Channels, Kraus, Choi, and Stinespring Representations

For all $x,y\in H$, [F2] expresses each inner product using the four displayed norm squares. Applying [F1] to $x+y$, $x-y$, $x+iy$, and $x-iy$ therefore gives $⟨Vx,Vy{⟩}_K=⟨x,y{⟩}_H$.

The adjoint exists by [F4]. By [F3] and step 1.1, $⟨x,(V^{†}V-I_H)y{⟩}_H=0$ for every $x,y\in H$. For each fixed $y$, conjugate symmetry and [F5] force $(V^{†}V-I_H)y=0$, hence $V^{†}V=I_H$.

If $V$ is surjective, choose for arbitrary $z\in K$ a vector $x\in H$ with $Vx=z$. Step 2.1 gives $V{V}^{†}z=V{V}^{†}Vx=Vx=z$, so $V{V}^{†}=I_K$. The proof includes zero-dimensional spaces and uses only pointwise preimages, not a choice axiom.

Put $S={\sum }_a{K}_a^{†}{K}_a\in \mathcal{L}(H)$. For every $X\in \mathcal{L}(H)$, [F1], trace linearity [F3], and rectangular cyclicity [F2] give ${\mathrm{tr}}_K(\mathrm{\Phi }(X))={\sum }_a{\mathrm{tr}}_K(K_{a}X{K}_a^{†})={\sum }_a{\mathrm{tr}}_H(K_a^{†}{K}_{a}X)={\mathrm{tr}}_H(SX)$. If $S=I_H$, this equals ${\mathrm{tr}}_H(X)$ for every $X$, so $\mathrm{\Phi }$ is trace-preserving by the Statement. This includes an empty family when $H=0$, where $I_H=0$.

Conversely, suppose $\mathrm{\Phi }$ is trace-preserving on all of $\mathcal{L}(H)$. The identity in step 1.1 gives ${\mathrm{tr}}_H((S-I_H)X)=0$ for every $X\in \mathcal{L}(H)$. Apply [F4] with $A=S-I_H$ to obtain $S-I_H=0$. Thus the normalization holds. In particular, if $H=0$ both sides are the unique zero endomorphism and the argument still applies.

For each $x\in H$, [F2] and $\mathrm{\mid }z\mathrm{\mid }=1$ give $\mathrm{\parallel }zUx\mathrm{\parallel }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\parallel }Ux\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$. The map $zU$ is surjective: for any $y\in K$, choose $x\in H$ with $Ux=\bar{z}y$, then $zUx=y$. Hence $zU$ is a surjective linear isometry and is unitary by [F1]; [F4] supplies its adjoint identities.

For every $X\in \mathcal{L}(H)$, [F1] and [F3] give ${\mathrm{Ad}}_{zU}(X)=(zU)X(zU{)}^{†}=z\bar{z}UX{U}^{†}=\mathrm{\mid }z{\mathrm{\mid }}^2{\mathrm{Ad}}_U(X)={\mathrm{Ad}}_U(X)$. This holds on every input operator, so the maps are equal. The proof concerns only a single conjugation map and yields no controlled-operation identity.

For a product basis vector $e_{v^{\mathrm{\prime }}}\otimes f_{t^{\mathrm{\prime }}}$, [F3] gives $(E_{uv}\otimes F_{st})(e_{v^{\mathrm{\prime }}}\otimes f_{t^{\mathrm{\prime }}})=e_u\otimes f_s$ when $v^{\mathrm{\prime }}=v$ and $t^{\mathrm{\prime }}=t$, and gives $0$ otherwise, by orthonormality in [F1]. Thus each tensor matrix unit has a single coordinate $1$ and zeros elsewhere in the product basis.

Every endomorphism $T$ of $H\otimes H^{\mathrm{\prime }}$ has a unique coordinate coefficient $c_{(u,s),(v,t)}$ for the image of each product basis vector $e_v\otimes f_t$. The finite sum ${\sum }_{u,s,v,t}{c}_{(u,s),(v,t)}{E}_{uv}\otimes F_{st}$ has the same image as $T$ on each such basis vector by step 1.1; [F4] therefore makes it equal to $T$. The same coordinate entries show that this expression is unique. If $n=0$ or $m=0$, the product space is zero by [F2] and both the basis and sum are empty.

