Pure and Mixed States, Ensembles, and Schmidt Decomposition

Fix $a\in \{1,\dots ,n\}$. The family identity gives $I_H-E_a={\sum }_{b\mathrm{\ne }a}{E}_b$. For every $x\in H$, each $⟨E_{b}x,x⟩$ is a nonnegative real by [F2]; their finite sum is therefore nonnegative, including the empty sum $0$ when $n=1$. The sum is self-adjoint because each $E_b$ is self-adjoint and the sum is finite. Thus $I_H-E_a$ is non-negative.

Since $E_a$ is non-negative by [F1] and its complement is non-negative by step 1.1, it is an effect by the definition in [F1]. The index $a$ was arbitrary.

Put $\rho =I_H\mathrm{/}d$. The scalar $1\mathrm{/}d$ is positive real. The identity is self-adjoint, and for every $x\in H$, $⟨\rho x,x⟩=(1\mathrm{/}d)⟨x,x⟩\ge 0$ by [F3] and [F7]. Thus $\rho$ is non-negative by [F2].

In a basis of $H$ with $d$ elements, the matrix of $\rho$ is $(1\mathrm{/}d)I_d$. By [F4]–[F6], its trace is $(1\mathrm{/}d){\sum }_{j=1}^{d}1=1$. Therefore $\rho$ is a density operator by [F1].

Let $u$ be unit and suppose $v=\alpha u$ is another unit representative of the same ray. Then $1=\mathrm{\parallel }v{\mathrm{\parallel }}^2=\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{\parallel }u{\mathrm{\parallel }}^2$, so $\mathrm{\mid }\alpha \mathrm{\mid }=1$. By [F2], $P_{v}z=⟨z,\alpha u⟩\alpha u=\bar{\alpha }\alpha ⟨z,u⟩u=P_{u}z$ for every $z$. Thus the projector does not depend on the unit representative.

By [F2], $P_u^{2}z=⟨z,u⟩⟨u,u⟩u=P_{u}z$. Its range is $\mathbb{C}u$, a one-dimensional subspace, so [F7] gives rank one. For every $z$, $P_{u}z\in \mathbb{C}u$ and $⟨z-P_{u}z,u⟩=⟨z,u⟩-⟨z,u⟩⟨u,u⟩=0,$ so the residual is in $(\mathbb{C}u{)}^{\perp }$; by the residual characterization in [F3], $P_u$ is the orthogonal projection onto $\mathbb{C}u$. Also, for all $z,w$, $⟨P_{u}z,w⟩=⟨z,u⟩⟨u,w⟩=⟨z,P_{u}w⟩$ and $⟨P_{u}z,z⟩=\mathrm{\mid }⟨z,u⟩{\mathrm{\mid }}^2\ge 0$. Hence $P_u$ is non-negative by [F4]. Choose an orthonormal basis $(e_j)$ by [F6] and write $u={\sum }_j{c}_j{e}_j$; then $⟨P_u{e}_j,e_j⟩=\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$, so [F5] gives $\mathrm{tr}(P_u)={\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2=⟨u,u⟩=1$ by orthonormality.

Conversely, let $P$ be a rank-one orthogonal projection onto $W$. By [F7], $\dim W=1$, so choose a unit $u\in W$. For every $z$, write $Pz=\alpha u$; [F3] gives $z-Pz\in W^{\perp }$, so $⟨z,u⟩=⟨Pz,u⟩=\alpha$. Therefore $Pz=⟨z,u⟩u=P_{u}z$, proving surjectivity. If $P_u=P_v$, their ranges are equal, so $u$ and $v$ span the same line and represent the same ray. This proves injectivity and the claimed bijection.

Choose ordered orthonormal bases $(e_i{)}_{i=1}^m$ of $H$ and $(f_k{)}_{k=1}^n$ of $K$ using [F4], and write $\mathrm{\Psi }={\sum }_{i,k}{c}_{ik}{e}_i\otimes f_k$ by [F1]. Define the complex-linear map $T_C:{\mathbb{C}}^n\to {\mathbb{C}}^m$ by $(T_{C}z{)}_i={\sum }_{k=1}^n{c}_{ik}{z}_k$; its standard matrix is exactly $C$. Apply the positive-family construction [F3] to $T_C$. If $q_j$ and $p_j$ are its right and positive left singular vectors, the identity there gives $T_{C}z={\sum }_{j=1}^r{s}_j⟨z,q_j⟩p_j$. Evaluating at each standard basis vector of ${\mathbb{C}}^n$ and using [F5], the $(i,k)$ entry of this identity is $c_{ik}={\sum }_{j=1}^r{s}_j(p_j{)}_i\overline{(q_j{)}_k}$. This is the coefficient formula needed below, obtained from the positive singular families without a codomain basis completion.

Define $u_j={\sum }_i(p_j{)}_i{e}_i$ and $v_j={\sum }_k\overline{(q_j{)}_k}{f}_k$. The coordinate columns of the $p_j$ are orthonormal by [F3]–[F5], so the $u_j$ are orthonormal. For the second family, $⟨v_j,v_{\mathrm{\ell }}⟩={\sum }_k\overline{(q_j{)}_k}(q_{\mathrm{\ell }}{)}_k=\overline{⟨q_j,q_{\mathrm{\ell }}⟩}={\delta }_{j\mathrm{\ell }}$, so it too is orthonormal. The coefficient of $e_i\otimes f_k$ in ${\sum }_j{s}_j{u}_j\otimes v_j$ is ${\sum }_j{s}_j(p_j{)}_i\overline{(q_j{)}_k}=c_{ik}$ by step 1.1, so this sum equals $\mathrm{\Psi }$. If $C=0$, then $\mathrm{\Psi }=0$ and $r=0$. The positive coefficients are precisely the positive singular values of $C$.

By [F1], the vectors $u_j\otimes v_j$ are orthonormal: their pairings are $⟨u_i,u_j⟩⟨v_i,v_j⟩={\delta }_{ij}$. Expanding the squared norm of the sum in step 2.1 therefore gives $\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}^2={\sum }_j{s}_j^2$, including the empty sum when $\mathrm{\Psi }=0$.

By [F1]–[F2], choose an orthonormal eigenbasis $e_1^{\mathrm{\prime }},\dots ,e_d^{\mathrm{\prime }}$ with $\rho e_k^{\mathrm{\prime }}={\mu }_k{e}_k^{\mathrm{\prime }}$ and ${\mu }_k\ge 0$. In this basis the representing matrix is diagonal, so [F1] and [F3] give ${\sum }_{k=1}^d{\mu }_k=\mathrm{tr}(\rho )=1$. Thus the index set $J=\{k:{\mu }_k>0\}$ is nonempty, and omitting its complement removes exactly the zero eigenvalues.

List the finite set $J$ in increasing order as $k_1,\dots ,k_m$ and put ${\lambda }_j={\mu }_{k_j}$ and $e_j=e_{k_j}^{\mathrm{\prime }}$. By [F4], each unit eigenvector has a rank-one projector $\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$; the eigenbasis gives $\rho ={\sum }_{j=1}^m{\lambda }_j\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$, with positive weights summing to one. The operator maps each $e_{k_j}^{\mathrm{\prime }}$ to the nonzero vector ${\mu }_{k_j}{e}_{k_j}^{\mathrm{\prime }}$ and maps every other eigenbasis vector to zero, so its image is exactly the span of the $e_{k_j}^{\mathrm{\prime }}$. Thus [F5] gives $\mathrm{rank}(\rho )=m$. Distinct basis vectors give orthogonal projectors.

Set $\bar{\rho }={\sum }_{a=1}^n{p}_a{\rho }_a$, the barycenter in [F1]. It is self-adjoint because it is a finite real linear combination of self-adjoint operators. For every $x\in H$, $⟨\bar{\rho }x,x⟩={\sum }_a{p}_a⟨{\rho }_{a}x,x⟩\ge 0$ by [F2]–[F3], so $\bar{\rho }$ is non-negative.

By trace linearity [F4], $\mathrm{tr}(\bar{\rho })={\sum }_a{p}_a\mathrm{tr}({\rho }_a)={\sum }_a{p}_a=1$. Thus $\bar{\rho }$ is a density operator by [F2].

