Tsallis Metrics on Density Matrices via the S-Divergence

Introduction

We continue the study of metric properties of symmetrized divergences on quantum states started in [19]. Square roots of symmetrized divergences sometimes turn out to be metrics, for instance Jensen–Shannon divergences for commuting states [6, 7]. For noncommuting states Lamberti et al. [10] conjectured the metric property, which was proved by both Sra [19] and Virosztek (2021) 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 $\mathbb{P}^{d}$ ($\mathbb{P}^{d}_{+}$) the set of $d\times d$ positive semidefinite (definite) Hermitian matrices; and let $\mathcal{D}_{d}:=\{\rho\in\mathbb{P}^{d}_{+}:\operatorname{tr}\rho=1\}$ be the density matrices, or states. For $\rho,\tau\in\mathcal{D}_{d}$, the Tsallis relative entropy [14] (Tsallis 1988) is

The symmetrized divergence obtained by computing the Tsallis centroid of $\rho$ and $\sigma$ is then:

where $\|X\|_{p}=(\operatorname{tr}|X|^{p})^{1/p}$ is the Schatten-$p$ norm. The equality in (2) is the quantum Sibson identity [9, 16, 4]; Proposition 3.1 verifies it for $0<\alpha<1$by Hölder’s inequality; the reverse Hölder inequality gives the case$\alpha>1 ({xref}Section 4`).

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 (2) is allowed to range over all $\mathbb{P}^{d}$, the minimum value is $2\alpha(1-h)$, where $h=\operatorname{tr}((\rho^{\alpha}+\sigma^{\alpha})/2)^{1/\alpha}$, and its square root is a metric for $1/2\leq\alpha<1$ [19 Theorem 4.1]. But in (2), we require the center to lie in $\mathcal{D}_{d}$, 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.

Theorem 1.1 also implies the previous results of Sra [19]; indeed, as $\alpha\uparrow1$, $d_\alpha^2/(2(1-\alpha)) \to\mathrm{QJSD}$, which gives the Jensen–Shannon metric (Corollary A.3), and the proof gives the cone metric of Sra [19]. At $\alpha= 1/2$, the distance is also of negative type (Proposition 3.5).

For $\alpha< 1$, with $p = 1/\alpha$, Theorem 1.1 follows from positivity of the trace-power kernel $\operatorname{tr}(A+B)^p-\operatorname{tr} A^p-\operatorname{tr} B^p$ on every three positive semidefinite matrices (Theorem 3.2). 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 $\mathbb{P}^d$, which turns our task into proving that $\left[-\log\det(I-X_iX_j)\right]_{i,j=1}^3$ is positive semidefinite. This task is nontrivial, and its proof uncovers a hitherto unknown property of the S-divergence (Sra [17]), denoted $\delta_S^2$ below, which leads to our second main result.

The S-divergence $\delta_S^2$ used in (4) is defined for any two matrices $P,Q \in\mathbb{P}_+^d$ as

It is a Jensen–Bregman divergence (Cherian et al. [3]) and $\delta_S$ is a metric (Sra [17]).

Recall that a symmetric function $N$ with $N(x,x)=0$ is 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 Sra [19]. However, the log-determinant inequality (3) 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 $\mathbb{P}_+^d$, and

Hence positivity of $\left[-\log\det(I-U_iU_j)\right]_{i,j=1}^3$ is the same as negative type of $\delta_S^2$ on the four matrices $I,P_1,P_2,P_3$. By Theorem 1.2 and Schoenberg’s criterion (Schoenberg [15]), every four-point subset of $(\mathbb{P}_+^d,\delta_S)$ embeds isometrically in $\mathbb{R}^3$. The number four is sharp when $d \ge2$: Sra [17] gave a numerical five-point obstruction in dimension two; Proposition 2.5 gives an analytic one.

