Step-by-Step Derivation

1) Ideal target state and its stabilizers

• The circuit acts (right-to-left) as H2 → CNOT21 → H0 → CNOT03 on |0000⟩, yielding the pure 4‑qubit state

This state is the [[4,2,2]]-encoded logical Bell (two‑qubit GHZ) state |Φ+⟩AB, i.e., (|00⟩AB+|11⟩AB)/√2, stabilized (in the logical basis) by X_A X_B and Z_A Z_B, exactly as a Bell state is stabilized by XX and ZZ. Using the given physical representatives X_A=XIXI, X_B=XXII, Z_A=ZZII, Z_B=ZIZI, we obtain physical logical stabilizers

Together with the code stabilizers XXXX and ZZZZ, these generate the 4‑qubit stabilizer group S of |Ψ⟩. In particular, S contains three weight‑2 Paulis on qubits (1,2): IXXI, IZZI, IYYI (up to phases), three weight‑2 Paulis on qubits (0,3): XIII X, ZIIZ, YIIY, and nine weight‑4 Paulis formed by their products (e.g., XXXX, ZZZZ, ZX XZ, XZZX, etc.). The Bell-state stabilizer fact is standard, and Clifford/logical replacement uses the given logical operators. (en.wikipedia.org)

2) Noise model located in the circuit

• After each CNOT gate we apply an independent two‑qubit depolarizing channel on that gate’s qubits:

i.e., with probability 1−p no error occurs; with probability p a uniformly random non‑identity 2‑qubit Pauli from the 15 possibilities is applied. One channel acts on qubits (2,1) after CNOT21; the other acts on qubits (0,3) after CNOT03. (A uniform Pauli mixture over all non‑identity n‑qubit Paulis is the “uniform Pauli error” form of the depolarizing channel.) (qiskit.github.io, qiskit-org-web-resources.s3.us.cloud-object-storage.appdomain.cloud)

• Because later gates act on disjoint qubits (H0 and CNOT03 do not touch qubits 1 or 2), the Pauli error drawn after CNOT21 remains a Pauli on qubits (1,2) until the end, and similarly for the error after CNOT03 on qubits (0,3). Clifford gates map Paulis to Paulis under conjugation; here, disjoint support also makes the propagation trivial. (en.wikipedia.org)

3) Fidelity reduction to counting stabilizer elements

• Let P03 be the (possibly identity) Pauli on qubits (0,3) drawn by the last channel, and P12 the (possibly identity) Pauli on (1,2) drawn by the first. The final state is a classical mixture of (P03⊗P12)|Ψ⟩. For a pure stabilizer state |Ψ⟩, the state fidelity with |Ψ⟩,

equals the total probability that the net Pauli P03⊗P12 lies in the stabilizer group S of |Ψ⟩ (since |Ψ⟩ is the unique +1 eigenstate of S; for P∉S the overlap vanishes). Thus, the fidelity is a counting problem over stabilizer elements reachable by the two channels. (zxcalc.github.io)

4) Counting cases that preserve the state

• Event types: 1) No error on either channel: probability (1−p)^2; the net Pauli is I and lies in S. 2) Exactly one two‑qubit error (on one pair only): probability 2(1−p)p. On a given pair, 3 of the 15 non‑identity two‑qubit Paulis are in S (for (1,2): IXXI, IZZI, IYYI; for (0,3): XIII X, ZIIZ, YIIY). Hence the conditional success probability is 3/15=1/5. 3) Two errors (one on each pair): probability p^2. There are 15×15=225 ordered pairs of non‑identity Paulis. Exactly 9 give a stabilizer element: all combinations where the (0,3) Pauli is one of {XX,YY,ZZ} and the (1,2) Pauli is one of {XX,YY,ZZ}, yielding the 9 weight‑4 stabilizers (e.g., XXXX= (XX)03⊗(XX)12, ZZZZ= (ZZ)03⊗(ZZ)12, etc.). Thus the conditional success probability is 9/225=1/25.

• Summing contributions:

Notes on ingredients cited: the stabilizers of Bell/|Φ+⟩ are generated by XX and ZZ; two‑qubit depolarizing as a uniform mixture over the 15 non‑identity Paulis is the standard “uniform Pauli error” form of an n‑qubit depolarizing channel; and Clifford gates map Pauli errors to Pauli errors under conjugation, which justifies treating each Pauli error at the end of the circuit. (en.wikipedia.org, qiskit.github.io, qiskit-org-web-resources.s3.us.cloud-object-storage.appdomain.cloud)

Final Answer: F(p) = 1 − (8/5)p + (16/25)p^2 = (1 − 4p/5)^2.