Let $B_1,\dots ,B_m$ partition the outcome indices and put $q_b={\sum }_{a\in B_b}{p}_a$. If $q_b>0$, the weights $p_a\mathrm{/}{q}_b$ within block $B_b$ sum to one, so the argument in steps 1.1–2.1, applied to this finite normalized subensemble, shows that ${\sigma }_b=q_b^{-1}{\sum }_{a\in B_b}{p}_a{\rho }_a$ is a density operator and $q_b{\sigma }_b={\sum }_{a\in B_b}{p}_a{\rho }_a$.

If $q_b=0$, nonnegativity of the weights forces every $p_a=0$ in $B_b$. Choose any density operator ${\sigma }_b$ using [F5]; then $q_b{\sigma }_b=0={\sum }_{a\in B_b}{p}_a{\rho }_a$. Summing the block identities gives ${\sum }_b{q}_b{\sigma }_b={\sum }_a{p}_a{\rho }_a$, so regrouping preserves the barycenter, including zero-weight blocks.

Use [F1] and let $K_0={\mathbb{C}}^r$ with the standard ordered basis relabelled $f_1,\dots ,f_r$, orthonormal by [F9]. Define ${\mathrm{\Psi }}_0={\sum }_{j=1}^r\sqrt{{\lambda }_j}{e}_j\otimes f_j$. The product pairing [F4] gives $\mathrm{\parallel }{\mathrm{\Psi }}_0{\mathrm{\parallel }}^2={\sum }_j{\lambda }_j=1$, and [F5] gives ${\mathrm{Tr}}_{K_0}(\mathrm{\mid }{\mathrm{\Psi }}_0⟩⟨{\mathrm{\Psi }}_0\mathrm{\mid })={\sum }_j{\lambda }_j\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }=\rho$. Thus ${\mathrm{\Psi }}_0$ is a purification on an auxiliary space of dimension $r=\mathrm{rank}(\rho )$ by [F12].

For each $i$, choose an orthonormal basis $(g_j^{(i)}{)}_{j<d_i}$ of $K_i$ by [F6], and put $h_j^{(i)}={\iota }_i{g}_j^{(i)}$. Since ${\iota }_i$ is linear and norm-preserving by [F11], each $h_j^{(i)}$ has norm one. For $j\mathrm{\ne }\mathrm{\ell }$, the four vectors $g_j^{(i)}\pm g_{\mathrm{\ell }}^{(i)}$ and $g_j^{(i)}\pm i{g}_{\mathrm{\ell }}^{(i)}$ have squared norm $2$, so their images do too. With the root inner product convention, for $z=⟨h_j^{(i)},h_{\mathrm{\ell }}^{(i)}⟩$ the differences $\mathrm{\parallel }{h}_j^{(i)}+h_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2-\mathrm{\parallel }{h}_j^{(i)}-h_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2=2(z+\bar{z})$ and $\mathrm{\parallel }{h}_j^{(i)}+i{h}_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2-\mathrm{\parallel }{h}_j^{(i)}-i{h}_{\mathrm{\ell }}^{(i)}{\mathrm{\parallel }}^2=2i(\bar{z}-z)$ both vanish. Thus $z+\bar{z}=0$ and $z=\bar{z}$, so $z=0$; the image vectors are orthonormal. Let $L_i=\mathrm{span}\{h_j^{(i)}:j<d_i\}$. By [F6], ${\mathbb{C}}^m=L_i\oplus L_i^{\perp }$ and $L_i^{\perp }$ is finite-dimensional; choose an orthonormal basis of $L_i^{\perp }$ by [F6] and append it to $(h_j^{(i)})$ to obtain an ordered orthonormal basis of ${\mathbb{C}}^m$. Choose an ordered orthonormal basis of $H$ by [F6]. The product basis result [F13] shows that $I_H\otimes {\iota }_i$ sends the orthonormal product basis of $H\otimes K_i$ to an orthonormal family, so the padded vector ${\tilde{\mathrm{\Psi }}}_i=(I_H\otimes {\iota }_i){\mathrm{\Psi }}_i$ is still a unit vector. Write ${\mathrm{\Psi }}_i={\sum }_{\alpha ,j}{c}_{\alpha j}{e}_{\alpha }\otimes g_j^{(i)}$ in this product basis. In the extended basis of ${\mathbb{C}}^m$, ${\tilde{\mathrm{\Psi }}}_i={\sum }_{\alpha ,j}{c}_{\alpha j}{e}_{\alpha }\otimes h_j^{(i)}$ has zero coefficients on every complement basis vector, so the partial-trace coordinate sum [F5] over the extended basis equals the original partial trace over $(g_j^{(i)})$, namely $\rho$. By [F3] write each ${\tilde{\mathrm{\Psi }}}_i$ in Schmidt form ${\sum }_j{s}_j^{(i)}{u}_j^{(i)}\otimes f_j^{(i)}$; [F5]–[F7] give its marginal ${\sum }_j(s_j^{(i)}{)}^2\mathrm{\mid }{u}_j^{(i)}⟩⟨u_j^{(i)}\mathrm{\mid }=\rho$. Therefore the vectors with $(s_j^{(i)}{)}^2=\lambda$ form an orthonormal basis of each positive $\lambda$-eigenspace of $\rho$.

