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

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 understand that the operations are applied from right to left, so the order is: $H_2$, $CNOT_{21}$, $H_0$, and finally $CNOT_{03}$. The task is 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: Analyze the circuit operation step by step

First, we apply $H_2$ to the initial state $|0000\rangle$. $H_2$ puts qubit 2 into a superposition state: $|0\rangle_2 \rightarrow \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)_2$. So, the state becomes $|000\rangle \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$.

Step 3: Continue with $CNOT_{21}$

Applying $CNOT_{21}$, we get $\frac{1}{\sqrt{2}}(|0000\rangle + |0100\rangle)$ because $CNOT_{21}$ flips qubit 1 if qubit 2 is $|1\rangle$.

Step 4: Apply $H_0$

$H_0$ acts on qubit 0, resulting in $\frac{1}{2}(|0000\rangle + |1000\rangle + |0100\rangle + |1100\rangle)$.

Step 5: Apply $CNOT_{03}$

Finally, $CNOT_{03}$ gives $\frac{1}{2}(|0000\rangle + |1001\rangle + |0100\rangle + |1101\rangle)$.

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 produces one of the 15 non-identity two-qubit Paulis with equal probability $p/15$. There are two $CNOT$ gates, so we need to consider the error probability for both.

Step 7: Calculate the probability of no error occurring after each $CNOT$ gate

For a single $CNOT$ gate followed by the error channel, the probability of no error is $1 - p$. Since there are two $CNOT$ gates, the overall probability of no error is $(1 - p)^2$.

Step 8: Calculate the fidelity of the final state

The fidelity of the final state is the probability that the state is prepared correctly. Given that the initial state is $|0000\rangle$, the correct final state is $\frac{1}{2}(|0000\rangle + |1001\rangle + |0100\rangle + |1101\rangle)$. Errors can occur after each $CNOT$ gate. We need to consider how these errors affect the state.

Step 9: Analyze the effect of errors on the state

For $CNOT_{21}$, an error can result in one of 15 possible two-qubit Pauli errors on qubits 1 and 2. For $CNOT_{03}$, an error affects qubits 0 and 3. The fidelity will be affected by the probability of these errors.

Step 10: Compute the fidelity

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 channel model, we directly calculate the effect of errors on the density matrix.

Step 11: Derive the expression for fidelity as a function of $p$

Let’s denote the ideal final state as $\rho_{ideal} = |\psi\rangle\langle\psi|$, where $|\psi\rangle = \frac{1}{2}(|0000\rangle + |1001\rangle + |0100\rangle + |1101\rangle)$. The actual state $\rho_{actual}$ is affected by the error channels. For a two-qubit depolarizing channel with error probability $p$, the output state after a $CNOT$ gate can be represented as $(1-p)\rho_{ideal,CNOT} + p\mathcal{E}(\rho_{ideal,CNOT})$, where $\mathcal{E}$ represents the depolarizing channel’s effect.

Step 12: Simplify the calculation by directly considering the effect of $p$ on the circuit

Since the circuit involves two $CNOT$ gates, each followed by a depolarizing error channel, we can simplify by directly applying the error model. The probability of an error on a single $CNOT$ is $p$, and there are two $CNOT$s, so the fidelity will be related to $(1-p)^2$ for the case of no errors on either $CNOT$. However, we must consider the overall effect on the 4-qubit state.

Step 13: Calculate the final state fidelity

To simplify, we directly analyze the impact of $p$ on the state. The initial state is $|0000\rangle$. After applying the circuit, the ideal state is achieved. The fidelity after errors is $F = \langle\psi|\rho_{actual}|\psi\rangle$. For a depolarizing error channel after each $CNOT$, the state’s fidelity is affected. The overall fidelity can be derived by considering the effect of errors on the density matrix.

Step 14: Explicitly derive $F(p)$

For a single $CNOT$ gate with error probability $p$, the fidelity of the operation is $1 - p$. With two $CNOT$ gates, the overall fidelity due to gate errors is $(1-p)^2$. However, we need to consider how these errors propagate through the circuit and affect the final state’s fidelity.

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

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

Step 1: Understand the task and the given circuit

We need to calculate the logical state fidelity of the final 2-qubit logical state after preparing a logical two-qubit GHZ state using the given circuit and post-selecting on all detectable errors in the [[4,2,2]] code.

Step 2: Recall the stabilizers and logical operators of the [[4,2,2]] code

The stabilizers are $XXXX$ and $ZZZZ$, and the logical operators are $X_A = XIXI$, $X_B = XXII$, $Z_A = ZZII$, $Z_B = ZIZI$. The logical GHZ state is $(|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}$.

Step 3: Analyze the circuit and its output

The circuit $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$ prepares a physical state. We need to understand how this state encodes the logical GHZ state.

Step 4: Understand the effect of post-selection on detectable errors

The [[4,2,2]] code can detect errors. Post-selecting on detectable errors means we only consider cases where no detectable error occurred.

Step 5: Determine the logical fidelity after post-selection

To calculate the logical fidelity, we need to consider how errors affect the logical state and how post-selection changes this.

Step 6: Analyze the types of errors that are detectable by the [[4,2,2]] code

The code detects errors that anti-commute with the stabilizers $XXXX$ and $ZZZZ$. Errors like $XIIX$, $IXXI$, $ZIIZ$, $IZIZ$, etc., are detectable if they are not part of the stabilizers or logical operators.

Step 7: Consider the effect of the depolarizing error channel on the logical fidelity

For each $CNOT$ gate, a depolarizing error channel applies. We need to consider how these errors affect the logical state and how post-selection on detectable errors influences the logical fidelity.

Step 8: Calculate the probability of undetectable errors after post-selection

Since the code is a [[4,2,2]] code, it can detect single-qubit errors but not necessarily correct them. The logical fidelity after post-selection will depend on the probability of undetectable errors.

Step 9: Derive the logical fidelity as a function of $p$

Given that the circuit involves two $CNOT$ gates, each with a depolarizing error channel, we need to analyze how $p$ affects the logical state fidelity after post-selection.

Step 10: Simplify the derivation by directly considering the impact of $p$ on logical fidelity

The logical fidelity after post-selection is related to the probability of errors that are not detectable by the code.

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

# Execute the function to get the final answer
F_logical = answer(p)
print("Final Answer: ", F_logical)

Checkpoint-2

Problem setup:

Checkpoints

Checkpoint 1

Step 1: Understand the given circuit and the task

Step 2: Analyze the circuit operation step by step

Step 3: Continue with \(CNOT_{21}\)

Step 4: Apply \(H_0\)

Step 5: Apply \(CNOT_{03}\)