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

Reasoning (model)

Analyzing Physical State Fidelity in a Noisy Circuit

As an expert in the field, I’m tasked with calculating the fidelity of a specific state preparation circuit, taking into account the presence of two-qubit depolarizing errors. My approach involves several key steps:

Understanding the Target Logical State: My goal is to create the logical GHZ state, which is a superposition of |00> and |11>. I need to determine the mapping of this logical state to a physical 4-qubit representation using the [[4,2,2]] code properties.
Circuit Analysis: The circuit comprises of a sequence of operations: $(CNOT_{03}) (H_0) (CNOT_{21}) (H_2)$. Since these are applied from right to left, I need to understand what state would be produced step-by-step.
Error Modeling: I’ll be modeling the errors in the circuit using the two-qubit depolarizing error channel applied after each CNOT gate.
Fidelity Calculation: To quantify the performance of this noisy circuit, I will be calculating the fidelity which is the overlap between the ideal target state and the noisy state. If the ideal state is pure $|\psi\rangle$, then the fidelity $F = \langle\psi|\rho_{noisy}|\psi\rangle$.

Decoding the Circuit and Error:

The code space is defined by two stabilizers: $S_1 = X_0 X_1 X_2 X_3$ and $S_2 = Z_0 Z_1 Z_2 Z_3$. I’ll use logical operators $X_A, X_B, Z_A, Z_B$.

Step 1. Encoding the Logical GHZ State:

I’ll proceed to determine the physical representation of the logical state through circuit application. The circuit will be applied to the state $|0000\rangle$, and then each step will be followed by the error channels, generating the noisy state. First I calculate the ideal state produced:

$H_2: |0000\rangle \rightarrow \frac{1}{\sqrt{2}}(|0000\rangle+|0010\rangle)$.
$CNOT_{21}: \frac{1}{\sqrt{2}}(|0000\rangle+|0010\rangle) \rightarrow \frac{1}{\sqrt{2}}(|0000\rangle+|0110\rangle)$.
$H_0: \frac{1}{\sqrt{2}}(|0000\rangle+|0110\rangle) \rightarrow \frac{1}{2}(|0000\rangle+|1000\rangle+|0110\rangle+|1110\rangle)$.
$CNOT_{03}: \frac{1}{2}(|0000\rangle+|1000\rangle+|0110\rangle+|1110\rangle) \rightarrow \frac{1}{2}(|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)$. So the ideal final state is: $\frac{1}{2}(|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)$. Verifying that this is a valid logical GHZ state is important. The logical GHZ must satisfy $X_A X_B |\psi_{GHZ}\rangle = |\psi_{GHZ}\rangle$ and $Z_A Z_B |\psi_{GHZ}\rangle = |\psi_{GHZ}\rangle$, and also the stabilizers $S_1, S_2$ must give the state back.

Step 2: Accounting for the Noise.

The fidelity calculation involves carefully tracking the state through the circuit and the depolarizing errors. The state undergoes:

$\rho_{noisy} = \mathcal{D}_{03}(CNOT_{03} H_0 \mathcal{D}_{21}(CNOT_{21} H_2 |0000\rangle\langle0000| H_2^\dagger) H_0^\dagger CNOT_{03}^\dagger)$.

Here I used $\mathcal{D}$ for the depolarizing error channels. Since the fidelity is $F = \langle\psi_{ideal}|\rho_{noisy}|\psi_{ideal}\rangle$, this can be simplified by repeatedly using the cyclic property of the trace and the channel structure. The fidelity can be rewritten to consider the expectation values of the Pauli operations.

To simplify the calculation of the noisy state, and since the

The problem asks us to calculate the physical state fidelity of a specific quantum circuit designed to prepare a logical GHZ state, given a model of two-qubit depolarizing errors following each CNOT gate.

The state preparation circuit, applied to the initial state $|0000\rangle$, is:

\[ U_{circ} = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2) \]

The operations are applied from right to left, starting with $|0000\rangle$.

1. Determine the ideal final physical state:

Let’s trace the initial state $|0000\rangle$ through the ideal circuit: * Initial state: $|\psi_{init}\rangle = |0000\rangle$. * After $H_2$ (on qubit 2): $|\psi_A\rangle = H_2 |0000\rangle = |00\rangle_0 \frac{|0\rangle_2+|1\rangle_2}{\sqrt{2}} |0\rangle_3 = \frac{1}{\sqrt{2}}(|0000\rangle+|0010\rangle)$. * After $CNOT_{21}$ (control qubit 2, target qubit 1): $CNOT_{21} |0000\rangle = |0000\rangle$. $CNOT_{21} |0010\rangle = |0110\rangle$. $|\psi_B\rangle = CNOT_{21} |\psi_A\rangle = \frac{1}{\sqrt{2}}(|0000\rangle+|0110\rangle)$. * After $H_0$ (on qubit 0): $H_0 |0000\rangle = \frac{1}{\sqrt{2}}(|0000\rangle+|1000\rangle)$. $H_0 |0110\rangle = \frac{1}{\sqrt{2}}(|0110\rangle+|1110\rangle)$. $|\psi_C\rangle = H_0 |\psi_B\rangle = \frac{1}{2}(|0000\rangle+|1000\rangle+|0110\rangle+|1110\rangle)$. * After $CNOT_{03}$ (control qubit 0, target qubit 3): $CNOT_{03} |0000\rangle = |0000\rangle$. $CNOT_{03} |1000\rangle = |1001\rangle$. $CNOT_{03} |0110\rangle = |0110\rangle$. $CNOT_{03} |1110\rangle = |1111\rangle$. The ideal final physical state is: $|\psi_{ideal}\rangle = CNOT_{03} |\psi_C\rangle = \frac{1}{2}(|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)$.

2. Model the noisy evolution and calculate the fidelity:

Let the initial state be $\rho_0 = |0000\rangle\langle0000|$. The circuit has two CNOT gates: $CNOT_{21}$ and $CNOT_{03}$. Each is followed by a two-qubit depolarizing channel $\mathcal{D}_{ij}$ with error rate $p$. The depolarizing channel on two qubits $i, j$ is defined as:

\[ \mathcal{D}_{ij}(\rho) = (1-p) \rho + \frac{p}{15} \sum_{P \in \mathcal{P}_{ij} \setminus \{I \otimes I\}} P \rho P \]

where $\mathcal{P}_{ij}$ is the set of 16 two-qubit Pauli operators acting on qubits $i,j$ and identity on other qubits.

Let $U_1=H_2$, $U_2=CNOT_{21}$, $U_3=H_0$, $U_4=CNOT_{03}$. The evolution of the density matrix is:

$\rho_1 = U_1 \rho_0 U_1^\dagger = |\psi_A\rangle\langle\psi_A|$.
$\rho_2 = U_2 \rho_1 U_2^\dagger = |\psi_B\rangle\langle\psi_B|$.
$\rho_{err,1} = \mathcal{D}_{21}(\rho_2)$. (Depolarizing error on qubits 2,1 after $CNOT_{21}$)
$\rho_3 = U_3 \rho_{err,1} U_3^\dagger$.
$\rho_4 = U_4 \rho_3 U_4^\dagger$.
$\rho_{noisy} = \mathcal{D}_{03}(\rho_4)$. (Depolarizing error on qubits 0,3 after $CNOT_{03}$)

The physical state fidelity is $F = \langle\psi_{ideal}|\rho_{noisy}|\psi_{ideal}\rangle = \text{Tr}(|\psi_{ideal}\rangle\langle\psi_{ideal}| \rho_{noisy})$. Let $\Pi_{ideal} = |\psi_{ideal}\rangle\langle\psi_{ideal}|$.

We can expand the fidelity using linearity:

\[ F = \text{Tr}\left(\Pi_{ideal} \left[ (1-p)\rho_4 + \frac{p}{15} \sum_{j=1}^{15} P_j^{(03)} \rho_4 P_j^{(03)} \right]\right) \]

\[ F = (1-p) \text{Tr}(\Pi_{ideal} \rho_4) + \frac{p}{15} \sum_{j=1}^{15} \text{Tr}(\Pi_{ideal} P_j^{(03)} \rho_4 P_j^{(03)}) \]

Let’s analyze $\text{Tr}(\Pi_{ideal} \rho_4)$:

\[ \text{Tr}(\Pi_{ideal} \rho_4) = \text{Tr}(\Pi_{ideal} U_4 \rho_3 U_4^\dagger) = \text{Tr}(U_4^\dagger \Pi_{ideal} U_4 \rho_3) \]

Since $\Pi_{ideal} = U_4 U_3 U_2 U_1 \rho_0 U_1^\dagger U_2^\dagger U_3^\dagger U_4^\dagger$, we have $U_4^\dagger \Pi_{ideal} U_4 = U_3 U_2 U_1 \rho_0 U_1^\dagger U_2^\dagger U_3^\dagger = |\psi_C\rangle\langle\psi_C|$. So, $\text{Tr}(\Pi_{ideal} \rho_4) = \langle\psi_C|\rho_3|\psi_C\rangle$.

Now expand $\rho_3 = U_3 \rho_{err,1} U_3^\dagger$:

\[ \langle\psi_C|\rho_3|\psi_C\rangle = \langle\psi_C|U_3 \rho_{err,1} U_3^\dagger|\psi_C\rangle = \text{Tr}(U_3^\dagger |\psi_C\rangle\langle\psi_C| U_3 \rho_{err,1}) \]

Since $U_3^\dagger |\psi_C\rangle\langle\psi_C| U_3 = |\psi_B\rangle\langle\psi_B|$, this becomes $\langle\psi_B|\rho_{err,1}|\psi_B\rangle$. Then, substituting $\rho_{err,1} = \mathcal{D}_{21}(\rho_2)$:

\[ \langle\psi_B|\rho_{err,1}|\psi_B\rangle = (1-p)\langle\psi_B|\rho_2|\psi_B\rangle + \frac{p}{15}\sum_{k=1}^{15}\langle\psi_B|P_k^{(21)}\rho_2 P_k^{(21)}|\psi_B\rangle \]

Since $\rho_2=|\psi_B\rangle\langle\psi_B|$, we have $\langle\psi_B|\rho_2|\psi_B\rangle = 1$. And $\langle\psi_B|P_k^{(21)}\rho_2 P_k^{(21)}|\psi_B\rangle = |\langle\psi_B|P_k^{(21)}|\psi_B\rangle|^2$.

So, the fidelity can be expressed as:

\[ F = (1-p)\left[ (1-p) + \frac{p}{15} \sum_{k=1}^{15} |\langle\psi_B|P_k^{(21)}|\psi_B\rangle|^2 \right] + \frac{p}{15} \sum_{j=1}^{15} \text{Tr}(P_j^{(03)} \Pi_{ideal} P_j^{(03)} U_4 \rho_3 U_4^\dagger) \]

The second term can be further simplified. Let $M_j = P_j^{(03)} \Pi_{ideal} P_j^{(03)}$. $\text{Tr}(M_j U_4 \rho_3 U_4^\dagger) = \text{Tr}(U_4^\dagger M_j U_4 \rho_3)$. $U_4^\dagger M_j U_4 = U_4^\dagger P_j^{(03)} U_4 U_4^\dagger \Pi_{ideal} U_4 U_4^\dagger P_j^{(03)} U_4 = (U_4^\dagger P_j^{(03)} U_4) |\psi_C\rangle\langle\psi_C| (U_4^\dagger P_j^{(03)} U_4)$. Let $\tilde{P}_j^{(03)} = U_4^\dagger P_j^{(03)} U_4$. Then the second sum is $\frac{p}{15} \sum_j \text{Tr}(\tilde{P}_j^{(03)} |\psi_C\rangle\langle\psi_C| \tilde{P}_j^{(03)} \rho_3)$. Using $\rho_3 = U_3 \rho_{err,1} U_3^\dagger$: $\text{Tr}(\tilde{P}_j^{(03)} |\psi_C\rangle\langle\psi_C| \tilde{P}_j^{(03)} U_3 \rho_{err,1} U_3^\dagger) = \text{Tr}(U_3^\dagger \tilde{P}_j^{(03)} |\psi_C\rangle\langle\psi_C| \tilde{P}_j^{(03)} U_3 \rho_{err,1})$. Let $\hat{P}_j^{(03)} = U_3^\dagger \tilde{P}_j^{(03)} U_3$. This becomes $\text{Tr}(\hat{P}_j^{(03)} |\psi_B\rangle\langle\psi_B| \hat{P}_j^{(03)} \rho_{err,1})$. Substituting $\rho_{err,1} = (1-p)\rho_2 + \frac{p}{15}\sum_k P_k^{(21)}\rho_2 P_k^{(21)}$: The terms contribute $p(1-p)\frac{1}{15} \sum_j |\langle\psi_B|\hat{P}_j^{(03)}|\psi_B\rangle|^2$ and $p^2\frac{1}{15^2} \sum_j\sum_k |\langle\psi_B|\hat{P}_j^{(03)}P_k^{(21)}|\psi_B\rangle|^2$.

