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:

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

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

Solution

We wish to prepare the logical GHZ state

\[ \bigl(|00\>\_{AB}+|11\>\_{AB}\bigr)/\sqrt2 \]

in the [[4,2,2]] code by the circuit

\[ \bigl(CNOT_{03}\bigr)\,\bigl(H_0\bigr)\,\bigl(CNOT_{21}\bigr)\,\bigl(H_2\bigr) \]

acting on the physical initial state $|0000\>$. Between each $CNOT$ and the next gate, we insert the two‐qubit depolarizing channel on its two qubit support.

1. Two‐qubit depolarizing channel

On qubits $(i,j)$, the error channel after $CNOT_{ij}$ is

\[ \mathcal D_{ij}(\rho) \;=\;(1-p)\,\rho \;+\;\frac p{15}\sum_{\substack{P\in\{I,X,Y,Z\}^{\!\otimes2}\\P\neq I\otimes I}} \bigl(P_{ij}\,\rho\,P_{ij}^\dagger\bigr)\,, \]

where $p$ is the gate’s “error probability.”

2. Ideal intermediate state after $H_2$ and $CNOT_{21}$

Start with $\rho_0=|0000\>\<0000|$. Apply

\[ H_2:\;|0\>_2\mapsto\frac{|0\>+|1\>}{\sqrt2}, \qquad CNOT_{21}:\;|x\>_2|y\>_1\mapsto|x\>_2\,|y\oplus x\>_1. \]

One finds the pure state

\[ |\psi_2\> \;=\; \frac1{\sqrt2}\bigl(|0000\>+|0110\>\bigr) \;=\; |00\>_{0,3}\;\otimes\;\frac{|00\>+|11\>}{\sqrt2}_{1,2}. \]

In particular, qubits $1,2$ carry the 2‐qubit GHZ state $\bigl(|00\>+|11\>\bigr)/\sqrt2$, while qubits $0,3$ remain in $|00\>$.

3. Fidelity loss from the first depolarizing channel

We now apply $\mathcal D_{21}$ on qubits $(2,1)$. Denote

\[ \rho_3=\mathcal D_{21}(|\psi_2\>\<\psi_2|). \]

The fidelity with the ideal $|\psi_2\>$ is

\[ F_1 =\<\psi_2|\rho_3|\psi_2\> =(1-p)\;+\;\frac p{15} \sum_{P\neq I\otimes I}\bigl|\<\psi_2|\,P_{21}\otimes I_{0,3}\,|\psi_2\>\bigr|^2. \]

But on the GHZ $\tfrac{|00\>+|11\>}{\sqrt2}$ of qubits $(1,2)$, exactly three non‐identity two‐qubit Paulis preserve the state (namely $X\otimes X,\;Y\otimes Y,\;Z\otimes Z$), each giving overlap magnitude 1, while all others give zero overlap. Hence

\[ \sum_{P\neq I\otimes I}\bigl|\<\psi_2|P_{21}|\psi_2\>\bigr|^2 =3, \]

and

\[ F_1 =(1-p)+\frac p{15}\,3 =1-p+\frac p5 =1-\frac{4p}5. \]

4. Remaining gates and second depolarizing channel

Next we apply the noiseless gates $H_0$ and $CNOT_{03}$. Since these unitaries act only on qubits $(0,3)$, they leave the (noisy) reduced state on $(1,2)$ unchanged and moreover map the ideal target to

\[ |\psi_{\rm id}\> \;=\; \frac1{2} \Bigl(|0000\>+|0110\>+|1001\>+|1111\>\Bigr) \;=\; \frac{|00\>+|11\>}{\sqrt2}_{1,2} \;\otimes\; \frac{|00\>+|11\>}{\sqrt2}_{0,3}. \]

Thus the final pure target factorizes into a GHZ on $(1,2)$ times a GHZ on $(0,3)$. We then apply $\mathcal D_{03}$ on qubits $(0,3)$. By exactly the same argument as above, this second depolarizing channel reduces the fidelity of the $(0,3)$‐GHZ by the same factor

\[ F_2 =1-\frac{4p}5. \]

5. Final physical fidelity

Because

the first noisy channel acted only on qubits $(1,2)$ preparing the $(1,2)$‐GHZ,
after that the state factorized across the two pairs,
the final channel acted only on the disjoint pair $(0,3)$,
and unitaries in between do not change state fidelity,

the overall fidelity is simply the product

\[ F_{\rm phys}(p) \;=\;F_1\,F_2 \;=\;\Bigl(1-\tfrac{4p}5\Bigr)\!\times\!\Bigl(1-\tfrac{4p}5\Bigr) \;=\;\Bigl(1-\tfrac{4p}5\Bigr)^2. \]

Final Answer:

\[ \boxed{F_{\rm phys}(p)=\Bigl(1-\tfrac{4p}{5}\Bigr)^2}. \]

