\rho_0 = |\psi_{init}\rangle\langle\psi_{init}|$ is:

$\rho_1 = U_1 \rho_0 U_1^\dagger$
$\rho_2 = C_1 \rho_1 C_1^\dagger$
$\rho_3 = \mathcal{E}_{21}(\rho_2) = (1-p)\rho_2 + \frac{p}{15}\sum_{P_1 \in \mathcal{P}_{21}^*} P_1 \rho_2 P_1$
$\rho_4 = U_2 \rho_3 U_2^\dagger$
$\rho_5 = C_2 \rho_4 C_2^\dagger$
$\rho_{final} = \mathcal{E}_{03}(\rho_5) = (1-p)\rho_5 + \frac{p}{15}\sum_{P_2 \in \mathcal{P}_{03}^*} P_2 \rho_5 P_2$

Let’s express $\rho_{final}$ in terms of $|\psi_{ideal}\rangle$. Let $U_{mid} = C_2 U_2$. The ideal state is $|\psi_{ideal}\rangle = U_{mid} C_1 U_1 |\psi_{init}\rangle$. The density matrix before the second error channel is:

\[ \rho_5 = U_{mid} \mathcal{E}_{21}(C_1 U_1 \rho_0 U_1^\dagger C_1^\dagger) U_{mid}^\dagger \]

\[ \rho_5 = (1-p) U_{mid} \rho_2 U_{mid}^\dagger + \frac{p}{15}\sum_{P_1} U_{mid} P_1 \rho_2 P_1 U_{mid}^\dagger \]

Since $\rho_2 = C_1 U_1 \rho_0 U_1^\dagger C_1^\dagger$, we have $U_{mid} \rho_2 U_{mid}^\dagger = |\psi_{ideal}\rangle\langle\psi_{ideal}|$. Let $\tilde{P}_1 = U_{mid} P_1 U_{mid}^\dagger$. Since $P_1$ acts on qubits {1,2} and $U_{mid}=CNOT_{03}H_0$ acts on qubits {0,3}, they commute, so $\tilde{P}_1 = P_1$.

\[ \rho_5 = (1-p) |\psi_{ideal}\rangle\langle\psi_{ideal}| + \frac{p}{15}\sum_{P_1 \in \mathcal{P}_{21}^*} P_1 |\psi_{ideal}\rangle\langle\psi_{ideal}| P_1 \]

Now we apply the second error channel $\mathcal{E}_{03}$:

\[ \rho_{final} = (1-p)\rho_5 + \frac{p}{15}\sum_{P_2 \in \mathcal{P}_{03}^*} P_2 \rho_5 P_2 \]

Substituting the expression for $\rho_5$:

\[ \rho_{final} = (1-p)^2 |\psi_{ideal}\rangle\langle\psi_{ideal}| + \frac{p(1-p)}{15}\sum_{P_1} P_1 |\psi_{ideal}\rangle\langle\psi_{ideal}| P_1 + \frac{p(1-p)}{15}\sum_{P_2} P_2 |\psi_{ideal}\rangle\langle\psi_{ideal}| P_2 + \frac{p^2}{15^2}\sum_{P_1,P_2} P_2 P_1 |\psi_{ideal}\rangle\langle\psi_{ideal}| P_1 P_2 \]

3. Calculate the fidelity

The physical state fidelity is $F = \langle\psi_{ideal}|\rho_{final}|\psi_{ideal}\rangle$. Let $|\psi\rangle = |\psi_{ideal}\rangle$.

\[ F = (1-p)^2 |\langle\psi|\psi\rangle|^2 + \frac{p(1-p)}{15}\sum_{P_1} |\langle\psi|P_1|\psi\rangle|^2 + \frac{p(1-p)}{15}\sum_{P_2} |\langle\psi|P_2|\psi\rangle|^2 + \frac{p^2}{15^2}\sum_{P_1,P_2} |\langle\psi|P_2 P_1|\psi\rangle|^2 \]

To evaluate the sums, we need to find which Pauli operators have non-zero expectation value with respect to $|\psi\rangle$. A Pauli operator $P$ has a non-zero expectation value $\langle\psi|P|\psi\rangle$ only if $|\psi\rangle$ is an eigenstate of $P$. Since $|\psi\rangle$ is a stabilizer state, this means $P$ must be an element of the stabilizer group $\mathcal{S}_\psi$ of $|\psi\rangle$ (or $-P \in \mathcal{S}_\psi$). For such a $P$, $|\langle\psi|P|\psi\rangle|^2=1$. Otherwise, it is 0.

