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

Reasoning (model)

Alright, here’s my understanding of the problem and the solution:

I’m being asked to calculate the fidelity of a 4-qubit state preparation circuit, focusing on the impact of two-qubit depolarizing errors after each CNOT gate. The circuit is defined in reverse order of operations, so I’ve carefully determined the correct sequence: applying a Hadamard on qubit 2, then a CNOT from qubit 2 to qubit 1, another Hadamard on qubit 0, and finally a CNOT from qubit 0 to qubit 3. The initial state is the computational basis state |0000>.

First, I traced the state evolution without any errors to find the ideal final state. I meticulously followed each gate’s operation, step by step, which led me to the ideal final state. I’ve verified that the ideal final state is equivalent to a logical two-qubit GHZ state encoded in the [[4,2,2]] code. I verified that each component of the final state is in the code space, meaning it is an eigenstate of all the stabilizer operators (XXXX and ZZZZ).

Next, I tackled the errors. I knew that after each CNOT gate, there’s a two-qubit depolarizing error channel acting on the two qubits involved in the CNOT. Each CNOT introduces an error from a set of 15 non-identity two-qubit Pauli operators with equal probability $p/15$. The Hadamard gates are assumed to be perfect. Using density matrix formalism, I carefully tracked the state evolution, considering the errors after each CNOT. This required applying the two-qubit depolarizing channel twice (after each CNOT gate).

The real challenge was dealing with the error channels. I broke down the problem, carefully applying the error channel formula after each gate. I then needed to consider the fact that the depolarizing channel is not a standard one, rather one that selects one of the 15 Pauli operators with equal probability. I used the fact that the sum of the Pauli operators applied to a pure state results in the state itself.

To simplify the math, I defined the ideal final state, and from there I calculated the fidelity: $F = \langle \psi_F | \rho_F | \psi_F \rangle$. After detailed calculations, I found the physical state fidelity is $F = 1 - \frac{8p}{5} + \frac{64p^2}{75}$, where p is the error probability of the two-qubit depolarizing channel.

This formula should correctly capture how fidelity decreases with increasing error rate. I’m confident in my work as I’ve double checked each step, verifying the application of the noise model. The derivation used the appropriate properties of the noise model.

The problem asks for the physical state fidelity of a specific 4-qubit state preparation circuit, considering two-qubit depolarizing errors after each CNOT gate. The fidelity is defined as $F = \langle \psi_F | \rho_F | \psi_F \rangle$, where $|\psi_F\rangle$ is the ideal final state and $\rho_F$ is the actual final density matrix after errors.

The circuit operations, applied from right to left, are:

$H_2$ (Hadamard on qubit 2)
$CNOT_{21}$ (Controlled-NOT with control qubit 2, target qubit 1)
$H_0$ (Hadamard on qubit 0)
$CNOT_{03}$ (Controlled-NOT with control qubit 0, target qubit 3)

We start with the initial state $|\psi_0\rangle = |0000\rangle$. Let $U_1=H_2$, $U_2=CNOT_{21}$, $U_3=H_0$, $U_4=CNOT_{03}$. The ideal final state is $|\psi_F\rangle = U_4 U_3 U_2 U_1 |0000\rangle$.

The error model states that each $CNOT_{ij}$ gate is followed by a two-qubit depolarizing error channel $\mathcal{E}_{ij}$. This channel applies one of the 15 non-identity two-qubit Pauli operators ($P \in \{I,X,Y,Z\}^{\otimes 2} \setminus \{II\}$) with equal probability $p/15$, or no error (identity operator) with probability $1-p$. Thus, for a general density matrix $\rho$ of 4 qubits, where the error acts on qubits $i$ and $j$, the channel is:

\[ \mathcal{E}_{ij}(\rho) = (1-p)\rho + \sum_{P_{ij} \in \mathcal{P}_{ij}, P_{ij} \ne I_4} \frac{p}{15} P_{ij} \rho P_{ij}^\dagger \]

Here, $\mathcal{P}_{ij}$ is the set of $4^2=16$ two-qubit Pauli operators acting on qubits $i,j$ (with identity on other qubits). The identity operator for the entire 4-qubit system is $I_4$. A key identity for depolarizing channels on a $D$-dimensional Hilbert space is $\sum_{P_k \in \mathcal{P}_D} P_k \rho P_k^\dagger = D \mathrm{Tr}(\rho) I_D$. For 2-qubit Paulis ($D=4$), and considering $\rho$ as a 4-qubit state (so $I_D$ is effectively $I_4$), this means:

\[ \sum_{P_{ij} \in \mathcal{P}_{ij}} P_{ij} \rho P_{ij}^\dagger = 4 I_4 \]

This is because $\mathrm{Tr}(\rho_{ij})$ (the partial trace over qubits not involved in $P_{ij}$) is 1, and the identity on the other qubits just propagates. Therefore, the sum over non-identity Paulis can be written as:

\[ \sum_{P_{ij} \in \mathcal{P}_{ij}, P_{ij} \ne I_4} P_{ij} \rho P_{ij}^\dagger = \left( \sum_{P_{ij} \in \mathcal{P}_{ij}} P_{ij} \rho P_{ij}^\dagger \right) - I_4 \rho I_4^\dagger = 4 I_4 - \rho \]

Substituting this into the channel definition:

\[ \mathcal{E}_{ij}(\rho) = (1-p)\rho + \frac{p}{15} (4 I_4 - \rho) \]

Now, let’s track the density matrix step-by-step:

Initial state: $\rho_0 = |0000\rangle\langle0000|$.
After $H_2$: $\rho_1 = U_1 \rho_0 U_1^\dagger = |\psi_1\rangle\langle\psi_1|$, where $|\psi_1\rangle = U_1|0000\rangle$.
After $CNOT_{21}$ (before error): $\rho_{2,no\_err} = U_2 \rho_1 U_2^\dagger = |\psi_2\rangle\langle\psi_2|$, where $|\psi_2\rangle = U_2|\psi_1\rangle$.
After error $\mathcal{E}_{21}$ (on qubits 1 and 2):

\[ \rho_2 = \mathcal{E}_{21}(\rho_{2,no\_err}) = (1-p)\rho_{2,no\_err} + \frac{p}{15} (4 I_4 - \rho_{2,no\_err}) \]

After $H_0$: $\rho_3 = U_3 \rho_2 U_3^\dagger$. Since $U_3$ is unitary ($U_3 I_4 U_3^\dagger = I_4$), this gives:

\[ \rho_3 = (1-p)U_3 \rho_{2,no\_err} U_3^\dagger + \frac{p}{15} (4 I_4 - U_3 \rho_{2,no\_err} U_3^\dagger) \]

Let $\rho_{3,no\_err} = U_3 \rho_{2,no\_err} U_3^\dagger = |\psi_3\rangle\langle\psi_3|$, where $|\psi_3\rangle = U_3|\psi_2\rangle$.

\[ \rho_3 = (1-p)\rho_{3,no\_err} + \frac{p}{15} (4 I_4 - \rho_{3,no\_err}) \]

After $CNOT_{03}$ (before error): $\rho_{4,no\_err\_pre} = U_4 \rho_3 U_4^\dagger$. Similarly:

\[ \rho_{4,no\_err\_pre} = (1-p)U_4 \rho_{3,no\_err} U_4^\dagger + \frac{p}{15} (4 I_4 - U_4 \rho_{3,no\_err} U_4^\dagger) \]

Let $\rho_{4,no\_err} = U_4 \rho_{3,no\_err} U_4^\dagger = |\psi_F\rangle\langle\psi_F|$, where $|\psi_F\rangle = U_4|\psi_3\rangle$ is the ideal final state.

\[ \rho_{4,no\_err\_pre} = (1-p)\rho_{4,no\_err} + \frac{p}{15} (4 I_4 - \rho_{4,no\_err}) \]

After error $\mathcal{E}_{03}$ (on qubits 0 and 3): This is the final density matrix $\rho_F$.

\[ \rho_F = \mathcal{E}_{03}(\rho_{4,no\_err\_pre}) = (1-p)\rho_{4,no\_err\_pre} + \frac{p}{15} (4 I_4 - \rho_{4,no\_err\_pre}) \]

The physical state fidelity is $F = \langle \psi_F | \rho_F | \psi_F \rangle$. Let’s substitute the expressions:

\[ F = \langle \psi_F | (1-p)\rho_{4,no\_err\_pre} + \frac{p}{15} (4 I_4 - \rho_{4,no\_err\_pre}) | \psi_F \rangle \]

Since $\langle \psi_F | I_4 | \psi_F \rangle = 1$ (as $|\psi_F\rangle$ is a normalized state), we can define $F' = \langle \psi_F | \rho_{4,no\_err\_pre} | \psi_F \rangle$:

\[ F = (1-p)F' + \frac{p}{15} (4 - F') \]

Now, let’s calculate $F'$:

\[ F' = \langle \psi_F | \left[ (1-p)\rho_{4,no\_err} + \frac{p}{15} (4 I_4 - \rho_{4,no\_err}) \right] | \psi_F \rangle \]

Since $\rho_{4,no\_err} = |\psi_F\rangle\langle\psi_F|$ and $\langle \psi_F | \psi_F\rangle = 1$:

\[ F' = (1-p)\langle \psi_F | \psi_F\rangle\langle\psi_F | \psi_F \rangle + \frac{p}{15} (4 \langle \psi_F | I_4 | \psi_F \rangle - \langle \psi_F | \psi_F\rangle\langle\psi_F | \psi_F \rangle) \]

\[ F' = (1-p)(1) + \frac{p}{15}(4 \times 1 - 1) = 1-p + \frac{3p}{15} = 1-p + \frac{p}{5} = 1 - \frac{4p}{5} \]

Finally, substitute $F'$ back into the expression for $F$:

\[ F = (1-p)\left(1 - \frac{4p}{5}\right) + \frac{p}{15}\left(4 - \left(1 - \frac{4p}{5}\right)\right) \]

\[ F = (1-p)\left(1 - \frac{4p}{5}\right) + \frac{p}{15}\left(3 + \frac{4p}{5}\right) \]

Expand the terms:

\[ F = \left(1 - \frac{4p}{5} - p + \frac{4p^2}{5}\right) + \left(\frac{3p}{15} + \frac{4p^2}{75}\right) \]

\[ F = \left(1 - \frac{9p}{5} + \frac{4p^2}{5}\right) + \left(\frac{p}{5} + \frac{4p^2}{75}\right) \]

Combine like terms:

\[ F = 1 - \frac{9p}{5} + \frac{p}{5} + \frac{4p^2}{5} + \frac{4p^2}{75} \]

\[ F = 1 - \frac{8p}{5} + \frac{60p^2}{75} + \frac{4p^2}{75} \]

\[ F = 1 - \frac{8p}{5} + \frac{64p^2}{75} \]

The final answer is $\boxed{1 - \frac{8p}{5} + \frac{64p^2}{75}}$.

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

The problem asks for the logical state fidelity of the final 2-qubit logical state, assuming post-selection on all detectable errors. This means we consider only the cases where the error syndrome is trivial (i.e., the state remains in the code space). The logical fidelity is then the probability that the final state is the ideal logical state, conditioned on not detecting an error.

The ideal logical state for the circuit is the physical state $|\psi_F\rangle$ derived in Subproblem 1:

\[ |\psi_F\rangle = \frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle) \]

This state is a valid codeword, being a +1 eigenstate of the stabilizers $S_1 = XXXX$ and $S_2 = ZZZZ$. We are seeking the fidelity $F_L$ with this state.

The errors occur after each CNOT gate. Let $E_1$ be the error (a 2-qubit Pauli operator) after $CNOT_{21}$ and $E_2$ be the error after $CNOT_{03}$. The errors are applied to the physical qubits. $E_1$ acts on qubits (1,2) and $E_2$ acts on qubits (0,3). Hadamard gates ($H_0, H_2$) are assumed perfect and do not introduce errors. The errors are chosen from a set of 15 non-identity two-qubit Paulis ($P_{ij} \ne I_{ij}$) with probability $p/15$, or no error (identity $I_{ij}$) with probability $1-p$. The total probability for $E_1$ (and $E_2$) is 1.

The ideal circuit transforms $|0000\rangle$ to $|\psi_F\rangle$. If an error $E_1$ (on $Q_1, Q_2$) occurs, followed by $E_2$ (on $Q_0, Q_3$), the final state is $E_2 U_4 E_1 U_3 U_2 U_1 |0000\rangle$. Since $E_1$ acts on qubits (1,2) and $U_4=CNOT_{03}$ acts on (0,3) and $U_3=H_0$ acts on (0), $E_1$ commutes with $U_4 U_3$. Therefore, the sequence of operations with errors applied is equivalent to applying $E_2 E_1$ to the ideal final state $|\psi_F\rangle$. The final (unnormalized) state is given by:

\[ \rho_F = \sum_{E_1, E_2} \mathrm{Pr}(E_1)\mathrm{Pr}(E_2) (E_2 E_1) |\psi_F\rangle\langle\psi_F| (E_2 E_1)^\dagger \]

where $E_1$ is a Pauli on $Q_1Q_2$ (identity on $Q_0Q_3$) and $E_2$ is a Pauli on $Q_0Q_3$ (identity on $Q_1Q_2$). $\mathrm{Pr}(I) = 1-p$, $\mathrm{Pr}(P \ne I) = p/15$.

1. Determine Undetectable Errors: An error $E$ is undetectable if it commutes with all stabilizers $S_k$. For Pauli errors, this means $S_k E S_k^\dagger = E$ for all $k$. The stabilizers are $S_1 = X_0X_1X_2X_3$ and $S_2 = Z_0Z_1Z_2Z_3$. Let $E_1 = P_1P_2$ where $P_1$ acts on $Q_1$ and $P_2$ acts on $Q_2$. $E_1$ is implicitly $I_0 P_1 P_2 I_3$. Let $E_2 = P_0P_3$ where $P_0$ acts on $Q_0$ and $P_3$ acts on $Q_3$. $E_2$ is implicitly $P_0 I_1 I_2 P_3$. Consider $S_1 E_1 S_1$. $S_1 = X_0(X_1X_2)X_3$. Since $E_1$ only acts on $Q_1Q_2$, and $S_1$ has parts on $Q_0Q_3$ and $Q_1Q_2$: $S_1 E_1 S_1 = (X_0X_3)(X_1X_2) E_1 (X_0X_3)(X_1X_2)$. Since $E_1$ acts only on $Q_1Q_2$, it commutes with $X_0X_3$. So $S_1 E_1 S_1 = (X_1X_2) E_1 (X_1X_2)$. An operator $P_A P_B$ (on $Q_A Q_B$) commutes with $X_A X_B$ if it anti-commutes an even number of times (0 or 2 times). An error $P_1P_2$ on $Q_1Q_2$ commutes with $X_1X_2$ if $A(P_1,X_1)+A(P_2,X_2)$ is even. Similarly, $P_1P_2$ commutes with $Z_1Z_2$ if $A(P_1,Z_1)+A(P_2,Z_2)$ is even. By examining all 16 2-qubit Pauli operators $P_1P_2$: - $A(P,X)=0$ for $P \in \{I,X\}$, $A(P,X)=1$ for $P \in \{Y,Z\}$. - $A(P,Z)=0$ for $P \in \{I,Z\}$, $A(P,Z)=1$ for $P \in \{X,Y\}$. The errors $E_1$ (on $Q_1Q_2$) that commute with $X_1X_2$ and $Z_1Z_2$ are $E_1 \in \{I_{12}, X_1X_2, Z_1Z_2, Y_1Y_2\}$. Let’s call this set $K_1$. Similarly, the errors $E_2$ (on $Q_0Q_3$) that commute with $X_0X_3$ and $Z_0Z_3$ are $E_2 \in \{I_{03}, X_0X_3, Z_0Z_3, Y_0Y_3\}$. Let’s call this set $K_2$. The combined error $E=E_2 E_1$ (since $E_1$ and $E_2$ act on disjoint qubits, they commute) commutes with $S_1$ if $S_1 E_2 E_1 S_1 = (S_1 E_2 S_1) (S_1 E_1 S_1)$ (incorrect step for $S_1$). $S_1 E_2 E_1 S_1 = S_1 E_2 S_1 E_1 S_1$ this step is incorrect because $E_1$ doesn’t commute with $S_1$. A simpler way: $E=E_2 E_1$ commutes with $S_1$ if $A(P_0,X_0)+A(P_1,X_1)+A(P_2,X_2)+A(P_3,X_3)$ is even. And it commutes with $S_2$ if $A(P_0,Z_0)+A(P_1,Z_1)+A(P_2,Z_2)+A(P_3,Z_3)$ is even.

Let $E_1=P_1P_2$ (on $Q_1Q_2$) and $E_2=P_0P_3$ (on $Q_0Q_3$). The undetectable errors $E=E_2E_1$ are those where $P_0P_1P_2P_3$ commutes with $XXXX$ and $ZZZZ$. This implies $E_1$ must be in $K_1$ and $E_2$ must be in $K_2$. There are $4 \times 4 = 16$ such undetectable combined errors $E_2 E_1$. Any other error is detectable.

2. Calculate the probability of not detecting an error ($P_{undetect}$): This is the sum of probabilities of all $16$ undetectable errors. - Case 1: $E_1=I, E_2=I$. Error is $I$. Prob: $(1-p)^2$. - Case 2: $E_1 \in K_1 \setminus \{I\}$ (3 errors), $E_2=I$. Error is $E_1$. Prob: $3 \times \frac{p}{15}(1-p)$. - Case 3: $E_1=I, E_2 \in K_2 \setminus \{I\}$ (3 errors). Error is $E_2$. Prob: $3 \times (1-p)\frac{p}{15}$. - Case 4: $E_1 \in K_1 \setminus \{I\}$ (3 errors), $E_2 \in K_2 \setminus \{I\}$ (3 errors). Error is $E_2E_1$. Prob: $3 \times 3 \times (\frac{p}{15})^2 = 9 \times (\frac{p}{15})^2$.

Summing these probabilities:

\[ P_{undetect} = (1-p)^2 + 6 \frac{p}{15}(1-p) + 9 \left(\frac{p}{15}\right)^2 \]

\[ P_{undetect} = (1-p)^2 + \frac{2p}{5}(1-p) + \frac{9p^2}{225} \]

\[ P_{undetect} = 1 - 2p + p^2 + \frac{2p}{5} - \frac{2p^2}{5} + \frac{p^2}{25} \]

\[ P_{undetect} = 1 - \frac{8p}{5} + p^2 \left(1 - \frac{2}{5} + \frac{1}{25}\right) = 1 - \frac{8p}{5} + p^2 \left(\frac{25-10+1}{25}\right) \]

\[ P_{undetect} = 1 - \frac{8p}{5} + \frac{16p^2}{25} \]

3. Calculate the probability of no logical error, given no detection: The logical fidelity $F_L$ is the probability that the final logical state is $|\psi_F\rangle$, given that no error was detected. This means the total error $E=E_2E_1$ must belong to the stabilizer group $G_S = \{I, XXXX, ZZZZ, YYYY\}$. We need to evaluate $|\langle\psi_F|E|\psi_F\rangle|^2$ for $E \in G_S$. For $E \in G_S$, $E|\psi_F\rangle = |\psi_F\rangle$ (since $S_1|\psi_F\rangle = |\psi_F\rangle$, $S_2|\psi_F\rangle = |\psi_F\rangle$, and $YYYY|\psi_F\rangle = S_1 S_2 |\psi_F\rangle = |\psi_F\rangle$). So, for $E \in G_S$, $|\langle\psi_F|E|\psi_F\rangle|^2 = 1$.

We list the $E_2E_1$ combinations that fall into $G_S$: - Case 1: $E_1=I, E_2=I$. $E=I$. Prob: $(1-p)^2$. This is in $G_S$. Contribution to numerator: $(1-p)^2 \times 1$. - Case 2: $E_1 \in K_1 \setminus \{I\}$, $E_2=I$. Error is $E_1 \in \{X_1X_2, Z_1Z_2, Y_1Y_2\}$. These are NOT in $G_S$. No contribution. - Case 3: $E_1=I, E_2 \in K_2 \setminus \{I\}$. Error is $E_2 \in \{X_0X_3, Z_0Z_3, Y_0Y_3\}$. These are NOT in $G_S$. No contribution. - Case 4: $E_1 \in K_1 \setminus \{I\}$, $E_2 \in K_2 \setminus \{I\}$. These are $3 \times 3=9$ errors: - $E_2E_1 = X_0X_3 X_1X_2 = X_0X_1X_2X_3 = XXXX = S_1$. Prob: $(p/15)^2$. This is in $G_S$. Contributes $1 \times (\frac{p}{15})^2$. - $E_2E_1 = Z_0Z_3 Z_1Z_2 = Z_0Z_1Z_2Z_3 = ZZZZ = S_2$. Prob: $(p/15)^2$. This is in $G_S$. Contributes $1 \times (\frac{p}{15})^2$. - $E_2E_1 = Y_0Y_3 Y_1Y_2 = Y_0Y_1Y_2Y_3 = YYYY = S_1S_2$. Prob: $(p/15)^2$. This is in $G_S$. Contributes $1 \times (\frac{p}{15})^2$. - The other 6 combinations (e.g., $X_0X_3 Z_1Z_2$) are in $N(G_S)$ but not in $G_S$. They represent logical errors. No contribution to $F_L$.

The numerator for $F_L$ (probability of an undetected physical error that results in the ideal logical state) is:

\[ N_{correct\_logical} = (1-p)^2 + 3 \left(\frac{p}{15}\right)^2 \]

\[ N_{correct\_logical} = 1 - 2p + p^2 + \frac{3p^2}{225} = 1 - 2p + p^2 + \frac{p^2}{75} = 1 - 2p + \frac{76p^2}{75} \]

4. Calculate Logical Fidelity: The logical fidelity is the ratio of $N_{correct\_logical}$ to $P_{undetect}$:

\[ F_L = \frac{N_{correct\_logical}}{P_{undetect}} = \frac{1 - 2p + \frac{76p^2}{75}}{1 - \frac{8p}{5} + \frac{16p^2}{25}} \]

To write this with a common denominator of 75 for the $p^2$ terms:

\[ F_L = \frac{1 - 2p + \frac{76p^2}{75}}{1 - \frac{24p}{15} + \frac{48p^2}{75}} \]

The final answer is $\boxed{\frac{1 - 2p + \frac{76p^2}{75}}{1 - \frac{8p}{5} + \frac{16p^2}{25}}}$.

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

Reasoning (model)

Analyzing Logical Fidelity with Depolarizing Errors

My objective is to determine the fidelity of a prepared logical $|00\rangle_{AB}$ state after a circuit with two-qubit depolarizing errors following each CNOT gate, and post-selection based on two conditions: (1) no detectable errors in the [[4,2,2]] code, and (2) the ancilla qubit measuring $|0\rangle$.

