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 $p$.

Challenge

Suppose that we prepare a logical two-qubit $|00\rangle_{AB}$ state in the [[4,2,2]] code. To do so, we introduce an ancilla qubit, qubit 4, and use the following state preparation circuit:

\[ M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1) \]

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 and $M$ is a single-qubit measurement. The ancilla is used to detect errors in the state preparation circuit and makes the circuit fault-tolerant. If the ancilla measurement is $|0\rangle$ ($|1\rangle$), the state preparation succeeds (fails).

What is the logical state fidelity of the final 2-qubit logical state at the end of the circuit as a function of two-qubit gate error rate $p$, assuming the state is post-selected on all detectable errors in the code and on the ancilla qubit measuring $|0\rangle$?

Assistant #2

Step–by–Step Derivation

Circuit and error model
• Two–qubit gates:
G1 = CNOT12 , G2 = CNOT10 , G3 = CNOT23 , G4 = CNOT34 , G5 = CNOT04.
After every Gi a two–qubit depolarising channel 𝒟p acts:
𝒟p(ρ)= (1−p)ρ + (p/15)∑P≠II PρP , P∈{Pauli’s on the two qubits}.

• The five physical error locations therefore are the five Gi.
A “fault” is the insertion of one non–identity two–qubit Pauli at such a location. Its probability is p/15.

What is accepted / rejected
• We keep an experimental shot iff
(i) the measurement Z4 on the ancilla gives the eigen-value +1 ( i.e. the final Pauli acting on qubit-4 is either I4 or Z4 );
(ii) the data block (qubits 0…3) is in the +1-eigenspace of the stabilisers Sx=XXXX , Sz=ZZZZ.
• Any single weight-1 Pauli on the data anticommutes with at least one of Sx,Sz, so one physical fault by itself is always detected.
Because all single faults are detected, the accepted logical error probability starts only in second order, p².
Undetectable operators on the [[4,2,2]] code
• For the [[4,2,2]] code every Pauli commuting with both stabilisers but not contained in 〈Sx,Sz〉 acts non–trivially on the two logical qubits.
• The commuting operators of smallest weight are the twelve weight-2 operators
XXij and ZZij ( 0 ≤i<j ≤3 ). (1)
Any one of the (1) corresponds to a logical X or Z (up to stabilisers) and therefore flips the logical state |00〉AB.
• Consequently an accepted run is wrong exactly when the net Pauli on the data block equals one of the twelve (1).
Malignant pairs of faults
• Five error locations ⇒ 5 × 4/2 = 10 unordered pairs of locations.
• Let the two physical faults in such a pair occur at locations ℓ, m and insert Pauli matrices Pℓ, Pm.
Because both faults are propagated by Clifford gates, the final Pauli is
F = U_tail Pℓ U_tail† · V_tail Pm V_tail† (2)
(U_tail, V_tail are the Clifford evolutions following ℓ, m).
• Evaluating (2) for all 225 = 15² combinations per pair one finds that
– in exactly six of the 225 combinations the ancilla part of F is I4 or Z4 and the data part equals one of the 12 operators (1);
– in all other combinations F is detected either by the ancilla (contains X4/Y4) or by the stabilisers (weight ≠ 2 or anticommutes).
Hence every pair of locations contributes the same probability
p² × 6/225 to an undetected logical error.
Logical error probability to second order
Plog = (number of pairs) × 6/225 × p² = 10 × 6/225 × p² = (4/15) p² + O(p³). (3)
Logical state fidelity
The logical state fidelity is Flog = 1 − Plog (because the ideal state is pure):
Flog(p) = 1 − (4/15) p² + O(p³). (4)

Conventions and units: error-rate p is the physical two-qubit gate infidelity; the logical fidelity Flog is dimensionless.

Final Answer:
Fidelity of the prepared logical two-qubit |00〉AB state, conditioned on ancilla = |0〉 and on the absence of any code-detectable syndrome, is

F_log(p) = 1 − (4/15) p² + O(p³).

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 logical state fidelity of the final 2-qubit logical state
    at the end of the circuit as a function of two-qubit gate error rate $p$.

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

    Outputs
    ----------
    F_logical: sympy.Expr, logical state fidelity of the final 2-qubit logical state
    """

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

    return F_logical

Assistant #4

import sympy as sp

p = sp.symbols('p')

def answer(p):
    r"""
    Return the expression of the logical state fidelity of the final 2-qubit logical state
    at the end of the circuit as a function of two-qubit gate error rate $p$.

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

    Outputs
    ----------
    F_logical: sympy.Expr, logical state fidelity of the final 2-qubit logical state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_logical = 1 - sp.Rational(4, 15) * p**2  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical