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

The problem asks for the logical state fidelity of the prepared two-qubit logical $|00\rangle_{AB}$ state as a function of the two-qubit gate error rate $p$. The fidelity is to be calculated assuming post-selection on the ancilla qubit measuring $|0\rangle$ and on the final state being in the code space (i.e., detectable errors in the code are discarded).

Let the initial state of the five qubits (0,1,2,3,4) be $|00000\rangle$. The circuit operations, applied from right to left as written in the problem, are: $U_{circuit} = M_4 (CNOT_{04}) (CNOT_{34}) (CNOT_{23}) (CNOT_{10}) (CNOT_{12}) (H_1)$

1. Ideal Circuit Evolution and Final State: Let’s first determine the ideal final state before any errors. Initial state: $|\psi_0\rangle = |00000\rangle$.

$H_1$: $|\psi_1\rangle = |0\rangle_0 \frac{1}{\sqrt{2}}(|0\rangle_1 + |1\rangle_1) |000\rangle_{234}$
$CNOT_{12}$: $|\psi_2\rangle = |0\rangle_0 \frac{1}{\sqrt{2}}(|00\rangle_{12} + |11\rangle_{12}) |00\rangle_{34}$
$CNOT_{10}$: $|\psi_3\rangle = \frac{1}{\sqrt{2}}(|000\rangle_{012} + |111\rangle_{012}) |00\rangle_{34}$
$CNOT_{23}$: $|\psi_4\rangle = \frac{1}{\sqrt{2}}(|0000\rangle_{0123} + |1111\rangle_{0123}) |0\rangle_4$
$CNOT_{34}$: $|\psi_5\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)$
$CNOT_{04}$: $|\psi_6\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$

Now, we perform the measurement $M_4$ on qubit 4. We post-select on measuring $|0\rangle_4$. The projector for this measurement outcome is $P_0^4 = I_{0123} \otimes |0\rangle_4\langle 0|_4$. The ideal state after post-selection is $P_0^4 |\psi_6\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$. This state is already normalized and has qubit 4 in $|0\rangle$. We denote this 5-qubit ideal state as $|\psi_{ideal\_5Q}\rangle$. The logical state $|\Psi_L\rangle$ is the 4-qubit component: $|\Psi_L\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)_{0123}$.

Let’s verify this is the logical $|00\rangle_{AB}$ state: Stabilizers are $S_1 = X_0X_1X_2X_3$ and $S_2 = Z_0Z_1Z_2Z_3$. $S_1|\Psi_L\rangle = \frac{1}{\sqrt{2}}(X_0X_1X_2X_3|0000\rangle + X_0X_1X_2X_3|1111\rangle) = \frac{1}{\sqrt{2}}(|1111\rangle + |0000\rangle) = |\Psi_L\rangle$. $S_2|\Psi_L\rangle = \frac{1}{\sqrt{2}}(Z_0Z_1Z_2Z_3|0000\rangle + Z_0Z_1Z_2Z_3|1111\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + (-1)^4|1111\rangle) = |\Psi_L\rangle$. Logical $Z_A = Z_0Z_1I_2I_3$ and $Z_B = Z_0I_1Z_2I_3$. $Z_A|\Psi_L\rangle = \frac{1}{\sqrt{2}}(Z_0Z_1|0000\rangle + Z_0Z_1|1111\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + (-1)^2|1111\rangle) = |\Psi_L\rangle$. $Z_B|\Psi_L\rangle = \frac{1}{\sqrt{2}}(Z_0Z_2|0000\rangle + Z_0Z_2|1111\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + (-1)^2|1111\rangle) = |\Psi_L\rangle$. The state $|\Psi_L\rangle$ is indeed the logical $|00\rangle_{AB}$ state.