If $\dim H=0$ or $\dim K=0$, the input or output operator space is zero, so $\mathrm{\Phi }=0$ and the empty family is a Kraus representation. Assume both dimensions are positive and choose an ordered orthonormal basis $(f_{\alpha }{)}_{\alpha <m}$ of $K$ by [F3]. If $J_{\mathcal{B}}(\mathrm{\Phi })=0$, its $(i,\alpha ),(j,\beta )$ product-basis entry is the $(\alpha ,\beta )$ entry of $\mathrm{\Phi }(E_{ij})$ by [L1], so every $\mathrm{\Phi }(E_{ij})=0$. For each $X\in \mathcal{L}(H)$, the finite sum ${\sum }_{i,j<n}{x}_{ij}{E}_{ij}$ whose coordinate entries are those of $X$ agrees with $X$ on each $e_j$; [F4] therefore makes it equal to $X$. Thus $\mathrm{\Phi }=0$ and the empty family works also in this case.

Suppose $J_{\mathcal{B}}(\mathrm{\Phi })\mathrm{\ne }0$. Let $S=J_{\mathcal{B}}(\mathrm{\Phi }{)}^{1\mathrm{/}2}$ be its non-negative square root from [F2]. By [F1], choose an orthonormal eigenbasis $(v_a{)}_{a<nm}$ of $S$ with eigenvalues $s_a\ge 0$. Then $J_{\mathcal{B}}(\mathrm{\Phi })=S^2={\sum }_{a:s_a>0}{s}_a^2{v}_a{v}_a^{†}$. For each $a$ with $s_a>0$, put $w_a=s_a{v}_a$ and write $w_a={\sum }_{i<n,\alpha <m}{w}_{a,i\alpha }{e}_i\otimes f_{\alpha }$. Define the linear map $K_a:H\to K$ on the basis by $K_a{e}_i={\sum }_{\alpha <m}{w}_{a,i\alpha }{f}_{\alpha }$.

For every $i,j<n$ and $\alpha ,\beta <m$, the $(i,\alpha ),(j,\beta )$ entry of [L1] is ${\sum }_{a:s_a>0}{w}_{a,i\alpha }\overline{w_{a,j\beta }}$. The $(\alpha ,\beta )$ entry of ${\sum }_a{K}_a{E}_{ij}{K}_a^{†}$ is ${\sum }_a(K_a{)}_{\alpha i}\overline{(K_a{)}_{\beta j}}$, which is the same sum because $(K_a{)}_{\alpha i}=w_{a,i\alpha }$ by construction and [F6] gives the conjugate-transpose matrix of $K_a^{†}$. Therefore $\mathrm{\Phi }(E_{ij})={\sum }_a{K}_a{E}_{ij}{K}_a^{†}$ for every matrix unit.

Every $X\in \mathcal{L}(H)$ is the finite coordinate combination ${\sum }_{i,j<n}{x}_{ij}{E}_{ij}$ established in step 1.1. By linearity, step 2.1 therefore gives $\mathrm{\Phi }(X)={\sum }_a{K}_{a}X{K}_a^{†}$ for every $X$. There are at most $nm=\dim (H)\dim (K)$ retained operators because the eigenbasis of $S$ has $nm$ elements; when $J=0$ the empty family was handled in step 1.1. This is a finite Kraus representation by [L2].

If $\dim H=0$, [F3] makes the Choi sum empty. If $\dim K=0$, the output operator space contains only the zero map, so every $\mathrm{\Phi }(E_{ij})=0$. In either case $J_{\mathcal{B}}(\mathrm{\Phi })=0$ is non-negative. Assume now both spaces are nonzero, write $\mathcal{B}=(e_i{)}_{i<n}$, let $R={\mathbb{C}}^n$ have its standard orthonormal basis $(c_i{)}_{i<n}$ by [F5], and set $\mathrm{\Omega }={\sum }_{i<n}{e}_i\otimes c_i$. Define $Q_{\mathrm{\Omega }}x=⟨x,\mathrm{\Omega }⟩\mathrm{\Omega }$. The root pairing is linear in its first argument, so $Q_{\mathrm{\Omega }}$ is linear. For all $x,y$, $⟨Q_{\mathrm{\Omega }}x,y⟩=⟨x,\mathrm{\Omega }⟩⟨\mathrm{\Omega },y⟩=⟨x,Q_{\mathrm{\Omega }}y⟩$, and $⟨Q_{\mathrm{\Omega }}x,x⟩=\mathrm{\mid }⟨x,\mathrm{\Omega }⟩{\mathrm{\mid }}^2\ge 0$. Thus $Q_{\mathrm{\Omega }}$ is non-negative by [F1].

For $i,j<n$, let $F_{ij}:R\to R$ be the matrix unit $F_{ij}{c}_k=0$ for $k\mathrm{\ne }j$ and $F_{ij}{c}_j=c_i$. On the product basis $(e_k\otimes c_l)$, the product pairing gives $Q_{\mathrm{\Omega }}(e_k\otimes c_l)=0$ if $k\mathrm{\ne }l$, and $Q_{\mathrm{\Omega }}(e_k\otimes c_k)=\mathrm{\Omega }$. The same action is given by ${\sum }_{i,j<n}{E}_{ij}\otimes F_{ij}$, so $Q_{\mathrm{\Omega }}={\sum }_{i,j<n}{E}_{ij}\otimes F_{ij}$ by [A2] and [A3]. The $R$-blocks of $Q_{\mathrm{\Omega }}$ are therefore $E_{ij}$; the actual block rule in [A1] gives the non-negative operator $Y={\mathrm{\Phi }}^{(n)}(Q_{\mathrm{\Omega }})={\sum }_{i,j<n}\mathrm{\Phi }(E_{ij})\otimes F_{ij}$ on $K\otimes R$.

Choose an ordered orthonormal basis $(f_{\alpha }{)}_{\alpha <m}$ of $K$ by [F2]. By [A2], both $(f_{\alpha }\otimes c_j{)}_{\alpha ,j}$ and $(e_j\otimes f_{\alpha }{)}_{j,\alpha }$ are orthonormal bases. Let $W:K\otimes R\to H\otimes K$ be the linear map sending $f_{\alpha }\otimes c_j$ to $e_j\otimes f_{\alpha }$. This bijection of orthonormal bases is a surjective linear isometry; [F6] gives $W^{†}W=I$ and $W{W}^{†}=I$. Applying the map and its inverse to product basis vectors shows $W(\mathrm{\Phi }(E_{ij})\otimes F_{ij})W^{†}=E_{ij}\otimes \mathrm{\Phi }(E_{ij})$ for each $i,j$. Hence $WY{W}^{†}=J_{\mathcal{B}}(\mathrm{\Phi })$ by [L1].

By [F4], $WY{W}^{†}$ is self-adjoint because $Y$ is self-adjoint. For every $z\in H\otimes K$, the adjoint identity gives $⟨WY{W}^{†}z,z⟩=⟨Y{W}^{†}z,W^{†}z⟩\ge 0$, since $Y$ is non-negative. Therefore $J_{\mathcal{B}}(\mathrm{\Phi })$ is non-negative.

Fix $r\ge 1$ and put $R={\mathbb{C}}^r$ with its standard ordered basis $(c_k{)}_{k<r}$, the basis used for the block amplification in [A1]. Choose ordered orthonormal bases $(e_i)$ of $H$ and $(f_j)$ of $K$ by [F6], allowing an empty basis. If $H=0$, the domain $\mathcal{L}(H\otimes R)$ is zero and the identity below holds for its sole operator. Otherwise let $E_{ij}$ be the matrix units of $H$ in the basis $(e_i)$ and let $F_{kl}$ be the matrix units of $R$ in its standard basis $(c_k)$. By [A2], $(e_i\otimes c_k)$ is a basis; the operators $E_{ij}\otimes F_{kl}$ are its matrix units, so they span $\mathcal{L}(H\otimes R)$ by [F4] and [F5]. For such a product matrix unit, its only nonzero ancilla block is block $(k,l)$, equal to $E_{ij}$. The block rule in [A1] therefore gives ${\mathrm{\Phi }}^{(r)}(E_{ij}\otimes F_{kl})=\mathrm{\Phi }(E_{ij})\otimes F_{kl}$. On the other hand, [A3], [A4], and [L1] give ${\sum }_a(K_a\otimes I_R)(E_{ij}\otimes F_{kl})(K_a^{†}\otimes I_R)={\sum }_a(K_a{E}_{ij}{K}_a^{†})\otimes F_{kl}=\mathrm{\Phi }(E_{ij})\otimes F_{kl}$. By linearity the two maps agree for every $X\in \mathcal{L}(H\otimes R)$.

