Finite-regulator entropy mathematics

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.

The matrix ${\rho }_N=I_{2^N}\mathrm{/}{2}^N$ is non-negative and has trace $2^N\mathrm{/}{2}^N=1$, so it is a density operator by [F2] and [F9]. Since the input factor is one-dimensional, the defining partial trace in [F1] has one summand and gives ${\rho }_{\mathrm{o}\mathrm{u}\mathrm{t},N}=I_{2^N}\mathrm{/}{2}^N$.

The spectrum of ${\rho }_{\mathrm{o}\mathrm{u}\mathrm{t},N}$ consists of $2^N$ copies of $2^{-N}$. For $N=1$, $\exp (N\log 2)=2$ by [F3]; if $\exp (N\log 2)=2^N$, [F6] gives $\exp ((N+1)\log 2)=\exp (N\log 2)\exp (\log 2)=2^{N+1}$. Induction yields $\exp (N\log 2)=2^N$ for every $N\ge 1$. By [F4], $\exp (-N\log 2)=1\mathrm{/}{2}^N=2^{-N}$, so [F3] and injectivity in [F7] give $\log (2^{-N})=-N\log 2$.

Substitution into [F1] gives $S_{\mathrm{o}\mathrm{u}\mathrm{t}}=-k_B{2}^N{2}^{-N}\log (2^{-N})=k_{B}N\log 2$. Also $\log 2>0$, because [F5] gives $\exp (0)=1<2=\exp (\log 2)$ and [F7] is strictly increasing.

Let $c=k_B\log 2>0$. For any real $M$, [F10] makes $\mathbb{R}$ a complete ordered field, so [F8] applied to $M\mathrm{/}c$ supplies a natural number $N_0\ge 1$ with $N_0>M\mathrm{/}c$. For every integer $N\ge N_0$, step 2.1 gives $S_{\mathrm{o}\mathrm{u}\mathrm{t}}=cN\ge c{N}_0>M$, which is exactly convergence to $+\mathrm{\infty }$.

Fix any real $a$ and any $\epsilon >0$. By [A1], choose $N_0$ so that $\mathrm{\mid }{x}_N-r_N-L\mathrm{\mid }<\epsilon$ for every $N\ge N_0$.

For every $N\ge N_0$, $\mathrm{\mid }{x}_N-r_N^{\mathrm{\prime }}-(L+a)\mathrm{\mid }=\mathrm{\mid }{x}_N-(r_N-a)-(L+a)\mathrm{\mid }=\mathrm{\mid }{x}_N-r_N-L\mathrm{\mid }<\epsilon$. Since $a$ and $\epsilon$ were arbitrary, the claimed limit is $L+a$ for every $a\in \mathbb{R}$.