Complete recovered-cycle bound with weighted trajectory branches

Isometric encoding and unitary conjugation are single-Kraus completely positive maps by [F7], and preserve trace by the diagonal-sum formula in [F11]: for an isometry $A$, $\mathrm{Tr}(AX{A}^{†})={\sum }_{i,j,k}{A}_{ij}{X}_{jk}\overline{A_{ik}}={\sum }_j{X}_{jj}$. The finite physical vector space has an ordered orthonormal basis by [F11], so is Banach by [F10]. The matrix $iH$ is bounded. Adjointing its exponential partial sums and taking limits in [F9] gives $(e^{iH\tau }{)}^{†}=e^{-iH\tau }$ because $H=H^{†}$; the group law in [F9] makes these exponentials mutual inverses. Thus the compensating matrix is unitary. Together with [F2] and [F6], this makes $C_{\tau }$ CPTP. Compositions preserve complete positivity at each finite extension and preserve trace. A finite reference extension of any channel here is itself a channel: a further identity extension is an extension by the product reference dimension, while trace is the sum of the traces of diagonal system blocks, each preserved by the channel. Thus [F5] applies to $R$ with its possibly unequal input and output dimensions and to every finite extension. Insert the two expansions [F3] between encoding and $R$. By [F2], $R\mathcal{D}\mathcal{V}=rI-(rI+rI)\mathrm{/}2=0$ and $R(-\{S,\mathcal{V}(\cdot )\}\mathrm{/}2)=-rI$. Encoding preserves trace norm by [F1], and $R$ contracts it by [F5], so the two remainders still have norm at most $K{\tau }^2$. This proves the full-cycle and null-branch estimates. A zero-dimensional reference gives only the zero input and trivial inequalities.

The physical vector space has a finite ordered orthonormal basis by [F11], so its Euclidean norm makes it Banach by [F10]; the finite matrix $G$ is bounded in spectral operator norm. Applying [F9] gives $\mathrm{\parallel }{e}^{uG}\mathrm{\parallel }\le e^{ug}$ for $u\ge 0$, where $g=\mathrm{\parallel }G\mathrm{\parallel }$. Applying the operator norm inequality successively to vectors gives submultiplicativity. The nonnegative propagation intervals in an $n$-jump amplitude in [F4] sum to $\tau$, so $\mathrm{\parallel }{K}_{a_1\cdots a_n}(s){\mathrm{\parallel }}^2\le e^{2g\tau }{\prod }_{k=1}^n\mathrm{\parallel }{L}_{a_k}{\mathrm{\parallel }}^2$. By [F1], the same constant bounds the trace norm of its congruence on an arbitrary reference-extended input of trace norm one. Apply [F8] in that finite output matrix space with its trace norm; the integrated congruence is bounded by $e^{2g\tau }{\prod }_k\mathrm{\parallel }{L}_{a_k}{\mathrm{\parallel }}^2{\tau }^n\mathrm{/}n!$. Summing the finite index family gives $e^{2g\tau }(j\tau {)}^n\mathrm{/}n!$, since $j={\sum }_a\mathrm{\parallel }{L}_a{\mathrm{\parallel }}^2$. This majorant has a convergent sum by the scalar series in [F9]; completeness in [F12] gives convergence in trace norm on each such input. Its sum agrees entrywise with [F4], so these are also the reference-extended trajectory branches of that channel.

Fix any positive integer reference dimension and a positive semidefinite extended input $Y$. Each trajectory congruence is completely positive by [F7]. Its entries are continuous functions of the ordered times by [F9] and finite matrix multiplication. By [F8], integrating its amplified value preserves Hermiticity and nonnegative quadratic forms: for each vector $v$, the quadratic form of the integral is the integral of the nonnegative scalar quadratic form. Thus every integrated $n$-jump map is completely positive. Finite sums have the same property. The convergent map series in [F4] converges on every system matrix unit, hence entrywise; after any fixed finite reference extension, each output block is still the image of one system block, so the series converges entrywise there too. Limits of Hermitian matrices are Hermitian and limits of their nonnegative quadratic forms remain nonnegative. It follows that every infinite jump tail, in particular the sum over $n\ge 1$, is completely positive. The null congruence is also completely positive by [F7]. Applying the same compensation, encoding and decoder to these maps preserves that property. By [F4], their sum is $C_{\tau }$ and the first tail is exactly $C_{\tau }^{+}=C_{\tau }-C_{\tau }^0$; this proves branch positivity by its trajectory representation, rather than by subtraction of arbitrary positive maps.

Subtract the two estimates of step 1.1 and use the triangle inequality in [F12] to obtain $\mathrm{\parallel }{C}_{\tau }^{+}-r\tau I\mathrm{\parallel }\le 2K{\tau }^2$, uniformly with references. For any reference-extended density input $\rho$, [F11] and [F12] give $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1=\mathrm{Tr}\rho =1$: its eigenvalues are nonnegative, so $\sqrt{{\rho }^{†}\rho }=\rho$. By step 2.1 and trace preservation of their sum, the two branch traces are nonnegative and sum to one. The inequality $\mathrm{\mid }\mathrm{Tr}X\mathrm{\mid }\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$ in [F12] bounds their deviations from $1-r\tau$ and $r\tau$ by $K{\tau }^2$ and $2K{\tau }^2$. Put $p_{\tau }=\mathrm{Tr}((C_{\tau }^{+}\otimes \mathrm{id})(\rho ))$. If $r>0$, for sufficiently small positive $\tau$ one has $p_{\tau }\ge r\tau \mathrm{/}2>0$, and the normalized jump-branch output obeys $\mathrm{\parallel }(C_{\tau }^{+}\otimes \mathrm{id})(\rho )\mathrm{/}{p}_{\tau }-\rho {\mathrm{\parallel }}_1\le (2K{\tau }^2+\mathrm{\mid }{p}_{\tau }-r\tau \mathrm{\mid })\mathrm{/}{p}_{\tau }\le 4K{\tau }^2\mathrm{/}{p}_{\tau }=O(\tau )$. This is a virtual trajectory branch; the decoder itself does not use a jump-count or jump-time record. Its unconditional error is already quadratic and requires no division by a vanishing probability. If $r=0$, the unconditional bounds apply without conditioning.

For a density input, the $n\ge 2$ tail is positive by step 2.1 and its trace is bounded by its trace norm by [F12]. Step 1.2 therefore bounds its probability by $e^{2g\tau }{\sum }_{n\ge 2}(j\tau {)}^n\mathrm{/}n!\le j^2{\tau }^2{e}^{(2g+j)\tau }\mathrm{/}2\le K_2{\tau }^2$ for $0\le \tau \le {\tau }_0$. The first inequality uses the scalar tail in [F9]; the last uses the nonnegative scalar exponential series and $K_2=j^2{e}^{(2g+j){\tau }_0}\mathrm{/}2$. Compensation and the decoder preserve every branch trace, including under reference extension, by step 1.1. Hence the same bound holds for the recovered cycle. The cases $j=0$ and $\tau =0$ give a zero tail by the same formulas and the degenerate integrals in [F8]. Together with steps 1.1 and 3.1, this establishes all the stated bounds and their interpretation using one decoder on both virtual branches.