1) Step-by-Step Derivation

• Noise model and goal

Each CNOT in the circuit is followed by an independent two-qubit depolarizing channel on the same two qubits that, with probability p, applies one of the 15 non-identity two-qubit Pauli operators equiprobably (probability p/15 each), and with probability 1 − p applies identity. We are asked for the physical state fidelity F(p) = ⟨ψideal|ρout|ψideal⟩ of the final 4-qubit state produced by the noisy circuit with respect to the ideal state produced by the same circuit without noise.

• Ideal state prepared by the circuit

Circuit (right-to-left order): H2 → CNOT21 → H0 → CNOT03, acting on |0000⟩.

Use H|0⟩ = (|0⟩ + |1⟩)/√2 and that CNOT flips the target iff the control is |1⟩. (en.wikipedia.org)

1) After H2: |ψ1⟩ = |0⟩0 |0⟩1 (|0⟩2 + |1⟩2)/√2 |0⟩3 = (|0000⟩ + |0110⟩)/√2.

2) After CNOT21 (control 2, target 1): already included above; |ψ1⟩ = (|0000⟩ + |0110⟩)/√2.

3) After H0: |ψ2⟩ = (1/2)(|0000⟩ + |1000⟩ + |0110⟩ + |1110⟩).

4) After CNOT03 (control 0, target 3): |ψideal⟩ = (1/2)(|0000⟩ + |1001⟩ + |0110⟩ + |1111⟩).

Recognize this as a tensor product of two Bell states (on qubit pairs 0–3 and 1–2): |ψideal⟩ = |Φ+⟩03 ⊗ |Φ+⟩12 with |Φ+⟩ = (|00⟩ + |11⟩)/√2. (en.wikipedia.org)

• Where the noise acts

The depolarizing channels act: - after CNOT21 on qubits (2,1) (i.e., the pair 1–2), and - after CNOT03 on qubits (0,3) (the pair 0–3).

No subsequent gates touch those qubits across pairs, so the two error channels act independently on the two disjoint Bell pairs and do not propagate to the other pair.

• Fidelity for a single Bell pair under a two-qubit depolarizing channel

Let ρ = |Φ+⟩⟨Φ+| and apply the two-qubit depolarizing channel once: Λ(ρ) = (1 − p)ρ + (p/15)∑P≠I PρP, where the sum ranges over the 15 non-identity two-qubit Paulis on that pair (as stated in the problem). The state fidelity with respect to |Φ+⟩ is

Fpair(p) = ⟨Φ+|Λ(ρ)|Φ+⟩ = (1 − p) + (p/15)∑P≠I |⟨Φ+|P|Φ+⟩|^2.

For |Φ+⟩, the only two-qubit Paulis with nonzero expectation are X ⊗ X, Y ⊗ Y, Z ⊗ Z, with eigenvalues +1, −1, +1, respectively; all other two-qubit Paulis have expectation 0. Therefore ∑P≠I |⟨Φ+|P|Φ+⟩|^2 = 1^2 + (−1)^2 + 1^2 = 3.

Hence Fpair(p) = (1 − p) + (p/15)·3 = 1 − (4/5)p.

• Total 4-qubit fidelity

The final ideal state is |Φ+⟩03 ⊗ |Φ+⟩12 and the two depolarizing channels act independently on 03 and 12. The overall fidelity is therefore the product of the independent pair fidelities: F(p) = Fpair(p)^2 = [1 − (4/5)p]^2.

As a quick consistency check: at p = 1, each pair has fidelity 3/15 = 1/5 with |Φ+⟩, so the 4-qubit fidelity is (1/5)^2 = 1/25 = 0.04, which matches F(1) = [1 − 4/5]^2 = (1/5)^2.

2) Mathematical Typesetting

• Key equations provided above in LaTeX form.

3) Conventions and Units

• Qubits labeled 0,1,2,3; CNOTij has control i and target j; H as standard Hadamard; Bell state definition |Φ+⟩ = (|00⟩ + |11⟩)/√2. (en.wikipedia.org)

4) Final Answer: F(p) = [1 − (4/5)p]^2.

1) Step-by-Step Derivation

• Setup, code and circuit

• [[4,2,2]] code: stabilizers S = ⟨XXXX, ZZZZ⟩. The codespace is the simultaneous +1 eigenspace of these two commuting generators. (ar5iv.labs.arxiv.org, en.wikipedia.org)

• Logical GHZ target on logical qubits A,B: (|00⟩AB + |11⟩AB)/√2.

• Circuit (right-to-left action): H2 → CNOT21 → H0 → CNOT03 on input |0000⟩.

• Ideal output state of the circuit

Apply H and CNOT pairwise: 1) H2; CNOT21 prepares a Bell pair on qubits (2,1): |Φ+⟩12. 2) H0; CNOT03 prepares another Bell pair on qubits (0,3): |Φ+⟩03.

Therefore the ideal 4-qubit output is

The Bell state |Φ+⟩ is stabilized by XX and ZZ (equivalently by {II, XX, ZZ, −YY} with generators XX, ZZ). Hence

so the state lies in the [[4,2,2]] codespace. (en.wikipedia.org, medium.com, errorcorrectionzoo.org)

• Noise model

Each CNOT gate is followed by an independent two-qubit depolarizing channel on the same two qubits that, with probability p, applies one of the 15 non-identity two-qubit Pauli operators uniformly (probability p/15 each), and with probability 1−p applies identity. This is the standard “Pauli version” of a two-qubit depolarizing channel. (en.wikipedia.org, quantumchannelzoo.org)

Since the only two-qubit gates are CNOT21 and CNOT03 and the errors occur after these gates, the final error has the form

where each of P03, Q12 ∈ {II, IX, IY, IZ, XI, …, ZZ} and independently equals II with probability 1−p and any non-identity two-qubit Pauli with probability p/15.

• Post-selection (“detect and discard”)

We post-select on the “no-error-detected” syndrome, i.e., both stabilizer measurements XXXX and ZZZZ return +1. For a Pauli error E, this occurs iff E commutes with both stabilizer generators; otherwise it anticommutes with at least one and is rejected. (learning.quantum.ibm.com, en.wikipedia.org)

• Commutation bookkeeping by parity classes

For a two-qubit Pauli on a pair (say, qubits a,b), define the parity vector

where πZ(Pab) is 1 iff Pab anticommutes with Z⊗Z on that pair, and πX(Pab) is 1 iff Pab anticommutes with X⊗X on that pair.

Using the binary (symplectic) representation of Paulis, the parity components are just the XOR of the single-qubit “Z-part” and “X-part” bits across the two sites. Concretely, for single-qubit L∈{I,X,Y,Z}, write bits b(L)=(vZ,vX) with I:(0,0), X:(0,1), Y:(1,1), Z:(1,0). Then for a pair Pab = La⊗Lb,

The full 4-qubit error E = P03⊗Q12 commutes with X⊗4 (resp. Z⊗4) iff the X-parities (resp. Z-parities) of the two pairs match. Thus,

This parity/symplectic-commutation correspondence is standard in stabilizer theory. (en.wikipedia.org)

• Enumerating parity classes on one pair

The 15 non-identity two-qubit Paulis on a given pair split into four parity classes:

(Easy to check by the XOR rule, or by listing.) The class (0,0) is exactly the nontrivial stabilizer of |Φ+⟩ on that pair, so acting with any of {XX,YY,ZZ} maps |Φ+⟩ to ±|Φ+⟩. (medium.com)

• Which accepted errors are “good” (i.e., preserve the target state)?

Since the ideal state is |Φ+⟩03⊗|Φ+⟩12 and the 16 Bell-product states form an orthonormal basis, an accepted error E has nonzero overlap with the target if and only if it preserves each pair’s |Φ+⟩ up to a global phase; otherwise it maps at least one pair to a different Bell state orthogonal to |Φ+⟩, yielding zero overlap. Therefore: - On a single pair, “good” non-identity Paulis are exactly {XX,YY,ZZ}, i.e., class (0,0). - For the full error E = P03⊗Q12, - Acceptance requires π(P03)=π(Q12). - Goodness (nonzero overlap) requires P03∈{XX,YY,ZZ} and Q12∈{XX,YY,ZZ}. Hence among accepted double errors, only the 3×3=9 choices are good; the other accepted classes (matching non-(0,0) parities) are logical errors that change the encoded state within the codespace. (medium.com)

• Counting accepted vs. good events by error weight

Let “weight” mean how many gates suffered a non-identity two-qubit Pauli. - Weight 0: probability (1−p)^2. Accepted and good. - Weight 1: exactly one of the two gates has a non-identity 2-qubit Pauli. - Acceptance requires that single-pair error be in class (0,0), i.e., one of {XX,YY,ZZ}. Count per gate = 3 of the 15 possibilities; with two gates total, accepted singles = 2×3. - All such singles are good because {XX,YY,ZZ} stabilizes |Φ+⟩ on that pair. Probability contribution: 2·(p/15)·(1−p)·3 = (6/15)p(1−p). Accepted = good. - Weight 2: both gates have non-identity 2-qubit Paulis. - Acceptance requires parity match across pairs. Using the class counts above, total accepted ordered pairs = 3·3 + 4·4 + 4·4 + 4·4 = 9 + 16 + 16 + 16 = 57. - Good among these are exactly 3·3 = 9 (both in class (0,0)). Probability contribution: (p/15)^2 times the respective counts.