4. Determine the stabilizer group of the ideal state

The state $|\psi\rangle = \frac{1}{2} (|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$ is a logical GHZ state in the [[4,2,2]] code. It is stabilized by the code’s stabilizers $S_1=XXXX$ and $S_2=ZZZZ$. It is also a logical Bell state, so it is an eigenstate of $X_A X_B$ and $Z_A Z_B$. $X_A X_B = (XIXI)(XXII) = I_0X_1X_2I_3$. $Z_A Z_B = (ZZII)(ZIZI) = I_0Z_1Z_2I_3$. One can verify that $|\psi\rangle$ is a +1 eigenstate of both $IXXI$ and $IZZI$. The stabilizer group $\mathcal{S}_\psi$ is generated by $\langle XXXX, ZZZZ, IXXI, IZZI \rangle$. A simpler set of generators can be found: * $g_1 = (XXXX)(IXXI) = X_0 I_1 I_2 X_3 = X_0X_3$ * $g_2 = (ZZZZ)(IZZI) = Z_0 I_1 I_2 Z_3 = Z_0Z_3$ * $g_3 = I_0 X_1 X_2 I_3 = X_1X_2$ * $g_4 = I_0 Z_1 Z_2 I_3 = Z_1Z_2$ So $\mathcal{S}_\psi = \langle X_0X_3, Z_0Z_3, X_1X_2, Z_1Z_2 \rangle$. This group consists of Pauli operators that are tensor products of a Pauli on {0,3} and a Pauli on {1,2}. Let $G_{03} = \langle X_0X_3, Z_0Z_3 \rangle = \{I, X_0X_3, Z_0Z_3, Y_0Y_3\}$ (up to phases). Let $G_{12} = \langle X_1X_2, Z_1Z_2 \rangle = \{I, X_1X_2, Z_1Z_2, Y_1Y_2\}$ (up to phases). Then $\mathcal{S}_\psi$ consists of products of elements from $G_{03}$ and $G_{12}$.

5. Evaluate the sums

Sum over $P_1 \in \mathcal{P}_{21}^*$: $P_1$ acts on qubits {1,2}. We need to count how many non-identity Paulis on {1,2} are in $\mathcal{S}_\psi$. These are the non-identity elements of $G_{12}$: $X_1X_2$, $Z_1Z_2$, and $Y_1Y_2$. There are 3 such operators.

\[ \sum_{P_1 \in \mathcal{P}_{21}^*} |\langle\psi|P_1|\psi\rangle|^2 = 1+1+1 = 3 \]

Sum over $P_2 \in \mathcal{P}_{03}^*$: $P_2$ acts on qubits {0,3}. We need to count how many non-identity Paulis on {0,3} are in $\mathcal{S}_\psi$. These are the non-identity elements of $G_{03}$: $X_0X_3$, $Z_0Z_3$, and $Y_0Y_3$. There are 3 such operators.

\[ \sum_{P_2 \in \mathcal{P}_{03}^*} |\langle\psi|P_2|\psi\rangle|^2 = 1+1+1 = 3 \]

Sum over $P_1 \in \mathcal{P}_{21}^*$ and $P_2 \in \mathcal{P}_{03}^*$: The product $P_2P_1$ is in $\mathcal{S}_\psi$ if and only if $P_1$ is one of the 3 non-identity elements in $G_{12}$ and $P_2$ is one of the 3 non-identity elements in $G_{03}$. This gives $3 \times 3 = 9$ pairs $(P_1, P_2)$.

\[ \sum_{P_1,P_2} |\langle\psi|P_2 P_1|\psi\rangle|^2 = 9 \]

6. Combine the terms to find the final fidelity

Substitute the values of the sums into the expression for $F$:

\[ F = (1-p)^2 + \frac{p(1-p)}{15}(3) + \frac{p(1-p)}{15}(3) + \frac{p^2}{15^2}(9) \]

\[ F = (1-p)^2 + \frac{6p(1-p)}{15} + \frac{9p^2}{225} \]

\[ F = (1-p)^2 + \frac{2p(1-p)}{5} + \frac{p^2}{25} \]

Now, expand and simplify the polynomial in $p$:

\[ F = (1 - 2p + p^2) + \left(\frac{2}{5}p - \frac{2}{5}p^2\right) + \frac{1}{25}p^2 \]

\[ F = 1 + \left(-2 + \frac{2}{5}\right)p + \left(1 - \frac{2}{5} + \frac{1}{25}\right)p^2 \]

\[ F = 1 - \frac{8}{5}p + \left(\frac{25 - 10 + 1}{25}\right)p^2 \]

\[ F = 1 - \frac{8}{5}p + \frac{16}{25}p^2 \]

This expression is a perfect square:

\[ F = \left(1 - \frac{4}{5}p\right)^2 \]

Final Answer: The physical state fidelity of the final 4-qubit state is $F(p) = \left(1 - \frac{4}{5}p\right)^2$.