Remark 1.3. Theorem 1.2 also implies the positivity of the Hua–Bellman matrices. For strict contractions $Z_1,Z_2,Z_3$, Sra [18] proves that the Hua–Bellman matrix (Hua [8]) $H_\gamma=\left[\det(I-Z_i^*Z_j)^{-\gamma}\right]_{i,j=1}^3$ is positive semidefinite for every $\gamma\ge0$. For Hermitian $Z_i$, Theorem 1.2 recovers this entire family (where $K$ is as in (3)) by the Schur product theorem:

where $K^{\circ m}$ denotes entrywise power. The converse direction gives less: differentiating $c^{\mathsf{T}}H_\gamma c$ at zero gives $c^{\mathsf{T}}Kc\ge0$ only when $\sum_i c_i=0$, whereas Theorem 1.2 gives $K\succeq0$.

Remark 1.4. Theorem 1.2 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–Rényi divergences of this form are of negative type on every four states for orders $1<q\le 2 and $0<q\le1$, respectively (Theorem B.1). This strengthens the metric results of Sra [19].

Below, we first prove Theorem 1.2 as it forms the basis of the proof of Theorem 1.1.

Proof of Theorem 1.2: four-point negative type of the S-divergence

For $\xi\in\mathbb{R}^{4}$ with $\sum_a \xi_a=0$ and $P_0,\ldots,P_3 \in\mathbb{P}_{+}^{d}$ put

so that (4) says $F_{\xi}\geq0$. 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.

Existence of minimizer

We start with three elementary properties of $\delta_{\mathrm{S}}$ that we will need.

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 2.1(a), we may take $P_0=I$. The domain $(\mathbb{P}_{+}^{d})^3$ 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 $\nu_{\xi}:=\min_{\varnothing\neq T\subsetneq\{0,1,2,3\}}\left|\sum_{a\in T}\xi_a\right|$, which is positive for the fixed $\xi$.

Proof. The values $\delta_{\mathrm{S}}^{2}(P_a,P_c)$ need not satisfy the triangle inequality, but by Lemma 2.1(b) and (c) they do after a shift: on $\{0,1,2,3\}$, the function $D(a,c)=\delta_{\mathrm{S}}^{2}(P_a,P_c)+d\log2$ for $a\neq c$, with $D(a,a)=0$, is a metric. Since $\sum_{a\neq c}\xi_a\xi_c=-\sum_a\xi_a^2$, the shift changes $F_{\xi}$ only by a constant:

Metrics on four points are conic combinations of cut metrics ([5 p. 422]), so,

where $\mathbf{1}_T$ is the indicator vector of $T$. Each cut contributes at least $2\nu_{\xi}^{2}w_T$, because $\sum_a\xi_a=0$ gives

and $\sum_Tw_T\geq\max_{a,c}D(a,c)\geq\max_i\delta_{\mathrm{S}}^{2}(I,P_i)$ as each cut takes values in $\{0,1\}$. Together, these prove the bound. Finally, if $F_{\xi}(I,P_1,P_2,P_3)\leq0$, the bound caps every $\delta_{\mathrm{S}}^{2}(I,P_i)$, so by Lemma 2.1(b) and $\log\cosh t\geq|t|-\log2$, 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 $(\mathbb{P}_{+}^{d})^3$. $\square$

Potential minimizers (critical points of $F_{\xi}$)

Since $(\mathbb{P}_{+}^{d})^3$ is open, the minimizer of Proposition 2.2 is a critical point. Lemma 2.3 describes these critical points.

Proof. Noting $\sum_a\xi_a=0$, and expanding $\delta_{\mathrm{S}}^{2}(P_a,P_c)$, we see that only joint terms survive, so that

Taking derivatives of $F_{\xi}$ with respect to the $P_i$ we obtain the criticality condition

The equation also holds for $i=0$. Indeed, by Lemma 2.1(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 (8) by $-\xi_0$ and rewrite it as the set of equations

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 $(P_i+P_j)^{-1}=\frac{1}{2}(I-U_j)(I-U_iU_j)^{-1}(I-U_i)$; so, the first equation becomes $\sum_j\lambda_j(I-U_iU_j)^{-1}=I$. □

Solving the critical-point equations

On a common eigenvector of the products $U_iU_j$, the system (7) becomes scalar, so we first show that these products commute.

