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

Reasoning (model)

Summary of my approach to calculating logical state fidelity:

As an expert in quantum information, I’ve been tasked with a challenging problem: determine the fidelity of a prepared |00⟩ state after it passes through a specific quantum circuit, taking into account the impact of depolarizing noise.

To tackle this, I’ve broken the problem into a few key areas. First, I need to deeply understand the [[4,2,2]] code, which is the cornerstone of protecting our logical qubits. This involves familiarizing myself with the stabilizers, logical operators, and, crucially, how the logical |00⟩ state is encoded and stabilized within the four physical qubits.

Next, I thoroughly reviewed the circuit itself. I carefully analyzed the sequence of Hadamard and CNOT gates, mapping each operation onto the state to see how it changes.

The noise model is a critical element, too. I fully grasped the depolarizing error channel that’s introduced after each CNOT gate, described by a specific probability p. The objective here is to determine how error detection works.

Post-selection is a key part, and here I understand that I am filtering for both successful ancilla measurements (|0⟩ on qubit 4) and ensuring the final state belongs to the |00⟩ logical subspace. This means I’ll effectively project the final state onto the logical |00⟩ state.

The core of the problem is about calculating the logical fidelity. The plan is to expand the fidelity to the first order of p, since the probability of errors is small. I must consider the outcome of the ideal circuit, which includes the ideal final state. Second, I need to account for a single error. Finally, I will determine what is produced if there are two errors. The circuit has five error-prone CNOT gates.

I start by building up the operations step by step. I use this approach to determine the ideal circuit outcome by mapping the qubits’ state along the ideal gates. After that, I include noise.

The strategy is to break the circuit into stages. I then trace how errors propagate forward and impact the output state. I calculated $F = (1 - \frac{28}{15}p)$.

To calculate the logical state fidelity, we need to determine the probability of successfully preparing the logical $|00\rangle_{AB}$ state and the probability of a successful ancilla measurement. The fidelity is the ratio of these two probabilities. We will work to first order in the physical infidelity $p$.

Let the initial state of the 5 qubits (4 physical, 1 ancilla) be $|\Psi_{in}\rangle = |00000\rangle$. The circuit consists of a Hadamard gate $H_1$ followed by 5 CNOT gates $C_{12}, C_{10}, C_{23}, C_{34}, C_{04}$. Let’s denote them as $G_1=H_1$, $G_2=C_{12}$, $G_3=C_{10}$, $G_4=C_{23}$, $G_5=C_{34}$, $G_6=C_{04}$. The order of application is $G_6 G_5 G_4 G_3 G_2 G_1$. The ideal final state, before ancilla measurement, is $|\Psi_{ideal}\rangle = G_6 G_5 G_4 G_3 G_2 G_1 |\Psi_{in}\rangle$. Let’s track the ideal state:

Initial: $|00000\rangle$
$H_1$: $H_1|00000\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)$. (Qubit order 01234)
$C_{12}$: $\frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)$.
$C_{10}$: $\frac{1}{\sqrt{2}}(|00000\rangle + |11100\rangle)$.
$C_{23}$: $\frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$. This is $|\text{GHZ}_4\rangle_Q |0\rangle_A$. The logical state is indeed $|00\rangle_{AB}$. It’s the $+1$ eigenstate of $XXXX$, $ZZZZ$, $ZZII$, and $ZIZI$. Let’s call the set of 16 Pauli operators on the 4 physical qubits that stabilize this state $G_{00}$.
$C_{34}$: $\frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)$.
$C_{04}$: $\frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$. So, the ideal final state is $|\Psi_{ideal}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)_Q |0\rangle_A$. In the ideal case, the ancilla qubit 4 is measured to be $|0\rangle$ with probability 1, and the logical state is the target $|00\rangle_{AB}$ state.

Each CNOT gate ($G_2, G_3, G_4, G_5, G_6$) is followed by a two-qubit depolarizing error channel $\mathcal{E}_{ij}$ with probability $p$. $\mathcal{E}_{ij}(\rho) = (1-p)\rho + \frac{p}{15} \sum_{P \in \text{Pauli}_2, P \ne I} P \rho P$. We are interested in the fidelity up to $O(p)$. This means we consider the case of no errors, and the case of a single error occurring after one of the CNOT gates.

Let $U_{ideal} = G_6 G_5 G_4 G_3 G_2 G_1$. The ideal final state is $|\Psi_{ideal}\rangle = U_{ideal}|\Psi_{in}\rangle$. If a Pauli error $P_{err}$ occurs after gate $G_k$, the state is $G_6 \dots G_{k+1} P_{err} G_k \dots G_1 |\Psi_{in}\rangle$. This is equivalent to applying an effective Pauli error $P_{eff}^{(k)}$ to the ideal state: $P_{eff}^{(k)} |\Psi_{ideal}\rangle$, where $P_{eff}^{(k)} = G_6 \dots G_{k+1} P_{err} G_{k+1}^\dagger \dots G_6^\dagger$.

We define three categories for each of the $5 \times 15 = 75$ possible single errors:

Ancilla Fails ($S=0$): If $P_{eff}^{(k)} |\Psi_{ideal}\rangle$ results in the ancilla qubit 4 being in a state orthogonal to $|0\rangle_A$. This means $P_{eff}^{(k)}$ contains an $X_4$ or $Y_4$ operator. These errors are detected by the ancilla measurement, and the state preparation fails.
Logical Fails ($S=1, F=0$): If the ancilla qubit is measured to be $|0\rangle_A$ (i.e., $P_{eff}^{(k)}$ contains $I_4$ or $Z_4$), but the 4-qubit physical state is no longer the target logical state $|\text{GHZ}_4\rangle$. This means the 4-qubit part of $P_{eff}^{(k)}$ is not in $G_{00}$. These errors are not detected by the ancilla and lead to logical infidelity.
Logical Success ($S=1, F=1$): If the ancilla qubit is measured to be $|0\rangle_A$ (i.e., $P_{eff}^{(k)}$ contains $I_4$ or $Z_4$), AND the 4-qubit physical state is still $|\text{GHZ}_4\rangle$ (i.e., the 4-qubit part of $P_{eff}^{(k)}$ is in $G_{00}$).

We need to list $P_{eff}^{(k)}$ for each of the 15 errors for each of the 5 CNOT gates ($G_2$ to $G_6$). Recall the Pauli transformation rules for $C_{ct}$: $C_{ct}(X_c I_t)C_{ct} = X_c X_t$ $C_{ct}(I_c X_t)C_{ct} = I_c X_t$ $C_{ct}(Z_c I_t)C_{ct} = Z_c I_t$ $C_{ct}(I_c Z_t)C_{ct} = Z_c Z_t$ The operators in $G_{00}$ (stabilizers of $|\text{GHZ}_4\rangle$) are the 16 elements of the group generated by $\{XXXX, ZZZZ, ZZII, ZIZI\}$.

Let $N_{ok}, N_{fail,logical}, N_{fail,ancilla}$ be the counts for each category for the 15 non-identity Pauli errors occurring after a specific CNOT.

1. Error after $G_6 = C_{04}$ (on qubits 0,4): $P_{eff}^{(6)} = P_{err}$. * $Q_A \in \{X_4, Y_4\}$: 8 errors (e.g., $I_0 X_4, X_0 X_4, Y_0 X_4, Z_0 X_4, I_0 Y_4, \dots$). $N_{fail,ancilla}=8$. * $Q_A \in \{I_4, Z_4\}$: 7 errors. * $I_0 Z_4$: $I_Q Z_A$. $I_Q \in G_{00}$. Logical Success. $N_{ok}=1$. * $X_0 I_4, Y_0 I_4, Z_0 I_4, X_0 Z_4, Y_0 Z_4, Z_0 Z_4$: (6 errors). Their logical parts ($X_0, Y_0, Z_0, X_0, Y_0, Z_0$) are not in $G_{00}$. Logical Fails. $N_{fail,logical}=6$. Summary for $G_6$: $(N_{ok}, N_{fail,logical}, N_{fail,ancilla}) = (1, 6, 8)$.

2. Error after $G_5 = C_{34}$ (on qubits 3,4): $P_{eff}^{(5)} = C_{04} P_{err} C_{04}$. * If $P_{err}$ has $X_4$ or $Y_4$ (8 errors, e.g., $I_3 X_4, \dots$): $C_{04}$ maps these to $I_Q X_4$ or $I_Q Y_4$. Ancilla Fails. $N_{fail,ancilla}=8$. * If $P_{err}$ has $I_4$ (3 errors: $X_3 I_4, Y_3 I_4, Z_3 I_4$): $C_{04}$ maps these to $P_3 I_4$. Their logical parts ($X_3, Y_3, Z_3$) are not in $G_{00}$. Logical Fails. $N_{fail,logical}=3$. * If $P_{err}$ has $Z_4$ (4 errors, e.g., $I_3 Z_4, \dots$): $C_{04}$ maps $P_3 Z_4$ to $P_3 Z_0 Z_4$. Their logical parts ($I_3 Z_0, X_3 Z_0, Y_3 Z_0, Z_3 Z_0$) are not in $G_{00}$. Logical Fails. $N_{fail,logical}=4$. Summary for $G_5$: $(N_{ok}, N_{fail,logical}, N_{fail,ancilla}) = (0, 7, 8)$.

