Step-by-Step Derivation
The goal is to compute the physical state fidelity \(F = \langle \psi_{ideal} | \rho_{actual} | \psi_{ideal} \rangle\) of the final state of a 4-qubit system prepared by a given circuit, subject to a specific error model.
1. Determine the Ideal Final State \(|\psi_{ideal}\rangle\)
The circuit operations are applied from right to left to the initial state \(|\psi_0\rangle = |0000\rangle\). The circuit is \(U = (CNOT_{03}) (H_0) (CNOT_{21}) (H_2)\).
- Initial state:
\[
|\psi_0\rangle = |0000\rangle
\]
- Apply \(H_2\) (Hadamard on qubit 2):
\[
|\psi_1\rangle = H_2 |\psi_0\rangle = |00\rangle_01 \otimes \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)_2 \otimes |0\rangle_3 = \frac{1}{\sqrt{2}}(|0000\rangle + |0010\rangle)
\]
- Apply \(CNOT_{21}\) (control qubit 2, target qubit 1):
The control qubit is qubit 2. The target qubit is qubit 1.
\[
CNOT_{21}|0000\rangle = |0000\rangle \quad (\text{control is } |0\rangle)
\]
\[
CNOT_{21}|0010\rangle = |0110\rangle \quad (\text{control is } |1\rangle, \text{flips target})
\]
Applying this to $|\psi_1\rangle$:
\[
|\psi_2\rangle = CNOT_{21} |\psi_1\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |0110\rangle)
\]
- Apply \(H_0\) (Hadamard on qubit 0):
\[
|\psi_3\rangle = H_0 |\psi_2\rangle = \frac{1}{\sqrt{2}} \left( \frac{|0\rangle+|1\rangle}{\sqrt{2}}|000\rangle + \frac{|0\rangle+|1\rangle}{\sqrt{2}}|110\rangle \right) = \frac{1}{2}(|0000\rangle + |1000\rangle + |0110\rangle + |1110\rangle)
\]
- Apply \(CNOT_{03}\) (control qubit 0, target qubit 3):
\[
CNOT_{03}|0000\rangle = |0000\rangle
\]
\[
CNOT_{03}|1000\rangle = |1001\rangle
\]
\[
CNOT_{03}|0110\rangle = |0110\rangle
\]
\[
CNOT_{03}|1110\rangle = |1111\rangle
\]
Applying this to $|\psi_3\rangle$ gives the ideal final state:
\[
|\psi_{ideal}\rangle = CNOT_{03} |\psi_3\rangle = \frac{1}{2}(|0000\rangle + |1001\rangle + |0110\rangle + |1111\rangle)
\]
2. Analyze the Error Model and Fidelity Calculation Strategy
The error model states that after each of the two \(CNOT\) gates, a two-qubit depolarizing error channel is applied. The channel for a state \(\rho\) is:
\[
\mathcal{E}(\rho) = (1-p)\rho + \frac{p}{15} \sum_{P \in \{I,X,Y,Z\}^{\otimes 2} \setminus \{II\}} P \rho P^\dagger
\]
where \(p\) is the total probability of an error.
We can calculate the fidelity by considering the four mutually exclusive events for the two \(CNOT\) gates:
* Event 0 (NN): No error on either gate.
* Event 1 (EN): Error on the first \(CNOT\) (\(CNOT_{21}\)), no error on the second (\(CNOT_{03}\)).
* Event 2 (NE): No error on the first \(CNOT\), error on the second.
* Event 3 (EE): Error on both \(CNOT\) gates.
The total fidelity is the sum of the fidelities from each event, weighted by the probability of that event:
\[
F = P(NN)F_{NN} + P(EN)F_{EN} + P(NE)F_{NE} + P(EE)F_{EE}
\]
The probability of no error on a gate is \((1-p)\), and the probability of an error is \(p\).
* \(P(NN) = (1-p)^2\)
* \(P(EN) = p(1-p)\)
* \(P(NE) = (1-p)p\)
* \(P(EE) = p^2\)
3. Calculate Fidelity for Each Event
-
Fidelity for Event 0 (NN):
If no errors occur, the final state is \(|\psi_{ideal}\rangle\). The fidelity is \(F_{NN} = \langle\psi_{ideal}|\psi_{ideal}\rangle = 1\).
Contribution to total fidelity: \((1-p)^2\).
