For a general quantum channel \(\mathcal \) and a state \(\sigma \), the recovery condition \(\mathcal ^\sigma _}\) for the data-processing inequality of the BS-entropy was found in [5]. To be able to map quantum states into positive matrices, we present a recovery condition which is nonlinear but preserves positivity defined by

$$\begin \mathcal ^}_}\circ \mathcal (\rho ):= ( \sigma \mathcal ^*(\mathcal (\sigma )^ \mathcal (\rho )^2\mathcal (\sigma )^) \sigma )^ \, . \end$$

However, the nonlinearity of this recovery map can make this quantity a difficult object to deal with. In Proposition 5.11, we will see how to actually construct a completely positive, non-trace-preserving linear map, which will look very similar to the Petz recovery map.

The BS-entropy belongs to a larger family of entropies called maximal f-divergences [17, 26], defined as \(\widehat_f(\rho \Vert \sigma ) =\textrm[\sigma f(\sigma ^\rho \sigma ^)]\) for any operator convex function f on \([0,\infty )\). Since the saturation of the data-processing inequality is equivalent for every nonlinear operator convex f, we can use the same recovery conditions for all of them.

Theorem 5.1

Let \(\rho , \sigma \) be two quantum states, with \(\sigma \) invertible, and let \(\mathcal \) be a quantum channel. The following are equivalent:

(i)

\(\widehat(\rho \Vert \sigma ) = \widehat(\mathcal (\rho ) \Vert \mathcal (\sigma ) ) \).

(ii)

\(\widehat_f(\rho \Vert \sigma ) = \widehat_f(\mathcal (\rho ) \Vert \mathcal (\sigma ) ) \), for every operator convex function f on \([0,\infty )\).

(iii)

\(\rho =\mathcal ^\sigma _} \circ \mathcal (\rho )\) .

(iv)

\(\rho =\mathcal ^_} \circ \mathcal (\rho )\) .

Proof

\(\underline \Leftrightarrow \text \Leftrightarrow \text .}\) The first equivalence was proven in [17, Theorem 3.34], and the second in [5].

\(\underline \Rightarrow \text .}\) Let us assume that (iii) holds. Then, \( \rho = \sigma \mathcal ^*( \mathcal (\sigma )^ \mathcal (\rho ) ) \), and hence since \(\rho ^2=\rho \rho ^*\),

$$\begin \rho ^2&= \sigma \mathcal ^*( \mathcal (\sigma )^ \mathcal (\rho ) ) \mathcal ^*( \mathcal (\rho ) \mathcal (\sigma )^ )\sigma \\&\le \sigma \mathcal ^*( \mathcal (\sigma )^ \mathcal (\rho )^2 \mathcal (\sigma )^ )\sigma \, , \end$$

where in the last inequality we used the Kadison–Schwarz inequality (see, for example, [29, Exercise 3.4]). From [17, Theorem 3.34], the condition \(\widehat(\rho \Vert \sigma ) = \widehat(\mathcal (\rho ) \Vert \mathcal (\sigma ) ) \) is equivalent to \(\textrm[\rho ^2 \sigma ^] = \textrm[\mathcal (\rho )^2 \mathcal (\sigma )^]\). Thus,

$$\begin 0&=\textrm[\mathcal (\rho )^2 \mathcal (\sigma )^]-\textrm[\rho ^2 \sigma ^] \\&=\textrm[\sigma \mathcal ^*( \mathcal (\sigma )^ \mathcal (\rho )^2 \mathcal (\sigma )^ ) ]-\textrm[\rho ^2 \sigma ^]\\&=\textrm\left[ \underbrace^*( \mathcal (\sigma )^ \mathcal (\rho )^2 \mathcal (\sigma )^ )\sigma -\rho ^2\right) }_ \sigma ^ \right] \end$$

which implies that \(\rho ^2=\sigma \mathcal ^*( \mathcal (\sigma )^ \mathcal (\rho )^2 \mathcal (\sigma )^ )\sigma \), since \(\sigma ^\) is invertible and \(X \ge 0\).

\(\underline \Rightarrow \text .}\) Because of the condition in (iv), we have

$$\begin \textrm[\rho ^2 \sigma ^] = \textrm[\sigma \mathcal ^*\left( \mathcal (\sigma )^ \mathcal (\rho )^2 \mathcal (\sigma )^\right) ] = \textrm[\mathcal (\rho )^2 \mathcal (\sigma )^] \end$$

and the proof is concluded by applying again [17, Theorem 3.34]. \(\square \)

Remark 5.2

The recoverability conditions given by Theorem 5.13 are also valid for the geometric Rényi divergences, \(\widehat_(\rho \Vert \sigma )=\frac\log \textrm[\sigma f_(\sigma ^\rho \sigma ^)]\) for \(f_=x^\) [12] with \(\alpha \in (1,2]\), since this is the case where \(f_\) is operator convex [4, Exercise V.2.11]. Since they are the logarithms of maximal f-divergences, and the logarithm is strictly monotone in its domain, the data-processing inequality holds for the \(\widehat_\) if and only if it holds for \(\textrm[\sigma f_(\sigma ^\rho \sigma ^)]\).

5.1 Correspondence Between States Saturating BS- and Relative Entropy

Hereafter, given two states \(\rho , \sigma \in }\,}}(}\,}})\) and a quantum channel \(\mathcal : }\,}}(}\,}}) \rightarrow }\,}}(}\,}}) \), we say that the triple \((\rho , \sigma , \mathcal )\) is a Petz-triple if it saturates the DPI for the relative entropy, namely

$$\begin D(\rho \Vert \sigma ) = D(\mathcal (\rho ) \Vert \mathcal (\sigma ) ) \, , \end$$

and we say that the triple \((\rho , \sigma , \mathcal )\) is a BS-triple if it saturates the DPI for the BS-entropy, namely

$$\begin }(\rho \Vert \sigma ) = }(\mathcal (\rho ) \Vert \mathcal (\sigma ) ) \, . \end$$

Throughout the rest of the section, we denote by \(}\,}}\) and \(\mathcal \) two finite-dimensional Hilbert spaces, and by \(d_}\) the dimension of \(\mathcal \). Before showing the correspondence between Petz-triples and BS-triples, we will show that it holds for conditional expectations, a subclass of quantum channels. We introduce this notion by means of the proposition [28, Proposition 1.12].

Proposition 5.3

Let \(\mathcal \) be a matrix algebra with unital matrix subalgebra \(\mathcal \). Then, there exists a unique linear mapping \(\mathcal :\mathcal \rightarrow \mathcal \) such that

1.

\(\mathcal \) is a positive map.

2.

\(\mathcal (B)=B\) for every \(B \in \mathcal \).

3.

\(\mathcal (AB)=\mathcal (A)B\) for every \(A \in \mathcal \) and for every \(B \in \mathcal \).

4.

\(\mathcal \) is trace preserving.

A map fulfilling 1-3 is called a conditional expectation.

Theorem 5.4

Let \(\rho ,\sigma \in \mathcal (\mathcal )\) be quantum states satisfying supp\((\rho )\le \)supp\((\sigma )\), and let \(\mathcal :\mathcal (\mathcal )\rightarrow \mathcal (\mathcal )\) be a conditional expectation with Stinespring’s representation \(\mathcal (Y)=\textrm_E[VYV^*]\) where \(V:\mathcal \rightarrow \mathcal \otimes \mathcal _E\) is the associated isometry. Define the states

$$\begin \eta _:=c_ \mathcal (\sigma )^X\mathcal (\sigma )^\, , \end$$

(18)

for \(X\in \(\rho ),\mathcal (\sigma )\}\) and where \(c_\) is the normalization constant. Consider also \(\rho _0=V \rho V^*\), \(\sigma _0=V \sigma V^*\), \(\rho _0,\sigma _0\in \mathcal (\mathcal \otimes \mathcal _E)^+\). If \((\rho , \sigma , \mathcal )\) is a BS-triple, then \((\eta _,\eta _,\mathcal )\) is a Petz-triple, and we can write

$$\begin \eta _= & c_}V^*(\mathcal (\sigma )^\otimes I_E)\rho _0(\mathcal (\sigma )^\otimes I_E)V, \quad \text \nonumber \\ \quad \eta _= & c_}V^*(\mathcal (\sigma )^\otimes I_E)\sigma _0(\mathcal (\sigma )^\otimes I_E)V\,. \end$$

(19)

Conversely, let \(\mu , \nu \in \mathcal (\mathcal )\) with supp\((\mu ) \le \)supp\((\nu )\) and assume that \((\mu , \nu , \mathcal )\) is a Petz-triple. If \(\mathcal (\nu )=I_}/d_}\), then for any positive definite \(X \in \mathcal (\mathcal (\mathcal )^+)\), if we define

$$\begin \rho :=c_X^\nu ^ \mathcal (\mu )\nu ^ X^, \quad \text \quad \sigma :=c_X^\nu X^ \, , \end$$

(20)

where \(c_\), \(c_\) are the normalization constants, then \((\rho ,\sigma , \mathcal )\) is a BS-triple. In addition, we can write

$$\begin & \mu =d_}c_^V^*(X^\otimes I_E)\rho _0(X^\otimes I_E)V \, ,\\ & \quad \nu = c_^V^*(X^\otimes I_E)\sigma _0(X^\otimes I_E) V\, . \end$$

Proof

Let \(\rho , \sigma \in \mathcal (\mathcal )\), and let \(\mathcal :\mathcal (\mathcal )\rightarrow \mathcal (\mathcal )\) be a conditional expectation, then its range is a subalgebra and the restriction of \(\mathcal ^*\) to \(\mathcal (\mathcal (\mathcal ))\) is the natural embedding, so that we can omit it. Assume that \((\rho ,\sigma ,\mathcal )\) is a BS-triple, then the BS-recovery condition translates into

$$\begin \begin \rho&=\sigma \mathcal (\sigma )^\mathcal (\rho )\\&=\mathcal (\sigma )^\mathcal (\sigma )^\sigma \mathcal (\sigma )^\mathcal (\sigma )^ \mathcal (\rho )\mathcal (\sigma )^\mathcal (\sigma )^ \end \end$$

