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)

Analysis of Physical State Fidelity in a [[4,2,2]] Code

My task is to calculate the fidelity of a prepared logical GHZ state in a [[4,2,2]] quantum error-correcting code, considering the effect of depolarizing errors on the circuit. First, I need to thoroughly understand the properties of this code, which involves four physical qubits, two logical qubits, and specific stabilizer and logical operators. The goal is to analyze a circuit intended to create a GHZ state and determine the impact of depolarizing errors on its fidelity.

The circuit starts with the initial state $|0000\rangle$, applies a Hadamard gate to qubit 2, followed by a CNOT gate between qubits 2 (control) and 1 (target). It then applies a Hadamard gate to qubit 0 and a CNOT gate between qubits 0 (control) and 3 (target). Ideally, the circuit’s output should represent the logical GHZ state $(|00\rangle_{AB} + |11\rangle_{AB})/\sqrt{2}$. My objective is to define the perfect output state based on the circuit operation, then quantify how much it deviates from it due to the errors.

I’ve determined the ideal state by simulating the circuit. The result, $|\psi_F^{\text{ideal}}\rangle = \frac{1}{2}(|0000\rangle + |0110\rangle + |1001\rangle + |1111\rangle)$, is a codeword in the code space, and importantly, an eigenstate of the operator $X_A X_B$ with eigenvalue +1 (where $X_A$ and $X_B$ are physical operators defined in the code). The state is also an eigenstate of $Z_A Z_B$ with eigenvalue +1. This, in turn, demonstrates that the state the circuit generates, in the absence of noise, represents the target logical GHZ state.

The core of the problem involves the error model, a two-qubit depolarizing channel applied after each CNOT gate. The fidelity calculation is based on the overlap between the ideal state and the final state after the error channels. The final state $\rho_{\text{final}}$ is determined by tracing the state evolution through each gate and error channel.

I am analyzing the error channels after each CNOT and how they impact the initial pure state. Because of the nature of the error channels, each Pauli operator in the expression of fidelity contributes an additional degree of freedom, but with the specific state $|\psi\rangle$, most are zero. I have determined that, of the 225 terms used in the calculation, only $P=I$ has an expectation value of unity.

Through a rigorous derivation, I’ve reduced the fidelity calculation to a simple expression. The final state fidelity $F$ should be $1 - \frac{8p}{5} + \frac{142p^2}{225}$, where $p$ is the depolarizing parameter. The analysis confirms that the fidelity decreases quadratically with increasing $p$.

The problem asks for the physical state fidelity of a prepared logical two-qubit GHZ state in the [[4,2,2]] code, subject to depolarizing errors on two-qubit gates.

First, let’s determine the ideal target state. The circuit operations are applied from right to left:

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

Starting from the initial state $|0000\rangle$:

Apply $H_2$ on qubit 2:

\[ |\psi_1\rangle = I_0 I_1 H_2 I_3 |0000\rangle = |00+0\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle) \]

Apply $CNOT_{21}$ (control qubit 2, target qubit 1):

\[ |\psi_2\rangle = CNOT_{21} |\psi_1\rangle = \frac{1}{\sqrt{2}}(CNOT_{21}|0000\rangle + CNOT_{21}|0010\rangle) \]

\[ CNOT_{21}|0000\rangle = |0000\rangle \]

\[ CNOT_{21}|0010\rangle = |0110\rangle \]

So,

\[ |\psi_2\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle) \]

Apply $H_0$ on qubit 0:

\[ |\psi_3\rangle = H_0 I_1 I_2 I_3 |\psi_2\rangle = \frac{1}{\sqrt{2}}(H_0|0000\rangle + H_0|0110\rangle) \]

\[ H_0|0000\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1000\rangle) \]

\[ H_0|0110\rangle = \frac{1}{\sqrt{2}}(|0110\rangle + |1110\rangle) \]

So,

\[ |\psi_3\rangle = \frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle) \]

Apply $CNOT_{03}$ (control qubit 0, target qubit 3):

\[ |\psi_{\text{target}}\rangle = CNOT_{03} |\psi_3\rangle = \frac{1}{2}(CNOT_{03}|0000\rangle + CNOT_{03}|1000\rangle + CNOT_{03}|0110\rangle + CNOT_{03}|1110\rangle) \]

\[ CNOT_{03}|0000\rangle = |0000\rangle \]

\[ CNOT_{03}|1000\rangle = |1001\rangle \]

\[ CNOT_{03}|0110\rangle = |0110\rangle \]

\[ CNOT_{03}|1110\rangle = |1111\rangle \]

The ideal final state is:

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

We should verify that this state is indeed a logical GHZ state and is in the code space. The stabilizers are $S_X = X_0X_1X_2X_3$ and $S_Z = Z_0Z_1Z_2Z_3$. For $S_X$: $X_0X_1X_2X_3(|0000\rangle + |0110\rangle + |1001\rangle + |1111\rangle) = (|1111\rangle + |1001\rangle + |0110\rangle + |0000\rangle)$, so $S_X|\psi_{\text{target}}\rangle = |\psi_{\text{target}}\rangle$. For $S_Z$: $Z_0Z_1Z_2Z_3(|0000\rangle + |0110\rangle + |1001\rangle + |1111\rangle) = (|0000\rangle + |0110\rangle + |1001\rangle + |1111\rangle)$ (each term maps to itself because an even number of Z gates act on 1-qubits), so $S_Z|\psi_{\text{target}}\rangle = |\psi_{\text{target}}\rangle$. Thus, $|\psi_{\text{target}}\rangle$ is a valid codeword.

The logical operators are $X_A = X_0X_2$, $X_B=X_0X_1$, $Z_A = Z_0Z_1$, $Z_B = Z_0Z_2$. The logical GHZ state $(|00\rangle_{AB}+|11\rangle_{AB})/\sqrt{2}$ satisfies $X_A X_B |\text{GHZ}\rangle = |\text{GHZ}\rangle$ and $Z_A Z_B |\text{GHZ}\rangle = |\text{GHZ}\rangle$. Let’s check this for $|\psi_{\text{target}}\rangle$: $X_A X_B = (X_0X_2)(X_0X_1) = X_0^2 X_1 X_2 = I_0X_1X_2I_3$. $I_0X_1X_2I_3|\psi_{\text{target}}\rangle = \frac{1}{2}(X_1X_2|0000\rangle + X_1X_2|0110\rangle + X_1X_2|1001\rangle + X_1X_2|1111\rangle)$ $= \frac{1}{2}(|0110\rangle + |0000\rangle + |1111\rangle + |1001\rangle) = |\psi_{\text{target}}\rangle$. $Z_A Z_B = (Z_0Z_1)(Z_0Z_2) = Z_0^2 Z_1 Z_2 = I_0Z_1Z_2I_3$. $I_0Z_1Z_2I_3|\psi_{\text{target}}\rangle = \frac{1}{2}(Z_1Z_2|0000\rangle + Z_1Z_2|0110\rangle + Z_1Z_2|1001\rangle + Z_1Z_2|1111\rangle)$ $= \frac{1}{2}((+1)|0000\rangle + (-1)(-1)|0110\rangle + (+1)(+1)|1001\rangle + (-1)(-1)|1111\rangle)$ $= \frac{1}{2}(|0000\rangle + |0110\rangle + |1001\rangle + |1111\rangle) = |\psi_{\text{target}}\rangle$. The state $|\psi_{\text{target}}\rangle$ is indeed a physical representative of the logical GHZ state.

Now, let’s incorporate the errors. Each CNOT gate is followed by a two-qubit depolarizing error channel. The Hadamard gates are assumed ideal. Let $\rho_0 = |0000\rangle\langle 0000|$. The circuit is $U_2 U_1$, where $U_1 = (CNOT_{21})(H_2)$ and $U_2 = (CNOT_{03})(H_0)$. The operations occur in this order: $H_2$, $CNOT_{21}$, $H_0$, $CNOT_{03}$. The errors are applied after $CNOT_{21}$ and $CNOT_{03}$.

Initial state $\rho_0 = |0000\rangle\langle 0000|$.
Apply $H_2$: $\rho_1 = H_2 \rho_0 H_2^\dagger = |\psi_1\rangle\langle\psi_1|$.
Apply $CNOT_{21}$: $\rho_2' = C_{21} \rho_1 C_{21}^\dagger = |\psi_2\rangle\langle\psi_2|$.
Apply depolarizing channel $\mathcal{E}_{12}$ on qubits (1,2) after $CNOT_{21}$:

\[ \rho_2 = (1-p)\rho_2' + \frac{p}{15} \sum_{k=1}^{15} P_k^{(12)} \rho_2' P_k^{(12)\dagger} \]

where $P_k^{(12)}$ are the 15 non-identity two-qubit Paulis acting on qubits 1 and 2.

Apply $H_0$: $\rho_3' = H_0 \rho_2 H_0^\dagger$. Since $H_0$ acts on qubit 0, it commutes with $P_k^{(12)}$.

\[ \rho_3' = (1-p)H_0\rho_2'H_0^\dagger + \frac{p}{15} \sum_{k=1}^{15} P_k^{(12)} (H_0\rho_2'H_0^\dagger) P_k^{(12)\dagger} \]

Let $\rho_3 = H_0|\psi_2\rangle\langle\psi_2|H_0^\dagger = |\psi_3\rangle\langle\psi_3|$. Then,

\[ \rho_3' = (1-p)\rho_3 + \frac{p}{15} \sum_{k=1}^{15} P_k^{(12)} \rho_3 P_k^{(12)\dagger} \]

Apply $CNOT_{03}$: $\rho_4' = C_{03} \rho_3' C_{03}^\dagger$. Since $C_{03}$ acts on qubits 0 and 3, it commutes with $P_k^{(12)}$.

\[ \rho_4' = (1-p)C_{03}\rho_3 C_{03}^\dagger + \frac{p}{15} \sum_{k=1}^{15} P_k^{(12)} (C_{03}\rho_3 C_{03}^\dagger) P_k^{(12)\dagger} \]

Let $\rho_{\text{target}} = C_{03}|\psi_3\rangle\langle\psi_3|C_{03}^\dagger = |\psi_{\text{target}}\rangle\langle\psi_{\text{target}}|$. Then,

\[ \rho_4' = (1-p)\rho_{\text{target}} + \frac{p}{15} \sum_{k=1}^{15} P_k^{(12)} \rho_{\text{target}} P_k^{(12)\dagger} \]

Apply depolarizing channel $\mathcal{E}_{03}$ on qubits (0,3) after $CNOT_{03}$:

\[ \rho_{\text{final}} = (1-p)\rho_4' + \frac{p}{15} \sum_{m=1}^{15} P_m^{(03)} \rho_4' P_m^{(03)\dagger} \]

where $P_m^{(03)}$ are the 15 non-identity two-qubit Paulis acting on qubits 0 and 3.

Substitute $\rho_4'$ into the expression for $\rho_{\text{final}}$:

\[ \rho_{\text{final}} = (1-p) \left[ (1-p)\rho_{\text{target}} + \frac{p}{15} \sum_{k=1}^{15} P_k^{(12)} \rho_{\text{target}} P_k^{(12)\dagger} \right] \]

\[ \qquad + \frac{p}{15} \sum_{m=1}^{15} P_m^{(03)} \left[ (1-p)\rho_{\text{target}} + \frac{p}{15} \sum_{k=1}^{15} P_k^{(12)} \rho_{\text{target}} P_k^{(12)\dagger} \right] P_m^{(03)\dagger} \]

The physical state fidelity is $F = \langle\psi_{\text{target}}| \rho_{\text{final}} |\psi_{\text{target}}\rangle$. Let $|\psi\rangle = |\psi_{\text{target}}\rangle$. Since $P_k^{(12)}$ acts on qubits 1,2 and $P_m^{(03)}$ acts on qubits 0,3, they commute ($P_m^{(03)} P_k^{(12)} = P_k^{(12)} P_m^{(03)}$). Also, $P_k^\dagger = P_k$.

\[ F = (1-p)^2 \langle\psi|\rho_{\text{target}}|\psi\rangle + (1-p)\frac{p}{15} \sum_{k=1}^{15} \langle\psi|P_k^{(12)}\rho_{\text{target}}P_k^{(12)}|\psi\rangle \]

\[ \qquad + (1-p)\frac{p}{15} \sum_{m=1}^{15} \langle\psi|P_m^{(03)}\rho_{\text{target}}P_m^{(03)}|\psi\rangle \]

\[ \qquad + \left(\frac{p}{15}\right)^2 \sum_{m=1}^{15} \sum_{k=1}^{15} \langle\psi|P_m^{(03)}P_k^{(12)}\rho_{\text{target}}P_k^{(12)}P_m^{(03)}|\psi\rangle \]

Since $\rho_{\text{target}} = |\psi\rangle\langle\psi|$ and $\langle\psi|\psi\rangle = 1$:

\[ F = (1-p)^2 + (1-p)\frac{p}{15} \sum_{k=1}^{15} |\langle\psi|P_k^{(12)}|\psi\rangle|^2 \]

\[ \qquad + (1-p)\frac{p}{15} \sum_{m=1}^{15} |\langle\psi|P_m^{(03)}|\psi\rangle|^2 \]

\[ \qquad + \left(\frac{p}{15}\right)^2 \sum_{m=1}^{15} \sum_{k=1}^{15} |\langle\psi|P_m^{(03)}P_k^{(12)}|\psi\rangle|^2 \]

Now we need to compute the expectation values $\langle\psi|P|\psi\rangle$ for various Pauli operators $P$. $|\psi\rangle = \frac{1}{2}(|0000\rangle + |0110\rangle + |1001\rangle + |1111\rangle)$.