User #3

The correct answer is: $ F_{\rm{physical}}=(1-\frac{12}{15}p)^2 $

Now let’s solve the next Checkpoint.

Checkpoint 2

Suppose that we wish to prepare a logical two-qubit GHZ state $(|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}$ in the [[4,2,2]] code. To do so, we use the following state preparation circuit:

\[ (CNOT_{03}) (H_0) (CNOT_{21}) (H_2). \]

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

Assistant #4

Reasoning (model)

Recap of My Strategy for Calculating Logical Fidelity

Okay, so I’m tasked with figuring out the logical state fidelity after some post-selection. I’ve already handled the physical fidelity in the first subproblem, so I’ve got that part down. Essentially, what I need to do is incorporate this post-selection step and then determine the fidelity of the final state with the ideal logical state.

Step-by-Step Breakdown:

Post-Selection Mechanics: The post-selection is based on detecting specific errors. My code here is a [[4,2,2]] code, which has stabilizers like XXXX and ZZZZ, meaning I have a subspace that is the codespace, C. The project onto the codespace is $\Pi_{\mathcal{C}}$. After measuring stabilizers (+1), I project the final state onto the codespace.
State After Post-Selection: After post-selection, I get a new state, $\rho_{ps} = \frac{\Pi_{\mathcal{C}} \rho_{final} \Pi_{\mathcal{C}}}{\text{Tr}(\Pi_{\mathcal{C}} \rho_{final} \Pi_{\mathcal{C}})}$. I want to calculate the logical fidelity, which is $\langle\psi_{ideal}| \rho_{ps} |\psi_{ideal}\rangle$.
Logical Fidelity is Simplified: Because the ideal state already lives in the codespace, the formula $\langle\psi_{ideal}| \rho_{ps} |\psi_{ideal}\rangle = \frac{\langle\psi_{ideal}| \rho_{final} |\psi_{ideal}\rangle}{\text{Tr}(\Pi_{\mathcal{C}} \rho_{final})}$ boils down to $F_{logical} = \frac{F_{physical}}{P_{success}}$ where $P_{success} = \text{Tr}(\Pi_{\mathcal{C}} \rho_{final})$. This also means, $F_{logical} = F_{physical} / P_{success}$.
Calculating Success Probability, P$_{success}$: This is the most involved part. I need to figure out what the probability is that the system survives post-selection. To do this I needed to classify the effects that the errors and stabilizers play on each other.

To figure out $P_{success}$, I break down the final density matrix $\rho_{final}$ into a sum of terms based on different Pauli errors.
I can also define detectable vs undetectable by the Pauli error and whether or not it commutes. Undetectable means no projection.
The math works out where errors that anticommute with stabilizers get projected out. This means that my total probability of success is a product of these two numbers (the sum of the terms that contribute to $\rho_{final}$ and $\Pi_\mathcal{C}$)

Putting it all Together: I have $F_{physical} = (1 - \frac{4}{5}p)^2$ and I worked my way through to get $P_{success} = 1 - \frac{8}{5}p + \frac{64}{75}p^2$. Therefore $F_{logical} = \frac{(1 - \frac{4}{5}p)^2}{1 - \frac{8}{5}p + \frac{64}{75}p^2}$ which is my final answer.

Step-by-Step Derivation

1. Define Logical Fidelity with Post-Selection

The logical state fidelity is the fidelity of the final state with the ideal logical state, conditioned on passing the error detection check. The error detection check consists of measuring the code stabilizers, $S_1 = XXXX$ and $S_2 = ZZZZ$, and accepting the state only if both measurement outcomes are +1. This corresponds to projecting the final state onto the codespace $\mathcal{C}$.

The projector onto the codespace is $\Pi_{\mathcal{C}} = \frac{1}{4}(I+S_1)(I+S_2)$. The final physical density matrix, derived in Subproblem 1, is $\rho_{final}$. After the stabilizer measurements, the unnormalized post-selected state is $\rho'_{ps} = \Pi_{\mathcal{C}} \rho_{final} \Pi_{\mathcal{C}}$. The probability of successfully passing the check is $P_{success} = \text{Tr}(\rho'_{ps}) = \text{Tr}(\Pi_{\mathcal{C}} \rho_{final})$. The normalized post-selected state is $\rho_{ps} = \rho'_{ps} / P_{success}$.

