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 $p$.

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

Step-by-Step Derivation

Noiseless target state produced by the circuit Apply the gates from right to left to |0000⟩:
H2 turns qubit 2 into |+⟩: |+⟩2|0⟩1|0⟩0|0⟩3.
CNOT21 (control 2, target 1) entangles (2,1) into a Bell pair |Φ+⟩21: |Φ+⟩21 = (|00⟩21 + |11⟩21)/√2.
H0 turns qubit 0 into |+⟩.
CNOT03 (control 0, target 3) entangles (0,3) into |Φ+⟩03.

Hence the ideal final state is

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

Stabilizer of the target state A Bell state |Φ+⟩ on a pair (a,b) is stabilized (up to phase) by {I⊗I, X_aX_b, Y_aY_b, Z_aZ_b}. Therefore

\[ \mathcal{S}_{03} = \langle X_0X_3,\;Z_0Z_3\rangle = \{II,\;X_0X_3,\;Y_0Y_3,\;Z_0Z_3\}, \]

\[ \mathcal{S}_{21} = \langle X_2X_1,\;Z_2Z_1\rangle = \{II,\;X_2X_1,\;Y_2Y_1,\;Z_2Z_1\}. \]

The total stabilizer (modulo overall phases) is the tensor product

\[ \mathcal{S} = \mathcal{S}_{21}\otimes\mathcal{S}_{03}. \]

Noise model and where it acts After each CNOT, a two-qubit depolarizing channel acts on the same two qubits:

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

where $\mathcal{P}_2\setminus\{II\}$ are the 15 non-identity 2-qubit Paulis on that pair.

In our circuit: - After CNOT21: noise $Q_{21}$ acts on qubits (2,1). - After CNOT03: noise $Q_{03}$ acts on qubits (0,3).

These two channels act on disjoint pairs and are independent, and the operations between them (H0 and CNOT03) do not affect qubits (2,1). Hence the final error operator is simply

\[ E_{\mathrm{tot}}=Q_{21}\otimes Q_{03}. \]

Fidelity for a stabilizer state under Pauli noise For a pure stabilizer state $|\psi_{\mathrm{id}}\rangle$, and a Pauli error $E$, the overlap satisfies

\[ 1,& E\in\mathcal{S}\\ 0,& E\notin\mathcal{S} \]

Therefore the physical state fidelity is the probability that the net Pauli error lies in the stabilizer:

\[ F=\Pr[E_{\mathrm{tot}}\in\mathcal{S}] = \Pr[Q_{21}\in\mathcal{S}_{21}]\;\Pr[Q_{03}\in\mathcal{S}_{03}], \]

where we used independence and the fact that $\mathcal{S}=\mathcal{S}_{21}\otimes\mathcal{S}_{03}$.

Counting “good” Pauli errors on each pair On any pair, the depolarizing channel outputs:
$II$ with probability $(1-p)$,
each of the 15 non-identity two-qubit Paulis with probability $p/15$.

Among the 15 non-identity Paulis, exactly 3 are in the Bell stabilizer: $XX,\ YY,\ ZZ$. Therefore

\[ \Pr[Q_{\text{pair}}\in\{II,XX,YY,ZZ\}] = (1-p) + 3\cdot\frac{p}{15} = 1-\frac{4p}{5}. \]

The two pairs are independent, so

\[ F(p) = \left(1-\frac{4p}{5}\right)^2. \]

Mathematical Typesetting - All equations above are typeset in LaTeX.
Conventions and Units - Standard qubit and Pauli operator conventions; no dimensional units apply.
Final Answer: F(p) = (1 - 4p/5)^2 = 1 - (8/5)p + (16/25)p^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

Step-by-Step Derivation

Ideal target state and its stabilizer The noiseless circuit prepares

\[ \]

A Bell state on a pair (a,b) is stabilized by {I, X_aX_b, Y_aY_b, Z_aZ_b}. Hence the 4-qubit target stabilizer is

\[ \mathcal{S}_{\mathrm{tar}}=\{S_{21}\otimes S_{03}\;|\;S_{21},S_{03}\in\{II,XX,YY,ZZ\}\}, \]

generated equivalently by the code stabilizers XXXX, ZZZZ together with the logical stabilizers X_AX_B = X_1X_2 and Z_AZ_B = Z_1Z_2.

Noise model and error structure Only the two CNOTs are noisy. After CNOT21 an error $E_{21}$ acts on qubits (2,1); after CNOT03 an error $E_{03}$ acts on (0,3). They are independent and uniformly distributed over the 15 non-identity Paulis with probability $p/15$ each; $II$ occurs with probability $1-p$ on each pair. The final error is

\[ E_{\text{tot}}=E_{21}\otimes E_{03}. \]

Post-selection (detectable error rejection) We post-select on measuring +1 for the code stabilizers XXXX and ZZZZ. A Pauli $E_{\text{tot}}$ is accepted iff it commutes with both XXXX and ZZZZ.

For a pair operator $E_{\text{pair}}$, define commutation parities with the pair stabilizers XX and ZZ:

\[ \]

Since XXXX = (X_2X_1)(X_0X_3) and ZZZZ = (Z_2Z_1)(Z_0Z_3), acceptance requires

\[ b_X(E_{21})=b_X(E_{03}),\qquad b_Z(E_{21})=b_Z(E_{03}). \]

The 16 two-qubit Paulis on a pair partition into four classes of size 4 by the pair (b_X,b_Z): - S: commute with both (b_X,b_Z)=(0,0): {II, XX, YY, ZZ}. - X-only: (1,0). - Z-only: (0,1). - Both: (1,1).

Under the depolarizing channel on a pair,

\[ \]

Therefore the acceptance probability is

\[ \]

Logical fidelity conditioned on acceptance For a pure stabilizer state and Pauli noise, the overlap squared is 1 iff the error lies in the state’s stabilizer, and 0 otherwise. Among accepted errors, the output equals the target iff $E_{21}\in$ S and $E_{03}\in$ S, i.e., $E_{\text{tot}}\in\mathcal{S}_{\mathrm{tar}}$. Thus the conditional logical fidelity is

\[ =\frac{\Pr[E_{21}\in \text{S}]\Pr[E_{03}\in \text{S}]} {P_{\mathrm{acc}}(p)} =\frac{\left(1-\frac{12}{15}p\right)^2} \]

Expanding numerator and denominator:

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

showing first-order error contributions are fully detected and removed by post-selection.

Mathematical Typesetting - All expressions are provided in LaTeX.
Conventions and Units - Standard qubit-Pauli conventions; no physical units.
Final Answer: $F_{\mathrm{logical}}(p)=\dfrac{\left(1-\frac{12}{15}p\right)^2}{\left(1-\frac{12}{15}p\right)^2+3\left(\frac{4}{15}p\right)^2}=\dfrac{1-\frac{8}{5}p+\frac{16}{25}p^2}{1-\frac{8}{5}p+\frac{64}{75}p^2}\,.$

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 ------------------
    s = 1 - sp.Rational(12, 15) * p
    a = sp.Rational(4, 15) * p
    F_logical = s**2 / (s**2 + 3 * a**2)  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical