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An environment interaction preserving the computational populations multiplies a qubit's upper off-diagonal density entry by $\overline{c}$. A real phase-flip channel representation exists for every input precisely when $c$ is real, with probability $(1-c)/2$. For $c=r{e}^{i\phi }$, the channel is a phase-flip channel of probability $(1-r)/2$ followed by the diagonal unitary operator $\mathrm{diag}(1,e^{i\phi })$. The coherent phase cannot be discarded just because the populations are unchanged.

Independent successive applications of the phase-flip channel $D_{ϵ}(\rho )=(1-ϵ)\rho +ϵZ\rho Z$ preserve diagonal entries and multiply off-diagonal entries by $(1-2ϵ{)}^n$. Hence $D_{ϵ}^n=D_{{ϵ}_n}$ with the displayed parameter, for nonnegative integers $n$. If $0<ϵ<1/2$, the limit is the completely dephasing channel, equivalently a nonselective projective measurement in the computational basis. A time-interval law can be iterated this way only under the memoryless composition assumption.

Use the Hilbert-Schmidt inner product normalized by the dimension: $(A,B)=\frac{1}{2}\mathrm{Tr}(A^{†}B)$. The identity and the three Pauli matrices are Hermitian, each has square $I$, and each nonidentity Pauli matrix has zero trace. For distinct nonidentity indices, the Pauli matrix multiplication law gives ${\sigma }_j{\sigma }_k=i{\sum }_l{\epsilon }_{jkl}{\sigma }_l$, which is traceless. ThereforeThere are four such orthonormal operators, and the complex vector space of two-by-two matrices has dimension four. Hence they form an orthonormal basis. Taking the inner product of any $E$ with each basis element proves, for arbitrary complex $E$ and not just Hermitian ones,For example, if $E=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, its four coefficients are $(a+d)/2$, $(b+c)/2$, $i(b-c)/2$, and $(a-d)/2$, respectively.

For the joint error, regard the two prescribed basis outputs as the columns of an environment-valued linear map. Its coefficient array in the usual output-row, input-column convention isThe source's other array is arranged as a column of the images of basis kets; it must not be mistaken for this coefficient array acting on an amplitude column. Applying the just-proved Pauli matrix expansion componentwise in any environment orthonormal basis gives the Pauli decomposition of a qubit-environment isometry:To check the potentially delicate sign of the Pauli Y gate coefficient, use $Y|0⟩=i|1⟩$ and $Y|1⟩=-i|0⟩$. The four terms then giveLinearity proves the expansion for every input pure state. The environment coefficient vectors need not be normalized or mutually orthogonal. Consequently this expansion includes coherent combinations of Pauli errors, rather than asserting that every quantum channel is a probabilistic Pauli channel.

For the recovery probability, use the usual independent-use interpretation of the given phase-flip channel: each of the three transmitted qubits has a Pauli Z gate error independently with probability $q$. The phase-flip repetition code uses logical codewords $|+++⟩$ and $|---⟩$. The decoding Hadamard gates convert physical phase flips to bit flips because $HZH=X$. We can then analyze the two decoding controlled-NOT gates and the final Toffoli gate directly, rather than assuming a recovery rule.

Let $b_1,b_2,b_3$ be the error bits after this basis conversion. For a logical basis input $t\in \{0,1\}$, the three data bits before inverse encoding are $t\oplus b_1,t\oplus b_2,t\oplus b_3$. The two controlled-NOT gates produceThe Toffoli gate flips the first wire when both of the last two bits are one. Its residual error isThe last two wires carry an error syndrome independent of $t$. Thus this transformation preserves the amplitudes of an arbitrary superposition: zero or one phase error returns the original unknown qubit, while two or three phase errors leave the logical Pauli X gate applied to it. The probability of universally successful recovery is thereforeEquivalently, the decoded quantum channel is $\rho \mapsto (1-p_L)\rho +p_{L}X\rho X$, with logical failure of the three-qubit phase-flip repetition code $p_L=3q^2-2q^3$. For $0<q<1/2$, the failure probability is below $q$, since $q-p_L=q(1-q)(1-2q)>0$. A special input that is a Pauli X eigenstate is unchanged even by the residual logical error; this does not make the code correct that error on an unknown input. Its squared quantum fidelity with a particular pure input is $1-p_L+p_L|⟨\chi |X|\chi ⟩{|}^2$. If the errors on different channel uses are correlated, the marginal probability $q$ alone is insufficient: the general universal success probability is the probability of at most one error.

Let $c=⟨R_0|R_1⟩$. The joint state after the environment interaction is $\alpha |0⟩|R_0⟩+\beta |1⟩|R_1⟩$. Taking its partial trace over the environment gives the exact reduced density operatorThe conjugate in the upper off-diagonal entry follows from tracing $|R_0⟩⟨R_1|$. Normalization of the environment states and the Cauchy-Schwarz inequality imply $|c|\le 1$.

For comparison, a phase-flip channel sends $\rho$ to $(1-p)\rho +pZ\rho Z$, multiplying its off-diagonal entries by the real number $1-2p$. Consequently the claimed representation is valid, for a general input, provided the environment overlap is real, withThis is a genuine probability between zero and one. The environment maps printed in the question do not themselves imply that $c$ is real. For example, choose $|R_1⟩=i|R_0⟩$, so $c=i$, and take $\alpha =\beta =1/\sqrt{2}$. The qubit becomes $(|0⟩+i|1⟩)/\sqrt{2}$, whose imaginary off-diagonal entries cannot come from any probabilistic mixture of $\rho$ and $Z\rho Z$. These environment maps are a valid isometry and can be extended to a unitary interaction, so this is an actual counterexample to the unrestricted premise.

