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 problem and the state preparation circuit.

The problem asks us to calculate the physical state fidelity of a 4-qubit state prepared using a specific circuit in the presence of two-qubit depolarizing errors after each $CNOT$ gate. The circuit is given by $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$, and the error model is a two-qubit depolarizing channel with error probability $p$.

Step 2: Analyze the state preparation circuit.

First, let’s understand the circuit. The operations are applied from right to left: $H_2$, then $CNOT_{21}$, then $H_0$, and finally $CNOT_{03}$. $H_i$ denotes a Hadamard gate on qubit $i$. The initial state is assumed to be $|0000\rangle$.

Step 3: Apply the operations in the circuit to the initial state.

$H_2$ on $|0000\rangle$ gives $|000\rangle + |001\rangle)/\sqrt{2} = (|00\rangle + |01\rangle)/\sqrt{2} \otimes |00\rangle$.
$CNOT_{21}$ gives $(|00\rangle + |01\rangle)/\sqrt{2} \otimes |00\rangle \rightarrow (|00\rangle|0\rangle + |01\rangle|1\rangle)/\sqrt{2} = (|000\rangle + |011\rangle)/\sqrt{2}$.
$H_0$ gives $((|0\rangle + |1\rangle)/\sqrt{2} \otimes |00\rangle + (|0\rangle - |1\rangle)/\sqrt{2} \otimes |11\rangle)/\sqrt{2} = (|000\rangle + |100\rangle + |011\rangle - |111\rangle)/2$.
$CNOT_{03}$ gives $(|0000\rangle + |1000\rangle + |0110\rangle - |1110\rangle)/2$.

Step 4: Simplify the final state without errors.

Let’s simplify the expression for the final state: $(|0000\rangle + |1000\rangle + |0110\rangle - |1110\rangle)/2$. This is the ideal state.

Step 5: Understand the effect of the depolarizing error channel on the $CNOT$ gates.

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 probability of no error is $1 - p$.

Step 6: Calculate the probability of the state being correct after the errors.

For the final state fidelity, we need to consider how the errors affect the state. The circuit has two $CNOT$ gates, so there are two opportunities for errors.

Step 7: Analyze the effect of errors on the state.

The error channel after each $CNOT$ gate can introduce errors. We need to consider the probability that the final state is correct (or equivalently, the fidelity of the final state).

Step 8: Compute the fidelity of the final state.

The fidelity $F$ of the final state can be related to the probability of no error or errors that do not affect the logical state. However, given the depolarizing noise model, we directly calculate the effect of errors on the state.

For a single $CNOT$ gate with depolarizing noise, the output density matrix $\rho'$ is given by $\rho' = (1-p)\rho + \frac{p}{15}\sum_{i=1}^{15} P_i \rho P_i$, where $P_i$ are the 15 non-identity two-qubit Paulis.

Step 9: Calculate the overall fidelity considering both $CNOT$ gates.

Let’s denote the initial state as $\rho_0 = |0000\rangle\langle0000|$. After applying $H_2$, $CNOT_{21}$, $H_0$, and $CNOT_{03}$ without errors, we get the ideal final state $\rho_{ideal}$. With errors, the state after the first $CNOT$ ($CNOT_{21}$) is $\rho_1 = (1-p)CNOT_{21}H_2\rho_0H_2CNOT_{21} + \frac{p}{15}\sum_{i=1}^{15} P_i CNOT_{21}H_2\rho_0H_2CNOT_{21}P_i$. After $H_0$ and the second $CNOT$ ($CNOT_{03}$), we apply a similar transformation.

Step 10: Simplify the calculation by directly considering the effect of $p$ on the fidelity.

The fidelity of the final state is given by $F = \langle\psi_{ideal}|\rho_{final}|\psi_{ideal}\rangle$, where $\rho_{final}$ is the density matrix of the final state with errors.

Step 11: Calculate the fidelity using the error model.

For two $CNOT$ gates, the probability of no error is $(1-p)^2$. The fidelity can be approximated by considering the effect of errors on the state. However, the exact fidelity calculation involves considering all possible error combinations and their effects.

Step 12: Derive the expression for the physical state fidelity as a function of $p$.

Given that each $CNOT$ gate is followed by a depolarizing channel, and there are two such gates, the overall fidelity can be related to the probability of error $p$. The fidelity $F = (1-p)^2$ because the errors are independent, and we are considering the probability that both gates 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 problem and the state preparation circuit.

The problem asks us to calculate the logical state fidelity of a 2-qubit logical state prepared using a specific circuit in the [[4,2,2]] code, with the state post-selected on all detectable errors. The circuit is given by $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$.

Step 2: Analyze the effect of the circuit on the logical qubits.

First, we need to understand how the given circuit affects the logical qubits. The initial state is $|00\rangle_{AB}$, and we aim to prepare $(|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}$.

Step 3: Understand the logical operations performed by the circuit.

$H_2$ acts on the physical qubit 2. $CNOT_{21}$ then acts on qubits 2 and 1. $H_0$ acts on qubit 0, and $CNOT_{03}$ acts on qubits 0 and 3.

Step 4: Translate the physical operations into logical operations.

To prepare the logical GHZ state, we need to understand how these physical operations translate into logical operations on qubits A and B.

Step 5: Determine the logical state after the circuit.

The logical $X$ operators are $X_A = XIXI$ and $X_B = XXII$. The circuit’s effect on the logical state can be understood by analyzing how it transforms the logical operators.

Step 6: Analyze the error detection capability of the [[4,2,2]] code.

The [[4,2,2]] code can detect single-qubit errors (among other errors) because its stabilizers are $XXXX$ and $ZZZZ$, allowing it to detect errors that anti-commute with these stabilizers.

Step 7: Understand the effect of post-selection on detectable errors.

Post-selecting on detectable errors means we discard states where errors are detected. This changes the probability distribution of the remaining states.

Step 8: Calculate the logical fidelity after post-selection.

The logical fidelity is the probability that the logical state is correct given that no detectable error is observed. We need to consider how the two-qubit depolarizing errors after each $CNOT$ gate affect the logical state and how post-selection changes this.

Step 9: Derive the expression for the logical state fidelity as a function of $p$.

To calculate the logical fidelity, we consider the probability of errors that are not detectable and affect the logical state. The [[4,2,2]] code detects single-qubit errors but not all two-qubit errors.

Step 10: Simplify the calculation by focusing on the relevant error probabilities.

For the logical GHZ state, errors that affect the logical state fidelity are those that are not detectable. We need to consider the probability of such errors.

Step 11: Compute the logical fidelity.

The probability of a two-qubit error (after a $CNOT$) is $p$. Some of these errors are detectable, and some are not. The code detects errors that anti-commute with $XXXX$ or $ZZZZ$.

Step 12: Calculate the final logical fidelity.

Let’s directly analyze the effect of $p$ on the logical fidelity. The circuit has two $CNOT$ gates. Errors after these gates can be detectable or not. Post-selection removes states with detectable errors.

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

F_logical = answer(p)
print(F_logical)

Checkpoint-2

Problem setup:

Checkpoint 1

Step 1: Understand the given problem and the state preparation circuit.

Step 2: Analyze the state preparation circuit.

Step 3: Apply the operations in the circuit to the initial state.

Step 4: Simplify the final state without errors.

Step 5: Understand the effect of the depolarizing error channel on the \(CNOT\) gates.

Step 6: Calculate the probability of the state being correct after the errors.

Step 7: Analyze the effect of errors on the state.

Step 8: Compute the fidelity of the final state.

Step 9: Calculate the overall fidelity considering both \(CNOT\) gates.

Step 10: Simplify the calculation by directly considering the effect of \(p\) on the fidelity.

Step 11: Calculate the fidelity using the error model.

Step 12: Derive the expression for the physical state fidelity as a function of \(p\).

Checkpoint 2

Step 1: Understand the given problem and the state preparation circuit.

Step 2: Analyze the effect of the circuit on the logical qubits.

Step 3: Understand the logical operations performed by the circuit.

Step 4: Translate the physical operations into logical operations.

Step 5: Determine the logical state after the circuit.

Step 6: Analyze the error detection capability of the [[4,2,2]] code.

Step 7: Understand the effect of post-selection on detectable errors.

Step 8: Calculate the logical fidelity after post-selection.

Step 9: Derive the expression for the logical state fidelity as a function of \(p\).

Step 10: Simplify the calculation by focusing on the relevant error probabilities.

Step 11: Compute the logical fidelity.

Step 12: Calculate the final logical fidelity.