Proof. The squares commute with all $U_i$. 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 (7) on the left by $U_i$ gives the first identity below, and the second is the equation of (7) for this $i$:

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

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 $k$,

Shared eigenbasis. By the first equation in (7), $U_3=-(\lambda_1U_1+\lambda_2U_2)/\lambda_3$, 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 (12), 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 $\lVert U_iv\rVert\lVert U_jv\rVert$. 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 $|z_i|=\lVert U_iv\rVert<1$. Thus $U_iU_j$ acts on $v$ as $\overline{z_i}z_j$.

The scalar moments vanish. On $v$, the second set of equations in (7) becomes

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

Let $\zeta_1,\ldots,\zeta_m$ 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 $n=1,\ldots,m$, 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, $U_1=U_2=U_3=0$.

Proof. of (4). We can now prove the four-point negative-type condition (4). By Proposition 2.2 and Lemmas 2.3 and 2.4, 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 (4) for the $\xi$ fixed above, and continuity in $\xi$ gives it for every zero-sum $\xi$; by the equivalence established via identity (6), the truth of (3) also follows.

Number of points

The following configuration gives the sharpness of both forms of Theorem 1.2 for $d=2$. The embeddings $X\mapsto X\oplus0$ and $P\mapsto P\oplus I$ extend it to every $d\ge2$.

Proof. (a) For $i,j\le2$, $X_iX_j$ has rank one and $\det(I-X_iX_j)=1-x^2(u_i^{\mathsf T}u_j)^2$. Since $(u_i^{\mathsf T}u_j)^2=\frac{1}{4}$ for $i\ne j$, this determinant is $\frac{1}{4}$ for $i=j$ and $\frac{13}{16}$ for $i\ne j$. Also $\det(I-X_kX_3)=1-x^2=\frac{1}{4}$ and $\det(I-X_3^2)=\frac{1}{16}$. With $L=\log4$ and $\ell=\log\frac{16}{13}$,

(b) The vector $\xi=(-\frac{3}{2},1,1,1,-\frac{3}{2})$ ($\xi^{\mathsf T}\mathbf{1}=0$) gives $\sum_{a,e=0}^{4}\xi_a\xi_e\delta_S^2(P_a,P_e)=-c^{\mathsf T}Kc>0$.

:::

Proof of Theorem 1.1 for $1/2\le\alpha<1$

We prove Theorem 1.1 for $\alpha<1$ on the positive unit sphere of a Schatten norm. The passage to the sphere rests on the equality in (2), which we prove first.

Reduction to a Schatten sphere

Proof. The scaled objective is $2-\operatorname{tr}(M\tau^{1-\alpha})$, and Schatten Hölder with exponents $1/\alpha$ and $1/(1-\alpha)$ gives $\operatorname{tr}(M\tau^{1-\alpha})\le\|M\|_{1/\alpha}\|\tau^{1-\alpha}\|_{1/(1-\alpha)}=\|M\|_{1/\alpha}$. The state (14) attains equality, and strict convexity of the Schatten $1/(1-\alpha)$-norm makes it the only maximizer.

For $1<p<\infty$, let $\mathcal{S}_p=\{A\in\mathcal{P}^d:\operatorname{tr}A^p=1\}$ and, for $A,B\in\mathcal{S}_p$, let $D_p(A,B)=(2-\|A+B\|_p)^{1/2}$. For $p=1/\alpha$, $\rho\mapsto\rho^\alpha$ is a bijection from $\mathcal{D}_d$ onto $\mathcal{S}_p$, and Proposition 3.1 gives $d_\alpha(\rho,\sigma)=D_p(\rho^\alpha,\sigma^\alpha)$. It therefore suffices to prove the triangle inequality for $1<p\le2$; Proposition 3.6 treats $p>2$.

An integral representation

We transfer the log-determinant form (3) of Theorem 1.2 to the trace-power kernel

and then use its three-point positivity on the sphere.