2. Error Model and Fidelity Calculation: There are 5 CNOT gates ($U_k$, $k=2 \dots 6$) in the circuit, each followed by a two-qubit depolarizing channel. The Hadamard gate $H_1$ (denoted $U_1$) is not followed by an error channel. The noisy evolution leads to a density matrix $\rho_{raw}$. We are interested in the fidelity of the post-selected state:

\[ F = \frac{\langle \psi_{ideal\_5Q} | \rho_{raw} | \psi_{ideal\_5Q} \rangle}{\text{Tr}(P_A \rho_{raw})} \]

where $P_A = P_S^{0-3} \otimes P_0^4$ is the projector onto the code space (qubits 0-3) and ancilla 4 being $|0\rangle$. $P_S^{0-3}$ is the projector onto the $+1$ eigenspace of $S_1$ and $S_2$. We calculate this fidelity to first order in $p$. $\rho_{raw} = (1-5p) |\psi_{ideal\_5Q}\rangle\langle\psi_{ideal\_5Q}| + \sum_{k=2}^6 \frac{p}{15} \sum_{G_k \in \mathcal{P}_{Q_k}^*} |\phi_{k,G_k}\rangle\langle\phi_{k,G_k}| + O(p^2)$, where $U_{final\_k} = U_6 \dots U_{k+1}$ and $U_{initial\_k} = U_k \dots U_1$. And $|\phi_{k,G_k}\rangle = U_{final\_k} G_k U_{initial\_k} |\psi_0\rangle = U_{final\_k} G_k U_{final\_k}^{-1} |\psi_{ideal\_5Q}\rangle$. Let $G_k^{\text{transformed}} = U_{final\_k} G_k U_{final\_k}^{-1}$. So $|\phi_{k,G_k}\rangle = G_k^{\text{transformed}} |\psi_{ideal\_5Q}\rangle$.

Numerator: $N = \langle \psi_{ideal\_5Q} | \rho_{raw} | \psi_{ideal\_5Q} \rangle$.

\[ N = (1-5p) + \sum_{k=2}^6 \frac{p}{15} \sum_{G_k \in \mathcal{P}_{Q_k}^*} |\langle\psi_{ideal\_5Q}| G_k^{\text{transformed}} |\psi_{ideal\_5Q}\rangle|^2 + O(p^2) \]

The term $|\langle\psi_{ideal\_5Q}| G_k^{\text{transformed}} |\psi_{ideal\_5Q}\rangle|^2$ is non-zero if and only if:

$G_k^{\text{transformed}}$ acts as $I_4$ on qubit 4 (because $Z_4$ anticommutes with one state, $X_4, Y_4$ flip the state).
$G_k^{\text{transformed}}$ on qubits 0-3 is a product of $Z$ operators ($P_{0123} = Z_0^{a_0}Z_1^{a_1}Z_2^{a_2}Z_3^{a_3}$) such that the total number of $Z$ operators $\sum a_i$ is even. (This is because $\langle\Psi_L| P_{0123} |\Psi_L\rangle = \frac{1}{2}(1 + (-1)^{\sum a_i})$). So, for $N$ contribution, $G_k^{\text{transformed}}$ must be $P_{0123} \otimes I_4$ and $P_{0123} \in \{I, Z_0Z_1, Z_0Z_2, Z_1Z_2, Z_2Z_3, Z_1Z_3, Z_0Z_3, Z_0Z_1Z_2Z_3\}$. All these 8 operators yield 1.

Denominator: $D = \text{Tr}(P_A \rho_{raw})$.

\[ D = (1-5p) + \sum_{k=2}^6 \frac{p}{15} \sum_{G_k \in \mathcal{P}_{Q_k}^*} \langle\psi_{ideal\_5Q}| (G_k^{\text{transformed}})^\dagger P_A G_k^{\text{transformed}} |\psi_{ideal\_5Q}\rangle + O(p^2) \]

The term $\langle\psi_{ideal\_5Q}| (G_k^{\text{transformed}})^\dagger P_A G_k^{\text{transformed}} |\psi_{ideal\_5Q}\rangle$ is non-zero if and only if:

