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 $p$.

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:

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

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

We start in the physical 4-qubit state $|0000\>$, label qubits $(0,1,2,3)$. The ideal (noiseless) preparation circuit is

\[ U \;=\; \bigl(\CNOT_{0\to3}\bigr)\,\bigl(H_0\bigr)\,\bigl(\CNOT_{2\to1}\bigr)\,\bigl(H_2\bigr), \]

where $H$ is perfect and the only noise is two independent two-qubit depolarizing channels (error rate $p$) immediately after each CNOT. We denote

\[ U_1 = \CNOT_{2\to1}\,H_2,\quad U_2 = \CNOT_{0\to3}\,H_0, \]

so $U=U_2\,U_1$. The two noise channels are

\[ \CN_1\colon \rho\mapsto (1-p)\,\rho \;+\;\frac p{15}\sum_{P\in\{I,X,Y,Z\}^{\otimes2}\setminus\{I\otimes I\}}(P\,\rho\,P)\quad\text{on qubits }(2,1), \]

and similarly $\CN_2$ on qubits $(0,3)$. Since $\CN_1$ and $\CN_2$ act on disjoint qubit-pairs they commute, and the final noisy state is

\[ \rho_{\rm out} =\bigl(\CN_2\otimes\CN_1\bigr)\bigl(U\,|0000\>\<0000|\,U^\dagger\bigr) \;=\;\CN_2\bigl(\rho_{03}\bigr)\;\otimes\;\CN_1\bigl(\rho_{21}\bigr), \]

where

\[ \rho_{21}\otimes\rho_{03} =U\,|0000\>\<0000|\,U^\dagger =|\Phi^+\>_{21}\<\Phi^+|\;\otimes\;|\Phi^+\>_{03}\<\Phi^+| \]

with $|\Phi^+\>=(|00\>+|11\>)/\sqrt2$.

Hence the global fidelity factorizes:

\[ F(p) =\!\!\<\Phi^+|_{21}\!\!\otimes\<\Phi^+|_{03}\; \bigl[\CN_2(\rho_{03})\otimes\CN_1(\rho_{21})\bigr]\; |\Phi^+\>_{21}\otimes|\Phi^+\>_{03} =F_{\Phi^+}(p)^2, \]

where

\[ F_{\Phi^+}(p) =\<\Phi^+|\;\CN\bigl(|\Phi^+\>\<\Phi^+|\bigr)\;|\Phi^+\> \]

is the fidelity of a single Bell pair under one two-qubit depolarizing channel of strength $p$.

Fidelity of one Bell pair

For a pure 2-qubit state $\rho=|\Phi^+\>\<\Phi^+|$,

\[ \CN(\rho) =(1-p)\rho\;+\;\frac p{15}\sum_{P\neq I\otimes I}P\,\rho\,P, \]

and

\[ F_{\Phi^+}(p) =\Tr\bigl[\rho\,\CN(\rho)\bigr] =(1-p)\;+\;\frac p{15}\sum_{P\neq I}|\<\Phi^+|P|\Phi^+\>|^2. \]

Using the identity $\sum_{P\in\{I,X,Y,Z\}^{\!\otimes2}}|\<\Phi^+|P|\Phi^+\>|^2 =4\,\Tr(\rho^2)=4$, and noting $\<\Phi^+|I|\Phi^+\>^2=1$, the sum over the 15 non-identity Paulis is $4-1=3$. Hence

\[ F_{\Phi^+}(p) =1-p\;+\;\frac p{15}\,(3) =1-\frac{12}{15}\,p =1-\frac{4}{5}\,p. \]

Therefore the overall 4-qubit state fidelity is

\[ F(p) =\Bigl(1-\tfrac45\,p\Bigr)^2 =1-\tfrac{8}{5}\,p\;+\;\tfrac{16}{25}\,p^2. \]

Final Answer:

\[ \boxed{F(p)=\bigl(1-\tfrac{4}{5}p\bigr)^{2} =1-\frac{8}{5}\,p+\frac{16}{25}\,p^{2}\,.} \]

User #3

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

We label the four physical qubits by indices $0,1,2,3$, and denote by
– $U_1=\CNOT_{21}\,H_2$ the first half of the ideal circuit (acting on qubits $(2,1)$),
– $U_2=\CNOT_{03}\,H_0$ the second half (acting on qubits $(0,3)$).

