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  \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}%
  \thanks{*\,TU Munich, Dept.~of Mathematics, School of CIT; Garching, Germany.
    Email: \texttt{s.sra@tum.de}.}}

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  their generous support via a Humboldt Professorship in AI.%
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\subjclass[2020]{15A42, 46L51, 47A63, 94A17}
\keywords{Tsallis relative entropy, quantum Jensen--Shannon divergence,
Sibson information radius, S-divergence, negative type, Schatten norm,
quantum R\'enyi entropy, log-determinant inequality}

\hypersetup{
  pdftitle={Tsallis metrics on density matrices via the S-divergence},
  pdfauthor={Suvrit Sra}
}

\makeatletter
\newcommand{\preprintslink}[2]{%
  \thanks{First public release: #2, as v1 in the author's preprint repository,
    \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every
    version is kept there with its timestamp.}}
\makeatother
\begin{document}
\title[Tsallis metrics on density matrices]{Tsallis Metrics on Density Matrices via the S-Divergence}
\aiauthor
\begin{abstract}
  We study symmetrized divergences induced by centroids of $\alpha$-Tsallis relative entropies between
  density matrices $\sigma$ and $\rho$; up to scaling, the induced divergence equals
  $|2-\pnorm{\rho^\alpha+\sigma^\alpha}{1/\alpha}|$.  Earlier work (S.~Sra. \darkblue{\emph{``Metrics
      induced by Jensen--Shannon and related divergences on positive definite matrices.''}}  Linear
  Algebra Appl., 2021) studied such divergences and metrics induced by them. That paper allowed
  centroids to range over the positive definite cone, while we require that the centroid itself be a
  density matrix. This (more natural) choice makes the metricity in this paper considerably stronger
  than that of the aforementioned paper. Specifically, we prove that
  $|2-\pnorm{\rho^\alpha+\sigma^\alpha}{1/\alpha}|^{1/2}$ is a metric on density matrices for
  $1/2\le\alpha\le2$, $\alpha\ne1$. For $\alpha<1$, the proof uses a new four-point theorem for the
  S-divergence that might be of independent interest. For $\alpha>1$, we first prove the endpoint
  $\alpha=2$, and then transfer it to the other orders by a tensoring trick. %
\end{abstract}
\preprintslink{tsallis}{13 October 2026}
\maketitle

\section{Introduction}\label{sec:intro}
We continue the study of metric properties of symmetrized divergences on quantum states started in~\citep{sra2021metrics}. Square roots of symmetrized divergences sometimes turn out to be metrics, for instance Jensen--Shannon divergences for commuting states \citep{endres2003new,fuglede2004jensen}. For noncommuting states \citet{lamberti2008metric} conjectured the metric property, which was proved by both~\citet{sra2021metrics} and
\citet{virosztek2021metric} independently.

We study herein a symmetrized version of Tsallis relative entropy obtained by computing
centroids, and ask whether the resulting divergence is a metric. We denote via $\pp^{d}$ ($\pp_+^d$) the set of $d\times d$ positive semidefinite (definite) Hermitian matrices; and let
$\Dc_d:=\{\rho\in\pp^d:\trace\rho=1\}$ be the density matrices, or states.  For $\rho,\tau\in\Dc_d$, the Tsallis relative entropy
\citep{tsallis1988possible,petz1986quasi} is
\begin{equation}\label{eq:tsallis-density}
T_\alpha(\rho,\tau):=
\frac{1-\trace\rho^\alpha\tau^{1-\alpha}}{1-\alpha},
\qquad \alpha>0,\quad\alpha\ne1.
\end{equation}
The symmetrized divergence obtained by computing the Tsallis centroid of $\rho$ and $\sigma$ is then:
\begin{equation}\label{eq:state-distance}
\begin{aligned}
d_\alpha^2(\rho,\sigma)
&:=|1-\alpha|\min_{\tau\in\Dc_d}
\bigl\{T_\alpha(\rho,\tau)+T_\alpha(\sigma,\tau)\bigr\}\quad=\quad\bigl|2-\pnorm{\rho^\alpha+\sigma^\alpha}{1/\alpha}\bigr|,
\end{aligned}
\end{equation}
where $\pnorm{X}{p}=(\trace|X|^p)^{1/p}$ is the Schatten-$p$
norm. %
The equality in~\eqref{eq:state-distance} is the quantum Sibson identity
\citep{konig2009strong,sharma2013fundamental,dalai2013lower}; Proposition~\ref{prop:centroid} verifies it
for $0<\alpha<1$ by H\"older's inequality; the reverse H\"older inequality gives the case $\alpha>1$
(Section~\ref{sec:alpha-large}).

It is evident that $d_\alpha$ is symmetric and nonnegative. The challenge is to show that it satisfies the triangle inequality. When the center $\tau$ in~\eqref{eq:state-distance} is allowed to range over all $\pp^d$, the minimum value is $2\alpha(1-h)$, where $h=\trace ((\rho^\alpha+\sigma^\alpha)/2)^{1/\alpha}$, and its square root is a
metric for $1/2\le\alpha<1$ \citep[Theorem~4.1]{sra2021metrics}. But in~\eqref{eq:state-distance}, we require the center to lie in $\Dc_d$, which changes the minimum to $d_\alpha^2=2(1-h^\alpha)$. It turns out to be much more difficult to establish that $d_\alpha$ is a metric in this case. The proof of this metricity is our first main result.

\begin{dashedbox}
\begin{theorem}\label{thm:tsallis}
For every $d$ and $1/2\le\alpha\le2$, $\alpha\ne1$, $d_\alpha$ is a metric on
$\Dc_d$.  %
\end{theorem}
\end{dashedbox}
Theorem~\ref{thm:tsallis} also implies the previous results of~\citet{sra2021metrics}; indeed, as
$\alpha\uparrow1$, $d_\alpha^2/(2(1-\alpha))\to \qjsd$, which gives the Jensen--Shannon metric
(Corollary~\ref{cor:qjsd}), and the proof gives the cone metric of
\citet[Theorem~4.1]{sra2021metrics}. At $\alpha=1/2$, the distance is also of negative type
(Proposition~\ref{prop:p2}).

For $\alpha<1$, with $p=1/\alpha$, Theorem~\ref{thm:tsallis} follows from positivity of the
trace-power kernel $\trace(A+B)^p-\trace A^p-\trace B^p$ on every three positive semidefinite matrices
(Theorem~\ref{thm:Lp}). For $1<p<2$, we write this kernel as a superposition of the kernels
$-\log\det(I-XY)$ on strict contractions $X$ and $Y$ in $\pp^d$, which turns our task into proving that
$[-\log\det(I-X_iX_j)]_{i,j=1}^3$ is positive semidefinite. This task is
nontrivial, and its proof uncovers a hitherto unknown property of the S-divergence~\citep{ssdiv}, denoted $\delta_S^2$ below, which leads to our second main result.
\begin{dashedbox}
  \begin{theorem}\label{thm:four}
    Let $U_1,U_2,U_3$ be Hermitian $d\times d$ matrices with
    $\pnorm{U_i}{\mathrm{op}}<1$.  Then,
    \begin{equation}\label{eq:logdet-psd}
      K := \bigl[-\log\det(I-U_iU_j)\bigr]_{i,j=1}^3\succeq0.
    \end{equation}
    Equivalently, if $P_0,P_1,P_2,P_3\in\pp_+^d$ and $\xi\in\reals^4$
    satisfies $\xi_0+\xi_1+\xi_2+\xi_3=0$, then
    \begin{equation}\label{eq:four-point}
      \sum_{a,c=0}^3\xi_a\xi_c\,\delta_S^2(P_a,P_c)\le0.
    \end{equation}
  \end{theorem}
\end{dashedbox}
\noindent The S-divergence $\delta_S^2$ used in~\eqref{eq:four-point} is defined for any two matrices $P, Q \in \pp_+^d$ as
\begin{equation}\label{eq:S-def}
\delta_S^2(P,Q):=\log\det\tfrac{P+Q}2-\tfrac12\log\det P-\tfrac12\log\det Q.
\end{equation}
It is a Jensen--Bregman divergence~\citep{cherian2012jensen} and $\delta_S$ is a metric \citep{ssdiv}.

Recall that a symmetric function $N$ with $N(x,x)=0$ is \emph{of negative type} if $\sum_{i,j}c_ic_jN(x_i,x_j)\le0$ for every finite family of points and real coefficients with $\sum_i c_i=0$.
On three points, negative type of $\delta_S^2$ is equivalent to the triangle
inequality for $\delta_S$, which is key to the results of \citet{sra2021metrics}. However, the log-determinant inequality~\eqref{eq:logdet-psd} needs four points. Indeed, using the Cayley transform
$P_i=(I+U_i)(I-U_i)^{-1}$ we map Hermitian strict contractions $U_i$ onto $\pp_+^d$, and
\begin{equation}\label{eq:polarization}
  -\log\det(I-U_iU_j)=\delta_S^2(P_i,I)+\delta_S^2(P_j,I)-\delta_S^2(P_i,P_j).
\end{equation}
Hence positivity of $[-\log\det(I-U_iU_j)]_{i,j=1}^3$ is the same as negative type of $\delta_S^2$ on
the four matrices $I,P_1,P_2,P_3$. By Theorem~\ref{thm:four} and Schoenberg's criterion~\citep{schoenberg1935remarks}, every four-point subset of $(\pp_+^d,\delta_S)$ embeds isometrically in $\reals^3$. The number four is sharp when $d\ge2$: \citet{ssdiv} gave a numerical five-point obstruction in dimension two; Proposition~\ref{prop:sharp}
gives an analytic one.

\begin{rmk}
  Theorem~\ref{thm:four} also implies the positivity of the Hua--Bellman matrices.
  For strict contractions $Z_1,Z_2,Z_3$, \citet[Corollary~3.2]{sra2021hua} proves that
  the Hua--Bellman matrix \citep{hua1955inequalities}
  $H_\gamma=[|\det(I-Z_i^*Z_j)|^{-\gamma}]_{i,j=1}^3$ is positive semidefinite
  for every $\gamma\ge0$.  For Hermitian $Z_i$, Theorem~\ref{thm:four}
  recovers this entire family (where $K$ is as in~(\ref{eq:logdet-psd})) by the Schur product theorem:
  \[
    H_\gamma=\sum_{m=0}^\infty\frac{\gamma^m}{m!}K^{\circ m}\succeq0
    \qquad(\gamma\ge0),
  \]
  where $K^{\circ m}$ denotes entrywise power. The converse direction gives less: differentiating $c^\top H_\gamma c$ at
  zero gives $c^\top Kc\ge0$ only when $\sum_i c_i=0$, whereas
  Theorem~\ref{thm:four} gives $K\succeq0$.
\end{rmk}

\begin{rmk}
  Theorem~\ref{thm:four} also applies to entropy-based Jensen divergences,
  which compare the entropy of $(\rho+\sigma)/2$ with the mean entropy of
  $\rho$ and $\sigma$.  The Jensen--Tsallis and Jensen--R\'enyi
  divergences of this form are of negative type on every four states for orders
  $0<q\le2$ and $0<q\le1$, respectively (Theorem~\ref{thm:jensen-four}).
  This strengthens the metric results of \citet{sra2021metrics}.
\end{rmk}

Below, we first prove Theorem~\ref{thm:four} as it forms the basis of the proof of Theorem~\ref{thm:tsallis}.
\section{Proof of \texorpdfstring{Theorem~\ref{thm:four}}{Theorem 1.2}: four-point negative type of the S-divergence}\label{sec:four-point}
For $\xi\in\reals^4$ with $\sum_a\xi_a=0$ and $P_0,\dots,P_3\in\pp_+^d$ put
\[
F_\xi(P_0,P_1,P_2,P_3):=-\nlsum_{a,c=0}^3\xi_a\xi_c\,\delta_S^2(P_a,P_c),
\]
so that \eqref{eq:four-point} says $F_\xi\ge0$.  Since $F_\xi$ vanishes when
the four matrices coincide, we show that zero is its minimum.  We prove that a minimizer exists and that the only critical
point is the trivial one.

\subsection{Existence of minimizer}\label{sec:normalization}
We start with three elementary properties of $\delta_S$ that we will need. %

\begin{lemma}\label{lem:S-basic}
Let $P,Q,R\in\pp_+^d$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item $\delta_S^2(GPG^*,GQG^*)=\delta_S^2(P,Q)$ for every invertible $G$.
\item $\delta_S^2(I,P)=\sum_{\ell=1}^d\log\cosh\bigl(\tfrac12\log\lambda_\ell(P)\bigr)$,
where $\lambda_1(P),\dots,\lambda_d(P)$ are the eigenvalues of $P$. 
\item $\delta_S^2(P,R)\le \delta_S^2(P,Q)+\delta_S^2(Q,R)+d\log2$.
\end{enumerate}
\end{lemma}


By continuity, it suffices to treat $\xi$ with nonzero coordinates
and a unique largest absolute value, say $|\xi_0|>|\xi_j|$ for $j=1,2,3$;
by Lemma~\ref{lem:S-basic}(a), we may take $P_0=I$. 
The domain $(\pp_+^d)^3$ of $F_\xi(I,\cdot)$ is not compact, so a minimizer
need not exist. A coercive lower bound rules this out. The bound depends on $\xi$ through
\(
\nu_\xi:=\min_{\varnothing\ne T\subsetneq\{0,1,2,3\}}
\Bigl|\nlsum_{a\in T}\xi_a\Bigr|,
\)
which is positive for the fixed $\xi$.


\begin{prop}\label{prop:coercive}
If $\nu_\xi>0$, then
$F_\xi(I,P_1,P_2,P_3)\ge2\nu_\xi^2\max_{1\le i\le3}\delta_S^2(I,P_i)
-d\log2\nlsum_a\xi_a^2.$
Consequently, $F_\xi(I,\cdot)$ attains its minimum on $(\pp_+^d)^3$.
\end{prop}

\begin{proof}
The values $\delta_S^2(P_a,P_c)$ need not satisfy the triangle inequality,
but by Lemma~\ref{lem:S-basic}(b) and~(c) they do after a shift: on
$\{0,1,2,3\}$, the function $D(a,c)=\delta_S^2(P_a,P_c)+d\log2$ for $a\ne c$,
with $D(a,a)=0$, is a metric.  Since $\sum_{a\ne c}\xi_a\xi_c=-\sum_a\xi_a^2$,
the shift changes $F_\xi$ only by a constant:
\[
-\nlsum_{a,c}\xi_a\xi_c\,D(a,c)=F_\xi(I,P_1,P_2,P_3)+d\log2\nlsum_a\xi_a^2.
\]
Metrics on four points are conic combinations of cut metrics
\citep[p.~422]{deza1997geometry}, so,
\[
D(a,c)=\nlsum_{\varnothing\ne T \subsetneq \{0,1,2,3\}} w_T\bigl(\one_T(a)-\one_T(c)\bigr)^2,\qquad w_T\ge0,
\]
where $\one_T$ is the indicator vector of $T$.  Each cut contributes at least
$2\nu_\xi^2w_T$, because $\sum_a\xi_a=0$ gives
\[
-\nlsum_{a,c}\xi_a\xi_c\bigl(\one_T(a)-\one_T(c)\bigr)^2
=2\Bigl(\nlsum_{a\in T}\xi_a\Bigr)^2\ge2\nu_\xi^2,
\]
and $\sum_Tw_T\ge\max_{a,c}D(a,c)\ge\max_i\delta_S^2(I,P_i)$ as each
cut takes values in $\{0,1\}$.  Together, these prove the bound.
Finally, if $F_\xi(I,P_1,P_2,P_3)\le0$, the bound caps every
$\delta_S^2(I,P_i)$, so by Lemma~\ref{lem:S-basic}(b) and
$\log\cosh t\ge|t|-\log2$, eigenvalues of each $P_i$ lie in a
compact interval that depends only on $\xi$ and $d$.  The
triples with this property form a compact set containing $(I,I,I)$, where
$F_\xi$ vanishes, and $F_\xi(I,\cdot)>0$ outside it.  Hence the minimum of
$F_\xi(I,\cdot)$ over this set is its minimum over $(\pp_+^d)^3$.
\end{proof}

\subsection{Potential minimizers (critical points of \texorpdfstring{$F_\xi$}{F})}\label{sec:cayley}
Since $(\pp_+^d)^3$ is open, the minimizer of
Proposition~\ref{prop:coercive} is a critical point.  Lemma~\ref{lem:critical} describes these critical points. %

\begin{lemma}\label{lem:critical}
For the $\xi$ fixed above, put $\lambda_j=-\xi_j/\xi_0$, so that
$\lambda_1+\lambda_2+\lambda_3=1$, $\lambda_j\ne0$, and $|\lambda_j|<1$.
If $(P_1,P_2,P_3)$ is a critical point of $F_\xi(I,\cdot)$, then the coordinates $U_j=(P_j-I)(P_j+I)^{-1}$ satisfy
\begin{equation}\label{eq:resolvent-system}
\nlsum_{j=1}^3\lambda_jU_j=0,
\qquad
\nlsum_{j=1}^3\lambda_j(I-U_iU_j)^{-1}=I\quad(i=1,2,3).
\end{equation}
\end{lemma}
\begin{proof}
Noting $\sum_a\xi_a=0$, and expanding $\delta_S^2(P_a,P_c)$, we see that only joint terms survive, so that
\[
F_\xi=-\nlsum_{a,c=0}^3\xi_a\xi_c\log\det(P_a+P_c).
\]
Taking derivatives of $F_\xi$ with respect to the $P_i$ we obtain the criticality condition
\begin{equation}\label{eq:stationary}
\nlsum_{a=0}^3\xi_a(P_i+P_a)^{-1}=0\qquad(i=1,2,3).
\end{equation}
The equation also holds for $i=0$. Indeed, by Lemma~\ref{lem:S-basic}(a), $F_\xi$ is unchanged when we replace $P_a$ by $e^{tH/2}P_ae^{tH/2}$.  The
derivative $(d/dt)F_\xi|_{t=0}$ along this curve is the sum of its four
partial derivatives.  At a critical point the last three vanish, so the
derivative in $P_0$ at $I$ in the direction $H$ also vanishes.

Now divide \eqref{eq:stationary} by $-\xi_0$ and rewrite it as the set of equations
\begin{equation}\label{eq:consensus}
\nlsum_{j=1}^3\lambda_j(P_i+P_j)^{-1}=(P_i+I)^{-1}\quad(i=1,2,3),
\qquad
\nlsum_{j=1}^3\lambda_j(I+P_j)^{-1}=\tfrac12I.
\end{equation}
The second equation in Cayley coordinates is $\sum_j\lambda_j(I-U_j)=I$, thus
$\sum_j\lambda_jU_j=0$.  Expanding $(I-U_i)(P_i+P_j)(I-U_j)$ with $(I-U_i)P_i=I+U_i$ and $P_j(I-U_j)=I+U_j$ and rearranging gives
$(P_i+P_j)^{-1}=\frac12(I-U_j)(I-U_iU_j)^{-1}(I-U_i)$; %
so, the first equation becomes $\sum_j \lambda_j(I-U_iU_j)^{-1}=I$. %
\end{proof}

\subsection{Solving the critical-point equations}\label{sec:rigidity}

On a common eigenvector of the products $U_iU_j$, the system
\eqref{eq:resolvent-system} becomes scalar, so we first show that
these products commute.

\begin{lemma}\label{lem:rigidity}
Let $\lambda_1,\lambda_2,\lambda_3$ be as in Lemma~\ref{lem:critical}, and let $U_1,U_2,U_3$ be Hermitian
strict contractions satisfying \eqref{eq:resolvent-system}.  Then
$U_1=U_2=U_3=0$.
\end{lemma}

\begin{proof}
\textbf{The squares commute with all $U_j$.}
Fix $i$, let $j,k$ be the other two indices, and write $X=U_i^2$,
$B_j=I-U_iU_j$ and $B_k=I-U_iU_k$.  Multiplying the first equation
in~\eqref{eq:resolvent-system} on the left by $U_i$ gives the first identity
below, and the second is the equation of~\eqref{eq:resolvent-system} for
this $i$:
\begin{equation}\label{eq:paired}
\lambda_i(I-X)+\lambda_jB_j+\lambda_kB_k=I,
\qquad
\lambda_i(I-X)^{-1}+\lambda_jB_j^{-1}+\lambda_kB_k^{-1}=I.
\end{equation}
Here $H=(1-\lambda_i)I+\lambda_iX$ is positive definite, since
$|\lambda_i|<1$.  With $B_j=HZ$, the first identity gives
$\lambda_kB_k=H(I-\lambda_jZ)$; substituting into the second and multiplying
on the right by $H$ yields
\begin{equation}\label{eq:rigidity-key}
\lambda_jZ^{-1}+\lambda_k^2(I-\lambda_jZ)^{-1}
=\bigl(I-\lambda_i(I-X)^{-1}\bigr)H=\varphi(X),
\qquad
\varphi(x):=(1-\lambda_i)^2-\lambda_i\frac{x^2}{1-x}.
\end{equation}
The left side commutes with $Z$.  Since $\lambda_i\ne0$ and $x^2/(1-x)$
increases strictly on $[0,1)$, $\varphi$ is injective there, so $X$ is a
polynomial in $\varphi(X)$ and commutes with $Z$, hence with $B_j=HZ$.
For $C=[U_i^2,U_j]$ this says $U_iC=[U_i^2,U_iU_j]=0$, and taking adjoints
gives $CU_i=0$.  So $C=PCP$ for the projection $P$ onto $\ker U_i$, and
$PCP=0$ because $U_i^2P=PU_i^2=0$.  By symmetry in $j$ and $k$,
\begin{equation}\label{eq:central-squares}
[U_i^2,U_j]=0\qquad\text{for all }i,j.
\end{equation}

\textbf{Shared eigenbasis.}
By the first equation in~\eqref{eq:resolvent-system},
$U_3=-(\lambda_1U_1+\lambda_2U_2)/\lambda_3$, so every product $U_iU_j$ lies
in the span of $U_1^2$, $U_2^2$, $U_1U_2$ and $U_2U_1$.  The first two commute
with all $U_j$ by~\eqref{eq:central-squares}, and $U_1U_2$ commutes with its
adjoint $U_2U_1$, since both products equal $U_1^2U_2^2$.  So these four
matrices are normal and commute, and have a common orthonormal eigenbasis.
Let $v$ be a vector in it.  The eigenvalue of $U_iU_j$ on $v$ is
$(U_iv)^*(U_jv)$, and since $(U_iU_j)(U_jU_i)=U_i^2U_j^2$, its modulus is
$\|U_iv\|\,\|U_jv\|$.  By the equality case of Cauchy--Schwarz, the vectors
$U_1v,U_2v,U_3v$ are parallel: $U_iv=z_ie$ for a unit vector $e$ and scalars
$z_i$ with $|z_i|=\|U_iv\|<1$.  Thus $U_iU_j$ acts on $v$ as
$\overline{z_i}\,z_j$.

\textbf{The scalar moments vanish.}
On $v$, the second set of equations in~\eqref{eq:resolvent-system} becomes
\[
\nlsum_{j=1}^3\frac{\lambda_j}{1-\overline{z_i}\,z_j}=1\qquad(i=1,2,3).
\]
Expand the denominators as geometric series, multiply by $\lambda_i$ and sum
over $i$.  Since $\sum_i\lambda_i=1$ and $|z_i|<1$, we obtain
\[
0=\nlsum_{n\ge1}\Bigl|\nlsum_{j=1}^3\lambda_jz_j^n\Bigr|^2.
\]
Let $\zeta_1,\dots,\zeta_m$ be the distinct nonzero values among the $z_j$,
with weights $w_\ell=\sum_{j:z_j=\zeta_\ell}\lambda_j$.  Then
$\sum_\ell w_\ell\zeta_\ell^n=0$ for $n=1,\dots,m$, and this Vandermonde
system forces every $w_\ell=0$.  But no nonempty subset of the $\lambda_j$
sums to zero: such a sum is $\lambda_j\ne0$, $1-\lambda_j>0$, or $1$.  Thus
$m=0$, so all $z_j=0$ and $U_jv=0$.  As $v$ runs over an orthonormal basis,
$U_1=U_2=U_3=0$.
\end{proof}

\noindent\textbf{\emph{Proof of~\eqref{eq:four-point}}}. We can now prove the four-point negative-type condition~\eqref{eq:four-point}.  By Proposition~\ref{prop:coercive} and Lemmas~\ref{lem:critical}
and~\ref{lem:rigidity}, the minimum of $F_\xi(I,\cdot)$ is attained at a
critical point, and every critical point has $U_1=U_2=U_3=0$, that is,
$P_1=P_2=P_3=I$.  The minimum is therefore $F_\xi(I,I,I,I)=0$.  This proves
\eqref{eq:four-point} for the $\xi$ fixed above, and continuity in $\xi$ gives it for every
zero-sum $\xi$; by the equivalence established via identity~\eqref{eq:polarization}, the truth of~\eqref{eq:logdet-psd} also follows.

\subsection{Number of points}\label{sec:sharp}
The following configuration gives the sharpness of both forms of
Theorem~\ref{thm:four} for $d=2$. The embeddings $X\mapsto X\oplus0$ and $P\mapsto P\oplus I$ extend it to
every $d\ge2$.

\begin{prop}\label{prop:sharp}
Let $u_k=(\cos\frac{k\pi}3,\sin\frac{k\pi}3)^\top$ for $k=0,1,2$, and let
$c=(1,1,1,-\frac32)^\top$.
\begin{enumerate}[label=\textup{(\alph*)}]
\item Let $x=\sqrt3/2$, $X_k=x\,u_ku_k^\top$ for $k=0,1,2$, and $X_3=xI$.  Then
$0\preceq X_k\prec I$, and the matrix
$K=[-\log\det(I-X_iX_j)]_{i,j=0}^3$ satisfies
$c^\top Kc=\log(2^{21}/13^6)<0$.
\item With $X_k$ as in \textup{(a)}, let $P_0=I$ and
$P_{k+1}=(I+X_k)(I-X_k)^{-1}$ for $k=0,\dots,3$.  Then $\delta_S^2$ is not of negative
type on $P_0,\dots,P_4$.
\end{enumerate}
\end{prop}

\begin{proof}
(a) For $i,j\le2$, $X_iX_j$ has rank one and
$\det(I-X_iX_j)=1-x^2(u_i^\top u_j)^2$.  Since $(u_i^\top u_j)^2=\frac14$ for
$i\ne j$, this determinant is $\frac14$ for $i=j$ and $\frac{13}{16}$ for $i\ne j$.  Also
$\det(I-X_kX_3)=1-x^2=\frac14$ and $\det(I-X_3^2)=\frac1{16}$.  With $L=\log4$
and $\ell=\log\frac{16}{13}$,
\[
K=\begin{pmatrix}L&\ell&\ell&L\\ \ell&L&\ell&L\\ \ell&\ell&L&L\\ L&L&L&2L\end{pmatrix},
\qquad
c^\top Kc=6\ell-\tfrac32L=\log\frac{2^{21}}{13^6}<0.
\]

(b) The vector
$\xi=(-\frac32,1,1,1,-\frac32)$ ($\xi^T1=0$) gives
$\sum_{a,e=0}^4\xi_a\xi_e\,\delta_S^2(P_a,P_e)=-c^\top Kc>0$.
\end{proof}


\section{Proof of \texorpdfstring{Theorem~\ref{thm:tsallis} for $1/2\le\alpha<1$}{Theorem 1.1 for 1/2 <= alpha < 1}}\label{sec:schatten}

We prove Theorem~\ref{thm:tsallis} for $\alpha<1$ on the positive unit
sphere of a Schatten norm.
The passage to the sphere rests on the equality in
\eqref{eq:state-distance}, which we prove first.

\subsection{Reduction to a Schatten sphere}\label{sec:reduction}

\begin{prop}\label{prop:centroid}
Let $0<\alpha<1$, let $\rho,\sigma\in\Dc_d$, and put
$M=\rho^\alpha+\sigma^\alpha$.  Then
\begin{equation}\label{eq:centroid-value}
(1-\alpha)\min_{\tau\in\Dc_d}
\bigl\{T_\alpha(\rho,\tau)+T_\alpha(\sigma,\tau)\bigr\}
=2-\pnorm{M}{1/\alpha},
\end{equation}
and the minimum is attained only at
\begin{equation}\label{eq:state-centroid}
\tau_\star=\frac{M^{1/\alpha}}{\trace M^{1/\alpha}}.
\end{equation}
\end{prop}

\begin{proof}
The scaled objective is $2-\trace(M\tau^{1-\alpha})$, and Schatten
H\"older with exponents $1/\alpha$ and $1/(1-\alpha)$ gives
$\trace(M\tau^{1-\alpha})\le\pnorm{M}{1/\alpha}\pnorm{\tau^{1-\alpha}}{1/(1-\alpha)}=\pnorm{M}{1/\alpha}$.
The state~\eqref{eq:state-centroid} attains equality, and strict convexity
of the Schatten $1/(1-\alpha)$-norm makes it the only maximizer.
\end{proof}

For $1<p<\infty$, let $\mathcal S_p=\{A\in\pp^d:\trace A^p=1\}$ and, for
$A,B\in\mathcal S_p$, let $D_p(A,B)=(2-\pnorm{A+B}{p})^{1/2}$.
For $p=1/\alpha$, $\rho\mapsto\rho^\alpha$ is a bijection from
$\Dc_d$ onto $\mathcal S_p$, and Proposition~\ref{prop:centroid}
gives $d_\alpha(\rho,\sigma)=D_p(\rho^\alpha,\sigma^\alpha)$.
It therefore suffices to prove the triangle inequality for $1<p\le2$;
Proposition~\ref{prop:p-large} treats $p>2$.

\subsection{An integral representation}\label{sec:mellin}

We transfer the log-determinant form \eqref{eq:logdet-psd} of
Theorem~\ref{thm:four} to the trace-power kernel
\begin{equation}\label{eq:Lp-definition}
L_p(A,B):=\trace(A+B)^p-\trace A^p-\trace B^p,
\qquad A,B\in\pp^d,
\end{equation}
and then use its three-point positivity on the sphere.

\begin{theorem}\label{thm:Lp}
Let $A_1,A_2,A_3\in\pp^d$.  For $1<p\le2$,
\begin{equation}\label{eq:Lp-psd}
\bigl[L_p(A_i,A_j)\bigr]_{i,j=1}^3\succeq0.
\end{equation}
\end{theorem}

\begin{proof}
At $p=2$, $L_2(A_i,A_j)=2\trace(A_iA_j)$ is a Gram matrix.
For $1<p<2$, put $\kappa_p=p\sin(\pi(p-1))/\pi>0$ and use the scalar
identity \citep{sra2021metrics}
\begin{equation}\label{eq:power-integral}
x^p=\kappa_p\int_0^\infty
\bigl(x/t-\log(1+x/t)\bigr)t^{p-1}\dd t\qquad(x\ge0).
\end{equation}
For $t>0$, set $X_A(t)=A(A+tI)^{-1}$ and
\begin{equation}\label{eq:Kt}
K_t(A,B):=\log\frac{\det(A+tI)\det(B+tI)}{t^d\det(A+B+tI)}
=-\log\det\bigl(I-X_A(t)X_B(t)\bigr).
\end{equation}
Applying \eqref{eq:power-integral} to $A+B$, $A$ and $B$ and taking traces
cancels the linear terms and gives
\begin{equation}\label{eq:mellin}
L_p(A,B)=\kappa_p\int_0^\infty t^{p-1}K_t(A,B)\dd t.
\end{equation}
Since $0\preceq X_{A_i}(t)\prec I$, Theorem~\ref{thm:four} gives
$[K_t(A_i,A_j)]\succeq0$, and integrating proves~\eqref{eq:Lp-psd}.
\end{proof}


\subsection{The triangle inequality}\label{sec:triangle}

Three-point positivity of $L_p$ places any three points of $\mathcal S_p$ on
a sphere, where $D_p$ becomes a function of an angle.  Let $1<p\le2$, let
$A_1,A_2,A_3\in\mathcal S_p$, and put
\begin{equation}\label{eq:gram}
G_{ij}:=2^{1-p}\trace(A_i+A_j)^p-1.
\end{equation}
Since $\trace A_i^p=1$,
\[
G=2^{1-p}\bigl[L_p(A_i,A_j)\bigr]_{i,j=1}^3+\bigl(2^{2-p}-1\bigr)\one\one^\top,
\]
which is positive semidefinite by Theorem~\ref{thm:Lp} and $2^{2-p}\ge1$.
Since $G_{ii}=1$, the matrix $G$ is the Gram matrix of three unit
vectors $u_1,u_2,u_3\in\reals^3$; write $G_{ij}=\cos\theta_{ij}$ with
$0\le\theta_{ij}\le\pi$.  These angles satisfy the spherical triangle
inequality
\begin{equation}\label{eq:spherical-triangle}
\theta_{13}\le\min\{\theta_{12}+\theta_{23},\pi\}.
\end{equation}

By \eqref{eq:gram}, $\pnorm{A_i+A_j}{p}=2\cos^{2/p}(\theta_{ij}/2)$, so
\begin{equation}\label{eq:fp-definition}
D_p(A_i,A_j)=f_p(\theta_{ij}),\qquad
f_p(\theta):=\bigl(2(1-\cos^{2/p}(\theta/2))\bigr)^{1/2}\quad(0\le\theta\le\pi).
\end{equation}
The triangle inequality for $D_p$ therefore reduces to a scalar property
of $f_p$.

\begin{lemma}\label{lem:fp}
For $1<p\le2$, the function $f_p$ in \eqref{eq:fp-definition}
is increasing and concave, $f_p(0)=0$, and
\begin{equation}\label{eq:fp-subadd}
f_p\bigl(\min\{x+y,\pi\}\bigr)\le f_p(x)+f_p(y)
\qquad(0\le x,y\le\pi).
\end{equation}
\end{lemma}

\begin{proof}
Since $\cos(\theta/2)$ decreases on $[0,\pi]$, the function $f_p$ is
increasing, and $f_p(0)=0$.  For $0<\theta<\pi$,
put $z=\cos^{2/p}(\theta/2)$.  Differentiation gives
\begin{equation}\label{eq:g-second}
f_p''(\theta)\le0
\quad\Longleftrightarrow\quad
(2-z)(1-z^p)\ge p(1-z).
\end{equation}
Since $t\mapsto t^{p-1}$ is concave on $[z,1]$,
\[
\frac{1-z^p}{p(1-z)}
=\frac1{1-z}\int_z^1t^{p-1}\dd t
\ge\frac{1+z^{p-1}}2
\ge\frac{1+z}2
\ge\frac1{2-z}.
\]
Thus $f_p$ is concave.

Being concave with $f_p(0)=0$, $f_p$ is subadditive; if $x+y\ge\pi$,
monotonicity gives $f_p(\pi)\le f_p(x)+f_p(\pi-x)\le f_p(x)+f_p(y)$.  This
proves~\eqref{eq:fp-subadd}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:tsallis} for $1/2\le\alpha<1$]
Minkowski's inequality and strict convexity of the Schatten $p$-norm
\citep{bhatia1997matrix} give $D_p\ge0$, with equality only when $A=B$.  For
the triangle inequality, let $A_1,A_2,A_3\in\mathcal S_p$,
with angles $\theta_{ij}$ as above.
By \eqref{eq:spherical-triangle}, the monotonicity of $f_p$, and
Lemma~\ref{lem:fp},
\[
D_p(A_1,A_3)=f_p(\theta_{13})
\le f_p\bigl(\min\{\theta_{12}+\theta_{23},\pi\}\bigr)
\le f_p(\theta_{12})+f_p(\theta_{23})
=D_p(A_1,A_2)+D_p(A_2,A_3).
\qedhere
\]
\end{proof}

The angles $\theta_{ij}$ give more than the triangle inequality: three states
lie on a sphere with chordal distance $d_\alpha$.

\begin{theorem}\label{thm:sphere}
Let $1<p\le2$ and $A_1,A_2,A_3\in\mathcal S_p$.  Then
\[
\bigl[\pnorm{(A_i+A_j)/2}{p}\bigr]_{i,j=1}^3\succeq0.
\]
In particular, for $1/2\le\alpha<1$, any three states with the metric
$d_\alpha$ embed isometrically in the unit sphere of $\reals^3$.
\end{theorem}

\begin{proof}
With the angles of \eqref{eq:gram}, $\pnorm{(A_i+A_j)/2}{p}=\cos\psi_p(\theta_{ij})$, where
$\psi_p(\theta):=\arccos\cos^{2/p}(\theta/2)$ is increasing on $[0,\pi]$, with
$\psi_p(0)=0$ and $\psi_p\le\pi/2$.  With $\beta=2/p$ and
$y=\cos^2(\theta/2)$,
$\psi_p'(\theta)^2=\frac{\beta^2}4\,y^{\beta-1}(1-y)/(1-y^\beta)$, and a brief
calculation shows that this increases in $y$ because
$(1-y^\beta)/(1-y)\le\beta$.  So $\psi_p$ is concave, and
\eqref{eq:spherical-triangle} with the argument of Lemma~\ref{lem:fp} gives
$\psi_p(\theta_{13})\le\psi_p(\theta_{12})+\psi_p(\theta_{23})$, and likewise
for the other two angles.  Three angles in $[0,\pi/2]$ with this property are
the angles between three unit vectors $v_1,v_2,v_3$, whose Gram matrix is
$[\cos\psi_p(\theta_{ij})]$.  For states, $A_i=\rho_i^\alpha$ gives
$\pnorm{(\rho_i^\alpha+\rho_j^\alpha)/2}{1/\alpha}=1-d_\alpha(\rho_i,\rho_j)^2/2$,
so these unit vectors satisfy $\|v_i-v_j\|=d_\alpha(\rho_i,\rho_j)$.
\end{proof}

\subsection{The case \texorpdfstring{$p=2$}{p=2}}\label{sec:p2}

At $p=2$, which corresponds to the endpoint $\alpha=1/2$ of
Theorem~\ref{thm:tsallis}, the distance
$D_2(A,B)=\bigl(2-(2+2\trace AB)^{1/2}\bigr)^{1/2}$
depends only on the Hilbert--Schmidt inner product $\trace AB$.
We can therefore work on the unit sphere of any real Hilbert space, where
the distance is also of negative type.

\begin{prop}\label{prop:p2}
On the unit sphere $\mathbb S$ of a real Hilbert space $\mathcal H$, put
$D(x,y)=(2-\|x+y\|)^{1/2}$.
Then $D$ is a metric of negative type on $\mathbb S$, so
$(\mathbb S,D^{1/2})$ embeds isometrically in a Hilbert space.
\end{prop}

\begin{proof}
For unit vectors at angle $\theta$, $D(x,y)=2\sin(\theta/4)=f_2(\theta)$, so
\eqref{eq:spherical-triangle} and Lemma~\ref{lem:fp} prove that $D$ is a
metric.

For negative type, write $D(x,y)=\varphi(\ip{x}{y})$, where
$\varphi(\cos\theta)=2\sin(\theta/4)$.  Then
$(1-t^2)\varphi''-t\varphi'+\varphi/16=0$, so the Taylor coefficients
$b_k$ of $\varphi$ at $0$, whose series converges on $|t|<1$, satisfy
\[
b_{k+2}=\frac{k^2-1/16}{(k+1)(k+2)}\,b_k\qquad(k\ge0).
\]
Since $b_0=2\sin(\pi/8)>0$ and $b_1=-\frac12\cos(\pi/8)<0$, the recurrence
gives $b_k<0$ for every $k\ge1$.
Let $x_1,\dots,x_n\in\mathbb S$, let $c\in\reals^n$ with $\sum_ic_i=0$,
and let $0<r<1$.  Since $\sum_ic_i=0$ and
$\ip{x_i}{x_j}^k=\ip{x_i^{\otimes k}}{x_j^{\otimes k}}$ in the $k$-fold tensor
power of $\mathcal H$,
\[
\nlsum_{i,j=1}^nc_ic_j\,\varphi\bigl(r\ip{x_i}{x_j}\bigr)
=\nlsum_{k\ge1}b_kr^k\Bigl\|\nlsum_{i=1}^nc_ix_i^{\otimes k}\Bigr\|^2\le0.
\]
Letting $r\uparrow1$ gives negative type, and Schoenberg's criterion gives
the embedding of $(\mathbb S,D^{1/2})$.
\end{proof}

For states, $d_{1/2}(\rho,\sigma)=D_2(\rho^{1/2},\sigma^{1/2})$ is a function of
the quantum affinity $\trace\rho^{1/2}\sigma^{1/2}$ \citep{luo2004informational},
so Proposition~\ref{prop:p2} shows that $d_{1/2}$ is of negative type.

\subsection{The range of \texorpdfstring{$p$}{p}}

For $p>2$, the triangle inequality fails even on diagonal matrices.

\begin{prop}\label{prop:p-large}
Let $p>2$ and $d=2$, and let $A=\diag(0,1)$, $C=\diag(1,0)$ and
$B_\varepsilon=\diag\bigl(\varepsilon,(1-\varepsilon^p)^{1/p}\bigr)$ for
$0<\varepsilon<1$.  Then $A,B_\varepsilon,C\in\mathcal S_p$, and
\[
D_p(C,B_\varepsilon)+D_p(B_\varepsilon,A)<D_p(C,A)
\]
for all sufficiently small $\varepsilon$.
\end{prop}

\begin{proof}
All three matrices are diagonal, so the norms are $\ell_p$ norms of the
diagonals.  As $\varepsilon\to0$, expanding
$(1-\varepsilon^p)^{1/p}=1-\varepsilon^p/p+O(\varepsilon^{2p})$ gives
$D_p(A,B_\varepsilon)^2=\frac{1-2^{1-p}}{p}\varepsilon^p+O(\varepsilon^{2p})$,
so $D_p(A,B_\varepsilon)=O(\varepsilon^{p/2})$.  On the other hand,
$\pnorm{C+B_\varepsilon}{p}^p=2+p\varepsilon+O(\varepsilon^2)$ because $p>2$, so
\[
D_p(C,B_\varepsilon)=D_p(C,A)-\frac{2^{1/p}}{4D_p(C,A)}\,\varepsilon+O(\varepsilon^2).
\]
Since $p/2>1$, the sum $D_p(C,B_\varepsilon)+D_p(B_\varepsilon,A)$ is
smaller than $D_p(C,A)$ for small $\varepsilon$.  Adjoining zero blocks gives
the same counterexample for every $d\ge2$.
\end{proof}

\section{Proof of \texorpdfstring{Theorem~\ref{thm:tsallis} for $1<\alpha\le2$}{Theorem 1.1 for 1 < alpha <= 2}}\label{sec:alpha-large}

For $\alpha>1$, the proof of Proposition~\ref{prop:centroid} with the reverse
H\"older inequality gives the equality in~\eqref{eq:state-distance}, with the
same minimizing state~\eqref{eq:state-centroid}.  We prove the triangle
inequality first at $\alpha=2$.  At a minimizing middle state,
the stationarity equation forces two auxiliary matrices to commute;
the trace-one constraints then reduce the problem to a scalar inequality.
A positive linear construction on tensor powers transfers this endpoint
to the remaining orders.

For $p=1/\alpha$ and $A_i=\rho_i^\alpha$, strict concavity of
$X\mapsto\trace X^p$ gives $\trace(A_i+A_j)^p\ge2^p$, with equality
only when $A_i=A_j$, so $d_\alpha$ separates states.  It remains to prove
the triangle inequality.

\subsection{The scalar inequality}

The trace-one constraints will give weighted scalar means of at least
one.  To use these inequalities directly, we subtract the tangent term
$(x+y)/\sqrt2$ before averaging.

\begin{lemma}\label{lem:weighted-endpoint}
Let $w_k,x_k,y_k\ge0$ satisfy
$\nlsum_k w_k=1$, $\nlsum_k w_kx_k\ge1$, and $\sum_k w_ky_k\ge1$.
If $r=\nlsum_k w_k\sqrt{1+x_k^2}$ and
$s=\nlsum_k w_k\sqrt{1+y_k^2}$, then
\begin{equation}\label{eq:weighted-endpoint}
\nlsum_k w_k\sqrt{x_k^2+y_k^2}
\le L(r,s):=\sqrt{2+\bigl(\sqrt{r^2-2}+\sqrt{s^2-2}\bigr)^2}.
\end{equation}
\end{lemma}
\begin{proof}
Jensen's inequality gives $r,s\ge\sqrt2$.  For $a,b\ge0$, put
\[
\delta(a,b)=\sqrt{a^2+b^2}-\frac{a+b}{\sqrt2},\qquad
D(a,b)=\sqrt{\delta(a,b)\bigl(\delta(a,b)+2\sqrt2\bigr)}.
\]
We first prove the pointwise inequality
\begin{equation}\label{eq:scalar-middle}
D(x,y)\le D(x,1)+D(1,y),\qquad x,y\ge0.
\end{equation}
For $a\le b$, both $\delta$ and $D$ are nonincreasing in $a$ and
nondecreasing in $b$.  Thus \eqref{eq:scalar-middle} holds when $x,y$
lie on the same side of $1$.
By symmetry and continuity, it remains to treat $0<x<1<y$.

We compare $-D_x(x,y)$ with its value at $y=1$.
For fixed $0<x<1$ and $y\ge1$, write
$H=\sqrt{2(x^2+y^2)}$, $S=x+y$, and $E=H-S>0$.
Then $D^2=E(E+4)/2$, and differentiation gives
\[
D_{xy}(x,y)=\frac{E^2}{H^3D(x,y)^3}
\bigl[6(S-xy)+(2-xy)E\bigr].
\]
Since $S-xy=x+(1-x)y>0$, the bracket is positive if $xy\le2$.  If $xy>2$, it has the sign of
\[
6-\frac{xy-2}{x+(1-x)y}E.
\]
Both factors in the subtracted product are positive and increasing in
$y$, since
\[
\frac{\partial}{\partial y}\frac{xy-2}{x+(1-x)y}
=\frac{x^2-2x+2}{(x+(1-x)y)^2}>0,\qquad
E_y=\frac{2y}{H}-1>0.
\]
Thus $D_{xy}$ can change sign only from positive to negative, so
$-D_x(x,y)$ has no interior maximum as a function of $y\ge1$.
Its limit at infinity is $1/\sqrt2$.

At $y=1$, rationalization gives
\[
D(x,1)=(1-x)h(x),\qquad
h(x)^2=\frac12+\frac{1-x}{\sqrt{2(1+x^2)}+x+1}.
\]
The fraction is positive and decreasing, so $h>1/\sqrt2$, $h'<0$, and
$-D_x(x,1)=h(x)-(1-x)h'(x)>1/\sqrt2$.
Hence $-D_x(x,y)\le-D_x(x,1)$ for $0<x<1$ and $y\ge1$, and
integrating proves \eqref{eq:scalar-middle}:
\[
D(x,y)-D(1,y)
=\int_x^1-D_a(a,y)\dd a
\le\int_x^1-D_a(a,1)\dd a=D(x,1).
\]

To average \eqref{eq:scalar-middle}, we use concavity of $L$ on
$[\sqrt2,\infty)^2$.  In the interior, put
$u=\sqrt{r^2-2}$, $v=\sqrt{s^2-2}$, $q=u+v$, and
$\xi=(r/u,s/v)^\top$.  Differentiation gives
\[
\nabla^2L=\frac2{L^3}
\left[\xi\xi^\top-qL^2\diag(u^{-3},v^{-3})\right]\preceq0,
\]
where the last inequality follows by weighted Cauchy--Schwarz from
$r^2u+s^2v=u^3+v^3+2q\le q(q^2+2)=qL^2$.
Also $L_r=qr/(uL)\ge1$, because
$q^2r^2-u^2L^2=2(q^2-u^2)\ge0$, and similarly $L_s\ge1$.
Integrating these derivative bounds gives
\begin{equation}\label{eq:L-increment}
L(r+a,s+b)\ge L(r,s)+a+b,
\qquad r,s\ge\sqrt2,\quad a,b\ge0.
\end{equation}

Let $(X,Y)$ take the values $(x_k,y_k)$ with probabilities $w_k$, and put
$t=\mathbb E\sqrt{X^2+Y^2}$.  Define
\[
\widehat R=\sqrt2+\delta(X,1),\qquad
\widehat S=\sqrt2+\delta(Y,1),\qquad
\widehat T=\sqrt2+\delta(X,Y).
\]
These variables are at least $\sqrt2$, and \eqref{eq:scalar-middle} is
exactly $\widehat T\le L(\widehat R,\widehat S)$.
The mean constraints give $a:=(\mathbb EX-1)/\sqrt2\ge0$ and
$b:=(\mathbb EY-1)/\sqrt2\ge0$, so $\mathbb E\widehat R=r-a$,
$\mathbb E\widehat S=s-b$ and $\mathbb E\widehat T=t-a-b$.
Concavity and \eqref{eq:L-increment} now give \eqref{eq:weighted-endpoint}:
\[
t-a-b\le\mathbb E L(\widehat R,\widehat S)
\le L(r-a,s-b)\le L(r,s)-a-b.\qedhere
\]
\end{proof}

\subsection{The endpoint \texorpdfstring{$\alpha=2$}{alpha=2}}

The scalar means in Lemma~\ref{lem:weighted-endpoint} arise from a trace
comparison.  Commutativity is needed only for the auxiliary matrices
in this comparison, not for the states.

\begin{lemma}\label{lem:commuting-trace}
Let $Z\succ0$, and let $R_1,\ldots,R_\ell$ be commuting Hermitian matrices.
If
\[
F^2:=\nlsum_jR_j^2-cI\succeq0,\qquad
C:=Z^{1/2}\left(\nlsum_jR_jZR_j-cZ\right)Z^{1/2}\succeq0,
\]
where $F\succeq0$ and $c\in\reals$, then
$\trace\sqrt C\le\trace(ZF)$.
\end{lemma}
\begin{proof}
For $W\succ0$, Hilbert--Schmidt Cauchy--Schwarz gives
$(\trace\sqrt C)^2\le\trace(CW^{-1})\trace W$.
Choose $W=Z^{1/2}(F+\varepsilon I)Z^{1/2}$.  Since each $R_j$ commutes
with $F$, cyclicity of trace gives
\[
\trace(CW^{-1})
=\trace\bigl[ZF^2(F+\varepsilon I)^{-1}\bigr]
\le\trace(ZF),\qquad
\trace W=\trace(ZF)+\varepsilon\trace Z.
\]
Let $\varepsilon\downarrow0$.
\end{proof}

\begin{prop}\label{prop:alpha-two}
The function $d_2$ satisfies the triangle inequality on $\Dc_d$
for every $d$.
\end{prop}
\begin{proof}
Write $T(X,Y)=\trace\sqrt{X^2+Y^2}$, so
$d_2(X,Y)=\sqrt{T(X,Y)^2-2}$ for states.  By continuity, it suffices to
treat distinct positive definite states $\rho,\sigma$.  Minimize
$f(z)=d_2(\rho,z)+d_2(z,\sigma)$ over $\Dc_d$.
A minimizer exists by compactness.  If it is an endpoint, there is
nothing to prove.  Otherwise put
\[
u=d_2(\rho,z)>0,\qquad v=d_2(\sigma,z)>0,\qquad
K_1=\sqrt{\rho^2+z^2},\quad K_2=\sqrt{\sigma^2+z^2},
\]
and $t_j=\trace K_j$, so $t_1^2=u^2+2$ and $t_2^2=v^2+2$.
The gradient is
\[
\nabla f(z)=\tfrac12(zH+Hz),\qquad
H=\frac{t_1}{u}K_1^{-1}+\frac{t_2}{v}K_2^{-1}\succ0.
\]
The minimizer is positive definite.  Indeed, if $e\in\ker z$ is a unit
vector, the feasible path $z_\varepsilon=(1-\varepsilon)z+\varepsilon ee^*$
has derivative $-\trace(z^2H)<0$.

Stationarity under $\trace z=1$ gives $zH+Hz=2\lambda I$, so
$H=\lambda z^{-1}$, that is,
\begin{equation}\label{eq:endpoint-stationarity}
\frac{t_1}{u}R_1^{-1}+\frac{t_2}{v}R_2^{-1}=\lambda I,\qquad
R_j:=z^{-1/2}K_jz^{-1/2}.
\end{equation}
Thus $R_1,R_2$ commute.  Operator monotonicity of the square root also
gives $K_j\succeq z$, so $R_j\succeq I$.
In a common eigenbasis $e_k$, write
\[
R_je_k=r_{jk}e_k,\qquad w_k=\langle e_k,ze_k\rangle,\qquad
x_k=\sqrt{r_{1k}^2-1},\quad y_k=\sqrt{r_{2k}^2-1}.
\]
Then $\sum_k w_k=1$ and
\[
t_1=\nlsum_k w_k\sqrt{1+x_k^2},\qquad
t_2=\nlsum_k w_k\sqrt{1+y_k^2}.
\]
Since $\rho^2=z^{1/2}(R_1zR_1-z)z^{1/2}$,
Lemma~\ref{lem:commuting-trace} gives
\[
1=\trace\rho\le\nlsum_k w_kx_k,\qquad
1=\trace\sigma\le\nlsum_k w_ky_k.
\]
Applying that lemma once more to
$\rho^2+\sigma^2=z^{1/2}(R_1zR_1+R_2zR_2-2z)z^{1/2}$ yields
\[
T(\rho,\sigma)\le\nlsum_k w_k\sqrt{x_k^2+y_k^2}
\le L(t_1,t_2)=\sqrt{2+(u+v)^2},
\]
where the second inequality is Lemma~\ref{lem:weighted-endpoint}.
Thus $d_2(\rho,\sigma)\le f(z)$, and minimality gives the triangle
inequality for every middle state.
\end{proof}

\subsection{Transfer to the other orders}

The transfer uses positive linear maps so that sums are preserved even
when the inputs do not commute.

\begin{lemma}\label{lem:power-transfer}
Let $1/2<p<1$, and let $A_1,A_2,A_3\in\pp^d$ satisfy
$\trace A_i^p=1$.  There are finite-dimensional density matrices
$B_i^{(N)}$ such that, for $i\ne j$,
\begin{equation}\label{eq:power-transfer}
\trace\sqrt{(B_i^{(N)})^2+(B_j^{(N)})^2}
\longrightarrow\bigl[\trace(A_i+A_j)^p\bigr]^{1/(2p)}.
\end{equation}
\end{lemma}
\begin{proof}
Let $S$ have density $p s^{-p-1}$ on $[1,\infty)$, and put
$\phi(t)=\mathbb E e^{-tS}$.  A change of variables gives
\[
1-\phi(t)=pt^p\int_t^\infty(1-e^{-v})v^{-p-1}\dd v
\sim\ell_p t^p,\qquad
1-\phi(t)\le\ell_p t^p,
\]
where $\ell_p=p\int_0^\infty(1-e^{-v})v^{-p-1}\dd v\in(0,\infty)$.

For independent copies $S_1,\ldots,S_N$, define the positive linear map
\[
\Phi_N(X)=N^{-1/p}\sum_{r=1}^N S_r X^{(r)}
\]
on $(\mathbb C^d)^{\otimes N}$, where $X^{(r)}$ acts in tensor factor $r$.
Writing $\tau_m=\trace/m$, independence and commutativity of the tensor
copies of each fixed $X\succeq0$ give
\[
\mathbb E\tau_{d^N}e^{-t\Phi_N(X)}
=\bigl[\tau_d\phi(tN^{-1/p}X)\bigr]^N
\longrightarrow e^{-\ell_p t^p\tau_dX^p}.
\]
Moreover,
\[
0\le1-\mathbb E\tau_{d^N}e^{-t\Phi_N(X)}
\le\min\{1,\ell_p t^p\tau_dX^p\}.
\]

Multiply by $t^{-3/2}$ and use
$\sqrt x=(2\sqrt\pi)^{-1}\int_0^\infty(1-e^{-tx})t^{-3/2}\dd t$.
The bound is integrable because $p>1/2$, so dominated convergence gives
\begin{equation}\label{eq:tensor-root-limit}
\mathbb E\tau_{d^N}\sqrt{\Phi_N(X)}
\longrightarrow\gamma_p(\tau_dX^p)^{1/(2p)},\qquad
\gamma_p:=\frac1{2\sqrt\pi}\int_0^\infty
(1-e^{-\ell_p t^p})t^{-3/2}\dd t\in(0,\infty).
\end{equation}

For each $N$, approximate these expectations simultaneously for the
six inputs $A_i$ and $A_i+A_j$ ($i<j$) by a finite weighted average,
with error at most $1/N$; such averages exist because the integrands are
continuous and dominated by
$\|X\|_{\mathrm{op}}^{1/2}N^{-1/(2p)}\sum_r\sqrt{S_r}$.
Let $s_\ell$ be the nodes and $q_\ell\ge0$ their weights, and set
\[
D_i^{(N)}=\bigoplus_\ell\frac{q_\ell}{d^N}
\sqrt{\Phi_{N,s_\ell}(A_i)}.
\]
Their traces are the chosen weighted averages.  Linearity gives
\[
\trace\sqrt{(D_i^{(N)})^2+(D_j^{(N)})^2}
=\sum_\ell q_\ell\tau_{d^N}
\sqrt{\Phi_{N,s_\ell}(A_i+A_j)}.
\]
Consequently, with $M=\gamma_p d^{-1/(2p)}>0$,
\[
m_i^{(N)}:=\trace D_i^{(N)}\longrightarrow M,\qquad
T_{ij}^{(N)}:=\trace\sqrt{(D_i^{(N)})^2+(D_j^{(N)})^2}
\longrightarrow M\bigl[\trace(A_i+A_j)^p\bigr]^{1/(2p)}.
\]

For sufficiently large $N$, normalize
$B_i^{(N)}=D_i^{(N)}/m_i^{(N)}$.  Operator monotonicity gives
\[
\frac{T_{ij}^{(N)}}{\max\{m_i^{(N)},m_j^{(N)}\}}
\le\trace\sqrt{(B_i^{(N)})^2+(B_j^{(N)})^2}
\le\frac{T_{ij}^{(N)}}{\min\{m_i^{(N)},m_j^{(N)}\}}.
\]
Both bounds have the limit in \eqref{eq:power-transfer}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:tsallis} for $1<\alpha\le2$]
Proposition~\ref{prop:alpha-two}
proves the triangle inequality at $\alpha=2$.  For $1<\alpha<2$,
set $p=1/\alpha$ and $A_i=\rho_i^\alpha$.
Lemma~\ref{lem:power-transfer} gives
\[
d_2(B_i^{(N)},B_j^{(N)})
\longrightarrow
\sqrt{\bigl[\trace(A_i+A_j)^p\bigr]^{1/p}-2}
=d_\alpha(\rho_i,\rho_j).
\]
The endpoint triangle inequality passes to this limit.
\end{proof}

\begin{aiusage}  %
The results in this paper were developed through extensive use of AI
assistants: OpenAI's GPT~Pro for the main line of argument, and Anthropic's
Claude for the endpoint results in Section~\ref{sec:p2} and Appendix~\ref{sec:qjsd},
Remark~\ref{rmk:free-metric}, Proposition~\ref{prop:p-large}, and the writing,
positioning, and literature search.  OpenAI assistants also developed the
four-point consequences in Appendix~\ref{sec:jensen} and assisted with
proof simplification and reorganization.  The arguments emerged through
sustained work involving failed attempts, counterexamples, numerical
checks, proof audits, and repairs.
\end{aiusage}

\aiacknowledgements

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\appendix

\section{Relation to the cone and Jensen--Shannon metrics}\label{sec:variational}

For $0<\alpha<1$, we relate $d_\alpha$ to the two earlier metrics of the introduction: the
cone metric of \citet{sra2021metrics}, which drops the trace constraint, and
the Jensen--Shannon metric, the limit of the rescaled $d_\alpha$ as
$\alpha\uparrow1$.

\subsection{Comparison with the cone minimum}

To let the center range over the cone, extend $T_\alpha$ to
$X,Z\in\pp^d$ by
\begin{equation}\label{eq:tsallis-general}
T_\alpha(X,Z):=\frac{\alpha\trace X+(1-\alpha)\trace Z-\trace X^\alpha Z^{1-\alpha}}{1-\alpha}.
\end{equation}
Minimizing over the cone gives the relative-entropy Jensen--Tsallis
divergence of \citet[Theorem~4.1]{sra2021metrics}; its commuting case was
studied by \citet{osterreicher2003new}.

\begin{prop}\label{prop:free}
Let $0<\alpha<1$ and $\rho,\sigma\in\Dc_d$, and put
\[
M_\alpha:=\Bigl(\frac{\rho^\alpha+\sigma^\alpha}2\Bigr)^{1/\alpha},
\qquad h:=\trace M_\alpha.
\]
The unique minimizer over $\pp^d$ is $M_\alpha$, and
\begin{equation}\label{eq:free-value}
(1-\alpha)\min_{Z\in\pp^d}
\bigl\{T_\alpha(\rho,Z)+T_\alpha(\sigma,Z)\bigr\}=2\alpha(1-h).
\end{equation}
Proposition~\ref{prop:centroid} gives the constrained center
$M_\alpha/h$ and
\begin{equation}\label{eq:constrained-value}
d_\alpha(\rho,\sigma)^2=2(1-h^\alpha).
\end{equation}
\end{prop}

\begin{proof}
Write $Z=s\tau$, where $s=\trace Z$ and $\tau\in\Dc_d$.
For fixed $s>0$, Proposition~\ref{prop:centroid} gives the unique optimal
state $\tau=M_\alpha/h$.  Since
$\pnorm{\rho^\alpha+\sigma^\alpha}{1/\alpha}=2h^\alpha$, the scaled objective
reduces to $2\alpha+2(1-\alpha)s-2h^\alpha s^{1-\alpha}$, a strictly convex
function of $s\ge0$ minimized at $s=h$, with value $2\alpha(1-h)$.  The minimizing matrix is $s\tau=M_\alpha$.
\end{proof}

\begin{rmk}\label{rmk:free-metric}
For $1/2\le\alpha<1$ and three states $\rho_i$, let $p=1/\alpha$ and take
$A_i=\rho_i^\alpha$ in the Gram matrix $G$ of \eqref{eq:gram}.  Then
$h_{ij}:=\trace((\rho_i^\alpha+\rho_j^\alpha)/2)^{1/\alpha}=(1+G_{ij})/2$.
Thus $H=[h_{ij}]=(\one\one^\top+G)/2$ is a Gram matrix of unit vectors
$v_i$, and the scaled cone minimum satisfies
$2\alpha(1-h_{ij})=\alpha\|v_i-v_j\|^2$.
Its square root satisfies the triangle inequality and separates states
by \eqref{eq:constrained-value} and Theorem~\ref{thm:tsallis}, recovering
the cone metric of \citet[Theorem~4.1]{sra2021metrics} on states.
In contrast, $d_\alpha(\rho_i,\rho_j)^2=2(1-h_{ij}^\alpha)$ by
\eqref{eq:constrained-value}, and its triangle inequality needs Lemma~\ref{lem:fp}.
\end{rmk}


\subsection{The limit \texorpdfstring{$\alpha\uparrow1$}{alpha->1}}\label{sec:qjsd}

For the Umegaki relative entropy $\trace\rho(\log\rho-\log\tau)$, the
optimal center is $m=(\rho+\sigma)/2$ and the least total is
$2\qjsd(\rho,\sigma)$, where
\begin{equation}\label{eq:qjsd}
\qjsd(\rho,\sigma):=\tfrac12\trace\rho(\log\rho-\log m)
+\tfrac12\trace\sigma(\log\sigma-\log m)
\end{equation}
is the quantum Jensen--Shannon divergence.

\begin{corr}\label{cor:qjsd}
For $\rho,\sigma\in\Dc_d$,
\[
\lim_{\alpha\uparrow1}\frac{d_\alpha(\rho,\sigma)^2}{2(1-\alpha)}
=\qjsd(\rho,\sigma).
\]
Consequently $\qjsd^{1/2}$ is a metric on $\Dc_d$.
\end{corr}

\begin{proof}
Put $Y_\alpha=\rho^\alpha+\sigma^\alpha$, $g(\alpha)=\trace Y_\alpha^{1/\alpha}$
and $f(\alpha)=g(\alpha)^\alpha=\pnorm{Y_\alpha}{1/\alpha}$.  Then $f(1)=2$, so
$d_\alpha(\rho,\sigma)^2=f(1)-f(\alpha)$ by Proposition~\ref{prop:centroid},
and the limit is $f'(1)/2$.  A brief calculation gives
\[
g'(1)=\trace(\rho\log\rho+\sigma\log\sigma)
-\trace\bigl((\rho+\sigma)\log(\rho+\sigma)\bigr)=2\qjsd(\rho,\sigma)-2\log2,
\]
so $(\log f)'(1)=\log g(1)+g'(1)/g(1)=\qjsd(\rho,\sigma)$ and
$f'(1)=2\qjsd(\rho,\sigma)$.  Since $d_\alpha/(2(1-\alpha))^{1/2}$ is a metric
for $1/2\le\alpha<1$ by Theorem~\ref{thm:tsallis}, its pointwise limit
$\qjsd^{1/2}$ satisfies the triangle inequality, and strict convexity of
$X\mapsto\trace X\log X$ makes it separate states.
\end{proof}

\section{Quantum Jensen--Tsallis and Jensen--R\'enyi divergences}\label{sec:jensen}

We apply Theorem~\ref{thm:four} to entropy-based Jensen divergences.
For $\rho,\sigma\in\Dc_d$ and $m=(\rho+\sigma)/2$, let
\begin{equation}\label{eq:jensen-tsallis}
\qjsd_q(\rho,\sigma)
:=\frac{\trace m^q-\tfrac12\trace\rho^q-\tfrac12\trace\sigma^q}{1-q}
\qquad(0<q\le2,\ q\ne1),
\end{equation}
and put $\qjsd_1=\qjsd$.  The quantum Jensen--R\'enyi divergence is
\begin{equation}\label{eq:jensen-renyi}
\qjrd_q(\rho,\sigma)
:=H_q(m)-\tfrac12H_q(\rho)-\tfrac12H_q(\sigma),
\qquad H_q(\rho):=\frac{\log\trace\rho^q}{1-q}
\quad(0<q<1),
\end{equation}
with $\qjrd_1=\qjsd$.  These divergences differ from both minima in
Proposition~\ref{prop:free}.

\begin{theorem}\label{thm:jensen-four}
On any four density matrices, $\qjsd_q$ is of negative type for
$0<q\le2$, and $\qjrd_q$ is of negative type for $0<q\le1$.
Their square roots are metrics, and every four states with either metric
embed isometrically in $\reals^3$.
\end{theorem}

The square-root metric properties were proved in
\citet[Corollaries~3.3--3.5 and Theorem~6.3]{sra2021metrics}.
More generally, the proof below gives four-point negative type for every
Jensen divergence that is a nonnegative integral of S-divergences of shifted
matrices, which includes the families of
\citet[Theorems~3.1 and~3.6]{sra2021metrics}.
For commuting states, $\qjsd_q$ is of negative type
on arbitrary finite sets \citep{briet2009properties,martins2009nonextensive}.

\begin{proof}
For $0<q<2$,
\begin{equation}\label{eq:jensen-integral}
\qjsd_q(\rho,\sigma)
=c_q\int_0^\infty t^{q-1}\delta_S^2(\rho+tI,\sigma+tI)\dd t,
\qquad
c_q=\frac{q\sin(\pi q)}{\pi(1-q)},\quad c_1=1.
\end{equation}
This follows from $c_q\int_0^\infty t^{q-1}(x+t)^{-2}\dd t=qx^{q-2}$ after
integrating twice in $x$ and taking traces: the midpoint combination cancels
the affine terms.  The integral converges since
$\delta_S^2(\rho+tI,\sigma+tI)$ is $O(1+|\log t|)$ at zero and $O(t^{-2})$ at
infinity.  Since $c_q>0$, Theorem~\ref{thm:four} proves four-point negative
type for $\qjsd_q$.  At $q=2$ the same conclusion follows from
$\qjsd_2(\rho,\sigma)=\tfrac14\pnorm{\rho-\sigma}{2}^2$.

For the R\'enyi case, we apply the logarithmic argument of
\citet[Theorem~6.3]{sra2021metrics} to four states.  Fix $0<q<1$ and
$\rho_0,\dots,\rho_3\in\Dc_d$, and put
$h_{ij}=\trace\bigl((\rho_i+\rho_j)/2\bigr)^q$, so that
$h_{ij}=(1-q)\qjsd_q(\rho_i,\rho_j)+(h_{ii}+h_{jj})/2$.  By the first part and
Schoenberg's theorem, $[e^{-s(1-q)\qjsd_q(\rho_i,\rho_j)}]\succeq0$ for every
$s>0$, and rescaling rows and columns by $e^{-sh_{ii}/2}$ gives
$[e^{-sh_{ij}}]\succeq0$.
For any $\xi\in\reals^4$ with $\sum_i\xi_i=0$, the identity
$\log r=\int_0^\infty(e^{-s}-e^{-sr})\dd s/s$ therefore gives
\[
\sum_{i,j}\xi_i\xi_j\log h_{ij}
=-\int_0^\infty\sum_{i,j}\xi_i\xi_j e^{-s h_{ij}}\,\frac{\dd s}{s}
\le0.
\]
Since
$(1-q)\qjrd_q(\rho_i,\rho_j)=\log h_{ij}-\frac12\log h_{ii}-\frac12\log h_{jj}$
and $\sum_i\xi_i=0$, this proves four-point negative type for $\qjrd_q$.

For distinct states, $\delta_S^2(\rho+tI,\sigma+tI)>0$, so
\eqref{eq:jensen-integral} gives $\qjsd_q(\rho,\sigma)>0$.
For $0<q<1$, this also gives
$h_{ij}>(h_{ii}+h_{jj})/2\ge\sqrt{h_{ii}h_{jj}}$, hence
$\qjrd_q(\rho_i,\rho_j)>0$ when $\rho_i\ne\rho_j$.
Schoenberg's criterion now gives the claimed Euclidean embeddings, so the
square roots are metrics.
\end{proof}

\end{document}