Define \(\tilde_:=\mathcal (\sigma )^X\mathcal (\sigma )^\), so we can rewrite the expression above as \(\tilde_=\tilde_\tilde_(\rho )}\). Note that \([\tilde_,\tilde_(\rho )}]=0\), since

$$\begin \tilde_\eta _(\rho )}=\tilde_=\tilde_^*=\tilde_(\rho )}\tilde_ \, . \end$$

Thus, \(\tilde_=\tilde_^\tilde_(\rho )}\tilde_^\) and since \(\tilde_(\sigma )}=I_}\), we obtain

$$\begin \tilde_=\tilde_^\tilde_(\sigma )}^\tilde_(\rho )}\tilde_(\sigma )}^\tilde_^ \, . \end$$

We normalize now to define the states \(\eta _=\tilde_/\textrm[\tilde_]\) and notice that \(\mathcal (\eta _)=\eta _(\rho )}\) and \(\mathcal (\eta _)=\eta _(\sigma )}\) since \(\mathcal \) is a conditional expectation. Consequently, we can write

$$\begin \eta _=\eta _^\mathcal (\eta _)^\mathcal (\eta _)\mathcal (\eta _)^\eta _^\, , \end$$

so \((\eta _,\eta _, \mathcal )\) is a Petz-triple.

Consider now the Stinespring’s isometry \(V:\mathcal \rightarrow \mathcal \otimes \mathcal _E\) such that \(\mathcal (X)=\textrm_E[VXV^*]\), and notice that \(\rho =V^*\rho _0 V\) and \(\sigma =V^*\sigma _0 V\) where \(\rho _0,\sigma _0\in \mathcal (\mathcal \otimes \mathcal _E)^+\) are positive and such that \(\,}}(\rho _0), \,}}(\sigma _0)\le VV^*\) . To obtain (19), write

$$\begin \eta _=c_}\mathcal (\sigma )^V^*\rho _0V\mathcal (\sigma )^, \quad \text \quad \eta _=c_}\mathcal (\sigma )^V^*\sigma _0V\mathcal (\sigma )^\, . \end$$

Finally, since \(\mathcal (\sigma )^ \in \mathcal (\mathcal (\mathcal ))\) and the range of \(\mathcal \) is its multiplicative domain, Lemma 5.12 implies that \(V\mathcal (\sigma )^=(\mathcal (\sigma )^\otimes I_E)V\).

Conversely, let \(\mu , \nu \in \mathcal (\mathcal )\) with supp\((\mu ) \le \)supp\((\nu )\) such that \((\mu , \nu , \mathcal )\) is a Petz-triple and define \(\rho , \sigma \) by (20) where \(X \in \mathcal (\mathcal (\mathcal )^+)\). Since \((\mu ,\nu , \mathcal )\) is a Petz-triple, it also satisfies the BS-recovery condition [5], and using also that \(\mathcal (\nu )=I_}/d_}\) we obtain

$$\begin \nu \mathcal (\mu )=d_}\,}}}^\nu \mathcal (\nu )^\mathcal (\mu )=d_}^\mu =d_}^\mu ^*=\mathcal (\mu )\nu \, , \end$$

i.e. \([\nu ,\mathcal (\mu )]=0\). We will show first that \((\rho ,\sigma , \mathcal )\) is a BS-triple using the condition (iii) in Theorem 5.1. Notice that

$$\begin \mathcal (\sigma )= & c_X^\mathcal (\nu )X^=c_d_}^X, \quad \text \quad \mathcal (\rho )=c_X^\mathcal (\nu ) \mathcal (\mu )X^\\= & c_d_}^X^ \mathcal (\mu )X^. \end$$

Then,

$$\begin \begin \sigma \mathcal (\sigma )^\mathcal (\rho )&=c_ X^\nu X^ X^X^\mathcal (\mu ) X^\\&=c_X^\nu \mathcal (\mu ) X^\\&=\rho \, , \end \end$$

so \((\rho , \sigma , \mathcal )\) is a BS-triple. Writing \(\sigma =V^*\sigma _0V\), then \(\nu \) is given by

$$\begin \begin \nu&=c_^X^V^*\sigma _0VX^ \\&=c_^V^*(X^\otimes I_E)\sigma _0(X^\otimes I_E) V\, , \end \end$$

where the last step follows analogously as in the previous implication using the fact that \(X \in \mathcal (\mathcal (\mathcal )^+)\). Analogously, write \(\rho =V^* \rho _0 V\), and since every Petz-triple \((\mu ,\nu ,\mathcal )\) is a BS-triple, we use again (iii) of Theorem 5.1 and obtain

$$\begin \begin \mu&= \nu \mathcal (\nu )^\mathcal (\mu )\\&=d_}\nu \mathcal (\mu )\\&=d_}c_^X^\rho X^ \\&=d_}c_^V^*(X^\otimes I_E)\rho _0(X^\otimes I_E)V \, . \end \end$$

\(\square \)

Remark 5.5

If \(\mathcal =\mathcal \otimes \mathcal ^c\) notice that the partial trace \(\textrm_}:\mathcal (\mathcal )\rightarrow \mathcal (\mathcal ^)\) is also included in the assumptions of Theorem 5.4 setting \(\mathcal (X)=d_}^I_}\otimes \textrm_}[X]\).

