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 $p$.

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

Ideal state prepared by the circuit:
Start from |0000⟩.
Apply H2: (|0⟩2 + |1⟩2)/√2 ⊗ |000⟩013.
Apply CNOT21: creates a Bell pair on qubits (2,1): |Φ+⟩21 = (|00⟩ + |11⟩)/√2. State is |Φ+⟩21 ⊗ |00⟩03.
Apply H0: puts qubit 0 in |+⟩.
Apply CNOT03: creates a Bell pair on qubits (0,3): |Φ+⟩03. Final ideal state

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

The stabilizers of |Φ+⟩ are generated by {X⊗X, Z⊗Z}; equivalently, the four elements {I⊗I, X⊗X, Y⊗Y, Z⊗Z} act as ±1 on |Φ+⟩.

Noise model:
After each CNOTij, a two-qubit depolarizing channel acts on qubits i and j:

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

where 𝒫2 = {I, X, Y, Z}⊗2.

There are two such channels: one on (2,1) after CNOT21 and one on (0,3) after CNOT03. The H gates are ideal. The first channel acts on qubits (2,1) and commutes with later gates (which act on qubits 0 and 3), so the total error at the end is a tensor product E03 ⊗ E21 of independent two-qubit Paulis on the two Bell pairs.
Fidelity for a single Bell pair under Λp:
For a pure target |Φ+⟩, the output fidelity after Λp is

\[ F_{\mathrm{pair}}(p) = \langle \Phi^+|\Lambda_p(|\Phi^+\rangle\langle\Phi^+|)|\Phi^+\rangle = (1-p) + \frac{p}{15}\sum_{P\neq II} |\langle \Phi^+|P|\Phi^+\rangle|^2. \]

Only P ∈ {X⊗X, Y⊗Y, Z⊗Z} give nonzero contribution, and for each of these, |⟨Φ+|P|Φ+⟩|^2 = 1. Hence

\[ F_{\mathrm{pair}}(p) = (1-p) + \frac{3p}{15} = 1 - \frac{4}{5}p. \]

Total 4-qubit fidelity:
The final state is a tensor product of two Bell pairs, and the two depolarizing channels act independently on disjoint pairs. Therefore the total fidelity is the product of the pair fidelities:

\[ F_{\mathrm{phys}}(p) = \big[F_{\mathrm{pair}}(p)\big]^2 = \left(1 - \frac{4}{5}p\right)^2. \]

Mathematical Typesetting

Done above.

Conventions and Units

Dimensionless probabilities; standard Pauli and CNOT conventions; depolarizing channel as specified.

Final Answer: F(p) = (1 − 4p/5)^2 = 1 − (8/5)p + (16/25)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-by-Step Derivation

Ideal output of the circuit
As in Subproblem 1, the noiseless circuit prepares a product of Bell pairs on disjoint pairs (03) and (21):

\[ \]

Noise model and factorization
A two-qubit depolarizing channel acts after each CNOT, once on qubits (2,1) and once on qubits (0,3). The channels are independent and act on disjoint pairs, so the final error is a tensor product $E_{03}\otimes E_{21}$ of two-qubit Paulis.
One-pair behavior under the depolarizing channel
The 16 two-qubit Paulis form four cosets of the stabilizer of $|\Phi^+\rangle$ on a pair. Consequently, after the channel on a given pair:
- Probability the pair remains in $|\Phi^+\rangle$:

\[ p_{\Phi^+}=(1-p) + 3\frac{p}{15} = 1-\frac{4}{5}p. \]

- Probability the pair is mapped to each of the other three Bell states $|\Phi^-\rangle,|\Psi^+\rangle,|\Psi^-\rangle$:

\[ p_{\text{other}}=\frac{4}{15}p. \]

What the post-selection accepts
The code stabilizers are

\[ \]

Each Bell state on a pair is a simultaneous eigenstate of $XX$ and $ZZ$ with eigenvalues $(\pm1,\pm1)$. Writing the eigenvalue pair on pair $03$ as $(a_{03},b_{03})$ for $(ZZ_{03},XX_{03})$ and similarly $(a_{21},b_{21})$ on pair $21$, the post-selection on $S_X=+1,S_Z=+1$ enforces

\[ a_{03}a_{21}=+1,\quad b_{03}b_{21}=+1 \;\;\Longleftrightarrow\;\; (a_{03},b_{03})=(a_{21},b_{21}). \]

So both pairs must end up in the same Bell state (both Φ+, or both Φ−, or both Ψ+, or both Ψ−).

Hence the acceptance probability is

\[ \]

Mapping to the logical 2-qubit state
Using the given logicals, observe

\[ \]

Thus the logical Bell labels on $(A,B)$ are exactly the $(XX,ZZ)$ eigenvalues on pair $(1,2)$. Because post-selection enforces pair $(0,3)$ to match $(1,2)$, the accepted 4-qubit state is one of the four orthonormal logical Bell states on $(A,B)$:
- Φ+⊗Φ+ → logical |Φ+⟩AB (the target),
- Φ−⊗Φ− → logical |Φ−⟩AB,
- Ψ+⊗Ψ+ → logical |Ψ+⟩AB,
- Ψ−⊗Ψ− → logical |Ψ−⟩AB.

Therefore, the logical state fidelity conditioned on acceptance is the conditional probability that both pairs are Φ+:

\[ \]

Simplifications and small-p expansion
Closed form:

\[ F_{\text{log}}(p)=\frac{75-120p+48p^2}{75-120p+64p^2}. \]

Series (small p):

\[ F_{\text{log}}(p)=1-\frac{16}{75}p^2+O(p^3), \]

showing the leading logical infidelity is second order in p due to post-selection.

Mathematical Typesetting

All expressions provided in LaTeX above.

Conventions and Units

Dimensionless probabilities; ideal single-qubit gates and stabilizer measurements; two-qubit depolarizing noise as specified.

Final Answer: F_log(p) = [(1 − 4p/5)^2] / [(1 − 4p/5)^2 + 3 (4p/15)^2] = (75 − 120 p + 48 p^2)/(75 − 120 p + 64 p^2).

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 = (75 - 120*p + 48*p**2) / (75 - 120*p + 64*p**2)
    # ---------------------------------------------------------------

    return F_logical