Step-by-Step Derivation
We label the two noisy CNOT gates as acting on qubit‐pairs
– first \( \mathrm{CNOT}_{21}\) with error \(P_1\) on qubits \((2,1)\),
– then \( \mathrm{CNOT}_{03}\) with error \(P_2\) on qubits \((0,3)\).
No errors occur on the Hadamards. Thus the total error is
\[
E \;=\; P_2^{(0,3)}\;P_1^{(2,1)},
\]
where \(P_i\) are drawn i.i.d. from the two-qubit depolarizing channel
\[
\Pr\bigl(P_i = I\otimes I\bigr)=1-p,
\quad
\Pr\bigl(P_i =\text{any one of the other }15\text{ Paulis}\bigr)
=\frac p{15}\,.
\]
- Post‐selection (“no‐detect”) condition.
We measure the two stabilizers
\[
S_X = X_0X_1X_2X_3,\qquad
S_Z = Z_0Z_1Z_2Z_3
\]
at the end and keep only those runs where both syndromes are “+1.”
Equivalently, we require
\[
[E,S_X]=0
\quad\text{and}\quad
[E,S_Z]=0.
\]
One checks that these two commutation conditions are equivalent to
requiring that the Pauli on \((0,3)\) and the Pauli on \((1,2)\) have the same
“\(Z/Y\)–parity” and the same “\(X/Y\)–parity.”
- Single‐pair parity classification.
For a single two‐qubit Pauli \(P\) on a pair we define
\[
a = \#\{\,\text{qubits where }P\in\{Z,Y\}\}\bmod2,\quad
b = \#\{\,\text{qubits where }P\in\{X,Y\}\}\bmod2.
\]
One finds that among the 16 Pauli operators on two qubits exactly
4 have parity \((a,b)=(0,0)\), and the other three parity‐classes
\((0,1),(1,0),(1,1)\) each also have 4 operators. Since the
depolarizing channel is uniform over the 15 non‐identity Paulis,
the probability that a random error on one pair lies in class \((a,b)\) is
\[
\pi(0,0)= (1-p)\;+\;3\frac p{15} \;=\;1-\frac{4p}{5},
\qquad
\pi(0,1)=\pi(1,0)=\pi(1,1)=4\frac p{15}\,.
\]
- Denominator: probability of no‐detection.
The two errors are independent, and commutation with both
stabilizers forces the two pairs to lie in the same parity‐class.
Hence
\[
\Pr[\text{no detect}]
= \sum_{(a,b)\in\{0,1\}^2}\;\pi(a,b)^2
= \Bigl(1-\tfrac{4p}{5}\Bigr)^2
\;+\;3\Bigl(4\tfrac p{15}\Bigr)^2
=1-\tfrac{8}{5}p+\tfrac{64}{75}p^2.
\]
- Numerator: probability of correct logical GHZ.
Conditioned on no‐detect, the remaining errors lie in the code’s
normalizer \(N(S)\). Those that act trivially on the logical GHZ are
exactly
– the four stabilizers \(\{I,XXXX,ZZZZ,YYYY\}\), and
– the four elements \(Z_AZ_B\cdot S\) (the coset which implements a
global logical \(Z_AZ_B\), which on \(\tfrac{|00\rangle+|11\rangle}{\sqrt2}\)
also has eigenvalue \(+1\)).
Of these eight “good” errors exactly four lie in the identity‐stabilizer
coset:
\((P_2,P_1)\in\{(I,I),(XX,XX),(ZZ,ZZ),(YY,YY)\}\),
and four lie in the \(Z_AZ_B\)–coset:
\(\bigl\{(I,ZZ),(ZZ,I),(XX,YY),(YY,XX)\bigr\}\).
Their probabilities are
– \(\Pr[P_2=P_1=I]=(1-p)^2\),
– each weight-2 pair \((XX,XX),(ZZ,ZZ),(YY,YY)\) has \(\,(p/15)^2\),
– the two “mixed” coset events \((I,ZZ)\) or \((ZZ,I)\) each have
\((1-p)(p/15)\),
– and the other two \((XX,YY),(YY,XX)\) each have \((p/15)^2\).
Summing gives
\[
\Pr[\text{logical‐good}]
= (1-p)^2
\;+\;2\,(1-p)\frac p{15}
\;+\;5\Bigl(\frac p{15}\Bigr)^2
=1-\frac{28}{15}\,p+\frac{8}{9}\,p^2.
\]
- Logical‐state fidelity.
Conditioning on no‐detection, the overlap of the logical state with the
ideal GHZ is the fraction of “good” over all no‐detect events:
\[
F_L(p)
=\frac{\Pr[\text{logical‐good}]}{\Pr[\text{no detect}]}
=\frac{1-\tfrac{28}{15}p+\tfrac{8}{9}p^2}
{1-\tfrac{8}{5}p+\tfrac{64}{75}p^2}\,.
\]
Final Answer:
\[
\boxed{
F_L(p)
\;=\;
\frac{1-\tfrac{28}{15}p+\tfrac{8}{9}p^2}
{1-\tfrac{8}{5}p+\tfrac{64}{75}p^2}
}.
\]