Ideal Circuit & State Evolution: First, I worked through the ideal circuit to determine the expected final state of the five qubits (four logical and one ancilla) in the absence of noise. Starting from $|00000\rangle$, I meticulously applied the Hadamard and CNOT gates. I found that the ideal final state of the circuit, before measurement, is a superposition of $|00000\rangle$ and $|11110\rangle$. This state, projected onto $|0\rangle$ ancilla, is indeed the logical $|00\rangle$ state encoded in the [[4,2,2]] code, as expected.

Error Model & Fidelity Calculation Strategy: The core of the problem lies in accounting for the two-qubit depolarizing noise that occurs after each of the five CNOT gates. I’m using the typical depolarizing channel representation. After each CNOT, the state is subject to a channel that, with probability ‘p’, replaces the state with a uniform mixture of Pauli operators. Given the complexity of tracking all possible error combinations, I recognized that calculating the exact fidelity for all possible errors is unwieldy.

Simplifying the Calculation (with small p approximation): I’m using the approximation $F = (1-\frac{16p}{15})^5$ because the formula for the fidelity is simplified, assuming small values of p, to include the main terms in the fidelity calculation. I can work out a recurrence relationship using the channel $\mathcal{E}(\rho) = (1-p)\rho + \frac{p}{15} (4I - \rho)$ $F_k = (1 - (16/15)p) F_{k-1} + (4/15) p$ Which is then $F_k = \frac{1}{4} + \frac{3}{4}(1-\frac{16p}{15})^k$ Given the post-selection, I need to account for (1) post-select on ancilla, and (2) ensure that any errors that lead to bad code states, we do not include in the overall value for calculating the final logical fidelity. I can then compute fidelity using the “general” approach: $F_L = \frac{P_{correct}}{P_{correct} + P_{logical\_error}}$. This has to include the probabilities.

Key Issues & Insights:

Fault-Tolerant vs. Non-Ideal Assumptions: I initially assumed that the code would detect all single errors. But it turns out that certain stabilizer and logical errors are undetectable.
Ancilla Post-Selection Effects: The ancilla measurement filters out certain error outcomes, but doesn’t guarantee a “pure” logical state. I have to think about what happens to the state, and also the probability.
The Importance of Post-Selection: The fidelity calculation requires understanding the effect of the post-selection on the ancilla and on the encoded qubit states and the probability of their respective states.
Error Propagation & Simplification: The key is to understand how the errors on the various qubits are transformed by the CNOT gates and their effect on the ancilla (and overall on the measurement outcomes).
The Role of the Depolarizing Channel: The depolarizing channel leads to states that aren’t ideal, which is important to the calculation.

Final Calculation I had to re-evaluate the calculation with $F_L = \frac{\frac{1}{4} + \frac{3}{4}(1-\frac{16p}{15})^5}{\frac{1}{8} + \frac{7}{8}(1-\frac{16p}{15})^5}$

The answer is $F_L = \frac{2 + 6A^5}{1 + 7A^5}$, where $A = 1 - 16p/15$.

Here is the correct approach to this problem.

To calculate the logical state fidelity, we need to consider the probability of observing the ancilla qubit 4 in state $|0\rangle$ and the remaining 4 physical qubits (0,1,2,3) being in the code space, and then, conditioned on this, calculate the probability that the logical state is the ideal logical $|00\rangle_{AB}$.

Let $N_{CNOT}=5$ be the number of CNOT gates in the circuit. The circuit operations, applied from right to left, are:

$H_1$ (Hadamard on qubit 1)
$CNOT_{12}$ (Control qubit 1, Target qubit 2)
$CNOT_{10}$ (Control qubit 1, Target qubit 0)
$CNOT_{23}$ (Control qubit 2, Target qubit 3)
$CNOT_{34}$ (Control qubit 3, Target qubit 4)
$CNOT_{04}$ (Control qubit 0, Target qubit 4)
$M_4$ (Measurement of qubit 4)

We start with the initial state $|\psi_0\rangle = |00000\rangle$.

1. Ideal Final State Derivation: Let’s trace the ideal state evolution without errors: * Initial state: $|\psi_{init}\rangle = |00000\rangle_{01234}$ * $H_1$: $|\psi_1\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01000\rangle)$ * $CNOT_{12}$: $|\psi_2\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |01100\rangle)$ * $CNOT_{10}$: $|\psi_3\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11100\rangle)$ * $CNOT_{23}$: $|\psi_4\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$ * $CNOT_{34}$: $|\psi_5\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11111\rangle)$ * $CNOT_{04}$: $|\psi_6\rangle = \frac{1}{\sqrt{2}}(|00000\rangle + |11110\rangle)$. This is the 5-qubit ideal state before measurement. Let this be $|\psi_{ideal\_5q}\rangle$.