$G_k^{\text{transformed}}$ does not flip qubit 4. So $G_k^{\text{transformed}}$ must not contain $X_4$ or $Y_4$. ($P_0^4 G_k^{\text{transformed}} |\psi_{ideal\_5Q}\rangle \neq 0$).
$G_k^{\text{transformed}}$ on qubits 0-3 must preserve the code space ($P_S^{0-3} G_k^{\text{transformed}} |\psi_{ideal\_5Q}\rangle \neq 0$). This means $G_k^{\text{transformed}}$ must commute with the stabilizers $S_1$ and $S_2$. Thus $P_{0123} \in N(S)$, the normalizer of the stabilizer group. So, for $D$ contribution, $G_k^{\text{transformed}}$ must be $P_{0123} \otimes I_4$ or $P_{0123} \otimes Z_4$, where $P_{0123} \in N(S)$. All such errors yield 1.

3. Pauli Error Propagation Analysis: Let’s analyze each of the 5 CNOT gates ($U_k$, $k=2 \dots 6$). $G_k$ is a two-qubit Pauli error on the qubits of $U_k$. $G_k^{\text{transformed}} = U_{final\_k} G_k U_{final\_k}^{-1}$.

CNOT Rules for $U_{c,t} P U_{c,t}^{-1}$:
- $X_c \to X_c X_t$
- $Z_c \to Z_c$
- $X_t \to X_t$
- $Z_t \to Z_c Z_t$
- $Y_c \to Y_c X_t$
- $Y_t \to Z_c Y_t$
- Paulis on qubits not involved commute.
List of $N(S)$ elements ($P_{0123}$): There are 256 such elements. These are products of $X_A, X_B, Z_A, Z_B, S_1, S_2$.

Case by Case Analysis:

$k=6$: $CNOT_{04}$ (qubits 0,4). $U_{final_6} = I$. So $G_6^{\text{transformed}} = G_6$. (15 errors)
- $N$: No error (non-identity) can satisfy the condition $P_0 \otimes I_4$ and $P_0=I_0$. So 0 contribution.
- $D$:
  - Errors with $X_4$ or $Y_4$ (8 errors like $I_0X_4, X_0X_4, \dots$): $0$ contribution as they flip ancilla.
  - Errors with $I_4$: $X_0I_4, Y_0I_4, Z_0I_4$. None of $X_0, Y_0, Z_0$ are in $N(S)$. $0$ contribution.
  - Errors with $Z_4$: $I_0Z_4, X_0Z_4, Y_0Z_4, Z_0Z_4$.
    - $I_0Z_4$: $I_0 \in N(S)$. This contributes 1 to $D$.
    - $X_0Z_4, Y_0Z_4, Z_0Z_4$: $X_0, Y_0, Z_0 \notin N(S)$. $0$ contribution.
- Total for $k=6$: $N=0$, $D=1$.
$k=5$: $CNOT_{34}$ (qubits 3,4). $U_{final_5} = CNOT_{04}$. (15 errors)
- $N$: No error can satisfy the condition $P_3 \otimes I_4$ and $P_3=I_3$. So 0 contribution.
- $D$:
  - Errors with $X_4$ or $Y_4$ (8 errors like $I_3X_4, X_3X_4, \dots$): $G_5^{\text{transformed}}$ will contain $X_4$ or $Y_4$. $0$ contribution.
  - Errors with $I_4$: $X_3I_4, Y_3I_4, Z_3I_4$. These map to themselves by $CNOT_{04}$. None of $X_3, Y_3, Z_3$ are in $N(S)$. $0$ contribution.
  - Errors with $Z_4$: $I_3Z_4, X_3Z_4, Y_3Z_4, Z_3Z_4$.
    - $I_3Z_4 \xrightarrow{C_{04}} I_3Z_0Z_4$. $Z_0 \notin N(S)$. $0$ contribution.
    - $X_3Z_4 \xrightarrow{C_{04}} X_3Z_0Z_4$. $X_3Z_0 \notin N(S)$. $0$ contribution.
    - $Y_3Z_4 \xrightarrow{C_{04}} Y_3Z_0Z_4$. $Y_3Z_0 \notin N(S)$. $0$ contribution.
    - $Z_3Z_4 \xrightarrow{C_{04}} Z_3Z_0Z_4$. $Z_3Z_0 \in N(S)$ (it’s $Z_A Z_B S_2$). This contributes 1 to $D$.
- Total for $k=5$: $N=0$, $D=1$.
$k=4$: $CNOT_{23}$ (qubits 2,3). $U_{final_4} = CNOT_{04} CNOT_{34}$. (15 errors)
- $N$: No $G_4$ error results in $P_{0123} \otimes I_4$ where $P_{0123}$ is an even-weight Z-type. $0$ contribution.
- $D$:
  - Errors with $Z_3$ (9 errors, e.g. $I_2Z_3, X_2Z_3, \dots$): $G_4$ containing $Z_3$ (e.g., $I_2Z_3$) will propagate to $I_2Z_3Z_4$ (via $CNOT_{34}$). So $G_4^{\text{transformed}}$ will contain $Z_4$.
    - $Z_2Z_3 \xrightarrow{C_{34}} Z_2Z_3Z_4 \xrightarrow{C_{04}} Z_2Z_3Z_4$. $Z_2Z_3 \in N(S)$ (it’s $Z_A S_2$). This contributes 1 to $D$.
    - Other 8 errors with $Z_3$: e.g. $I_2Z_3 \to Z_3Z_4$, $Z_3 \notin N(S)$. Check other $X,Y$ combinations, e.g. $X_2Z_3 \to X_2Z_3Z_4$, $X_2Z_3 \notin N(S)$. So 1 from $Z_2Z_3$.
  - Errors without $Z_3$ (6 errors): $I_2X_3, I_2Y_3, X_2I_3, X_2X_3, X_2Y_3, Y_2I_3, Y_2X_3, Y_2Y_3$. These propagate to $P_{0123} \otimes I_4$.
    - $X_2X_3 \xrightarrow{U_{final_4}} X_2X_3$. $X_2X_3 \in N(S)$. This contributes 1 to $D$.
    - $Y_2Y_3 \xrightarrow{U_{final_4}} Y_2Y_3$. $Y_2Y_3 \in N(S)$. This contributes 1 to $D$.
    - Others: $X_2, Y_2, X_3, Y_3, X_2Y_3, Y_2X_3$ are not in $N(S)$. $0$ contribution.
- Total for $k=4$: $N=0$, $D=3$.
$k=3$: $CNOT_{10}$ (qubits 1,0). $U_{final_3} = CNOT_{04} CNOT_{34} CNOT_{23}$. (15 errors)
- $N$: No $G_3$ error results in $P_{0123} \otimes I_4$ where $P_{0123}$ is an even-weight Z-type. $0$ contribution.
- $D$:
  - Errors with $Z_0$ (9 errors): $G_3$ containing $Z_0$ (e.g., $I_1Z_0$) will propagate to $Z_0Z_4$ (via $CNOT_{04}$).
    - $Z_1Z_0 \xrightarrow{U_{final_3}} Z_1Z_0Z_4$. $Z_1Z_0 \in N(S)$ (it’s $Z_A$). This contributes 1 to $D$.
    - Other 8 errors with $Z_0$: E.g. $X_1Z_0 \to X_1Z_0Z_4$. $X_1Z_0 \notin N(S)$. $0$ contribution.
  - Errors without $Z_0$ (6 errors): $I_1X_0, I_1Y_0, X_1I_0, X_1X_0, X_1Y_0, Y_1I_0, Y_1X_0, Y_1Y_0$. These propagate to $P_{0123} \otimes I_4$.
    - $X_1X_0 \xrightarrow{U_{final_3}} X_1X_0$. $X_1X_0 \in N(S)$ (it’s $X_B$). This contributes 1 to $D$.
    - $Y_1Y_0 \xrightarrow{U_{final_3}} Y_1Y_0$. $Y_1Y_0 \in N(S)$. This contributes 1 to $D$.
    - Others (4 errors): not in $N(S)$. $0$ contribution.
- Total for $k=3$: $N=0$, $D=3$.
$k=2$: $CNOT_{12}$ (qubits 1,2). $U_{final_2} = CNOT_{04} CNOT_{34} CNOT_{23} CNOT_{10}$. (15 errors)
- $N$:
  - $G_2=Z_1Z_2 \xrightarrow{U_{final_2}} Z_1Z_2$. $Z_1Z_2 \in N(S)$ and is even-weight $Z$-type. This contributes 1 to $N$.
  - No other error contributes to $N$.
- $D$: No error originating from $G_2$ will propagate to $Z_0$ or $Z_3$ (which would then become $Z_0Z_4$ or $Z_3Z_4$). So all 15 errors effectively become $P_{0123} \otimes I_4$.
  - $Z_1Z_2 \xrightarrow{U_{final_2}} Z_1Z_2$. $Z_1Z_2 \in N(S)$. This contributes 1 to $D$.
  - $X_1X_2 \xrightarrow{U_{final_2}} X_1X_2$. $X_1X_2 \in N(S)$. This contributes 1 to $D$.
  - $Y_1Y_2 \xrightarrow{U_{final_2}} Y_1Y_2$. $Y_1Y_2 \in N(S)$. This contributes 1 to $D$.
  - Other 12 errors (e.g. $I_1X_2, Z_1I_2$): Not in $N(S)$. $0$ contribution.
- Total for $k=2$: $N=1$, $D=3$.

Summary of Contributions: Number of CNOT gates $N_{CNOT}=5$. Sum of $N$ contributions from errors: $0+0+0+0+1 = 1$. Sum of $D$ contributions from errors: $1+1+3+3+3 = 11$.

Substituting these into the fidelity formula:

\[ N = (1-5p) + \frac{p}{15} \times 1 = 1 - 5p + \frac{p}{15} = 1 - \frac{75p-p}{15} = 1 - \frac{74p}{15} \]

\[ D = (1-5p) + \frac{p}{15} \times 11 = 1 - 5p + \frac{11p}{15} = 1 - \frac{75p-11p}{15} = 1 - \frac{64p}{15} \]

The logical state fidelity is:

\[ F = \frac{1 - \frac{74p}{15}}{1 - \frac{64p}{15}} \]

For small $p$, we can use the approximation $\frac{1}{1-x} \approx 1+x$:

\[ F \approx \left(1 - \frac{74p}{15}\right) \left(1 + \frac{64p}{15}\right) = 1 + \left(\frac{64}{15} - \frac{74}{15}\right)p + O(p^2) = 1 - \frac{10p}{15} + O(p^2) = 1 - \frac{2p}{3} + O(p^2) \]

The final answer is $\boxed{\frac{1 - \frac{74p}{15}}{1 - \frac{64p}{15}}}$.

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)

Okay, here’s what I’m thinking:

The user wants me to take that final, elegant answer I just derived - the one with all the fractions, $\frac{1 - \frac{74p}{15}}{1 - \frac{64p}{15}}$ - and translate it directly into Python code. Specifically, they want me to plug it into the F_logical variable in their script. I know they’re using sympy since that’s a prerequisite to work in this domain. So, it’s just a matter of transforming that mathematical expression into a valid sympy representation. Let’s keep the variable ‘p’, I’m comfortable in this context and know the significance. Shouldn’t be too hard, it’s really just a matter of representing the numerator and the denominator correctly and then dividing. Should be good to go.

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 - (74*p)/15) / (1 - (64*p)/15)
    # ---------------------------------------------------------------

    return F_logical