Finite jump expansion and ensemble equation

Under [A1], put $D={\sum }_a{L}_a^{\ast }{L}_a$ and $K=-iH\mathrm{/}\mathrm{\hslash }-D\mathrm{/}2$. Then $K+K^{\ast }=-D$. The matrix space is complete by [F3], so [F4] differentiates $R(u)=e^{uK}$. For every vector $v$, $\frac{d}{du}\mathrm{\parallel }R(u)v{\mathrm{\parallel }}^2=⟨(K+K^{\ast })R(u)v,R(u)v⟩=-{\sum }_a\mathrm{\parallel }{L}_{a}R(u)v{\mathrm{\parallel }}^2\le 0$. Since $R(0)=I$, the continuous derivative is integrable by [F8], and [F9] gives $\mathrm{\parallel }R(u)v\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$ for $u\ge 0$, hence $R(u)$ is a contraction.

On $M_d(\mathbb{C})$ define $\mathcal{N}(X)=KX+X{K}^{\ast }$, $\mathcal{J}(X)={\sum }_a{L}_{a}X{L}_a^{\ast }$, and ${\mathcal{T}}_u=e^{u\mathcal{L}}$. By [F5], these finite sums of left and right matrix multiplications are bounded in trace norm. Expanding [F1] gives $\mathcal{L}=\mathcal{N}+\mathcal{J}$. Fix $t>0$. Integrate finite-dimensional matrix and superoperator entries by their real and imaginary components, using the repeated Riemann convention in [F2]. The curve $s\mapsto e^{(t-s)\mathcal{N}}{\mathcal{T}}_s$ has continuously differentiable entries by [F4]; its real and imaginary derivative components are continuous and hence integrable by [F8]. Differentiation gives $\frac{d}{ds}(e^{(t-s)\mathcal{N}}{\mathcal{T}}_s)=e^{(t-s)\mathcal{N}}\mathcal{J}{\mathcal{T}}_s$, since $\mathcal{L}=\mathcal{N}+\mathcal{J}$. Applying [F9] entrywise yields ${\mathcal{T}}_t=e^{t\mathcal{N}}+{\int }_0^t{e}^{(t-s)\mathcal{N}}\mathcal{J}{\mathcal{T}}_{s}ds$. This finite identity follows by induction on $N$: the base case is the displayed variation-of-constants identity, and substituting that identity for the final ${\mathcal{T}}_{s_1}$ in the remainder adds the next ordered-simplex term and a remainder with one more innermost time. Only finite linearity of the nested integrals is used; no integration order is exchanged. Thus, for each $N\ge 0$, ${\mathcal{T}}_t={\sum }_{n=0}^N{S}_n(t)+Q_{N+1}(t)$, where $S_0(t)=e^{t\mathcal{N}}$, $S_n(t)={\int }_{0<s_1<\cdots <s_n<t}{e}^{(t-s_n)\mathcal{N}}\mathcal{J}{e}^{(s_n-s_{n-1})\mathcal{N}}\cdots \mathcal{J}{e}^{s_1\mathcal{N}}d\mathbf{s}$ for $n\ge 1$, and $Q_m(t)={\int }_{0<s_1<\cdots <s_m<t}{e}^{(t-s_m)\mathcal{N}}\mathcal{J}\left({\prod }_{j=m-1,\dots ,1}{e}^{(s_{j+1}-s_j)\mathcal{N}}\mathcal{J}\right){\mathcal{T}}_{s_1}d\mathbf{s}$ for $m=N+1$, where the product is ordered from left to right with descending $j$, and the empty product is the identity. For $m=1$ the intermediate exponential factors are absent: $Q_1(t)={\int }_0^t{e}^{(t-s_1)\mathcal{N}}\mathcal{J}{\mathcal{T}}_{s_1}d{s}_1$, exactly the displayed variation-of-constants remainder. In particular there is no extra $e^{s_1\mathcal{N}}$ before ${\mathcal{T}}_{s_1}$. All these integrands are continuous in finite-dimensional coordinates on the closed ordered simplex; [F11] proves that every inner integral is continuous in its remaining parameters and hence the full repeated integral exists. The insertion of one further nested integral in the induction is therefore legitimate.

Left multiplication by $K$ and right multiplication by $K^{\ast }$ commute; multiplying their absolutely norm-convergent exponential series and grouping by total degree with the binomial formula yields $e^{u\mathcal{N}}(X)=e^{uK}X{e}^{u{K}^{\ast }}=R(u)XR(u{)}^{\ast }$. This establishes the no-jump superoperator formula used in the norm estimates below.

