Liouville matrix and Choi reshuffling conventions

Let $H_{\mathrm{i}\mathrm{n}}$ and $H_{\mathrm{o}\mathrm{u}\mathrm{t}}$ be finite-dimensional complex inner-product spaces with ordered orthonormal bases $(e_i{)}_{i<n}$ and $(f_k{)}_{k<m}$. Use the convention that the inner product is linear in its first argument. Write $\mathcal{L}(H)$ for the complex-linear endomorphisms of $H$. Define input matrix units $E_{ij}x=⟨x,e_j⟩e_i$ and output matrix units $F_{kl}y=⟨y,f_l⟩f_k$. A superoperator here means a complex-linear map $\mathcal{N}:\mathcal{L}(H_{\mathrm{i}\mathrm{n}})\to \mathcal{L}(H_{\mathrm{o}\mathrm{u}\mathrm{t}})$.

The Liouville matrix of $\mathcal{N}$ in these ordered matrix-unit bases is the $m^2\times n^2$ matrix $S$ whose entries satisfy $\mathcal{N}(E_{ij})={\sum }_{k,l<m}{S}_{(k,l),(i,j)}{F}_{kl}$. Pairs are ordered lexicographically. The input-first Choi matrix is $J={\sum }_{i,j<n}{E}_{ij}\otimes \mathcal{N}(E_{ij})$, represented in the ordered product basis $(e_i\otimes f_k{)}_{(i,k)}$. Its entries obey $J_{(i,k),(j,l)}=S_{(k,l),(i,j)}$. This exchange of index groupings is called reshuffling; the two matrices generally have different shapes and represent different linear maps.

If $H_{\mathrm{i}\mathrm{n}}=H_{\mathrm{o}\mathrm{u}\mathrm{t}}=H$ with the same ordered basis and $X=\mathcal{L}(H)$ is equipped with a Banach-space norm for which a linear endomorphism $\mathcal{G}:X\to X$ is bounded, write $L$ for the Liouville matrix of $\mathcal{G}$. For real $t$, the Liouville matrix of $e^{t\mathcal{G}}={\sum }_{r=0}^{\mathrm{\infty }}{t}^r{\mathcal{G}}^r\mathrm{/}r!$ is $S=e^{tL}={\sum }_{r=0}^{\mathrm{\infty }}{t}^r{L}^r\mathrm{/}r!$. Reshuffling this $S$ gives the Choi matrix of $e^{t\mathcal{G}}$.

Given: The finite-dimensional spaces, ordered orthonormal bases and linear maps of the Definition, with the stated Banach norm and bounded generator when the exponential is used.

1.1 For any $A\in \mathcal{L}(H_{\mathrm{i}\mathrm{n}})$, expand each $A{e}_j$ in the input basis. The resulting coefficients give $A={\sum }_{i,j}{a}_{ij}{E}_{ij}$; evaluating a vanishing such sum at each $e_j$ shows all coefficients vanish. Thus the $E_{ij}$ form a basis, and the same argument gives the output basis $F_{kl}$. Unique coordinates therefore make $S$ well defined. The input-first Choi definition and the tensor-map action give $J(e_j\otimes f_l)={\sum }_{i,k}{S}_{(k,l),(i,j)}{e}_i\otimes f_k$, establishing the stated entries in the ordered product basis. These arguments also cover an empty basis, with empty sums and the unique zero-space maps. Linear map between vector spaces over the same field A finite list v : n → V is an ordered basis if and only if every x ∈ V equals ∑_i<n λᵢ vᵢ for exactly one λ : n → F; those scalars are the coordinates of x in that ordered basis Choi matrix with input factor first Tensor products of finitely many linear maps Ordered tensor registers, their inner product, and wire permutations. [given, algebra]

1.2 For endomorphisms $\mathcal{P},\mathcal{Q}$ of $X=\mathcal{L}(H)$, let $A,B$ be their Liouville matrices and abbreviate a matrix-unit index pair by $\alpha$, $\beta$ or $\gamma$. Linearity gives $(\mathcal{P}\circ \mathcal{Q})(E_{\beta })={\sum }_{\alpha }({\sum }_{\gamma }{A}_{\alpha \gamma }{B}_{\gamma \beta })E_{\alpha }$. The inner sum is exactly $(AB{)}_{\alpha \beta }$ by the matrix-product definition. The identity map has matrix $I_{n^2}$. Thus, defining zeroth powers as these identities, repeated composition gives the matrix $L^r$ for ${\mathcal{G}}^r$, and linearity gives ${\sum }_{r=0}^N{t}^r{L}^r\mathrm{/}r!$ for the partial sum $T_N:={\sum }_{r=0}^N{t}^r{\mathcal{G}}^r\mathrm{/}r!$. This includes $n=0$, where $I_0$ and all matrices concerned are empty. Linear map between vector spaces over the same field Rectangular matrix multiplication and the identity matrix Iₙ, including zero-sized shapes. [given, step 1.1, algebra]

2.1 Under the stated Banach and boundedness hypotheses the operator exponential converges in operator norm. Write $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ for the operator norm on bounded endomorphisms of $X$ and $T=e^{t\mathcal{G}}$. For every matrix unit the operator-norm bound gives $\mathrm{\parallel }(T_N-T)E_{\beta }\mathrm{\parallel }\le \mathrm{\parallel }{T}_N-T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }{E}_{\beta }\mathrm{\parallel }\to 0$. Thus evaluation on each fixed matrix unit respects this limit. Banach space A bounded linear operator between normed spaces The operator norm as the least bound and as the unit-sphere or unit-ball supremum The exponential series of a bounded operator. [given, step 1.2]

2.2 The coordinate-map theorem applied to the normed space $X$ and its ordered matrix-unit basis gives a bounded coordinate map $c:X\to {\mathbb{C}}^{n^2}$, where the coordinate space has its ${\mathrm{\ell }}^1$ norm. In particular, there is a constant $C\ge 0$ such that ${\sum }_{\alpha }\mathrm{\mid }c(x{)}_{\alpha }\mathrm{\mid }\le C\mathrm{\parallel }x\mathrm{\parallel }$ for all $x\in X$. For each $\beta$, applying this to $(T_N-T)E_{\beta }$ and using step 2.1 shows that every matrix entry of $T_N$ converges to the corresponding entry of the Liouville matrix of $T$. By step 1.2 these entries are the partial sums of ${\sum }_{r=0}^{\mathrm{\infty }}{t}^r{L}^r\mathrm{/}r!$, proving the claimed equality $S=e^{tL}$. The Choi entry identity from step 1.1 then gives the Choi matrix by reshuffling. If $n=0$, both matrix series and the Choi matrix are empty and the same conclusions hold. A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space A finite list v : n → V is an ordered basis if and only if every x ∈ V equals ∑_i<n λᵢ vᵢ for exactly one λ : n → F; those scalars are the coordinates of x in that ordered basis. [given, step 1.1, step 1.2, step 2.1] ∎