May 27, 2026
Motivated by several open problems in the Afham–Ferrie theory of generalized fidelity, we study polar fidelities, Holevo bases, and unitary factors on \(\mathbb{P}_d\), the cone of \(d\times d\) complex positive definite matrices. We prove that, for every fixed pair \(P,Q\in\mathbb{P}_d\), the one-sided polar fidelities \(x\mapsto F_{P^x}(P,Q)\) and \(x\mapsto F_{Q^x}(P,Q)\), as well as their symmetrization, are nondecreasing on \((-\infty,1]\) and nonincreasing on \([1,\infty)\). Hence the polar paths realize exactly the interval \([F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)]\) on \([-1,1]\), yielding pointwise, generally pair-dependent realizations of all \(z\)-fidelity values with \(z\ge1/2\) and of the Log-Euclidean fidelity as generalized fidelities. We also show that for \(d\ge2\) and \(0<z<1/2\), such realization fails in general, even for interior fidelities. We further solve the fixed-pair Holevo-base equation \(F_R(P,Q)=F^{\mathrm H}(P,Q)\), classify all Holevo bases, and classify exactly which unitary factors of generalized fidelity can arise from a base \(R\). These are precisely the unitaries \(W\) for which \(P^{-1/2}Q^{1/2}W\) is similar to a positive definite matrix. This recovers the special-unitary constraint and disproves the global reverse inclusion \(SU(d)\) for \(d\ge2\).
Fix a dimension \(d\in\mathbb{N}\), and let \(\mathbb{P}_d\) denote the cone of \(d\times d\) complex positive definite matrices. For \(P,Q,R\in\mathbb{P}_d\), Afham and Ferrie introduced the generalized fidelity \[F_R(P,Q):= \operatorname{Tr}\!\left[\bigl(R^{1/2}PR^{1/2}\bigr)^{1/2}R^{-1} \bigl(R^{1/2}QR^{1/2}\bigr)^{1/2}\right],\] a base-dependent fidelity whose special choices of \(R\) recover several standard quantum fidelities [1]. The relevant specializations will be recalled below. In general, \(F_R(P,Q)\) need not be real-valued; by contrast, the one-sided polar quantities studied below are positive real.
The same framework gives the one-sided polar paths \(x\mapsto F_{P^x}(P,Q)\) and \(x\mapsto F_{Q^x}(P,Q)\), together with the symmetrized family \[F^{\mathrm{pol}}_{x}(P,Q):=\frac{F_{P^x}(P,Q)+F_{Q^x}(P,Q)}{2}, \qquad x\in\mathbb{R}.\] Since \(P^x,Q^x\in\mathbb{P}_d\) for every real \(x\), this family is well defined on all of \(\mathbb{R}\). On \([-1,1]\), it recovers Matsumoto, Holevo, and Uhlmann fidelities at \(x=-1,0,1\), respectively.
This should be distinguished from other recent \(\alpha\)-\(z\)-fidelity frameworks [2]–[4]. We do not introduce a new \(\alpha\)-\(z\) fidelity. Rather, we study base selection and fixed-pair structure inside the Afham–Ferrie generalized-fidelity framework, especially along the polar paths and in the equality problem \(F_R(P,Q)=F^{\mathrm H}(P,Q)\).
We now spell out how the results of this paper relate to the open problems in [1]. The status is as follows.
Open Problem 4: monotonicity of polar fidelities. We answer this problem affirmatively and in a stronger form. For every fixed pair \(P,Q\in\mathbb{P}_d\), the one-sided polar paths \(x\mapsto F_{P^x}(P,Q)\) and \(x\mapsto F_{Q^x}(P,Q)\) are nondecreasing on \((-\infty,1]\) and nonincreasing on \([1,\infty)\). Consequently, the symmetrized map \(x\mapsto F^{\mathrm{pol}}_{x}(P,Q)\) has the same two-range monotonicity, attains a global maximum at \(x=1\), and restricts on \([-1,1]\) to a monotone bridge between \(F^{\mathrm M}\), \(F^{\mathrm H}\), and \(F^{\mathrm U}\).
Open Problem 3: recovery of \(z\)-fidelities and Log-Euclidean fidelity. We give a pointwise positive-definite answer. For every \(z\ge\frac{1}{2}\), and also for the Log-Euclidean fidelity, each fixed-pair value is realized as a generalized fidelity along each of the one-sided polar paths \(R=P^x\) and \(R=Q^x\), with generally pair-dependent and nonunique parameters. Conversely, for \(d\ge2\) and \(0<z<\frac{1}{2}\), such pointwise realization fails in general, even if one allows interior fidelities. Thus the pointwise realization problem has an affirmative answer for \(z\ge\frac{1}{2}\) and a negative answer in general below this threshold. We do not claim a universal base, nor a canonical rule \(R=R_z(P,Q)\), valid for all pairs.
Open Problem 5: other bases for Holevo fidelity. We solve the fixed-pair Holevo-base problem. First, we classify all bases satisfying the natural sufficient polar condition \[\operatorname{Pol}(R^{1/2}P^{1/2})=\operatorname{Pol}(R^{1/2}Q^{1/2}). \label{eq:holevo-polar-condition}\tag{1}\] We then remove this sufficient condition and solve the full fixed-pair equation \[F_R(P,Q)=F^{\mathrm H}(P,Q)\] as an equality of complex numbers. The polar condition turns out to describe only the distinguished stratum \(W=I\) inside the full solution set. As an additional global consequence, the identity is, up to positive scaling, the unique universal Holevo base.
Open Problem 7: unitary factors and \(SU(d)\). We answer the reverse-inclusion question negatively. More precisely, for each fixed pair \(P,Q\in\mathbb{P}_d\), we classify exactly which unitary factors of generalized fidelity can arise from some base \(R\): a unitary \(W\) can arise if and only if \[P^{-1/2}Q^{1/2}W\sim\mathbb{P}_d.\] This recovers the known special-unitary constraint and shows that, for \(d\ge2\), not every element of \(SU(d)\) can arise as a unitary factor of generalized fidelity.
Open Problems 1, 2, and 6. We do not solve the full data-processing problem for generalized Bures distance (Open Problem 1), nor the convexity and SDP questions (Open Problems 2 and 6). We only prove a partial fidelity-form data-processing result for the symmetrized polar fidelities under commutative-interface hypotheses.
Two mechanisms behind the main results are worth emphasizing. First, the monotonicity theorem yields exact realization of the full interval \[[F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)]\] by the one-sided and symmetrized polar paths on \([-1,1]\); this is the mechanism behind the positive part of the recovery theorem for \(z\)-fidelities and for the Log-Euclidean fidelity. Second, the same fixed-pair parameterization that describes Holevo bases also controls the unitary factors of generalized fidelity.
All structural results below are stated for strictly positive definite matrices. Negative powers and invertible polar decomposition are essential throughout. Positive semidefinite matrices enter only as an auxiliary limiting device in the proof of Theorem 7(iii); no semidefinite extension of generalized fidelity or of the \(x\)-polar family is claimed here.
The paper is organized as follows. Section 2 collects notation and basic identities for generalized fidelity and polar decomposition. Section 3 proves the two-range monotonicity of the one-sided and symmetrized \(x\)-polar families. Section 4 proves exact realization of the interval \([F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)]\), derives the pointwise recovery of \(z\)-fidelities and of the Log-Euclidean fidelity, and establishes the commutative-interface data-processing result. Section 5 studies Holevo bases and unitary factors: first the polar slice of Holevo bases, then the full fixed-pair Holevo-base equation, and finally the unitary-factor classification and its special-unitary consequences.
For \(R,P,Q\in\mathbb{P}_d\), the generalized fidelity of Afham–Ferrie is \[F_R(P,Q):= \operatorname{Tr}\!\left[\bigl(R^{1/2}PR^{1/2}\bigr)^{1/2}R^{-1} \bigl(R^{1/2}QR^{1/2}\bigr)^{1/2}\right]. \label{eq:generalized-fidelity}\tag{2}\] An equivalent unitary-factor formula is \[F_R(P,Q)=\operatorname{Tr}\!\bigl[Q^{1/2}U_QU_P^*P^{1/2}\bigr], \label{eq:generalized-fidelity-unitary}\tag{3}\] where \[U_P:=\operatorname{Pol}(P^{1/2}R^{1/2}), \qquad U_Q:=\operatorname{Pol}(Q^{1/2}R^{1/2}).\] The special choices \(R=P,Q\), \(R=I\), and \(R=P^{-1},Q^{-1}\) recover the standard named fidelities: Uhlmann fidelity [5]–[8], Holevo fidelity [9], [10], and Matsumoto fidelity [11], [12]. More precisely, \[\begin{align} F^{\mathrm U}(P,Q)&=F_P(P,Q)=F_Q(P,Q),\\ F^{\mathrm H}(P,Q)&=F_I(P,Q),\\ F^{\mathrm M}(P,Q)&=F_{P^{-1}}(P,Q)=F_{Q^{-1}}(P,Q). \end{align}\] For \(z>0\), we also write \[F_z(P,Q):=\operatorname{Tr}\!\left[\left(P^{\frac{1}{2z}}Q^{\frac{1}{2z}}\right)^z\right],\] where the power is the principal matrix power. We shall write \[T\sim\mathbb{P}_d\] to mean that the invertible matrix \(T\) is similar to a positive definite matrix, i.e., there exists \(S\in GL_d(\mathbb{C})\) such that \(T=SDS^{-1}\) for some \(D\in\mathbb{P}_d\). Equivalently, \(T\) is diagonalizable and has strictly positive real spectrum. We shall also use the notation \[S^{-*}:=(S^{-1})^* \qquad (S\in GL_d(\mathbb{C})).\]
The limiting Log-Euclidean fidelity, obtained as the \(z\to\infty\) limit of the \(z\)-fidelities, is \[F^{\mathrm{LE}}(P,Q):=\operatorname{Tr}\!\left[\exp\!\left(\frac{\log P+\log Q}{2}\right)\right],\] see [1], [13]. We shall use the following standard positive representative of the \(z\)-fidelity.
Lemma 1. For every \(z>0\) and every \(P,Q\in\mathbb{P}_d\), \[F_z(P,Q) = \operatorname{Tr}\!\left[\left(P^{\frac{1}{4z}}Q^{\frac{1}{2z}}P^{\frac{1}{4z}}\right)^z\right] = \operatorname{Tr}\!\left[\left(Q^{\frac{1}{4z}}P^{\frac{1}{2z}}Q^{\frac{1}{4z}}\right)^z\right].\] In particular, \(F_z(P,Q)\) is a positive real number. Moreover, the displayed positive formulas define the unique continuous extension of \(F_z\) to positive semidefinite \(P,Q\).
Proof. This is the \(\alpha=\frac{1}{2}\) instance of the standard positive trace functional underlying the \(\alpha\)-\(z\) Rényi divergences [2]. Equivalently, \(P^{1/(2z)}Q^{1/(2z)}\) is similar to \(P^{1/(4z)}Q^{1/(2z)}P^{1/(4z)}\), and principal powers are preserved under similarity. The positive-semidefinite extension follows from the displayed positive formulas and continuity of the functional calculus \(X\mapsto X^t\) for \(t>0\). ◻
We shall also use the standard variational formula for Uhlmann fidelity and its immediate generalized-fidelity consequence.
