Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 51 / 4

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.

Since $0<1-2ϵ<1$, the off-diagonal multiplier of the iterated phase-flip channel tends to zero. ThusLet $P_0=|0⟩⟨0|$ and $P_1=|1⟩⟨1|$ be the computational basis measurement projectors. The nonselective projective measurement channel iswhich is exactly the displayed limit. Its diagonal entries retain the original outcome probabilities; the measurement outcome is ignored and the coherences are erased. The limiting evolution is therefore the completely dephasing channel.

Number the three data qubits from top to bottom. It suffices initially to follow a pure input $|\psi ⟩=a|0⟩+b|1⟩$; linearity will extend the result to every density operator. The first two controlled-NOT gates create $a|000⟩+b|111⟩$. The first layer of Hadamard gates then encodes this asthe three-qubit phase-flip repetition code. Let $e_i$ indicate whether a Pauli Z gate occurred on data qubit $i$ during storage. The next layer of Hadamard gates uses $HZH=X$ and givesThe two initially zero syndrome quantum ancillas receive the adjacent parities. In the diagram's top-to-bottom order, their error syndrome isBoth terms of the logical superposition have these same parities, so the syndrome is independent of $a,b$. Measuring it reveals no logical-state information.

At the marked position of $R$ in the printed circuit, there has already been another layer of Hadamard gates on the data. There is also a layer immediately after $R$. Thus a desired correction $X_i$ in the parity-extraction frame must be implemented as $R=Z_i$ at the actual recovery location, since $H^{\otimes 3}{Z}_i{H}^{\otimes 3}=X_i$. Choose the recovery according toThe two patterns on each row share a syndrome because complementing all three bits does not change either parity. For a weight-zero or one error, the recovery removes the entire error in the parity-extraction frame. The remaining state is $a|000⟩+b|111⟩$, and the two final controlled-NOT gates give $|\psi ⟩|00⟩$.

For a weight-two or three error, minimum-weight recovery instead leaves all three bits flipped, giving $a|111⟩+b|000⟩$. The final controlled-NOT gates send this to $(a|1⟩+b|0⟩)|00⟩=X|\psi ⟩|00⟩$. Consequently, for every input density operator,The logical failure is a Pauli X gate after decoding, even though the stored physical errors were Pauli Z gates: $Z_1{Z}_2{Z}_3$ exchanges the two encoded codewords. This is the logical failure of the three-qubit phase-flip repetition code.

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.

Take a normalized environment quantum stateThe Controlled-Z gate is symmetric between the two qubits, so it can be written with the environment as the controlling system:Expanding the joint density operator after the interaction givesIn the partial trace, the two off-diagonal environment operators have trace zero, while each diagonal projector has trace one. HenceEquivalently, the Kraus operators are $K_0=\sqrt{1-ϵ}I$ and $K_1=\sqrt{ϵ}Z$, and $K_0^{†}{K}_0+K_1^{†}{K}_1=I$. This explicit Stinespring dilation describes the system's dephasing by discarding information in the environment. A fresh environment in this state at each storage interval also realizes the memoryless iteration used in part (a).