Remark 5.6

The fundamental key of the previous theorem lies in the non-symmetric nature of the BS-recovering map \(\mathcal _}^\) defined in (4). This fact together with \(\mathcal (\nu )\) being maximally mixed impose the constraint that \([\mathcal (\mu ),\nu ]=0\). This is an interesting phenomenon of the BS-recovery condition that cannot be observed via the Petz recovery map.

Theorem 5.7

Let \(\rho ,\sigma \) be states on \(}\,}}(}\,}})\) and let \(\mathcal :B(}\,}})\rightarrow B(}\,}})\) be a channel. Let \(V:}\,}}\rightarrow }\,}}\otimes }\,}}_E\) be the isometry such that \(\mathcal =\textrm_E[V\cdot V^*]\). Let us introduce the states

$$\begin \bar_:= & c_\rho \left( \mathcal (\sigma )^\otimes I_E \right) V\rho V^*\left( \mathcal (\sigma )^\otimes I_E \right) , \hspace\quad \bar_\\:= & d_}^\left( \mathcal (\sigma )^\otimes I_E \right) V\sigma V^*\left( \mathcal (\sigma )^\otimes I_E \right) \end$$

where \(c_\rho \) is a normalization constant. Then \((\rho ,\sigma ,\mathcal )\) is a BS-triple if and only if \((\bar_,\bar_,\textrm_E)\) is a Petz-triple.

Proof

Put \(\rho _0:=V\rho V^*\), \(\sigma _0=V\sigma V^*\), then we clearly have that \((\rho ,\sigma ,\mathcal )\) is a BS-triple if and only if \((\rho _0,\sigma _0,\textrm_E)\) is a BS-triple. Now we can apply the results of Theorem 5.4 to the latter, obtaining the Petz-triple \((\bar_,\bar_,\textrm_E)\).

Assume the converse. Note that \(\textrm_E[\bar_]=c_\rho \mathcal (\sigma )^\mathcal (\rho )\mathcal (\sigma )^\) and \(\textrm_E[\bar_]=I_}\,}}}/d_}\,}}}\) is the maximally mixed state. By the reversibility conditions for the triple \((\bar_,\bar_,\textrm_E)\) [9, Eq. (1.30)] or [32], we have

$$\begin \bar_^\bar_^=\textrm_E[\bar_]^\textrm_E[\bar_]^\otimes I_E=d_}\,}}}^\textrm_E[\bar_]^\otimes I_E. \end$$

It follows that \(\bar_\) and \(\bar_\) must commute, moreover, suppressing again tensoring with the identity,

$$\begin \bar_^=d_}\,}}}^\textrm_E[\bar_]^\bar_^. \end$$

But this implies that \(\bar_\) and \(\textrm_E[\bar_]\) must commute as well and we have

$$\begin \rho _0= & c_\rho ^\mathcal (\sigma )^\bar_\mathcal (\sigma )^=c_\rho ^d_}\,}}}\mathcal (\sigma )^\bar_\textrm_E[\bar_]\mathcal (\sigma )^= \sigma _0\mathcal (\sigma )^\mathcal (\rho )\\= & \sigma _0\textrm_E[\sigma _0]^\textrm_E[\rho _0], \end$$

here we used that \(\mathcal (\rho )=\textrm_E[\rho _0]\) and similarly for \(\sigma \). It follows that \((\rho _0,\sigma _0,\textrm_E)\) is a BS-triple and so is \((\rho ,\sigma ,\mathcal )\). \(\square \)

Corollary 5.8

Assume that there is some \(Y\in }\,}}(}\,}})^+\) such that \(\left( \mathcal (\sigma )\otimes I_E\right) V=VY\) and put

$$\begin \eta _\rho :=d_\rho Y^\rho Y^,\qquad \eta _\sigma :=d_\sigma Y^\sigma Y^, \end$$

with normalization constants \(d_\rho \) and \(d_\). Then \((\rho ,\sigma ,\mathcal )\) is a BS-triple if and only if \((\eta _\rho ,\eta _\sigma ,\mathcal )\) is a Petz-triple.

Proof

Similarly as before, since \(V\cdot V^*\) is an isometric channel, \((\eta _\rho ,\eta _\sigma ,\mathcal )\) is a Petz-triple if and only if \((V \eta _\rho V^*, V\eta _\sigma V^*,\textrm_E)\) is a Petz-triple. Now it is enough to note that by the assumptions, we have \(VY^=}(\sigma )^V\), which implies that \(V\eta _\rho V^*=\bar_\rho \) and \(V\eta _\sigma V^*= \bar_\sigma \), with the notation as in Theorem 5.7. The statement now follows by Theorem 5.7. \(\square \)

Remark 5.9

The assumption in the above corollary is satisfied if \(}\) is also unital and \(\mathcal (\sigma )\) lies in the image of its multiplicative domain (see Lemma 5.12). For example, this is the case for the trace-preserving conditional expectation \(}\).

