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

Reasoning (model)

My analysis of the quantum error detection circuit and its fidelity begins with a thorough understanding of the problem’s setup. I’m given a [[4,2,2]] code designed to protect two logical qubits using four physical qubits, and I need to calculate the fidelity of preparing the logical $|00\rangle_{AB}$ state under a specific error model.

I start by deconstructing the problem: I identify the stabilizers ($S_1$ and $S_2$), the logical operators ($X_A, Z_A, X_B, Z_B$), and the circuit composed of a Hadamard gate and five CNOT gates, with the initial state $|00000\rangle$. I also note the two-qubit depolarizing error model applied after each CNOT gate, the post-selection on detectable errors (projecting onto the code space and requiring a $|0\rangle$ measurement on the ancilla), and the final fidelity calculation.

Next, I analyze the ideal (error-free) circuit. I step-by-step follow the circuit’s operations, transforming the initial state to the logical $|00\rangle_{AB}$ state, defined as the simultaneous +1 eigenstate of $Z_A$ and $Z_B$ (as well as $S_1$ and $S_2$). I meticulously verify that the circuit indeed prepares the intended logical state $|L_{00}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)$, and it commutes with all the operators, confirming this state is the logical target.

Then, I delve into the error analysis, focusing on single Pauli errors to the first order in the error probability (p). I recognize that errors can propagate through the CNOT gates, potentially leading to errors on the logical qubits. An error is “undetectable” if it’s both consistent with an ancilla $|0\rangle$ outcome and commutes with both stabilizers. I break down the 15 possible two-qubit errors after each CNOT to understand their impact. I note a crucial factor here: if we find an error that is “undetectable,” it also has to contain $X$ or $Y$. If not, the logical state will not change, and the result is benign. The ideal fidelity remains 1.

I methodically propagate errors through the circuit, gate-by-gate, examining 15 distinct errors at each CNOT gate. For each error scenario, I first propagate the Pauli error operator. Then, I analyze the ancilla measurement and the stabilizer properties to characterize each propagated error as detectable or undetectable/benign. Undetectable errors that have no impact on the fidelity are logical $Z$ operators or stabilizers. I label the logical $Z$ operator that doesn’t change fidelity “benign”. I carefully apply commutation to check if each propagated operator is a logical one. I notice that for the ancilla measurement to return a $|0\rangle$, the ancilla error must be either $I$ or $Z$.

My error analysis continues for all 75 potential errors to show that any bad errors are caught or are benign. As I perform this exhaustive analysis, I come to a surprising conclusion: to the first order in the error probability, none of the errors lead to a loss of fidelity. This finding implies the circuit is robust and fault-tolerant to the first order in $p$ against single-gate errors and post-selection is working.

After several iterations of verification, my result is that the fidelity of preparing the logical $|00\rangle_{AB}$ state is 1 in the limit of first-order errors. This makes sense considering that a single error is detectable by the ancilla or the code’s stabilizers, or is benign. The fidelity can be written as $1 - c p^2$, with no first-order term. It can also be written as a function to first order, where the fidelity is 1.

This problem asks for the logical state fidelity of a prepared logical $|00\rangle_{AB}$ state using a specific circuit, subject to a depolarizing error model on the CNOT gates. The fidelity calculation must account for post-selection based on both code stabilizers and an ancilla qubit measurement. We will solve this by analyzing the effect of a single two-qubit Pauli error at each of the five CNOT gates in the circuit, working to first order in the error probability $p$.

Step-by-Step Derivation

1. The Ideal State and Circuit

First, let’s trace the evolution of the initial state $|0000\rangle_{0123}|0\rangle_4$ through the error-free circuit. The operations are applied from right to left: $|\psi_0\rangle = |00000\rangle$

$H_1$: The state becomes $\frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)$.
$CNOT_{12}$: Acts on the second term, flipping qubit 2. State: $\frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)$.
$CNOT_{10}$: Acts on the second term, flipping qubit 0. State: $\frac{1}{\sqrt{2}}(|00000\rangle + |11100\rangle)$.
$CNOT_{23}$: Acts on the second term, flipping qubit 3. State: $\frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$.

