On the continuity of quantum correlation quantifiers

Abstract

We show an equivalence relation between different types of continuity of the generalized discord function (GDF) that leads to the continuity of the generalized quantum discord (GQD) in the finite-dimensional case. We extend the definition of the GQD to the case where the GDF is optimized over the set of all states with zero quantum discord and prove its continuity by showing that this set is topologically compact. However, for an unmeasured subsystem with infinite dimension, we find that this set is no longer compact while the set of locally measured states is shown to maintain this property in the space of Hilbert-Schmidt (HS) operators. This allows us to prove the continuity of the GQD when the GDF is jointly continuous in the infinite case. As an application, we obtain that the geometric discord is continuous (HS topology) and has the zero set given by the zero quantum discord set in the infinite-dimensional case as a consequence of our previous results.

Appendices

Appendix

Lemma 1

Let $C\subset \mathbb {R}^{p}$ be a compact convex set. The function $F:C\rightarrow \mathbb {R}$ is continuous if, and only if, there exists a continuous function $g:[0,1]\rightarrow \mathbb {R}$, with $g(0)=0$, such that $|F(x)-F((1-r)x+ry)|\le g(r)$ for all $x,y\in C$ with $r\in [0,1]$.

Proof

First, we suppose that F is continuous. Let $g:[0,1]\rightarrow \mathbb {R}$ be defined as $g(r)\equiv \sup _{(x,y)\in C\times C}|F(x)-F((1-r)x+ry)|$. Set $H(x,y,r)\equiv x$, $J(x,y,r)\equiv (1-r)x+ry$ and $G(x,y,r)=|F\circ H(x,y,r)-F\circ J(x,y,r)|$. Then $G=|.|\circ (F\circ H-F\circ J)$. Using, as a metric on $C\times C\times [0,1]$, the function $d((x,y,r),(x',y',r'))\equiv \Vert x-x'\Vert +\Vert y-y'\Vert +|r-r'|$, it is easy to see that H and J are continuous, implying that G is continuous on $C\times C\times [0,1]$. The set $C\times C$ is compact in the product topology induced by the metric $d((x,y),(x',y'))\equiv \Vert x-x'\Vert +\Vert y-y'\Vert$. Then $g(r)=\sup _{(x,y)\in C\times C}G(x,y,r)$ is a continuous function. Thus for all $x,y\in C$ and $r\in [0,1]$,

$$\begin{aligned} |F(x)-F((1-r)x+ry)|\le g(r). \end{aligned}$$

(23)

Now, suppose that Eq. (23) is valid with a continuous function $g:[0,1]\rightarrow \mathbb {R}$ such that $g(0)=0$. First, we prove the continuity of F on the relative interior of C, relint(C), and then prove the continuity on the relative boundary of C: relbd(C). Here, we define a metric on ${{\,\mathrm{Aff}\,}}(C)\subseteq \mathbb {R}^{n}$ by the function $d(x,y)\equiv \Vert x-y\Vert$, where $\Vert \cdot \Vert$ is the norm on $\mathbb {R}^{n}$. As the relative interior of any nonempty convex set is nonempty, $\exists x\in$ relint(C). Then, $\exists \delta '>0$ such that $B_{\delta '}(x)\subseteq C$, where $B_{\delta '}(x)\equiv \{w\in {{\,\mathrm{Aff}\,}}(C):\Vert w-x\Vert <\delta '\}$. As g is continuous and $g(0)=0$, for any $\epsilon >0$, there is $r\in (0,1)$ such that $g(r)<\epsilon$. Defining $\delta \equiv r\delta '<\delta '$, for any $y\in B_{\delta }(x)\in C$ we have $z\equiv (1-r^{-1})x+r^{-1}y\in {{\,\mathrm{Aff}\,}}(C)$ and $\Vert z-x\Vert =r^{-1}\Vert x-y\Vert <\delta '$, implying that $z\in B_{\delta '}(x)$. This means that for any $y\in B_{\delta }(x)$, we can find $z\in C$ such that $y=(1-r)x+rz$. By the continuity of g and $g(0)=0$, for any $\epsilon >0$ there is $r\in (0,1)$ such that $g(r)<\epsilon$. If $x\in$ relint(C), by the previous considerations, there is $\delta >0$ such that $B_{\delta }(x)\subset C$ and for any $y\in B_{\delta }(x)$ we can find $z\in C$ where $y=(1-r)x+rz$. Then,

$$\begin{aligned} \forall \epsilon>0,x\in {{\,\mathrm{relint}\,}}(C),\exists \delta >0: y\in B_{\delta }(x)\subset C\Rightarrow |F(x)-F(y)|\le g(r)<\epsilon . \end{aligned}$$

