Fast-interval decay approaches a correctable coherent error

Let $H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$ be the full finite model space in [F1], let $H_{\mathrm{s}\mathrm{e}\mathrm{l}}$ be the direct sum of the blocks labelled by $S$ in [F3], and let $\iota :H_{\mathrm{s}\mathrm{e}\mathrm{l}}\to H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$ be its isometric inclusion. Regard $V$ as mapping into $H_{\mathrm{s}\mathrm{e}\mathrm{l}}$ and put $P=V{V}^{\ast }$ there. Choose a finite spanning list $E_1,\dots ,E_m$ for the selected coherent tensor span. By [F3], $P{E}_i^{\ast }{E}_{j}P=c_{ij}P$. Each $c_{ii}\ge 0$, and $c={\sum }_i{c}_{ii}>0$ because the span contains the identity and the logical space is nonzero. Define $A_i=\iota E_{i}P\mathrm{/}\sqrt{c}$ and $A_0=\iota (I_{H_{\mathrm{s}\mathrm{e}\mathrm{l}}}-P)$, all with input $H_{\mathrm{s}\mathrm{e}\mathrm{l}}$ and output $H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$. Since ${\iota }^{\ast }\iota =I_{H_{\mathrm{s}\mathrm{e}\mathrm{l}}}$, their adjoint-square sum is $P+(I_{H_{\mathrm{s}\mathrm{e}\mathrm{l}}}-P)=I_{H_{\mathrm{s}\mathrm{e}\mathrm{l}}}$, so [F7] makes their finite operator sum a completely positive, trace-preserving channel. Their scalar pair products on the code are $c_{ij}\mathrm{/}c$, and all such compressed products involving $A_0$ vanish.

Let $B_{\tau }^{\mathrm{e}\mathrm{x}}$ be the exact unnormalized checked decay output, and let $d=\dim H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}\ge 1$. For any endomorphism $X$ of this space, [F9] gives singular values $s_j$ and unit vectors with $X{e}_j=s_j{f}_j$, so $s_j\le \mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. By the trace-norm definition in [F8], $\mathrm{\parallel }X{\mathrm{\parallel }}_1={\sum }_j{s}_j\le d\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. Thus [F2] gives ${\epsilon }_{\tau }:=\mathrm{\parallel }{B}_{\tau }^{\mathrm{e}\mathrm{x}}-B_{\tau }{\mathrm{\parallel }}_1=O({\tau }^2)$ uniformly for density inputs; their operator norm is at most one, since their nonnegative eigenvalues sum to one. By [F8], $\mathrm{\mid }\mathrm{Tr}(B_{\tau }^{\mathrm{e}\mathrm{x}}-B_{\tau })\mathrm{\mid }\le {\epsilon }_{\tau }$. Taking traces gives $p_{\tau }=\mathrm{Tr}{B}_{\tau }^{\mathrm{e}\mathrm{x}}=\gamma \tau +O({\tau }^2)$. Since $\gamma >0$, for all sufficiently small positive $\tau$ we have $p_{\tau }\ge \gamma \tau \mathrm{/}2>0$.

Apply [F4] to this channel with output space $H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$. It supplies a channel $\mathcal{R}:\mathcal{L}(H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}})\to \mathcal{L}(H_L)$, where $H_L$ is the logical space and $\mathcal{L}(K)$ denotes the linear operators on $K$. Its syndrome ranges lie in $\iota H_{\mathrm{s}\mathrm{e}\mathrm{l}}$, and its fixed-state completion includes the whole orthogonal complement of $\iota H_{\mathrm{s}\mathrm{e}\mathrm{l}}$ in $H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$. Thus the decoder is defined even on outputs in other model blocks. The off-diagonal error formula in [F4], after undoing the factor $1\mathrm{/}\sqrt{c}$, gives $\mathcal{R}(\iota E_{i}VX{V}^{\ast }{E}_j^{\ast }{\iota }^{\ast })=c_{ji}X$ for every logical operator $X$. By sesquilinearity this holds for all pairs in the selected error span. This extends the selected-space decoder without changing its action on selected-space inputs.

On the selected input block, the three jumps in [F1] are coherent rank-one components into its downward neighbor, hence belong to the selected error span after restriction to the code. By [F2], $B_{\tau }\mathrm{/}(\gamma \tau )={\sum }_{\delta }{L}_{\delta }V\rho V^{\ast }{L}_{\delta }^{\ast }\mathrm{/}\gamma$ has trace one. The full-space decoder of step 2.1 sends this normalized mixture to $\rho$: its cross-term formula applies to each jump, and the trace-one normalization makes the sum of the scalar coefficients one. This proves exact recovery of the first-order detected-decay state for either named code.