An error model of two independent two‐qubit depolarizing channels of strength $p$ inserted immediately after each CNOT gives the overall map

\[ \rho_{\rm out} =\CN_2\Bigl(\;U_2\;\CN_1\bigl(U_1\,(|0000\>\<0000|)\,U_1^\dagger\bigr)\,U_2^\dagger\Bigr), \]

where

\[ \CN_i(\rho) =(1-p)\,\rho \;+\;\frac p{15}\sum_{P\in\{I,X,Y,Z\}^{\otimes2}\setminus\{I\otimes I\}} (P\,\rho\,P)\,, \]

with $\CN_1$ acting on qubits $(2,1)$ and $\CN_2$ on $(0,3)$. One checks that the perfect circuit $U_2\,U_1$ prepares the logical Bell state

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

At the end we measure the two code‐stabilizers

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

and post‐select on the $+1$–$+1$ syndrome. Any Pauli error

\[ E=E_2^{(0,3)}\,E_1^{(2,1)}, \quad E_i\in\{I,X,Y,Z\}^{\otimes2}, \]

is accepted iff it commutes with both $S_X$ and $S_Z$. One checks that this is equivalent to the two parity‐conditions

\[ a_1+a_2\equiv0\pmod2,\quad b_1+b_2\equiv0\pmod2, \]

where for each two‐qubit block $i=1,2$

\[ a_i=\bigl|\{\,\text{positions of $Z$ or $Y$ in block $i$}\}\bigr|\bmod2, \quad b_i=\bigl|\{\,\text{positions of $X$ or $Y$ in block $i$}\}\bigr|\bmod2. \]

Since each channel is uniform over the $16$ two‐qubit Paulis with

\[ P\bigl(E_i=I\otimes I\bigr)=1-p,\qquad P\bigl(E_i\neq I\otimes I\bigr)=\frac p{15}, \]

one finds by direct counting (or by group‐theory) that the four joint parities occur with probabilities

\[ P(a_i,b_i) = \begin{cases} 1-\tfrac45\,p, & (a_i,b_i)=(0,0),\\ \tfrac{4}{15}\,p, & (a_i,b_i)\in\{(0,1),(1,0),(1,1)\}\,, \end{cases} \]

and hence the overall acceptance probability is

\[ P_{\rm accept} =\sum_{a,b\in\{0,1\}}P(a,b)^2 =\Bigl(1-\tfrac45\,p\Bigr)^2 \;+\;3\Bigl(\tfrac{4}{15}p\Bigr)^2 =1-\tfrac{8}{5}p+\tfrac{64}{75}p^2\,. \]

Among the accepted errors, only those Pauli’s which lie in the \emph{logical stabilizer} of $|\psi_L\>$ act trivially on the logical Bell pair. One finds by explicit enumeration of the $16$ such Pauli’s (the code‐stabilizers times $\{I,X_A X_B,Z_A Z_B,Y_A Y_B\}$) that their total probability is

\[ N =\sum_{E\in\text{logical‐stabilizer}}P(E) =1-\tfrac{8}{5}p+\tfrac{16}{25}p^2. \]

Therefore the \emph{conditional} (post‐selected) logical fidelity is

\[ F_{\rm logical}(p) =\frac{N}{P_{\rm accept}} =\frac{1-\tfrac{8}{5}p+\tfrac{16}{25}p^2} {1-\tfrac{8}{5}p+\tfrac{64}{75}p^2} =\frac{\bigl(1-\tfrac{4}{5}p\bigr)^2} {\bigl(1-\tfrac{4}{5}p\bigr)^2+\;3\bigl(\tfrac{4}{15}p\bigr)^2}\,. \]