The ideal state $|\psi_{ideal}\rangle = \frac{1}{2} (|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$ is a codeword, so it lies within the codespace, meaning $\Pi_{\mathcal{C}}|\psi_{ideal}\rangle = |\psi_{ideal}\rangle$.

The logical fidelity is the fidelity of the post-selected state with the ideal state:

\[ F_{logical} = \langle\psi_{ideal}| \rho_{ps} |\psi_{ideal}\rangle = \frac{\langle\psi_{ideal}| \Pi_{\mathcal{C}} \rho_{final} \Pi_{\mathcal{C}} |\psi_{ideal}\rangle}{P_{success}} \]

Using $\Pi_{\mathcal{C}}|\psi_{ideal}\rangle = |\psi_{ideal}\rangle$, the numerator simplifies to:

\[ \langle\psi_{ideal}| \rho_{final} |\psi_{ideal}\rangle = F_{physical} \]

where $F_{physical}$ is the physical fidelity calculated in Subproblem 1. Thus, we have:

\[ F_{logical} = \frac{F_{physical}}{P_{success}} \]

From Subproblem 1, we have $F_{physical} = \left(1 - \frac{4}{5}p\right)^2$. We now need to compute $P_{success}$.

2. Calculate the Success Probability ($P_{success}$)

The success probability is $P_{success} = \text{Tr}(\Pi_{\mathcal{C}} \rho_{final})$. The final density matrix is:

\[ \rho_{final} = (1-p)^2 |\psi\rangle\langle\psi| + \frac{p(1-p)}{15}\sum_{P_1} P_1 |\psi\rangle\langle\psi| P_1 + \frac{p(1-p)}{15}\sum_{P_2} P_2 |\psi\rangle\langle\psi| P_2 + \frac{p^2}{225}\sum_{P_1,P_2} P_2 P_1 |\psi\rangle\langle\psi| P_1 P_2 \]

where $|\psi\rangle = |\psi_{ideal}\rangle$, $P_1 \in \mathcal{P}_{21}^*$ (non-identity Paulis on qubits {1,2}), and $P_2 \in \mathcal{P}_{03}^*$ (non-identity Paulis on qubits {0,3}).

The trace of a generic term $P|\psi\rangle\langle\psi|P$ is $\text{Tr}(\Pi_{\mathcal{C}} P|\psi\rangle\langle\psi|P) = \langle\psi|P\Pi_{\mathcal{C}}P|\psi\rangle$. An error operator $P$ is detectable if it anticommutes with at least one code stabilizer ($S_1$ or $S_2$). If an error $P$ is detectable, the state $P|\psi\rangle$ is in an error space orthogonal to the codespace, so $\Pi_{\mathcal{C}} P|\psi\rangle = 0$. Consequently, for a detectable error $P$, $\text{Tr}(\Pi_{\mathcal{C}} P|\psi\rangle\langle\psi|P) = 0$. An error $P$ is undetectable if it commutes with all code stabilizers. For such a $P$, the state $P|\psi\rangle$ remains in the codespace. Thus, $\text{Tr}(\Pi_{\mathcal{C}} P|\psi\rangle\langle\psi|P) = \text{Tr}(P|\psi\rangle\langle\psi|P) = 1$.

So, to calculate $P_{success}$, we need to count the number of undetectable errors for each term in $\rho_{final}$.

Classification of single CNOT errors: A two-qubit Pauli $P_{ij}$ is undetectable if it commutes with both $S_1=XXXX$ and $S_2=ZZZZ$. This occurs if $P_{ij}$ is one of $\{X_iX_j, Y_iY_j, Z_iZ_j\}$. There are 3 undetectable errors out of 15 non-identity Paulis. The remaining 12 are detectable.
Classification of two CNOT errors: An error $P = P_1P_2$ is undetectable if it commutes with both $S_1$ and $S_2$. Since $P_1$ and $P_2$ act on disjoint sets of qubits, $P$ commutes with a stabilizer $S$ if either both $P_1, P_2$ commute with $S$ or both anticommute with $S$. Let’s classify the 15 errors on a given pair of qubits based on their commutation with $(S_1, S_2)$:
- Commute with both (C,C): 3 operators ($\{XX, YY, ZZ\}$)
- Commute with $S_1$, anticommute with $S_2$ (C,A): 4 operators ($\{IX, XI, YZ, ZY\}$)
- Anticommute with $S_1$, commute with $S_2$ (A,C): 4 operators ($\{IZ, ZI, XY, YX\}$)
- Anticommute with both (A,A): 4 operators ($\{IY, YI, XZ, ZX\}$) The total number of pairs $(P_1, P_2)$ for which $P_1P_2$ is undetectable is the sum of products of counts for matching types:

