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:

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

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 solve this problem, we need to analyze the given state preparation circuit and understand how quantum errors impact the fidelity of the final prepared state. The circuit is designed to prepare a logical two-qubit GHZ state $(|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}$ in a [[4,2,2]] quantum code. Let’s go through the derivation step by step.

Step-by-Step Derivation:

Circuit Description:

The circuit provided is:

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

This indicates the quantum operations are applied in this order: - Apply Hadamard $H$ on qubit 2. - Apply $CNOT_{21}$ with qubit 2 as the control and qubit 1 as the target. - Apply Hadamard $H$ on qubit 0. - Apply $CNOT_{03}$ with qubit 0 as the control and qubit 3 as the target.

Error Model:

After each $CNOT$ gate, the circuit is followed by a two-qubit depolarizing channel, which introduces one of the 15 non-identity two-qubit Pauli errors with probability $p/15$. The depolarizing channel for two qubits is given by:

\[ \mathcal{E}(\rho) = (1-p) \rho + \frac{p}{15} \sum_{P \in \{X,Y,Z\}^2} P \rho P^\dagger, \]

where the sum runs over all combinations of two-qubit Paulis, excluding the identity.

State Preparation without Errors:

Consider the initial state $|0000\rangle$. The operations transform this as follows: - $H_2$: Converts $|0\rangle_2 \to \frac{|0\rangle_2 + |1\rangle_2}{\sqrt{2}}$. - $CNOT_{21}$: Entangles qubits 2 and 1:

\[ \frac{1}{\sqrt{2}}(|00\rangle_{21} + |11\rangle_{21})|00\rangle_{03}. \]

$H_0$: Converts $|0\rangle_0 \to \frac{|0\rangle_0 + |1\rangle_0}{\sqrt{2}}$.
$CNOT_{03}$: Further entangles qubits 0 and 3:

\[ \frac{1}{2}(|0000\rangle + |1001\rangle + |0100\rangle + |1111\rangle). \]

State with Errors:

Each $CNOT$ introduces an error $\mathcal{E}$, so the sequence $CNOT_{21}$ and $CNOT_{03}$ each may add a depolarizing channel. At this point, there are two main $CNOT$ operations contributing to errors: - (1-p) represents the probability of no error after both gates. - $2p/15$ for a single error occurring in the first gate sequence. - $p^2/225$ for errors in both $CNOT$ operations.

Fidelity Calculation:

Fidelity between expected $|\psi_{ideal}\rangle$ and error-induced $\rho$ is calculated by:

\[ F(\rho, |\psi_{ideal}\rangle) = \langle \psi_{ideal} | \rho | \psi_{ideal}\rangle. \]

For small $p$, assuming independence of error sources: - Fidelity contributions for each $CNOT$ give:

\[ F \approx 1 - 2\frac{p}{15}. \]

Final Answer:

The physical state fidelity $ F $ of the final physical 4-qubit state with respect to the target logical two-qubit GHZ state is approximately:

\[ F \approx 1 - \frac{4p}{15}. \]

This formula hints at the degradation of fidelity due to errors introduced by the $CNOT$ gates in the presence of a depolarizing channel following each such operation.

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 determine the logical state fidelity of the final 2-qubit logical state after running the circuit while post-selecting on all detectable errors, we analyze how the quantum error detection properties of the [[4,2,2]] code handle errors introduced by the circuit.

Step-by-Step Derivation:

State Preparation Circuit:

The circuit sequence is:

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

The operations execute in reverse order, affecting qubits as follows: - Initial state: $|0000\rangle$. - $H_2$: Transforms qubit 2:

\[ |0000\rangle \to \frac{1}{\sqrt{2}}\left(|0000\rangle + |0010\rangle\right). \]

$CNOT_{21}$: Entangles qubits 2 (control) and 1 (target):

\[ \to \frac{1}{\sqrt{2}}\left(|0000\rangle + |0011\rangle\right). \]

$H_0$: Acts on qubit 0:

\[ \to \frac{1}{2}\left(|0000\rangle + |1000\rangle + |0011\rangle + |1011\rangle\right). \]

$CNOT_{03}$: Entangles qubits 0 (control) and 3 (target):

\[ \to \frac{1}{2}\left(|0000\rangle + |1001\rangle + |0011\rangle + |1010\rangle\right). \]

Error Model with Post-Selection:

Each $CNOT$ is followed by a depolarizing channel. However, since we are post-selecting on no errors being detected by the code, the stabilizers $XXXX$ and $ZZZZ$ must return $+1$ when employed.

Post-Selection: - Only errors not detectable by the stabilizers are post-selected. - Detectable errors causing syndrome changes are discarded. - For one error in either $CNOT$, an error might be detected by messing up a stabilizer condition.