Lemma 2. For every \(A,B\in\mathbb{P}_d\), \[F^{\mathrm U}(A,B)=\max_{U\in U(d)} \left|\operatorname{Tr}\!\left(B^{1/2}UA^{1/2}\right)\right|.\] Moreover, for every \(R\in\mathbb{P}_d\), \[|F_R(A,B)|\le F^{\mathrm U}(A,B).\]
Proof. The first formula is the standard trace-norm variational characterization of Uhlmann fidelity [8]. The second bound is precisely [1]. ◻
Lemma 3. For all \(c,a,b>0\) and all \(P,Q,R\in\mathbb{P}_d\), \[F_{cR}(P,Q)=F_R(P,Q), \qquad F_R(aP,bQ)=\sqrt{ab}\,F_R(P,Q).\] Consequently, for every \(\mathcal{F}\in\{F^{\mathrm M},F^{\mathrm H},F^{\mathrm U},F_z,F^{\mathrm{LE}}\}\), \[\mathcal{F}(aP,bQ)=\sqrt{ab}\,\mathcal{F}(P,Q).\]
Proof. The identities for \(F_R\) are precisely the scaling property of generalized fidelity [1]. The homogeneity of \(F^{\mathrm U},F^{\mathrm H},F^{\mathrm M}\) follows from the standard reductions to named fidelities [1]. The homogeneity of \(F_z\) and \(F^{\mathrm{LE}}\) is immediate from their definitions. ◻
Remark 1. Let \(P,Q\in\mathbb{P}_d\), set \[p:=\operatorname{Tr}P,\qquad q:=\operatorname{Tr}Q,\qquad \rho:=P/p,\qquad \sigma:=Q/q.\] Then, for every \(\mathcal{F}\in\{F^{\mathrm M},F^{\mathrm H},F^{\mathrm U},F_z,F^{\mathrm{LE}}\}\), \[\mathcal{F}(P,Q)=\sqrt{pq}\,\mathcal{F}(\rho,\sigma).\] Thus pointwise inequalities and data-processing inequalities proved for faithful states extend immediately to arbitrary \(P,Q\in\mathbb{P}_d\), as long as the output matrices remain positive definite.
In this paper, by a normalized monotone quantum fidelity on faithful states we mean a function \(G\) which is positive, symmetric, satisfies \(G(\rho,\rho)=1\), agrees with the classical fidelity on commuting faithful states, and is monotone increasing under quantum channels whenever the two output states are faithful.
Lemma 4. Let \(G\) be a normalized monotone quantum fidelity on faithful states. Then, for all faithful states \(\rho,\sigma\), \[F^{\mathrm M}(\rho,\sigma)\le G(\rho,\sigma)\le F^{\mathrm U}(\rho,\sigma).\]
Proof. This is the extremal characterization of Matsumoto and Uhlmann fidelities among monotone quantum fidelities extending the classical fidelity: \(F^{\mathrm M}\) is the minimal such extension and \(F^{\mathrm U}\) is the maximal one [11], [12], [14]. Applying this characterization to \(G\) gives the stated inequalities. ◻
For \(P,Q\in\mathbb{P}_d\), define \[\begin{align} \Phi_P(x) &:=F_{P^x}(P,Q) =\operatorname{Tr}\!\left[P^{\frac{1-x}{2}} \left(P^{\frac{x}{2}}QP^{\frac{x}{2}}\right)^{1/2}\right],\\ \Phi_Q(x) &:=F_{Q^x}(P,Q) =\operatorname{Tr}\!\left[Q^{\frac{1-x}{2}} \left(Q^{\frac{x}{2}}PQ^{\frac{x}{2}}\right)^{1/2}\right]. \end{align}\]
Remark 2. In general the generalized fidelity \(F_R(P,Q)\) need not be real-valued. By contrast, for every \(x\in\mathbb{R}\), both \(\Phi_P(x)\) and \(\Phi_Q(x)\) are positive real numbers, because for \(A,B\in\mathbb{P}_d\), \[\operatorname{Tr}(AB)=\operatorname{Tr}(A^{1/2}BA^{1/2})>0.\]
The symmetrized \(x\)-polar fidelity is \[F^{\mathrm{pol}}_{x}(P,Q):=\frac{\Phi_P(x)+\Phi_Q(x)}{2}, \qquad x\in\mathbb{R}. \label{eq:x-polar-def}\tag{4}\] In particular, \(F^{\mathrm{pol}}_{x}(P,Q)\) is defined for every real \(x\). The main result below shows that the one-sided and symmetrized polar families are nondecreasing on \((-\infty,1]\) and nonincreasing on \([1,\infty)\). For the interpolation statements, we will mainly use the increasing branch on \([-1,1]\). Thus \[F^{\mathrm{pol}}_{-1}(P,Q)=F^{\mathrm M}(P,Q), \qquad F^{\mathrm{pol}}_{0}(P,Q)=F^{\mathrm H}(P,Q), \qquad F^{\mathrm{pol}}_{1}(P,Q)=F^{\mathrm U}(P,Q).\] We shall use the following elementary facts about polar decomposition and products of positive definite matrices.
Lemma 5. Let \(X\in GL_d(\mathbb{C})\) and \(U,V\in U(d)\). Then \[\operatorname{Pol}(UXV)=U\,\operatorname{Pol}(X)\,V, \qquad \operatorname{Pol}(X^*)=\operatorname{Pol}(X)^*.\] Moreover, for every unitary \(W\in U(d)\), \[\operatorname{Pol}(X)=W \iff W^*X\in\mathbb{P}_d \iff XW^*\in\mathbb{P}_d.\] Finally, for \(A,B\in\mathbb{P}_d\), \[AB\in\mathbb{P}_d\iff AB=BA.\]
Proof. The polar identities follow immediately from uniqueness of the polar decomposition for invertible matrices. The last equivalence follows from the standard fact that, for Hermitian \(A,B\), the product \(AB\) is Hermitian if and only if \(AB=BA\) [15]. ◻
Lemma 6. Let \(A,B\in\mathbb{P}_d\) and let \(f:(0,\infty)\to[0,\infty)\) be nondecreasing. Then \[\operatorname{Tr}\bigl[f(A)A^{1/2}B^{1/2}\bigr] \le \operatorname{Tr}\Bigl[f(A)\bigl(A^{1/2}BA^{1/2}\bigr)^{1/2}\Bigr].\]
Proof. This is the case \(s=\frac{1}{2}\) of [16]. ◻
Lemma 7. Let \(A,B\in\mathbb{P}_d\) and let \(g:(0,\infty)\to\mathbb{R}\) be such that the function \[t\longmapsto t^{1/2}g(t)\] is nonnegative and nonincreasing on \((0,\infty)\). Then \[\operatorname{Tr}\Bigl[g(A)\bigl(A^{1/2}BA^{1/2}\bigr)^{1/2}\Bigr] \le \operatorname{Tr}\bigl[g(A)A^{1/2}B^{1/2}\bigr].\]
Proof. This is the case \(s=\frac{1}{2}\) of [16]. ◻
The main result of this section is the following two-range monotonicity theorem for the one-sided polar paths.
Theorem 3. Let \(P,Q\in\mathbb{P}_d\). Then the maps \[x\longmapsto \Phi_P(x)=F_{P^x}(P,Q), \qquad x\longmapsto \Phi_Q(x)=F_{Q^x}(P,Q)\] are nondecreasing on \((-\infty,1]\) and nonincreasing on \([1,\infty)\). Equivalently, for every \(y\le x\le 1\), \[\Phi_P(y)\le \Phi_P(x), \qquad \Phi_Q(y)\le \Phi_Q(x),\] while for every \(1\le y\le x\), \[\Phi_P(y)\ge \Phi_P(x), \qquad \Phi_Q(y)\ge \Phi_Q(x).\] In particular, each one-sided polar fidelity is nondecreasing up to \(x=1\), nonincreasing after \(x=1\), and attains a global maximum at \(x=1\), where \[\Phi_P(1)=\Phi_Q(1)=F^{\mathrm U}(P,Q).\]
Proof. We prove the claim for \(\Phi_P\); the proof for \(\Phi_Q\) is similar after swapping \(P\) and \(Q\).
Fix \(y<x\), and set \[\Delta:=x-y>0, \qquad A:=P^\Delta, \qquad B:=P^{y/2}QP^{y/2}, \qquad \beta:=\frac{1-x}{2\Delta}. \label{eq:one-sided-monotonicity-setup}\tag{5}\] Then \[\begin{align} \operatorname{Tr}\bigl[A^\beta A^{1/2}B^{1/2}\bigr] &= \operatorname{Tr}\!\left[P^{\beta\Delta}P^{\Delta/2}\bigl(P^{y/2}QP^{y/2}\bigr)^{1/2}\right] \notag\\ &= \operatorname{Tr}\!\left[P^{\frac{1-x}{2}}P^{\frac{x-y}{2}}\bigl(P^{y/2}QP^{y/2}\bigr)^{1/2}\right] \notag \\ &= \operatorname{Tr}\!\left[P^{\frac{1-y}{2}}\bigl(P^{y/2}QP^{y/2}\bigr)^{1/2}\right] \notag\\ &=\Phi_P(y), \label{eq:one-sided-monotonicity-left} \end{align}\tag{6}\] while \[\begin{align} \operatorname{Tr}\Bigl[A^\beta\bigl(A^{1/2}BA^{1/2}\bigr)^{1/2}\Bigr] &= \operatorname{Tr}\!\left[P^{\beta\Delta} \bigl(P^{\Delta/2}P^{y/2}QP^{y/2}P^{\Delta/2}\bigr)^{1/2}\right] \notag \\ &= \operatorname{Tr}\!\left[P^{\frac{1-x}{2}}\bigl(P^{x/2}QP^{x/2}\bigr)^{1/2}\right] \notag \\ &=\Phi_P(x). \label{eq:one-sided-monotonicity-right} \end{align}\tag{7}\]
It remains to compare these two expressions in the two monotonicity ranges stated in the theorem. First assume that \(y<x\le 1\). Then \(\beta\ge 0\), so the function \(f(t):=t^\beta\) is nonnegative and nondecreasing on \((0,\infty)\). By Lemma 6, \[\operatorname{Tr}\bigl[A^\beta A^{1/2}B^{1/2}\bigr] \le \operatorname{Tr}\Bigl[A^\beta\bigl(A^{1/2}BA^{1/2}\bigr)^{1/2}\Bigr].\] Using 6 –7 , we obtain \[\Phi_P(y)\le \Phi_P(x).\]
Next assume that \(1\le y<x\). Then \(\beta\le 0\). Set \(g(t):=t^\beta\). Then \[t^{1/2}g(t)=t^{\beta+1/2}=t^{\frac{1-y}{2\Delta}},\] which is nonnegative and nonincreasing on \((0,\infty)\) because \(y\ge 1\). Therefore Lemma 7 gives \[\operatorname{Tr}\Bigl[A^\beta\bigl(A^{1/2}BA^{1/2}\bigr)^{1/2}\Bigr] \le \operatorname{Tr}\bigl[A^\beta A^{1/2}B^{1/2}\bigr].\] Again using 6 –7 , we conclude that \[\Phi_P(x)\le \Phi_P(y).\]
The case \(x=y\) is trivial. This proves the theorem. ◻
Corollary 1. Let \(P,Q\in\mathbb{P}_d\). Then the map \[x\longmapsto F^{\mathrm{pol}}_{x}(P,Q)\] is nondecreasing on \((-\infty,1]\) and nonincreasing on \([1,\infty)\). In particular, \(F^{\mathrm{pol}}_{x}(P,Q)\) is nondecreasing up to \(x=1\), nonincreasing after \(x=1\), attains a global maximum at \(x=1\), and \[F^{\mathrm{pol}}_{x}(P,Q)\le F^{\mathrm{pol}}_{1}(P,Q)=F^{\mathrm U}(P,Q) \qquad (x\in\mathbb{R}).\] Its restriction to \([-1,1]\) is therefore nondecreasing.