Step-by-Step Derivation

1) What “post-selecting on all detectable errors” means and how to compute logical fidelity

• The [[4,2,2]] code has stabilizer generators S = ⟨XXXX, ZZZZ⟩. Measuring these two generators and keeping only +1 outcomes projects onto the code space. A Pauli error P is “detectable” (and thus rejected) if it anticommutes with at least one generator; equivalently, it is accepted iff it commutes with both generators, i.e., P∈N(S) (the normalizer/centralizer of S). Errors in N(S)\S act as nontrivial logical Paulis; errors in S act trivially. (en.wikipedia.org, leftasexercise.com)

• The target logical state is the logical Bell state |Φ+⟩AB = (|00⟩AB+|11⟩AB)/√2. In the logical basis it is stabilized by XAXB and ZAZB, just as a 2‑qubit Bell state is stabilized by XX and ZZ. Hence, among accepted (normalizer) errors, those that are equivalent (mod S) to I, XAXB, ZAZB, or YAYB leave the logical |Φ+⟩ unchanged; any other logical Pauli maps it to an orthogonal Bell state. Therefore the logical fidelity, conditioned on acceptance, is Flogical(p) = P{P ∈ Sstate}/P{P ∈ N(S)}, where Sstate is the full stabilizer group of the prepared 4‑qubit codeword (generated by XXXX, ZZZZ together with physical representatives of XAXB and ZAZB). (zxcalc.github.io)

2) Noise model and independence across pairs

• The circuit has two CNOTs; we place an independent two‑qubit depolarizing channel after each CNOT. For a 2‑qubit system (d=4), the depolarizing channel is the uniform mixture over the 15 non‑identity 2‑qubit Paulis, so each non‑identity Pauli occurs with probability p/15 and “no error” occurs with probability 1−p. One channel acts on qubits (2,1) after CNOT21; the other acts on (0,3) after CNOT03. Subsequent gates act on disjoint qubits, so each drawn Pauli remains on its pair until the end. (en.wikipedia.org)

• Denote the (possibly identity) Pauli on pair (1,2) by P12 and on (0,3) by P03. The net (final) Pauli is P = P03 ⊗ P12.

3) Classifying a 2‑qubit Pauli by its commutation with XXXX and ZZZZ

• For any Pauli P, P commutes with ZZZZ iff the total number of X or Y factors in P is even, and P commutes with XXXX iff the total number of Z or Y factors is even. For a 2‑qubit Pauli Q acting on a specific pair, define its “syndrome parity” s(Q) = (sX(Q), sZ(Q)) with sX(Q) = parity of {Z,Y} in Q, sZ(Q) = parity of {X,Y} in Q. For the two disjoint pairs, acceptance requires the total parity over all 4 qubits to be even, i.e., s(P12) ⊕ s(P03) = (0,0). (en.wikipedia.org)

• Enumerating the 15 non‑identity 2‑qubit Paulis on one pair and grouping by s(Q): • s=(0,0): {XX, YY, ZZ} → 3 elements, • s=(1,0): {ZI, IZ, XY, YX} → 4 elements, • s=(0,1): {XI, IX, YZ, ZY} → 4 elements, • s=(1,1): {YI, IY, XZ, ZX} → 4 elements. (Here I, X, Y, Z act on the two qubits of that pair.)

4) Probability the run is accepted (passes both stabilizers)

Break into mutually exclusive cases:

• No error on either channel: probability (1−p)^2, accepted.

• Exactly one non‑identity error (on one pair only): probability 2(1−p)p. Acceptance requires that that single 2‑qubit Pauli commute with both generators, i.e., s=(0,0). There are 3 such Paulis out of 15, so conditional acceptance probability = 3/15 = 1/5.

• Non‑identity errors on both pairs: probability p^2. Acceptance requires s(P12)=s(P03). With counts above, the number of accepted ordered pairs is 3^2+4^2+4^2+4^2 = 57 out of 15×15 = 225. Hence conditional acceptance probability = 57/225 = 19/75.

Therefore the total acceptance probability is

5) Probability the prepared 4‑qubit state equals the ideal (i.e., is in the state stabilizer)

• The run produces the exact target 4‑qubit state iff the net Pauli P lies in the stabilizer Sstate of that pure state. Counting (as in Subproblem 1): identity contributes (1−p)^2; a single non‑identity on one pair yields a stabilizer element only for the 3 Paulis {XX,YY,ZZ} → probability 2(1−p)p·(3/15); with non‑identity on both pairs, being in Sstate occurs for the 3×3=9 ordered choices in {XX,YY,ZZ}×{XX,YY,ZZ} → probability p^2·(9/225). Thus

Because |Φ+⟩AB is stabilized by XX and ZZ in the logical basis, P∈Sstate is equivalent to “no logical error” on |Φ+⟩AB. (zxcalc.github.io)

6) Logical fidelity conditioned on acceptance

• By definition,

Two useful checks: • Small‑p expansion: \(F_{\mathrm{logical}}(p)=1-\frac{16}{75}p^2+O(p^3)\), i.e., post‑selection removes all O(p) infidelity, as expected for a distance‑2 detector. • Worst case p=1: \(F_{\mathrm{logical}}(1)=\frac{1/25}{19/75}=\frac{3}{19}\).

Final Answer: The logical state fidelity of the post‑selected logical Bell state is

with acceptance probability \(P_{\mathrm{acc}}(p)=1-\frac{8}{5}p+\frac{64}{75}p^2\). (en.wikipedia.org, zxcalc.github.io)