Let \(\mathcal{H}\) be a finite-dimensional Hilbert space, \(\mathfrak{T}(\mathcal{H})\) the Banach space of operators on \(\mathcal{H}\) equipped with trace norm, \(\mathfrak{S}=\mathfrak{S}(\mathcal{H})\) the compact convex set of quantum states equipped with trace-norm distance. An ensemble  is a probability measure \(\pi(dS)\) on the set \(\mathfrak{S}\), cf. [2], and the set of all such measures is denoted \(\mathcal{P}(\mathfrak{S})\).

Let \(f\) be a continuous concave function on the compact convex set \(\mathfrak{S}\). The proof of theorem below uses very little of other special properties of \(f\) and \(\mathfrak{S}\). Consider the functional

$$F(\pi)=\int\limits_{\mathfrak{S}}f(S)\pi(dS),$$
(1)

on \(\mathcal{P}(\mathfrak{S})\). From the definition, it is a continuous affine functional. We are interested in minimization of this functional on the closed convex subset \(\mathcal{P}_{\overline{S}}\) of probability measures \(\pi\) with the fixed given barycenter

$$\overline{S}=\int\limits_{\mathfrak{S}}S\,\pi(dS).$$
(2)

Under mild additional conditions the functional \(F(\pi)\) attains its minimum \(\mathcal{F}(\overline{S})\) on the compact set \(\mathcal{P}_{\overline{S}}\). The resulting function \(\mathcal{F}(S),S\in\mathfrak{S},\) is convex, in fact it is equal to the convex closure of \(f(S)\) i.e. the greatest lower semicontinuous convex function majorized by \(f(S)\) [7]. Concavity of \(f\) implies then that we can choose the minimizing measure \(\pi\) to be supported by the pure states since we can always make spectral decompositions of all the density operators \(S\) into pure states without changing the barycenter and without increasing the value \(F(\pi)\).

This problem is relevant to a number of issues in quantum information theory.

1. Computation of the \(\chi\)-capacity [2] of a quantum channel \(\Phi\) defined as

$$C_{\chi}(\Phi)=\sup_{\pi}\left[H\left(\Phi\left[\int\limits_{\mathfrak{S}}S\,\pi(dS)\right]\right)-\int\limits_{\mathfrak{S}}H(\Phi[S])\pi(dS)\right]$$
$${}=\sup_{\overline{S}}\left[H\left(\Phi\left[\overline{S}\right]\right)-\inf_{\pi\in\mathcal{P}_{\overline{S}}}\int\limits_{\mathfrak{S}}H(\Phi[S])\pi(dS)\right].$$

Here the second term in the squared brackets is the convex closure of the channel output entropy, with

$$f(S)=H(\Phi[S])=\textrm{Tr\,}K(S)S,$$
$$K(S)=-\,\Phi^{\ast}\left[\log\Phi[S]\right].$$

2. A similar case is the classical capacity of quantum-classical channel (observable) given by the map \(M:S\rightarrow p_{S}(z)=\textrm{Tr\,}Sm(z),\) where \(m(z)\) is a uniformly bounded positive-operator-valued function of \(z\in Z,\) such that \(\int m(z)dz=I\) (the unit operator). Here \((Z,dz)\) is a measure space. Then \(p_{S}(z)\) is the probability density of the outcomes of the quantum measurement described by \(m(z)\). In this case the classical capacity of the channel is equal to the \(\chi\)-capacity (the channel is entanglement-breaking) and

$$C(M)=C_{\chi}(M)=\sup_{S}\left[h\left(p_{S}\right)-\inf_{\pi:\overline{S}_{\pi}=S}\int\limits_{\mathfrak{S}}h\left(p_{S}\right)\pi(dS)\right],$$

where \(h\left(p_{S}\right)=-\int p_{S}(z)\log p_{S}(z)dz\) is the differential entropy of this probability density and

$$f(S)=h\left(p_{S}\right)=\textrm{Tr\,}\,K(S)S,$$
$$K(S)=-\int m(z)\log p_{S}(z)dz.$$