3. Error after $G_4 = C_{23}$ (on qubits 2,3): $P_{eff}^{(4)} = C_{04} C_{34} P_{err} C_{34} C_{04}$. Let $P_{err}$ be $P_2 P_3$. * If $P_3$ is $X$ or $Y$ (8 errors, e.g., $I_2 X_3, \dots$): $C_{34}$ maps $P_2 P_3$ to $P_2 P_3 X_4$. These lead to Ancilla Fails. $N_{fail,ancilla}=8$. * If $P_3$ is $I$ (3 errors: $X_2 I_3, Y_2 I_3, Z_2 I_3$): $C_{34}$ maps $P_2 I_3$ to $P_2 I_3 I_4$. $C_{04}$ keeps it $P_2 I_3 I_4$. Their logical parts ($X_2, Y_2, Z_2$) are not in $G_{00}$. Logical Fails. $N_{fail,logical}=3$. * If $P_3$ is $Z$ (4 errors: $I_2 Z_3, X_2 Z_3, Y_2 Z_3, Z_2 Z_3$): $C_{34}$ maps $P_2 Z_3$ to $P_2 Z_3 I_4$. $C_{04}$ keeps it $P_2 Z_3 I_4$. * $Z_2 Z_3$: Logical part $I_0 I_1 Z_2 Z_3 \in G_{00}$. Logical Success. $N_{ok}=1$. * $I_2 Z_3, X_2 Z_3, Y_2 Z_3$: Logical parts ($I_2 Z_3, X_2 Z_3, Y_2 Z_3$) are not in $G_{00}$. Logical Fails. $N_{fail,logical}=3$. Summary for $G_4$: $(1, 6, 8)$.

4. Error after $G_3 = C_{10}$ (on qubits 1,0): $P_{eff}^{(3)} = C_{04} C_{34} C_{23} P_{err} C_{23} C_{34} C_{04}$. Let $P_{err}$ be $P_1 P_0$. These do not act on qubits 2,3,4. So $C_{23}$ and $C_{34}$ don’t change them. $P_{eff}^{(3)} = C_{04} P_{10} I_4 C_{04}$. * If $P_0$ is $X$ or $Y$ (8 errors): $C_{04}$ maps $P_1 P_0 I_4$ to $P_1 P_0 X_4$. Ancilla Fails. $N_{fail,ancilla}=8$. * If $P_0$ is $I$ (3 errors: $X_1 I_0, Y_1 I_0, Z_1 I_0$): $C_{04}$ maps $P_1 I_0 I_4$ to $P_1 I_0 I_4$. Logical parts ($X_1, Y_1, Z_1$) are not in $G_{00}$. Logical Fails. $N_{fail,logical}=3$. * If $P_0$ is $Z$ (4 errors: $I_1 Z_0, X_1 Z_0, Y_1 Z_0, Z_1 Z_0$): $C_{04}$ maps $P_1 Z_0 I_4$ to $P_1 Z_0 Z_4$. Logical parts ($I_1 Z_0, X_1 Z_0, Y_1 Z_0, Z_1 Z_0$) are not in $G_{00}$. Logical Fails. $N_{fail,logical}=4$. Summary for $G_3$: $(0, 7, 8)$.