Moreover, if \([P,Q]=0\), then for every \(x\in\mathbb{R}\), \[\Phi_P(x)=\Phi_Q(x)=F^{\mathrm{pol}}_{x}(P,Q)=\operatorname{Tr}(P^{1/2}Q^{1/2}).\] Hence the maximizer at \(x=1\) need not be unique; the monotonicity assertions above are weak in general.
Proof. The two-range monotonicity follows immediately by averaging the corresponding inequalities for \(\Phi_P\) and \(\Phi_Q\) in Theorem 3. Since \(\Phi_P(1)=\Phi_Q(1)=F^{\mathrm U}(P,Q)\), we have \(F^{\mathrm{pol}}_{1}(P,Q)=F^{\mathrm U}(P,Q)\), and the asserted maximum and upper bound follow.
If \([P,Q]=0\), then \[\Phi_P(x) = \operatorname{Tr}\!\left[ P^{\frac{1-x}{2}} \left(P^{\frac{x}{2}}QP^{\frac{x}{2}}\right)^{1/2} \right] = \operatorname{Tr}\!\left[ P^{\frac{1-x}{2}}(P^xQ)^{1/2} \right] = \operatorname{Tr}(P^{1/2}Q^{1/2}).\] The same computation with \(P\) and \(Q\) interchanged gives \[\Phi_Q(x)=\operatorname{Tr}(P^{1/2}Q^{1/2}).\] Thus \[\Phi_P(x)=\Phi_Q(x)=F^{\mathrm{pol}}_{x}(P,Q)=\operatorname{Tr}(P^{1/2}Q^{1/2})\] for every \(x\in\mathbb{R}\). This proves the final assertion. ◻
Corollary 2. Let \(P,Q\in\mathbb{P}_d\). For every \(x\in[-1,1]\), \[F^{\mathrm M}(P,Q)=F^{\mathrm{pol}}_{-1}(P,Q) \le F^{\mathrm{pol}}_{x}(P,Q) \le F^{\mathrm{pol}}_{1}(P,Q)=F^{\mathrm U}(P,Q),\] and \[F^{\mathrm{pol}}_{0}(P,Q)=F^{\mathrm H}(P,Q).\] Hence \[F^{\mathrm M}(P,Q)\le F^{\mathrm H}(P,Q)\le F^{\mathrm U}(P,Q)\] is refined by the monotone bridge \(x\mapsto F^{\mathrm{pol}}_{x}(P,Q)\).
Proof. By Corollary 1, the map \[x\longmapsto F^{\mathrm{pol}}_{x}(P,Q)\] is nondecreasing on \([-1,1]\). Using \[F^{\mathrm{pol}}_{-1}(P,Q)=F^{\mathrm M}(P,Q),\qquad F^{\mathrm{pol}}_{0}(P,Q)=F^{\mathrm H}(P,Q),\qquad F^{\mathrm{pol}}_{1}(P,Q)=F^{\mathrm U}(P,Q),\] the stated inequalities follow immediately. ◻
Remark 4. Theorem 3 answers Open Problem 4 of [1] in the affirmative and proves a stronger global statement. Indeed, for every fixed \(P,Q\in\mathbb{P}_d\), the one-sided polar paths \(x\mapsto F_{P^x}(P,Q)\) and \(x\mapsto F_{Q^x}(P,Q)\) are nondecreasing on \((-\infty,1]\) and nonincreasing on \([1,\infty)\). Averaging the two one-sided inequalities gives the same two-range monotonicity for the symmetrized family \(x\mapsto F^{\mathrm{pol}}_{x}(P,Q)\). In particular, on the interval \([-1,1]\), these polar families provide a monotone interpolation from Matsumoto fidelity at \(x=-1\) to Uhlmann fidelity at \(x=1\), passing through Holevo fidelity at \(x=0\).
The next result records the exact realization of the extremal interval \([F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)]\) on \([-1,1]\), together with the corresponding pointwise recovery statement along the one-sided and symmetrized polar paths.
Proposition 5. Let \(P,Q\in\mathbb{P}_d\). For \(x\in[-1,1]\), recall that \[F^{\mathrm{pol}}_{x}(P,Q):=\frac{F_{P^x}(P,Q)+F_{Q^x}(P,Q)}{2}.\] Then the following statements hold.
The restrictions to \([-1,1]\) realize exactly the extremal interval: \[\begin{align} \{F_{P^x}(P,Q):x\in[-1,1]\} &= \{F_{Q^x}(P,Q):x\in[-1,1]\} \\ &= \{F^{\mathrm{pol}}_{x}(P,Q):x\in[-1,1]\} \\ &= [F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)]. \end{align}\] In particular, for every \(x\in[-1,1]\), \[F^{\mathrm M}(P,Q)\le F^{\mathrm{pol}}_{x}(P,Q)\le F^{\mathrm U}(P,Q).\]
Let \(\mathcal{F}:\mathbb{P}_d\times\mathbb{P}_d\to\mathbb{R}\) be any function such that \[F^{\mathrm M}(P,Q)\le \mathcal{F}(P,Q)\le F^{\mathrm U}(P,Q) \qquad (P,Q\in\mathbb{P}_d).\] Then for every fixed pair \(P,Q\in\mathbb{P}_d\), there exist generally nonunique parameters \[\theta_P,\theta_Q,\vartheta\in[-1,1]\] (depending on \(\mathcal{F},P,Q\)) such that \[\mathcal{F}(P,Q) = F_{P^{\theta_P}}(P,Q) = F_{Q^{\theta_Q}}(P,Q) = F^{\mathrm{pol}}_{\vartheta}(P,Q).\]
Proof. By Theorem 3, the maps \[x\mapsto F_{P^x}(P,Q), \qquad x\mapsto F_{Q^x}(P,Q)\] are nondecreasing on \([-1,1]\). They are continuous there by functional calculus and continuity of the square-root map. Since their endpoint values are \[F_{P^{-1}}(P,Q)=F_{Q^{-1}}(P,Q)=F^{\mathrm M}(P,Q), \qquad F_P(P,Q)=F_Q(P,Q)=F^{\mathrm U}(P,Q),\] each of their ranges on \([-1,1]\) is exactly \([F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)]\).
The same argument applies to \(x\mapsto F^{\mathrm{pol}}_{x}(P,Q)\), using Corollary 1 and the endpoint values \[F^{\mathrm{pol}}_{-1}(P,Q)=F^{\mathrm M}(P,Q), \qquad F^{\mathrm{pol}}_{1}(P,Q)=F^{\mathrm U}(P,Q).\] This proves (i). For (ii), fix \(P,Q\in\mathbb{P}_d\). By assumption, \[\mathcal{F}(P,Q)\in[F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)].\] The required parameters then follow from the range statement in (i). ◻
Remark 6. Proposition 5(ii) is a preparatory realization result. It shows that any function whose values lie pointwise between \(F^{\mathrm M}\) and \(F^{\mathrm U}\) can already be realized, for each fixed pair \(P,Q\in\mathbb{P}_d\), as a genuine generalized-fidelity value along either one-sided polar path \(R=P^x\) or \(R=Q^x\). The realizing parameter is generally pair-dependent and generally nonunique; in particular, if \([P,Q]=0\), then all named fidelities coincide and every parameter in \([-1,1]\) gives the same value.
The connection with \(z\)-fidelities and the Log-Euclidean fidelity is made in Theorem 7, where this realization mechanism is applied after establishing the bounds \[F^{\mathrm M}(P,Q)\le F_z(P,Q)\le F^{\mathrm U}(P,Q) \quad (z\ge 1/2), \qquad F^{\mathrm M}(P,Q)\le F^{\mathrm{LE}}(P,Q)\le F^{\mathrm U}(P,Q).\]
For the next result, we use the term interior fidelity in the sense of Afham–Ferrie: a finite convex combination of generalized fidelities, \[\sum_{i=1}^n \mu_i F_{R_i}(P,Q),\] where \(\mu=(\mu_1,\dots,\mu_n)\) is a probability vector and \(R_1,\dots,R_n\in\mathbb{P}_d\).
Theorem 7. The following hold.
For every \(P,Q\in\mathbb{P}_d\) and every \(z\in[1/2,\infty)\), there exist generally nonunique parameters \[\theta_P,\theta_Q\in[-1,1] \qquad\text{(depending on \(z,P,Q\))}\] such that \[F_z(P,Q) = F_{P^{\theta_P}}(P,Q) = F_{Q^{\theta_Q}}(P,Q).\] In particular, for every \(z\ge 1/2\), every \(z\)-fidelity value admits a pointwise representation as a generalized-fidelity value with pair-dependent base.
For every \(P,Q\in\mathbb{P}_d\), there exist generally nonunique parameters \[\theta_P,\theta_Q\in[-1,1] \qquad\text{(depending on \(P,Q\))}\] such that \[F^{\mathrm{LE}}(P,Q) = F_{P^{\theta_P}}(P,Q) = F_{Q^{\theta_Q}}(P,Q).\] In particular, every Log-Euclidean fidelity value also admits a pointwise representation as a generalized-fidelity value with pair-dependent base.
Assume \(d\ge 2\). For every \(z\in(0,1/2)\), there exist faithful states \(\rho,\sigma\in\mathbb{P}_d\) such that \(F_z(\rho,\sigma)\) cannot be represented either as a generalized fidelity \(F_R(\rho,\sigma)\) for any \(R\in\mathbb{P}_d\), or even as an interior fidelity \[\sum_{i=1}^n \mu_i F_{R_i}(\rho,\sigma).\] In particular, even allowing pair-dependent bases, \(F_z\) is in general neither a generalized fidelity nor an interior fidelity on \(\mathbb{P}_d\).
Proof. We begin with (i). Let \(z\in[1/2,\infty)\) and fix \(P,Q\in\mathbb{P}_d\). Write \[p:=\operatorname{Tr}P,\qquad q:=\operatorname{Tr}Q,\qquad \rho:=P/p,\qquad \sigma:=Q/q.\] By Lemma [lem:joint-homogeneity], \[F_z(P,Q)=\sqrt{pq}\,F_z(\rho,\sigma),\qquad F^{\mathrm M}(P,Q)=\sqrt{pq}\,F^{\mathrm M}(\rho,\sigma),\qquad F^{\mathrm U}(P,Q)=\sqrt{pq}\,F^{\mathrm U}(\rho,\sigma).\] If \([\rho,\sigma]=0\), then \[F_z(\rho,\sigma) = \operatorname{Tr}\!\left[\left(\rho^{\frac{1}{2z}}\sigma^{\frac{1}{2z}}\right)^z\right] = \operatorname{Tr}(\rho^{1/2}\sigma^{1/2}),\] so \(F_z\) agrees with the classical fidelity on commuting faithful states. By Lemma 1, \(F_z\) is symmetric and positive, and \[F_z(\rho,\rho)=\operatorname{Tr}\rho=1.\] Moreover, \[F_z(\rho,\sigma) = \operatorname{Tr}\!\left[\left(\sigma^{\frac{1}{4z}}\rho^{\frac{1}{2z}} \sigma^{\frac{1}{4z}}\right)^z\right]\] is the \(\alpha=\frac{1}{2}\) instance of the standard \(Q_{\alpha,z}\)-quantity associated with the \(\alpha\)-\(z\) Rényi divergence. Since \[D_{\alpha,z}(\rho\|\sigma) = \frac{1}{\alpha-1}\log Q_{\alpha,z}(\rho\|\sigma),\] the data-processing inequality for \(D_{\alpha,z}\) at \(\alpha=\frac{1}{2}\) is equivalent to monotone increase of \(Q_{\alpha,z}\), hence of \(F_z\), under quantum channels. Therefore \(F_z\) is monotone for \(z\ge\frac{1}{2}\) by [17].