(24)

Thus F is continuous on ${{\,\mathrm{relint}\,}}(C)$. Now, we prove the continuity of F on relbd(C). For any $\epsilon \in (0,1)$ (we can put an upper bound on $\epsilon$ w.l.o.g), there exists $0<r<\epsilon /3$ such that $g(r)<\epsilon /3$. With such r, for any $x\in$ relbd(C), we can choose an arbitrary $z\in$ relint(C) and define $z_{x}\equiv (1-r)x+rz\in$ intrel(C). As F is continuous on ${{\,\mathrm{relint}\,}}(C)$,

$$\begin{aligned} \exists \delta ''>0:\Vert z_{x}-w\Vert<\delta '',w\in C\Rightarrow |F(z_{x})-F(w)|< \epsilon /3. \end{aligned}$$

(25)

For any $y\in B_{\delta ''}(x)\cap C$ and defining $z_{y}\equiv (1-r)y+rz\in C$, we have

$$\begin{aligned} \Vert z_{x}-z_{y}\Vert =(1-r)\Vert x-y\Vert<\delta ''\Rightarrow |F(z_{x})-F(z_{y})|<\epsilon /3. \end{aligned}$$

(26)

Using Proposition (26),

$$\begin{aligned} |F(x)-F(y)|&=|F(x)-F(z_{x})+F(z_{x})-F(z_{y})+F(z_{y})-F(y)| \nonumber \\&\le |F(x)-F((1-r)x+rz)|+|F(y)\nonumber \\&\quad -\,F((1-r)y+rz)|+|F(z_{x})-F(z_{y})|\nonumber \\&< 2g(r)+\epsilon /3<\epsilon . \end{aligned}$$

(27)

Then

$$\begin{aligned} \forall \epsilon>0,x\in {{\,\mathrm{relbd}\,}}(C), \exists \delta ''>0:y\in B_{\delta ''}(x)\cap C\Rightarrow |F(x)-F(y)|<\epsilon , \end{aligned}$$

(28)

which proves that F is continuous on ${{\,\mathrm{relbd}\,}}(C)\cup {{\,\mathrm{relint}\,}}(C)=C$.$\square$

Appendix

Lemma 2

The set $\mathcal {P}^{B}\subset H^{\oplus n}$ is compact.

Proof

Let’s define the set $\tilde{\mathcal {P}}^{B}$. The operator $\tilde{X}^{B}\in H_{B}^{\oplus n}$, where $H_{B}$ is the space of hermitian operators on $\mathcal {H}_{B}$, is an element of $\tilde{\mathcal {P}}^{B}$ if, and only if, Eq. (29) is satisfied for any k, l:

$$\begin{aligned} F_{k}(\tilde{X}^{B})&\equiv (\tilde{\Pi }_{k}(\tilde{X}^{B}))^2-\tilde{\Pi }_{k}(\tilde{X}^{B})=0,\quad G_{k,l}(\tilde{X}^{B})\equiv {{\,\mathrm{Tr}\,}}(\tilde{\Pi }_{k}(\tilde{X}^{B})\tilde{\Pi }_{l}(\tilde{X}^{B}))-\delta _{k,l}=0,\nonumber \\ J(\tilde{X}^{B})&\equiv \sum _{k}\tilde{\Pi }_{k}(\tilde{X}^{B})-\mathbb {1}^{B}=0, \end{aligned}$$

(29)

