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 (no noise)
Start from |0000⟩ on physical qubits 0,1,2,3.
The subcircuit H2 then CNOT21 prepares a Bell state on the pair (2,1): H on the control followed by CNOT maps |00⟩ → |Φ+⟩ = (|00⟩+|11⟩)/√2. Hence qubits (2,1) are in |Φ+⟩21. The subsequent gates act on other qubits, so this entanglement is unaffected. (en.wikipedia.org)
The subcircuit H0 then CNOT03 prepares a Bell state on the pair (0,3), giving |Φ+⟩03. (en.wikipedia.org)
Therefore, the ideal 4-qubit output state is the product of two Bell pairs:

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

Noise model applied
After each CNOTij, a 2-qubit depolarizing Pauli channel acts on that pair: with probability 1−p do nothing; with probability p choose uniformly one of the 15 non-identity 2-qubit Paulis P∈{IX,IY,IZ, … , ZZ} and apply P (probability p/15 for each). This is the standard “uniform Pauli” description (the 16 two-qubit Paulis are II plus the 15 non-identity strings). (qiskit.github.io, qiskit.qotlabs.org)
In our circuit, the two such channels act on disjoint pairs (2,1) and (0,3), and there are no subsequent gates touching those qubits after each error is applied; thus the final state is

\[ \rho_{\mathrm{out}}=\mathcal{E}_{21}\!\left(|\Phi^+\rangle\!\langle\Phi^+|_{21}\right)\;\otimes\;\mathcal{E}_{03}\!\left(|\Phi^+\rangle\!\langle\Phi^+|_{03}\right), \]

i.e., a tensor product of two independently–noisy Bell pairs.

Fidelity for a single noisy Bell pair
For a target pure state |ψ⟩, the fidelity with a state ρ is F=⟨ψ|ρ|ψ⟩ (we use the “squared” fidelity convention). For our tensor-product target |ψideal⟩ the total fidelity factorizes as the product of the two pairwise fidelities (see next bullet). (en.wikipedia.org)
Because ρout is a product and |ψideal⟩ is a product, the fidelity factorizes:

\[ \]

This follows from ⟨a⊗b| (A⊗B) |a⊗b⟩ = ⟨a|A|a⟩ ⟨b|B|b⟩.

Evaluate Fpair for one pair subject to the 2-qubit Pauli channel:

\[ F_{\text{pair}}=\sum_{P} \Pr(P)\,|\langle\Phi^+|P|\Phi^+\rangle|^2, \]

where P runs over {II}∪{15 non-identity Paulis}.
- The Bell state |Φ+⟩ is a simultaneous eigenstate of X⊗X and Z⊗Z with eigenvalue +1, and of Y⊗Y with eigenvalue −1:

  Hence these three two-qubit Paulis leave |Φ+⟩ invariant up to a global phase. All other non-identity two-qubit Paulis map |Φ+⟩ to one of the other three (orthogonal) Bell states, giving zero overlap. ([chem.libretexts.org](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_%28Rioux%29/08%3A_Quantum_Teleportation/8.53%3A_Bell_State_Exercises?utm_source=openai))
- Therefore,

\[ \]

Total physical-state fidelity
Using Fphys = Fpair^2,

\[ F_{\mathrm{phys}}(p)=\Big(1-\frac{4}{5}p\Big)^2. \]

2) Mathematical Identities Used

Bell-state creation by H on the control followed by CNOT: CNOT(H⊗I)|00⟩ = (|00⟩+|11⟩)/√2. (en.wikipedia.org)
The 2-qubit Pauli set has 16 elements (II and the 15 non-identity strings); “uniform Pauli” depolarizing distributes weight equally over those non-identity Paulis. (qiskit.qotlabs.org, qiskit.github.io)
Bell states are eigenstates of X⊗X, Z⊗Z (and Y⊗Y), so XX, YY, ZZ stabilize |Φ+⟩ up to a phase; the four Bell states form an orthonormal basis. (chem.libretexts.org, en.wikipedia.org)
Fidelity with a pure target equals F = ⟨ψ|ρ|ψ⟩. (en.wikipedia.org)

3) Conventions and Units

Fidelity is taken as F(|ψ⟩,ρ) = ⟨ψ|ρ|ψ⟩ (not the root-fidelity). (en.wikipedia.org)
Error rate p is the probability that a given CNOT is followed by a non-identity 2-qubit Pauli error; each of the 15 non-identity Paulis occurs with probability p/15. This channel acts once per CNOT, on that gate’s two qubits. (qiskit.github.io, qiskit.qotlabs.org)

