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)

Analysis of a [[4,2,2]] Quantum Error Correction Code

Alright, let’s break this down. The user wants me to analyze this [[4,2,2]] code, and I know the basics: four physical qubits (n=4), two logical qubits (k=2), and a distance of two (d=2). This means it can detect single-qubit errors but can’t correct them. I’m labeling the physical qubits 0, 1, 2, and 3, and the logical qubits A and B.

First, I need to understand the stabilizers and the logical operators. The stabilizers are pretty straightforward: S1 = XXXX and S2 = ZZZZ. A state in the code space has to be a +1 eigenstate of both. Now, the trickier part is defining the logical operators. They’re like the “qubit flip” and “phase flip” for our logical qubits.

Here’s what I have: XA = XIXI, XB = XXII, ZA = ZZII, and ZB = ZIZI. I need to make sure these behave correctly. Specifically, they should anti-commute within a logical qubit (like XA and ZA) and commute with each other (like XA and ZB). I have to remember that “up to multiplication by stabilizers” means I might see some extra stabilizer operators.

I’ve carefully checked the commutation relations. It looks like everything works out correctly, including the anti-commutation. The target state, the logical |00⟩ state, should be a simultaneous eigenstate of all the positive logical operator expectation values (i.e. +1 eigenvalues) of ZA and ZB (i.e., +1), as well as both stabilizers, which has to be a symmetric superposition of all-zeros and all-ones: |00⟩L = 1/sqrt(2) (|0000> + |1111>).

Next, I look at how this state is prepared using a quantum circuit. I can see the circuit has a series of CNOT gates and a Hadamard gate. I’ve checked the evolution of the qubits step-by-step, including the ancilla, which should measure $|0\rangle$. If this is the case, the data qubits are left in the $|00⟩_L$ state.

Now, the user introduces a nasty twist: errors! Each CNOT gate is followed by a two-qubit depolarizing channel, so I have to carefully examine the effect of these errors on the overall fidelity. To first order, I only have to look at single-qubit errors. My strategy is to analyze how each of the 15 possible two-qubit Pauli errors (remember, it’s a two-qubit depolarizing channel) after each CNOT propagates through the rest of the circuit.

There’s a key point here. Errors are detectable if they are anti-commute with either stabilizer or Z (measure of ancilla qubit) or $Z$ measurement that should result in $+1$ (i.e. $|0\rangle$). If an error leaves the ancilla in $|0\rangle$ and commutes with all stabilizers, it’s a logical error, meaning the state gets changed without being detected. Otherwise, the code detects an error, and the user can attempt recovery.

I have analyzed the effect of each error and found that there are nine single Pauli errors that, when propagated through the entire circuit, lead to a logical error. These are the errors that I need to worry about.

To calculate the fidelity, I must count the number of X/Y errors that lead to a non-recoverable state. I computed the total count of detectable error events, i.e., the state changes or the ancilla changes from $|0\rangle$. After making the calculation, it can be stated that the logical state fidelity $F_L$ is $1 - p/5$.

Answering this research-level problem requires a detailed step-by-step analysis of error propagation through the quantum circuit. We will determine the logical fidelity by calculating the probability of uncorrectable errors that alter the quantum state, conditioned on passing all error detection checks.

Step 1: Define the Code and the Target State

The problem specifies the [[4,2,2]] quantum error detection code with the following properties: - Physical qubits: 0, 1, 2, 3. - Logical qubits: A, B. - Stabilizers: $S_1 = X_0X_1X_2X_3$ and $S_2 = Z_0Z_1Z_2Z_3$. Any valid codeword $|\psi\rangle_L$ must satisfy $S_1 |\psi\rangle_L = |\psi\rangle_L$ and $S_2 |\psi\rangle_L = |\psi\rangle_L$. - Logical operators: - $X_A = X_0 \otimes I_1 \otimes X_2 \otimes I_3$ - $X_B = X_0 \otimes X_1 \otimes I_2 \otimes I_3$ - $Z_A = Z_0 \otimes Z_1 \otimes I_2 \otimes I_3$ - $Z_B = Z_0 \otimes I_1 \otimes Z_2 \otimes I_3$

The target state is the logical $|00\rangle_{AB}$. This state is the unique +1 eigenstate of the stabilizers $S_1, S_2$ and the logical Z operators $Z_A, Z_B$. A state satisfying these conditions is:

\[ |00\rangle_L = \frac{1}{\sqrt{2}} (|0000\rangle + |1111\rangle) \]

Step 2: Analyze the Ideal State Preparation Circuit

The circuit prepares the state $|00\rangle_L$ on qubits 0-3, starting from $|0000\rangle$ and using an ancilla qubit 4, initially in $|0\rangle$. The sequence of operations (right-to-left) is:

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

Let’s trace the evolution of the initial state $|\psi_{in}\rangle = |00000\rangle$:

$H_1$: $|\psi_1\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)$
$CNOT_{12}$: $|\psi_2\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)$
$CNOT_{10}$: $|\psi_3\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11100\rangle)$
$CNOT_{23}$: $|\psi_4\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle) = |00\rangle_L \otimes |0\rangle_4$
$CNOT_{34}$: $|\psi_5\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)$
$CNOT_{04}$: $|\psi_6\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle) = |00\rangle_L \otimes |0\rangle_4$
$M_4$: Measurement of qubit 4. Since qubit 4 is in the state $|0\rangle$, the measurement outcome is ‘0’ with probability 1. The data qubits are left in the desired state $|00\rangle_L$.