The exact fidelity expression is:

\[ F = (1-p)^2 + (1-p)\frac{p}{15} \sum_{k=1}^{15} |\langle\psi_B|P_k^{(21)}|\psi_B\rangle|^2 + (1-p)\frac{p}{15} \sum_{j=1}^{15} |\langle\psi_{ideal}|P_j^{(03)}|\psi_{ideal}\rangle|^2 + \frac{p^2}{15^2} \sum_{j=1}^{15}\sum_{k=1}^{15} |\langle\psi_{ideal}|P_j^{(03)} U_4 U_3 P_k^{(21)} U_2 U_1 |0000\rangle|^2 \]

Let’s evaluate each sum:

Term 1: $\sum_{k=1}^{15} |\langle\psi_B|P_k^{(21)}|\psi_B\rangle|^2$ $|\psi_B\rangle = \frac{1}{\sqrt{2}}(|0000\rangle+|0110\rangle)$. This can be written as $|0\rangle_0 \otimes \frac{1}{\sqrt{2}}(|00\rangle_{12}+|11\rangle_{12}) \otimes |0\rangle_3$. Let $|\chi\rangle_{12} = \frac{1}{\sqrt{2}}(|00\rangle_{12}+|11\rangle_{12})$. The operator $P_k^{(21)}$ acts on qubits 1 and 2, so $P_k^{(21)} = I_0 \otimes P'_{12} \otimes I_3$. So, $\langle\psi_B|P_k^{(21)}|\psi_B\rangle = \langle\chi|_{12} P'_{12} |\chi\rangle_{12}$. For the Bell state $|\chi\rangle_{12}$, the expectation values of two-qubit Paulis are non-zero only for $P'_{12} \in \{II, XX, YY, ZZ\}$. $\langle\chi|XX|\chi\rangle = 1$ $\langle\chi|YY|\chi\rangle = -1$ $\langle\chi|ZZ|\chi\rangle = 1$ The sum is over 15 non-identity Paulis. So, we exclude $II$. $\sum_{k=1}^{15} |\langle\psi_B|P_k^{(21)}|\psi_B\rangle|^2 = |1|^2 + |-1|^2 + |1|^2 = 3$.

Term 2: $\sum_{j=1}^{15} |\langle\psi_{ideal}|P_j^{(03)}|\psi_{ideal}\rangle|^2$ $|\psi_{ideal}\rangle = \frac{1}{2}(|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)$. The operator $P_j^{(03)}$ acts on qubits 0 and 3, so $P_j^{(03)} = P'_0 \otimes I_{12} \otimes P'_3$. Let’s check the expectation values for $P'_0 P'_3$: $\langle\psi_{ideal}|X_0 X_3|\psi_{ideal}\rangle = 1$ $\langle\psi_{ideal}|Y_0 Y_3|\psi_{ideal}\rangle = -1$ $\langle\psi_{ideal}|Z_0 Z_3|\psi_{ideal}\rangle = 1$ All other 12 two-qubit Paulis on qubits 0 and 3 (e.g., $X_0 I_3$, $I_0 Z_3$, $X_0 Z_3$, etc.) yield an expectation value of 0 due to the structure of $|\psi_{ideal}\rangle$. For example, $\langle\psi_{ideal}|X_0 I_3|\psi_{ideal}\rangle = \frac{1}{4} (\langle0000|+\langle1001|+\langle0110|+\langle1111|) (X_0 I_3) (|0000\rangle+|1001\rangle+|0110\rangle+|1111\rangle)$. $X_0 I_3 |\psi_{ideal}\rangle = \frac{1}{2}(|1000\rangle+|0001\rangle+|1110\rangle+|0111\rangle)$. This is orthogonal to $|\psi_{ideal}\rangle$, so the inner product is 0. Thus, $\sum_{j=1}^{15} |\langle\psi_{ideal}|P_j^{(03)}|\psi_{ideal}\rangle|^2 = |1|^2 + |-1|^2 + |1|^2 = 3$.