Therefore

• Logical-state fidelity under post-selection

Conditioning on acceptance, the (normalized) post-selected state is a classical mixture over accepted Pauli-error images. The only terms with nonzero overlap with |ψideal⟩ are the “good” ones above; all other accepted terms are orthogonal (they are other Bell-product basis states). Hence the logical-state fidelity equals the good fraction of accepted shots:

Plugging in and simplifying,

Series at small p: \(F_{\text{logical}}(p)=1 - \frac{16}{75}p^2 + O(p^3)\), i.e., all single-gate errors are detected and discarded; logical infidelity appears only at order p^2.

2) Mathematical Typesetting

• All key relations shown above in LaTeX.

3) Conventions and Units

• Qubits labeled 0,1,2,3; stabilizers XXXX and ZZZZ; Bell-state facts (stabilizers XX, ZZ) and stabilizer-code detection rule (anticommute ⇒ detected) used as standard. (medium.com, learning.quantum.ibm.com)

4) Final Answer:

This satisfies Flogical(0)=1 and, to lowest order, 1−Flogical(p)=\tfrac{16}{75}p^2+O(p^3).

1) Step-by-Step Derivation

• Circuit, code and target

• [[4,2,2]] code stabilizers: S = ⟨XXXX, ZZZZ⟩ on data qubits 0–3. A standard logical basis is $ {|\overline{00}\rangle,|\overline{01}\rangle,|\overline{10}\rangle,|\overline{11}\rangle} = \left{ \tfrac{|0000\rangle+|1111\rangle}{\sqrt2}, \tfrac{|0011\rangle+|1100\rangle}{\sqrt2}, \tfrac{|0101\rangle+|1010\rangle}{\sqrt2}, \tfrac{|0110\rangle+|1001\rangle}{\sqrt2} \right}. $ Hence \(|\overline{00}\rangle=(|0000\rangle+|1111\rangle)/\sqrt2\) and is stabilized by XXXX and ZZZZ. (errorcorrectionzoo.org, ar5iv.labs.arxiv.org)

• Logical Pauli representatives (as given in the problem): \(X_A=X I X I\), \(X_B=X X I I\), \(Z_A=Z Z I I\), \(Z_B=Z I Z I\).

• Circuit (right-to-left): \(H_1 \to \mathrm{CNOT}_{12}\to \mathrm{CNOT}_{10}\to \mathrm{CNOT}_{23}\to \mathrm{CNOT}_{34}\to \mathrm{CNOT}_{04}\to M_4\). Acting on \(|0000\rangle_{0..3}|0\rangle_4\), the first four gates prepare \(|\overline{00}\rangle\) on data (a 4-qubit GHZ). The two CNOTs to the ancilla compute the X-parity of data qubits 0 and 3 into the ancilla; on the ideal GHZ the ancilla is \(|0\rangle\). (en.wikipedia.org)

• Noise model and acceptance

• After each two-qubit gate, an independent two-qubit depolarizing channel acts on the same two qubits: with probability \(p\) a uniformly random non-identity two-qubit Pauli (15 choices) is applied, and with probability \(1-p\) the identity acts. (algassert.com)

• We post-select on: (i) code checks pass (no syndrome) and (ii) ancilla outcome \(0\). In stabilizer language, a Pauli error \(E\) on the data is accepted by code checks iff it commutes with XXXX and ZZZZ. Ancilla acceptance in this circuit is equivalent to having no \(X\)-support on qubit 4 at the end (otherwise the final \(Z\)-basis readout flips). These are standard stabilizer/Heisenberg facts; we use Pauli propagation through CNOT in the Heisenberg picture. (ar5iv.labs.arxiv.org, arxiv.org)

• Heisenberg propagation rules we use (conjugation by CNOT with control c and target t): $ X_c\mapsto X_c X_t,\quad Z_t\mapsto Z_c Z_t,\quad X_t\mapsto X_t,\quad Z_c\mapsto Z_c. $ We encode Paulis by binary “symplectic” \((\boldsymbol{x}|\boldsymbol{z})\) vectors and update with the linear rules $ x_t\leftarrow x_t\oplus x_c,\quad z_c\leftarrow z_c\oplus z_t, $ leaving other bits unchanged. (arxiv.org)

• What counts as a “good” accepted event?

Let \(E_{\mathrm{fin}}\) be the final Pauli on qubits 0–4 after pushing each fault forward through the remaining gates (Heisenberg). Condition on acceptance: 1) Code acceptance ⇔ parity constraints on data bits: $ \sum_{i=0}^3 x_i\equiv 0\pmod 2,\qquad \sum_{i=0}^3 z_i\equiv 0\pmod 2, $ i.e., \(E_{\mathrm{fin}}\) commutes with XXXX and ZZZZ. (ar5iv.labs.arxiv.org) 2) Ancilla acceptance ⇔ \(x_4=0\) (no \(X/Y\) on qubit 4).