Thus \(F_z\) satisfies the hypotheses of Lemma 4, and so \[F^{\mathrm M}(\rho,\sigma)\le F_z(\rho,\sigma)\le F^{\mathrm U}(\rho,\sigma).\] Multiplying by \(\sqrt{pq}\) yields \[F^{\mathrm M}(P,Q)\le F_z(P,Q)\le F^{\mathrm U}(P,Q).\] Now apply Proposition 5(ii) with \(\mathcal{F}=F_z\). This gives generally nonunique parameters \(\theta_P,\theta_Q\in[-1,1]\) such that \[F_z(P,Q)=F_{P^{\theta_P}}(P,Q)=F_{Q^{\theta_Q}}(P,Q).\] This proves (i).
For (ii), fix \(P,Q\in\mathbb{P}_d\). By the Lie–Trotter product formula, \[F^{\mathrm{LE}}(P,Q)=\lim_{z\to\infty}F_z(P,Q);\] see, for example, [18]. Passing to the limit in \[F^{\mathrm M}(P,Q)\le F_z(P,Q)\le F^{\mathrm U}(P,Q) \qquad (z\ge 1/2)\] gives \[F^{\mathrm M}(P,Q)\le F^{\mathrm{LE}}(P,Q)\le F^{\mathrm U}(P,Q).\] Applying Proposition 5(ii) with \(\mathcal{F}=F^{\mathrm{LE}}\), we obtain generally nonunique parameters \(\theta_P,\theta_Q\in[-1,1]\) such that \[F^{\mathrm{LE}}(P,Q)=F_{P^{\theta_P}}(P,Q)=F_{Q^{\theta_Q}}(P,Q).\] This proves (ii).
We now prove (iii). Fix \(z\in(0,1/2)\). To show failure in general, it suffices to exhibit one faithful-state counterexample.
Choose unit vectors \(\psi,\phi\in\mathbb{C}^d\) with \[0<c:=|\langle\psi,\phi\rangle|<1,\] and define the rank-one projections \[P_0:=|\psi\rangle\langle\psi|, \qquad Q_0:=|\phi\rangle\langle\phi|.\] Since \(P_0^t=P_0\) and \(Q_0^t=Q_0\) for every \(t>0\), Lemma 1 gives \[F_z(P_0,Q_0) = \operatorname{Tr}\!\left[\left(P_0^{\frac{1}{4z}}Q_0^{\frac{1}{2z}}P_0^{\frac{1}{4z}}\right)^z\right] = \operatorname{Tr}\!\left[(P_0Q_0P_0)^z\right].\] But \(P_0Q_0P_0=c^2P_0\), hence \[F_z(P_0,Q_0)=\operatorname{Tr}\!\left[(c^2P_0)^z\right]=c^{2z}.\] On the other hand, \[P_0^{1/2}Q_0P_0^{1/2}=P_0Q_0P_0=c^2P_0,\] and therefore \[F^{\mathrm U}(P_0,Q_0)=\operatorname{Tr}\!\left[(c^2P_0)^{1/2}\right]=c.\] Because \(0<z<1/2\) and \(0<c<1\), we obtain \[c^{2z}>c, \qquad\text{i.e.}\qquad F_z(P_0,Q_0)>F^{\mathrm U}(P_0,Q_0).\]
We now regularize this singular pair to a full-rank pair. For \(\varepsilon\in(0,1)\), set \[P_\varepsilon:=(1-\varepsilon)P_0+\varepsilon I/d, \qquad Q_\varepsilon:=(1-\varepsilon)Q_0+\varepsilon I/d.\] Then \(P_\varepsilon,Q_\varepsilon\in\mathbb{P}_d\), and \[P_\varepsilon\to P_0, \qquad Q_\varepsilon\to Q_0 \qquad (\varepsilon\downarrow0).\] Moreover, \[\operatorname{Tr}P_\varepsilon=(1-\varepsilon)\operatorname{Tr}P_0+\varepsilon\operatorname{Tr}(I/d)=1, \qquad \operatorname{Tr}Q_\varepsilon=(1-\varepsilon)\operatorname{Tr}Q_0+\varepsilon\operatorname{Tr}(I/d)=1,\] so \(P_\varepsilon\) and \(Q_\varepsilon\) are faithful states.
The standard formula \[F^{\mathrm U}(A,B)=\operatorname{Tr}\!\left[\left(A^{1/2}BA^{1/2}\right)^{1/2}\right]\] extends \(F^{\mathrm U}\) continuously to positive semidefinite pairs. Likewise, by Lemma 1, the formula \[F_z(A,B) = \operatorname{Tr}\!\left[\left(A^{\frac{1}{4z}}B^{\frac{1}{2z}}A^{\frac{1}{4z}}\right)^z\right]\] extends \(F_z\) continuously to positive semidefinite pairs. Hence \[F_z(P_\varepsilon,Q_\varepsilon)\to c^{2z}, \qquad F^{\mathrm U}(P_\varepsilon,Q_\varepsilon)\to c.\] Since \(c^{2z}-c>0\), for all sufficiently small \(\varepsilon>0\) we still have \[F_z(P_\varepsilon,Q_\varepsilon)>F^{\mathrm U}(P_\varepsilon,Q_\varepsilon).\] Fix such an \(\varepsilon\).
If the value \(F_z(P_\varepsilon,Q_\varepsilon)\) admitted a generalized-fidelity representation, there would exist \(R\in\mathbb{P}_d\) such that \[F_z(P_\varepsilon,Q_\varepsilon)=F_R(P_\varepsilon,Q_\varepsilon).\] But Lemma 2 gives \[|F_R(P_\varepsilon,Q_\varepsilon)| \le F^{\mathrm U}(P_\varepsilon,Q_\varepsilon),\] contradicting the strict inequality above. The same argument also rules out a representation of \(F_z(P_\varepsilon,Q_\varepsilon)\) as \(\operatorname{Re}F_R(P_\varepsilon,Q_\varepsilon)\), since \[\operatorname{Re}F_R(P_\varepsilon,Q_\varepsilon) \le |F_R(P_\varepsilon,Q_\varepsilon)| \le F^{\mathrm U}(P_\varepsilon,Q_\varepsilon).\]
Likewise, if the value \(F_z(P_\varepsilon,Q_\varepsilon)\) admitted an interior-fidelity representation, then there would exist a probability vector \(\mu=(\mu_1,\dots,\mu_n)\) and bases \(R_1,\dots,R_n\in\mathbb{P}_d\) such that \[F_z(P_\varepsilon,Q_\varepsilon) = \sum_{i=1}^n \mu_i F_{R_i}(P_\varepsilon,Q_\varepsilon).\] Since \(F_z(P_\varepsilon,Q_\varepsilon)\) is a positive real number, we may take absolute values and estimate \[F_z(P_\varepsilon,Q_\varepsilon) = \left| \sum_{i=1}^n \mu_i F_{R_i}(P_\varepsilon,Q_\varepsilon) \right| \le \sum_{i=1}^n \mu_i\,|F_{R_i}(P_\varepsilon,Q_\varepsilon)| \le F^{\mathrm U}(P_\varepsilon,Q_\varepsilon),\] again a contradiction. Thus, taking \(\rho:=P_\varepsilon\) and \(\sigma:=Q_\varepsilon\), we obtain the faithful states asserted in (iii). In particular, \(F_z\) is, in general, neither a generalized fidelity nor an interior fidelity on \(\mathbb{P}_d\). This proves (iii). ◻
Corollary 3. Let \(P,Q\in\mathbb{P}_d\). Then for every \(z\in[1/2,\infty)\), there exists a generally nonunique parameter \(\theta_z(P,Q)\in[-1,1]\) such that \[F_z(P,Q)=F^{\mathrm{pol}}_{\theta_z(P,Q)}(P,Q).\] Likewise, there exists a generally nonunique parameter \(\theta_{\mathrm{LE}}(P,Q)\in[-1,1]\) such that \[F^{\mathrm{LE}}(P,Q)=F^{\mathrm{pol}}_{\theta_{\mathrm{LE}}(P,Q)}(P,Q).\]
Proof. This is Proposition 5(ii) applied with \(\mathcal{F}=F_z\) and \(\mathcal{F}=F^{\mathrm{LE}}\), using the interval bounds proved in Theorem 7(i)–(ii). ◻
Remark 8. Theorem 7 gives the pointwise positive-definite answer to Open Problem 3 of [1]. For \(z\ge 1/2\), every \(z\)-fidelity value is realized as a generalized-fidelity value along each of the one-sided polar paths \(R=P^x\) and \(R=Q^x\), with a parameter depending in general on the pair \((P,Q)\). The same holds for the Log-Euclidean fidelity. Conversely, when \(d\ge2\) and \(0<z<1/2\), such a pointwise realization fails in general, even if one allows finite convex combinations of generalized fidelities, i.e. interior fidelities.
Corollary 3 records the corresponding realization through the symmetrized family \(F^{\mathrm{pol}}_{x}\). The one-sided statement in Theorem 7(i)–(ii) is stronger: the same value is obtained as a genuine generalized-fidelity value along each one-sided polar path, namely with bases \(R=P^{\theta_P}\) and \(R=Q^{\theta_Q}\).
The positive result is pointwise and existential. It realizes individual values and does not assert the existence of a universal base, nor of a canonical rule \(R=R_z(P,Q)\), valid for all pairs. The proof ultimately combines the interval bounds for \(F_z\) and \(F^{\mathrm{LE}}\) with the exact interval-realization statement in Proposition 5(ii), whose proof relies on the one-sided monotonicity of \(\Phi_P\) and \(\Phi_Q\).
We close this section with a partial data-processing statement for the symmetrized \(x\)-polar fidelities.
Proposition 9. Fix \(x\in[-1,1]\), let \(P,Q\in\mathbb{P}_d\), and let \(\Lambda\) be a quantum channel such that \(\Lambda(P),\Lambda(Q)\in\mathbb{P}_d\). Then \[F^{\mathrm{pol}}_{x}(\Lambda(P),\Lambda(Q))\ge F^{\mathrm{pol}}_{x}(P,Q)\] whenever one of the following conditions is satisfied:
\([P,Q]=0\);
\([\Lambda(P),\Lambda(Q)]=0\);
\(\Lambda=\Gamma\circ M\), where \(M\) is a quantum channel with commuting range, \(\Gamma\) is a quantum channel, and \(M(P),M(Q)\in\mathbb{P}_d\).
Proof. Fix \(x\in[-1,1]\). Here a quantum channel means a completely positive trace-preserving linear map on matrices. Although such maps are usually stated on states, we apply them to positive definite matrices as linear maps; the state case extends to this setting by joint homogeneity and trace preservation, as in Remark 1. The displayed \(F^{\mathrm{pol}}_{x}\)-terms are defined by the hypotheses; in case (c) this includes \(M(P),M(Q)\in\mathbb{P}_d\).