3. Another case of interest described e.g. in Sec. 7.5 of [1] is the Entanglement of Formation. Let \(S_{12}\) be a state in the tensor product of two Hilbert spaces \(\mathcal{H}_{1}\otimes\mathcal{H}_{2}.\) Entanglement of Formation of the state \(S_{12}\) is the convex closure of \(H(S_{1})\), where \(S_{1}=\) \(\textrm{Tr}_{2}S_{12}\) is the partial state:

$$E_{F}(S_{12})=\inf_{\pi:\overline{S}_{\pi}=S_{12}}\int\limits_{\mathfrak{S}(\mathcal{H}_{1}\otimes\mathcal{H}_{2})}H(\textrm{Tr}_{2}S)\,\pi(dS),$$

minimization is over all probability measures \(\pi(dS)\) on \(\mathfrak{S}(\mathcal{H}_{1}\otimes\mathcal{H}_{2})\) satisfying \(\int_{\mathfrak{S}(\mathcal{H}_{1}\otimes\mathcal{H}_{2})}S\,\pi(dS)=S_{12}.\) In that case \(f(S_{12})=H(S_{1}),K(S_{12})=-\left(\log S_{1}\otimes I_{2}\right).\)

In all these cases we are looking for the solution of the convex programming problem

$$F(\pi)\equiv\int\limits_{\mathfrak{S}}\textrm{Tr\,}\,K(S)S\,\pi(dS)\longrightarrow\min$$
$$\pi\in\mathcal{P}(\mathfrak{S})$$
$$\int\limits_{\mathfrak{S}}S\,\pi(dS)=\overline{S},$$
(3)

where \(\overline{S}\) is a fixed density operator. We will give the duality relation and necessary and sufficient conditions for optimality basing on the results obtained in the monograph [5].

We start by introducing an equivalent but more convenient definition: we now call ensemble a measure \(\Pi(dS)\) on \(\mathfrak{S}\) with values in the positive cone of \(\mathfrak{T}(\mathcal{H}),\) such that \(\Pi(\mathfrak{S})\in\mathfrak{S}.\) We call \(\Pi(\mathfrak{S})\equiv\overline{S}_{\Pi}\) the average state of the ensemble. The equivalence with the initial definition is established by the relations

$$\Pi(dS)=S\pi(dS);\quad\overline{S}_{\Pi}=\int\limits_{\mathfrak{S}}S\pi(dS),$$

where \(\pi(A)=\textrm{Tr\,}\Pi(A),\,A\) is any Borel subset of \(\mathfrak{S}.\) Then the minimized functional can be rewritten as the scalar integral of the operator-valued function \(K(S)\) with respect to the operator-valued measure \(\Pi(dS),\)  the construction of which was given in the infinite-dimensional case in Ch. I of [5] (see also [4]):

$$F(\pi)=\int\limits_{\mathfrak{S}}\left\langle K(S),\Pi(dS)\right\rangle.$$

In what follows we assume that \(\mathcal{H}\) is finite-dimensional. In that case case we can just assume that \(K(S)\) is measurable function with values in the cone of positive operators on \(\mathcal{H}\) and the scalar integral can be understood via the expression \(\int_{\mathfrak{S}}\textrm{Tr\,}\,K(S)S\,\pi(dS)\). Then the optimization problem

$$\int\limits_{\mathfrak{S}}\left\langle K(S),\Pi(dS)\right\rangle\longrightarrow\min$$
$$\Pi(A)\geq 0\text{ for any Borel }A\subseteq\mathfrak{S},$$
$$\Pi(\mathfrak{S})=\overline{S},$$

becomes similar to a generalization of the Bayes problem [4, 5]. Combination of Theorem 2.1 and Theorem 2.2 from Ch. II of [5] implies

Theorem. The problem dual to (3) is

$$\max\left\{\textrm{Tr\,}\overline{S}\Lambda:\Lambda^{\ast}=\Lambda,\,\Lambda\leq K(S)\text{ for all }S\in\mathfrak{S}\right\}.$$
(4)

The following statements are equivalent:

(i) \(\Pi_{0}(dS)\) is the solution of the problem (3); \(\Lambda_{0}\) is the solution of the problem (4);

(ii) a. \(\Lambda_{0}\leq K(S)\) for all \(S\in\mathfrak{S};\)

b. \(\int_{A}\left[K(S)-\Lambda_{0}\right]\,\Pi_{0}(dS)=0\) for any Borel subset \(A\subseteq\mathfrak{S}.\)

The condition (ii.b) can be rewritten as

$$\left[K(S)-\Lambda_{0}\right]\,S=0\quad(\textrm{mod\,}\pi_{0}),$$

which means that the equality holds a.e. with respect to the measure \(\pi_{0}.\) By integrating, we obtain

$$\int\limits_{\mathfrak{S}}K(S)\,S\,\pi_{0}(dS)=\Lambda_{0}\overline{S},$$

which gives equation for determination of \(\Lambda_{0}\). Note that \(\Lambda_{0}\) must be Hermitean operator.

