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 calculate the physical state fidelity of the 4-qubit state prepared by the given circuit, we need to account for the errors introduced by the depolarizing channel after each $CNOT$ gate.

Step-by-Step Derivation:

Circuit Gates and Errors:

The circuit, in sequence from right to left, executes: - $H_2$: Hadamard gate on qubit 2 - $CNOT_{21}$: $CNOT$ with qubit 2 as control and qubit 1 as target - $H_0$: Hadamard gate on qubit 0 - $CNOT_{03}$: $CNOT$ with qubit 0 as control and qubit 3 as target

Each $CNOT$ gate ($CNOT_{21}$ and $CNOT_{03}$) is followed by a depolarizing error on the two qubits involved with probability $p$.

Depolarizing Channel:

After a $CNOT$ gate, consider a two-qubit depolarizing channel where one of the 15 non-identity Pauli operators is applied with probability $p/15$. The identity (no error) is applied with probability $1-p$.

State Fidelity Calculation:

Since the depolarizing channel is symmetric and acts the same on each pair, we can represent the effect on the state fidelity as an average effect of decoherence introduced by both gates:

Each depolarizing error changes the fidelity of any two-qubit entangled state it acts upon by a factor dependent on $p$.
Two $CNOT$ gates introduce two error points, each propagating error with probability $p$.

The fidelity reduction of the entangled state is represented by:

\[ \text{Fidelity} = (1 - p)^{2}, \]

where the exponent 2 accounts for the two depolarizing errors (post $CNOT_{21}$ and $CNOT_{03}$).

Combine Errors:

We assume errors are independent. The compounded error probability from the two depolarizing channels is multiplicative due to their independence: - The probability that no error occurs in either gate is $(1 - p)$ per gate, squared for two $CNOT$ gates:

\[ \text{Overall Probability of No Error} = (1 - p)^2 \]

The fidelity of the state is effectively reduced by this compounded error probability.

Physical State Fidelity:

The physical fidelity of the final 4-qubit state, therefore, remains:

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

This fidelity reflects the probability that no error affected the GHZ state preparation circuit.

Final Answer:

The physical state fidelity of the final physical 4-qubit state at the end of the circuit is given by:

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

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 the described state preparation circuit, we need to understand how logical errors propagate in the [[4,2,2]] code when post-selecting on all detectable errors.

Step-by-Step Derivation:

Error Detection with the [[4,2,2]] Code:

The [[4,2,2]] code can detect a single arbitrary error on any of the qubits, but it cannot correct them. Errors are identified through stabilizer measurements:

The stabilizers $XXXX$ and $ZZZZ$ can detect:
- Any $X$-type or $Z$-type error on a single qubit.
- No detection for $Y$ (causing simultaneous $X$ and $Z$ errors) or Pauli combinations affecting two qubits, if they preserve codewords.

Logical GHZ State and Circuit:

The logical GHZ state should ideally be:

\[ |\psi_{\text{ideal}}\rangle_{\text{logical}} = \frac{1}{\sqrt{2}}(|00\rangle_{AB} + |11\rangle_{AB}). \]

The preparation circuit $ (CNOT_{03})(H_0)(CNOT_{21})(H_2) $ prepares the logical state by acting on the qubits $0$ to $3$. Each $CNOT$ introduces errors with probability $p$.

Effect of Errors:

After each $CNOT_{ij}$, introduce a depolarizing channel that outputs any non-identity Pauli on qubits $i$ and $j$ with probability $p/15$:

Each error introduces a probability cost to state fidelity.
However, since all detectable errors are post-selected out, the remaining states after projection are error-free at logical level.

Post-Selection and Logical Fidelity:

When post-selecting on no detectable error: - Errors detected by stabilizers are discarded, improving fidelity. - The remaining errors are those that either: - Do not commute with stabilizers, thus detectable. - Commute but still preserve the logical GHZ state upon correction.

Probability No Error Occurs:

Unwanted errors that do not affect logical state appear when: - Both CNOT gates succeed (apply correctly), probability remains $(1-p)^2$. - Post-selected Circuit Success Probability: - We post-select on the state not flipping due to detectable errors (only undetectable logical errors affect this case).

Logical Fidelity Formula:

With post-selection, the logical state fidelity is maintained as if no error occurred:

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

This probability reflects post selection of states consistent with no alterations to logical encoding.

Final Answer:

The logical state fidelity of the final 2-qubit logical state, assuming detection of all errors, is:

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

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_logical = (1 - p)**2  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical