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 $p$.

Challenge

Suppose that we prepare a logical two-qubit $|00\rangle_{AB}$ state in the [[4,2,2]] code. To do so, we introduce an ancilla qubit, qubit 4, and use the following state preparation circuit:

\[ M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1) \]

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 and $M$ is a single-qubit measurement. The ancilla is used to detect errors in the state preparation circuit and makes the circuit fault-tolerant. If the ancilla measurement is $|0\rangle$ ($|1\rangle$), the state preparation succeeds (fails).

What is the logical state fidelity of the final 2-qubit logical state at the end of the circuit as a function of two-qubit gate error rate $p$, assuming the state is post-selected on all detectable errors in the code and on the ancilla qubit measuring $|0\rangle$?

Assistant #2

This problem asks for the logical state fidelity of a prepared logical $|00\rangle_{AB}$ state in the [[4,2,2]] code, considering a specific circuit and error model. The fidelity is calculated to first order in the physical two-qubit gate error rate $p$.

Step-by-Step Derivation

1. Ideal State Preparation

First, let’s analyze the circuit in the absence of errors. The initial state is $|00000\rangle = |0\rangle_0|0\rangle_1|0\rangle_2|0\rangle_3|0\rangle_4$. The circuit operations are applied from right to left as given in the problem statement: $M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1)$.

Initial state: $|\psi_0\rangle = |00000\rangle$.
After $H_1$: $|\psi_1\rangle = H_1 |\psi_0\rangle = |0\rangle_0 (\frac{|0\rangle_1+|1\rangle_1}{\sqrt{2}}) |0\rangle_2 |0\rangle_3 |0\rangle_4 = \frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)$.
After $CNOT_{12}$: $|\psi_2\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)$.
After $CNOT_{10}$: $|\psi_3\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11100\rangle)$.
After $CNOT_{23}$: $|\psi_4\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$.
After $CNOT_{34}$: The state has two components. For $|00000\rangle$, $d_3=0$, so the ancilla is unchanged. For $|11110\rangle$, $d_3=1$, so the ancilla is flipped. $|\psi_5\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)$.
After $CNOT_{04}$: For $|00000\rangle$, $d_0=0$, ancilla is unchanged. For $|11111\rangle$, $d_0=1$, ancilla is flipped from $|1\rangle$ back to $|0\rangle$. $|\psi_6\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) \otimes |0\rangle_4$.

The state of the four data qubits (0-3) is $|\psi_L\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)$. This state is an eigenstate of the stabilizers $S_1=XXXX$ and $S_2=ZZZZ$ with eigenvalue +1, so it is a valid codeword. Let’s assume this is the logical $|00\rangle_L$ state. The ancilla qubit is in the state $|0\rangle_4$.

Measurement $M_4$: The ancilla qubit (4) is measured in the Z-basis. In the ideal case, the outcome is $|0\rangle$ with probability 1. The post-measurement state of the data qubits is $|\psi_L\rangle$.

2. Error Model and Analysis Framework

Each of the 5 CNOT gates is followed by a two-qubit depolarizing channel, which introduces one of the 15 non-identity two-qubit Pauli errors $P$ with probability $p/15$. We are interested in the logical fidelity to first order in $p$. This means we only need to consider a single error event occurring at one of the 5 CNOT locations.

The final fidelity is given by $F_L = 1 - P(\text{logical error})$. The probability of a logical error is $P(\text{logical error}) = \sum_{k=1}^5 \sum_{P} P(\text{error } P \text{ at location } k) \times I(P,k)$, where $I(P,k)=1$ if the error is a logical error and 0 otherwise. To first order in $p$, $P(\text{error } P \text{ at location } k) = p/15$. So, $F_L = 1 - \frac{p}{15} N_{log, total}$, where $N_{log, total}$ is the total number of single-error events that result in a logical error.

An error is classified as: * Detected by ancilla: If the error causes the final state of the ancilla to be $|1\rangle_4$ (or have a non-zero component along $|1\rangle_4$). The post-selection on measuring $|0\rangle_4$ discards these runs. * Detected by stabilizers: If the error evolves the data qubits to a state outside the codespace. Post-selection on detectable errors means we assume the final state is in the codespace. * Trivial: If the error is not detected and the final logical state is the correct $|00\rangle_L$. This includes errors that are equivalent to a stabilizer or the identity on the codespace. * Logical: If the error is not detected and the final logical state is incorrect (e.g., $|01\rangle_L, |10\rangle_L, |11\rangle_L$).

We need to count the number of logical errors, $N_{log,k}$, for each of the 5 gate locations $k$. Let an error $P_k$ occur after gate $U_k$. Its effect on the final state is equivalent to applying a propagated error $E_k = U_{>k} P_k U_{>k}^\dagger$ at the end of the circuit, where $U_{>k}$ is the product of gates after $U_k$.