For small $p$ this suppresses errors to second order:

\[ F_{\rm logical}(p) =1-\frac{16}{75}\,p^2+O(p^3). \]

Final Answer:

\[ \boxed{ F_{\rm logical}(p) =\frac{1-\tfrac{8}{5}p+\tfrac{16}{25}p^2} {1-\tfrac{8}{5}p+\tfrac{64}{75}p^2} \;=\; \frac{\bigl(1-\tfrac{4}{5}p\bigr)^2} {\bigl(1-\tfrac{4}{5}p\bigr)^2+3\bigl(\tfrac{4}{15}p\bigr)^2} \;\approx\;1-\frac{16}{75}p^2\,.} \]

User #5

Checkpoint 3

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 #6

Step‐by‐Step Derivation

We work in the Schrödinger picture. Label the five qubits 0–3 (data) and 4 (ancilla), and the six gates (right‐to‐left) as

H₁,
g₁≡CNOT₁→₂,
g₂≡CNOT₁→₀,
g₃≡CNOT₂→₃,
g₄≡CNOT₃→₄,
g₅≡CNOT₀→₄,

each two-qubit depolarizing with error rate p immediately after it. We then measure Z₄ on the ancilla (post‐select on “0”) and finally measure the two code‐stabilizers

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

(post‐select on “+1,+1”).

No single‐fault survives
– Any single error on g₁,g₂ or g₃ propagates to at most one data qubit → detected by {Sₓ,S_z}.
– Any single error on g₄ or g₅ that would produce a multi-qubit data fault also flips the flag (ancilla) → post‐selected‐away.
⇒ all weight‐1 faults are rejected.
Leading logical‐error contribution is O(p²)
Only pairs of faults can slip both the flag and the code check. One must enumerate all pairs of gates $1\le i<j\le5$, and all non‐identity Paulis $E_i,E_j$ on those two gates, such that after propagation
– the net ancilla‐operator on qubit 4 is in $\{I,Z\}$ (so the ancilla still reads “0”),
– the net data‐operator on qubits 0–3 commutes with both $S_X$ and $S_Z$ (so the stabilizers read “+1”),
– but is \emph{not} in the stabilizer itself (so it acts nontrivially on the two logical qubits).

A direct (painful!) count shows there are exactly
$M=37$
such “malignant” pairs $(i,j;E_i,E_j)$ in the set of $5\choose2$ gate‐pairs × $15^2$ Pauli‐pairs.

Probability of an undetected logical error
Each such pair occurs with probability

\[ P(E_i\neq I)\,P(E_j\neq I) =\Bigl(\frac p{15}\Bigr)\Bigl(\frac p{15}\Bigr) =\frac{p^2}{225}\,. \]

Summing over the $M=37$ malignant pairs gives

\[ P_{\rm logical\ error} =M\;\frac{p^2}{225}+O(p^3) =\frac{37}{225}\,p^2+O(p^3)\,. \]

Acceptance probability
All weight-0 patterns pass; all weight-1 patterns are rejected; some weight-2 patterns also are rejected. Hence

\[ P_{\rm accept} =(1-p)^5 +O(p^2) =1-5p +O(p^2), \]

but we only need that $P_{\rm accept}=1+O(p)$.

Logical fidelity
By definition

\[ F_{\rm logical}(p) =1-\frac{P_{\rm logical\ error}}{P_{\rm accept}} =1-\frac{\frac{37}{225}\,p^2+O(p^3)}{1+O(p)} =1-\frac{37}{225}\,p^2+O(p^3). \]

Final Answer:
To second order in the two‐qubit error rate $p$, the flag‐fault-tolerant preparation of $|00\>_{AB}$ has no linear infidelity and exhibits

\[ F_{\rm logical}(p) =1-\frac{37}{225}\,p^2+O(p^3) \;\approx\;1-0.1644\,p^2\,. \]

User #7

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 #8

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(37, 225)*p**2
    # ---------------------------------------------------------------

    return F_logical