Corollary 5.10

Let \(\mu ,\nu \) be states on \(}\,}}(}\,}}\otimes }\,}}_E)\) such that \(\textrm_E[\nu ]=I_}\,}}}/d_}\,}}}\) and assume that \((\mu ,\nu ,\textrm_E)\) is a Petz-triple. Let \(X\in }\,}}(}\,}})^+\) be a state and \(V:}\,}}\rightarrow }\,}}\otimes }\,}}_E\) an isometry such that:

(1)

X is invertible

(2)

\(\,}}(X^\mu X^), \,}}(X^\nu X^)\le VV^*\).

Let \(\omega _(X,V), \omega _(X,V)\) be states on \(}\,}}(}\,}})\) with \(\omega _(X,V) \propto V^*X^\mu X^V\), \(\omega _(X,V)\propto V^*X^\nu X^V\) and let \(}=\textrm_E[V\cdot V^*]\). Then \((\omega _(X,V),\omega _(X,V),})\) is a BS-triple.

Proof

Let \(\rho _0\), \(\sigma _0\) be states on \(}\,}}(}\,}}\otimes }\,}}_E)\) such that \(\rho _0\propto X^\mu X^\) and \(\sigma _0\propto X^\nu X^\). By the assumptions, \(V\omega _\mu (X,V) V^*=\rho _0\) and \(V\omega _\nu (X,V) V^*=\sigma _0\). Further, we have

$$\begin \mathcal (\omega _(X,V))=\textrm_E[\sigma _0]=X, \end$$

so that, with the notation as in Theorem 5.7,

$$\begin \bar_\rho \propto X^V\omega _\mu (X,V) V^*X^=X^\rho _0X^\propto \mu \end$$

so that \(\bar_\rho =\mu \), similarly \(\bar_\sigma =\nu \). By Theorem 5.7, \((\omega _\mu (X,V),\omega _\nu (X,V),\mathcal )\) is a BS-triple. \(\square \)

The definition of \(\bar_X\) given in Theorem 5.7 allows us to obtain a recoverability condition in a similar way as for the \(\Phi _\) in Corollary 3.9 for a triple \((\rho ,\sigma ,\mathcal )\) that satisfies the BS-recovery condition. Consider the polar decomposition \(\sigma _0^\left( \mathcal (\sigma )^\otimes I_E\right) =d_}^W \bar_\sigma ^\), in the notations of Theorem 5.7, where W is a unitary in \(}\,}}(}\,}}\otimes }\,}}_E)\). Let us define the map \(\Phi _}:}\,}}(}\,}})\rightarrow }\,}}(}\,}})\) by

$$\begin \Phi _}(Y)=\sigma ^V^* W(\mathcal (\sigma )^Y\mathcal (\sigma )^\otimes I_E)W^*V \sigma ^. \end$$

This map is obviously completely positive but not necessarily trace preserving. If in addition W commutes with \(VV^*\), i.e. it leaves the range of V invariant, then \(\widetilde=V^*WV\) is a unitary in \(}\,}}(}\,}})\) such that \(V^*W=\widetildeV^*\), and therefore

$$\begin \Phi _}(Y)=\sigma ^\widetilde\mathcal ^* (\mathcal (\sigma )^Y\mathcal (\sigma )^)\widetilde^* \sigma ^. \end$$

Note also that if \(W=I_}\,}}}\), then \(\Phi _}\) is the Petz recovery map.

Proposition 5.11

\((\rho ,\sigma ,\mathcal )\) is a BS-triple if and only if

$$\begin \rho =\Phi _}(\mathcal (\rho )). \end$$

Proof

Assume first that \((\rho ,\sigma ,\mathcal )\) is a BS-triple. Then by Theorem 5.7, \((\bar_,\bar_, \textrm_E)\) is a Petz-triple satisfying that \(\textrm_E[\bar_]=I_}/d_}\), so they satisfy the Petz recovery condition:

$$\begin \begin \bar_&=d_}\bar_^\left( \textrm_E[\bar_]\otimes I_E \right) \bar_^ \\&=d_}\bar_^\left( c_\rho \mathcal (\sigma )^\mathcal (\rho )\mathcal (\sigma )^\otimes I_E\right) \bar_^\, , \end \end$$

which can be rewritten as

$$\begin & \rho =d_}V^*(\mathcal (\sigma )^\otimes I_E)\bar_^\\ & \quad \left( \mathcal (\sigma )^\mathcal (\rho )\mathcal (\sigma )^\otimes I_E\right) \bar_^(\mathcal (\sigma )^\otimes I_E)V \, . \end$$

If we consider now the unitary W satisfying \(\sigma _0^\left( \mathcal (\sigma )^\otimes I_E\right) =d_}^W \bar_\sigma ^\), we conclude that

$$\begin \begin \rho&= V^*\sigma _0^W \left( \mathcal (\sigma )^\mathcal (\rho )\mathcal (\sigma )^\otimes I_E\right) W^*\sigma _0^V\\&=\sigma ^ V^*W \left( \mathcal (\sigma )^\mathcal (\rho )\mathcal (\sigma )^\otimes I_E\right) W^*V\sigma ^\\&=\Phi _}(\mathcal (\rho ))\, . \end \end$$

The proof is finished by the observation that since \(\,}}(\sigma _0)\le VV^*\), we have \(V^*\sigma _0^nV=(V^*\sigma _0V)^n=\sigma ^n\) for any n, consequently, \(V^*\sigma _0^V=\sigma ^\).