The state of the four code qubits at this point is $|\psi_{code}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)$. Let’s verify this is the logical $|00\rangle_{AB}$ state, denoted $|L_{00}\rangle$. A logical state must be a +1 eigenstate of the stabilizers, and $|L_{00}\rangle$ must be a +1 eigenstate of the logical $Z_A$ and $Z_B$ operators. * $S_1 = XXXX$: $XXXX|\psi_{code}\rangle = \frac{1}{\sqrt{2}}(|1111\rangle + |0000\rangle) = |\psi_{code}\rangle$. Correct. * $S_2 = ZZZZ$: $ZZZZ|\psi_{code}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) = |\psi_{code}\rangle$. Correct. * $Z_A = ZZII$: $ZZII|\psi_{code}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) = |\psi_{code}\rangle$. Correct. * $Z_B = ZIZI$: $ZIZI|\psi_{code}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) = |\psi_{code}\rangle$. Correct.

The state before the final two CNOTs is $|L_{00}\rangle \otimes |0\rangle_4$.

$CNOT_{34}$: The state becomes $\frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)$.
$CNOT_{04}$: The state becomes $\frac{1}{\sqrt{2}}(|00000\rangle + CNOT_{04}|11111\rangle) = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$.
The final state before measurement is $|\psi_{final}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) \otimes |0\rangle_4 = |L_{00}\rangle \otimes |0\rangle_4$.

Measurement $M_4$ of the ancilla qubit yields $|0\rangle$ with probability 1. The resulting state of the code qubits is $|L_{00}\rangle$.

2. Error Model and Fidelity Calculation

Each of the 5 CNOT gates is followed by a depolarizing channel $\mathcal{E}(\rho) = (1-p)\rho + p \sum_{P \in \mathcal{P}_{ij}\setminus\{II\}} \frac{1}{15} P\rho P^\dagger$. We are interested in the fidelity $F = \langle L_{00} | \rho_{final} | L_{00} \rangle$ to first order in $p$.

The final density matrix $\rho_{final}$ is the state after the circuit and post-selection. An error can be: a) Detected by the ancilla: If the ancilla measurement outcome is $|1\rangle$. b) Detected by the stabilizers: If the error moves the state out of the codespace (i.e., the propagated error on the code qubits anti-commutes with at least one stabilizer). c) Undetectable: If the ancilla measures $|0\rangle$ and the propagated error commutes with all stabilizers. Undetectable errors are logical operators or stabilizers.

An undetectable error causes infidelity if it is a “bad” logical operator, i.e., one that changes the state $|L_{00}\rangle$. These are logical operators containing $X$ or $Y$ factors (e.g., $X_A, Y_A, X_B, ...$). An undetectable error is “benign” if it preserves the state (e.g., stabilizers, $Z_A, Z_B, Z_AZ_B$).

To find the fidelity to first order in $p$, we count the number of single two-qubit Pauli errors (out of $5 \times 15 = 75$ total possibilities) that are “bad”. Let this number be $N_{bad}$. The probability of one specific bad error occurring is $p/15$. The total probability of a bad outcome is $P_{bad} = N_{bad} \frac{p}{15}$.

The fidelity to first order is $F(p) \approx 1 - P_{bad} = 1 - N_{bad} \frac{p}{15}$.

3. Error Propagation Analysis

Let the CNOTs be $U_1=CNOT_{12}, U_2=CNOT_{10}, U_3=CNOT_{23}, U_4=CNOT_{34}, U_5=CNOT_{04}$. An error $E^{(k)}$ occurs after $U_k$. The subsequent gates $U_{k+1}, ..., U_5$ propagate this error. The final propagated error is $E' = (U_5 \dots U_{k+1}) E^{(k)} (U_5 \dots U_{k+1})^\dagger$.

We analyze the effect of $E'$ on the final ideal state $|L_{00}\rangle \otimes |0\rangle_4$. * Ancilla Detection: The ancilla measures $|1\rangle$ if the part of $E'$ acting on the ancilla, $E'_4$, is $X_4$ or $Y_4$. * Stabilizer Detection: If the ancilla part is $I_4$ or $Z_4$, the code qubit part $E'_{0123}$ must be checked. It is detected if it anti-commutes with $S_1=XXXX$ or $S_2=ZZZZ$.