For $P_k^{(12)}$ (Paulis on qubits 1 and 2): The set $\mathcal{P}_{12}$ consists of 15 non-identity Paulis of the form $I \otimes P_1 \otimes P_2 \otimes I$. The expectation value $\langle\psi|P_k^{(12)}|\psi\rangle$ is non-zero only for specific correlators:
- $\langle\psi|X_1X_2|\psi\rangle = \frac{1}{4} \langle (0000) + (0110) + (1001) + (1111) | X_1X_2 | (0000) + (0110) + (1001) + (1111) \rangle$ $= \frac{1}{4} (\langle 0000|X_1X_2|0110\rangle + \langle 0110|X_1X_2|0000\rangle + \langle 1001|X_1X_2|1111\rangle + \langle 1111|X_1X_2|1001\rangle)$ $= \frac{1}{4} (1+1+1+1) = 1$. So $|\langle\psi|X_1X_2|\psi\rangle|^2 = 1$.
- $\langle\psi|Y_1Y_2|\psi\rangle$: $Y_1Y_2|0000\rangle = -|0110\rangle$, $Y_1Y_2|0110\rangle = -|0000\rangle$, $Y_1Y_2|1001\rangle = -|1111\rangle$, $Y_1Y_2|1111\rangle = -|1001\rangle$. So $\langle\psi|Y_1Y_2|\psi\rangle = \frac{1}{4} (-1-1-1-1) = -1$. So $|\langle\psi|Y_1Y_2|\psi\rangle|^2 = 1$.
- $\langle\psi|Z_1Z_2|\psi\rangle = \frac{1}{4} (\langle 0000|Z_1Z_2|0000\rangle + \langle 0110|Z_1Z_2|0110\rangle + \langle 1001|Z_1Z_2|1001\rangle + \langle 1111|Z_1Z_2|1111\rangle)$ $= \frac{1}{4} (1 \cdot 1 + (-1) \cdot (-1) + 1 \cdot 1 + (-1) \cdot (-1)) = \frac{1}{4}(1+1+1+1) = 1$. So $|\langle\psi|Z_1Z_2|\psi\rangle|^2 = 1$. For all other 12 Paulis in $\mathcal{P}_{12}$ (e.g., $X_1I_2$, $I_1X_2$, $X_1Y_2$), the expectation value is 0. Thus, $\sum_{k=1}^{15} |\langle\psi|P_k^{(12)}|\psi\rangle|^2 = 1+1+1 = 3$.
For $P_m^{(03)}$ (Paulis on qubits 0 and 3): Similarly, the set $\mathcal{P}_{03}$ consists of 15 non-identity Paulis of the form $P_0 \otimes I_1 \otimes I_2 \otimes P_3$. The expectation values are non-zero for correlators $X_0X_3, Y_0Y_3, Z_0Z_3$:
- $\langle\psi|X_0X_3|\psi\rangle = \frac{1}{4} (\langle 0000|X_0X_3|1001\rangle + \langle 0110|X_0X_3|1111\rangle + \langle 1001|X_0X_3|0000\rangle + \langle 1111|X_0X_3|0110\rangle) = 1$. So $|\langle\psi|X_0X_3|\psi\rangle|^2 = 1$.
- $\langle\psi|Y_0Y_3|\psi\rangle = -1$. So $|\langle\psi|Y_0Y_3|\psi\rangle|^2 = 1$.
- $\langle\psi|Z_0Z_3|\psi\rangle = 1$. So $|\langle\psi|Z_0Z_3|\psi\rangle|^2 = 1$. For all other 12 Paulis in $\mathcal{P}_{03}$, the expectation value is 0. Thus, $\sum_{m=1}^{15} |\langle\psi|P_m^{(03)}|\psi\rangle|^2 = 1+1+1 = 3$.
For $P_m^{(03)}P_k^{(12)}$ (product of Paulis from disjoint sets): These are 4-qubit Paulis where $P_m \in \mathcal{P}_{03}$ and $P_k \in \mathcal{P}_{12}$. An expectation value $\langle\psi|P_m P_k|\psi\rangle$ is non-zero only if $P_m$ and $P_k$ are two-qubit Paulis (not single qubit Paulis like $X_0 I_3$). This means $P_m \in \{X_0X_3, Y_0Y_3, Z_0Z_3\}$ (3 terms) and $P_k \in \{X_1X_2, Y_1Y_2, Z_1Z_2\}$ (3 terms). There are $3 \times 3 = 9$ such combinations. Let $A = P_m^{(03)}$ and $B = P_k^{(12)}$.
- $\langle\psi|X_0X_3 X_1X_2|\psi\rangle = \langle\psi|X_0X_1X_2X_3|\psi\rangle = \langle\psi|S_X|\psi\rangle = 1$. So $1^2=1$.
- $\langle\psi|X_0X_3 Y_1Y_2|\psi\rangle = 0$. So $0^2=0$.
- $\langle\psi|X_0X_3 Z_1Z_2|\psi\rangle = \frac{1}{4} (\langle 0000|X_0Z_1Z_2X_3|1001\rangle + \langle 0110|X_0Z_1Z_2X_3|1111\rangle + \langle 1001|X_0Z_1Z_2X_3|0000\rangle + \langle 1111|X_0Z_1Z_2X_3|0110\rangle) = 1$. So $1^2=1$.
- $\langle\psi|Y_0Y_3 X_1X_2|\psi\rangle = 0$. So $0^2=0$.
- $\langle\psi|Y_0Y_3 Y_1Y_2|\psi\rangle = \langle\psi|Y_0Y_1Y_2Y_3|\psi\rangle = 1$. So $1^2=1$.
- $\langle\psi|Y_0Y_3 Z_1Z_2|\psi\rangle = \frac{1}{4} (\langle 0000|Y_0Z_1Z_2Y_3|1001\rangle + \langle 0110|Y_0Z_1Z_2Y_3|1111\rangle + \langle 1001|Y_0Z_1Z_2Y_3|0000\rangle + \langle 1111|Y_0Z_1Z_2Y_3|0110\rangle) = 1$. So $1^2=1$.
- $\langle\psi|Z_0Z_3 X_1X_2|\psi\rangle = \frac{1}{4} (\langle 0000|Z_0X_1X_2Z_3|0110\rangle + \langle 0110|Z_0X_1X_2Z_3|0000\rangle + \langle 1001|Z_0X_1X_2Z_3|1111\rangle + \langle 1111|Z_0X_1X_2Z_3|1001\rangle) = 1$. So $1^2=1$.
- $\langle\psi|Z_0Z_3 Y_1Y_2|\psi\rangle = \frac{1}{4} (\langle 0000|Z_0Y_1Y_2Z_3|0110\rangle + \langle 0110|Z_0Y_1Y_2Z_3|0000\rangle + \langle 1001|Z_0Y_1Y_2Z_3|1111\rangle + \langle 1111|Z_0Y_1Y_2Z_3|1001\rangle) = -1$. So $(-1)^2=1$.
- $\langle\psi|Z_0Z_3 Z_1Z_2|\psi\rangle = \langle\psi|Z_0Z_1Z_2Z_3|\psi\rangle = \langle\psi|S_Z|\psi\rangle = 1$. So $1^2=1$. Thus, $\sum_{m=1}^{15} \sum_{k=1}^{15} |\langle\psi|P_m^{(03)}P_k^{(12)}|\psi\rangle|^2 = 1+0+1+0+1+1+1+1+1 = 7$.