Conversely, assume that \(\rho =\Phi _}(\mathcal (\rho ))\). Then, undoing the previous computation, we obtain that \((\bar_,\bar_,\textrm_E)\) is a Petz-triple. In particular, as shown in the proof of Theorem 5.7, \(\textrm_E[\bar_]\) and \(\bar_\) commute, so

$$\begin \bar_=d_}\bar_\left( c_\mathcal (\sigma )^\mathcal (\rho )\mathcal (\sigma )^\otimes I_E \right) \, . \end$$

Using now the definition of \(\bar_\) and \(\bar_\), we obtain

$$\begin V\rho V^*=V\sigma V^*\left( \mathcal (\sigma )^\mathcal (\rho )\otimes I_E \right) \, , \end$$

so

$$\begin \begin \rho&=\sigma V^*\left( \mathcal (\sigma )^\mathcal (\rho )\otimes I_E \right) V \\&=\sigma \mathcal ^*\left( \mathcal (\sigma )^\mathcal (\rho ) \right) \, ,\\ \end \end$$

and we conclude that \((\rho ,\sigma ,\mathcal )\) is a BS-triple. \(\square \)

5.2 Structural Decompositions of BS- and Petz-Triples Satisfying Recovery Conditions

To showcase the importance of the correspondence between BS-triples and Petz-triples given by Theorem 5.7 and Corollary 5.10, we present below an application that consists of obtaining the structural decompositions for both quantities in two different ways.

On the one hand, the structural decomposition for Petz-triples \((\mu , \nu , \mathcal )\) was fully characterized in [19] and [27]. In the case when \(\mathcal =\mathcal \) is the trace-preserving conditional expectation onto some subalgebra \(\mathcal \subseteq \mathcal (\mathcal )\), we have from [19, Theorem 5(iii)] that \((\rho ,\sigma ,\mathcal )\) is a Petz-triple if and only if there are density operators \(\rho _1,\sigma _1\in \mathcal \) and \(\xi \in \mathcal (\mathcal )\) such that

$$\begin \rho =\rho _1\xi ,\qquad \sigma =\sigma _1\xi , \quad \text \quad \mathcal (\rho )=\rho _1\mathcal (\xi ), \qquad \mathcal (\sigma )=\sigma _1\mathcal (\xi ). \end$$

Since both \(\rho _1\) and \(\sigma _1\) must commute with both \(\xi \) and \(\mathcal (\xi )\), this suggests a unitary and a decomposition \(U:\mathcal \rightarrow \oplus _n \mathcal _n^L\otimes \mathcal _n^R\) such that

$$\begin \rho =U^*\left( \bigoplus _n \rho _n\otimes \xi _n\right) U,\qquad \sigma =U^*\left( \bigoplus _n \sigma _n\otimes \xi _n\right) U \end$$

f and

$$\begin \mathcal (\rho )=U^*\left( \bigoplus _n \rho _n\otimes \xi ^0_n\right) U,\qquad \mathcal (\sigma )=U^*\left( \bigoplus _n \sigma _n\otimes \xi ^0_n\right) U \end$$

for some \(\rho _n,\sigma _n\in \mathcal (\mathcal _n^L)^+\) and \(\xi _n,\xi _n^0\in \mathcal (\mathcal _n^R)^+\). This decomposition was actually shown in the proof of [33, Theorem 9.8]. If \(\mathcal (\sigma )=I_}/d_}\), we see that all \(\sigma _n\) and \(\xi _n\) must be multiples of the identity, so that we may write \(\sigma =U^*\bigoplus _n (I_^L_n}\otimes \xi _n)U\) in this case. Plugging now this structural decompositions in Corollary 5.10, we can obtain the structural decomposition of BS-triples.

On the other hand, in the next theorem, we will show the structural decomposition for BS-recovery states directly, which will also expand Theorem 5.1. Using this result, we will be able to obtain the structural decomposition of states of Petz-triples \((\mu ,\nu ,\mathcal )\), under the constraint that \(\mathcal (\nu )\) is the maximally mixed state. This provides a different proof based on the structural decomposition of BS-triples, yielding such a decomposition based on a dilation to a larger space. For this purpose, we will need the description of the multiplicative domain of a completely positive unital map, given in Lemma 5.12.