After measuring qubit 4, the ideal outcome is $|0\rangle$. The state on qubits 0,1,2,3 is: $|\psi_{L00}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)$. This is the logical $|00\rangle_{AB}$ state, which is the target state. We confirm this state is in the code space: $XXXX |\psi_{L00}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) = |\psi_{L00}\rangle$. $ZZZZ |\psi_{L00}\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) = |\psi_{L00}\rangle$. It is also a logical $|00\rangle_{AB}$ state: $Z_A = ZZII$, $Z_B = ZIZI$. $Z_A |\psi_{L00}\rangle = \frac{1}{\sqrt{2}}(ZZII|0000\rangle + ZZII|1111\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) = |\psi_{L00}\rangle$. $Z_B |\psi_{L00}\rangle = \frac{1}{\sqrt{2}}(ZIZI|0000\rangle + ZIZI|1111\rangle) = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle) = |\psi_{L00}\rangle$.

2. Error Model and Final Density Matrix: Each CNOT gate is followed by a two-qubit depolarizing error channel $\mathcal{E}_{ij}$. For a 5-qubit system (dimension $D=2^5=32$), if the error channel acts on 2 qubits $i,j$, it is specified as:

\[ \mathcal{E}_{ij}(\rho) = (1-p)\rho + \sum_{P_{ij} \in \mathcal{P}_{ij}, P_{ij} \ne I_{ij}} \frac{p}{15} (P_{ij} \otimes I_{rest}) \rho (P_{ij} \otimes I_{rest})^\dagger \]

where $I_{rest}$ is the identity on the other 3 qubits. Using the identity $\sum_{P_{ij} \in \mathcal{P}_{ij}} (P_{ij} \otimes I_{rest}) \rho (P_{ij} \otimes I_{rest})^\dagger = 4 I_5$, where $I_5$ is the 5-qubit identity, we get:

\[ \mathcal{E}_{ij}(\rho) = (1-p)\rho + \frac{p}{15} (4 I_5 - \rho) = \left(1-p-\frac{p}{15}\right)\rho + \frac{4p}{15}I_5 = \left(1-\frac{16p}{15}\right)\rho + \frac{4p}{15}I_5 \]

Let $A = 1 - \frac{16p}{15}$. So $\mathcal{E}(\rho) = A\rho + \frac{4p}{15}I_5$. There are $N_{CNOT}=5$ CNOT gates. Let $\rho_k$ be the density matrix after $k$ noisy CNOT gates, with $F_k = \langle \psi_{ideal\_k} | \rho_k | \psi_{ideal\_k} \rangle$. The Hadamard gate $H_1$ is assumed perfect. So $\rho_0 = |\psi_{init}\rangle\langle\psi_{init}|$. Let $\rho_0' = H_1 \rho_0 H_1^\dagger$. Then $\rho_1 = \mathcal{E}_{12}(CNOT_{12} \rho_0' CNOT_{12}^\dagger)$. The fidelity $F_k = \langle \psi_{ideal\_k} | \rho_k | \psi_{ideal\_k} \rangle$ evolves as: $F_k = A F_{k-1} + \frac{4p}{15} \langle \psi_{ideal\_k} | I_5 | \psi_{ideal\_k} \rangle = A F_{k-1} + \frac{4p}{15}$. This is a linear recurrence relation. With $F_0=1$ (fidelity with the initial ideal state being 1). The general solution is $F_k = \frac{1}{4} + \frac{3}{4} A^k$. For $N_{CNOT}=5$, the final physical fidelity of the 5-qubit state with the ideal 5-qubit state is:

\[ P_{GC} = F_5 = \frac{1}{4} + \frac{3}{4}\left(1-\frac{16p}{15}\right)^5 \]

This represents the probability that the final (noisy) 5-qubit state is identical to the ideal 5-qubit state $|\psi_{ideal\_5q}\rangle$. Since $|\psi_{ideal\_5q}\rangle$ has $Q_4=|0\rangle$ and $Q_{0123}$ in the correct logical $|00\rangle_{AB}$ state (which is a member of the code space), this quantity $P_{GC}$ serves as the numerator for the logical fidelity.

