Uniform expansion of calibrated finite channels and null branches

A zero-dimensional reference gives zero input space and trivial inequalities, so fix a reference of dimension $r\ge 1$. The system vector space ${\mathbb{C}}^d$ has its ordered coordinate basis. The original and reference-extended matrix spaces have ordered matrix-unit bases $E_{pq}$: each matrix is uniquely the sum of its entries times these units, ordered lexicographically. Thus [F4] applies to the system vector space with its Euclidean norm and to those matrix spaces with their trace norms. In particular all spaces on which we use [F3] are Banach. On matrix maps write $\mathrm{\parallel }T{\mathrm{\parallel }}_{1\to 1}={\sup }_{\mathrm{\parallel }X{\mathrm{\parallel }}_1=1}\mathrm{\parallel }T(X){\mathrm{\parallel }}_1$, suppressing the subscript below; applying this inequality successively gives $\mathrm{\parallel }PQ\mathrm{\parallel }\le \mathrm{\parallel }P\mathrm{\parallel }\mathrm{\parallel }Q\mathrm{\parallel }$. Set $A(X)=i[H,X]$, $B_0(X)=GX+X{G}^{†}$ and $B_1=\mathcal{L}$, with identity reference factors suppressed. For any system matrix $T$, conjugating and transposing its entries twice gives $(T^{†}{)}^{†}=T$. By [F5], $\mathrm{\parallel }{T}^{†}{\mathrm{\parallel }}^2=\mathrm{\parallel }T{T}^{†}\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }{T}^{†}\mathrm{\parallel }$. If $\mathrm{\parallel }{T}^{†}\mathrm{\parallel }>0$, division gives $\mathrm{\parallel }{T}^{†}\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$; the same inequality is immediate if $\mathrm{\parallel }{T}^{†}\mathrm{\parallel }=0$. Repeating this argument with $T^{†}$ gives the reverse inequality, hence $\mathrm{\parallel }{T}^{†}\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$. This equality and [F5] give $\mathrm{\parallel }S\mathrm{\parallel }\le {\sum }_a\mathrm{\parallel }{L}_a^{†}\mathrm{\parallel }\mathrm{\parallel }{L}_a\mathrm{\parallel }=j$ and $\mathrm{\parallel }G\mathrm{\parallel }\le h+j\mathrm{/}2$. Applying [F1] term by term now yields $\mathrm{\parallel }A\mathrm{\parallel }\le 2h$, $\mathrm{\parallel }{B}_0\mathrm{\parallel }\le 2h+j$ and $\mathrm{\parallel }{B}_1\mathrm{\parallel }\le 2h+2j$, independently of $r$. These bounds also establish the boundedness required in [F3].

For a fixed system matrix $D$, left multiplication by $D$ and right multiplication by $D^{†}$ commute, also with reference factors. Their sum $T_D(X)=DX+X{D}^{†}$ is bounded by $2\mathrm{\parallel }D\mathrm{\parallel }$ by [F1]. The finite binomial identity gives $T_D^n(X)={\sum }_{p+q=n}\left(\genfrac{}{}{0px}{}{n}{p}\right)D^{p}X(D^{†}{)}^q$. The square partial products of the two matrix exponential series converge to $e^{\tau D}X(e^{\tau D}{)}^{†}$ by [F3] and the congruence continuity estimate in [F1]; adjointing the finite sums and taking limits identifies the right exponential. Their difference from the total-degree-$N$ triangular sum consists only of terms with $p+q>N$, whose induced trace-norm sum is at most ${\sum }_{n>N}(2\tau \mathrm{\parallel }D\mathrm{\parallel }{)}^n\mathrm{/}n!$, which tends to zero. The triangular sums are precisely the exponential partial sums of $T_D$ by the binomial identity. Therefore ${\mathrm{Ad}}_{e^{\tau D}}=e^{\tau T_D}$. Apply this with $D=iH$ and $D=G$ to obtain ${\mathcal{U}}_{\tau }=e^{\tau A}$ and ${\mathrm{Ad}}_{e^{\tau G}}=e^{\tau B_0}$. Likewise the powers of $B_1={\mathrm{id}}_R\otimes \mathcal{L}$ are the reference extensions of the powers of $\mathcal{L}$; the exponential series acts on each reference matrix block, giving $e^{\tau B_1}={\mathrm{id}}_R\otimes {\mathcal{N}}_{\tau }$.

Let $a=\mathrm{\parallel }A\mathrm{\parallel }$, $c=\mathrm{\parallel }{B}_i\mathrm{\parallel }$ for either $i\in \{0,1\}$, and $Q_T=e^{\tau T}-I-\tau T$. In addition to the quadratic remainder in [F3], the exponential series gives $\mathrm{\parallel }{e}^{\tau B_i}-I\mathrm{\parallel }\le {\sum }_{n\ge 1}(\tau c{)}^n\mathrm{/}n!\le \tau c{e}^{\tau c}$, since $n!\ge (n-1)!$. The exact algebraic identity $e^{\tau A}{e}^{\tau B_i}-I-\tau (A+B_i)=Q_A{e}^{\tau B_i}+\tau A(e^{\tau B_i}-I)+Q_{B_i}$ needs no commutation of $A$ and $B_i$. Taking norms and using [F3] bounds its right side by ${\tau }^2(a^2{e}^{\tau (a+c)}\mathrm{/}2+ac{e}^{\tau c}+c^2{e}^{\tau c}\mathrm{/}2)\le {\tau }^2(a+c{)}^2{e}^{\tau (a+c)}\mathrm{/}2\le K{\tau }^2$, because $a+c\le b$ and $0\le \tau \le {\tau }_0$. This also covers $b=0$, when the bound is zero.

By the displayed definitions, $A+B_1=\mathcal{D}$ and $A+B_0=-\{S,\cdot \}\mathrm{/}2$, with the identity reference understood. Steps 2.1 and 2.2 therefore give both claimed first-order expansions and a remainder of induced trace norm at most $K{\tau }^2$ on every reference-extended matrix space. Applying the induced norm bound to an arbitrary $X$ gives each stated inequality. The constants depend only on the fixed finite data and ${\tau }_0$, as required.