Step 3: Error Analysis Framework

We consider a two-qubit depolarizing error channel after each of the 5 CNOT gates. This channel applies one of the 15 non-identity two-qubit Pauli operators with probability $p/15$. We work to first order in $p$, meaning we only consider single error events.

An error event leads to a logical failure if it satisfies three conditions:

Ancilla post-selection: The ancilla qubit must be measured as $|0\rangle$. An error $E$ occurring at some point in the circuit will propagate to a final error operator $E_{final}$ acting on the 5-qubit state just before measurement. The ancilla measurement is not triggered if $E_{final}$ commutes with the ancilla’s Z-operator, $Z_4$. This means $E_{final}$ must not have an $X_4$ or $Y_4$ component.
Code post-selection: The error must not be detected by the code stabilizers. This means the data-qubit part of the final error, $E_{data}$, must commute with both $S_1=XXXX$ and $S_2=ZZZZ$. An error satisfying this condition is, by definition, a logical operator (or a stabilizer).
State change: The logical operator must change the ideal state $|00\rangle_L$. The ideal state is a +1 eigenstate of $Z_A$ and $Z_B$. Therefore, any logical Z-type error ($Z_A, Z_B, Z_A Z_B$) will preserve the state. Only logical X and Y-type errors will change it to an orthogonal state, causing a loss of fidelity.

We will systematically analyze the 15 possible Pauli errors after each of the 5 CNOT gates. Let’s label the gates chronologically from $G_2$ to $G_6$.

Step 4: Detailed Error Propagation and Classification

Let’s find the number of single-error events that cause a logical X/Y-type error, $N_{LE,XY}$, and those that cause a logical Z-type error, $N_{LE,Z}$. We also count errors detected by the ancilla ($N_{det,anc}$), by the code ($N_{det,code}$), and those equivalent to the identity ($N_I$). The total for each gate location must be 15.

A key tool is tracking how Pauli operators are transformed by CNOT gates. For a $CNOT_{c,t}$ gate: - $X_c \to X_c X_t$, $Z_t \to Z_c Z_t$. - $X_t \to X_t$, $Z_c \to Z_c$.