-
Fidelity for Event 2 (NE): (This is simpler than Event 1, so we do it first)
The circuit is error-free until after the final \(CNOT_{03}\). The state is \(|\psi_{ideal}\rangle\). An error \(P_{03} \in \{I,X,Y,Z\}^{\otimes 2} \setminus \{II\}\) then occurs on qubits 0 and 3 with probability \(p/15\). The final state is \(P_{03}|\psi_{ideal}\rangle\). The fidelity, conditioned on error \(P_{03}\) occurring, is \(|\langle\psi_{ideal}|P_{03}|\psi_{ideal}\rangle|^2\).
We need to average this over the 15 possible errors:
\[
F_{NE} = \frac{1}{15} \sum_{P_{03} \neq II} |\langle\psi_{ideal}|P_{03}|\psi_{ideal}\rangle|^2
\]
The state $|\psi_{ideal}\rangle$ is an eigenstate of the code's stabilizers $S_1=XXXX$ and $S_2=ZZZZ$. An operator $P$ that anti-commutes with a stabilizer $S$ of a state $|\psi\rangle$ will have zero expectation value: $\langle\psi|P|\psi\rangle = \langle\psi|SPS|\psi\rangle = -\langle\psi|P|\psi\rangle = 0$. We only need to sum over Paulis $P_{03}$ that commute with both $S_1$ and $S_2$. These are $\{II, X_0X_3, Y_0Y_3, Z_0Z_3\}$.
Let's compute the expectation values:
\[
\langle\psi_{ideal}|X_0X_3|\psi_{ideal}\rangle = \frac{1}{4}(\langle0000|+\dots)(X_0X_3)(|0000\rangle+\dots) = \frac{1}{4}(\langle0000|+\dots)(|1001\rangle+|0000\rangle+|1111\rangle+|0110\rangle) = 1
\]
\[
\langle\psi_{ideal}|Y_0Y_3|\psi_{ideal}\rangle = \frac{1}{4}(\langle0000|+\dots)(Y_0Y_3)(|0000\rangle+\dots) = \frac{1}{4}(\langle0000|+\dots)(-1|1001\rangle-1|0000\rangle-1|0110\rangle-1|1111\rangle) = -1
\]
\[
\langle\psi_{ideal}|Z_0Z_3|\psi_{ideal}\rangle = \frac{1}{4}(\langle0000|+\dots)(|0000\rangle-|1001\rangle-|0110\rangle+|1111\rangle) = \frac{1}{4}(1-1-1+1) = 0
\]
The sum is $1^2 + (-1)^2 + 0^2 = 2$.
\[
F_{NE} = \frac{2}{15}
\]
Contribution to total fidelity: $p(1-p)\frac{2}{15}$.
- Fidelity for Event 1 (EN):
An error \(Q_{21} \neq II\) occurs after \(CNOT_{21}\). The state becomes \(Q_{21}|\psi_2\rangle\). The rest of the circuit \(U_{rem} = CNOT_{03}H_0\) is applied, resulting in the state \(U_{rem}Q_{21}|\psi_2\rangle\). The ideal final state is \(|\psi_{ideal}\rangle = U_{rem}|\psi_2\rangle\).
The fidelity, conditioned on error \(Q_{21}\), is \(|\langle\psi_{ideal}|U_{rem}Q_{21}|\psi_2\rangle|^2 = |\langle\psi_2|U_{rem}^\dagger U_{rem}Q_{21}|\psi_2\rangle|^2 = |\langle\psi_2|Q_{21}|\psi_2\rangle|^2\).
We need to average this over the 15 possible errors \(Q_{21}\):
\[
F_{EN} = \frac{1}{15} \sum_{Q_{21} \neq II} |\langle\psi_2|Q_{21}|\psi_2\rangle|^2
\]
The state is $|\psi_2\rangle = \frac{1}{\sqrt{2}}(|0000\rangle+|0110\rangle) = |0\rangle_0 \otimes \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)_{12} \otimes |0\rangle_3$. Let $|\Phi^+\rangle_{12} = \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)_{12}$. The operator $Q_{21}$ acts on qubits 1 and 2, so $\langle\psi_2|Q_{21}|\psi_2\rangle = \langle\Phi^+_{12}|Q_{12}|\Phi^+_{12}\rangle$.
For a Bell state $|\Phi^+\rangle$, the non-zero expectation values for Pauli operators $P \neq II$ are:
\[
\langle\Phi^+|XX|\Phi^+\rangle = 1, \quad \langle\Phi^+|YY|\Phi^+\rangle = -1, \quad \langle\Phi^+|ZZ|\Phi^+\rangle = 1
\]
The sum is $1^2 + (-1)^2 + 1^2 = 3$.
\[
F_{EN} = \frac{3}{15} = \frac{1}{5}
\]
Contribution to total fidelity: $p(1-p)\frac{3}{15}$.
- Fidelity for Event 3 (EE):
Error \(Q_{21}\) on the first gate, error \(P_{03}\) on the second. The final state is \(P_{03} U_{rem} Q_{21} |\psi_2\rangle\).
Fidelity is \(|\langle\psi_{ideal}|P_{03} U_{rem} Q_{21}|\psi_2\rangle|^2 = |\langle\psi_2|U_{rem}^\dagger P_{03} U_{rem} Q_{21}|\psi_2\rangle|^2\).
The state \(|\psi_2\rangle = |00\rangle_{03} \otimes |\Phi^+\rangle_{12}\). The operator \(U_{rem}^\dagger P_{03} U_{rem}\) acts on qubits 0,3 and \(Q_{21}\) acts on qubits 1,2. The expectation value factorizes:
\[
\langle\psi_2|U_{rem}^\dagger P_{03} U_{rem} Q_{21}|\psi_2\rangle = \langle 00|_{03} U_{rem}^\dagger P_{03} U_{rem} |00\rangle_{03} \cdot \langle\Phi^+_{12}|Q_{21}|\Phi^+_{12}\rangle
\]
The first term is $\langle (U_{rem}|00\rangle_{03}) | P_{03} | (U_{rem}|00\rangle_{03}) \rangle$. Let's compute $U_{rem}|00\rangle_{03}$:
\[
U_{rem}|00\rangle_{03} = CNOT_{03}H_0|00\rangle_{03} = CNOT_{03}\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)_0|0\rangle_3 = \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)_{03} = |\Phi^+\rangle_{03}
\]
So, $\langle\psi_2|\dots|\psi_2\rangle = \langle\Phi^+_{03}|P_{03}|\Phi^+_{03}\rangle \cdot \langle\Phi^+_{12}|Q_{21}|\Phi^+_{12}\rangle$.
The average fidelity is:
\[
F_{EE} = \frac{1}{15^2} \sum_{P_{03}\neq II, Q_{21}\neq II} |\langle\Phi^+_{03}|P_{03}|\Phi^+_{03}\rangle \cdot \langle\Phi^+_{12}|Q_{21}|\Phi^+_{12}\rangle|^2
\]
\[
F_{EE} = \left(\frac{1}{15}\sum_{P\neq II}|\langle\Phi^+|P|\Phi^+\rangle|^2\right) \left(\frac{1}{15}\sum_{Q\neq II}|\langle\Phi^+|Q|\Phi^+\rangle|^2\right) = \left(\frac{3}{15}\right)\left(\frac{3}{15}\right) = \frac{9}{225} = \frac{1}{25}
\]
Contribution to total fidelity: $p^2 \frac{9}{225}$.
4. Assemble the Total Fidelity
Summing the contributions from all four events:
\[
F = (1-p)^2 \cdot 1 + p(1-p) \cdot \frac{3}{15} + p(1-p) \cdot \frac{2}{15} + p^2 \cdot \frac{9}{225}
\]
\[
F = (1-p)^2 + p(1-p)\left(\frac{3+2}{15}\right) + \frac{9p^2}{225}
\]
\[
F = (1-p)^2 + p(1-p)\frac{5}{15} + \frac{p^2}{25}
\]
\[
F = (1-2p+p^2) + \frac{p-p^2}{3} + \frac{p^2}{25}
\]
\[
F = 1 + (-2+\frac{1}{3})p + (1-\frac{1}{3}+\frac{1}{25})p^2
\]
\[
F = 1 - \frac{5}{3}p + \left(\frac{75-25+3}{75}\right)p^2
\]
\[
F = 1 - \frac{5}{3}p + \frac{53}{75}p^2
\]
Final Answer:
The physical state fidelity as a function of the two-qubit gate error rate \(p\) is:
\[
F(p) = 1 - \frac{5}{3}p + \frac{53}{75}p^2
\]