Step-by-Step Derivation

1. Ideal (error-free) output state
   • Start in the computational ground state

• Apply the circuit (right-to-left order)

a) \(H_2\)

b) \(CNOT_{21}\) (control 2, target 1)

c) \(H_0\)

d) \(CNOT_{03}\) (control 0, target 3)

e) Rearranging qubits pairwise,

   where $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$.
   Hence the ideal state is a tensor product of two Bell pairs:
   • qubits 2–1,  
   • qubits 0–3.

2. Stabilizers of each Bell pair
   For a Bell state \(|\Phi^+\rangle\) on qubits \(i,j\),

Thus the set $\{XX,YY,ZZ\}$ (and $II$) stabilizes each pair.

3. Two-qubit depolarizing channel after every CNOT
   After each CNOT the noise map on its two qubits is

•  After $CNOT_{21}$ the noise acts only on qubits 2–1.  
•  After $CNOT_{03}$ the noise acts only on qubits 0–3.  
No subsequent gates mix these qubit pairs, so the final error operators are simply Pauli products on the corresponding pair(s).

4. Fidelity for a single Pauli error
   Let \(P\) be any of the 15 non-identity two-qubit Paulis on a given pair.

   Fidelity contribution

Therefore, averaged over the depolarizing distribution

5. Fidelity for two independent Pauli errors
   With Pauli operators \(P_{21}\) on qubits (2,1) and \(Q_{03}\) on (0,3),

The magnitude is 1 only if **both** $P_{21}$ and $Q_{03}$
are in $\{XX,YY,ZZ\}$; otherwise it is 0.  
Hence

6. Combine the mutually exclusive error cases
   • no errors: probability \((1-p)^2\), fidelity \(1\)
   • exactly one error: probability \(2p(1-p)\), fidelity \(1/5\)
   • two errors: probability \(p^2\), fidelity \(1/25\)

   Altogether

Final Answer:
The physical four-qubit state fidelity after the preparation circuit, as a function of the two-qubit gate error rate \(p\), is

$$

Step-by-Step Derivation

1. Error model
   • After each of the two CNOTs a two–qubit depolarising channel acts on the same two qubits as the gate

  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 $CNOT_{03}$).

2. Which two–qubit Paulis are undetectable?
   An error \(E\) is undetectable iff it commutes with both stabilisers

For a two–qubit operator supported on some qubit pair
only the three elements

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.

3. Syndrome pattern produced by a single two-qubit Pauli

   For every error we record the pair of parity bits

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)   | $Y,\;Y,\;XZ,\;ZX$                 | $p_{11}= \dfrac{4p}{15}$ |

4. 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

contains at least one ‘1’.  
Consequently the run is **kept** when simultaneously

Because the two channels are independent,

5. Logical state fidelity conditioned on keeping the run

   The ideal encoded GHZ state obtained by the circuit is

For a single Bell pair

Hence after the two independent errors

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

Expanding for small $p$,

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

Step-by-Step Derivation

(The numbering of the five physical qubits is
0,1,2,3 (data) and 4 (ancilla). A two–qubit depolarising channel of strength \(p\) hits immediately after every CNOT.)

1. Gates, noise locations and notation

Executed (right-to-left) the circuit is

1. \(H_1\)
2. \(CNOT_{12}\) (control 1, target 2) ⟶ noise \(E_{12}\)
3. \(CNOT_{10}\) (control 1, target 0) ⟶ noise \(E_{10}\)
4. \(CNOT_{23}\) (control 2, target 3) ⟶ noise \(E_{23}\)
5. \(CNOT_{34}\) (control 3, target 4) ⟶ noise \(E_{34}\)
6. \(CNOT_{04}\) (control 0, target 4) ⟶ noise \(E_{04}\)
7. measure qubit 4 in the \(Z\) basis.

• Each \(E_{ij}\) is a Pauli picked from \(\{II\}\cup\mathcal P_{ij}\) with
\(\Pr(II)=1-p,\quad \Pr(P)=p/15\;(P\ne II)\).
• We keep the run iff
(a) the ancilla is read out in \(\ket{0}_4\) and
(b) the four data qubits are in the simultaneous \(+1\) eigenspace of the stabilisers \(XXXX\) and \(ZZZZ\).

2. Which single–gate Paulis can survive the post-selection?

• The last two CNOTs (acting on qubit 4) form a flag:
any Pauli on \(\{3,4\}\) or \(\{0,4\}\) that contains \(X_4\) or \(Y_4\) flips the ancilla and is rejected.
Every Pauli on those pairs that does not contain \(X_4\) or \(Y_4\) anticommutes with at least one stabiliser and is also rejected.
⇒ the only error–free choice on those two gates is the identity.

• For \(E_{23}\) the only undetected non-identity Pauli is \(ZZ_{23}\); the others either propagate an \(X_4/Y_4\) onto the ancilla or trigger a stabiliser. Thus

    good set on $(2,3)$ : $\{II,\,ZZ\}$,

    with probability $g_3 = 1-\dfrac{14}{15}p$.

• For the first two CNOTs (pairs \((1,2)\) and \((1,0)\)) the familiar set

    $\{II,\,XX,\,YY,\,ZZ\}$

 commutes with both stabilisers, so

    good probability $g_1 = 1-\dfrac45p$,  
    bad probability  $b_1 = \dfrac{4}{15}p$   (each of the three
    syndrome-raising classes).

3. Probability that the run is kept

Because the two last CNOTs must be error-free, their joint survival probability contributes a factor \((1-p)^2\) that cancels out later and will be dropped from now on.

On the first three noise locations we must have overall syndrome \((a,b)=(0,0)\) (XOR of the three individual syndromes).

Up to second order in \(p\) the only ways this can happen are

a) no error at all on the three locations
probability \(g_1^2\,g_3\),

b) the same non-trivial syndrome on both of the first two locations (and none on the third)
probability \(\,3\,b_1^2\,g_3\)
(there are three possible non-trivial syndromes).

Hence, to \(\mathcal O(p^2)\),

4. Probability that the logical state is correct and the run is kept

The only way to pass and still agree with the target \(\ket{00}_{AB}\) is that all three early locations are of the good, stabiliser-type \((0,0)\):

5. Logical fidelity conditioned on acceptance

Expanding,

i.e. all first-order contributions are eliminated by the fault-tolerant preparation and the double post-selection.

Final Answer:

The logical two-qubit fidelity after post-selection on both
(i) the ancilla outcome \(\ket{0}_4\) and
(ii) the code stabilisers being \(+1\) is

Thus the first non-vanishing term is quadratic in the physical gate error rate, confirming that the circuit is first-order fault-tolerant for preparing the logical state \(\ket{00}_{AB}\).