\[ N_{undetect} = N_{CC}^2 + N_{CA}^2 + N_{AC}^2 + N_{AA}^2 = 3^2 + 4^2 + 4^2 + 4^2 = 9 + 16 + 16 + 16 = 57 \]

Calculating $P_{success}$: We sum the probabilities of all events that result in an overall undetectable error.
1. No error (Identity operator, is undetectable): $(1-p)^2$
2. One error on first CNOT, none on second: $p(1-p) \times \frac{3}{15}$ (3 undetectable errors)
3. One error on second CNOT, none on first: $(1-p)p \times \frac{3}{15}$ (3 undetectable errors)
4. Errors on both CNOTs: $p^2 \times \frac{57}{225}$ (57 undetectable pairs)

\[ P_{success}(p) = (1-p)^2 + \frac{3p(1-p)}{15} + \frac{3p(1-p)}{15} + \frac{57p^2}{225} \]

\[ P_{success}(p) = (1-p)^2 + \frac{6p(1-p)}{15} + \frac{57p^2}{225} \]

\[ P_{success}(p) = (1-2p+p^2) + \frac{2}{5}(p-p^2) + \frac{19}{75}p^2 \]

\[ P_{success}(p) = 1 + (-2+\frac{2}{5})p + (1-\frac{2}{5}+\frac{19}{75})p^2 \]

\[ P_{success}(p) = 1 - \frac{8}{5}p + \left(\frac{75-30+19}{75}\right)p^2 = 1 - \frac{8}{5}p + \frac{64}{75}p^2 \]

3. Final Calculation of Logical Fidelity

Now we combine our results for $F_{physical}$ and $P_{success}$.

\[ F_{physical} = \left(1 - \frac{4}{5}p\right)^2 = 1 - \frac{8}{5}p + \frac{16}{25}p^2 \]

\[ F_{logical}(p) = \frac{F_{physical}(p)}{P_{success}(p)} = \frac{1 - \frac{8}{5}p + \frac{16}{25}p^2}{1 - \frac{8}{5}p + \frac{64}{75}p^2} \]

The logical infidelity $1-F_{logical}$ to leading order in $p$ is:

\[ 1 - F_{logical} = 1 - \frac{1 - \frac{8}{5}p + \frac{16}{25}p^2}{1 - \frac{8}{5}p + \frac{64}{75}p^2} = \frac{(1 - \frac{8}{5}p + \frac{64}{75}p^2) - (1 - \frac{8}{5}p + \frac{16}{25}p^2)}{1 - \frac{8}{5}p + \frac{64}{75}p^2} \]

\[ 1 - F_{logical} = \frac{(\frac{64}{75} - \frac{16}{25})p^2}{1 - \frac{8}{5}p + \frac{64}{75}p^2} = \frac{(\frac{64-48}{75})p^2}{1 - O(p)} = \frac{16}{75}p^2 + O(p^3) \]

This confirms that logical errors are a second-order effect, as single undetectable errors from one CNOT happen to be stabilizers of the final state and thus cause no logical error.

Final Answer: The logical state fidelity is given by the function:

\[ F_{logical}(p) = \frac{1 - \frac{8}{5}p + \frac{16}{25}p^2}{1 - \frac{8}{5}p + \frac{64}{75}p^2} \]

User #5

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 logical state fidelity of the final 2-qubit logical state
    at the end of the circuit as a function of the two-qubit gate error rate $p$ in Sympy format.

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

    Outputs
    ----------
    F_logical: sympy.Expr, the logical state fidelity as a function of $p$
    """

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

    return F_logical

Assistant #6

import sympy as sp

p = sp.symbols('p')

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

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

    Outputs
    ----------
    F_logical: sympy.Expr, the logical state fidelity as a function of $p$
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_physical = (1 - 8/5 * p + 16/25 * p**2)
    P_success = (1 - 8/5 * p + 64/75 * p**2)
    F_logical = F_physical / P_success
    # ---------------------------------------------------------------

    return F_logical