Logical Operators and Error Detection Abilities:

Logical State with Ideal Code: $(|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}$.
Logical Errors and Undetected Errors: The crucial undetectable error combinations which don’t trigger stabilizers are either no error or very rare uncorrelated double-qubit errors considering the stabilizer constraints of the code.

Fidelity Calculation:

Undetected Errors: Error modalities that don’t trigger a detectable change in the syndrome are logical errors that occur less often than detectable errors.
Probability of Detectable Error: Given the error model for a single $CNOT$, detectable errors precipitate a logical error with less than a 2-fold probability reduction for each 2-qubit gate.
The probability of not detecting an error, which scapegoats contribution from undetected errors outweighs full fidelity degradation to logical state:

\[ F_{\text{logical}}(\rho, |\psi_{\text{ideal}}\rangle) \approx 1 - \frac{2p}{15}, \]

where additional negligible contributions originate from unmonitored stabilizer-breach events after single round undetected stabilization, reflecting the coherence preservation.

Final Answer:

The logical state fidelity $F_{\text{logical}}$ of the final 2-qubit logical state post selection is approximately:

\[ F_{\text{logical}} \approx 1 - \frac{2p}{15}. \]

This reflects the code’s ability to post-select the final state while discarding detectable errors, retaining a fidelity higher than simple physical fidelity compromised by errors.

User #5

Checkpoint 3

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 #6

To address Subproblem 3, we need to analyze how the provided state preparation circuit, which includes an ancilla qubit, prepares a fault-tolerant logical $|00\rangle_{AB}$ state in the [[4,2,2]] code. The circuit capitalizes on post-selection based on the ancilla measurement.

Step-by-Step Derivation:

Circuit Description:

The operations execute in reverse from the given order: - Initial state: $|00000\rangle$ (including ancilla qubit 4 initialized to $|0\rangle$). - $H_1$: Applies Hadamard to qubit 1, entangling it:

\[ |00000\rangle \to \frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle). \]

$CNOT_{12}$: Transfers control from qubit 1 to qubit 2:

\[ \to \frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle). \]

$CNOT_{10}$: Further control from qubit 1 to qubit 0:

\[ \to \frac{1}{\sqrt{2}}(|00000\rangle + |01110\rangle). \]

$CNOT_{23}$: Twin control between qubits 2 and 3:

\[ \to \frac{1}{\sqrt{2}}(|00000\rangle + |01111\rangle). \]

$CNOT_{34}$: Uses qubit 3 as control for the ancilla:

\[ \to \frac{1}{\sqrt{2}}(|00000\rangle + |01111\rangle). \]

$CNOT_{04}$: Final control from qubit 0 to ancilla (4):

\[ \to \frac{1}{\sqrt{2}}(|00000\rangle + |01111\rangle). \]

Measurement $M_4$: Ancilla is measured. The correct operation ensures:
- Measured $|0\rangle$, state collapses to desired:

\[ \to \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle). \]

Error Model with Ancilla and Post-Selection:

Each $CNOT$ gate is followed by a depolarizing channel:

\[ \mathcal{E}(\rho) = (1-p)\rho + \frac{p}{15}\sum_{P \in \{X,Y,Z\}^2} P\rho P^\dagger. \]

Post-Selection on Ancilla: - Successful post-selection occurs if the ancilla yields $|0\rangle$, accommodating only cases where no logical errors are deposited into detectable forms or mitigated by measured $|0\rangle$.

Logical State Fidelity Calculation:

Consider: - Detectable Errors Removal: Post-selection on the ancilla discards significant error forms measured as $|1\rangle$. - Undetected Errors Contribution Reduction: Suppose induced logical errors beyond ancilla post-selection are implausible: - Key contribution from the gates $CNOT_{23}$, $CNOT_{34}$, and $CNOT_{04}$. - Fidelity enhancement as follows:

The logical state fidelity $F_{\text{logical}}$ expectantly improves due to successful ancilla detection, leading to:

\[ F_{\text{logical}} \approx 1 - \frac{4p}{15}, \]

founded on significant mitigation of logical errors by ancilla utility.

Final Answer:

The logical state fidelity $F_{\text{logical}}$ of the prepared logical $|00\rangle_{AB}$ state, conditioned on the ancilla measuring $|0\rangle$ and post-selecting detectable errors, is approximately:

\[ F_{\text{logical}} \approx 1 - \frac{4p}{15}. \]

This reflects the error resilience due to the ancilla ensuring correct error post-selection and discarding logical contaminations that trigger unwanted measurements.

User #7

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 #8

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 - (4 * p / 15)
    # ---------------------------------------------------------------

    return F_logical