We first record a simple commuting-case identity. If \([A,B]=0\), then \[\begin{align} F_{A^x}(A,B) &=\operatorname{Tr}\!\left[A^{\frac{1-x}{2}} \Bigl(A^{\frac{x}{2}}BA^{\frac{x}{2}}\Bigr)^{1/2}\right] \\ &=\operatorname{Tr}\!\left[A^{\frac{1-x}{2}}(A^xB)^{1/2}\right] \\ &=\operatorname{Tr}\!\left[A^{\frac{1-x}{2}}A^{\frac{x}{2}}B^{1/2}\right] \\ &=\operatorname{Tr}\!\left[A^{1/2}B^{1/2}\right]. \end{align}\] Similarly, \[F_{B^x}(A,B)=\operatorname{Tr}\!\left[A^{1/2}B^{1/2}\right].\] Consequently, whenever \([A,B]=0\), \[F^{\mathrm{pol}}_{x}(A,B)=\operatorname{Tr}\!\left[A^{1/2}B^{1/2}\right]=F^{\mathrm M}(A,B)=F^{\mathrm U}(A,B).\] In the estimates below we use the standard data-processing inequalities for \(F^{\mathrm M}\) and \(F^{\mathrm U}\), extended from faithful states to \(\mathbb{P}_d\) by the homogeneity reduction just noted.
Assume first that \([P,Q]=0\). By the commuting-case identity just proved, \[F^{\mathrm{pol}}_{x}(P,Q)=F^{\mathrm M}(P,Q).\] Using Proposition 5(i) at the output pair and the DPI of \(F^{\mathrm M}\), we obtain \[F^{\mathrm{pol}}_{x}(\Lambda(P),\Lambda(Q)) \ge F^{\mathrm M}(\Lambda(P),\Lambda(Q)) \ge F^{\mathrm M}(P,Q) =F^{\mathrm{pol}}_{x}(P,Q).\]
Assume next that \([\Lambda(P),\Lambda(Q)]=0\). Again by the commuting-case identity, \[F^{\mathrm{pol}}_{x}(\Lambda(P),\Lambda(Q))=F^{\mathrm U}(\Lambda(P),\Lambda(Q)).\] Using the DPI of \(F^{\mathrm U}\) and Proposition 5(i), we get \[F^{\mathrm{pol}}_{x}(\Lambda(P),\Lambda(Q)) = F^{\mathrm U}(\Lambda(P),\Lambda(Q)) \ge F^{\mathrm U}(P,Q) \ge F^{\mathrm{pol}}_{x}(P,Q).\]
Finally assume that \(\Lambda=\Gamma\circ M\), where \(M\) is a quantum channel with commuting range and \(M(P),M(Q)\in\mathbb{P}_d\). Since \(M(P)\) and \(M(Q)\) commute, the first part just proved, applied to the channel \(\Gamma\) and the pair \(M(P),M(Q)\), gives \[F^{\mathrm{pol}}_{x}\bigl(\Gamma(M(P)),\Gamma(M(Q))\bigr) \ge F^{\mathrm{pol}}_{x}(M(P),M(Q)).\] Moreover, the second part just proved, applied to the channel \(M\), gives \[F^{\mathrm{pol}}_{x}(M(P),M(Q))\ge F^{\mathrm{pol}}_{x}(P,Q),\] because the output pair \(M(P),M(Q)\) is commuting. Combining the two inequalities and using \(\Lambda=\Gamma\circ M\), we obtain \[F^{\mathrm{pol}}_{x}(\Lambda(P),\Lambda(Q))\ge F^{\mathrm{pol}}_{x}(P,Q).\] This proves the proposition. ◻
Remark 10. Proposition 9 is independent of the recovery question in Open Problem 3. It gives a positive data-processing result for the symmetrized polar fidelity under commutative-interface hypotheses: \[F^{\mathrm{pol}}_{x}(\Lambda(P),\Lambda(Q))\ge F^{\mathrm{pol}}_{x}(P,Q), \qquad x\in[-1,1].\] Since \(\Lambda\) is trace-preserving, this is equivalently the contraction of the associated Hellinger-type quantity \[\operatorname{Tr}(P+Q)-2F^{\mathrm{pol}}_{x}(P,Q).\] Consequently, any counterexample to full data processing for an interior parameter \(x\in(-1,1)\) must be genuinely noncommutative: both the input pair and the output pair must be noncommuting, and the channel cannot factor through a commuting range in the positive-definite regime considered here.
We first classify the bases satisfying the sufficient polar condition 1 .
Theorem 11. Let \(P,Q\in\mathbb{P}_d\) and define \[H:=P^{-1/4}Q^{1/2}P^{-1/4}\in\mathbb{P}_d.\] For \(R\in\mathbb{P}_d\), the following are equivalent:
\[\operatorname{Pol}\bigl(R^{1/2}P^{1/2}\bigr)=\operatorname{Pol}\bigl(R^{1/2}Q^{1/2}\bigr).\]
There exists \(B\in\mathbb{P}_d\) such that \([B,H]=0\) and \[R=P^{-1/4}BP^{1/2}BP^{-1/4}.\]
Moreover, whenever these equivalent conditions hold, one has \[F_R(P,Q)=F^{\mathrm H}(P,Q)=\operatorname{Tr}\bigl[P^{1/2}Q^{1/2}\bigr].\]
Proof. We first prove \((ii)\Rightarrow(i)\). Assume that \(B\in\mathbb{P}_d\) commutes with \(H=P^{-1/4}Q^{1/2}P^{-1/4}\) and define \[C:=P^{1/4}BP^{-1/4}.\] Then \[R=C^*C=P^{-1/4}BP^{1/2}BP^{-1/4}.\] Also, \[CP^{1/2}=P^{1/4}BP^{1/4}>0.\] Since \(Q^{1/2}=P^{1/4}HP^{1/4}\), we also have \[CQ^{1/2}=P^{1/4}BH P^{1/4}=P^{1/4}HB P^{1/4}>0,\] because \(B\) commutes with \(H\) and both are positive definite.
Now let \(C=WR^{1/2}\) be the polar decomposition of \(C\), so that \(W=\operatorname{Pol}(C)\) is unitary. Then \[R^{1/2}P^{1/2}=W^*(CP^{1/2}), \qquad R^{1/2}Q^{1/2}=W^*(CQ^{1/2}).\] Since \(CP^{1/2}\) and \(CQ^{1/2}\) are positive definite, both matrices on the right-hand side have the same polar factor \(W^*\). Hence \[\operatorname{Pol}\bigl(R^{1/2}P^{1/2}\bigr)=W^*=\operatorname{Pol}\bigl(R^{1/2}Q^{1/2}\bigr),\] which proves \((i)\).
We now prove \((i)\Rightarrow(ii)\). Assume \[\operatorname{Pol}\bigl(R^{1/2}P^{1/2}\bigr)=\operatorname{Pol}\bigl(R^{1/2}Q^{1/2}\bigr)=:U.\] Define \[C:=U^*R^{1/2}.\] Then \[CP^{1/2}=U^*R^{1/2}P^{1/2}>0, \qquad CQ^{1/2}=U^*R^{1/2}Q^{1/2}>0,\] because the common polar factor has been removed. Let \[A:=CP^{1/2}\in\mathbb{P}_d.\] Then \(C=AP^{-1/2}\), and therefore \[R=C^*C=P^{-1/2}A^2P^{-1/2}.\] Moreover, \[AP^{-1/2}Q^{1/2}=CQ^{1/2}>0.\] Now define \[B:=P^{-1/4}AP^{-1/4}\in\mathbb{P}_d.\] Using \(H=P^{-1/4}Q^{1/2}P^{-1/4}\), we get \[AP^{-1/2}Q^{1/2}=P^{1/4}BHP^{1/4}>0.\] Conjugating by \(P^{-1/4}\) yields \[BH=P^{-1/4}\bigl(AP^{-1/2}Q^{1/2}\bigr)P^{-1/4}\in\mathbb{P}_d.\] By Lemma [lem:product-positive-iff-commute], this implies \[BH=HB.\] Thus \([B,H]=0\). Finally, \[R=P^{-1/2}A^2P^{-1/2} =P^{-1/2}(P^{1/4}BP^{1/4})(P^{1/4}BP^{1/4})P^{-1/2} =P^{-1/4}BP^{1/2}BP^{-1/4}.\] This proves \((ii)\).
It remains to prove that either condition implies \(F_R(P,Q)=F^{\mathrm H}(P,Q)\). Recall that \[F_R(P,Q)=\operatorname{Tr}\bigl[Q^{1/2}U_QU_P^*P^{1/2}\bigr],\] where \[U_P:=\operatorname{Pol}(P^{1/2}R^{1/2}), \qquad U_Q:=\operatorname{Pol}(Q^{1/2}R^{1/2}).\] By Lemma 5, \(\operatorname{Pol}(X^*)=\operatorname{Pol}(X)^*\) for every invertible matrix \(X\). Hence condition \((i)\) implies \[U_P=\operatorname{Pol}(R^{1/2}P^{1/2})^*=\operatorname{Pol}(R^{1/2}Q^{1/2})^*=U_Q.\] Therefore, \[F_R(P,Q)=\operatorname{Tr}\bigl[Q^{1/2}P^{1/2}\bigr]=F^{\mathrm H}(P,Q).\] This completes the proof. ◻
Corollary 4. Let \(P,Q\in\mathbb{P}_d\) and \(H:=P^{-1/4}Q^{1/2}P^{-1/4}\). If \(B\in\mathbb{P}_d\) satisfies \([B,H]=0\), then \[R_B:=P^{-1/4}BP^{1/2}BP^{-1/4}\] satisfies \[F_{R_B}(P,Q)=F^{\mathrm H}(P,Q).\] In particular, if \(f:\sigma(H)\to(0,\infty)\) is any positive function on the spectrum, then \[R_f:=P^{-1/4}f(H)P^{1/2}f(H)P^{-1/4}\] is a Holevo base. Hence, for every \(t\in\mathbb{R}\), \[R_t:=P^{-1/4}H^tP^{1/2}H^tP^{-1/4}\] is a valid base for Holevo fidelity.
If \(d\ge2\), the commutant family above contains infinitely many distinct bases satisfying the polar condition.
Proof. The first assertion is exactly Theorem 11. The functional-calculus subfamily follows by taking \(B=f(H)\), and the power family follows by taking \(f(\lambda)=\lambda^t\).
It remains only to justify the final infinitude statement. The map \[B\longmapsto R_B=P^{-1/4}BP^{1/2}BP^{-1/4}\] is injective. Indeed, \[P^{1/2}R_B P^{1/2} = (P^{1/4}BP^{1/4})^2,\] and therefore \[(P^{1/2}R_B P^{1/2})^{1/2}=P^{1/4}BP^{1/4},\] which determines \(B\) uniquely.
If \(H\) is not a scalar matrix, then the matrices \(B_t=e^{tH}\), \(t\in\mathbb{R}\), give infinitely many distinct positive definite matrices commuting with \(H\). If \(H\) is scalar, then every positive definite \(B\) commutes with \(H\), and since \(d\ge2\) there are again infinitely many choices of \(B\). By injectivity, these give infinitely many distinct bases. ◻
Remark 12. The functional subfamily \(B=H^t\) in Corollary 4 contains several simple examples: \[R_0=I, \qquad R_1=P^{-1/2}QP^{-1/2}, \qquad R_{-1}=Q^{-1/2}PQ^{-1/2}.\] When \(H\) is scalar, this one-parameter power subfamily consists only of scalar multiples of \(I\), and hence is trivial up to the scaling invariance of the base. The full commutant family in Corollary 4, however, remains infinite in every dimension \(d\ge2\).
Proposition 13. Let \(R\in\mathbb{P}_d\) be fixed. If \[F_R(P,Q)=F^{\mathrm H}(P,Q) \qquad\text{for every }P,Q\in\mathbb{P}_d,\] then \(R=cI\) for some \(c>0\). Conversely, every base \(R=cI\) has this universal Holevo property. In particular, up to the scaling invariance of the base, the identity is the unique universal Holevo base, even without imposing the polar condition 1 .
