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

Step-by-Step Derivation

Ideal (error–free) preparation
• The circuit

\[ \bigl(CNOT_{03}\bigr)\, H_{0}\, \bigl(CNOT_{21}\bigr)\, H_{2} \]

 acting on the initial product state $|0000\rangle_{0123}$ produces

\[ \]

 (Qubit ordering is $0,1,2,3$.)  
 The encoded state $|\psi_{\text{ideal}}\rangle$ lies in the
 $+1$–eigenspace of the stabilizers $XXXX$ and $ZZZZ$ and represents the logical
 two-qubit GHZ state  
 $(|00\rangle_{AB}+|11\rangle_{AB})/\sqrt2$.

Error model for a single two–qubit gate
• After every $CNOT_{ij}$ we apply an independent two-qubit depolarising channel on the same pair $(i,j)$:

\[ (1-p)\,\rho+\frac{p}{15}\sum_{P\in\mathcal P_{ij}\setminus\{II\}}\! \]

 where $\mathcal P_{ij}$ is the $4\times4$ two-qubit Pauli group on
 qubits $i$ and $j$.

• Only the $CNOT_{21}$ and $CNOT_{03}$ gates are noisy, so two such channels act in the whole circuit.

Fidelity after one noisy two-qubit gate
Let $\rho=|\psi\rangle\langle\psi|$ be the input to the channel and $|\psi\rangle$ any pure state. The output state is $\rho'=\mathcal D_{ij}(\rho)$, and its fidelity with $\rho$ is

\[ =(1-p)+\frac{p}{15}\sum_{P\neq II} \]

For the particular state $|\psi_{\text{ideal}}\rangle$ we must evaluate the overlaps $\langle\psi_{\text{ideal}}|P|\psi_{\text{ideal}}\rangle$ for the 15 non-identity Paulis acting on the relevant two qubits.

• Qubits 2 & 1 (after $CNOT_{21}$)
Among the 15 Paulis on qubits 2 and 1 only the three operators

\[ XX_{21},\;YY_{21},\;ZZ_{21} \]

 leave $|\psi_{\text{ideal}}\rangle$ invariant up to a global phase
 (all others map the state to an orthogonal subspace).
 Hence

\[ \]

• Qubits 0 & 3 (after $CNOT_{03}$)
Exactly the same situation holds; the only invariants are

\[ XX_{03},\;YY_{03},\;ZZ_{03}, \]

 so the sum of squared overlaps is again 3.

Therefore, for either noisy $CNOT$ gate,

\[ =(1-p)+\frac{p}{15}\times 3 \]

Two independent noisy $CNOT$ gates
The two noise channels act on disjoint pairs of qubits and are statistically independent.
Writing $r\equiv 1-\dfrac{4p}{5}$ for the fidelity retention factor of one gate, the overall process is

\[ \]

The final fidelity with the ideal state is then

\[ =\operatorname{Tr}\!\bigl(|\psi_{\text{ideal}}\rangle\langle\psi_{\text{ideal}}| \,\rho_{\text{final}}\bigr) =r^{2} \]

Expanding,

\[ F_{\text{phys}}(p)=1-\frac{8p}{5}+\frac{16p^{2}}{25}. \]

Conventions and units: All quantities are dimensionless probabilities. No numerical approximation was required.

Final Answer:
The physical four-qubit state fidelity after the complete preparation circuit is

\[ \]

User #3

The correct answer is: $ F_{\rm{physical}}=(1-\frac{12}{15}p)^2 $

Now let’s solve the next Checkpoint.

Checkpoint 2

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:

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

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

Assistant #4

Step-by-Step Derivation

Error filtering by the code
• After each $CNOT$ a two–qubit depolarising channel acts on the same two physical qubits.
• We subsequently measure the two stabilisers

\[ S_1=XXXX,\qquad S_2=ZZZZ \]

   and keep the run only when the $+1$ syndrome is obtained (“post-
   selection on all detectable errors’’).

•  For a two–qubit depolarising channel the probabilities on a single
   gate are  
   – identity (no error)                $1-p$  
   – one of 15 non-identity Paulis      $p/15$ each.

   Of those 15 Paulis, exactly the three

\[ \{\,XX,\;YY,\;ZZ\,\} \]

   commute with both stabilisers on that pair of qubits and are therefore
   {\it undetectable}; the remaining 12 are detected and the
   corresponding runs are discarded.

Undetectable errors and their action in the logical subspace
• For the $CNOT_{21}$ gate (qubits 2 & 1) we have

\[ XX_{21}=X_2X_1=X_A X_B, \]

\[ ZZ_{21}=Z_2Z_1=Z_A Z_B, \]

\[ YY_{21}=Y_2Y_1=i^2\,X_A X_B Z_A Z_B=Y_A Y_B, \]

   up to an irrelevant overall phase.

•  For the $CNOT_{03}$ gate (qubits 0 & 3) the undetectable Paulis are
   related to the same logical operators by a stabiliser,
   e.g.

\[ \]

   and similarly for $YY_{03}$.  
   Multiplying by a stabiliser does not change the encoded state.

Logical GHZ state is an eigenstate of the surviving logical Paulis
The target logical state

\[ |\text{GHZ}\rangle_{AB}=\frac{|00\rangle_{AB}+|11\rangle_{AB}}{\sqrt2} \]

is simultaneously a $+1$ eigenstate of
$X_A X_B,\;Z_A Z_B$, and of their product $Y_A Y_B$:

\[ Z_A Z_B|\text{GHZ}\rangle=|\text{GHZ}\rangle,\; \]

where the $-$ sign for $Y_A Y_B$ is a global phase and hence physically
irrelevant.  Therefore {\it every undetectable error that survives
post-selection acts only by (at most) a global phase on the logical
state}.  The encoded state is unchanged.

Logical fidelity conditioned on successful post-selection
Because the post-selection discards all detectable errors and every remaining (undetectable) error leaves the logical state invariant, the conditional logical density operator equals the ideal projector:

\[ \rho_{\text{logical,\,post}}=|\text{GHZ}\rangle\langle\text{GHZ}|. \]

Consequently the logical state fidelity is unity for any physical error
rate $p$,

\[ F_{\text{logical}}(p)=\;1. \]

Conventions and units: Dimensionless probabilities; no numerical approximation required.

Final Answer:
After discarding all runs in which the [[4,2,2]] code detects an error, the 2-qubit logical state is {\it always} the intended
$\bigl(|00\rangle_{AB}+|11\rangle_{AB}\bigr)/\sqrt2$. Hence

\[ F_{\rm logical}(p)=1\qquad\text{for all }p. \]

User #5

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

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

    Outputs
    ----------
    F_logical: sympy.Expr, the logical state fidelity as a function of $p$
    """

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

    return F_logical

Assistant #6

import sympy as sp

p = sp.symbols('p')

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

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

    Outputs
    ----------
    F_logical: sympy.Expr, the logical state fidelity as a function of $p$
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_logical = sp.Integer(1)  # a SymPy expression of inputs
    # ---------------------------------------------------------------

    return F_logical