Let’s analyze the error locations from last to first.

Error after $U_5 = CNOT_{04}$: $E' = E^{(5)}$ acts on qubits $(0,4)$. For the ancilla to measure $|0\rangle$, $E^{(5)}$ must be of the form $P_0 \otimes I_4$ or $P_0 \otimes Z_4$. The error on the code is $P_0 \otimes I_{123}$. Any non-identity single-qubit Pauli error on qubit 0 (i.e., $X_0, Y_0, Z_0$) is detected by the code stabilizers. Thus, no undetectable logical errors occur from this location.
Error after $U_4 = CNOT_{34}$: $E' = CNOT_{04} E^{(4)} CNOT_{04}^\dagger$. $E^{(4)}$ acts on $(3,4)$. Let $E^{(4)}=P_3 \otimes Q_4$. The gate $CNOT_{04}$ does not act on qubit 3. The error on qubit 4 propagates as $CNOT_{04}(I_0 \otimes Q_4)CNOT_{04}^\dagger$.
- If $Q_4 \in \{I,Z\}$, the ancilla part remains $\{I,Z\}$ and the code error is $P_3$. This is detected by stabilizers.
- If $Q_4 \in \{X,Y\}$, the ancilla part becomes $\{Z_0 \otimes X_4, Z_0 \otimes Y_4\}$. The ancilla measures $|1\rangle$. No undetectable logical errors occur from this location.
Error after $U_3 = CNOT_{23}$: $E' = CNOT_{04}CNOT_{34} E^{(3)} CNOT_{34}^\dagger CNOT_{04}^\dagger$. $E^{(3)}=P_2 \otimes Q_3$.
- The gate $CNOT_{34}$ propagates $Q_3 \in \{X,Y\}$ to $Q_3 \otimes X_4$.
- The gate $CNOT_{04}$ further propagates the $X_4$ term to $Z_0 \otimes X_4$.
- An error $Q_3 \in \{X,Y\}$ will always result in the ancilla measuring $|1\rangle$.
- If $Q_3 \in \{I,Z\}$, no error propagates to the ancilla. The code error is $P_2 \otimes Q_3$. Any such two-qubit error on qubits $(2,3)$ is detected by the stabilizers. No undetectable logical errors occur from this location.
Error after $U_2 = CNOT_{10}$: $E' = (U_5 U_4 U_3) E^{(2)} (U_5 U_4 U_3)^\dagger$. $E^{(2)}=Q_0 \otimes P_1$. The gates $U_3, U_4$ do not act on these qubits. The only propagation is through $U_5=CNOT_{04}$.
- Let $E^{(2)} = Q_0 \otimes P_1$. Then $E' = CNOT_{04} (Q_0 \otimes P_1) CNOT_{04}^\dagger$.
- If $Q_0 \in \{X,Y\}$, the error propagates to $Q_0 \otimes P_1 \otimes X_4$. Detected by the ancilla.
- If $Q_0 \in \{I,Z\}$, the error on the ancilla is $I_4$. The code error is $Q_0 \otimes P_1$. We check for undetectable logical errors.
  - $E^{(2)} = Z_0 \otimes Z_1$: $E'_{code} = Z_0 Z_1 = Z_A$. This is a logical operator. It’s benign as $Z_A|L_{00}\rangle = |L_{00}\rangle$.
  - All other 7 errors in this class (e.g., $Z_0 \otimes X_1, I_0 \otimes X_1, \dots$) are detected by the stabilizers. For example, $Z_0X_1$ anti-commutes with $S_2=ZZZZ$. One benign undetectable error ($Z_0Z_1$) occurs from this location.
Error after $U_1 = CNOT_{12}$: $E' = (U_5 U_4 U_3 U_2) E^{(1)} (U_5 U_4 U_3 U_2)^\dagger$. Let $E^{(1)}=P_1 \otimes Q_2$. This is the most complex case. A detailed analysis shows that an error propagates to the ancilla unless specific cancellations occur. The cancellation happens when an $X$ or $Y$ error on qubit 1 propagates to qubit 0 (via $CNOT_{10}$), and an $X$ or $Y$ error on qubit 2 propagates to qubit 3 (via $CNOT_{23}$) and then to the ancilla (via $CNOT_{34}$). The $X_0$ from the first path and the $X_4$ from the second path cancel their effect on the ancilla when propagated through $CNOT_{04}$. This occurs for the 4 initial errors where $P_1, Q_2 \in \{X, Y\}$. The propagated code error is $E'_{code} = X_0 P_1 Q_2 X_3$.
1. $E^{(1)}=X_1X_2 \implies E'_{code} = X_0X_1X_2X_3 = S_1$. Benign (stabilizer).
2. $E^{(1)}=Y_1Y_2 \implies E'_{code} = X_0Y_1Y_2X_3 = S_1 (Z_1Z_2)$. This is a logical Z-type operator, so it is benign.
3. $E^{(1)}=X_1Y_2 \implies E'_{code} = X_0X_1Y_2X_3$. Anti-commutes with both $S_1$ and $S_2$. Detected.
4. $E^{(1)}=Y_1X_2 \implies E'_{code} = X_0Y_1X_2X_3$. Anti-commutes with both $S_1$ and $S_2$. Detected.
Another class of errors that does not get detected by the ancilla is when $P_1, Q_2 \in \{I, Z\}$. The propagated error is $E'_{code} = P_1 Q_2$. 1. $E^{(1)}=Z_1Z_2$: Commutes with stabilizers. It is a logical Z-type operator. Benign. 2. Other errors like $Z_1I_2, I_1Z_2$ are detected by stabilizers.

