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, mathematical identities or numerical codes.
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. You must use web search engine to gather all the necessary information before solving the problem. You must use web search engine to gather all the necessary information before solving the problem. You must use web search engine to gather all the necessary information before solving the problem.

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

1) Step-by-Step Derivation

Ideal state prepared by the circuit
The circuit is applied right-to-left:
- H2, then CNOT21 creates a Bell pair on qubits (2,1): |Φ+>21 = (|00>+|11>)/√2.
- H0, then CNOT03 creates a Bell pair on qubits (0,3): |Φ+>03 = (|00>+|11>)/√2.
The noiseless 4-qubit output is therefore the product state

\[ |\psi_{\mathrm{id}}\rangle = |\Phi^+\rangle_{21}\otimes|\Phi^+\rangle_{03}. \]

Noise model per CNOT
After each CNOT, a two-qubit depolarizing channel acts on the two involved qubits:

\[ \mathcal{D}^{(2)}(\rho)=(1-p)\rho+\frac{p}{15}\sum_{P\in\mathcal{P}_2\setminus\{II\}} P\rho P, \]

where the sum runs over the 15 non-identity two-qubit Pauli operators. ([quantumcomputing.stackexchange.com](https://quantumcomputing.stackexchange.com/questions/25869/how-can-i-apply-2-qubit-pauli-operator-error-instead-of-cnot-operator-in-stim-co?utm_source=openai), [en.wikipedia.org](https://en.wikipedia.org/wiki/Quantum_depolarizing_channel?utm_source=openai))

Stabilizers of a Bell state
The Bell state |Φ+> is a +1 eigenstate of XX and ZZ; equivalently, it is stabilized by ⟨XX, ZZ⟩ (hence also by −YY, so Y⊗Y maps |Φ+> to −|Φ+>, which is the same physical state up to global phase). Thus, among the 15 non-identity two-qubit Paulis, exactly three—XX, YY, ZZ—leave |Φ+> unchanged up to a phase; all others map it to an orthogonal Bell state. (qubit.guide, docs.quantum.ibm.com)
Fidelity for one Bell pair going through one two-qubit depolarizing channel
For a pure target |Φ+>, the state fidelity after the channel is

\[ F_{\text{pair}}(p)=\langle\Phi^+|\mathcal{D}^{(2)}(|\Phi^+\rangle\!\langle\Phi^+|)|\Phi^+\rangle. \]

Using the channel form and the stabilizer property,

\[ \]

Factorization for the 4-qubit product state
The two noise channels act on disjoint pairs: CNOT21’s noise acts only on qubits (2,1) and CNOT03’s noise only on (0,3). Since the ideal state is a tensor product across these pairs,

\[ \]

2) Mathematical Typesetting

See the display equations above.

3) Conventions and Units

Qubits are labeled 0,1,2,3.
Only the two-qubit gates (CNOT21, CNOT03) are followed by two-qubit depolarizing errors; H gates are ideal.
Physical state fidelity is F(|ψid>, ρ) = ⟨ψid|ρ|ψid⟩.
No units are involved; p is a dimensionless probability per two-qubit gate.

4) Final Answer: Fphys(p) = (1 − 4p/5)^2.

User #3

The correct answer is: $ F_{\rm{physical}}=(1-\frac{12}{15}p)^2 $

Now let’s solve the next Checkpoint.

Checkpoint 2

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:

\[ (CNOT_{03}) (H_0) (CNOT_{21}) (H_2). \]

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

Assistant #4

1) Step-by-Step Derivation

Noiseless target and where noise acts
The circuit is applied right-to-left: H2 → CNOT21 → H0 → CNOT03.
In the ideal (noise-free) case, H2 and CNOT21 prepare a Bell pair on qubits (2,1), and H0 and CNOT03 prepare a Bell pair on qubits (0,3):

\[ |\psi_{\rm id}\rangle=|\Phi^+\rangle_{21}\otimes|\Phi^+\rangle_{03}. \]

Only the two CNOTs are noisy. After each CNOT, a two-qubit depolarizing channel acts on that gate’s two qubits:

\[ \mathcal D^{(2)}(\rho)=(1-p)\rho+\frac{p}{15}\!\!\sum_{P\in\mathcal P_2\setminus\{II\}}\!P\rho P, \]

i.e., one of the 15 non-identity two-qubit Paulis is applied uniformly at random with probability p/15. ([en.wikipedia.org](https://en.wikipedia.org/wiki/Quantum_depolarizing_channel?utm_source=openai), [quantumchannelzoo.org](https://quantumchannelzoo.org/channel/depolarizing?utm_source=openai))

