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-by-Step Derivation

Circuit and error model
The circuit is (in time order right→left): apply H2, then CNOT21, then H0, then CNOT03.
Only the two CNOTs are noisy. After each CNOTij a 2-qubit depolarizing channel acts on qubits i,j:

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

where the 15 non-identity two-qubit Paulis are equiprobable.

Ideal (noiseless) output state and its stabilizer
H2 creates |+⟩ on qubit 2; CNOT21 then prepares a Bell pair |Φ+⟩ on qubits (2,1).
H0 creates |+⟩ on qubit 0; CNOT03 then prepares a Bell pair |Φ+⟩ on qubits (0,3).
Hence the ideal 4-qubit output is

\[ \]

The stabilizer of a Bell pair |Φ+⟩ij is generated by X_iX_j and Z_iZ_j. Therefore the ideal 4-qubit state is the unique stabilizer state with generators

\[ X_1X_2,\; Z_1Z_2,\; X_0X_3,\; Z_0Z_3. \]

Equivalently, this is the logical Bell state (|00⟩AB+|11⟩AB)/√2, since
X_A X_B = IXXI and Z_A Z_B = IZZI together with the code stabilizers XXXX and ZZZZ generate the same group.

Location and propagation of errors
The error after CNOT21 acts only on qubits (2,1) and commutes with the later gates H0 and CNOT03 (which act on disjoint qubits). Thus it reaches the end unchanged as some P21.
The error after CNOT03 acts only on qubits (0,3) and is the last operation, so it is Q03 at the end.
Therefore the final state is a classical mixture over Pauli errors of the form

\[ E = Q_{03}\otimes P_{21},\qquad P_{21},Q_{03}\in\mathcal{P}_2. \]

Fidelity reduction to stabilizer membership
With Pauli errors on a pure stabilizer state, the output density matrix is

\[ \rho_{\text{out}}=\sum_{P,Q} \Pr(P)\Pr(Q)\,(Q\otimes P)\,|\psi\rangle\langle\psi|\,(Q\otimes P), \]

and the state fidelity is

\[ F=\langle\psi|\rho_{\text{out}}|\psi\rangle=\sum_{P,Q}\Pr(P)\Pr(Q)\,|\langle\psi|(Q\otimes P)|\psi\rangle|^2. \]

For a stabilizer state, a Pauli E has overlap 1 iff E is in its stabilizer group S, and 0 otherwise. Because S factorizes here as

\[ S = \{II,\,X_0X_3,\,Z_0Z_3,\,Y_0Y_3\}\ \otimes\ \{II,\,X_1X_2,\,Z_1Z_2,\,Y_1Y_2\}, \]

we have

\[ F = \Pr(Q_{03}\in S_{03})\cdot \Pr(P_{21}\in S_{12}). \]

Probability a 2-qubit depolarizing error lands in the pairwise stabilizer
For a given pair (i,j), the stabilizer of |Φ+⟩ij contains exactly the 4 Paulis {II, XX, ZZ, YY} on that pair.
Under the depolarizing channel:
- Identity II occurs with probability 1−p.
- Among the 15 non-identity Paulis, exactly 3 belong to the stabilizer (XX, YY, ZZ). Hence the additional probability is (3/15)p = (1/5)p.
Therefore for either pair,

\[ \Pr(\text{pairwise error} \in \{II,XX,YY,ZZ\}) = (1-p) + \frac{p}{5} = 1-\frac{4}{5}p. \]

Independence across the two CNOTs yields

\[ F(p)=\Big(1-\frac{4}{5}p\Big)^2. \]

Mathematical Typesetting - Provided inline and display equations above.
Conventions and Units - Dimensionless error rate p. No units involved.
Final Answer: F(p) = (1 - 4p/5)^2 = 1 - (8/5)p + (16/25)p^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