Proof. If \(R=cI\), then \(F_R=F_I=F^{\mathrm H}\) by Lemma 3. Conversely, assume that \(F_R(P,Q)=F^{\mathrm H}(P,Q)\) for all \(P,Q\in\mathbb{P}_d\). Put \(P=I\). Then \[U_P=\operatorname{Pol}(I^{1/2}R^{1/2})=I, \qquad U_Q=\operatorname{Pol}(Q^{1/2}R^{1/2}),\] and the unitary-factor formula gives \[\operatorname{Tr}\!\left[Q^{1/2}U_Q\right]=\operatorname{Tr}(Q^{1/2}) \qquad (Q\in\mathbb{P}_d).\] We claim that this forces \(U_Q=I\). Indeed, diagonalize \(Q^{1/2}=V\operatorname{diag}(\lambda_1,\ldots,\lambda_d)V^*\), with \(\lambda_i>0\), and set \(Z=V^*U_QV\). Taking real parts in \[\sum_i\lambda_i Z_{ii}=\sum_i\lambda_i\] gives \(\sum_i\lambda_i\operatorname{Re}Z_{ii}=\sum_i\lambda_i\). Since \(\operatorname{Re}Z_{ii}\le1\) and all \(\lambda_i>0\), we have \(Z_{ii}=1\) for every \(i\). Unitarity of \(Z\) then gives \(Z=I\), hence \(U_Q=I\).
Therefore \[\operatorname{Pol}(Q^{1/2}R^{1/2})=I \qquad(Q\in\mathbb{P}_d),\] so \(Q^{1/2}R^{1/2}\in\mathbb{P}_d\) for every \(Q\in\mathbb{P}_d\). By Lemma [lem:product-positive-iff-commute], \(Q^{1/2}\) commutes with \(R^{1/2}\) for every \(Q\in\mathbb{P}_d\). Hence \(R^{1/2}\), and therefore \(R\), commutes with every positive definite matrix. Thus \(R=cI\) for some \(c>0\). ◻
Theorem 11 classifies the bases satisfying the sufficient polar condition 1 . The next subsection removes the remaining gap by classifying all bases \(R\in\mathbb{P}_d\) for which \[F_R(P,Q)=F^{\mathrm H}(P,Q)\] for a fixed pair \((P,Q)\), without assuming the polar condition.
Throughout this subsection, the equality \[F_R(P,Q)=F^{\mathrm H}(P,Q)\] is understood as equality of complex numbers. Since \(F^{\mathrm H}(P,Q)=\operatorname{Tr}(P^{1/2}Q^{1/2})\) is a positive real number, this is stronger than equality after taking real parts.
Theorem 14. Let \(P,Q\in\mathbb{P}_d\) be fixed, and set \[M:=P^{-1/2}Q^{1/2}.\] For each \(A\in\mathbb{P}_d\), define \[R_A:=P^{-1/2}A^2P^{-1/2}, \qquad W_A:=\operatorname{Pol}(AM)^*.\] Then the map \(A\mapsto R_A\) is a bijection from \(\mathbb{P}_d\) onto \(\mathbb{P}_d\), with inverse \[R\longmapsto (P^{1/2}RP^{1/2})^{1/2}.\] Moreover, for every \(A\in\mathbb{P}_d\), \[F_{R_A}(P,Q)=\operatorname{Tr}\!\bigl[Q^{1/2}W_AP^{1/2}\bigr] =\operatorname{Tr}\!\Bigl[Q^{1/2}\operatorname{Pol}\bigl(AP^{-1/2}Q^{1/2}\bigr)^*P^{1/2}\Bigr].\] Consequently, the full set of bases for which generalized fidelity equals Holevo fidelity is \[\mathcal{H}(P,Q) :=\{R\in\mathbb{P}_d: F_R(P,Q)=F^{\mathrm H}(P,Q)\}\] and admits the exact description \[\mathcal{H}(P,Q) = \Bigl\{P^{-1/2}A^2P^{-1/2}: A\in\mathbb{P}_d,\; \operatorname{Tr}\!\Bigl[Q^{1/2}\operatorname{Pol}\bigl(AP^{-1/2}Q^{1/2}\bigr)^*P^{1/2}\Bigr] =\operatorname{Tr}(P^{1/2}Q^{1/2})\Bigr\}.\] Equivalently, \[R_A\in\mathcal{H}(P,Q) \iff \operatorname{Tr}\!\Bigl[Q^{1/2}\bigl(\operatorname{Pol}(AP^{-1/2}Q^{1/2})^*-I\bigr)P^{1/2}\Bigr]=0.\]
Proof. Fix \(A\in\mathbb{P}_d\), and define \[C:=AP^{-1/2}.\] Then \[C^*C=P^{-1/2}A^2P^{-1/2}=R_A.\] Write the polar decomposition of \(C\) as \[C=V_A^*R_A^{1/2}, \qquad\text{equivalently}\qquad R_A^{1/2}=V_AC.\] Since \(A>0\), we have \[R_A^{1/2}P^{1/2}=V_ACP^{1/2}=V_AA,\] whose positive factor is \(A\). Hence \[\operatorname{Pol}(R_A^{1/2}P^{1/2})=V_A.\] Likewise, \[R_A^{1/2}Q^{1/2}=V_ACQ^{1/2}=V_AAP^{-1/2}Q^{1/2}=V_AAM,\] so, by Lemma 5, \[\operatorname{Pol}(R_A^{1/2}Q^{1/2})=V_A\operatorname{Pol}(AM).\] Now let \[U_P:=\operatorname{Pol}(P^{1/2}R_A^{1/2}), \qquad U_Q:=\operatorname{Pol}(Q^{1/2}R_A^{1/2}).\] By Lemma 5, \[U_P=\operatorname{Pol}(R_A^{1/2}P^{1/2})^*=V_A^*, \qquad U_Q=\operatorname{Pol}(R_A^{1/2}Q^{1/2})^* =\operatorname{Pol}(AM)^*V_A^*.\] Therefore \[U_QU_P^*=\operatorname{Pol}(AM)^*.\] Invoking the unitary-factor representation of generalized fidelity, \[F_{R_A}(P,Q)=\operatorname{Tr}\!\bigl[Q^{1/2}U_QU_P^*P^{1/2}\bigr],\] we get \[F_{R_A}(P,Q)=\operatorname{Tr}\!\bigl[Q^{1/2}\operatorname{Pol}(AM)^*P^{1/2}\bigr].\] This proves the stated formula.
Next, the map \(A\mapsto R_A\) is clearly injective. It is also surjective, since for every \(R\in\mathbb{P}_d\), choosing \[A:=(P^{1/2}RP^{1/2})^{1/2}\] gives \[P^{-1/2}A^2P^{-1/2}=R.\] Thus \(A\mapsto R_A\) is a bijection with the stated inverse.
Finally, \(F^{\mathrm H}(P,Q)=\operatorname{Tr}(P^{1/2}Q^{1/2})\), so the characterization of \(\mathcal{H}(P,Q)\) follows immediately from the formula just proved. ◻
Corollary 5. With the notation of Theorem 14, the following are equivalent:
\[\operatorname{Pol}(R_A^{1/2}P^{1/2})=\operatorname{Pol}(R_A^{1/2}Q^{1/2}).\]
\[\operatorname{Pol}(AM)=I.\]
\[AM>0.\]
Hence the previously obtained family of Holevo bases corresponds exactly to the slice \(W_A=I\) inside the full classification.
Proof. From the proof of Theorem 14, \[\operatorname{Pol}(R_A^{1/2}P^{1/2})=V_A, \qquad \operatorname{Pol}(R_A^{1/2}Q^{1/2})=V_A\operatorname{Pol}(AM).\] Thus (i) holds if and only if \(\operatorname{Pol}(AM)=I\), proving (i)\(\Leftrightarrow\)(ii). For an invertible matrix \(X\), one has \(\operatorname{Pol}(X)=I\) if and only if \(X>0\), so (ii)\(\Leftrightarrow\)(iii). ◻
Example 1. Let \[P=\begin{pmatrix}4&0\\0&1\end{pmatrix}, \qquad Q=\begin{pmatrix}2&1\\1&2\end{pmatrix}, \qquad R_r:=\begin{pmatrix}r&0\\0&1\end{pmatrix} \quad (r>0).\] Then \[F^{\mathrm H}(P,Q)=\operatorname{Tr}(P^{1/2}Q^{1/2})=\frac{3}{2}(1+\sqrt3).\] Moreover, \[F_{R_r}(P,Q)= \frac{(4+\sqrt3)\sqrt r+2(1+\sqrt3)}{\sqrt{2r+2+2\sqrt{3r}}}.\] Therefore \(F_{R_r}(P,Q)=F^{\mathrm H}(P,Q)\) if and only if, writing \(u:=\sqrt r\), \[(u-1)\bigl((1-\sqrt3)u+(2+\sqrt3)\bigr)=0.\] Besides the trivial solution \(u=1\) (that is, \(R=I\)), there is a second solution \[u=\frac{2+\sqrt3}{\sqrt3-1}=\frac{5+3\sqrt3}{2}, \qquad r=u^2=13+\frac{15\sqrt3}{2}.\] Hence the base \[R_*=\begin{pmatrix}13+\frac{15\sqrt3}{2}&0\\0&1\end{pmatrix} \neq I\] satisfies \[F_{R_*}(P,Q)=F^{\mathrm H}(P,Q).\] However, \[R_*^{1/2}P^{1/2}=\begin{pmatrix}2\sqrt r&0\\0&1\end{pmatrix}>0,\] so \[\operatorname{Pol}(R_*^{1/2}P^{1/2})=I.\] On the other hand, \[Q^{1/2}=\frac{1}{2} \begin{pmatrix} 1+\sqrt3 & \sqrt3-1\\ \sqrt3-1 & 1+\sqrt3 \end{pmatrix},\] and therefore \[R_*^{1/2}Q^{1/2} =\frac{1}{2} \begin{pmatrix} \sqrt r(1+\sqrt3) & \sqrt r(\sqrt3-1)\\ \sqrt3-1 & 1+\sqrt3 \end{pmatrix},\] which is not Hermitian because \(r\ne1\). Hence \(R_*^{1/2}Q^{1/2}\) is not positive definite, so \[\operatorname{Pol}(R_*^{1/2}Q^{1/2})\ne I.\] Thus \(R_*\) is a Holevo base that does not satisfy the sufficient polar condition \[\operatorname{Pol}(R^{1/2}P^{1/2})=\operatorname{Pol}(R^{1/2}Q^{1/2}).\]
Proof of the formula in Example 1. Since \(R_r\) commutes with \(P\), \[(R_r^{1/2}PR_r^{1/2})^{1/2}=\begin{pmatrix}2\sqrt r&0\\0&1\end{pmatrix}.\] Also, \[R_r^{1/2}QR_r^{1/2}=\begin{pmatrix}2r&\sqrt r\\ \sqrt r&2\end{pmatrix}, \qquad \det(R_r^{1/2}QR_r^{1/2})=3r.\] For a positive definite \(2\times2\) matrix \[X=\begin{pmatrix}a&c\\c&b\end{pmatrix},\] one has \[X^{1/2}=\frac{1}{\sqrt{a+b+2\sqrt{ab-c^2}}} \begin{pmatrix} a+\sqrt{ab-c^2}&c\\ c&b+\sqrt{ab-c^2} \end{pmatrix}.\] Applying this with \(a=2r\), \(b=2\), \(c=\sqrt r\) gives \[(R_r^{1/2}QR_r^{1/2})^{1/2} = \frac{1}{\sqrt{2r+2+2\sqrt{3r}}} \begin{pmatrix} 2r+\sqrt{3r}&\sqrt r\\ \sqrt r&2+\sqrt{3r} \end{pmatrix}.\] Hence \[\begin{align} F_{R_r}(P,Q) &=\operatorname{Tr}\!\left[(R_r^{1/2}PR_r^{1/2})^{1/2}R_r^{-1}(R_r^{1/2}QR_r^{1/2})^{1/2}\right]\\ &= \frac{2}{\sqrt r}\cdot\frac{2r+\sqrt{3r}}{\sqrt{2r+2+2\sqrt{3r}}} + \frac{2+\sqrt{3r}}{\sqrt{2r+2+2\sqrt{3r}}}\\ &= \frac{(4+\sqrt3)\sqrt r+2(1+\sqrt3)}{\sqrt{2r+2+2\sqrt{3r}}}. \end{align}\] Finally, \[Q^{1/2}=\frac{1}{2} \begin{pmatrix} 1+\sqrt3 & \sqrt3-1\\ \sqrt3-1 & 1+\sqrt3 \end{pmatrix},\] so \[F^{\mathrm H}(P,Q)=\operatorname{Tr}(P^{1/2}Q^{1/2})=2\cdot\frac{1+\sqrt3}{2}+\frac{1+\sqrt3}{2} =\frac{3}{2}(1+\sqrt3).\] Setting \(u=\sqrt r>0\), the equality \(F_{R_r}(P,Q)=F^{\mathrm H}(P,Q)\) is an equality between positive quantities, so squaring is equivalent. A short simplification yields \[(u-1)\bigl((1-\sqrt3)u+(2+\sqrt3)\bigr)=0,\] and hence \[u=1 \qquad\text{or}\qquad u=\frac{2+\sqrt3}{\sqrt3-1}=\frac{5+3\sqrt3}{2}.\] ◻
Remark 15. Theorem 14 solves the full equality problem \[F_R(P,Q)=F^{\mathrm H}(P,Q)\] for fixed \(P,Q\in\mathbb{P}_d\): every such base \(R\) is obtained uniquely from a positive matrix \(A=(P^{1/2}RP^{1/2})^{1/2}\), and the equality is equivalent to a single explicit trace constraint involving the polar factor of \(AP^{-1/2}Q^{1/2}\). Corollary 5 shows that the earlier family obtained from the sufficient polar condition is only the special slice where this polar factor is the identity.