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

Step-by-Step Derivation

We wish to prepare the logical two-qubit GHZ state

\[ \lvert\psi_L\>=\frac{\lvert00\>_{AB}+\lvert11\>_{AB}}{\sqrt2} \]

in the $[[4,2,2]]$ code via

\[ \bigl(CNOT_{03}\bigr)\;\bigl(H_0\bigr)\;\bigl(CNOT_{21}\bigr)\;\bigl(H_2\bigr) \]

with two independent two-qubit depolarizing errors (rate $p$) acting immediately after each CNOT. We then measure the two stabilizers $\;S_X=X^{\otimes4},\;S_Z=Z^{\otimes4}$, post-select on the “$+1,+1$” outcome, and ask for the fidelity of the logical 2-qubit state (i.e.\ the overlap with $\lvert\psi_L\>$) in that post-selected ensemble.

Error model and final error operator
• After $CNOT_{21}$ an error $E_{21}\in\{I\otimes I,P\}/\{P\neq I\}$ occurs on qubits $(2,1)$, with

\[ \Pr[E_{21}=I\otimes I]=1-p,\quad \Pr[E_{21}=P]=\frac p{15}\;(P\neq I\otimes I). \]

• After $CNOT_{03}$ an independent error $E_{03}$ of the same form acts on qubits $(0,3)$.
• Because all intervening gates either act on $(0,3)$ or on $(2,1)$, the two errors commute and never spread between the two pairs. Hence the total physical error on the final codeword is

\[ E \;=\; E_{03}\,\otimes\,E_{21}\,, \]

acting on disjoint qubit‐pairs $(0,3)$ and $(2,1)$.

Post-selection condition (“undetectable” errors)
We measure $S_X=X^{\otimes4},\;S_Z=Z^{\otimes4}$ on the final 4-qubit state and keep only the “$+1,+1$” outcomes. Equivalently, we keep only those Pauli errors $E$ which commute with both $S_X$ and $S_Z$. The set of all Pauli operators that commute with every stabilizer is the normalizer $N$. Among these, the stabilizer subgroup $S\subset N$ acts trivially on the code space (i.e.\ no logical error), while the remaining $N\setminus S$ are non-trivial logical errors.

Thus, if we let

\[ \mathcal{P}_2=\{\,\text{all 2-qubit Paulis}\} \quad\text{and}\quad q(P)= \begin{cases} 1-p,&P=I\otimes I,\\ \tfrac p{15},&P\neq I\otimes I, \end{cases} \]

then the joint law is $\Pr[E_{21}=P_{21},E_{03}=P_{03}]=q(P_{21})\,q(P_{03})$. We define

• Denominator (probability undetected):

\[ D =\sum_{\substack{P_{21},P_{03}\\(P_{21}\otimes P_{03})\in N}} q(P_{21})\,q(P_{03})\,. \]

• Numerator (probability undetected & no logical error):

\[ N =\sum_{\substack{P_{21},P_{03}\\(P_{21}\otimes P_{03})\in S}} q(P_{21})\,q(P_{03})\,. \]

Then the logical fidelity (post-selected on no syndrome) is

\[ F_{\rm logical} =\frac{N}{D}\,. \]

Characterizing $N$ vs.\ $S$ for disjoint pairs
The stabilizer group of the $[[4,2,2]]$ code is generated by

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

and has four elements $\{I,\,X^{\otimes4},\,Z^{\otimes4},\,Y^{\otimes4}\}$.
Because $E=E_{03}\otimes E_{21}$ acts on disjoint pairs, $E\in S$ if and only if

\[ E_{03}\otimes E_{21}\;\in\;\{\,I,I\},\;\{X_0X_3,X_1X_2\},\;\{Z_0Z_3,Z_1Z_2\},\;\{Y_0Y_3,Y_1Y_2\}. \]

Hence there are exactly four joint events in the numerator:

\[ \bigl(P_{03},P_{21}\bigr) = (I\!,I),\;(X\!X,X\!X),\;(Z\!Z,Z\!Z),\;(Y\!Y,Y\!Y), \]

each occurring with probability

\[ q(P)^2 = \begin{cases} (1-p)^2,&P=I\!,\\ \bigl(\tfrac p{15}\bigr)^2,&P\neq I. \end{cases} \]

Thus

\[ N =(1-p)^2 \;+\;3\;\Bigl(\frac p{15}\Bigr)^{\!2} =1-2p+p^2\;+\;\frac{3p^2}{225} =1-2p+\frac{76\,p^2}{75}\,. \]