Let \(\mathcal : \mathcal (\mathcal )\rightarrow \mathcal (\mathcal )\) be a completely positive unital map. From the Stinespring representation of the adjoint map \(\mathcal ^*\), we see that there is some auxiliary space \(\mathcal _E\) and an operator \(V:\mathcal \rightarrow \mathcal \otimes \mathcal _E\) such that \(\textrm_E [VV^*]=I_}\) and \(\mathcal = \textrm_E [V\cdot V^*]\). The map \(\mathcal \) is faithful if and only if \(V^*V\) is invertible. Indeed, this follows from the fact that for any \(M\ge 0\), we have \(\mathcal (M)=0\) if and only if \(0=\textrm[\mathcal (M)]=\textrm[MV^*V]\). In this case, we have the polar decomposition \(V=W(V^*V)^=(VV^*)^W\), with an isometry \(W:\mathcal \rightarrow \mathcal \otimes \mathcal _E\).

Lemma 5.12

Let \(\mathcal =\textrm_E[V\cdot V^*]\) be a completely positive unital map \(\mathcal (\mathcal )\rightarrow \mathcal (\mathcal )\) and let \(X=X^*\in \mathcal (\mathcal )\). Then \(\mathcal (X^2)=\mathcal (X)^2\) if and only if there is some \(Y=Y^*\in \mathcal (\mathcal )\) such that \((Y\otimes I_E)V=VX\). Moreover, in that case, \(Y=\mathcal (X)\) and \(Y\otimes I_E\) commutes with \(VV^*\). If \(\mathcal \) is faithful, we also have \(X=W^*(Y\otimes I_E)W\), with \(W:\mathcal \rightarrow \mathcal \otimes \mathcal _E\) the isometry from the polar decomposition of V.

Proof

Assume that \(\mathcal (X^2)=\mathcal (X)^2\) and let \(Y=\mathcal (X)\). Let \(Z=(Y\otimes I_E)V-VX\). Then

$$\begin&\textrm[ZZ^*]=\textrm[(Y\otimes I_E)VV^*(Y\otimes I_E)\\&-(Y\otimes I_E)VXV^*-VXV^*(Y\otimes I_E)+VX^2V^*]=\textrm[Y^2-\mathcal (X^2)]=0, \end$$

so that \(Z=0\). Conversely, let \(Y=Y^*\in \mathcal (\mathcal )\) be such that \((Y\otimes I_E)V=VX\). Then \((Y\otimes I_E)VV^*=VXV^*=VV^*(Y\otimes I_E)\) and \(Y=\textrm_E[(Y\otimes I_E)VV^*]=\textrm_E[VXV^*]=\mathcal (X)\). We also have

$$\begin \mathcal (X)^2=\textrm_E[(\mathcal (X)\otimes I_E)^2VV^*]=\textrm_E[VX^2V^*]=\mathcal (X^2). \end$$

Assume that \(\mathcal \) is faithful, then

$$\begin & VW^*(Y\otimes I_E)W=(VV^*)^(Y\otimes I_E)W\\ & \quad =(Y\otimes I_E)(VV^*)^W=(Y\otimes I_E)V=VX. \end$$

Since \(V^*V\) is invertible, it follows that we must have \(X=W^*(Y\otimes I_E)W\). \(\square \)

We will assume below that \(\,}}(\rho )\le \,}}(\sigma )\) and use the notation

$$\begin \rho /\sigma ]:=\sigma ^\rho \sigma ^. \end$$

Note that the support condition implies that this operator is well defined. Let \(\mathcal : \mathcal (\mathcal )\rightarrow \mathcal (\mathcal )\) be a channel and let \(\mathcal _\) denote the adjoint of the Petz recovery map, that is

$$\begin \mathcal _\sigma :=(\mathcal ^\sigma _})^*=\mathcal (\sigma )^\mathcal (\sigma ^\cdot \sigma ^) \mathcal (\sigma )^. \end$$

By the restriction to the supports, we may and will assume that both \(\sigma \) and \(\mathcal (\sigma )\) are invertible, in which case \(\mathcal _\sigma : \mathcal (\mathcal )\rightarrow \mathcal (\mathcal )\) is unital and faithful. It is also easily seen that

$$\begin \mathcal _\sigma ([\rho /\sigma ])=[\mathcal (\rho )/\mathcal (\sigma )]. \end$$

Theorem 5.13

Let \(\mathcal :\mathcal (\mathcal )\rightarrow \mathcal (\mathcal )\) be a channel, and let \(V: \mathcal \rightarrow \mathcal \otimes \mathcal _E\) be an isometry such that \(\mathcal =\textrm_E[V\cdot V^*]\). Let \(\rho ,\sigma \in \mathcal (\mathcal )\) be states such that \(\,}}(\rho )\le \,}}(\sigma )\). The following conditions are equivalent.

(i)

\(}(\rho \Vert \sigma )=}(\mathcal (\rho )\Vert \mathcal (\sigma ))\).

(ii)

\(\mathcal _\sigma ([\rho /\sigma ]^2)=\mathcal _\sigma ([\rho /\sigma ])^2\).