3. Post-selection and Logical Fidelity: We are post-selecting on two conditions: (a) The ancilla qubit (Q4) measures $|0\rangle$. Let $P_0 = I_{0123} \otimes |0\rangle_4\langle0|_4$. (b) The remaining 4 qubits (Q0123) are in the code space. Let $P_{code} = \frac{1}{4}(I+XXXX)(I+ZZZZ)$ be the projector onto the code space. The combined projector for the desired subspace is $S = P_{code} \otimes P_0$. The final density matrix is $\rho_F = \mathcal{E}_5(U_6 \mathcal{E}_4(U_5 \mathcal{E}_3(U_4 \mathcal{E}_2(U_3 \mathcal{E}_1(U_2 U_1 \rho_0 U_1^\dagger U_2^\dagger) U_3^\dagger) U_4^\dagger) U_5^\dagger) U_6^\dagger)$. The state $\rho_F$ can be expressed in terms of the ideal final state and identity due to the depolarizing channels:

\[ \rho_F = \left(1-\frac{16p}{15}\right)^5 |\psi_{ideal\_5q}\rangle\langle\psi_{ideal\_5q}| + \frac{1-\left(1-\frac{16p}{15}\right)^5}{4} I_5 \]

The logical fidelity $F_L$ is given by $F_L = \frac{P_{correct}}{P_{good}}$, where $P_{correct}$ is the probability that the final state is $|\psi_{L00}\rangle$ after post-selection, and $P_{good}$ is the total probability of post-selection success (ancilla measures $|0\rangle$ and the state is in the code space). The numerator $P_{correct}$ is $\langle \psi_{L00} | \left( \mathrm{Tr}_4(P_0 \rho_F P_0) \times P_{code} \right) | \psi_{L00} \rangle$. Since $|\psi_{ideal\_5q}\rangle = |\psi_{L00}\rangle \otimes |0\rangle_4$ and $P_{code}|\psi_{L00}\rangle = |\psi_{L00}\rangle$, this simplifies to:

\[ P_{correct} = \langle \psi_{ideal\_5q} | \rho_F | \psi_{ideal\_5q} \rangle = F_5 \]

So, $P_{correct} = P_{GC} = \frac{1}{4} + \frac{3}{4}\left(1-\frac{16p}{15}\right)^5$.

The denominator $P_{good}$ is the trace of the post-selected state:

\[ P_{good} = \mathrm{Tr}(S \rho_F) \]

\[ P_{good} = \mathrm{Tr}\left(S \left[ \left(1-\frac{16p}{15}\right)^5 |\psi_{ideal\_5q}\rangle\langle\psi_{ideal\_5q}| + \frac{1-\left(1-\frac{16p}{15}\right)^5}{4} I_5 \right]\right) \]

Since $S |\psi_{ideal\_5q}\rangle = |\psi_{ideal\_5q}\rangle$, the first term is $\left(1-\frac{16p}{15}\right)^5 \mathrm{Tr}(|\psi_{ideal\_5q}\rangle\langle\psi_{ideal\_5q}|) = \left(1-\frac{16p}{15}\right)^5$. For the second term, $\mathrm{Tr}(S I_5)$ is the dimension of the subspace projected by $S$. The dimension of the code space for 2 logical qubits is $2^2=4$. The dimension for ancilla $|0\rangle_4$ is $1$. So $\mathrm{Tr}(S I_5) = 4 \times 1 = 4$. Thus, the second term is $\frac{1-\left(1-\frac{16p}{15}\right)^5}{4} \times \frac{\mathrm{Tr}(S I_5)}{\mathrm{dim}(\mathcal{H})} = \frac{1-\left(1-\frac{16p}{15}\right)^5}{4} \times \frac{4}{32} = \frac{1-\left(1-\frac{16p}{15}\right)^5}{32}$. So $P_{good} = \left(1-\frac{16p}{15}\right)^5 + \frac{1-\left(1-\frac{16p}{15}\right)^5}{32}$. Let $X = \left(1-\frac{16p}{15}\right)^5$.

\[ P_{good} = X + \frac{1-X}{32} = \frac{32X + 1 - X}{32} = \frac{1+31X}{32} \]

The logical fidelity $F_L = \frac{P_{correct}}{P_{good}}$ is:

\[ F_L = \frac{\frac{1}{4} + \frac{3}{4}X}{\frac{1}{32} + \frac{31}{32}X} \]

To simplify, multiply numerator and denominator by 32:

\[ F_L = \frac{8 + 24X}{1 + 31X} \]

Substituting $X = \left(1-\frac{16p}{15}\right)^5$:

\[ F_L = \frac{8 + 24\left(1-\frac{16p}{15}\right)^5}{1 + 31\left(1-\frac{16p}{15}\right)^5} \]

The final answer is $\boxed{\frac{8 + 24\left(1-\frac{16p}{15}\right)^5}{1 + 31\left(1-\frac{16p}{15}\right)^5}}$.

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 ------------------
    X = (1 - (16*p)/15)**5
    F_logical = (8 + 24*X) / (1 + 31*X)
    # ---------------------------------------------------------------

    return F_logical