Fix an eigenvalue $\lambda >0$. By step 1.2, its two lists of Schmidt vectors are orthonormal bases of the same eigenspace. Write $u_{\lambda ,j}^{(2)}={\sum }_k(V_{\lambda }{)}_{kj}{u}_{\lambda ,k}^{(1)}$. Orthonormality gives ${\sum }_k(V_{\lambda }{)}_{kj}\overline{(V_{\lambda }{)}_{k\mathrm{\ell }}}={\delta }_{j\mathrm{\ell }}$; conjugating this identity gives $V_{\lambda }^{\ast }{V}_{\lambda }=I$, so the square matrix $V_{\lambda }$ is unitary. Prescribe $f_{\lambda ,j}^{(2)}\mapsto {\sum }_k\overline{(V_{\lambda }{)}_{kj}}{f}_{\lambda ,k}^{(1)}$. For indices $j,\mathrm{\ell }$, the inner product of the two images is ${\sum }_k\overline{(V_{\lambda }{)}_{kj}}(V_{\lambda }{)}_{k\mathrm{\ell }}={\delta }_{j\mathrm{\ell }}$; distinct eigenvalue blocks remain orthogonal. Thus this basis map extends linearly by [F10] to an onto isometry between the Schmidt-support subspaces.

By [F6], each padded auxiliary space is the orthogonal direct sum of its Schmidt-support subspace and its orthogonal complement. Step 1.2 shows that each support has dimension $r$, so [F8] gives dimension $m-r$ for both complements. Choose orthonormal bases for them by [F6] and prescribe a bijection between those bases. The complement map and the support isometry from step 2.1 together send an orthonormal basis of ${\mathbb{C}}^m$ to an orthonormal basis of ${\mathbb{C}}^m$; [F10] extends this map uniquely and linearly, and the basis map is surjective and preserves norms. Hence [F11] makes it a unitary $U$. In each eigenvalue block, ${\sum }_j(V_{\lambda }{)}_{kj}\overline{(V_{\lambda }{)}_{\mathrm{\ell }j}}={\delta }_{k\mathrm{\ell }}$, and therefore ${\sum }_j\sqrt{\lambda }{u}_{\lambda ,j}^{(2)}\otimes U{f}_{\lambda ,j}^{(2)}={\sum }_{k=1}^{m_{\lambda }}\sqrt{\lambda }{u}_{\lambda ,k}^{(1)}\otimes f_{\lambda ,k}^{(1)}\mathrm{.}$ Summing these equalities over the positive eigenvalues gives the displayed equivalence after padding.

For any pure-state ray, [F1] gives a unit representative and [F3] makes its projector non-negative with trace one. By [F2] it is a density operator.

Let $\rho$ be a density operator whose image has dimension one by [F5]. By [F4] its spectral ensemble has exactly one positive term, so $\rho =\lambda \mathrm{\mid }u⟩⟨u\mathrm{\mid }$ for a unit $u$ and $\lambda >0$. The trace-one condition in [F2] gives $1=\mathrm{tr}(\rho )=\lambda \mathrm{tr}(\mathrm{\mid }u⟩⟨u\mathrm{\mid })=\lambda$, by [F3] and linearity [F6]. Hence $\rho$ is the projector of $[u]$, and the bijection in [F3] makes that ray unique.