Let $X$ be non-negative and put $T_a=K_a\otimes I_R$. Each $T_{a}X{T}_a^{†}$ is self-adjoint because $(T_{a}X{T}_a^{†}{)}^{†}=T_a{X}^{†}{T}_a^{†}=T_{a}X{T}_a^{†}$ by [F2]. For every $y$, the adjoint identity in [F2] gives $⟨T_{a}X{T}_a^{†}y,y⟩=⟨X{T}_a^{†}y,T_a^{†}y⟩\ge 0$ by [F1]. Thus each summand and their finite sum are non-negative. If $A$ is empty, the sum is zero and has the same property.

Since $r\ge 1$ was arbitrary, [A1] and step 2.1 show that every finite ancilla block amplification preserves non-negativity. This is precisely complete positivity by [A1].

Choose ordered orthonormal bases $(f_{\alpha })$ of $K$ and $(u_a)$ of $E$ by finite-dimensional orthonormal-basis existence. For each $a$, define the linear map $P_a:K\otimes E\to K$ on the product basis by $P_a(f_{\alpha }\otimes u_b)=f_{\alpha }$ if $b=a$ and $0$ otherwise. Its adjoint satisfies $P_a^{†}{f}_{\alpha }=f_{\alpha }\otimes u_a$: both sides have the same pairing with every product basis vector by [F2] and [F4]. Linearity extends the equality of pairings to every first argument, and [F5] then makes the vectors equal. For any $R\in \mathcal{L}(K\otimes E)$, the $(\alpha ,\beta )$ entry of ${\sum }_a{P}_{a}R{P}_a^{†}$ is ${\sum }_a{R}_{(\alpha ,a),(\beta ,a)}$, which is the defining coordinate of ${\mathrm{Tr}}_{E}R$ by [F3]. Thus ${\mathrm{Tr}}_{E}R={\sum }_a{P}_{a}R{P}_a^{†}$.

Put $L_a=P_{a}V:H\to K$. By [F4], $L_a^{†}=V^{†}{P}_a^{†}$. Substituting $R=VX{V}^{†}$ into step 1.1 gives $\mathrm{\Phi }(X)={\sum }_a{P}_{a}VX{V}^{†}{P}_a^{†}={\sum }_a{L}_{a}X{L}_a^{†}$ for every $X$. This is a finite Kraus representation, so [F6] makes $\mathrm{\Phi }$ completely positive.

On each product basis vector $f_{\alpha }\otimes u_b$, the sum ${\sum }_a{P}_a^{†}{P}_a$ acts as the identity; hence ${\sum }_a{P}_a^{†}{P}_a=I_{K\otimes E}$. By [F4] and [F1], ${\sum }_a{L}_a^{†}{L}_a=V^{†}({\sum }_a{P}_a^{†}{P}_a)V=V^{†}V=I_H$. Rectangular trace cyclicity [F7] and trace linearity [F8] then give, for every $X\in \mathcal{L}(H)$, ${\mathrm{tr}}_K\mathrm{\Phi }(X)={\sum }_a{\mathrm{tr}}_K(L_{a}X{L}_a^{†})={\mathrm{tr}}_H(X{\sum }_a{L}_a^{†}{L}_a)={\mathrm{tr}}_{H}X$. The complete positivity from step 2.1 and this trace identity are exactly the channel conditions [F9]. Empty bases and sums give the same identities when $E=0$; the isometry equation then forces $H=0$.

Suppose $\mathrm{\Phi }$ is completely positive. By [F1], its Choi matrix is non-negative. Suppose instead that its Choi matrix is non-negative. By [F2], $\mathrm{\Phi }$ has a finite Kraus representation. These give the first two implications in the cycle.

If $\mathrm{\Phi }$ has a finite Kraus representation in the sense of [L2], [F3] implies that $\mathrm{\Phi }$ is completely positive. This closes the cycle, so each of the three stated conditions implies the other two. All statements use the same fixed input basis and the input-first/output-second Choi convention of [L1].