Example 1 shows that this slice is proper: there exist nontrivial bases yielding Holevo fidelity for which the two polar factors are not equal.
We now turn to the unitary factors that can arise from generalized fidelity. For fixed \(P,Q\in\mathbb{P}_d\), we say that a unitary \(W\in U(d)\) arises as a unitary factor of generalized fidelity if there exists \(R\in\mathbb{P}_d\) such that \[W=U_QU_P^*, \qquad U_P:=\operatorname{Pol}(P^{1/2}R^{1/2}),\quad U_Q:=\operatorname{Pol}(Q^{1/2}R^{1/2}).\]
Theorem 16. Fix \(P,Q\in\mathbb{P}_d\), and define \[M:=P^{-1/2}Q^{1/2}.\] For a unitary matrix \(W\in U(d)\), the following are equivalent.
The unitary \(W\) arises as a unitary factor of generalized fidelity for the pair \((P,Q)\).
There exists \(A\in\mathbb{P}_d\) such that \(A M W\in\mathbb{P}_d.\)
The matrix \(MW\) is similar to a positive definite matrix.
The matrix \(MW\) is diagonalizable and all of its eigenvalues are strictly positive real numbers.
Moreover, if \(MW=SDS^{-1}\) for some \(S\in GL_d(\mathbb{C})\) and some \(D\in\mathbb{P}_d\), then the set of all \(A\in\mathbb{P}_d\) satisfying (ii) is exactly \[\mathcal{A}_W(P,Q) := \{S^{-*}CS^{-1}: C\in\mathbb{P}_d,\;[C,D]=0\}.\] Consequently, the set of all bases \(R\in\mathbb{P}_d\) whose unitary factor is \(W\) is exactly \[\mathcal{R}_W(P,Q) := \{P^{-1/2}A^2P^{-1/2}: A\in\mathcal{A}_W(P,Q)\}.\]
Proof. We first prove \((i)\Leftrightarrow(ii)\). Let \(R\in\mathbb{P}_d\) be arbitrary and set \[A:=(P^{1/2}RP^{1/2})^{1/2}\in\mathbb{P}_d.\] Let \[V:=\operatorname{Pol}(R^{1/2}P^{1/2})\in U(d).\] Then \[R^{1/2}P^{1/2}=VA, \qquad \text{hence} \qquad R^{1/2}=VAP^{-1/2}.\] Therefore, \[R^{1/2}Q^{1/2}=VAP^{-1/2}Q^{1/2}=VAM.\] By Lemma 5, \[\operatorname{Pol}(R^{1/2}P^{1/2})=V, \qquad \operatorname{Pol}(R^{1/2}Q^{1/2})=V\operatorname{Pol}(AM),\] and hence \[U_P=\operatorname{Pol}(P^{1/2}R^{1/2})=V^*, \qquad U_Q=\operatorname{Pol}(Q^{1/2}R^{1/2})=\operatorname{Pol}(AM)^*V^*.\] Therefore \[U_QU_P^*=\operatorname{Pol}(AM)^*.\] It follows that \[U_QU_P^*=W \iff \operatorname{Pol}(AM)=W^*.\] Now for any invertible matrix \(Z\) and unitary \(U\), \[\operatorname{Pol}(Z)=U \iff U^*Z\in\mathbb{P}_d,\] because \(Z=U|Z|\) if and only if \(U^*Z=|Z|\in\mathbb{P}_d\). Applying this with \(Z=AM\) and \(U=W^*\), we obtain \[\operatorname{Pol}(AM)=W^* \iff WAM\in\mathbb{P}_d.\] Since \[AMW=W^*(WAM)W,\] unitary conjugation shows that \[WAM\in\mathbb{P}_d \iff AMW\in\mathbb{P}_d.\] Hence \[U_QU_P^*=W \iff AMW\in\mathbb{P}_d.\] This proves that, for every \(R\in\mathbb{P}_d\) and the associated matrix \[A:=(P^{1/2}RP^{1/2})^{1/2},\] one has \[U_QU_P^*=W \iff AMW\in\mathbb{P}_d.\] Conversely, if \(A\in\mathbb{P}_d\) is given and satisfies \(AMW\in\mathbb{P}_d\), define \[R:=P^{-1/2}A^2P^{-1/2}\in\mathbb{P}_d.\] Applying the preceding computation to this \(R\), whose associated positive matrix is precisely the chosen \(A\), yields \(U_QU_P^*=W\). Hence \((i)\Leftrightarrow(ii)\). We next prove \((ii)\Rightarrow(iii)\). If \(AMW\in\mathbb{P}_d\), then \[A^{1/2}(MW)A^{-1/2}=A^{-1/2}(AMW)A^{-1/2}\in\mathbb{P}_d,\] so \(MW\) is similar to a positive definite matrix.
The implication \((iii)\Rightarrow(iv)\) is immediate, because a positive definite matrix is Hermitian with strictly positive spectrum, and similarity preserves both spectrum and diagonalizability.
Conversely, assume (iv). Then \(MW=SDS^{-1}\) for some \(S\in GL_d(\mathbb{C})\) and some positive diagonal matrix \(D\), hence in particular \(D\in\mathbb{P}_d\). Therefore \(MW\) is similar to the positive definite matrix \(D\), which proves \((iv)\Rightarrow(iii)\).
It remains to characterize all \(A\) satisfying (ii). Fix a decomposition \(MW=SDS^{-1}\) with \(D\in\mathbb{P}_d\), and let \(A\in\mathbb{P}_d\). Set \[B:=S^*AS\in\mathbb{P}_d.\] Then \[AMW=A(SDS^{-1})=S^{-*}BDS^{-1}.\] Since congruence by an invertible matrix preserves positive definiteness, \[AMW\in\mathbb{P}_d \iff BD\in\mathbb{P}_d.\] Since \(B,D\in\mathbb{P}_d\), the matrix \(BD\) is positive definite if and only if it is Hermitian, and this is equivalent to \[BD=(BD)^*=DB.\] Therefore \[AMW\in\mathbb{P}_d \iff [B,D]=0.\] Equivalently, \[A\in\mathbb{P}_d\text{ satisfies (ii)} \iff A=S^{-*}CS^{-1} \text{ for some }C\in\mathbb{P}_d\text{ with }[C,D]=0.\] This proves the formula for \(\mathcal{A}_W(P,Q)\), and the description of \(\mathcal{R}_W(P,Q)\) follows from the identity \(R=P^{-1/2}A^2P^{-1/2}\). ◻
The next result combines the trace criterion from Theorem 14 with the unitary-factor classification above and gives the full fixed-pair stratification of Holevo bases by unitary strata.
Theorem 17. Fix \(P,Q\in\mathbb{P}_d\), and again write \[X:=P^{1/2}Q^{1/2}, \qquad M:=P^{-1/2}Q^{1/2}.\] Define the set of Holevo bases \[\mathcal{H}(P,Q):= \{R\in\mathbb{P}_d: F_R(P,Q)=F^{\mathrm H}(P,Q)\}.\] Then \(\mathcal{H}(P,Q)\) decomposes as the disjoint union \[\mathcal{H}(P,Q) = \bigcup_{W\in\mathcal{U}_H(P,Q)} \mathcal{R}_W(P,Q),\] where \[\mathcal{U}_H(P,Q) := \{W\in U(d): MW\sim\mathbb{P}_d,\;\operatorname{Tr}(WX)=\operatorname{Tr}(X)\},\] and \(\mathcal{R}_W(P,Q)\) is the \(W\)-stratum described in Theorem 16. Equivalently, \(R\in\mathcal{H}(P,Q)\) if and only if there exist a unitary \(W\in U(d)\) and a positive matrix \(A\in\mathbb{P}_d\) such that \[AMW\in\mathbb{P}_d, \qquad \operatorname{Tr}(WX)=\operatorname{Tr}(X), \qquad R=P^{-1/2}A^2P^{-1/2}.\] More explicitly, if \(MW=SDS^{-1}\) with \(D\in\mathbb{P}_d\), then the corresponding Holevo bases in the \(W\)-stratum are precisely \[R=P^{-1/2}A^2P^{-1/2}, \qquad A=S^{-*}CS^{-1}, \qquad C\in\mathbb{P}_d, \;[C,D]=0,\] with the sole additional requirement that \[\operatorname{Tr}(WP^{1/2}Q^{1/2})=\operatorname{Tr}(P^{1/2}Q^{1/2}).\]
Proof. Let \(R\in\mathbb{P}_d\), and denote its unitary factor by \[W_R:=U_QU_P^*.\] By the equivalent form of generalized fidelity, \[F_R(P,Q)=\operatorname{Tr}\bigl[Q^{1/2}W_RP^{1/2}\bigr]=\operatorname{Tr}(W_RX).\] On the other hand, \[F^{\mathrm H}(P,Q)=\operatorname{Tr}(P^{1/2}Q^{1/2})=\operatorname{Tr}(X).\] Hence \[F_R(P,Q)=F^{\mathrm H}(P,Q) \iff \operatorname{Tr}(W_RX)=\operatorname{Tr}(X).\] By Theorem 16, the possible values of \(W_R\) are exactly those unitaries \(W\) for which \(MW\) is similar to a positive definite matrix, and for each such \(W\) the corresponding bases are precisely the elements of \(\mathcal{R}_W(P,Q)\). Intersecting these strata with the single scalar constraint \(\operatorname{Tr}(WX)=\operatorname{Tr}(X)\) yields the claimed description of \(\mathcal{H}(P,Q)\). The union is disjoint because the unitary factor \(W_R=U_QU_P^*\) is uniquely determined by \(R\). ◻
Corollary 6. Fix \(P,Q\in\mathbb{P}_d\), and set \(M:=P^{-1/2}Q^{1/2}\). The condition 1 is exactly the \(W=I\) stratum in Theorem 17. Equivalently, \[\operatorname{Pol}(R^{1/2}P^{1/2})=\operatorname{Pol}(R^{1/2}Q^{1/2}) \iff R=P^{-1/2}A^2P^{-1/2} \text{ for some }A\in\mathbb{P}_d\text{ with }AM\in\mathbb{P}_d.\] Thus the family obtained from 1 is one stratum of the full solution, but not in general the whole solution set.
