Finite matrix C-star algebra and density-state duality

Choose an ordered orthonormal basis $e_1,\dots ,e_D$. Its matrix units form an ordered basis of the complex endomorphism space, since $A={\sum }_{i,j}{A}_{ij}{E}_{ij}$ uniquely. The operator norm is a norm: its triangle inequality and homogeneity follow by applying the vector norm to $Av$ for unit $v$, and norm zero implies $A{e}_j=0$ for all $j$. Matrix composition is associative and complex bilinear, and conjugate transpose is conjugate-linear, involutive and reverses products. By [F2] the norm is submultiplicative, complete, and satisfies $\mathrm{\parallel }{A}^{\ast }A\mathrm{\parallel }=\mathrm{\parallel }A{\mathrm{\parallel }}^2$. Thus all axioms of [F4] hold. The identity is nonzero and has norm one because $D\ge 1$. The C*-identity and submultiplicativity give $\mathrm{\parallel }A{\mathrm{\parallel }}^2\le \mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }\mathrm{\parallel }A\mathrm{\parallel }$, and applying this also to $A^{\ast }$ gives $\mathrm{\parallel }{A}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }A\mathrm{\parallel }$, including $A=0$.

For every vector $v$, $⟨(\mathrm{\parallel }A{\mathrm{\parallel }}^{2}I-A^{\ast }A)v,v⟩=\mathrm{\parallel }A{\mathrm{\parallel }}^2\mathrm{\parallel }v{\mathrm{\parallel }}^2-\mathrm{\parallel }Av{\mathrm{\parallel }}^2\ge 0$. This Hermitian matrix is positive semidefinite, so [F1] gives its positive square root. The same spectral calculus shows that matrix positivity is equivalent to being $B^{\ast }B$, agreeing with the algebraic positivity used in [F4].

Let $\omega$ be complex linear and positive, and define ${\rho }_{ij}:=\omega (E_{ji})$. Finite expansion gives $\omega (A)={\sum }_{i,j}{A}_{ij}\omega (E_{ij})=\mathrm{Tr}(\rho A)$. For a column vector $z$, take $B=e_1{z}^{\ast }$; then $B^{\ast }B=z{z}^{\ast }$, so $z^{\ast }\rho z=\mathrm{Tr}(\rho z{z}^{\ast })=\omega (B^{\ast }B)\ge 0$ is real. Substitution of $z=e_i$, $e_i+e_j$, and $e_i+i{e}_j$ shows that the diagonal is real and ${\rho }_{ji}=\overline{{\rho }_{ij}}$; hence $\rho$ is Hermitian and its nonnegative quadratic forms make it positive semidefinite. Taking $A=I$ gives $\mathrm{Tr}\rho =\omega (I)$. The formula on the matrix units forces every entry of any representing matrix, proving uniqueness.

Conversely let $\rho \ge 0$ and diagonalize it by [F1] as $\rho ={\sum }_j{p}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$, $p_j\ge 0$. Then ${\omega }_{\rho }(A)={\sum }_j{p}_j⟨A{u}_j,u_j⟩$ is complex linear and ${\omega }_{\rho }(B^{\ast }B)={\sum }_j{p}_j\mathrm{\parallel }B{u}_j{\mathrm{\parallel }}^2\ge 0$. For every complex $A$, vector Cauchy–Schwarz gives $\mathrm{\mid }{\omega }_{\rho }(A)\mathrm{\mid }\le {\sum }_j{p}_j\mathrm{\parallel }A{u}_j\mathrm{\parallel }\le ({\sum }_j{p}_j)\mathrm{\parallel }A\mathrm{\parallel }$. Thus its functional norm is at most $\mathrm{Tr}\rho$, and evaluating at the norm-one identity gives the reverse inequality. This proves the norm formula, including $\rho =0$. The preceding reconstruction and this converse give the full bijection; imposing norm one is exactly imposing trace one by [F4], with no faithfulness requirement.