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 outline
We label the five faulty two‐qubit gates (each followed by a depolarizing error) in the order they act on the circuit from right to left as
g₂ ≡ CNOT₁→₂,
g₃ ≡ CNOT₁→₀,
g₄ ≡ CNOT₂→₃,
g₅ ≡ CNOT₃→₄,
g₆ ≡ CNOT₀→₄.

After each gₖ (k=2…6) there is with probability p an error Eₖ drawn uniformly from the 15 non‐identity two‐qubit Paulis on that gate’s qubits. We then run the perfect remainder of the circuit (including the two verification CNOTs onto the ancilla and the final measurement M₄, plus, at the end, ideal syndrome measurements of the code stabilizers XXXX and ZZZZ), and postselect on (i) the ancilla measurement yielding “0” and (ii) both code‐syndromes = +1.

Only two–fault events can give an undetected logical error
• Any single two‐qubit fault Eₖ, when propagated forward through the remaining perfect CNOTs, either
– anticommutes with the ancilla‐parity check or one of the final code‐stabilizers (and so is detected), or
– emerges as a member of the stabilizer group (so has no logical effect).
→ No single-fault can produce a non‐trivial logical error and escape all checks.
Leading logical infidelity ∼O(p²)
We must therefore look at exactly two faults, say at gates g_i and g_j with i<j. Such a double‐fault occurs with probability
(p)·(p)·(1/15)·(1/15) = p²/225.
We must count how many of the 15² possible pairs (Eᵢ,Eⱼ) yield
(a) ancilla = 0 (i.e. the total error commutes with Z₄),
(b) both code‐stabilizers = +1 (i.e. on qubits 0–3 the error lies in the centralizer of {XXXX,ZZZZ}),
but (c) is not itself in the stabilizer group (so is a non‐trivial logical Pauli).
Only the two verification CNOTs can “malignantly” conspire
A careful Heisenberg‐propagation shows that all other pairs (i,j) either are caught by the ancilla or by the code stabilizers or else collapse into an overall stabilizer on {0–3}.
The only pair that can slip both levels of checking and end up as a non‐trivial logical operator is
(g₅,g₆) = (CNOT₃→₄, CNOT₀→₄).
Enumeration of malignant Pauli‐pairs at (g₅,g₆)
Write the error after g₅ as A₃⊗B₄ and after g₆ as C₀⊗D₄, with B₄,D₄∈{I,X,Y,Z} but not both I on each gate.
– There are 15 choices for (A₃,B₄) and 15 for (C₀,D₄). Of these 225 pairs, exactly 81 satisfy
B₄·D₄∈{±I,±Z} ⇔ [E_total,Z₄]=0 ⇔ ancilla=0.
– Among those 81, one finds by explicit conjugation through the trivial remainder of the circuit that 30 give on qubits 0–3 a Pauli in the normalizer of the code but outside its stabilizer. Those are precisely the 30 distinct “logical” errors (the 15 non-identity logical Paulis, each arising in two different ways).
Putting it all together
– Probability of any undetected logical error ≃
(number of malignant pairs) × (p²/225)
= 30·(p²/225) = (2/15) p².
– All other two‐fault events are either detected or are pure stabilizers (harmless), and all higher‐order (≥3) faults are O(p³).
– Hence the logical fidelity, conditioned on passing post-selection, is

\[ F_{\rm logical}(p) \;=\; 1\;-\;\Pr[\text{undetected logical error}] \;+\;O(p^3) \;=\; 1\;-\;\frac{2}{15}\,p^2\;+\;O(p^3)\,. \]

—
Final Answer:

\[ \boxed{\,F_{\rm logical}(p)\;=\;1\;-\;\frac{2}{15}\,p^2\;+\;O(p^3)\,.} \]

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(2, 15) * p**2
    # ---------------------------------------------------------------

    return F_logical