Error after $G_6=CNOT_{04}$: The error $E_{04}$ is not propagated. Ancilla detection fails if $E_{04}$ contains $X_4$ or $Y_4$. This accounts for 8 of the 15 errors. The remaining 7 errors are of the form $P_0 \otimes \{I_4, Z_4\}$. Since the ancilla is in state $|0\rangle_4$, errors with $Z_4$ act as if they were $I_4$. All non-identity data-qubit errors ($X_0, Y_0, Z_0$) are detected by the code stabilizers. The error $I_0 \otimes Z_4$ acts as the identity on the final state.
- $N_{LE,XY}=0, N_{LE,Z}=0$.
Error after $G_5=CNOT_{34}$: An error $E_{34} = P_3 \otimes P_4$ propagates through $CNOT_{04}$. Ancilla detection fails if the propagated error has an $X_4$ or $Y_4$ component. This happens if $E_{34}$ has $X_4$ or $Y_4$ (8 errors). The remaining 7 errors ($P_3 \otimes \{I_4, Z_4\}$) are not detected by the ancilla.
- $E_{34} = P_3 \otimes I_4$: The final error on data is $P_3$. $X_3, Y_3, Z_3$ are all detected by the code.
- $E_{34} = P_3 \otimes Z_4$: The final data error is $Z_0P_3$.
  - $I_3Z_4 \to Z_0$: Detected by code.
  - $X_3Z_4 \to Z_0X_3$: Logical error. It anti-commutes with both $X_A$ and $X_B$, so it’s a $Z_AZ_B$ type error.
  - $Y_3Z_4 \to Z_0Y_3$: Detected by code.
  - $Z_3Z_4 \to Z_0Z_3$: Logical error ($Z_AZ_B$ type).
- $N_{LE,XY}=0, N_{LE,Z}=2$.
Error after $G_4=CNOT_{23}$: An error $E_{23}=P_2 \otimes P_3$ propagates through $CNOT_{34}$ and $CNOT_{04}$. Ancilla detection fails if $P_3$ is $X_3$ or $Y_3$ (8 errors). The remaining 7 errors ($P_2 \otimes \{I_3, Z_3\}$) pass ancilla check.
- $E_{23}=P_2 \otimes I_3$: Data error $P_2$, all detected.
- $E_{23}=P_2 \otimes Z_3$: Data error $Z_0P_2Z_3$.
  - $I_2Z_3 \to Z_0Z_3$: Logical ($Z_AZ_B$ type).
  - $X_2Z_3 \to Z_0X_2Z_3$: Detected.
  - $Y_2Z_3 \to Z_0Y_2Z_3$: Logical. Anti-commutes with $Z_B$, not $Z_A$. X/Y-type.
  - $Z_2Z_3 \to Z_0Z_2Z_3$: Logical ($Z_B$ type).
- $N_{LE,XY}=1, N_{LE,Z}=2$.
Error after $G_3=CNOT_{10}$: An error $E_{10}=P_1 \otimes P_0$ propagates through $CNOT_{23}, CNOT_{34}, CNOT_{04}$. Ancilla detection fails if $P_0$ is $X_0$ or $Y_0$ (8 errors). The remaining 7 errors ($P_1 \otimes \{I_0, Z_0\}$) pass ancilla check.
- $E_{10}=P_1 \otimes I_0$: Data error $P_1$, all detected.
- $E_{10}=P_1 \otimes Z_0$: Data error $Z_0P_1$.
  - $I_1Z_0 \to Z_0$: Detected.
  - $X_1Z_0 \to Z_0X_1$: Detected.
  - $Y_1Z_0 \to Z_0Y_1$: Logical. Anti-commutes with $Z_A$, not $Z_B$. X/Y-type.
  - $Z_1Z_0 \to Z_0Z_1$: Logical ($Z_A$ type).
- $N_{LE,XY}=1, N_{LE,Z}=1$.
Error after $G_2=CNOT_{12}$: An error $E_{12}=P_1 \otimes P_2$ propagates through $CNOT_{10}, CNOT_{23}, CNOT_{34}, CNOT_{04}$. An $X_1$ or $Y_1$ component on $P_1$ creates an $X_0$ term after $CNOT_{10}$, which then triggers ancilla detection after $CNOT_{04}$ (8 errors). The remaining 7 errors ($I_1 \otimes P_2, Z_1 \otimes P_2$) pass ancilla check.
- $E_{12}=I_1 \otimes P_2$: Data error $P_2$, all detected.
- $E_{12}=Z_1 \otimes P_2$: Data error $Z_1P_2$.
  - $Z_1I_2 \to Z_1$: Detected.
  - $Z_1X_2 \to Z_1X_2$: Detected.
  - $Z_1Y_2 \to Z_1Y_2$: Logical. Anti-commutes with $Z_B$, not $Z_A$. X/Y-type.
  - $Z_1Z_2 \to Z_1Z_2$: Logical ($Z_AZ_B$ type).
- $N_{LE,XY}=1, N_{LE,Z}=1$.

Step 5: Calculate Logical Fidelity

We sum the counts of logical errors over all 5 gate locations: - Total Z-type logical errors: $N_{LE,Z} = 0 + 2 + 2 + 1 + 1 = 6$. - Total X/Y-type logical errors: $N_{LE,XY} = 0 + 0 + 1 + 1 + 1 = 3$. - Total logical errors: $N_{LE} = 6 + 3 = 9$.

The logical state fidelity $F_L$ is the probability that the final state, conditioned on passing all checks, is the ideal state $|00\rangle_L$. A Z-type logical error preserves the state $|00\rangle_L$ (since it is a +1 eigenstate of all logical Z operators). An X/Y-type logical error maps $|00\rangle_L$ to an orthogonal logical state. Therefore, the infidelity is the conditional probability of an X/Y-type error occurring.

$1 - F_L = P(\text{X/Y error} | \text{not detected})$

To first order in $p$, this is:

\[ 1 - F_L \approx \frac{P(\text{X/Y error})}{P(\text{not detected})} \]

The total probability of an X/Y-type logical error is:

\[ P(\text{X/Y error}) = N_{LE,XY} \times \frac{p}{15} = 3 \times \frac{p}{15} = \frac{p}{5} \]

The probability of the measurement not being discarded (a “successful” run) is the probability of no error, an identity-equivalent error, or any logical error. To first order, this is $P(\text{not detected}) \approx 1$. A more precise calculation gives: $P(\text{not detected}) = 1 - P(\text{detected error}) = 1 - (N_{det,anc}^{tot} + N_{det,code}^{tot})\frac{p}{15}$. From our analysis, the total number of detected errors is $(5 \times 15 - N_{LE} - N_{I}) = 75 - 9 - 1 = 65$. So, $P(\text{not detected}) \approx 1 - \frac{65}{15}p = 1 - \frac{13}{3}p$.

The infidelity is:

\[ 1 - F_L = \frac{p/5}{1 - \frac{13}{3}p} \approx \frac{p}{5} (1 + \frac{13}{3}p) \approx \frac{p}{5} \]

Therefore, the logical state fidelity to first order in $p$ is:

\[ F_L = 1 - \frac{p}{5} \]

Final Answer: The logical state fidelity of the final 2-qubit logical state is $F_L(p) = 1 - \frac{p}{5} + O(p^2)$.

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 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 - p / 5
    # ---------------------------------------------------------------

    return F_logical