Finite-regulator outside density and entropy

For each basis vector $f_b$, prescribe ${\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}{f}_b:={\sum }_{a=1}^{d_{\mathrm{o}\mathrm{u}\mathrm{t}}}({\sum }_{i=1}^{d_{\mathrm{i}\mathrm{n}}}{\rho }_{(i,a),(i,b)})f_a$. The product basis in [F3] supplies the matrix entries of $\rho$, and [F4]–[F6] give a unique linear extension from the prescribed images; its matrix is exactly the displayed formula.

Since $\rho$ is self-adjoint by [F1]–[F2], its matrix satisfies $\overline{{\rho }_{(i,b),(i,a)}}={\rho }_{(i,a),(i,b)}$, so ${\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ is self-adjoint. For every $y\in H_{\mathrm{o}\mathrm{u}\mathrm{t}}$, expansion in $(f_a)$ gives $⟨{\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}y,y⟩={\sum }_i⟨\rho (e_i\otimes y),e_i\otimes y⟩\ge 0$ by [F2]–[F3]. Its trace is ${\sum }_a({\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}{)}_{aa}={\sum }_{i,a}{\rho }_{(i,a),(i,a)}=\mathrm{tr}\rho =1$ by [F1] and [F7]; hence ${\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ is a density operator by [F1].

For every endomorphism $B$ of $H_{\mathrm{o}\mathrm{u}\mathrm{t}}$, define $I_{\mathrm{i}\mathrm{n}}\otimes B$ on the product basis by $(I_{\mathrm{i}\mathrm{n}}\otimes B)(e_i\otimes f_b):=e_i\otimes B{f}_b$; [F3]–[F6] give its unique linear extension. For arbitrary $x,y$, expansion in the chosen bases shows this map sends $x\otimes y$ to $x\otimes By$, so it is independent of those bases. Finite matrix multiplication gives $\mathrm{tr}({\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}B)={\sum }_{i,a,b}{\rho }_{(i,a),(i,b)}{B}_{ba}=\mathrm{tr}(\rho (I_{\mathrm{i}\mathrm{n}}\otimes B))$. The right side is therefore basis-independent. If a second choice produced ${\tilde{\rho }}_{\mathrm{o}\mathrm{u}\mathrm{t}}$, then $\mathrm{tr}(({\tilde{\rho }}_{\mathrm{o}\mathrm{u}\mathrm{t}}-{\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}})B)=0$ for all $B$; taking $B$ to be the matrix unit with sole nonzero entry in row $b$, column $a$ gives $\mathrm{tr}(({\tilde{\rho }}_{\mathrm{o}\mathrm{u}\mathrm{t}}-{\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}})B)=({\tilde{\rho }}_{\mathrm{o}\mathrm{u}\mathrm{t}}-{\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}{)}_{ab}$, so every entry of the difference is zero. Thus the endomorphism is basis-independent.

By [F9], ${\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}}$ has an orthonormal eigenbasis with nonnegative real eigenvalues; [F7] and step 1.2 show their sum is $\mathrm{tr}({\rho }_{\mathrm{o}\mathrm{u}\mathrm{t}})=1$, so each lies in $[0,1]$. The finite spectral sum in [F8] is therefore real, with zero eigenvalues contributing $0$ by convention. Its eigenvalue multiset is intrinsic to the endomorphism, and step 1.3 proves that endomorphism is basis-independent; multiplying by the fixed real $k_B$ preserves finiteness and basis-independence.