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, mathematical identities or numerical codes.
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-by-Step Derivation

Assumptions and noise model
Initial state is the computational basis state on four qubits: $|0000\rangle$.
The state preparation circuit is applied right-to-left: 1) $H_2$, 2) $CNOT_{21}$, 3) $H_0$, 4) $CNOT_{03}$.
After each $CNOT_{ij}$, a two-qubit depolarizing channel acts on $(i,j)$:

\[ \mathcal{D}_p(\rho)=(1-p)\rho+\frac{p}{15}\sum_{P\in\mathcal{P}_2\setminus\{II\}} P\rho P, \]

where $\mathcal{P}_2=\{I,X,Y,Z\}^{\otimes 2}$ on the two involved qubits.

Ideal final state and its stabilizer
Apply the circuit to $|0000\rangle$:
- $H_2$ followed by $CNOT_{21}$ prepares a Bell pair on qubits $(2,1)$:

\[ |{\Phi^+}\rangle_{21}=\frac{|00\rangle_{21}+|11\rangle_{21}}{\sqrt{2}}. \]

- $H_0$ followed by $CNOT_{03}$ prepares a Bell pair on qubits $(0,3)$:

\[ |{\Phi^+}\rangle_{03}=\frac{|00\rangle_{03}+|11\rangle_{03}}{\sqrt{2}}. \]

Hence the ideal four-qubit state is

\[ |\psi\rangle=|{\Phi^+}\rangle_{03}\otimes|{\Phi^+}\rangle_{21}. \]

The stabilizer generators (conjugating $Z_i$ through the circuit) are

\[ X_0X_3,\quad Z_0Z_3,\quad Z_2Z_1,\quad X_2X_1, \]

confirming $|\psi\rangle$ is a product of two Bell states (each Bell state has stabilizer $\langle X\otimes X,\,Z\otimes Z\rangle$).

Location of errors and their propagation
Errors occur only after $CNOT_{21}$ (on qubits $(2,1)$) and after $CNOT_{03}$ (on qubits $(0,3)$).
The gates between these errors ($H_0$ and $CNOT_{03}$ after the first error, and no further gate after the second error) act on disjoint qubits from $(2,1)$; hence the first error does not propagate to $(0,3)$ and vice versa.
Therefore, the net error at the end is a tensor product

\[ E_{\text{tot}}=E_{21}\otimes E_{03}, \]

with $E_{21}$ a two-qubit Pauli on $(2,1)$ and $E_{03}$ a two-qubit Pauli on $(0,3)$, independently drawn from the depolarizing distributions.

Fidelity contribution from a single pair
For a single Bell pair $|{\Phi^+}\rangle$, the two-qubit Paulis that leave the density matrix invariant (map the state to itself up to a global phase) are

\[ \{II,\ XX,\ YY,\ ZZ\}. \]

Indeed, $XX$ and $ZZ$ stabilize $|{\Phi^+}\rangle$ with eigenvalue $+1$, while $YY=-(XX)(ZZ)$ acts with eigenvalue $-1$, which still leaves the projector $|\Phi^+\rangle\langle\Phi^+|$ unchanged.

Under the depolarizing channel on that pair:
- With probability $(1-p)$, $II$ occurs (good).
- With probability $p$, a uniformly random non-identity Pauli from the 15 occurs; among these, exactly 3 are good: $XX,YY,ZZ$.
Thus, the probability that the pair remains in $|{\Phi^+}\rangle$ is

\[ f_{\text{pair}}=(1-p)+\frac{3}{15}p=1-\frac{12}{15}p=1-\frac{4}{5}p. \]

Overall four-qubit fidelity
Because the two error channels act on disjoint pairs and independently, the total state is preserved if and only if both pairs are preserved. Hence the total physical-state fidelity is

\[ F(p)=\big(f_{\text{pair}}\big)^2=\left(1-\frac{4}{5}p\right)^2. \]

Mathematical Typesetting

All key steps have been expressed in LaTeX above.

Conventions and Units

Qubits labeled as specified: physical qubits 0,1,2,3; pairs affected by the two CNOTs are (2,1) and (0,3).
Noise model is the stated two-qubit depolarizing channel with parameter $p$.

Final Answer: F(p) = (1 - 4p/5)^2.

User #3

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 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 $p$.

    Inputs
    ----------
    p: sympy.Symbol, two-qubit gate error rate $p$

    Outputs
    ----------
    F_physical: sympy.Expr, the physical state fidelity of the final physical 4-qubit state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_physical = ...  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_physical

Assistant #4

import sympy as sp

p = sp.symbols('p')

def answer(p):
    r"""
    Return the expression of 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 $p$.

    Inputs
    ----------
    p: sympy.Symbol, two-qubit gate error rate $p$

    Outputs
    ----------
    F_physical: sympy.Expr, the physical state fidelity of the final physical 4-qubit state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_physical = (1 - sp.Rational(4, 5)*p)**2  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_physical