From [F5] and [F7], $\mathrm{\parallel }\mathcal{J}(X){\mathrm{\parallel }}_1\le ({\sum }_a\mathrm{\parallel }{L}_a{\mathrm{\parallel }}^2)\mathrm{\parallel }X{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }\mathcal{N}(X){\mathrm{\parallel }}_1\le 2\mathrm{\parallel }K\mathrm{\parallel }\mathrm{\parallel }X{\mathrm{\parallel }}_1$, so the induced $1\to 1$ norms are finite. By step 2.1 and the contraction in step 1.1, [F5] gives $\mathrm{\parallel }{e}^{u\mathcal{N}}{\mathrm{\parallel }}_{1\to 1}\le 1$ for $u\ge 0$. The ordered-simplex volume is $t^n\mathrm{/}n!$: the zero-variable volume is $1$; if the $n$-variable volume is $s^n\mathrm{/}n!$, then [F10] gives the primitive $s^{n+1}\mathrm{/}(n+1)!$, and [F8] makes the integrand Riemann integrable, so [F9] evaluates the next repeated integral as $t^{n+1}\mathrm{/}(n+1)!$. Induction proves the volume formula. The continuous-integrand norm bound in [F11], applied in the finite-dimensional superoperator space with norm $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{1\to 1}$, now bounds each integral by that volume times its pointwise norm bound. Thus $\mathrm{\parallel }{S}_n(t){\mathrm{\parallel }}_{1\to 1}\le (t\mathrm{\parallel }\mathcal{J}{\mathrm{\parallel }}_{1\to 1}{)}^n\mathrm{/}n!$. By [F4], $\mathrm{\parallel }{\mathcal{T}}_s{\mathrm{\parallel }}_{1\to 1}\le e^{s\mathrm{\parallel }\mathcal{L}{\mathrm{\parallel }}_{1\to 1}}$, so $\mathrm{\parallel }{Q}_{N+1}(t){\mathrm{\parallel }}_{1\to 1}\le e^{t\mathrm{\parallel }\mathcal{L}{\mathrm{\parallel }}_{1\to 1}}(t\mathrm{\parallel }\mathcal{J}{\mathrm{\parallel }}_{1\to 1}{)}^{N+1}\mathrm{/}(N+1)!\to 0$. Replacing $t$ by any bound $T$ gives summable bounds for all $0\le t\le T$ and a remainder tending uniformly to zero. Thus the series ${\sum }_n{S}_n(t)$ converges in induced trace norm to ${\mathcal{T}}_t$, uniformly on bounded time intervals.

If $t=0$, then ${\mathcal{T}}_0=\mathrm{id}$ and $R(0)=I$; all positive-jump simplexes are empty, and the sole branch has weight $\mathrm{Tr}\rho =1$. For $t>0$, applying the operator identity from step 3.1 to $\rho$ and expanding each occurrence of $\mathcal{J}$ in $S_n(t)$ sums over labels $a_1,\dots ,a_n$. Step 2.1 turns each factor $e^{u\mathcal{N}}$ into the no-jump conjugation, so each integrand is $W_{\mathbf{a},\mathbf{s}}(t)\rho W_{\mathbf{a},\mathbf{s}}(t{)}^{\ast }$ by [F2]. This matrix is positive since $⟨W\rho W^{\ast }v,v⟩=⟨\rho W^{\ast }v,W^{\ast }v⟩\ge 0$ for every vector $v$; this argument uses only $\rho \ge 0$, proving branch positivity for all positive inputs. All label sums at fixed $n$ are finite, and [F11] allows moving those sums and the finite-dimensional linear trace through the nested integrals. Trace-norm convergence from step 3.1 and the trace bound in [F5] then allow taking the trace of the infinite series. Each integrated branch weight is nonnegative: its continuous real trace integrand is pointwise nonnegative, and [F11] preserves this at every successive integration. For each vector $v$, the finite-dimensional linear functional $X\mapsto ⟨Xv,v⟩$ likewise commutes with these integrals by [F11]; its nonnegative integrand proves that every integrated branch state is positive. This positivity and convergence argument has not assumed the complete positivity of the final ensemble map. By [F6], $\mathrm{Tr}(e^{t\mathcal{L}}\rho )=\mathrm{Tr}\rho =1$, so the nonnegative integrated branch weights sum to one. If $r=0$, then $\mathcal{J}=0$, all $S_n$ with $n\ge 1$ vanish, and only the no-jump term remains.