where $\tilde{\Pi _{k}}$ projects $\tilde{X}^{B}=\bigoplus _{l}\tilde{X}^{B}_{l}$ on its kth component $\tilde{\Pi _{k}}(\tilde{X}^{B})\equiv \tilde{X}^{B}_{k}$. Eq. (29) implies that the elements of $\tilde{\mathcal {P}}^{B}$ have the form $\tilde{P}^{B}=\bigoplus _{l} \tilde{P}^{B}_{l}$ such that $\{\tilde{P}^{B}_{l}:l\in \{1,\ldots ,n\}\}$ is a set of rank one orthogonal projections with $\sum _{k}\tilde{P}^{B}_{l}=\mathbb {1}^{B}$. As $\Vert \tilde{P}^{B}_{l}\Vert _{2}=1$, we have $\Vert \tilde{P}^{B}\Vert =\sum _{l}\Vert \tilde{P}^{B}_{l}\Vert _{2}=n$, implying that $\tilde{\mathcal {P}}^{B}$ is a bounded set. Suppose $(\tilde{P}^{B,m})_{m\in \mathbb {N}}$ is a convergent sequence in $H_{B}^{\oplus n}$ such that $\tilde{P}^{B,m}\in \tilde{\mathcal {P}}^{B}$ for any m. If $\lim _{m\rightarrow \infty }\tilde{P}^{B,m}=\tilde{X}^{B}$, by the continuity of the functions $F_{k}$, $G_{k,l}$ and J, we have $F_{k}(\tilde{X}^{B})=0$, $G_{k,l}(\tilde{X}^{B})=0$ and $J(\tilde{X}^{B})=0$ by Eq. (29), which implies that $\tilde{X}^{B}\in \tilde{\mathcal {P}}^{B}$. It means that $\tilde{\mathcal {P}}^{B}$ is a closed set and, as it is also bounded, we conclude that $\tilde{\mathcal {P}}^{B}$ is compact. Now, consider the function $L:H_{B}^{\oplus n}\rightarrow H^{\oplus n}$ given by $L(\tilde{X}^{B})\equiv X^{B}=\bigoplus _{l}\mathbb {1}^{A}\otimes \tilde{X}^{B}_{l}$. As L is a linear function, it is continuous, which implies that $L(\tilde{\mathcal {P}}^{B})=\mathcal {P}^{B}$ is a compact set.$\square$

Appendix

Lemma 3

The function $\tilde{\mathcal {K}}_{k}:H\times H^{\oplus n}\rightarrow \mathbb {R}$ defined as $\tilde{\mathcal {K}}_{k}(X,Y)\equiv \Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X$ $\Pi _{k}(Y))\Vert _{1}$ $\times \Arrowvert {{\,\mathrm{Tr}\,}}_{B}(X)-{{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))/\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))\Vert _{1}\Arrowvert ^{2}_{2}$ if $\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)$ $X\Pi _{k}(Y))\Vert _{1}\ne 0$ and $\tilde{\mathcal {K}}_{k}(X,Y)=0$ if $\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))\Vert _{1}=0$, is jointly continuous in the product topology of $H\times H^{\oplus n}$.

Proof

Defining $f_{B}(X,Y)\equiv {{\,\mathrm{Tr}\,}}_{B}(YXY)$ and $\Pi ^{\beta }_{k}(X,Y)\equiv (X,\Pi _{k}(Y))$, we have $f_{B}\circ \Pi ^{\beta }_{k}(X,$ Y) $={{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))$. As $f_{B}$ is jointly continuous and $\Pi _{k}$ is linear, then $f_{B}\circ \Pi ^{\beta }_{k}$ is jointly continuous. For any point $(X,Y)\in H\times H^{\oplus n}$ such that $\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))\Vert _{1}\ne 0$, $1/\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))\Vert _{1}$ is jointly continuous. As $\tilde{\mathcal {K}}_{k}$ is composition of jointly continuous functions, $\tilde{\mathcal {K}}_{k}$ also has this property for any point $(X,Y)\in H\times H^{\oplus n}$ such that $\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))\Vert _{1}\ne 0$.

Now, suppose that (X, Y) is a point such that $\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y)X\Pi _{k}(Y))\Vert _{1}=0$. For arbitrary $\delta _{x}>0$ and $\delta _{y}>0$, suppose that $\Vert X-X'\Vert _{1}<\delta _{x}$ and $\Vert Y-Y'\Vert _{1}<\delta _{y}$, where $X,X'\in H$ and $Y,Y'\in H^{\oplus n}$. Define $\Delta X\equiv X'-X$ and $\Delta Y\equiv Y'-Y$. By $|\tilde{\mathcal {K}}_{k}(X+\Delta X,Y+\Delta Y)-\tilde{\mathcal {K}}_{k}(X,Y)|=|\tilde{\mathcal {K}}_{k}(X+\Delta X,Y+\Delta Y)|$, we have

