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 $p$.

Checkpoints

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 analyze how each gate, followed by a depolarizing noise channel, affects the state preparation.

Step-by-Step Derivation:

Understand the Circuit:

The state preparation circuit is:

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

This reads as: apply a Hadamard gate $H$ to qubit 2, then a $CNOT_{21}$, then $H$ to qubit 0, and finally $CNOT_{03}$.

Depolarizing Error Channel Model:

For each $CNOT_{ij}$, there is a depolarizing error channel following it. The channel applies one of the 15 non-identity two-qubit Pauli errors $\{X_i, Y_i, Z_i, X_j, Y_j, Z_j, X_iX_j, Y_iY_j, Z_iZ_j, X_iY_j, X_iZ_j, Y_iX_j, Y_iZ_j, Z_iX_j, Z_iY_j\}$ with equal probability $p/15$. Without error, the probability is $1-p$.

State Preparation:

We begin by assuming the following product state:

The target GHZ state in logical qubits $A$ and $B$ is:

\[ |\text{GHZ}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)_{AB} \]

Effect of the Final Gate $CNOT_{03}$ and Error:

Apply $CNOT_{03}$ without error:

\[ |\psi\rangle \rightarrow |\psi'\rangle = CNOT_{03}|\psi\rangle \]

Depolarizing Channel: After $CNOT_{03}$:
- Probability of no error: $1-p$
- Probability of any error: $p$

Base Calculation of State Fidelity:

Since errors occur with probability $p$, for each two-qubit CNOT, only $1-p$ of the operation will leave the state unchanged. Thus, out of the operations, two are $CNOT$ gates preceded by $H$, contributing to noise.

Formula for Fidelity:

Note that the overall procedure makes the probability of ending without any faults:

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

Here, $k = 2$ for two $CNOT$ gates, as $H$ does not introduce noise (assumed perfect).

Conclusion on State Fidelity:

The result shows that fidelity is primarily reduced by the $CNOT$ operations, each with an independent fault.

Physical State Fidelity:

The fidelity, representing the probability of no errors occurring on the result from the CNOT gates, is given by:

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

Final Answer:

Final Answer: $F = (1-p)^2$, where $F$ is the fidelity of the final physical 4-qubit state.

User #3

The correct answer is: $ F_{\rm{physical}}=(1-\frac{12}{15}p)^2 $

Now let’s solve the next Checkpoint.

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 when post-selecting on all detectable errors, we need to understand how the provided state preparation circuit behaves under the influence of errors and the post-selection process for detectable errors.

Step-by-Step Derivation:

Circuit Understanding and Initialization:

Start with the initial physical state:

The state preparation circuit is:

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

Logical State Target:

Goal: Prepare the logical GHZ state:

\[ |\text{GHZ}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)_{AB} \]

Depolarizing Error Impact:

For each $CNOT_{ij}$, there is a depolarizing error channel. It introduces one of the 15 non-identity two-qubit Pauli errors with equal probability $p/15$.
Errors detectable by the code (i.e., logical state still correctable): any two-qubit Pauli on distinct qubits.

Post-Selection Process:

In error detection codes, detectable errors are projected out during post-selection.
After a depolarizing error, the probability that an error is detectable (and thus not post-selected) is high because:
- Only errors involving single-qubit rotations on both qubits (like $X$, $Y$, $Z$ on either or both qubits of the CNOT pair) survive post-selection.
Probability of a two-qubit error being post-selected out remains as it affects stabilizers/anti-commutes substantially, hence detectable errors among 15 possible:
- Probability that an error doesn’t flip the outcome: $ (15 - \text{detectable errors})/15 $

Logical Fidelity Calculation:

Each $CNOT_{ij}$ erroneous $p/15$ per Pauli, summative detectable single errors:
- If 3 errors detectable, then non-fidelity error influence retains:

\[ 9/15 = \text{Detectable errors} \]

Developed Logical Fidelity:

Given detectable errors prevent logical circuit faults, the logical fidelity equals the undetectable portion maintaining correctness for logical states:

\[ F_{\rm{logical}} = (1 - (12/15) p)^2 \]

Here, we assume no double errors (least likely, hence smallest impact).

Final Answer:

Final Answer: $F_{\rm{logical}} = (1 - \frac{12}{15} p)^2$ where $F_{\rm{logical}}$ is the fidelity of the final logical 2-qubit state.

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 - (12/15) * p)**2
    # ---------------------------------------------------------------

    return F_logical