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

Ideal output of the circuit (no noise)
Start from |0000⟩ on qubits (0,1,2,3).
The rightmost subcircuit H2 then CNOT21 prepares a Bell state on qubits (2,1): CNOT(H⊗I)|00⟩ = |Φ+⟩ = (|00⟩+|11⟩)/√2. Hence qubits (2,1) are in |Φ+⟩21. The later gates act only on (0,3). (en.wikipedia.org)
The subcircuit H0 then CNOT03 similarly prepares |Φ+⟩03 on (0,3). Therefore the ideal 4-qubit output state is

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

Noise model and its localization
After each CNOT, a two-qubit depolarizing Pauli channel acts on that gate’s two qubits: with probability 1−p nothing happens, and with probability p one of the 15 non-identity two-qubit Paulis is applied uniformly (probability p/15 each). The two noisy channels act on disjoint pairs (2,1) and (0,3). (qiskit.github.io, quantumchannelzoo.org)
How a two-qubit Pauli acts on a Bell state
The Bell states are simultaneous eigenstates of X⊗X and Z⊗Z; in particular,

\[ |\Phi^+\rangle \text{ has } (X\!\otimes\!X)=+1,\ (Z\!\otimes\!Z)=+1. \]

The only two-qubit Paulis that map |Φ+⟩ to itself (up to phase) are XX, YY, ZZ; all the other 12 map it to one of the other three Bell states {Φ−, Ψ+, Ψ−}. Consequently, for one noisy pair,

\[ \Pr(\Phi^+) = 1-\tfrac{12}{15}p,\qquad \Pr(\Phi^-)=\Pr(\Psi^+)=\Pr(\Psi^-)=\tfrac{4}{15}p. \]

The two pairs are independent. ([qubit.guide](https://qubit.guide/7.3-single-stabiliser-states?utm_source=openai))

Post-selection on the [[4,2,2]] code stabilizers
The code stabilizers are S = ⟨XXXX, ZZZZ⟩. On the product form, XXXX = (X0X3)(X2X1) and ZZZZ = (Z0Z3)(Z2Z1). Measuring XXXX (resp. ZZZZ) on the product of two Bell states returns the product of the XX (resp. ZZ) eigenvalues of the two pairs. We post-select runs with +1 for both stabilizers, which is equivalent to requiring that the two pairs have identical (XX, ZZ) eigenvalue pairs, i.e., they are the same Bell state. (ar5iv.labs.arxiv.org)
Therefore the acceptance probability is

\[ =(1-\tfrac{12}{15}p)^2+3\Big(\tfrac{4}{15}p\Big)^2 \]

Logical state produced upon acceptance and its fidelity
Conditioning on acceptance, the four-qubit state is stabilized by XXXX, ZZZZ and also by IXXI and IZZI with eigenvalues inherited from the (2,1)-pair. Thus the accepted state is exactly the logical Bell state whose (XX, ZZ) logical stabilizers match the pair’s eigenvalues: if both pairs are Φ+, we get the target logical |Φ+⟩AB; if both are Φ−, Ψ+, or Ψ−, we obtain the corresponding other logical Bell state, orthogonal to the target. Hence the logical-state fidelity equals the conditional probability that both pairs are Φ+:

\[ =\frac{\Pr(\Phi^+)^2}{P_{\rm acc}(p)} \]

Equivalently,

\[ \]

showing first-order error suppression under post-selection (distance-2 detection). ([qubit.guide](https://qubit.guide/7.3-single-stabiliser-states?utm_source=openai), [ar5iv.labs.arxiv.org](https://ar5iv.labs.arxiv.org/html/1907.11157v1?utm_source=openai))

2) Mathematical identities and checks used

Factorization of stabilizer measurements across the two Bell pairs: XXXX = (X0X3)(X2X1), ZZZZ = (Z0Z3)(Z2Z1). The eigenvalue measured is the product of pairwise eigenvalues. This matches the Bell-state eigenvalue table for XX and ZZ. (qubit.guide)
Two-qubit depolarizing channel chooses uniformly among the 15 non-identity two-qubit Pauli operators. (qiskit.github.io)

3) Conventions and Units

Fidelity with a pure target state |ψ⟩ is F = ⟨ψ|ρ|ψ⟩.
Error rate p is the probability that a given CNOT is followed by a non-identity 2-qubit Pauli error (each of the 15 non-identity Paulis occurs with probability p/15). (qiskit.github.io)

4) Final Answer:

\[ =\frac{(1-\frac{4}{5}p)^2}{1-\frac{8}{5}p+\frac{64}{75}p^2} \]

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