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

1. Step-by-Step Derivation

Ideal circuit action

• Start from the computational vacuum

\[ |0000\rangle_{0123}. \]

• Apply the gates in the (right-to-left) order given in the problem

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

• After the first two operations

\[ H_2 :\; |0000\rangle \;\longrightarrow\; \frac{|0\rangle_2+|1\rangle_2}{\sqrt2}\otimes|000\rangle_{013}, \]

\[ CNOT_{21} :\; \frac{|0\rangle_2|0\rangle_1+|1\rangle_2|1\rangle_1}{\sqrt2}\otimes|00\rangle_{03} = |\Phi^{+}\rangle_{21}\otimes|00\rangle_{03}, \]

where $ |\Phi^{+}\rangle\equiv (|00\rangle+|11\rangle)/\sqrt2 $.

• After the last two operations

\[ H_0:\; |00\rangle_{03}\;\longrightarrow\; \frac{|0\rangle_0+|1\rangle_0}{\sqrt2}\otimes|0\rangle_3 , \]

\[ CNOT_{03}:\; \frac{|0\rangle_0|0\rangle_3 + |1\rangle_0|1\rangle_3}{\sqrt2} = |\Phi^{+}\rangle_{03}. \]

• Hence the ideal final 4-qubit state

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

It is a stabilizer state whose independent generators can be chosen as

\[ \bigl\{\,Z_2Z_1,\;X_2X_1,\;Z_0Z_3,\;X_0X_3\,\bigr\}. \]

Error model

• After each CNOT gate a two-qubit depolarising channel acts on the same two qubits:

\[ \mathcal{D}_{ij}^{(p)}(\rho)= (1-p)\,\rho+\frac{p}{15}\sum_{P\neq I\otimes I}P_{ij}\,\rho\,P_{ij}. \]

• There are two such channels: – $\mathcal{D}_{21}^{(p)}$ immediately after $CNOT_{21}$
– $\mathcal{D}_{03}^{(p)}$ immediately after $CNOT_{03}$.

• Denote the Pauli error applied by the first channel by $E_{21}$
(acting only on qubits 2 and 1) and the second by $E_{03}$
(acting only on qubits 0 and 3).

• Because the gates following $E_{21}$ touch only qubits 0 and 3, the two error operators act on disjoint pairs and therefore commute. The final physical state is

\[ \rho(p)= \sum_{E_{21},E_{03}} p(E_{21})\,p(E_{03})\; E_{03}\,E_{21}\; |\psi_{\text{ideal}}\rangle\langle\psi_{\text{ideal}}|\; E_{21}\,E_{03}, \]

 where

\[ p(E)= \begin{cases} 1-p, & E=I\otimes I,\\[4pt] p/15,& \text{any of the 15 non-identity Paulis.} \end{cases} \]

When does a Pauli error leave the state invariant?

• For any stabilizer state $|\psi\rangle$, a Pauli operator $P$ satisfies
$\langle\psi|P|\psi\rangle=\pm1$ if $P$ is in the stabilizer group and $0$ otherwise, hence

\[ F=\langle\psi_{\text{ideal}}|\rho(p)|\psi_{\text{ideal}}\rangle =\sum_{E_{21},E_{03}} p(E_{21})\,p(E_{03}) \,\delta\bigl(E_{21}E_{03}\in\mathcal{S}\bigr). \]

• Because the stabilizer factors into the two Bell-pair stabilizers, $E_{21}E_{03}\in\mathcal{S}$ ⇔ $E_{21}\in\{I,X_2X_1,Z_2Z_1,Y_2Y_1\}$ and $E_{03}\in\{I,X_0X_3,Z_0Z_3,Y_0Y_3\}$.

• For one depolarising channel

\[ P_{\text{pair}}(E\in\text{local stabilizer}) =(1-p)+\frac{3p}{15}=1-\frac{4p}{5}. \]

Physical state fidelity

• The two error channels are independent, so the total fidelity is the product of the two probabilities:

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

2. Mathematical Typesetting

All equations above are typeset in LaTeX.

3. Conventions and Units

The error rate $p$ is dimensionless. Fidelity is dimensionless and lies in [0\le F\le1.]

4. Final Answer

Final Answer:

\[ \boxed{\,F(p)=\Bigl(1-\dfrac{4p}{5}\Bigr)^2\,} \]

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 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 $p$.

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

    Outputs
    ----------
    F_physical: sympy.Expr, the physical state fidelity of the final physical 4-qubit state
    """

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

    return F_physical

Assistant #4

import sympy as sp

p = sp.symbols('p')

def answer(p):
    r"""
    Return the expression of 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 $p$.

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

    Outputs
    ----------
    F_physical: sympy.Expr, the physical state fidelity of the final physical 4-qubit state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_physical = (1 - sp.Rational(4, 5) * p) ** 2  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_physical