Substitute these values back into the fidelity expression:

\[ F = (1-p)^2 + (1-p)\frac{p}{15}(3) + (1-p)\frac{p}{15}(3) + \left(\frac{p}{15}\right)^2(7) \]

\[ F = (1-2p+p^2) + \frac{6p(1-p)}{15} + \frac{7p^2}{225} \]

\[ F = (1-2p+p^2) + \frac{2p(1-p)}{5} + \frac{7p^2}{225} \]

\[ F = 1 - 2p + p^2 + \frac{2p}{5} - \frac{2p^2}{5} + \frac{7p^2}{225} \]

Group terms by powers of $p$:

\[ F = 1 + p\left(-2 + \frac{2}{5}\right) + p^2\left(1 - \frac{2}{5} + \frac{7}{225}\right) \]

\[ F = 1 + p\left(-\frac{10}{5} + \frac{2}{5}\right) + p^2\left(\frac{225}{225} - \frac{90}{225} + \frac{7}{225}\right) \]

\[ F = 1 - \frac{8p}{5} + p^2\left(\frac{225 - 90 + 7}{225}\right) \]

\[ F = 1 - \frac{8p}{5} + \frac{142p^2}{225} \]

The final answer is $\boxed{1 - \frac{8p}{5} + \frac{142p^2}{225}}$.

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*p)/5 + (142*p**2)/225
    # ---------------------------------------------------------------

    return F_physical