$$\begin{aligned}&|\tilde{\mathcal {K}}_{k}(X+\Delta X,Y+\Delta Y)|\nonumber \\&\quad ={{\,\mathrm{Tr}\,}}_{A}\left( |{{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y+\Delta Y)(X+\Delta X)\Pi _{k}(Y+\Delta Y))|\right) \Big \Vert {{\,\mathrm{Tr}\,}}_{B}(X\nonumber \\&\qquad +\,\Delta X)-\frac{{{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y+\Delta Y)(X+\Delta X)\Pi _{k}(Y+\Delta Y))}{{{\,\mathrm{Tr}\,}}_{A}\left( |{{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y+\Delta Y)(X+\Delta X)\Pi _{k}(Y+\Delta Y))|\right) }\Big \Vert _{2}^{2} \end{aligned}$$

(30)

By the equivalence of norms, there exists a constant c such that, for any $Z\in H_{A}$, $\Vert Z\Vert _{2}\le c\Vert Z\Vert _{1}$. Defining $Z^{B}\equiv {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y+\Delta Y)(X+\Delta X)\Pi _{k}(Y+\Delta Y))$, we have

$$\begin{aligned} \frac{\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y+\Delta Y)(X+\Delta X)\Pi _{k}(Y+\Delta Y))\Vert _{2}}{{{\,\mathrm{Tr}\,}}_{A}\left( |{{\,\mathrm{Tr}\,}}_{B}(\Pi _{k}(Y+\Delta Y)(X+\Delta X)\Pi _{k}(Y+\Delta Y))|\right) }=\frac{\Vert Z^{B}\Vert _{2}}{\Vert Z^{B}\Vert _{1}}\le c. \end{aligned}$$

(31)

Equations (30) and (31) imply that

$$\begin{aligned} |\tilde{\mathcal {K}}_{k}(X+\Delta X,Y+\Delta Y)|&\le \left( \Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi ^{B}(Y)X\Pi ^{B}(Y))\Vert _{1}+\ldots \right) (\Vert {{\,\mathrm{Tr}\,}}_{B}(X+\Delta X)\Vert _{2}+c)^{2}. \end{aligned}$$

(32)

As $Tr_{B}$ is a bounded linear operator, there exists $M>0$ such that $\Vert {{\,\mathrm{Tr}\,}}_{B}(XYZ)\Vert _{1}\le M\Vert XYZ\Vert _{1}\le M\Vert X\Vert _{1}\Vert Y\Vert _{1}\Vert Z\Vert _{1}$. By the same argument, there exists $N>0$ such that $\Vert \Pi _{k}(Y)\Vert _{1}\le N\Vert Y\Vert _{1}$. In Inequality (32), the “$\ldots$” represents several terms with the form $\Vert {{\,\mathrm{Tr}\,}}_{B}(X'Y'Z')\Vert _{1}$, such that at least one variable is $\Delta X$ or $\Pi _{k}(\Delta Y)$. By these considerations and knowing that $\Vert {{\,\mathrm{Tr}\,}}_{B}(\Pi ^{B}(Y)X\Pi ^{B}(Y))\Vert _{1}=0$, there exists a constant R such that

$$\begin{aligned} |\tilde{\mathcal {K}}_{k}(X+\Delta X,Y+\Delta Y)|&\le R\max \{\delta _{x},\delta _{y},\delta _{x}\delta _{y},\delta _{x}\delta _{y}^2\}(\Vert {{\,\mathrm{Tr}\,}}_{B}(X+\Delta X)\Vert _{2}+c)^{2}\nonumber \\&\le R\max \{\delta _{x},\delta _{y},\delta _{x}\delta _{y},\delta _{x}\delta _{y}^2\}(\Vert {{\,\mathrm{Tr}\,}}_{B}(X)\Vert _{2}+S\delta _{x}+c)^2, \end{aligned}$$

(33)

where $S>0$ is a constant. Clearly, Inequality (33) implies that $\tilde{\mathcal {K}}_{k}$ is jointly continuous.$\square$

Lemma 3 implies that:

Corollary 3

The function $\tilde{\mathcal {K}}:D(\mathcal {H})\times \mathcal {P}^{B}\rightarrow \mathbb {R}_{\ge 0}$ defined as $\tilde{\mathcal {K}}(\rho ,P^{B})\equiv \sum _{k}\tilde{\mathcal {K}}_{k}(\rho ,P^{B})$, where $D(\mathcal {H})\times \mathcal {P}^{B}\subset H\times H^{\oplus n}$, is jointly continuous.