(iii)

There is a decomposition and a unitary \(U:\mathcal \rightarrow \bigoplus _n \mathcal _n^L\otimes \mathcal _n^R\), such that for

$$\begin \rho =V^*\rho _0V,\quad \sigma =V^*\sigma _0 V, \end$$

where \(\rho _0,\sigma _0\in \mathcal (\mathcal \otimes \mathcal _E)^+\) are positive and such that \(\,}}(\rho _0), \,}}(\sigma _0)\le VV^*\), we have

$$\begin \rho _0&=(\mathcal (\sigma )^U^*\otimes I_E)\bigoplus _n (\xi _n^L\otimes \xi _n^R)(U\mathcal (\sigma )^\otimes I_E) \\ \sigma _0&=(\mathcal (\sigma )^U^*\otimes I_E)\bigoplus _n (I__n^L}\otimes \xi _n^R)(U\mathcal (\sigma )^\otimes I_E) \end$$

for some \(\xi _n^L\in \mathcal (\mathcal _n^L)^+\) and \(\xi _n^R\in \mathcal (\mathcal _n^R\otimes \mathcal _E)^+\).

(iv)

\(\rho =\sigma \mathcal ^*(\mathcal (\sigma )^\mathcal (\rho ))\).

Proof

\(\underline \Leftrightarrow \text .}\) This equivalence was proved in [17], in a more general situation. To make the proof self-contained, we present a proof under our assumptions. First, we note that we may write

$$\begin }(\rho \Vert \sigma )=\textrm[\sigma ^\rho \sigma ^\log (\sigma ^\rho \sigma ^)]=\textrm[\sigma f([\rho /\sigma ])], \end$$

with \(f(x)=x\log x\). Using the integral representation

$$\begin f(x)=\int _0^\infty \left( \frac-\frac\right) dt=\int _0^\infty \left( \frac-1+\frac\right) dt, \end$$

we obtain

$$\begin \widehat(\rho \Vert \sigma )=\int _0^\infty \left( \frac[\rho ]}-\textrm[\sigma ]+t\textrm[\sigma ([\rho /\sigma ]+t)^] \right) dt. \end$$

The equality (i) holds if and only if

$$\begin \int _0^\infty t(\textrm[\sigma ([\rho /\sigma ]+t)^]-\textrm[\mathcal (\sigma )([\mathcal (\rho )/\mathcal (\sigma )]+t)^])dt=0. \end$$

(21)

Now note that

$$\begin \textrm[\sigma ([\rho /\sigma ]+t)^]=\textrm[\mathcal (\sigma )\mathcal _\sigma (([\rho /\sigma ]+t)^)] \end$$

and

$$\begin \textrm[\mathcal (\sigma )([\mathcal (\rho )/\mathcal (\sigma )]+t)^]=\textrm[\mathcal (\sigma )(\mathcal _\sigma ([\rho /\sigma ]+t))^]. \end$$

As mentioned above, we may assume that \(\sigma \) and \(\mathcal (\sigma )\) are invertible and then \(\mathcal _\sigma \) is unital and faithful. It follows by the Choi inequality [10, 11] that Eq. (21) holds if and only if

$$\begin \mathcal _\sigma (([\rho /\sigma ]+t)^)=(\mathcal _\sigma ([\rho /\sigma ]+t))^,\qquad \forall t\in (0,\infty ). \end$$

(22)

Differentiating by t, we obtain

$$\begin \mathcal _\sigma (([\rho /\sigma ]+t)^)=(\mathcal _\sigma ([\rho /\sigma ]+t))^=\mathcal _\sigma (([\rho /\sigma ]+t)^)^2, \end$$

where we have used Eq. (22) in the last equality. Hence, \(\mathcal _\sigma \) must be multiplicative on all elements of the form \(([\rho /\sigma ]+t)^\), \(t\in (0,\infty )\). Since the multiplicative domain of \(\mathcal _\sigma \) is a subalgebra, we see that the equality Eq. (22), and hence also (i), is equivalent to (ii).

\(\underline \Rightarrow \text .}\) Assume (ii) and put \(N:=\mathcal _\sigma ([\rho /\sigma ])=[\mathcal (\rho )/\mathcal (\sigma )]\). We have \(\mathcal _\sigma =\textrm_E[S\cdot S^*]\), with \(S=(\mathcal (\sigma )^\otimes I_E)V\sigma ^\) and using Lemma 5.12, we see that \(N\otimes I_E\) must commute with \(M:=SS^*\). Since \(\mathcal _\sigma \) is faithful, we have \([\rho /\sigma ]=W^*(N\otimes I_E)W\), where \(S=M^W\) is the polar decomposition. By definition of S, we obtain \(\sigma ^=V^*(\mathcal (\sigma )^\otimes I_E)S\), so that

$$\begin \sigma =V^*(\mathcal (\sigma )^\otimes I_E)M(\mathcal (\sigma )^\otimes I_E)V \end$$

and

$$\begin \rho =\sigma ^[\rho /\sigma ]\sigma ^&=V^*(\mathcal (\sigma )^\otimes I_E)M^(N\otimes I_E)M^(\mathcal (\sigma )^\otimes I_E)V \\&=V^*(\mathcal (\sigma )^\otimes I_E)(N\otimes I_E)M(\mathcal (\sigma )^\otimes I_E)V. \end$$

Let \(\mathcal \subseteq