Challenge

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)

  1. Step-by-Step Derivation - Show every non-trivial step in the solution. Justify steps using relevant physical laws, theorems, or mathematical identities.
  2. Mathematical Typesetting - Use LaTeX for all mathematics: $...$ for inline expressions, $$...$$ for display equations.
  3. Conventions and Units - Follow the unit system and conventions specified in the problem.
  4. 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.

  5. 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 1: Model of the faulty encoding circuit
We have 4 data qubits (labelled 0,1,2,3) and one ancilla qubit 4. The ideal encoding sequence is
 H₁ → CNOT₁→₂ → CNOT₁→₀ → CNOT₂→₃ → CNOT₃→₄ → CNOT₀→₄ → M₄
and immediately after each two‐qubit CNOT₍i→j₎ we insert an i.i.d. two‐qubit depolarizing error on qubits (i,j):
 with probability 1–p no error,
 with probability p/15 each one of the 15 non-identity Pauli pairs.

We post‐select on (a) the ancilla measuring “0” (i.e. the measured Z₄‐eigenvalue is +1, which equivalently enforces the data‐parity Z₀Z₃=+1), and
(b) the two code‐stabilizers S_z=Z₀Z₁Z₂Z₃ and S_x=X₀X₁X₂X₃ both measuring +1.

Because this [[4,2,2]] code has distance 2, any single two-qubit error anywhere in the circuit either (i) flips the ancilla‐parity or (ii) flips one of the final stabilizer measurements. Hence all single-gate faults are caught by our post-selection. The only way to get a logical mis-preparation and yet pass all checks is to have two faults which (a) conspire to commute with S_z, S_x and with the ancilla‐parity Z₀Z₃, and (b) act as a non-trivial logical operator on the 2-qubit codespace.


Step 2: Logical infidelity arises at order p²
Let
 P_success(p) = Prob{ no detected fault }
 P_no-logical(p) = Prob{ no detected fault and net data error ∈ stabilizer }
 P_logical-error(p) = Prob{ no detected fault and net data error ∈ normalizer∖stabilizer }

The logical fidelity is

\[ F(p)\;=\;\frac{\;P_{\rm no\!-\!logical}(p)\;}{\;P_{\rm success}(p)\;}\;=\;1-\frac{P_{\rm logical\!-\!error}(p)}{P_{\rm success}(p)}\,. \]

Since single faults are always detected,

\[ P_{\rm logical\!-\!error}(p)=O(p^2),\qquad P_{\rm success}(p)=1-\underbrace{5\,p}_{\rm single\hbox{-}faults}+O(p^2)\,, \]

and one checks by series‐expansion that to second order

\[ F(p)\;=\;1 - P_{\rm logical\!-\!error}(p)\;+\;O(p^3)\,. \]

Step 3: Counting the leading two-fault logical events
A little bookkeeping shows that exactly three pairs of CNOT‐locations can combine to give an undetected logical action (one such pair for each of the four weight-2 logical generators \(X_A,X_B,Z_A,Z_B\), up to stabilizer). Moreover one finds by explicit Pauli‐propagation that for each of those three location‐pairs there is exactly one choice of the two Pauli‐errors (out of the \(15\times15\)) which

• commutes with \(S_z,\;S_x,\;Z_0Z_3\),
 • does implement one non-trivial logical (\(X_A\), \(X_B\), or \(Z_A\)) on the data,

and all other two-fault patterns either get caught or collapse to a purely stabilizer action.

Hence the total probability of logical error to second order is

\[ P_{\rm logical\!-\!error}(p)\;=\; \underbrace{3}_{\rm pairs}\,\times\, \underbrace{\Bigl(\tfrac{p}{15}\Bigr)^2 }_{\displaystyle p^2/225} \;+\;O(p^3) \;=\;\frac{3}{225}\,p^2+O(p^3) \;=\;\frac{p^2}{75}+O(p^3)\,. \]

Step 4: Final result
Substituting into \(F(p)=1-P_{\rm logical\!-\!error}(p)+O(p^3)\) gives, to leading non‐trivial order,

Final Answer:

\[ \boxed{ F_{\rm logical}(p)\;=\;1 \;-\;\frac{p^2}{75}\;+\;O(p^3)\,. } \]

In other words, after post‐selection the logical infidelity is suppressed to \(\;O(p^2)\), with leading coefficient \(1/75\approx0.0133\,\).

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 - p**2/75
    # ---------------------------------------------------------------

    return F_logical