Entropy inequalities, recovery boundaries and thermal entanglement

All Hilbert spaces in this item are nonzero and finite-dimensional over $\mathbb{C}$. A density operator $\rho$ on $H$ is a Hermitian positive semidefinite endomorphism with $\mathrm{Tr}\rho =1$. Its dimensionless von Neumann entropy, in nats, is

$S(\rho ):=-\mathrm{Tr}(\rho \log \rho )=-{\sum }_{j:{\lambda }_j>0}{\lambda }_j\log {\lambda }_j,$

where ${\lambda }_j$ are the eigenvalues of $\rho$, the logarithm is natural, and $0\log 0:=0$. For any positive semidefinite endomorphisms $X,Y$ on one finite-dimensional $H$, let $\mathrm{supp}X:=(\ker X{)}^{\perp }$. Define

$D(X\mathrm{\parallel }Y):=\mathrm{Tr}(X\log X)-\mathrm{Tr}(X\log Y)$

when $\mathrm{supp}X\subseteq \mathrm{supp}Y$, taking each logarithm spectrally on its positive support and setting zero-eigenvalue contributions to zero; set $D(X\mathrm{\parallel }Y):=+\mathrm{\infty }$ otherwise. For density operators this is the finite-dimensional quantum relative entropy in nats.

For a density operator ${\rho }_{AB}$ on the ordered product $H_A\otimes H_B$, define ${\rho }_A:={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B:={\mathrm{Tr}}_A{\rho }_{AB}$, where in orthonormal product bases $({\mathrm{Tr}}_{B}R{)}_{i{i}^{\mathrm{\prime }}}={\sum }_j{R}_{(i,j),(i^{\mathrm{\prime }},j)}$ and the first factor is retained. Define

$I(A:B{)}_{\rho }:=S({\rho }_A)+S({\rho }_B)-S({\rho }_{AB})\mathrm{.}$

For a density operator ${\rho }_{ABC}$ on $H_A\otimes H_B\otimes H_C$, define

${\rho }_{AB}:={\mathrm{Tr}}_C{\rho }_{ABC},{\rho }_{BC}:={\mathrm{Tr}}_A{\rho }_{ABC},{\rho }_B:={\mathrm{Tr}}_A({\mathrm{Tr}}_C{\rho }_{ABC}),$

where each partial trace contracts the indicated tensor factor and retains the other factors in their displayed order.

$I(A:C\mid B{)}_{\rho }:=S({\rho }_{AB})+S({\rho }_{BC})-S({\rho }_B)-S({\rho }_{ABC}),$

using the partial traces in the displayed factor order. For any fixed positive conversion factor $k_B$ with units J/K, the corresponding physical entropy is $S_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(\rho ):=k_{B}S(\rho )$; no numerical value of $k_B$ is specified here.

Density operator

Finite-dimensional von Neumann entropy

Finite-dimensional quantum relative entropy

Partial trace over a finite tensor factor

Partial trace over the first tensor factor

Partial traces of a finite bipartite state are states

Let $H_A,H_B,H_C$ be nonzero finite-dimensional complex inner-product spaces and let ${\rho }_{ABC}$ be a density operator on $H_A\otimes H_B\otimes H_C$. With dimensionless entropy $S$ and conditional mutual information $I(A:C\mid B)$ defined by the conventions below,

$I(A:C\mid B{)}_{\rho }=S({\rho }_{AB})+S({\rho }_{BC})-S({\rho }_B)-S({\rho }_{ABC})\ge 0.$

Also let $d,e\ge 1$ and let $\mathrm{\Phi }:M_d(\mathbb{C})\to M_e(\mathbb{C})$ be a finite completely positive trace-preserving map. For all positive semidefinite $X,Y\in M_d(\mathbb{C})$,

$D(\mathrm{\Phi }(X)\mathrm{\parallel }\mathrm{\Phi }(Y))\le D(X\mathrm{\parallel }Y),$

where $D$ is the support-sensitive finite-dimensional quantum relative entropy: it equals $\mathrm{Tr}(X\log X)-\mathrm{Tr}(X\log Y)$ when $\mathrm{supp}X\subseteq \mathrm{supp}Y$, and $+\mathrm{\infty }$ otherwise. These are finite-dimensional mathematical statements. The channel inequality is one-sided and asserts no equality condition, recovery map, or infinite-dimensional extension.

Dimensionless entropy, mutual and conditional information

Finite completely positive trace-preserving maps

Finite-dimensional quantum relative entropy

Strong subadditivity of finite-dimensional quantum entropy

Partial-trace data processing for finite quantum relative entropy

Complete finite Choi-Kraus equivalence

diagram failed to render:
\begin{tikzcd}
M_d(\mathbb C)_+ \arrow[r, "X\mapsto VXV^*"] \arrow[dr, "\Phi"'] & \operatorname{End}(\mathbb C^e\otimes\mathcal E)_+ \arrow[d, "\operatorname{Tr}_{\mathcal E}"] \\
& M_e(\mathbb C)_+
\end{tikzcd}

Here $M_d(\mathbb{C}{)}_{+}$ denotes positive semidefinite matrices, $\mathcal{E}$ is the finite environment constructed in step 1.2, and $V$ is its isometry. The triangle commutes because step 1.2 computes ${\mathrm{Tr}}_{\mathcal{E}}(VX{V}^{\ast })=\mathrm{\Phi }(X)$ entrywise. The proof uses that computation and the stated partial-trace inequality, not the diagram.

