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.

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 $p$.

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

Step-by-Step Derivation

Ideal state prepared by the circuit
Start from |0000⟩.
Apply H2 then CNOT21: this prepares a Bell pair on qubits (2,1),

\[ |\Phi^+\rangle_{21}=\frac{|00\rangle_{21}+|11\rangle_{21}}{\sqrt{2}}. \]

Apply H0 then CNOT03: this prepares a Bell pair on qubits (0,3),

\[ |\Phi^+\rangle_{03}=\frac{|00\rangle_{03}+|11\rangle_{03}}{\sqrt{2}}. \]

Since the two halves of the circuit act on disjoint qubit pairs, the ideal final 4-qubit state is

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

Noise model and independence across pairs
After each CNOT, a two-qubit depolarizing channel acts on that CNOT’s qubit pair:

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

where $\mathcal{P}_2=\{I,X,Y,Z\}^{\otimes 2}$.

The two channels act on disjoint pairs (first on qubits 2&1, second on 0&3) and are independent. Because no subsequent gates couple these pairs, the final state factorizes:

\[ \rho_{\text{out}}=\mathcal{D}_p^{(21)}\!\big(|\Phi^+\rangle\langle\Phi^+|_{21}\big)\;\otimes\;\mathcal{D}_p^{(03)}\!\big(|\Phi^+\rangle\langle\Phi^+|_{03}\big). \]

Fidelity for one Bell pair under two-qubit depolarizing noise
For $|\Phi^+\rangle$, the (density-matrix) stabilizer up to sign is

\[ \{II,\;XX,\;YY,\;ZZ\}. \]

For these Paulis $P$, one has $P|\Phi^+\rangle=\pm|\Phi^+\rangle$, which leaves the projector $|\Phi^+\rangle\langle\Phi^+|$ unchanged.

Under $\mathcal{D}_p$, the event “no error” occurs with probability $(1-p)$ and yields fidelity 1. If an error occurs, it is uniformly one of the 15 non-identity two-qubit Paulis. Exactly 3 of these (XX, YY, ZZ) preserve $|\Phi^+\rangle\langle\Phi^+|$.
Hence the single-pair fidelity is

\[ F_{\text{pair}}(p)=(1-p)+\frac{3}{15}p=1-\frac{4}{5}p. \]

Total 4-qubit fidelity
Since the two subsystems are independent and the ideal state is a product, the total fidelity factorizes:

\[ F_{\text{total}}(p)=F_{\text{pair}}(p)\times F_{\text{pair}}(p)=\left(1-\frac{4}{5}p\right)^2. \]

Mathematical Typesetting - Already included above.
Conventions and Units - Two-qubit depolarizing channel after each CNOT with error probability $p$ as specified.
Final Answer:

\[ F_{\text{phys}}(p)=\left(1-\frac{4}{5}p\right)^2=1-\frac{8}{5}p+\frac{16}{25}p^2. \]

User #3

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

Step-by-Step Derivation

Ideal encoded target state
The circuit prepares two independent Bell pairs on disjoint qubit pairs:
- On qubits (2,1): after $(H_2)$ and $(CNOT_{21})$ we have $|\Phi^+\rangle_{21}=(|00\rangle+|11\rangle)/\sqrt{2}$.
- On qubits (0,3): after $(H_0)$ and $(CNOT_{03})$ we have $|\Phi^+\rangle_{03}$.
The ideal 4-qubit state is

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

This state lies in the codespace (stabilized by $XXXX$ and $ZZZZ$) and encodes the logical Bell (GHZ) state $(|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}$.
Noise model on each CNOT and its effect on one Bell pair
After each CNOT, a 2-qubit depolarizing channel acts on that pair:

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

For a single Bell pair $|\Phi^+\rangle$, classify the 15 non-identity two-qubit Paulis by their commutation with the pair stabilizers $XX$ and $ZZ$:
- Commute with both $XX$ and $ZZ$ (leave $|\Phi^+\rangle$ invariant up to phase): exactly 3 elements $\{XX,YY,ZZ\}$.
- Anticommute with exactly one of them: 4 elements each, mapping $|\Phi^+\rangle$ to one of the other three Bell states:
- Flip $XX$ only: $|\Phi^+\rangle\to|\Phi^-\rangle$.
- Flip $ZZ$ only: $|\Phi^+\rangle\to|\Psi^+\rangle$.
- Flip both: $|\Phi^+\rangle\to|\Psi^-\rangle$.
Thus for one pair, the probabilities for the “Bell-type label” $c\in\{00,10,01,11\}$ (whether the $XX$ and $ZZ$ eigenvalues flip) are

\[ \]

Post-selection on the [[4,2,2]] code
The code checks are $M_X=XXXX=(XX)_{21}\otimes(XX)_{03}$ and $M_Z=ZZZZ=(ZZ)_{21}\otimes(ZZ)_{03}$.
A run is accepted iff the products of the pair eigenvalues are $+1$ for both $M_X$ and $M_Z$. Equivalently, the two pairs must have the same Bell-type label:

\[ \text{accept} \iff c_{21}=c_{03}\in\{00,10,01,11\}. \]

Therefore the acceptance probability is

\[ \]

Logical state after post-selection and its fidelity
Conditioned on acceptance, the two pairs end up in the same Bell state:
- If $c=00$, both are $|\Phi^+\rangle$, which is exactly the target logical Bell state.
- If $c\in\{10,01,11\}$, both pairs are in the same other Bell state ($|\Phi^-\rangle$, $|\Psi^+\rangle$, or $|\Psi^-\rangle$), which corresponds to a nontrivial logical Pauli on $A$ or $B$ and is orthogonal to the target logical state.
Hence the logical fidelity conditioned on acceptance is the conditional probability of $c=00$:

\[ \]

Simplify:

\[ \]

For small $p$: $F_{\text{log}}(p)=1-\frac{16}{75}p^2+O(p^3)$, showing first-order errors are entirely post-selected away.

Mathematical Typesetting - All mathematical expressions are provided in LaTeX.
Conventions and Units - Two-qubit depolarizing channel with error probability $p$ acts once after each of the two CNOTs. - Post-selection keeps only runs with $+1$ outcomes for both stabilizers $XXXX$ and $ZZZZ$ (perfect stabilizer measurement assumed).
Final Answer:

\[ \]

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 = (75 - 120*p + 48*p**2) / (75 - 120*p + 64*p**2)
    # ---------------------------------------------------------------

    return F_logical