3. Counting Logical Errors Location by Location

Let $E_k = E_{D,k} \otimes E_{A,k}$ be the propagated error on the data (D) and ancilla (A) spaces. - Ancilla Detection: An error is detected by the ancilla if $E_{A,k}$ contains an $X_4$ or $Y_4$ term, which would flip the ancilla from $|0\rangle_4$. - Stabilizer Detection: An error is detected by stabilizers if $[E_{D,k}, S_1] \neq 0$ or $[E_{D,k}, S_2] \neq 0$.

Location 5 (after $CNOT_{04}$): $P_{04}$ on qubits {0,4}. $U_{>5}=I$. $E_5=P_{04}$.
- Ancilla detection: Any $P_0 \otimes P_4$ with $P_4 \in \{X_4, Y_4\}$ is detected. This accounts for $3 \times 2 = 6$ errors, plus $I_0 \otimes X_4, I_0 \otimes Y_4$, for a total of 8 errors.
- Remaining 7 errors have $P_4 \in \{I_4, Z_4\}$. The data part of the error is $P_0$.
- $P_0 \in \{X_0, Y_0, Z_0\}$. These are single-qubit errors, all detectable by stabilizers. For instance, $[X_0, S_2] \ne 0$ and $[Z_0, S_1] \ne 0$.
- The error $I_0 Z_4$ is trivial.
- Thus, all 15 errors are either detected or trivial.
- $N_{log,5} = 0$.
Location 4 (after $CNOT_{34}$): $P_{34}$ on {3,4}. $U_{>4}=CNOT_{04}$. $E_4 = CNOT_{04} P_{34} CNOT_{04}^\dagger$.
- Propagating $P_4$ through $CNOT_{04}$ gives: $X_4 \to X_0 X_4$, $Y_4 \to -Z_0 Y_4$, $Z_4 \to Z_0 Z_4$.
- Ancilla detection: Errors with $P_4 \in \{X_4, Y_4\}$ are detected (8 errors).
- Remaining 7 errors have $P_4 \in \{I_4, Z_4\}$.
- If $P_4=I_4$, data error is $P_3$. $X_3, Y_3, Z_3$ are all detectable by stabilizers.
- If $P_4=Z_4$, data error is $P_3 Z_0$.
  - $I_3 Z_0 = Z_0$: Detectable.
  - $X_3 Z_0, Y_3 Z_0, Z_3 Z_0$: These are weight-2 Pauli operators that commute with both stabilizers $S_1, S_2$. They are logical errors.
- $N_{log,4} = 3$.
Location 3 (after $CNOT_{23}$): $P_{23}$ on {2,3}. $U_{>3}=CNOT_{04} CNOT_{34}$.
- $E_3 = CNOT_{04} CNOT_{34} P_{23} CNOT_{34}^\dagger CNOT_{04}^\dagger$. $P_{23}$ commutes with $CNOT_{04}$.
- $E_3 = CNOT_{34} P_{23} CNOT_{34}^\dagger$.
- Ancilla detection: $P_{23} = P_2 \otimes Y_3 \to P_2 \otimes Y_3 X_4$. Detected. (4 errors).
- Remaining errors:
  - $P_2 \otimes X_3 \to E_D=P_2 X_3$: $X_3$ (det), $X_2 X_3$ (log), $Y_2 X_3$ (log), $Z_2 X_3$ (log). (3 logical errors).
  - $P_2 \otimes Z_3 \to E_D=P_2 Z_3$: $Z_3$ (det), $X_2 Z_3$ (log), $Y_2 Z_3$ (log), $Z_2 Z_3$ (log). (3 logical errors).
  - $P_2 \otimes I_3 \to E_D=P_2$: all detectable.
- $N_{log,3} = 3+3=6$.
Location 2 (after $CNOT_{10}$): $P_{10}$ on {1,0}. $U_{>2}=CNOT_{04} CNOT_{34} CNOT_{23}$. $P_{10}$ commutes with $CNOT_{34}, CNOT_{23}$.
- $E_2 = CNOT_{04} P_{10} CNOT_{04}^\dagger$. This is symmetric to location 4.
- Ancilla detection: Errors $P_1 \otimes Y_0$ are detected (4 errors).
- Remaining errors:
  - $P_1 \otimes X_0 \to E_D = P_1 X_0$: $X_0$ (det), $X_1 X_0$ (log), $Y_1 X_0$ (log), $Z_1 X_0$ (log). (3 logical).
  - $P_1 \otimes Z_0 \to E_D = P_1 Z_0$: $Z_0$ (det), $X_1 Z_0$ (log), $Y_1 Z_0$ (log), $Z_1 Z_0$ (log). (3 logical).
  - $P_1 \otimes I_0 \to E_D = P_1$: all detectable.
- $N_{log,2} = 3+3=6$.
Location 1 (after $CNOT_{12}$): $P_{12}$ on {1,2}. $U_{>1}=CNOT_{04} CNOT_{34} CNOT_{23} CNOT_{10}$.
- A detailed analysis shows that no error $P_{12}$ creates an $X_4$ or $Y_4$ component on the ancilla. No errors are detected by the ancilla.
- We must trace the propagation of each of the 15 errors and classify the resulting data error $E_D$.
- Let’s classify the 15 data errors propagated from $P_{12}$: 1. $X_1 I_2 \to X_0 X_1$ (Logical) 2. $Y_1 I_2 \to Y_1 X_0$ (Detectable by $S_2$) 3. $Z_1 I_2 \to Z_0 Z_1$ (Logical) 4. $I_1 X_2 \to X_2 X_3$ (Logical) 5. $I_1 Y_2 \to Y_2 X_3$ (Detectable by $S_2$) 6. $I_1 Z_2 \to Z_2 Z_3$ (Logical) 7. $X_1 X_2 \to X_0 X_1 X_2 X_3 = S_1$ (Trivial) 8. $X_1 Y_2 \to X_0 X_1 Y_2 X_3$ (Detectable by $S_2$) 9. $X_1 Z_2 \to X_0 X_1 Z_2 Z_3$ (Logical) 10. $Y_1 X_2 \to Y_1 X_0 X_2 X_3$ (Detectable by $S_2$) 11. $Y_1 Y_2 \to Y_1 X_0 Y_2 X_3$ (Detectable by $S_1$) 12. $Y_1 Z_2 \to Y_1 X_0 Z_2 Z_3$ (Detectable by $S_1$) 13. $Z_1 X_2 \to Z_0 Z_1 X_2 X_3$ (Logical) 14. $Z_1 Y_2 \to Z_0 Z_1 Y_2 X_3$ (Detectable by $S_1$) 15. $Z_1 Z_2 \to Z_0 Z_1 Z_2 Z_3 = S_2$ (Trivial)
- Counting the logical errors from the list above gives $N_{log,1} = 6$.

