Step-by-Step Derivation
- Error model
• After each of the two CNOTs a two–qubit depolarising channel acts on the
same two qubits as the gate
\[
(1-p)\rho+\frac{p}{15}\!\!
\]
Denote by
• $P_{21}$ the Pauli error on qubits $(2,1)$ (coming from $CNOT_{21}$),
• $Q_{03}$ the Pauli error on qubits $(0,3)$ (coming from $CNOT_{03}$).
- Which two–qubit Paulis are undetectable?
An error \(E\) is undetectable iff it commutes with both stabilisers
\[
S_X = XXXX,\qquad S_Z = ZZZZ.
\]
For a two–qubit operator supported on some qubit pair
only the three elements
\[
\{\,XX,\;YY,\;ZZ\}\,
\]
commute with **both** $S_X$ and $S_Z$ (identity of course also does).
Therefore, for either pair of qubits
• “good” operators (no syndrome): $I I,\,XX,\,YY,\,ZZ$
• “bad” operators (trigger syndrome): the remaining 12 Paulis.
-
Syndrome pattern produced by a single two-qubit Pauli
For every error we record the pair of parity bits
\[
0 & \text{commutes with }S_X,\\
1 & \text{anticommutes with }S_X,
\end{cases}\qquad
b=\begin{cases}
0 & \text{commutes with }S_Z,\\
1 & \text{anticommutes with }S_Z .
\]
A direct enumeration gives for **each** depolarising channel
| $(a,b)$ | operators | probability |
|---------|-----------|-------------|
| (0,0) | $II,XX,YY,ZZ$ | $p_{00}=1-p+\dfrac{3p}{15}=1-\dfrac{4p}{5}$ |
| (1,0) | $Z,\;Z,\;XY,\;YX$ | $p_{10}= \dfrac{4p}{15}$ |
| (0,1) | $X,\;X,\;YZ,\;ZY$ | $p_{01}= \dfrac{4p}{15}$ |
| (1,1) | $Y,\;Y,\;XZ,\;ZX$ | $p_{11}= \dfrac{4p}{15}$ |
-
Probability that no error is detected (post-selection kept)
Let \((a_1,b_1)\) be the parity bits of \(P_{21}\) and \((a_2,b_2)\) those of
\(Q_{03}\).
After both channels the combined error is detected
iff
\[
a=a_1\oplus a_2,\; b=b_1\oplus b_2
\]
contains at least one ‘1’.
Consequently the run is **kept** when simultaneously
\[
a_1=a_2,\qquad b_1=b_2.
\]
Because the two channels are independent,
\[
P_{\text{keep}}
=\sum_{(a,b)\in\{0,1\}^2} p_{ab}^2
=p_{00}^2+3\,p_{10}^2
=(1-\tfrac{4p}{5})^{\!2}+3\Bigl(\tfrac{4p}{15}\Bigr)^{\!2}
=1-\frac{8p}{5}+\frac{64p^{2}}{75}.
\]
-
Logical state fidelity conditioned on keeping the run
The ideal encoded GHZ state obtained by the circuit is
\[
|\Psi_{\mathrm{enc}}\rangle
=|\Phi^{+}\rangle_{21}\otimes|\Phi^{+}\rangle_{03},
\qquad
|\Phi^{+}\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}.
\]
\[
\begin{cases}
\;\;1 & P\in\{II,XX,YY,ZZ\},\\[4pt]
\;\;0 & \text{otherwise}.
\]
Hence after the two independent errors
\[
\langle\Psi_{\mathrm{enc}}|Q_{03}P_{21}|\Psi_{\mathrm{enc}}\rangle
=\langle\Phi^{+}|P_{21}|\Phi^{+}\rangle\,
\langle\Phi^{+}|Q_{03}|\Phi^{+}\rangle
=
\begin{cases}
1 & P_{21},Q_{03}\in\{II,XX,YY,ZZ\},\\
0 & \text{otherwise}.
\end{cases}
\]
Therefore the (squared) overlap with the target logical state equals 1
**iff** both individual errors are of the $(a,b)=(0,0)$ type.
The probability for that event is $p_{00}^2=(1-\tfrac{4p}{5})^{2}$.
The logical fidelity **conditioned on acceptance** is then
\[
F_{\mathrm{logical}}(p)=
\frac{p_{00}^2}{P_{\text{keep}}}
=\frac{(1-\frac{4}{5}p)^{2}}
{(1-\frac{4}{5}p)^{2}+3\left(\frac{4}{15}p\right)^{2}}
=\frac{(1-\frac{4}{5}p)^{2}}
{1-\frac{8}{5}p+\frac{64}{75}p^{2}}.
\]
\[
F_{\mathrm{logical}}(p)=1-\frac{64}{75}p^{2}+O(p^{3}),
\]
i.e. all first-order error contributions are removed by the
post-selection.
Final Answer:
The logical two-qubit state fidelity after post-selection on all detectable
errors is
\[
\boxed{\,F_{\mathrm{logical}}(p)=
\frac{\bigl(1-\tfrac{4}{5}p\bigr)^{2}}
{1-\tfrac{8}{5}p+\tfrac{64}{75}p^{2}}\,}.
\]