Assume the Axiom of Choice. Let ${\mathcal{H}}_A,{\mathcal{H}}_B$ be nonzero finite-dimensional complex inner-product spaces. Let $H_A=H_A^{\ast }$ act on ${\mathcal{H}}_A$, let $H_B=H_B^{\ast }$ act on ${\mathcal{H}}_B$, and let $H_{\mathrm{\partial }}=H_{\mathrm{\partial }}^{\ast }$ act on ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$. Set

$H:=H_A\otimes I_B+I_A\otimes H_B+H_{\mathrm{\partial }}\mathrm{.}$

For $\beta >0$, define $Z_{\beta }:=\mathrm{Tr}(e^{-\beta H})$ and the finite Gibbs density ${\rho }_{\beta }:=e^{-\beta H}\mathrm{/}{Z}_{\beta }$. Then

$I(A:B{)}_{{\rho }_{\beta }}\le 2\beta \mathrm{\parallel }{H}_{\mathrm{\partial }}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}},$

where $I(A:B)$ uses dimensionless entropy in nats and $\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}:={\sup }_{\mathrm{\parallel }v\mathrm{\parallel }=1}\mathrm{\parallel }Kv\mathrm{\parallel }$ is the operator norm.

For the optional lattice-family conclusion, fix $m\ge 1$. For each member $s$ of a specified family, take a finite site set ${\mathrm{\Lambda }}_s\subseteq {\mathbb{Z}}^m$, a partition ${\mathrm{\Lambda }}_s=A_s\bigsqcup B_s$, and nonzero finite-dimensional complex inner-product spaces ${\mathcal{K}}_{s,x}$ for $x\in {\mathrm{\Lambda }}_s$. Use the lexicographic order on ${\mathbb{Z}}^m$ in ${\mathcal{H}}_{s,X}:={⨂}_{x\in X}{\mathcal{K}}_{s,x}$, with the empty tensor product defined as $\mathbb{C}$, and identify ${\mathcal{H}}_{s,{\mathrm{\Lambda }}_s}$ with ${\mathcal{H}}_{s,A_s}\otimes {\mathcal{H}}_{s,B_s}$ in the displayed factor order. Let $J_s$ be a finite set of interaction labels with nonempty supports $X_{s,j}\subseteq {\mathrm{\Lambda }}_s$ and self-adjoint terms $h_{s,j}$ on ${\mathcal{H}}_{s,X_{s,j}}$. For $Y\in \{{\mathrm{\Lambda }}_s,A_s,B_s\}$ containing $X_{s,j}$, write ${\widehat{h}}_{s,j}^Y$ for the extension of $h_{s,j}$ to ${\mathcal{H}}_{s,Y}$ by the identity on every factor in $Y\setminus X_{s,j}$. The crossing-label set is

$J_{\mathrm{\partial },s}:=\{j\in J_s:X_{s,j}\cap A_s\mathrm{\ne }\mathrm{\varnothing }\text{ and }{X}_{s,j}\cap B_s\mathrm{\ne }\mathrm{\varnothing }\},\mathrm{\mid }\mathrm{\partial }{A}_s\mathrm{\mid }:=\mathrm{\mid }{J}_{\mathrm{\partial },s}\mathrm{\mid }\mathrm{.}$

Let $J_{A,s}:=\{j\in J_s:X_{s,j}\subseteq A_s\}$ and $J_{B,s}:=\{j\in J_s:X_{s,j}\subseteq B_s\}$. Set $H_{A_s}:={\sum }_{j\in J_{A,s}}{\widehat{h}}_{s,j}^{A_s}$, $H_{B_s}:={\sum }_{j\in J_{B,s}}{\widehat{h}}_{s,j}^{B_s}$, and $H_{\mathrm{\partial },s}:={\sum }_{j\in J_{\mathrm{\partial },s}}{\widehat{h}}_{s,j}^{{\mathrm{\Lambda }}_s}$. Under the stated identification these give the displayed Hamiltonian decomposition. For each $s$, assume ${\beta }_s>0$ and define ${\rho }_{{\beta }_s}$ by the Gibbs formula above. If one specified $C\ge 0$ satisfies $\mathrm{\parallel }{H}_{\mathrm{\partial },s}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le C\mathrm{\mid }\mathrm{\partial }{A}_s\mathrm{\mid }$ for every family member, the bound gives $I(A_s:B_s{)}_{{\rho }_{{\beta }_s}}\le 2{\beta }_{s}C\mathrm{\mid }\mathrm{\partial }{A}_s\mathrm{\mid }$ memberwise. No finite-range condition or additional geometric scaling assumption is used. When $H_{\mathrm{\partial },s}$ and each interaction term are measured in joules, ${\beta }_s$ has units J${}^{-1}$ and $C$ has units J, so the bound is dimensionless.

Dimensionless entropy, mutual and conditional information

Finite Gibbs free-energy and entropy optimization

The Axiom of Choice