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 the task

The given circuit to prepare a logical two-qubit GHZ state is $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$. We need to calculate the physical state fidelity of the final 4-qubit state as a function of the two-qubit gate error rate $p$.

Step 2: Break down the circuit into individual operations

First, we apply $H_2$, then $CNOT_{21}$, followed by $H_0$, and finally $CNOT_{03}$. The initial state is $|0000\rangle$.

Step 3: Apply the first operation, $H_2$, to the initial state $|0000\rangle$

Applying $H_2$ to $|0000\rangle$ gives $\frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)$.

Step 4: Apply $CNOT_{21}$ to the state $\frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)$

$CNOT_{21}$ results in $\frac{1}{\sqrt{2}}(|0000\rangle + |0011\rangle)$ because $CNOT_{21}$ flips the state of qubit 1 if qubit 2 is $|1\rangle$.

Step 5: Apply $H_0$ to $\frac{1}{\sqrt{2}}(|0000\rangle + |0011\rangle)$

$H_0$ gives $\frac{1}{2}(|0000\rangle + |1000\rangle + |0011\rangle + |1011\rangle)$.

Step 6: Apply $CNOT_{03}$ to $\frac{1}{2}(|0000\rangle + |1000\rangle + |0011\rangle + |1011\rangle)$

$CNOT_{03}$ results in $\frac{1}{2}(|0000\rangle + |1001\rangle + |0011\rangle + |1010\rangle)$.

Step 7: Understand the effect of the two-qubit depolarizing error channel

Each $CNOT$ gate is followed by a two-qubit depolarizing error channel that produces one of the 15 non-identity two-qubit Paulis with equal probability $p/15$. We need to consider the errors after $CNOT_{21}$ and $CNOT_{03}$.

Step 8: Calculate the probability of error for each $CNOT$ gate

For a $CNOT$ gate, the probability of no error is $1-p$, and the probability of an error is $p$. The error can be any of the 15 non-identity two-qubit Paulis.

Step 9: Analyze the effect of errors on the state fidelity

To calculate the physical state fidelity, we need to understand how errors affect the prepared state. The ideal final state is $\frac{1}{2}(|0000\rangle + |1001\rangle + |0011\rangle + |1010\rangle)$. Errors will cause deviations from this state.

Step 10: Compute the fidelity of the final state

The fidelity $F$ of the final state with respect to the ideal state is given by the overlap between the ideal state and the actual state produced by the noisy circuit. For a depolarizing channel after each $CNOT$, the fidelity can be affected by the errors.

Step 11: Calculate the fidelity after the first $CNOT_{21}$ and its associated error channel

After $CNOT_{21}$, the state is $\frac{1}{\sqrt{2}}(|0000\rangle + |0011\rangle)$. The error channel following $CNOT_{21}$ acts on qubits 1 and 2. The probability that no error occurs is $1-p$.

Step 12: Calculate the fidelity after $H_0$ and $CNOT_{03}$ with their associated error channels

$H_0$ is a single-qubit gate and does not have a two-qubit error channel associated with it in this model. $CNOT_{03}$ is followed by an error channel acting on qubits 0 and 3.

Step 13: Derive the expression for the fidelity of the final state

Let’s denote the ideal state as $|\psi\rangle = \frac{1}{2}(|0000\rangle + |1001\rangle + |0011\rangle + |1010\rangle)$. The fidelity $F = \langle\psi|\rho|\psi\rangle$, where $\rho$ is the density matrix of the final state.

Step 14: Analyze the effect of the depolarizing error channels on the fidelity

For each $CNOT$, the error channel can cause errors. The overall fidelity will be affected by the probability of no error occurring after both $CNOT$ gates.

Step 15: Compute the final state fidelity as a function of $p$

The probability of no error after both $CNOT$ gates is $(1-p)^2$. When an error occurs, it can be one of 15 possible two-qubit Pauli errors. The fidelity will decrease as $p$ increases.

The final answer is:

Step 1: Understand that the initial state is $|0000\rangle$ and the circuit applies a series of gates to it.

The circuit is $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$.

The $CNOT$ gates are followed by depolarizing error channels.

Step 3: Calculate the effect of the error channels on the fidelity.

