Challenge

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)

  1. Step-by-Step Derivation - Show every non-trivial step in the solution. Justify steps using relevant physical laws, theorems, or mathematical identities.
  2. Mathematical Typesetting - Use LaTeX for all mathematics: $...$ for inline expressions, $$...$$ for display equations.
  3. Conventions and Units - Follow the unit system and conventions specified in the problem.
  4. 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.

  5. 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

Step 1: Understand the given problem and the circuit for state preparation

The problem involves preparing a logical two-qubit \(|00\rangle_{AB}\) state in the [[4,2,2]] code using a specific circuit and assessing its logical fidelity as a function of the physical two-qubit gate error rate \(p\). The circuit given is \(M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1)\), where the operations are applied from right to left.

Step 2: Analyze the circuit and its operation

First, let’s break down the circuit:

  1. \(H_1\) applies a Hadamard gate to qubit 1.
  2. \(CNOT_{12}\) applies a CNOT with control on qubit 1 and target on qubit 2.
  3. \(CNOT_{10}\) applies a CNOT with control on qubit 1 and target on qubit 0.
  4. \(CNOT_{23}\) applies a CNOT with control on qubit 2 and target on qubit 3.
  5. \(CNOT_{34}\) applies a CNOT with control on qubit 3 and target on qubit 4 (ancilla).
  6. \(CNOT_{04}\) applies a CNOT with control on qubit 0 and target on qubit 4.
  7. \(M_4\) measures qubit 4.

Step 3: Determine the effect of the circuit on the initial state

Initially, all qubits are in \(|0\rangle\). After \(H_1\), qubit 1 is in \(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\). The sequence of CNOTs and \(H_1\) prepares the state. Let’s analyze the circuit’s effect step by step.

Step 4: Calculate the state after each operation

  1. After \(H_1\): \(|0\rangle_0 \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)_1 \otimes |0\rangle_2 \otimes |0\rangle_3 \otimes |0\rangle_4\)
  2. After \(CNOT_{12}\): \(|0\rangle_0 \otimes \frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{12} \otimes |0\rangle_3 \otimes |0\rangle_4\)
  3. After \(CNOT_{10}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{01} \otimes |0\rangle_2 \otimes |0\rangle_3 \otimes |0\rangle_4\)
  4. After \(CNOT_{23}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{01} \otimes |0\rangle|0\rangle)_{23} \otimes |0\rangle_4\)
  5. After \(CNOT_{34}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle + |1\rangle|1\rangle)_{01} \otimes |0\rangle|0\rangle)_{23} \otimes |0\rangle|0\rangle)_{34}\)
  6. After \(CNOT_{04}\): \(\frac{1}{\sqrt{2}}(|0\rangle|0\rangle|0\rangle + |1\rangle|1\rangle|1\rangle)_{013} \otimes |0\rangle)_{2} \otimes |0\rangle)_{4}\), simplifying the effect on qubits 0, 1, 3, and 4.

Step 5: Understand the effect of measurement \(M_4\)

The measurement \(M_4\) projects qubit 4 onto \(|0\rangle\) or \(|1\rangle\). For the state preparation to succeed, \(M_4\) must yield \(|0\rangle\). This implies that the overall state before measurement must be such that the probability of measuring \(|1\rangle\) on qubit 4 is zero or negligible for a successful preparation.

Step 6: Analyze the error model and its impact on the circuit

Each \(CNOT\) gate is followed by a two-qubit depolarizing error channel with probability \(p\) of producing one of the 15 non-identity two-qubit Paulis. We need to consider how these errors affect the state preparation and the measurement outcome.

Step 7: Calculate the probability of successful state preparation and its fidelity

To calculate the logical state fidelity, we must consider the probability of errors that are detectable by the code and the ancilla measurement. The [[4,2,2]] code detects single-qubit errors but cannot correct them. The ancilla helps in detecting errors during state preparation.

Step 8: Assess the impact of errors on the logical fidelity

Given that the circuit involves 5 CNOT gates, each with a probability \(p\) of error, we need to consider how these errors propagate and affect the logical state.

Step 9: Derive the expression for logical fidelity as a function of \(p\)

The logical fidelity is related to the probability that no undetectable error occurs. For the [[4,2,2]] code, errors are detectable if they are not in the stabilizer or logical operator set. The probability of an error in a single CNOT is \(p\), and there are 5 CNOTs.

Step 10: Calculate the probability of success and logical fidelity

The probability that a CNOT gate has no error is \(1-p\). Thus, the probability that all 5 CNOT gates have no error is \((1-p)^5\). However, we need to consider the effect of errors and the post-selection on \(|0\rangle\) for qubit 4.

Step 11: Simplify the expression for logical fidelity

Since the state is post-selected on the ancilla being \(|0\rangle\), and given that detectable errors will cause the preparation to fail or be rejected, we focus on the cases where the state is accepted.

The final answer is: \(\boxed{1-p}\)

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 = 1 - p  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical

# Print the final answer
print("Final Answer: ", answer(p))