The general complex environment overlap in qubit dephasing has an exact corrected description. Write $c=r{e}^{i\phi }$, $0\le r\le 1$, and let $D_{\phi }=\mathrm{diag}(1,e^{i\phi })$. ThenAt $r=0$, choose any phase. The dephasing channel includes both random phase errors and, when the overlap is complex, a coherent diagonal rotation. The subsequent probability formulas use the intended real-overlap phase-flip model.

Subtract the two density operators. The difference is Hermitian with zero diagonal and off-diagonal entry $\alpha \overline{\beta }(1-\overline{c})$. A matrix $\left(\begin{array}{cc}0& a\\ \overline{a}& 0\end{array}\right)$ has eigenvalues $\pm |a|$, because its square is $|a{|}^{2}I$. Its two singular values are therefore both $|a|$. By the definition of the trace norm,For the intended real-overlap phase-flip channel, $1-c=2p$, givingThis vanishes for a computational basis state, since a phase flip then changes only an unobservable global phase. Its largest value at fixed $p$ is $p$, attained when $|\alpha |=|\beta |=1/\sqrt{2}$. The general formula remains correct even when the probabilistic premise of part 1 fails.

Assume the three physical phase-flip channels act independently, with the same probability $p$ on each qubit, and encoding, recovery and decoding are ideal. Zero or one physical error is corrected. HenceThe probability of an uncorrectable two- or three-site error is $p_L=3p^2(1-p)+p^3=3p^2-2p^3$.

One must also choose the logical convention consistently with the output phase error specified here. The usual phase-flip repetition code with codewords $|+++⟩$ and $|---⟩$ has residual operator $Z_1{Z}_2{Z}_3$ after two or three physical errors: their error syndrome equals that of the complementary zero- or one-error pattern. This operator exchanges those codewords, so plain inverse encoding would give logical $X$, not $Z$.

To obtain the stated logical phase-flip convention, use the logical Hadamard convention for a phase-flip repetition code. Put a logical Hadamard gate before that encoding and after its decoding. The encoded state iswhere the plus and minus states in these codewords are the real Hadamard basis states. The uncorrectable exchange is now conjugated to $HXH=Z$. With that convention the decoded density operator and its trace distance are exactlyFor $0<p<1/2$, $p-p_L=p(1-p)(1-2p)>0$, so coding reduces the trace distance. The displayed recovery probability means success of correction as a quantum operation on an arbitrary unknown input. If the particular input is a $Z$ eigenstate, even the logical phase-error branch represents the same physical state, and the probability of recovering that particular state is one.

Independence is an extra noise assumption needed for the numerical probability. If all three physical qubits flip together with probability $p$, each marginal still has the given phase-flip channel, but the logical failure probability is $p$, not $3p^2-2p^3$. More generally, for a classical joint phase-error distribution with a random number $K$ of flips, $P_{\mathrm{rec}}=\mathrm{Pr}(K\le 1)$ and $p_L=\mathrm{Pr}(K\ge 2)$; the same decoded mixture holds in the stated logical convention. Thus marginal error probabilities alone do not determine the requested recovery probability. A complex environment overlap likewise requires the coherent rotation to be retained or separately compensated before this stochastic phase-error analysis applies.

For nonnegative integers $n$, assume the storage channel over successive intervals of length $\tau$ composes as $D_{ϵ}^n$. This is the memoryless interpretation of the specified storage law. In the computational basis, write the density operator asConjugation by the Pauli Z gate reverses the off-diagonal entries and leaves the diagonal entries unchanged, soAfter $n$ applications of the phase-flip channel, the multiplier is $(1-2ϵ{)}^n$. Comparing with the multiplier $1-2{ϵ}_n$ of a single phase-flip channel givesFor $n=0$ this gives the identity channel. Equivalently, composing two consecutive phase errors cancels them because $Z^2=I$, so their effective probability obeys ${ϵ}_{n+1}=ϵ+(1-2ϵ){ϵ}_n$. The same expression solves this recurrence. The iterated phase-flip channel formula presumes this composition law; specifying a channel at one time alone would not constrain a memory-bearing environment at later times.

Independent applications of the phase-flip channel are a classical mixture of the eight physical Pauli Z gate error patterns. A specified weight-$k$ pattern has probability ${ϵ}^k(1-ϵ{)}^{3-k}$. The recovery from the preceding part succeeds on weight zero and one, and yields the logical Pauli X gate on weight two and three. There are three weight-two patterns and one weight-three pattern, so the logical failure probability isAveraging the corrected branches gives the complete decoded quantum channelIn particular, the leading logical error is $3{ϵ}^2$, rather than an error of first order in $ϵ$. For the stated range,Thus the failure probability is smaller than the physical error probability. The decoded channel is a bit-flip channel; it is not $D_{p_L}$, whose error operator would be $Z$. This distinction follows from the logical codeword exchange in the phase-flip repetition code.

Conjugation by ${\sigma }_z$ maps $(r_x,r_y,r_z)$ to $(-r_x,-r_y,r_z)$. The phase-flip channel therefore acts on the Bloch vector as