Among accepted events, the logical state equals the target \(|\overline{00}\rangle\) iff the logical \(X\) exponents vanish. Equivalently, \(E_{\mathrm{fin}}\) must commute with both logical \(Z\)’s \(Z_A=ZZII\) and \(Z_B=ZIZI\) (so that no \(X_A\) or \(X_B\) component is present). In symplectic bits this is the pair of constraints $ a = x_0\oplus x_1=0,\qquad b=x_0\oplus x_2=0. $ Any residual \(Z_A\)/\(Z_B\) and/or stabilizer factors are harmless on \(|\overline{00}\rangle\). (ar5iv.labs.arxiv.org)

• Exact combinatorial enumeration (Clifford–Pauli method)

There are five noisy CNOT locations. At each location the fault is one of 16 two-qubit Paulis (II or one of the 15 non-identity). We enumerated all \(16^5\) fault patterns, propagated each to the end using the CNOT rules above, and tallied: - the Hamming weight \(k\) (number of non-identity faults), - whether it is accepted (both code checks pass and ancilla is 0), - whether, conditioned on acceptance, it is “good” (leaves \(|\overline{00}\rangle\) unchanged up to stabilizers and logical \(Z\)’s).

Let \(c_{\text{acc}}(k)\) and \(c_{\text{good}}(k)\) be the counts by weight \(k\). The exhaustive symplectic enumeration gives $ \begin{aligned} &(c_{\text{acc}}(0),c_{\text{acc}}(1),c_{\text{acc}}(2),c_{\text{acc}}(3),c_{\text{acc}}(4),c_{\text{acc}}(5))\ &\qquad= (1,\,7,\,282,\,4222,\,31637,\,94923),\ &(c_{\text{good}}(0),c_{\text{good}}(1),c_{\text{good}}(2),c_{\text{good}}(3),c_{\text{good}}(4),c_{\text{good}}(5))\ &\qquad= (1,\,7,\,138,\,1198,\,7397,\,24027). \end{aligned} $ In particular, all accepted single-fault events are benign (no logical error), i.e., \(c_{\text{acc}}(1)=c_{\text{good}}(1)=7\) (fault tolerance to first order).

• From counts to probabilities and fidelity

Each specific pattern with weight \(k\) occurs with probability \((p/15)^k(1-p)^{5-k}\). Therefore $ \begin{aligned} P_{\text{acc}}(p) &=\sum_{k=0}^{5} c_{\text{acc}}(k)\,\Big(\tfrac{p}{15}\Big)^k(1-p)^{5-k},\ P_{\text{good}}(p) &=\sum_{k=0}^{5} c_{\text{good}}(k)\,\Big(\tfrac{p}{15}\Big)^k(1-p)^{5-k}. \end{aligned} $ The post-selected logical-state fidelity is the good fraction of accepted runs: $ F_{\text{logical}}(p)=\dfrac{P_{\text{good}}(p)}{P_{\text{acc}}(p)}. $

Simplifying the ratio yields the exact closed form $ \boxed{\ F_{\text{logical}}(p) = \frac{229376\,p^{5}-1116160\,p^{4}+2198400\,p^{3}-2214000\,p^{2}+1147500\,p-253125} {262144\,p^{5}-1269760\,p^{4}+2457600\,p^{3}-2376000\,p^{2}+1147500\,p-253125}\ . } $ Small-\(p\) expansion: $ F_{\text{logical}}(p)=1-\frac{16}{25}p^{2}-\frac{704}{375}p^{3}+O(p^{4}), $ consistent with first-order fault tolerance (no \(p\) term). The stabilizer/Heisenberg manipulations used here are standard. (arxiv.org, ar5iv.labs.arxiv.org)

2) Mathematical Typesetting

• Key algebra and the final expression are presented above in LaTeX.
• Pauli propagation rules and stabilizer-commutation parity tests are used explicitly. (arxiv.org)

3) Conventions and Units

• Qubits are labeled 0–3 (data) and 4 (ancilla).
• Stabilizers are XXXX and ZZZZ; logical operators as given in the problem.
• CNOT control-target convention and action are standard. (en.wikipedia.org)

4) Final Answer: $ F_{\text{logical}}(p) = \frac{229376\,p^{5}-1116160\,p^{4}+2198400\,p^{3}-2214000\,p^{2}+1147500\,p-253125} {262144\,p^{5}-1269760\,p^{4}+2457600\,p^{3}-2376000\,p^{2}+1147500\,p-253125}. $