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

```markdown Step-by-Step Derivation

Ideal encoded GHZ state in the [[4,2,2]] code.
The circuit

\[ |0000\rangle \xrightarrow{H_2} \xrightarrow{\rm CNOT_{2\to1}} \xrightarrow{H_0} \xrightarrow{\rm CNOT_{0\to3}} |\,\psi\>\;=\;\frac{1}{2}\bigl(|0000\> + |0110\> + |1001\> + |1111\>\bigr) \]

prepares the logical GHZ

\[ \bigl(|00\>_{AB}+|11\>_{AB}\bigr)/\sqrt2 \;=\;\frac1{2}\bigl(|0000\>+|0110\>+|1001\>+|1111\>\bigr)\,. \]

Error model.
Each CNOT gate is followed by an i.i.d.\ two-qubit depolarizing channel

\[ \mathcal D(\rho)=(1-p)\,\rho\;+\;\frac{p}{15}\sum_{E\neq I\otimes I}E\,\rho\,E\,. \]

The Hadamards are assumed perfect.

Final noisy state as a Pauli‐mixture.
Let the first faulty gate (CNOT$_{2\to1}$) insert error $E_1$ on qubits $\{1,2\}$, drawn from

\[ \Pr[E_1=I]=1-p,\quad \Pr[E_1=P]=\tfrac p{15}\quad(P\in\{X,Y,Z\}^{\otimes2}\setminus\{II\}). \]

Likewise the second faulty gate (CNOT$_{0\to3}$) inserts $E_2$ on qubits $\{0,3\}$ with the same law.

Fidelity as sum over “harmless” errors.
The final density matrix is

\[ \rho=\sum_{E_1,E_2}\Pr[E_1]\Pr[E_2]\;\bigl(E_2\,U\,E_1\,V\,|0000\>\<0000|\,V^\dagger\,E_1\,U^\dagger\,E_2\bigr)\,, \]

where $V=H_2$, $U={\rm CNOT}_{2\to1}\,H_0\,{\rm CNOT}_{0\to3}$. Since the second half of the circuit, $H_0$ and CNOT$_{0\to3}$, acts only on qubits $\{0,3\}$, it commutes with any Pauli on $\{1,2\}$. Hence

\[ E_1\;\xrightarrow{\;U\;} \;E_1,\qquad E_2\;\xrightarrow{\;\cdot\;} \;E_2, \]

and the net error before comparison with $|\psi\>$ is $E=E_2\otimes E_1$.

Which errors leave $|\psi\>$ invariant?
The stabilizer group of $|\psi\>$ (the logical‐GHZ) is generated by the four independent Pauli stabilizers

\[ S_1=X_0X_1X_2X_3,\quad S_2=Z_0Z_1Z_2Z_3,\quad S_3=X_1X_2,\quad S_4=Z_1Z_2. \]

One checks that the entire stabilizer group is exactly

\[ G=\bigl\{A_{0,3}\otimes B_{1,2}\,\bigm|\,A,B\in\{II,\;XX,\;ZZ,\;YY\}\bigr\}. \]

Any $E\in G$ satisfies $E\,|\psi\>=\pm|\psi\>$, so contributes unit fidelity overlap.

Compute the fidelity.
Since $E_2$ lives on $\{0,3\}$ and $E_1$ on $\{1,2\}$, they are independent. The total fidelity is

\[ F(p) =\sum_{\substack{E_2\in\{II,\,XX,\,YY,\,ZZ\}\\E_1\in\{II,\,XX,\,YY,\,ZZ\}}} \Pr[E_2]\,\Pr[E_1] \;=\;\Bigl(\Pr[II]+\!3\,\tfrac p{15}\Bigr)^2 =\Bigl((1-p)+3\frac p{15}\Bigr)^2 =\bigl(1-\tfrac{4p}{5}\bigr)^2. \]

Final Answer:

\[ \boxed{F_{\rm phys}(p)\;=\;\bigl(1 - \tfrac{4}{5}\,p\bigr)^2\,.} \]

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