In the case of measurement channel, this equation reduces to

$$-\int\limits_{\mathfrak{S}}\int m(z)\log p_{S}(z)dz\,S\,\pi_{0}(dS)=\Lambda_{0}\overline{S}.$$

For completeness, we give the proof of the Theorem, taking into account simplifications due to finite dimensionality of \(\mathcal{H}.\) An infinite-dimensional generalization of the Theorem would have important applications to ensemble optimization problems.

Proof. In what follows we use the notation \(\left\langle X,Y\right\rangle=\textrm{Tr\,}X\,Y.\) Let us fix \(S_{0}\in\mathfrak{S}\) and show that

$$\inf\left\{\int\limits_{\mathfrak{S}}\left\langle K(S),\Pi(dS)\right\rangle:\Pi(\mathfrak{S})=\overline{S}\right\}$$
$${}=\inf\left\{\int\limits_{\mathfrak{S}}\left\langle K(S)-K(S_{0}),\Pi(dS)\right\rangle:\Pi(\mathfrak{S})\leq\overline{S}\right\}+\left\langle K(S_{0}),\overline{S}\right\rangle.$$

It is sufficient to show that

$$\inf\left\{\int\limits_{\mathfrak{S}}\left\langle K(S)-K(S_{0}),\Pi(dS)\right\rangle:\Pi(\mathfrak{S})=\overline{S}\right\}$$
$${}=\inf\left\{\int\limits_{\mathfrak{S}}\left\langle K(S)-K(S_{0}),\Pi(dS)\right\rangle:\Pi(\mathfrak{S})\leq\overline{S}\right\}.$$
(5)

Let \(\Pi\) be such that \(\Pi(\mathfrak{S})\leq\overline{S}.\) Defining

$$\overline{\Pi}(A)=\Pi(A)+\left(\overline{S}-\Pi(\mathfrak{S})\right)1_{A}(S_{0}),$$

we get \(\overline{\Pi}(\mathfrak{S})=\overline{S}\) and equality in (5).

Denote \(G(\Pi)=\Pi(\mathfrak{S})-\overline{S}\) and consider the problem of minimizing the functional

$$F(\Pi)=\int\limits_{\mathfrak{S}}\left\langle K(S)-K(S_{0}),\Pi(dS)\right\rangle+\left\langle K(S_{0}),\overline{S}\right\rangle$$

over the convex set of \(\Pi\)’s satisfying \(G(\Pi)\leq 0.\) For this we compute the dual functional

$$\varphi(S)=\inf\left\{F(\Pi)+\left\langle S,G(\Pi)\right\rangle\right\},$$

where the infimum is over the set of all positive \(\mathfrak{S}-\)valued measures. We show that

$$\varphi(S)=\begin{cases}\left\langle K(S_{0}),\overline{S}\right\rangle-\left\langle S,\overline{S}\right\rangle,\quad\text{if}\quad S\geq K(S_{0})-K(S^{\prime}),\quad\,\forall S^{\prime};\\ -\infty\quad\text{otherwise}.\end{cases}$$
(6)

Let \(S\) be such that for some \(S^{\prime}\) and some \(X\geq 0\)

$$\left\langle S,X\right\rangle<\left\langle K(S_{0})-K(S^{\prime}),X\right\rangle.$$

Defining \(\Pi_{n}(A)=nX\,1_{A}(S^{\prime}),\) we have

$$\varphi(S)\leq n\left[\left\langle S,X\right\rangle-\left\langle K(S_{0})-K(S^{\prime}),X\right\rangle\right]+\left\langle K(S_{0}),\overline{S}\right\rangle-\left\langle S,\overline{S}\right\rangle$$

whence \(\varphi(S)=-\infty.\)

Let now

$$S\geq K(S_{0})-K(S^{\prime})\quad\,\forall S^{\prime}.$$
(7)

By letting \(\Pi(A)\equiv 0,\) we obtain \(\varphi(S)\leq\left\langle K(S_{0}),\overline{S}\right\rangle-\left\langle S,\overline{S}\right\rangle.\) The converse inequality follows from

$$\left\langle S,\Pi(\mathfrak{S})\right\rangle\geq\int\limits_{\mathfrak{S}}\left\langle K(S_{0})-K(S^{\prime}),\Pi(dS^{\prime})\right\rangle,$$
(8)

which is obtained from (7) by integration (see Lemma 2.1 in Ch. I of [5]). This implies the first line in (6).