For a single $CNOT$ gate followed by a depolarizing error channel, the fidelity of the operation is $1-p$ because $p$ is the probability of an error.

Step 4: Since there are two $CNOT$ gates, the overall probability of no error is $(1-p)^2$.

This is because both $CNOT_{21}$ and $CNOT_{03}$ must not have an error.

Given that the initial state is $|0000\rangle$ and the operations are applied, the ideal final state is achieved with probability $(1-p)^2$.

Step 6: Consider the effect of $H_0$ and $H_2$.

$H_0$ and $H_2$ are single-qubit gates. The problem does not specify an error model for single-qubit gates, so we assume they are ideal.

Step 7: Compute the final state fidelity.

The final state fidelity $F = (1-p)^2$ because this is the probability that both $CNOT$ operations are error-free.

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 the task

The given circuit to prepare a logical two-qubit GHZ state is $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$. We need to calculate the logical state fidelity of the final 2-qubit logical state as a function of the two-qubit gate error rate $p$, assuming the state is post-selected on all detectable errors in the code.

Step 2: Identify the logical states and the effect of the circuit on them

The logical GHZ state is $(|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}$. We need to understand how the given circuit prepares this state and how errors affect it.

Step 3: Analyze the circuit’s operation on the physical qubits

First, $H_2$ is applied to $|0\rangle$, resulting in $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. Then, $CNOT_{21}$ is applied, followed by $H_0$, and finally $CNOT_{03}$.

Step 4: Understand the encoding and the logical operations

The [[4,2,2]] code encodes two logical qubits into four physical qubits. The logical $X$ and $Z$ operators are given. We need to relate the physical operations to logical operations.

Step 5: Consider the effect of errors on the logical state

Errors that are detectable by the code will be post-selected out. The code can detect errors that anti-commute with the stabilizers $XXXX$ and $ZZZZ$.

Step 6: Analyze the probability of undetectable errors

For the [[4,2,2]] code, undetectable errors are those that commute with both $XXXX$ and $ZZZZ$. We need to consider how the two-qubit depolarizing error channels after $CNOT$ gates affect the logical state.

Step 7: Compute the probability of undetectable errors after the $CNOT$ gates

The error channels after $CNOT_{21}$ and $CNOT_{03}$ can introduce errors. We need to identify which of these errors are undetectable.

Step 8: Calculate the logical fidelity given post-selection on detectable errors

The logical fidelity after post-selection is related to the probability of undetectable errors. We need to calculate this probability as a function of $p$.

Step 9: Derive the expression for logical fidelity

Let’s consider the errors that can occur after $CNOT_{21}$ and $CNOT_{03}$. For $CNOT_{21}$, errors on qubits 1 and 2 can be $I$, $X$, $Y$, $Z$ on each qubit, resulting in 16 possible two-qubit Paulis. Similarly, for $CNOT_{03}$, errors on qubits 0 and 3 are considered.

Step 10: Understand that post-selection removes states with detectable errors

The [[4,2,2]] code detects single-qubit errors and some two-qubit errors. Errors that are not detected are those that commute with $XXXX$ and $ZZZZ$, such as $II$, $XX$, $YY$, $ZZ$ on all four qubits or certain combinations that effectively act as identity or logical operations on the encoded qubits.

Step 11: Simplify the calculation by focusing on the relevant error probabilities

The probability that an error is undetectable is related to the probability of certain error combinations occurring.

Step 12: Compute the logical fidelity

Given that the state is post-selected on detectable errors, the logical fidelity is related to the cases where errors are not detected.

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
    # ---------------------------------------------------------------

    return F_logical

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

Checkpoint-2

Problem setup:

Checkpoint 1

Step 1: Understand the given circuit and the task

Step 2: Break down the circuit into individual operations

Step 3: Apply the first operation, \(H_2\), to the initial state \(|0000\rangle\)

Step 4: Apply \(CNOT_{21}\) to the state \(\frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)\)

Step 5: Apply \(H_0\) to \(\frac{1}{\sqrt{2}}(|0000\rangle + |0011\rangle)\)

Step 6: Apply \(CNOT_{03}\) to \(\frac{1}{2}(|0000\rangle + |1000\rangle + |0011\rangle + |1011\rangle)\)

