Four-level syndrome recovery

PDF p. 2: Four-level system, text after Eq. (2)

Use the four-level model with ordered orthonormal basis $\mathrm{\mid }0⟩,\mathrm{\mid }1⟩,\mathrm{\mid }2⟩,\mathrm{\mid }3⟩$. Its code space is $\mathcal{C}=\mathrm{span}\{\mathrm{\mid }0⟩,\mathrm{\mid }1⟩\}$ and its decay space is $\mathcal{D}=\mathrm{span}\{\mathrm{\mid }2⟩,\mathrm{\mid }3⟩\}$. Put $P=\mathrm{\mid }0⟩⟨0\mathrm{\mid }+\mathrm{\mid }1⟩⟨1\mathrm{\mid }$, $Q=\mathrm{\mid }2⟩⟨2\mathrm{\mid }+\mathrm{\mid }3⟩⟨3\mathrm{\mid }$, and $U=\mathrm{\mid }0⟩⟨2\mathrm{\mid }+\mathrm{\mid }1⟩⟨3\mathrm{\mid }$. Here $P$ and $Q$ project onto the code and decay spaces, respectively, and $U$ maps $\mathrm{\mid }2⟩$ to $\mathrm{\mid }0⟩$ and $\mathrm{\mid }3⟩$ to $\mathrm{\mid }1⟩$.

The input is a density operator $\rho$ on the four-level space, in particular an output of the postulated noise channel. A density operator is positive semidefinite with trace one. Measure the two-outcome syndrome with projectors $P$ and $Q$: the outcome labelled null has unnormalized state $P\rho P$, and the outcome labelled decay has unnormalized state $Q\rho Q$. Leave the null branch unchanged; on the decay branch apply $U$. Thus the recovered, unnormalized branch outputs are ${\mathcal{I}}_{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}}(\rho )=P\rho P$ and ${\mathcal{I}}_{\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}}(\rho )=U\rho U^{†}$. Their traces are the outcome probabilities. When an outcome has positive probability, divide its recovered output by that probability to obtain its conditional recovered state; when its probability is zero, no conditional state is assigned.

If the outcome is discarded, the physical recovery is the channel ${\mathcal{R}}_4(X)=PXP+UX{U}^{†}$, whose output is supported on $\mathcal{C}$. For a logical two-dimensional space with isometric encoding $V\mathrm{\mid }{0}_L⟩=\mathrm{\mid }0⟩$ and $V\mathrm{\mid }{1}_L⟩=\mathrm{\mid }1⟩$, the corresponding decoder is $X\mapsto V^{†}{\mathcal{R}}_4(X)V$. This procedure specifies no compensation for an unknown jump time and makes no claim that the decay branch restores an arbitrary input exactly.

Four-level spontaneous-decay model

Finite operator and ket conventions

Code restriction and decoded channel types

Finite instruments and conditional states

Finite projective measurements

For A∈ M_m× n(F) and B∈ M_n× m(F), tr(AB)=tr(BA)

Instrument outcome weights normalize

Positive-probability instrument branches are density operators

Given: The stated definitions, hypotheses and cited interfaces.

Goal: Justify the stated construction and its scope.

1.1 By orthonormality and the outer-product action in [F1] and [F2], $P^{†}=P$, $P^2=P$, $Q^{†}=Q$, $Q^2=Q$, $PQ=QP=0$, and $P+Q=I$. Also $U^{†}U=Q$, $UP=0$, and $UQ=U$. The four-level space is nonzero and finite-dimensional, and $\{\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l},\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\}$ is finite and nonempty. Thus $(P,Q)$ meets every projector hypothesis in [F5] and defines the stated projective syndrome measurement. [given, F1, F2, F5]

2.1 The specified syndrome branch operations are $X\mapsto PXP$ and $X\mapsto QXQ$. Applying $U$ after the latter gives $U(QXQ)U^{†}=UX{U}^{†}$ by $UQ=U$. By [F6], for any input density operator $\rho$, $\mathrm{tr}(P\rho P)=\mathrm{tr}(\rho P)$ and $\mathrm{tr}(U\rho U^{†})=\mathrm{tr}(\rho U^{†}U)=\mathrm{tr}(\rho Q)$. These are exactly the syndrome weights defined in [F5], so the decay correction preserves its outcome weight. [F2, F5, F6, step 1.1]

3.1 Each recovered branch map is complex-linear and completely positive: for $A=P$ or $A=U$, its amplification is conjugation by $A\otimes I$; this preserves self-adjointness and, for positive semidefinite $Z$, its quadratic form at $z$ is $((A^{†}\otimes I)z{)}^{†}Z((A^{†}\otimes I)z)\ge 0$. For every $X$, [F6] and step 1.1 give $\mathrm{tr}(PXP)+\mathrm{tr}(UX{U}^{†})=\mathrm{tr}(XP)+\mathrm{tr}(XQ)=\mathrm{tr}(X)$, where the last equality follows by adding the finite diagonal sums and using $P+Q=I$. Hence the two recovered branch maps form a finite instrument from the four-level space to itself by [F4]. Their sum is linear, completely positive and trace-preserving, hence a channel by [F3]. The ranges of $P$ and $U$ lie in $\mathcal{C}$, so its outputs are code-supported. The displayed $V$ sends an orthonormal logical basis to an orthonormal basis of $\mathcal{C}$ and is therefore an isometric encoding; the decoder rule now follows from [F3]. [F1, F2, F3, F4, F6, step 1.1, step 2.1]

4.1 Apply [F7] to the finite instrument established in step 3.1 and the density input $\rho$ specified in the Definition: its two recovered branch traces are probabilities summing to one. For either branch whose trace is positive, [F8] applies to the same instrument and input and shows that the stated normalized output is a density operator. At zero trace, [F4] assigns no conditional state. Thus the output and conditional-state rules have their asserted meanings without asserting exact correction of an arbitrary decay-branch input. [given, F4, F7, F8, step 3.1] ∎