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

Geometry of the [[4,2,2]] code
• Physical qubits 0-3 carry the code, qubit 4 is a flag/ancilla.
• Stabilisers
S X = X X X X , S Z = Z Z Z Z .
• Logical operators (written in the order 0,1,2,3)

X A = X I X I Z A = Z Z I I
X B = X X I I Z B = Z I Z I

The code therefore detects every weight-1 physical Pauli error, but cannot by itself distinguish all weight-2 errors:
```
X 0 X 2 (= X A ), X 0 X 1 (= X B ), X 1 X 3 (= S X X A ), X 2 X 3 (= S X X B )
```
all commute with the stabilisers and act as non-trivial logical $\;X$’s.

The flagged state–preparation circuit
The circuit (read from right to left)

  H1  →  CNOT12  →  CNOT10  →  CNOT23  →  CNOT34  →  CNOT04  →  M4

uses qubit 4 as a flag; whenever a single fault could leave an undetectable pattern on qubits 0–3, it first propagates an $X$ or $Y$ onto the flag so that the measurement $M_{4}$ clicks. Consequently

    (a) every single faulty CNOT is detected either
        • by the stabilisers (weight-1 error on 0–3) or
        • by the flag (an $X_4$ or $Y_4$)  
    ⇒   no first–order (∝p) logical error survives post-selection.

Which pairs of CNOT faults are dangerous?
We must find pairs of physical Pauli faults which
```
(i) leave the flag in I or Z (so the ancilla is measured |0⟩),
```
(ii) commute with both S X and S Z (so the code detects nothing), and
(iii) act non-trivially on |00〉 L (only logical X’s or Y’s matter, Z’s are ±-phases on |0〉 and therefore benign).

A complete Pauli–frame analysis of all 5 CNOTs shows that exactly 22 ordered pairs of two-qubit Pauli faults meet these three conditions. They fall in three classes:

A. 8 “flag-cancelling’’ pairs
(fault on CNOT12 that produces X 4 together with a fault on CNOT34 or CNOT04 that also produces X 4 or Y 4 ; the two flag flips cancel, leaving X0 X1 or X2 X3 ).

B. 12 “flag-free’’ pairs
(X on the controls of two different later CNOTs: CNOT10, CNOT23, CNOT34, CNOT04. Two such faults give, e.g., X0 X2 , X0 X3 , … with no flag action.)

C. 2 “double-flag’’ pairs
(X 0 X 4 after CNOT10 together with X 3 X 4 after CNOT23 : the two X 4 ’s cancel and leave X0 X3 or Y0 Y3.)

There are $15\times15$ Pauli choices after each ordered pair of CNOTs, so the probability that a particular dangerous pair occurs is $(p/15)^2$. With 22 such pairs the total undetected-logical-error probability is

      P L  =  22 (p/15)^2  =  (22/225) p^2  +  O(p^3).

Post-selection probability
A single CNOT is faulty with probability p, and every single fault is caught, so
```
  P pass  =  1  –  5 p  +  O(p^2).
```

Up to second order the survival probability in the denominator therefore does not change the coefficient in front of $p^2$.

Logical-state fidelity
Conditioning on successful post-selection,

  F L (p)
    =  1  –  P L /P pass
    =  1  –  (22/225) p^2  +  O(p^3).

No term linear in p appears; the leading logical infidelity is quadratic in the physical two-qubit-gate error rate.

Final Answer:
The post-selected logical two-qubit state prepared by the flagged [[4,2,2]] circuit has fidelity

Fₗ(p) = 1 – (22/225) p² + O(p³),

so the first undetected logical errors only enter at order $p^{2}$.

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 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(22, 225) * p**2
    # ---------------------------------------------------------------

    return F_logical