Computing $D$ by parity classes
A Pauli $P\in\mathcal P_4$ commutes with $S_X$ iff it has an even number of factors in $\{Y,Z\}$, and commutes with $S_Z$ iff it has an even number of factors in $\{X,Y\}$. Restricting to each 2-qubit block $\mathcal P_2$, these two parity conditions split the 16 two-qubit Paulis into four equally-sized “parity classes” of size 4. Label these classes by $(r,s)\in\{0,1\}^2$. Denote

\[ p_{r,s}\;=\;\sum_{\substack{P\in\mathcal P_2\\\text{class}(P)=(r,s)}} q(P)\,. \]

By inspection one finds

\[ p_{0,0}=(1-p)+3\,\frac{p}{15} =1-\frac{4p}{5}, \quad p_{0,1}=p_{1,0}=p_{1,1} =4\,\frac{p}{15} =\frac{4p}{15}\,. \]

Since $E_{03},E_{21}$ must lie in the same parity class in order for $E_{03}\otimes E_{21}\in N$, the total undetected probability is

\[ D =\sum_{(r,s)\in\{0,1\}^2} p_{r,s}^{\,2} =\Bigl(1-\tfrac{4p}{5}\Bigr)^2 +3\;\Bigl(\tfrac{4p}{15}\Bigr)^2 =1-\frac{8p}{5}+\frac{64\,p^2}{75}\,. \]

Assembling the logical fidelity

\[ F_{\rm logical} =\frac{N}{D} =\frac{\,1-2p+\tfrac{76\,p^2}{75}\,} {\,1-\tfrac{8p}{5}+\tfrac{64\,p^2}{75}\,} \;=\; \frac{75-150p+76p^2}{75-120p+64p^2}\,. \]

Final Answer:

\[ \boxed{ F_{\rm logical}(p) = \frac{\,1-2p+\tfrac{76}{75}\,p^2\,} {\,1-\tfrac{8}{5}\,p+\tfrac{64}{75}\,p^2\,} \;=\; \frac{75-150p+76\,p^2}{75-120p+64\,p^2}\,.} \]

User #5

The correct answer is: $ F_{\rm{logical}}=1 - \frac{\frac{16}{75}p^2}{1-\frac{8}{5}p + \frac{64}{75}p^2} $

Now let’s solve the next Checkpoint.

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 label the five two‐qubit CNOT gates (in time‐order, just before measurement) as
g₂ = CNOT₁₂, g₃ = CNOT₁₀, g₄ = CNOT₂₃, g₅ = CNOT₃₄, g₆ = CNOT₀₄,
each followed by an i.i.d. two‐qubit depolarizing error with probability p. We then measure the ancilla (qubit 4) in the Z–basis and post‐select on “0,” and (separately) measure the two code stabilizers S_X and S_Z on the data qubits 0–3 and post‐select on “+1” for both.

By fault‐tolerance of this flagged scheme, every single‐gate fault (any one gi failing) either (i) is caught by the ancilla (ancilla flip) or (ii) yields a single–data‐qubit error which the code–syndrome will detect. Hence no logical error survives if there is only one fault. The leading surviving logical‐error events come from two faults whose combined Pauli on the data (after propagation through the remaining CNOTs)

(a) commutes with both stabilizers (so is undetected by S_X,S_Z)
(b) acts as a nontrivial logical (i.e. lies in N∖S)
(c) induces no X–flip on qubit 4 (so ancilla Z–measurement still reads “0”).

One checks by explicit propagation that there are exactly 8 distinct two‐fault patterns $(g_i,g_j)$ (with $2\le i<j\le6$)—each involving a $\;ZZ$-error on the two data qubits of one data‐only CNOT and a single $\;Z$-error on the ancilla‐gate CNOT—that slip both detection checks and implement one of the four weight‐2 logical Z–operators. Since each two‐fault pattern occurs with probability

\[ \Bigl(\frac p{15}\Bigr)\times\Bigl(\frac p{15}\Bigr) \;=\;\frac{p^2}{225}\,, \]

the total logical‐error probability given acceptance (no flag, no syndrome) to leading nonzero order is

\[ P_{\rm err\;\&\;accept} \;=\; 8\;\frac{p^2}{225} \;+\;O(p^3)\,. \]

Meanwhile the acceptance probability itself is

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

(since any one of the 5 gates fails ⇒ detection and rejection). Hence the logical fidelity, to second order in p, is

\[ F_{\rm logical} \;=\; 1 \;-\;\frac{P_{\rm err\;\&\;accept}}{P_{\rm accept}} \;=\; 1 \;-\;\frac{8}{225}\,p^2 \;+\;O(p^3)\,. \]

Final Answer:
To leading nonzero order in the two‐qubit error rate $p$, the logical fidelity of the post–selected $|00\>_{AB}$ is

\[ \boxed{ F_{\rm logical}(p) \;=\; 1 \;-\;\frac{8}{225}\,p^2 \;+\;O(p^3)\,.} \]

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

    return F_logical

Checkpoint-3