Step 7: Understand the effect of the two-qubit depolarizing error channel

Step 8: Calculate the probability of error for each \(CNOT\) gate

Step 9: Analyze the effect of errors on the state fidelity

Step 10: Compute the fidelity of the final state

Step 11: Calculate the fidelity after the first \(CNOT_{21}\) and its associated error channel

Step 12: Calculate the fidelity after \(H_0\) and \(CNOT_{03}\) with their associated error channels

Step 13: Derive the expression for the fidelity of the final state

Step 14: Analyze the effect of the depolarizing error channels on the fidelity

Step 15: Compute the final state fidelity as a function of \(p\)

Step 1: Understand that the initial state is \(|0000\rangle\) and the circuit applies a series of gates to it.

Step 3: Calculate the effect of the error channels on the fidelity.

Step 4: Since there are two \(CNOT\) gates, the overall probability of no error is \((1-p)^2\).

Step 6: Consider the effect of \(H_0\) and \(H_2\).

Step 7: Compute the final state fidelity.

Checkpoint 2

Step 1: Understand the given circuit and the task

Step 2: Identify the logical states and the effect of the circuit on them

Step 3: Analyze the circuit’s operation on the physical qubits

Step 4: Understand the encoding and the logical operations

Step 5: Consider the effect of errors on the logical state

Step 6: Analyze the probability of undetectable errors

Step 7: Compute the probability of undetectable errors after the \(CNOT\) gates

Step 8: Calculate the logical fidelity given post-selection on detectable errors

Step 9: Derive the expression for logical fidelity

Step 10: Understand that post-selection removes states with detectable errors

Step 11: Simplify the calculation by focusing on the relevant error probabilities

Step 12: Compute the logical fidelity

Problem setup:

Checkpoint 1

Step 1: Understand the given circuit and the task

Step 2: Break down the circuit into individual operations

Step 3: Apply the first operation, \(H_2\), to the initial state \(|0000\rangle\)

Step 4: Apply \(CNOT_{21}\) to the state \(\frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)\)

Step 5: Apply \(H_0\) to \(\frac{1}{\sqrt{2}}(|0000\rangle + |0011\rangle)\)

Step 6: Apply \(CNOT_{03}\) to \(\frac{1}{2}(|0000\rangle + |1000\rangle + |0011\rangle + |1011\rangle)\)

Step 7: Understand the effect of the two-qubit depolarizing error channel

Step 8: Calculate the probability of error for each \(CNOT\) gate

Step 9: Analyze the effect of errors on the state fidelity

Step 10: Compute the fidelity of the final state

Step 11: Calculate the fidelity after the first \(CNOT_{21}\) and its associated error channel

Step 12: Calculate the fidelity after \(H_0\) and \(CNOT_{03}\) with their associated error channels

Step 13: Derive the expression for the fidelity of the final state

Step 14: Analyze the effect of the depolarizing error channels on the fidelity

Step 15: Compute the final state fidelity as a function of \(p\)

Step 1: Understand that the initial state is \(|0000\rangle\) and the circuit applies a series of gates to it.

Step 2: Recognize that the fidelity of the final state is related to the probability of errors in the \(CNOT\) gates.

Step 3: Calculate the effect of the error channels on the fidelity.

Step 4: Since there are two \(CNOT\) gates, the overall probability of no error is \((1-p)^2\).

Step 5: The fidelity of the final state with respect to the ideal state is directly related to the probability of no error.

Step 6: Consider the effect of \(H_0\) and \(H_2\).

Step 7: Compute the final state fidelity.

Checkpoint 2

Step 1: Understand the given circuit and the task

Step 2: Identify the logical states and the effect of the circuit on them

Step 3: Analyze the circuit’s operation on the physical qubits

Step 4: Understand the encoding and the logical operations

Step 5: Consider the effect of errors on the logical state

Step 6: Analyze the probability of undetectable errors

Step 7: Compute the probability of undetectable errors after the \(CNOT\) gates

Step 8: Calculate the logical fidelity given post-selection on detectable errors

Step 9: Derive the expression for logical fidelity

Step 10: Understand that post-selection removes states with detectable errors

Step 11: Simplify the calculation by focusing on the relevant error probabilities

Step 12: Compute the logical fidelity