By [F1]–[F2], the barycenter $\rho$ is a density operator, as required for the purification hypothesis. By [F3], write $\mathrm{\Psi }={\sum }_{j=1}^r\sqrt{{\lambda }_j}{e}_j\otimes f_j$ with ${\lambda }_j>0$ and orthonormal families; $r\ge 1$ because $\mathrm{\Psi }$ is a unit vector. Set $S=\mathrm{span}\{e_1,\dots ,e_r\}$ and $L=\mathrm{span}\{f_1,\dots ,f_r\}$. By [F13]–[F14], the complements $S^{\perp }$ and $L^{\perp }$ are finite-dimensional and have orthonormal bases; adjoining those bases completes the Schmidt families to orthonormal bases of $H$ and $K$. The basis-independent partial-trace formula [F5]–[F6], together with [F7], gives $\rho ={\mathrm{Tr}}_K(\mathrm{\mid }\mathrm{\Psi }⟩⟨\mathrm{\Psi }\mathrm{\mid })={\sum }_j{\lambda }_j\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$. This operator is positive with eigenvalue ${\lambda }_j$ on $e_j$ and zero on $S^{\perp }$; the orthogonal decomposition shows $\ker \rho =S^{\perp }$, so the support definition [F9] and kernel definition [F18] give $S=\mathrm{supp}(\rho )$. The expansion of $\mathrm{\Psi }$ shows that its auxiliary support is $L$.

If $p_a>0$ and $x\in \ker \rho$ as defined in [F18], then $0=⟨\rho x,x⟩={\sum }_b{p}_b⟨{\sigma }_{b}x,x⟩$. Every summand is nonnegative by [F9], so $⟨{\sigma }_{a}x,x⟩=0$. Writing $R_a=\sqrt{{\sigma }_a}$ by [F10], self-adjointness gives $⟨{\sigma }_{a}x,x⟩=⟨R_{a}x,R_{a}x⟩=0$, hence ${\sigma }_{a}x=0$. Self-adjointness then gives $⟨{\sigma }_{a}y,x⟩=0$ for every $y$, so ${\sigma }_{a}y\in S$.

On $S$, define $R{e}_j={\lambda }_j^{-1\mathrm{/}2}{e}_j$ and $A_a=p_{a}R({\sigma }_a{\mathrm{\mid }}_S)R$ when $p_a>0$; put $A_a=0$ when $p_a=0$. Each $A_a$ is self-adjoint because $R$ and ${\sigma }_a{\mathrm{\mid }}_S$ are self-adjoint, and it is non-negative because its quadratic form is $p_a⟨{\sigma }_{a}Rx,Rx⟩\ge 0$. Since ${\sum }_a{p}_a{\sigma }_a=\rho$, one has ${\sum }_a{A}_a=R(\rho {\mathrm{\mid }}_S)R=I_S$.

By [F13], $K=L\oplus L^{\perp }$; choose an orthonormal basis of $L^{\perp }$ by [F14], so $(f_j)$ together with that basis is an orthonormal basis of $K$. On $L$, define $E_a$ by the matrix $A_a^T$ in $(f_j)$ and extend it by zero on $L^{\perp }$. Since each $A_a$ is self-adjoint, $A_a^T$ is self-adjoint; for every coordinate vector $z$, [F17] gives $z^{\ast }{A}_a^{T}z=\overline{(\stackrel{ˉ}{z}{)}^{\ast }{A}_a\stackrel{ˉ}{z}}\ge 0$, so this block is non-negative. Their sum is identity on $L$ and zero on $L^{\perp }$. Choose $a_0$ with $p_{a_0}>0$, which exists since ${\sum }_a{p}_a=1$, and add the projection $0\oplus I_{L^{\perp }}$ to $E_{a_0}$. The resulting operators are non-negative and sum to $I_K$, so [F11]–[F12] make them a complete effect family. In particular, $E_a=0$ when $p_a=0$.

Let $\mathrm{\Lambda }{e}_j=\sqrt{{\lambda }_j}{e}_j$ and $S_a=\sqrt{E_a}$. Each $E_a$ is block diagonal on $L\oplus L^{\perp }$. The direct sum of the non-negative square roots of its two blocks is a non-negative square root of $E_a$, so uniqueness in [F10] shows that $S_a$ is block diagonal. If $B_a$ is the matrix of $S_a{\mathrm{\mid }}_L$ in $(f_j)$, then $B_a$ is self-adjoint and $B_a^2=A_a^T$.