Positivity and the trace-norm definition in [F8] give $\mathrm{\parallel }{B}_{\tau }{\mathrm{\parallel }}_1=\mathrm{Tr}{B}_{\tau }=\gamma \tau$. Splitting the normalized difference as $(B_{\tau }^{\mathrm{e}\mathrm{x}}-B_{\tau })\mathrm{/}{p}_{\tau }+B_{\tau }(1\mathrm{/}{p}_{\tau }-1\mathrm{/}(\gamma \tau ))$, the triangle inequality and step 1.2 give $\mathrm{\parallel }{B}_{\tau }^{\mathrm{e}\mathrm{x}}\mathrm{/}{p}_{\tau }-B_{\tau }\mathrm{/}(\gamma \tau ){\mathrm{\parallel }}_1\le ({\epsilon }_{\tau }+\mathrm{\mid }{p}_{\tau }-\gamma \tau \mathrm{\mid })\mathrm{/}{p}_{\tau }\le 2{\epsilon }_{\tau }\mathrm{/}{p}_{\tau }=O(\tau )$. Both normalized branches are density operators on $H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$, using $\iota$ to embed the first-order branch. Apply [F6] to the full-space channel $\mathcal{R}$ from step 2.1 and these two states. Step 3.1 then yields the stated decoded conditional error, uniformly over logical density operators.

Write a known diagonal calibration as $U_h=\mathrm{diag}(e^{ih{d}_x})$ on $H_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$, where the finitely many $d_x$ are real fixed lift values, $0\le h\le \tau$, and either sign convention is absorbed into $d_x$. Each $h{d}_x$ is real, so [F11] gives $\mathrm{\mid }{e}^{ih{d}_x}\mathrm{\mid }=1$. Entrywise multiplication of the diagonal matrices therefore gives $U_h^{\ast }{U}_h=U_h{U}_h^{\ast }=I$, proving that $U_h$ is unitary. Put $M={\max }_x\mathrm{\mid }{d}_x\mathrm{\mid }$. If $M=0$, then $U_h=I$. Otherwise choose $M\tau \le 1\mathrm{/}2$. For $\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}2$, the finite partial sums of [F10] satisfy $\mathrm{\mid }{\sum }_{n=1}^N{z}^n\mathrm{/}n!\mathrm{\mid }\le {\sum }_{n=1}^N\mathrm{\mid }z{\mathrm{\mid }}^n\le \mathrm{\mid }z\mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }z\mathrm{\mid })\le 2\mathrm{\mid }z\mathrm{\mid }$; the middle inequality follows by multiplying the finite geometric sum by $1-\mathrm{\mid }z\mathrm{\mid }$. Passing to the convergent limit gives $\mathrm{\mid }{e}^z-1\mathrm{\mid }\le 2\mathrm{\mid }z\mathrm{\mid }$. Applying this to $z=ih{d}_x$ and using the diagonal basis gives $\mathrm{\parallel }{U}_h-I{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 2M\tau$. Since $U_h$ is unitary, $\mathrm{\parallel }{U}_h^{\ast }-I{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }{U}_h-I{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ and $\mathrm{\parallel }{U}_h^{\ast }{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$. For every density operator $\sigma$, [F8] gives $\mathrm{\parallel }\sigma {\mathrm{\parallel }}_1=1$ and, from $U_h\sigma U_h^{\ast }-\sigma =(U_h-I)\sigma U_h^{\ast }+\sigma (U_h^{\ast }-I)$, the bound $\mathrm{\parallel }{U}_h\sigma U_h^{\ast }-\sigma {\mathrm{\parallel }}_1\le 2\mathrm{\parallel }{U}_h-I{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 4M\tau$. Apply [F6] to $U_h\sigma U_h^{\ast }$ and $\sigma$ for the decoder $\mathcal{R}$; together with step 4.1 this preserves its $O(\tau )$ conditional error when calibration precedes decoding. If the calibration is instead on the decoded logical space, the same estimate there applies directly.

Multiplying the decoded conditional difference by $p_{\tau }=O(\tau )$ gives the $O({\tau }^2)$ unnormalized branch bound, uniformly over logical density operators. Under [F5], an unconditional repeated-cycle claim would additionally require the other outcomes and a complete composition budget; neither is supplied by this single-branch estimate.