5. Error after $G_2 = C_{12}$ (on qubits 1,2): $P_{eff}^{(2)} = C_{04} C_{34} C_{23} C_{10} P_{err} C_{10} C_{23} C_{34} C_{04}$. This is the most complex one. Let’s trace it carefully. * Errors with $X_1$ or $Y_1$ (8 errors: $X_1 I_2, X_1 X_2, \dots, Y_1 Z_2$). $C_{10}(X_1 P_2)C_{10}$ yields $X_1 X_0 P_2$. $C_{10}(Y_1 P_2)C_{10}$ yields $Y_1 X_0 P_2$. The $X_0$ component then means $C_{04}$ will map these to terms with $X_4$. All 8 of these lead to Ancilla Fails. $N_{fail,ancilla}=8$. * Errors with $Z_1$ (4 errors: $Z_1 I_2, Z_1 X_2, Z_1 Y_2, Z_1 Z_2$). $C_{10}(Z_1 P_2)C_{10}$ yields $Z_1 I_0 P_2$. These have $I_0$ so $C_{04}$ won’t flip $q_4$ from $q_0$’s action. * $Z_1 I_2 \to Z_1 I_0 I_2 I_3 I_4$. Logical part ($I_0 Z_1 I_2 I_3$) not in $G_{00}$. Logical Fail. * $Z_1 X_2 \to Z_1 I_0 X_2 X_3 X_4$. Ancilla Fails. * $Z_1 Y_2 \to Z_1 I_0 Y_2 X_3 X_4$. Ancilla Fails. * $Z_1 Z_2 \to Z_1 I_0 Z_2 I_3 I_4$. Logical part ($I_0 Z_1 Z_2 I_3$) is $I_0 Z_1 Z_2 I_3 |\text{GHZ}_4\rangle = |\text{GHZ}_4\rangle$. So it’s in $G_{00}$. Logical Success. $N_{ok}=1$. Sub-summary for $Z_1$: $(1, 1, 2)$. * Errors with $I_1$ (3 errors: $I_1 X_2, I_1 Y_2, I_1 Z_2$). $C_{10}(I_1 P_2)C_{10}$ yields $I_1 P_2$. * $I_1 X_2 \to I_1 X_2 X_3 X_4$. Ancilla Fails. * $I_1 Y_2 \to I_1 Y_2 X_3 X_4$. Ancilla Fails. * $I_1 Z_2 \to I_1 Z_2 I_3 I_4$. Logical part ($I_0 I_1 Z_2 I_3$) not in $G_{00}$. Logical Fail. Sub-summary for $I_1$: $(0, 1, 2)$. Summary for $G_2$: $(N_{ok}, N_{fail,logical}, N_{fail,ancilla}) = (1, 1+1, 8+2+2) = (1, 2, 12)$.

Total counts for 75 errors: Total $N_{ok} = 1+0+1+0+1 = 3$. Total $N_{fail,logical} = 6+7+6+7+2 = 28$. Total $N_{fail,ancilla} = 8+8+8+8+12 = 44$. Total errors = $3+28+44 = 75$.

The probability of successful ancilla measurement ($P_{succ}$) to $O(p)$: $P_{succ} = (1-p)^5 \cdot 1 + \frac{p}{15} (1-p)^4 (N_{ok} + N_{fail,logical}) + O(p^2)$ $P_{succ} \approx (1-5p) + \frac{p}{15} (3+28) = 1 - 5p + \frac{31p}{15} = 1 - (\frac{75-31}{15})p = 1 - \frac{44}{15}p$.

The probability of successful ancilla measurement AND correct logical state ($\mathcal{F}$) to $O(p)$: $\mathcal{F} = (1-p)^5 \cdot 1 + \frac{p}{15} (1-p)^4 N_{ok} + O(p^2)$ $\mathcal{F} \approx (1-5p) + \frac{p}{15} \cdot 3 = 1 - 5p + \frac{p}{5} = 1 - (\frac{25-1}{5})p = 1 - \frac{24}{5}p$.

The logical state fidelity $F_{logical}$ is the ratio of $\mathcal{F}$ to $P_{succ}$:

\[ F_{logical} = \frac{\mathcal{F}}{P_{succ}} = \frac{1 - \frac{24}{5}p}{1 - \frac{44}{15}p} \]

Using the approximation $\frac{1}{1-x} \approx 1+x$ for small $x$:

\[ F_{logical} \approx \left(1 - \frac{24}{5}p\right) \left(1 + \frac{44}{15}p\right) \]

\[ F_{logical} \approx 1 + \left(\frac{44}{15} - \frac{24}{5}\right)p + O(p^2) \]

\[ F_{logical} \approx 1 + \left(\frac{44 - 3 \times 24}{15}\right)p + O(p^2) \]

\[ F_{logical} \approx 1 + \left(\frac{44 - 72}{15}\right)p + O(p^2) \]

\[ F_{logical} \approx 1 - \frac{28}{15}p + O(p^2) \]

The final answer is $\boxed{1 - \frac{28}{15}p}$.

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

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(28, 15)) * p
    # ---------------------------------------------------------------

    return F_logical