Proof. Let \[U_P:=\operatorname{Pol}(P^{1/2}R^{1/2}), \qquad U_Q:=\operatorname{Pol}(Q^{1/2}R^{1/2}), \qquad W_R:=U_QU_P^*.\] By Lemma 5, \[U_P=\operatorname{Pol}(P^{1/2}R^{1/2}) =\operatorname{Pol}\!\bigl((R^{1/2}P^{1/2})^*\bigr) =\operatorname{Pol}(R^{1/2}P^{1/2})^*,\] and similarly \[U_Q=\operatorname{Pol}(Q^{1/2}R^{1/2}) =\operatorname{Pol}\!\bigl((R^{1/2}Q^{1/2})^*\bigr) =\operatorname{Pol}(R^{1/2}Q^{1/2})^*.\] Therefore, \[\operatorname{Pol}(R^{1/2}P^{1/2}) =\operatorname{Pol}(R^{1/2}Q^{1/2}) \iff U_P=U_Q \iff W_R=I.\] The rest follows from Theorem 16 with \(W=I\). ◻
Corollary 7. If \([P,Q]=0\), then \[\mathcal{U}_H(P,Q)=\{I\}, \qquad \mathcal{H}(P,Q)=\mathcal{R}_I(P,Q).\] Equivalently, for commuting pairs the polar slice exhausts all Holevo bases.
Proof. If \([P,Q]=0\), then \[X=P^{1/2}Q^{1/2}\in\mathbb{P}_d.\] Let \(W\in\mathcal{U}_H(P,Q)\). By definition, \[\operatorname{Tr}(WX)=\operatorname{Tr}(X).\] Choose a unitary \(U\) such that \[X=U\Lambda U^*, \qquad \Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_d), \qquad \lambda_i>0,\] and set \(V:=U^*WU\). Then \[\sum_{i=1}^d \lambda_i V_{ii} =\operatorname{Tr}(V\Lambda) =\operatorname{Tr}(WX) =\operatorname{Tr}(X) =\sum_{i=1}^d \lambda_i.\] Taking real parts gives \[\sum_{i=1}^d \lambda_i \Re(V_{ii})=\sum_{i=1}^d \lambda_i.\] Since \(|V_{ii}|\le 1\), one has \(\Re(V_{ii})\le 1\) for every \(i\). Because each \(\lambda_i>0\), the last equality forces \(\Re(V_{ii})=1\) for all \(i\), hence \(V_{ii}=1\) for all \(i\). For each row \(i\), \[1=\sum_{j=1}^d |V_{ij}|^2 \ge |V_{ii}|^2=1,\] so \(V_{ij}=0\) for every \(j\ne i\). Hence \(V=I\), and therefore \(W=I\). The conclusion follows from Theorem 17. ◻
Remark 18. Theorem 17 gives the strongest fixed-pair form of the Holevo-base classification. Indeed, the freedom in the base splits into two clean pieces:
a unitary stratum \(W\), constrained only by the two conditions \[MW\sim \mathbb{P}_d, \qquad \operatorname{Tr}(WX)=\operatorname{Tr}(X),\] where \(MW\sim \mathbb{P}_d\) means that \(MW\) is similar to a positive definite matrix;
within each such stratum, the positive part is completely classified by the commutant of the positive matrix \(D\) in any decomposition \(MW=SDS^{-1}\).
In particular, there are generally Holevo bases with \(W\neq I\), so the polar condition 1 is sufficient but not necessary.
Corollary 8. For \(P,Q\in\mathbb{P}_d\), set \(M:=P^{-1/2}Q^{1/2}\) and define the set of unitary factors arising from generalized fidelity by \[\mathcal{U}_{\mathrm{att}}(P,Q) := \{W\in U(d): MW \text{ is similar to a positive definite matrix}\}.\] Then \[\mathcal{U}_{\mathrm{att}}(P,Q)\subseteq SU(d).\] By Theorem 16, \(\mathcal{U}_{\mathrm{att}}(P,Q)\) is exactly the set of unitary factors \(U_QU_P^*\) arising from generalized fidelity for the fixed pair \((P,Q)\). In particular, every unitary factor \(U_Q U_P^*\) arising from generalized fidelity has determinant \(1\). Moreover, \[\mathcal{U}_H(P,Q)\subseteq \mathcal{U}_{\mathrm{att}}(P,Q).\]
Proof. Let \(W\in\mathcal{U}_{\mathrm{att}}(P,Q)\). Then \(MW\) is similar to some \(D\in\mathbb{P}_d\), so \[\det(M)\det(W)=\det(MW)=\det(D)>0.\] Also, \[\det(M)=\det(P^{-1/2})\det(Q^{1/2})>0\] because \(P^{-1/2}\) and \(Q^{1/2}\) are positive definite. Therefore \(\det(W)\) is a positive real number. Since \(W\) is unitary, \(|\det(W)|=1\), hence \(\det(W)=1\). The final inclusion is immediate from the definition of \(\mathcal{U}_H(P,Q)\). ◻
Corollary 9. Let \(d\ge2\), and let \(\omega\ne1\) be a \(d\)-th root of unity. Then \[\omega I\in SU(d),\] but \(\omega I\) cannot arise as a unitary factor of generalized fidelity for any triple \(P,Q,R\in\mathbb{P}_d\). Consequently, \[\bigcup_{P,Q\in\mathbb{P}_d}\mathcal{U}_{\mathrm{att}}(P,Q) \subsetneq SU(d).\]
Proof. Fix arbitrary \(P,Q\in\mathbb{P}_d\), and write \[M:=P^{-1/2}Q^{1/2}.\] Then \(M\) is similar to a positive definite matrix. Indeed, with \[H:=P^{-1/4}Q^{1/2}P^{-1/4}\in\mathbb{P}_d,\] one has \[M=P^{-1/4}HP^{1/4}.\] Hence all eigenvalues of \(M\) are strictly positive real numbers.
Now take \(W=\omega I\). Then \[MW=\omega M,\] so the eigenvalues of \(MW\) are \(\omega\lambda_1,\dots,\omega\lambda_d\), where \(\lambda_1,\dots,\lambda_d>0\) are the eigenvalues of \(M\). Since \(\omega\ne1\), these eigenvalues are not strictly positive real numbers. Therefore \(MW\) is not similar to a positive definite matrix.
By Theorem 16, \(W=\omega I\) cannot arise as a unitary factor of generalized fidelity for the fixed pair \(P,Q\). Since \(P,Q\) were arbitrary, \(\omega I\) cannot arise in this way for any triple \(P,Q,R\in\mathbb{P}_d\).
Finally, \(\omega I\in SU(d)\) because \(\det(\omega I)=\omega^d=1\). Thus the global set of unitary factors arising from generalized fidelity is a proper subset of \(SU(d)\). ◻
Remark 19. Afham and Ferrie proved that every generalized-fidelity unitary factor belongs to \(SU(d)\) and asked whether the reverse inclusion holds. Corollary 8 recovers the known inclusion in the present fixed-pair parameterization, while Corollary 9 shows that the reverse inclusion is false for every \(d\ge2\). Thus the correct global problem is not equality with \(SU(d)\), but rather the finer characterization of \[\bigcup_{P,Q\in\mathbb{P}_d}\mathcal{U}_{\mathrm{att}}(P,Q) = \bigcup_{P,Q\in\mathbb{P}_d} \{W\in U(d): P^{-1/2}Q^{1/2}W\sim\mathbb{P}_d\}.\] The fixed-pair classification in Theorem 16 provides the corresponding local answer.
We have proved monotonicity of the one-sided and symmetrized \(x\)-polar families on the two ranges \((-\infty,1]\) and \([1,\infty)\). We have also shown that, for each fixed positive definite pair, every \(z\)-fidelity value in the data-processing regime \(z\ge \frac{1}{2}\), as well as every Log-Euclidean fidelity value, admits a pointwise generalized-fidelity representation with pair-dependent base. In addition, we obtained a structural classification of Holevo bases for a fixed pair of positive definite matrices, and a fixed-pair classification of all unitary factors arising from generalized fidelity.
The monotonicity theorem strengthens the numerical observation of Afham–Ferrie from \([-1,1]\) to the whole real line: the one-sided polar paths and the symmetrized family are nondecreasing on \((-\infty,1]\) and nonincreasing on \([1,\infty)\). Together with the exact realization of the interval \([F^{\mathrm M}(P,Q),F^{\mathrm U}(P,Q)]\), this gives a positive-definite, pointwise, and pair-dependent answer to Open Problem 3 of [1] for all \(z\ge \frac{1}{2}\) and for the Log-Euclidean fidelity. On the other hand, Theorem 7(iii) shows that such a pointwise generalized-fidelity recovery fails in general in dimensions \(d\ge2\) for \(0<z<\frac{1}{2}\), even at the level of interior fidelities.
The Holevo-base results solve the fixed-pair equality problem \[F_R(P,Q)=F^{\mathrm H}(P,Q),\] showing that the sufficient polar condition captures only one distinguished stratum of the full solution set. The polar slice corresponds exactly to the unitary stratum \(W=I\), while Example 1 shows that nonpolar Holevo bases do occur.
The classification of unitary factors also settles the special-unitary reverse-inclusion question in the negative in every dimension \(d\ge2\). Although every generalized-fidelity unitary factor lies in \(SU(d)\), not every element of \(SU(d)\) can arise as such a unitary factor when \(d\ge2\). In fact, for every \(d\ge2\), every nontrivial central special unitary \(\omega I\), with \(\omega^d=1\) and \(\omega\ne1\), is excluded by the fixed-pair criterion \[W\text{ can arise as a unitary factor for }(P,Q) \iff P^{-1/2}Q^{1/2}W\sim\mathbb{P}_d.\] Thus Open Problem 7 of [1] has a negative answer in every dimension \(d\ge2\).
The main questions that remain open are therefore of a different nature. On the polar-fidelity side, the principal unresolved issue is full data-processing monotonicity beyond commutative interfaces, together with related convexity questions for generalized Bures distance. On the generalized-fidelity side, the global set of arising unitary factors \[\bigcup_{P,Q\in\mathbb{P}_d} \{W\in U(d):P^{-1/2}Q^{1/2}W\sim\mathbb{P}_d\}\] now appears as the natural object to characterize more explicitly. It would also be interesting to determine whether semidefinite-program formulations are available for the real part of generalized fidelity or for the associated optimization problems.