4. Final Fidelity Calculation

The total number of single-error events that result in a logical error is:

\[ N_{log, total} = N_{log,1} + N_{log,2} + N_{log,3} + N_{log,4} + N_{log,5} \]

\[ N_{log, total} = 6 + 6 + 6 + 3 + 0 = 21 \]

There are 5 CNOT gates, each with 15 possible errors, for a total of $5 \times 15 = 75$ single-error events. The probability of any specific single error event is $p/15$.

The total probability of a logical error, to first order in $p$, is the sum of probabilities of each event that leads to a logical error:

\[ P(\text{logical error}) = N_{log, total} \times \frac{p}{15} = 21 \times \frac{p}{15} = \frac{7}{5}p = 1.4p \]

\[ F_L = 1 - P(\text{logical error}) = 1 - 1.4p \]

Final Answer: The logical state fidelity of the final 2-qubit logical state is:

\[ F_L(p) = 1 - \frac{7}{5}p = 1 - 1.4p \]

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

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

    Outputs
    ----------
    F_logical: sympy.Expr, logical state fidelity of the final 2-qubit logical state
    """

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

    return F_logical

Assistant #4

Reasoning (model)

My Reasoning for the Python Code Block

Okay, so I need to present my final answer as a Python code block, specifically using SymPy. The answer itself is quite straightforward: $F_L(p) = 1 - 1.4p$. But, because it’s a symbolic expression, I need to make sure I represent it properly using SymPy. The variable p is already defined as a SymPy symbol, which is convenient. Thus, I can express the fidelity as 1 - 1.4*p. However, to be more precise and avoid floating-point representation, I’ll use 1 - sp.Rational(7, 5) * p.

Now, let’s fill in the provided template. I’ll define a function answer(p) that takes the gate error rate p as input. I’ll add a docstring that clearly explains the function’s purpose, the input, and the output. Inside the function, I’ll place the SymPy expression that I just determined, which is 1 - sp.Rational(7, 5) * p, and return it. This approach ensures that the code is well-documented and produces the correct symbolic output using precise rational numbers. I’ve also added comments to guide the user in identifying where the solution is placed. I’m satisfied that this is the best and cleanest representation for the result.

import sympy as sp

p = sp.symbols('p')

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

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

    Outputs
    ----------
    F_logical: sympy.Expr, logical state fidelity of the final 2-qubit logical state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_logical = 1 - sp.Rational(7, 5) * p
    # ---------------------------------------------------------------

    return F_logical