Term 3: $\frac{p^2}{15^2} \sum_{j=1}^{15}\sum_{k=1}^{15} |\langle\psi_{ideal}|P_j^{(03)} U_4 U_3 P_k^{(21)} U_2 U_1 |0000\rangle|^2$ Let $U_{DC} = U_4 U_3 = CNOT_{03} H_0$. The term inside the sum is $|\langle\psi_{ideal}|P_j^{(03)} U_{DC} P_k^{(21)} |\psi_B\rangle|^2$. Using $|\psi_{ideal}\rangle = U_{DC} |\psi_B\rangle$, this becomes: $|\langle U_{DC} \psi_B | P_j^{(03)} U_{DC} P_k^{(21)} |\psi_B\rangle|^2 = |\langle \psi_B | U_{DC}^\dagger P_j^{(03)} U_{DC} P_k^{(21)} |\psi_B\rangle|^2$. Let $M_j = U_{DC}^\dagger P_j^{(03)} U_{DC} = H_0 CNOT_{03} P_j^{(03)} CNOT_{03} H_0$. The operator $P_j^{(03)}$ acts only on qubits 0 and 3. The conjugation by $CNOT_{03}$ keeps the operator localized to qubits 0 and 3. The conjugation by $H_0$ (only on qubit 0) also keeps the operator localized to qubits 0 and 3. So $M_j$ acts only on qubits 0 and 3. $M_j = M_{j(03)} \otimes I_{12}$. The operator $P_k^{(21)}$ acts only on qubits 1 and 2. $P_k^{(21)} = I_0 \otimes P'_{k(12)} \otimes I_3$. The state $|\psi_B\rangle = |0\rangle_0 \otimes |\chi\rangle_{12} \otimes |0\rangle_3$. So, $\langle\psi_B | M_j P_k^{(21)} |\psi_B\rangle = \langle 0_0 \chi_{12} 0_3 | (M_{j(03)} \otimes I_{12}) (I_0 \otimes P'_{k(12)} \otimes I_3) | 0_0 \chi_{12} 0_3 \rangle$. $= \langle 0_0 0_3 | M_{j(03)} | 0_0 0_3 \rangle \langle \chi_{12} | P'_{k(12)} | \chi_{12} \rangle$. This allows us to factor the sum: $\sum_{j,k} |\langle\psi_B | M_j P_k^{(21)} |\psi_B\rangle|^2 = \left( \sum_j |\langle 0_0 0_3 | M_{j(03)} | 0_0 0_3 \rangle|^2 \right) \left( \sum_k |\langle \chi_{12} | P'_{k(12)} | \chi_{12} \rangle|^2 \right)$.

The second factor is 3, as calculated for Term 1. For the first factor: $\langle 0_0 0_3 | M_{j(03)} | 0_0 0_3 \rangle = \langle 0_0 0_3 | H_0 CNOT_{03} P_j^{(03)} CNOT_{03} H_0 | 0_0 0_3 \rangle$. Applying $H_0$ to $|0\rangle_0$ gives $|+\rangle_0 = \frac{1}{\sqrt{2}}(|0\rangle_0+|1\rangle_0)$. So, this is $\langle +_0 0_3 | CNOT_{03} P_j^{(03)} CNOT_{03} | +_0 0_3 \rangle$. The action of $CNOT_{03}$ on $|+\rangle_0 |0\rangle_3$ is $CNOT_{03} \frac{1}{\sqrt{2}}(|00\rangle_{03}+|10\rangle_{03}) = \frac{1}{\sqrt{2}}(|00\rangle_{03}+|11\rangle_{03}) = |\Phi^+\rangle_{03}$. So, this term is $\langle \Phi^+_{03} | P_j^{(03)} | \Phi^+_{03} \rangle$. The sum $\sum_j |\langle \Phi^+_{03} | P_j^{(03)} | \Phi^+_{03} \rangle|^2$ is over 15 non-identity Paulis on qubits 0 and 3, in the Bell state $|\Phi^+\rangle_{03}$. This is the same form as Term 2, so the sum is 3.

Therefore, the third term’s contribution is $\frac{p^2}{15^2} \cdot (3 \times 3) = \frac{9p^2}{225} = \frac{p^2}{25}$.

3. Combine terms for total fidelity:

\[ F = (1-p)^2 + (1-p)\frac{p}{15}(3) + (1-p)\frac{p}{15}(3) + \frac{p^2}{25} \]

\[ F = (1-2p+p^2) + \frac{3p(1-p)}{15} + \frac{3p(1-p)}{15} + \frac{p^2}{25} \]

\[ F = (1-2p+p^2) + \frac{2p(1-p)}{5} + \frac{p^2}{25} \]

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

Collect terms by powers of $p$:

\[ F = 1 + \left(-2 + \frac{2}{5}\right)p + \left(1 - \frac{2}{5} + \frac{1}{25}\right)p^2 \]

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

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

This expression can also be written as $F = \left(1 - \frac{4}{5}p\right)^2$.

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

User #3

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 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 $p$.

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

    Outputs
    ----------
    F_physical: sympy.Expr, the physical state fidelity of the final physical 4-qubit state
    """

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

    return F_physical

Assistant #4

import sympy as sp

p = sp.symbols('p')

def answer(p):
    r"""
    Return the expression of 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 $p$.

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

    Outputs
    ----------
    F_physical: sympy.Expr, the physical state fidelity of the final physical 4-qubit state
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    F_physical = 1 - (8/5)*p + (16/25)*p**2
    # ---------------------------------------------------------------

    return F_physical