According to the general Lagrange duality theorem (see Appendix)

$$\inf_{G(\Pi)\leq 0}F(\Pi)=\max\left\{\varphi(S):S\geq 0\right\},$$

i.e. taking into account (5), (6)

$$\inf\left\{\int\limits_{\mathfrak{S}}\left\langle K(S),\Pi(dS)\right\rangle:\Pi(\mathfrak{S})=\overline{S}\right\}$$
$${}=\max\left\{\left\langle K(S_{0}),\overline{S}\right\rangle-\left\langle S,\overline{S}\right\rangle:\quad\ S\geq K(S_{0})-K(S^{\prime})\quad\,\forall S^{\prime}\right\}.$$

Denoting \(\Lambda=K(S_{0})-S,\) we come to (4).

Let (i) be fulfilled, (4) implies

$$\int\limits_{\mathfrak{S}}\left\langle K(S),\Pi_{0}(dS)\right\rangle=\left\langle\Lambda_{0},\overline{S}\right\rangle.$$
(9)

The inequality (iia): \(\Lambda_{0}\leq K(S)\), \(S\in\mathfrak{S}\) holds by (4). It follows for any Borel \(A\subseteq\mathfrak{S}:\)

$$\left\langle\Lambda_{0},\Pi_{0}(A)\right\rangle=\int\limits_{A}\left\langle\Lambda_{0},\Pi_{0}(dS)\right\rangle\leq\int\limits_{A}\left\langle K(S),\Pi_{0}(dS)\right\rangle.$$
(10)

By (9) it should be equality here, i.e. \(\int_{A}\left\langle K(S)-\Lambda_{0},\Pi_{0}(dS)\right\rangle=0.\) But since \(K(S)-\Lambda_{0}\geq 0,\) this implies (iib) (for detail see Proposition 3.3 from Ch. I of [5]).

Conversely, for arbitrary \(\Pi\) satisfying \(\Pi(\mathfrak{S})=\overline{S}\) we have by (iia)

$$\int\limits_{\mathfrak{S}}\left\langle K(S),\Pi(dS)\right\rangle\geq\left\langle\Lambda_{0},\overline{S}\right\rangle$$

and taking \(A=\mathfrak{S}\) in (iib) we obtain (9) whence (i) follows. \(\Box\)

Appendix. Let \(F\) be a convex functional on a convex subset \(\mathfrak{S}\) of a linear subspace \(L\) and \(G\) be a convex map of \(\mathfrak{S}\) into partially ordered Banach space \(L_{1}.\)

Consider the optimization problem

$$F(x)\longrightarrow\inf,\quad x\in\mathfrak{S};\quad G(x)\leq 0.$$
(11)

The following duality theorem holds (see e.g. [6], pp. 217, 224):

Theorem. Assume that the positive cone of \(L_{1}\)  contains an inner point, and there exists \(x_{1}\in S\)  such that \(\left\langle\lambda,G(x_{1})\right\rangle<0\)  for all \(\lambda\in L_{1}^{\ast},\lambda>0.\)  Then if the quantity (11) is finite,

$$\inf\left\{F(x):\quad x\in\mathfrak{S;\quad}G(x)\leq 0\right\}=\max\left\{\varphi(\lambda):\quad\lambda\in L_{1}^{\ast},\quad\lambda\geq 0\right\},$$
(12)

where

$$\varphi(\lambda)=\inf_{x\in\mathfrak{S}}\left\{F(x)+\left\langle\lambda,G(x)\right\rangle\right\}$$

is the dual functional.

Let \(\lambda_{0}\) be a solution of the dual problem in the right-hand side of (12). If the infimum in the left-hand side of (12) is attained on \(x_{0},\) then \(x_{0}\) is a solution of the problem \(\min\left\{F(x)+\left\langle\lambda_{0},G(x)\right\rangle;x\in\mathfrak{S}\right\}\) and \(\left\langle\lambda_{0},G(x_{0})\right\rangle=0.\)