Bell-pair stabilizers and how two-qubit Paulis move Bell states
The Bell state $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$ is stabilized by $XX$ and $ZZ$; its stabilizer group is $\{II,XX,ZZ,-YY\}$. Thus $XX,YY,ZZ$ leave $|\Phi^+\rangle$ unchanged up to a phase, while other two-qubit Paulis map it to one of the other three Bell states $|\Phi^-\rangle,|\Psi^+\rangle,|\Psi^-\rangle$. (medium.com)
Equivalently, each two-qubit Pauli $P$ has a “parity” with respect to the pair-stabilizers $(XX,ZZ)$: • It commutes with a stabilizer if it flips that stabilizer’s eigenvalue an even number of times and anticommutes otherwise. Errors with parities (commute, commute), (anticommute, commute), (commute, anticommute), (anticommute, anticommute) map $|\Phi^+\rangle$ to $|\Phi^+\rangle,|\Phi^-\rangle,|\Psi^+\rangle,|\Psi^-\rangle$, respectively. Counting the 15 non-identity two-qubit Paulis gives 3, 4, 4, 4 in these four classes, respectively. Hence, for one Bell pair after one two-qubit depolarizing channel:

\[ \]

What “post-select on all detectable errors” means here
The [[4,2,2]] code has stabilizers $S=\langle XXXX,\;ZZZZ\rangle$; measuring them and keeping only the $+1,+1$ outcomes projects onto the codespace. A Pauli error is discarded if it anticommutes with at least one stabilizer and kept if it commutes with both. (ar5iv.labs.arxiv.org, en.wikipedia.org)
Note $XXXX=XX_{21}\otimes XX_{03}$ and $ZZZZ=ZZ_{21}\otimes ZZ_{03}$. The four-qubit syndrome is $+1$ precisely when the two pairs (21) and (03) have the same $(XX,ZZ)$ eigenvalues. Equivalently, we accept if and only if both pairs are in the same Bell state (Φ+, Φ−, Ψ+, or Ψ−). This is because the stabilizer eigenvalues on four qubits are products of the pair eigenvalues. (ar5iv.labs.arxiv.org)
Acceptance probability and accepted-state composition
Let $a:=\Pr[\Phi^+ \text{ on a pair}]=1-\frac{4p}{5}=1-\frac{12}{15}p$ and $b:=\Pr[\text{any one of }\Phi^-,\Psi^+,\Psi^-]=\frac{4}{15}p$ (for each of the three). The two pairs are independent channels on disjoint qubits.
Acceptance requires the same Bell label on both pairs, so

\[ P_{\rm acc}=a^2+3b^2. \]

Logical state after post-selection and its fidelity
The codespace of the [[4,2,2]] code can be viewed as the span of the four “identical-Bell” states $\{|\Phi^+\rangle_{21}\!\otimes\!|\Phi^+\rangle_{03},|\Phi^-\rangle_{21}\!\otimes\!|\Phi^-\rangle_{03},|\Psi^+\rangle_{21}\!\otimes\!|\Psi^+\rangle_{03},|\Psi^-\rangle_{21}\!\otimes\!|\Psi^-\rangle_{03}\}$, and the given logical operators (e.g., $\bar X_A=XIXI$, $\bar X_B=XXII$, etc.) act within this space as single-qubit Paulis on the two logical qubits. In particular, the ideal logical target is the logical Bell state corresponding to the physical state $|\Phi^+\rangle_{21}\otimes|\Phi^+\rangle_{03}$. (ar5iv.labs.arxiv.org)
Conditioned on acceptance, the logical two-qubit state is the Bell state whose label both pairs share. Therefore, the logical fidelity with the intended logical $|\Phi^+\rangle$ equals the conditional probability that both pairs are in $\Phi^+$:

\[ \]

Small-$p$ expansion (useful check): since post-selection removes all first-order contributions, the infidelity starts at $O(p^2)$:

\[ F_{\rm logical}(p)=1-\frac{16}{75}p^2+O(p^3). \]

2) Mathematical Typesetting

All expressions above are in LaTeX.

3) Conventions and Units

Qubits are labeled 0,1,2,3; logical qubits A,B.
Stabilizers $XXXX$ and $ZZZZ$; logicals as in the problem statement (consistent with literature). (ar5iv.labs.arxiv.org)
Only two-qubit gates are noisy, with a two-qubit depolarizing channel of rate $p$ applied after each CNOT. (en.wikipedia.org)
Fidelity is $F(|\psi\rangle,\rho)=\langle\psi|\rho|\psi\rangle$; we post-select on the $+1,+1$ stabilizer outcomes (accept only undetected errors). (en.wikipedia.org)

4) Final Answer:

\[ =\frac{\big(1-\frac{4}{5}p\big)^{2}}{\big(1-\frac{4}{5}p\big)^{2}+\frac{48}{225}p^{2}} \]

User #5

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

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

    Outputs
    ----------
    F_logical: sympy.Expr, the logical state fidelity as a function of $p$
    """

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

    return F_logical

Assistant #6

import sympy as sp

p = sp.symbols('p')

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

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

    Outputs
    ----------
    F_logical: sympy.Expr, the logical state fidelity as a function of $p$
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_logical = ((1 - sp.Rational(12, 15)*p)**2) / ((1 - sp.Rational(12, 15)*p)**2 + 3*(sp.Rational(4, 15)*p)**2)
    # ---------------------------------------------------------------

    return F_logical