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

Step 1: Understand the given circuit and its operations

The given state preparation circuit is $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$. To understand the physical state fidelity, we first need to analyze the circuit’s operation. The circuit operates from right to left, so the first operation is $H_2$, followed by $CNOT_{21}$, then $H_0$, and finally $CNOT_{03}$.

Step 2: Analyze the initial state and the effect of each gate

The initial state of the 4 qubits is $|0000\rangle$. The first operation is $H_2$, which puts qubit 2 into a superposition state: $|0\rangle_2 \rightarrow \frac{1}{\sqrt{2}}(|0\rangle_2 + |1\rangle_2)$. Thus, the state becomes $|0000\rangle \rightarrow |00\rangle_0|0\rangle_1\frac{1}{\sqrt{2}}(|0\rangle_2 + |1\rangle_2)|0\rangle_3$.

Step 3: Apply $CNOT_{21}$ to the state

$CNOT_{21}$ flips qubit 1 if qubit 2 is $|1\rangle$. So, $|00\rangle_0|0\rangle_1\frac{1}{\sqrt{2}}(|0\rangle_2 + |1\rangle_2)|0\rangle_3 \rightarrow |00\rangle_0\frac{1}{\sqrt{2}}(|0\rangle_1|0\rangle_2 + |1\rangle_1|1\rangle_2)|0\rangle_3 = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)$.

Step 4: Apply $H_0$ to the state

$H_0$ puts qubit 0 into a superposition: $\frac{1}{\sqrt{2}}(|0\rangle_0 + |1\rangle_0)$. So, $\frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle) \rightarrow \frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle)$.

Step 5: Apply $CNOT_{03}$ to the state

$CNOT_{03}$ flips qubit 3 if qubit 0 is $|1\rangle$. So, $\frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle) \rightarrow \frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$.

Step 6: Understand the effect of the two-qubit depolarizing error channel

Each $CNOT$ gate is followed by a two-qubit depolarizing error channel that applies one of the 15 non-identity two-qubit Paulis with probability $p/15$. The identity operation is applied with probability $1-p$.

Step 7: Calculate the probability of error for each $CNOT$ gate

For $CNOT_{21}$ and $CNOT_{03}$, the probability of an error is $p$. The error channel affects the qubits involved in the $CNOT$ gate. We need to consider how these errors propagate through the circuit.

Step 8: Determine the ideal final state

The ideal final state after the circuit is $\frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$. This is a superposition that corresponds to the logical state $(|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}$ in the [[4,2,2]] code.

Step 9: Calculate the fidelity of the final physical state

To calculate the physical state fidelity, we need to consider the effect of errors on the final state. The fidelity $F$ is given by the probability that the final state is the ideal state.

Step 10: Analyze the effect of errors on $CNOT_{21}$ and $CNOT_{03}$

For $CNOT_{21}$, an error affects qubits 1 and 2. For $CNOT_{03}$, an error affects qubits 0 and 3. We need to consider how these errors affect the final state’s fidelity.

Step 11: Compute the fidelity as a function of $p$

Let’s denote the ideal output state as $\rho_{ideal} = |\psi_{ideal}\rangle\langle\psi_{ideal}|$ where $|\psi_{ideal}\rangle = \frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$. The actual output state $\rho_{actual}$ is affected by the depolarizing channels. The fidelity is $F = \langle\psi_{ideal}|\rho_{actual}|\psi_{ideal}\rangle$.

Step 12: Calculate the probability of no error and the effect of errors

The probability that no error occurs in either $CNOT$ gate is $(1-p)^2$. For each $CNOT$ gate, the probability of an error is $p$, and there are 15 possible errors, each occurring with probability $p/15$.

Step 13: Simplify the calculation by directly considering the impact of $p$ on fidelity

Since the circuit involves two $CNOT$ gates, and each has a depolarizing error channel, we can directly analyze the impact of $p$ on the fidelity. The fidelity after one $CNOT$ gate with error probability $p$ can be related to the probability of no error or errors that do not affect the logical state.

Step 14: Calculate the final fidelity

Given that there are two $CNOT$ gates, the overall fidelity $F$ can be expressed in terms of $p$. For a depolarizing channel following a $CNOT$, the fidelity of the operation is $1-p$ because $1-p$ is the probability that the identity operation (or an operation that does not change the state) occurs.

The final answer is: $\boxed{(1-p)^2}$

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

Step 1: Understand the given circuit and its operations

The given state preparation circuit is $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$. The circuit operates from right to left.

Step 2: Recall the definition of logical state fidelity

The logical state fidelity is the fidelity between the ideal logical state and the actual logical state after post-selection on detectable errors.

Step 3: Understand the [[4,2,2]] code and its error detection capability

The [[4,2,2]] code encodes 2 logical qubits into 4 physical qubits. It can detect errors but not correct them. The stabilizers are $XXXX$ and $ZZZZ$, indicating that any single-qubit error or certain two-qubit errors can be detected.

Step 4: Analyze the effect of the depolarizing error channel on the $CNOT$ gates

Each $CNOT$ gate is followed by a depolarizing error channel with error probability $p$. This means that with probability $p$, one of the 15 non-identity two-qubit Paulis is applied.

Step 5: Consider the post-selection on detectable errors

The state is post-selected on all detectable errors. This means we only consider cases where no error or undetectable errors occur.

Step 6: Determine the probability of undetectable errors

For the [[4,2,2]] code, errors that are not detectable are those that are proportional to the stabilizers ($XXXX$ and $ZZZZ$) or commute with both. We need to consider how the depolarizing channels affect the detectability of errors.

Step 7: Calculate the probability of no error or undetectable error

The probability that a $CNOT$ gate has no error is $1-p$. For two $CNOT$ gates, the probability of no error is $(1-p)^2$. Errors that are undetectable are related to the stabilizers.

Step 8: Understand that the logical fidelity after post-selection depends on the probability of undetectable errors

Since we post-select on detectable errors, the logical fidelity is related to the cases where errors are either absent or undetectable.

Step 9: Analyze the specific errors that are undetectable in the [[4,2,2]] code

Undetectable errors are those that are in the stabilizer group or have the same syndrome as the stabilizers. For two-qubit errors following $CNOT$ gates, we need to check which errors are undetectable.

Step 10: Compute the logical fidelity as a function of $p$

To compute the logical fidelity after post-selection, we consider the probability of having an undetectable error versus the total probability of not being detected (which includes having no error).

Step 11: Simplify the expression for logical fidelity

The logical fidelity $F_L$ after post-selection can be simplified by considering the ratio of the probability of being in the correct logical state (or an undetectable error) to the total probability of not being rejected by the error detection.

The final answer is: $\boxed{1}$

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  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical

# Example usage:
print(answer(p))

Checkpoint-2

Problem setup:

Checkpoint 1

Step 1: Understand the given circuit and its operations

Step 2: Analyze the initial state and the effect of each gate

Step 3: Apply \(CNOT_{21}\) to the state

Step 4: Apply \(H_0\) to the state

Step 5: Apply \(CNOT_{03}\) to the state

Step 6: Understand the effect of the two-qubit depolarizing error channel

Step 7: Calculate the probability of error for each \(CNOT\) gate

Step 8: Determine the ideal final state

Step 9: Calculate the fidelity of the final physical state

Step 10: Analyze the effect of errors on \(CNOT_{21}\) and \(CNOT_{03}\)

Step 11: Compute the fidelity as a function of \(p\)

Step 12: Calculate the probability of no error and the effect of errors

Step 13: Simplify the calculation by directly considering the impact of \(p\) on fidelity

Step 14: Calculate the final fidelity

Checkpoint 2

Step 1: Understand the given circuit and its operations

Step 2: Recall the definition of logical state fidelity

Step 3: Understand the [[4,2,2]] code and its error detection capability

Step 4: Analyze the effect of the depolarizing error channel on the \(CNOT\) gates

Step 5: Consider the post-selection on detectable errors

Step 6: Determine the probability of undetectable errors

Step 7: Calculate the probability of no error or undetectable error

Step 8: Understand that the logical fidelity after post-selection depends on the probability of undetectable errors

Step 9: Analyze the specific errors that are undetectable in the [[4,2,2]] code

Step 10: Compute the logical fidelity as a function of \(p\)

Step 11: Simplify the expression for logical fidelity