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 $p$.

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

To determine the physical state fidelity of the final physical 4-qubit state, we need to understand how errors from each $CNOT$ gate contribute to the overall fidelity. Here, we will derive the fidelity through each step of the preparation circuit.

Step 1: Circuit Operations and Error Model

Circuit Description: - The circuit from right to left is: $H_2$, $CNOT_{21}$, $H_0$, $CNOT_{03}$.
Hadamard Gate Operations: - Before analyzing errors, consider how the $H$ transforms the state.
- $H_0$ converts $|0\rangle$ to $(|0\rangle + |1\rangle)/\sqrt{2}$.
- $H_2$ converts $|0\rangle$ to $(|0\rangle + |1\rangle)/\sqrt{2}$.
- Apply $H_2$ first, then $CNOT_{21}$, followed by $H_0$, and finally $CNOT_{03}$.
Error Model for $CNOT$ gates: - After each $CNOT_{ij}$, there’s a two-qubit depolarizing error channel which introduces one of 15 non-identity Pauli errors with equal probability $p/15$.

Step 2: Calculate the Fidelity

Initial State: - Begin with the product state of 4 qubits: $|0000\rangle_{0123}$.
Ideal Final State: - The target is the logical GHZ state:

\[ |\psi_{\text{ideal}}\rangle = \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB}) \rightarrow \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) \]

Multiple $CNOT$ gates and $H$ create entangled states acting upon physical qubits.

Fidelity Calculation: - Chain of Operations: Apply each operation, with its associated errors. - Fidelity decreases with each $CNOT$ due to depolarizing error. Assume perfect single-qubit gates (ideal $H$). - Total fidelity $F$ is calculated as the product of fidelity contributions from each $CNOT$:

\[ F = (1 - p)^2 \]

Explain the formula: With each $CNOT$, successful operation probability is $(1-p)$; square this since two $CNOT$ gates are involved:

\[ \text{Fidelity due to } CNOT_{21}: (1-p) \]

\[ \text{Fidelity due to } CNOT_{03}: (1-p) \]

Final Answer:

The physical state fidelity $F$ of the final 4-qubit state after the circuit given the two-qubit gate error rate $p$ is:

\[ F = (1 - p)^2 \]

This quantifies the efficiency of executing this specific circuit under the error model described.

User #3

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

To calculate the logical state fidelity of the final 2-qubit logical state after post-selecting on all detectable errors using the [[4,2,2]] code, we’ll consider how errors within the circuit affect the detection via stabilizers and impact the logical qubits.

Step 1: Understand Error Detection Capabilities

Error Detection by Stabilizers: - The stabilizers for the [[4,2,2]] code are $XXXX$ and $ZZZZ$. These stabilizers can detect any single-qubit error. - The errors that occur due to the depolarizing channel on the $CNOT$ gates will be detected if they involve a single-qubit error. The code cannot detect correlated two-qubit errors on different pairs of qubits simultaneously if they do not commute with stabilizers.
Circuit Errors: - Each $CNOT$ introduces a two-qubit depolarizing error with probability $p$, introducing one of the 15 non-identity Paulis on those qubits. - The probability of introducing an undetectable error is smaller, as most single errors will be detected.

Step 2: Logical Fidelity Calculation

Initial State: - Start with $|0000\rangle_{0123}$.
Logical State Preparation: - The desired logical GHZ state is $\frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB})$.
Effect of $H$ and $CNOT$ with Errors: - $H$ gates are assumed error-free. - Each $CNOT$ gate is followed by an error model, producing non-identity two-qubit Pauli errors with probability $p/15$ for each specific pair.
Calculate Post-Selection Probability: - The probability of having no detectable error after both $CNOT$ operations (corresponding to undetectable errors) is calculated assuming independence:
- Odds of introducing any detectable error per $CNOT$: $14/15 \times p$ for 14 detected two-qubit Pauli errors.
- Probability of no detectable error (post-selected):

\[ (1 - 14/15 \cdot p)^2 \]

Logical Fidelity: - Post-selection eliminates states with detectable errors: Trivial physical errors are detected. - Remaining errors can still affect logical state fidelity. - Assuming perfect correction of all detectable single-qubit errors: logical errors are due to undetected errors not causing detectable syndrome results.

Logical fidelity of state, approximates:

\[ F_{\text{logical}} = (1 + (1-p)^2)/2 \]

Dominated by the lack of error formation or error correction.

Conclusion:

The logical state fidelity $F_{\text{logical}}$ of the final 2-qubit logical state in the circuit after post-selection, as a function of the error rate $p$, is approximated as:

Final Answer:

\[ F_{\text{logical}} \approx (1 + (1-p)^2) / 2 \]

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_logical = (1 + (1 - p)**2) / 2
    # ---------------------------------------------------------------

    return F_logical