In total, from this location we find 3 benign undetectable errors: $X_1X_2$, $Y_1Y_2$, and $Z_1Z_2$.

4. Final Fidelity Calculation

Let’s summarize the undetectable errors: * From $U_1$ ($CNOT_{12}$): 3 benign errors ($X_1X_2, Y_1Y_2, Z_1Z_2$). * From $U_2$ ($CNOT_{10}$): 1 benign error ($Z_0Z_1$). * From $U_3, U_4, U_5$: 0 undetectable errors.

The total number of single Pauli errors that result in a “bad” logical operator is $N_{bad}=0$. All single errors are either detected or benign.

The state of the system, to first order in $p$, after post-selection is a mixture of the ideal state (from the no-error and benign-error branches). Since none of these branches produce a state orthogonal to $|L_{00}\rangle$, the final state is pure.

Let $P_{succ}$ be the probability of passing post-selection. $P(\text{no error}) = (1-p)^5 \approx 1-5p$. $N_{benign} = 3+1=4$. $P(\text{any benign error}) = 4 \times \frac{p}{15}$. $P_{succ} = P(\text{no error}) + P(\text{any benign error}) \approx 1 - 5p + \frac{4p}{15} = 1 - \frac{71p}{15}$.

The probability of obtaining the correct state $|L_{00}\rangle$ upon successful post-selection is: $P(\text{correct} | \text{success}) = \frac{P(\text{correct and success})}{P_{succ}} = \frac{P(\text{no error}) + P(\text{any benign error})}{P_{succ}} = \frac{1 - 71p/15 + O(p^2)}{1 - 71p/15 + O(p^2)} \approx 1$.

The fidelity is $F = \langle L_{00} | \rho_{final} | L_{00} \rangle = P(\text{correct} | \text{success})$. To first order in $p$, there is no infidelity term. The infidelity is of order $p^2$. Thus, the fidelity is: $F(p) = 1 + O(p^2)$.

The linear term in the infidelity is zero, demonstrating the fault-tolerant nature of this state preparation circuit against this error model. The question asks for the fidelity as a function of $p$. To first order, this function is constant.

Final Answer: The logical state fidelity $F$ as a function of the two-qubit gate error rate $p$, to first order in $p$, is:

\[ F(p) = 1 \]

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

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 = sp.Integer(1)  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical