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 $CNOT_{03}$.

Step 2: Apply the operations in the given order to understand the circuit’s effect

First, apply $H_2$ to the state. Assuming the initial state is $|0000\rangle$, $H_2$ gives $\frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)$.

Step 3: Continue with the application of $CNOT_{21}$

Applying $CNOT_{21}$ to $\frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)$ results in $\frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)$.

Step 4: Apply $H_0$

$H_0$ on $\frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)$ gives $\frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle)$.

Step 5: Apply $CNOT_{03}$

Finally, applying $CNOT_{03}$ results in $\frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$.

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 $p/15$. There are two $CNOT$ gates, so we need to consider the error channels for both $CNOT_{21}$ and $CNOT_{03}$.

Step 7: Calculate the probability of no error occurring after both $CNOT$ gates

The probability that no error occurs after a $CNOT$ gate is $1 - p$. Since there are two $CNOT$ gates, the probability of no error occurring after both 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 correctly prepared. For a depolarizing channel, the fidelity after one $CNOT$ gate is $1 - p$. For two $CNOT$ gates, it’s $(1 - p)^2$ because the errors are independent.

Step 9: Consider the effect of errors on the state preparation

However, we must directly calculate or reason about the fidelity considering the specific circuit and error model.

Step 10: Analyze the circuit and error model to determine the state fidelity

The initial state is $|0000\rangle$. After applying $H_2$, $CNOT_{21}$, $H_0$, and $CNOT_{03}$, we get $\frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$. The $CNOT$ gates are subject to depolarizing noise. The fidelity of the state is related to the probability of no error or errors that do not affect the logical state.

Step 11: Compute the fidelity considering the error channels

Let’s directly compute the effect of the error channels on the fidelity. The state after the circuit without errors is $\frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)$. We need to consider how the depolarizing error channels after $CNOT_{21}$ and $CNOT_{03}$ affect this state.

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

For a two-qubit depolarizing channel with error probability $p$, the fidelity after one $CNOT$ is $1 - p$. After two $CNOT$s, it becomes $(1-p)^2$. However, we need to directly apply this understanding to our specific circuit.

Step 13: Calculate the final state fidelity as a function of $p$

The final state’s fidelity is directly related to the probability of successful gate operations. Given that $H$ gates are assumed to be error-free or their errors are not considered, we focus on $CNOT$ gates. The probability of an error-free operation for both $CNOT$ gates is $(1-p)^2$.

Step 14: Derive the expression for the physical state fidelity

Since the $H$ gates are not subject to the depolarizing error model described, and assuming they are perfect, the fidelity of the final state is primarily affected by the $CNOT$ gates. Thus, the fidelity $F = (1-p)^2$ because there are two $CNOT$ gates.

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 given problem and the circuit

The problem asks us to find the logical state fidelity of the final 2-qubit logical state after preparing a logical two-qubit GHZ state using the given circuit, assuming post-selection 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 [[4,2,2]] code has stabilizers $XXXX$ and $ZZZZ$, and logical operators $X_A = XIXI$, $X_B = XXII$, $Z_A = ZZII$, $Z_B = ZIZI$. The logical GHZ state is $(|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}$.

Step 3: Understand the effect of the circuit on the logical state

The circuit $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$ prepares a physical state that encodes the logical GHZ state.

Step 4: Analyze the errors detectable by the [[4,2,2]] code

The [[4,2,2]] code can detect single-qubit errors because any single-qubit Pauli error will anti-commute with at least one of the stabilizers ($XXXX$ or $ZZZZ$), thus being detectable.

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

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

Step 6: Determine the conditions under which errors are detectable

Errors are detectable if they cause the state to be outside the codespace or if they anti-commute with the stabilizers.

Step 7: Understand the post-selection process

The state is post-selected on all detectable errors, meaning that if an error is detected, the state is discarded.

Step 8: Calculate the probability of undetectable errors

To have an undetectable error, the error must be a product of the stabilizers or commute with both $XXXX$ and $ZZZZ$. For two-qubit depolarizing errors after $CNOT$ gates, we need to consider how these errors propagate.

Step 9: Analyze the logical fidelity after post-selection

The logical fidelity after post-selection is related to the probability that the errors are either detected or do not affect the logical state.

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

Given that the code detects single-qubit errors and some two-qubit errors, we need to consider how the two-qubit depolarizing error channels affect the logical state fidelity.

Step 11: Simplify the problem by directly considering the effect of $p$ on logical fidelity

For the [[4,2,2]] code, certain errors are detectable. The logical fidelity is related to the probability of having no error or an undetectable error that does not affect the logical state.

Step 12: Derive the expression for the logical state fidelity

The probability of an undetectable error (which could be a combination of errors that commute with the stabilizers) needs to be considered. However, given the post-selection, the fidelity is related to the cases where no undetectable logical error occurs.

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

Checkpoint-2