Finite tensor density, observable and measurement language — Examples

Take $q=2$, $e_1=0$, $e_2=ϵ>0$, and choose finite $\beta >0$ with $\beta ϵ=\log \sqrt{2}$. By [F1], $p_2=\frac{e^{-\beta ϵ}}{1+e^{-\beta ϵ}}=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1.$ This number is irrational: if $\sqrt{2}$ were rational, writing it as a reduced fraction $a\mathrm{/}b$ gives $a^2=2b^2$, so $a$ is even, then $b$ is even, contradicting reduction. Hence $p_2$ is irrational.

For any integer $N\ge 1$, an $N$-site label has energy $mϵ$ for some integer $m$, whereas the canonical mean is $NE=Nϵp_2$. The zero-width band condition forces $mϵ=Nϵp_2$, hence $m=N{p}_2$. Since $p_2$ is irrational, $N{p}_2$ is not an integer. Thus no label satisfies the condition, ${\mathrm{\Gamma }}_N=\mathrm{\varnothing }$, and ${\mathrm{\Pi }}_N\mathrm{/}\mathrm{\mid }{\mathrm{\Gamma }}_N\mathrm{\mid }$ is undefined. This shows why a nonemptiness or positive width hypothesis cannot be omitted.

On $V={\mathbb{C}}^2$, the supplier [F3] identifies the stated canonical one-site density as $\gamma =I\mathrm{/}2$ and its $N$-site density as ${\gamma }_N=(I\mathrm{/}2{)}^{\otimes N}$. For each $N\ge 1$, set ${\gamma }_N=(I\mathrm{/}2{)}^{\otimes N}$ and ${\tau }_N=\mathrm{\mid }1⟩⟨1\mathrm{\mid }\otimes (I\mathrm{/}2{)}^{\otimes (N-1)}$. Both are positive of trace one, so are finite tensor densities. Their positive eigenvalues are respectively $2^{-N}$ with multiplicity $2^N$ and $2^{-(N-1)}$ with multiplicity $2^{N-1}$. By [F1], $s_N({\gamma }_N)=N\log 2$ and $s_N({\tau }_N)=(N-1)\log 2$; therefore $\mathrm{\mid }{s}_N({\tau }_N)-s_N({\gamma }_N)\mathrm{\mid }\mathrm{/}N=(\log 2)\mathrm{/}N\to 0$.

By [F2], the one-site reductions are $\gamma$ and $\mathrm{\mid }1⟩⟨1\mathrm{\mid }$. Their difference in the ordered basis is $\mathrm{diag}(1\mathrm{/}2,-1\mathrm{/}2)$, whose eigenvalues have absolute sum one. Thus $\mathrm{\parallel }{\tau }_{N,1}-\gamma {\mathrm{\parallel }}_1=1$ for every $N$. Entropy-density agreement alone therefore does not imply local trace-norm convergence.

For $N\ge 2$, index each binary factor's ordered basis by $\{0,1\}$; a string $x=(x_1,\dots ,x_N)$ has even parity when ${\sum }_j{x}_j$ is even. Let ${\pi }_N$ be the diagonal matrix with entry $2^{-(N-1)}$ on even strings and zero on odd strings, and let ${\gamma }_N=I\mathrm{/}{2}^N$. There are $2^{N-1}$ even strings: the first $N-1$ bits can be arbitrary and the last bit is uniquely determined. Thus both matrices are nonnegative of trace one. Fix $0\le k<N$ coordinates and their values. Among the $N-k\ge 1$ unassigned coordinates, exactly half of the $2^{N-k}$ completions have the required parity, namely $2^{N-k-1}$. Summing by [F1], the marginal diagonal entry of each fixed $k$-bit string is $2^{N-k-1}\mathrm{/}{2}^{N-1}=2^{-k}$; all off-diagonal entries vanish. Thus $({\pi }_N{)}_k=I\mathrm{/}{2}^k=({\gamma }_N{)}_k$ for every proper block.

On an even string the diagonal difference ${\pi }_N-{\gamma }_N$ is $2^{-(N-1)}-2^{-N}=2^{-N}$; on an odd string it is $-2^{-N}$. There are $2^N$ strings in total, so the absolute eigenvalues sum to $2^N{2}^{-N}=1$. Therefore $\mathrm{\parallel }{\pi }_N-{\gamma }_N{\mathrm{\parallel }}_1=1$ for every $N$.

Let $P_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ be the diagonal matrix with entry one on even strings and zero on odd strings. Both $P_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ and $I-P_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}$ are nonnegative and sum to $I$, so they form a two-outcome POVM by [F2]. Their first-effect trace pairings are $\mathrm{Tr}({\pi }_N{P}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}})=1$ and $\mathrm{Tr}({\gamma }_N{P}_{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}})=1\mathrm{/}2$, a gap of $1\mathrm{/}2$. Hence equality of all proper local marginals does not entail global trace-norm convergence or equality of global POVM trace pairings. This is a mathematical separation; no physical probability interpretation is adopted.