```markdown Step-by-Step Derivation

Error propagation and final Pauli errors
The circuit

\[ |0000\rangle \xrightarrow{H_2} \xrightarrow{\rm CNOT_{2\to1}} \xrightarrow{H_0} \xrightarrow{\rm CNOT_{0\to3}} \;|\psi\>\;=\;\tfrac1{2}\bigl(|0000\>+|0110\>+|1001\>+|1111\>\bigr) \]

is followed by independent two‐qubit depolarizing errors on $\{2,1\}$ (after CNOT$_{2\to1}$) and on $\{0,3\}$ (after CNOT$_{0\to3}$).
Denote these Pauli errors by

\[ E_1\in\{\!I,X,Y,Z\}^{\otimes2}\quad\text{on qubits 2,1}, \]

\[ E_2\in\{\!I,X,Y,Z\}^{\otimes2}\quad\text{on qubits 0,3}, \]

with

\[ \Pr[E_i=II]=1-p,\;\Pr[E_i=P\neq II]=\tfrac p{15}. \]

Because $H_0$ and CNOT$_{0\to3}$ act only on $\{0,3\}$, $E_1$ and $E_2$ propagate trivially to the end and remain on disjoint qubit‐pairs. Thus the total Pauli error at the end is

\[ E = E_2\otimes E_1. \]

Post‐selection (error detection)
We post‐select on “no detectable error,” i.e.\ we keep only those runs for which the syndrome of the code’s stabilizers

\[ S_X = X_0X_1X_2X_3,\quad S_Z = Z_0Z_1Z_2Z_3 \]

is trivial. A Pauli $E$ is undetected exactly if $[E,S_X]=[E,S_Z]=0$. Since

\[ [E_2\otimes E_1,\;S_X]=0 \;\Longleftrightarrow\; [E_2,\;X_0X_3]=0 \quad\text{and}\quad [E_1,\;X_1X_2]=0, \]

and similarly for $S_Z$, one finds that the undetectable errors on each pair are

\[ \{II,\;X\,X,\;Y\,Y,\;Z\,Z\} \]

(on the respective qubit‐pair). Hence the total probability to pass post‐selection is \begin{align} P_{\rm accept} &=\biggl[\sum_{E_2\in{II,XX,YY,ZZ}}\Pr[E_2]\biggr]\, \biggl[\sum_{E_1\in{II,XX,YY,ZZ}}\Pr[E_1]\biggr]\ &=\Bigl((1-p)+3\frac p{15}\Bigr)^2 =\Bigl(1-\tfrac{12}{15}p\Bigr)^2 =\Bigl(1-\tfrac{4}{5}p\Bigr)^2. \end{align}

Logical errors vs.\ stabilizer errors
Among the undetectable errors, those that lie in the stabilizer group of the code

\[ G=\{I,\;X_0X_1X_2X_3,\;Z_0Z_1Z_2Z_3,\;Y_0Y_1Y_2Y_3\} \]

act trivially on the logical qubits and therefore do not harm the logical GHZ. In terms of $(E_2,E_1)$ these four are

\[ (II,II),\;(XX,XX),\;(ZZ,ZZ),\;(YY,YY). \]

Their total probability is

\[ P_{\rm stab} =\Pr[E_2=II]\Pr[E_1=II] +3\Bigl(\tfrac p{15}\Bigr)^2 =(1-p)^2+3\,\frac{p^2}{225}. \]

Conditional (logical) fidelity
The logical fidelity given that no detectable error occurred is

\[ F_{\rm logical}(p) =\frac{P_{\rm stab}}{P_{\rm accept}} =\frac{(1-p)^2 \;+\;3\,\dfrac{p^2}{225}} {\bigl(1-\tfrac{4}{5}p\bigr)^2} =\frac{(1-p)^2 + \tfrac{3}{225}\,p^2} {\bigl(1-\tfrac{4}{5}p\bigr)^2}. \]

Final Answer:

\[ F_{\rm logical}(p) =\frac{(1-p)^2 \;+\;\tfrac{3}{225}\,p^2} {\bigl(1-\tfrac{4}{5}p\bigr)^2}\,. \]

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 = ((1 - p)**2 + sp.Rational(3, 225)*p**2) / (1 - sp.Rational(4, 5)*p)**2
    # ---------------------------------------------------------------

    return F_logical