Proof. At $p=2$, $L_2(A_i,A_j)=2\operatorname{tr}(A_iA_j)$ is a Gram matrix. For $1<p<2$, put $\kappa_p=p\sin(\pi(p-1))/\pi>0$ and use the scalar identity [19]

For $t>0$, set $X_A(t)=A(A+tI)^{-1}$ and

Applying (3.5) to $A+B$, $A$ and $B$ and taking traces cancels the linear terms and gives

Since $0\preceq X_{A_i}(t)\prec I$, Theorem 1.2 gives $\left[K_t(A_i,A_j)\right]\succeq0$, and integrating proves (3.4). □

The triangle inequality

Three-point positivity of $L_p$ places any three points of $S_p$ on a sphere, where $D_p$ becomes a function of an angle. Let $1<p\le2$, let $A_1,A_2,A_3\in S_p$, and put

Since $\operatorname{tr}A_i^p=1$,

which is positive semidefinite by Theorem 3.2 and $2^{2-p}\ge1$. Since $G_{ii}=1$, the matrix $G$ is the Gram matrix of three unit vectors $u_1,u_2,u_3\in\mathbb{R}^3$; write $G_{ij}=\cos\theta_{ij}$ with $0\le\theta_{ij}\le\pi$. These angles satisfy the spherical triangle inequality

By (3.8), $\lVert A_i+A_j\rVert_p=2\cos^{2/p}(\theta_{ij}/2)$, so

The triangle inequality for $D_p$ therefore reduces to a scalar property of $f_p$.

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

Since $t\mapsto t^{p-1}$ is concave on $[z,1]$,

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 (3.11). □

Proof. of Theorem 1.1 for $1/2 \le\alpha< 1$. Minkowski’s inequality and strict convexity of the Schatten $p$-norm ([1]) 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 (3.9), the monotonicity of $f_p$, and Lemma 3.3,

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

Proof. With the angles of (3.8), $\left\|(A_i+A_j)/2\right\|_p=\cos\psi_p(\theta_{ij})$, 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 $y=\cos^2(\theta/2)$, $\psi_p''(\theta)=\frac{\beta^2}{4}y^{\beta-1}(1-y)/(1-y^\beta)$, and a brief calculation shows that this increases in $y$ because $(1-y^\beta)/(1-y)\le\beta$. So $\psi_p$ is concave, and (3.9) with the argument of Lemma 3.3 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 $\left\|(\rho_i^\alpha+\rho_j^\alpha)/2\right\|_{1/\alpha}=1-d_\alpha(\rho_i,\rho_j)^2/2$, so these unit vectors satisfy $\|v_i-v_j\|=d_\alpha(\rho_i,\rho_j)$.

The case $p=2$

At $p=2$, which corresponds to the endpoint $\alpha=1/2$ of Theorem 1.1, the distance $D_2(A,B)=(2-(2+2\operatorname{tr}AB)^{1/2})^{1/2}$ depends only on the Hilbert–Schmidt inner product $\operatorname{tr}AB$. We can therefore work on the unit sphere of any real Hilbert space, where the distance is also of negative type.

Proof. For unit vectors at angle $\theta$, $D(x,y)=2\sin(\theta/4)=f_2(\theta)$, so (3.9) and Lemma 3.3 prove that $D$ is a metric.

For negative type, write $D(x,y)=\varphi(\langle x,y\rangle)$, 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

Since $b_0=2\sin(\pi/8)>0$ and $b_1=-\frac{1}{2}\cos(\pi/8)<0$, the recurrence gives $b_k<0$ for every $k\ge1$. Let $x_1,\ldots,x_n\in\mathbb{S}$, let $c\in\mathbb{R}^n$ with $\sum_i c_i=0$, and let $0<r<1$. Since $\sum_i c_i=0$ and $\langle x_i,x_j\rangle^k=\langle x_i^{\otimes k},x_j^{\otimes k}\rangle$ in the $k$-fold tensor power of $\mathcal{H}$,

Letting $r\uparrow1$ gives negative type, and Schoenberg’s criterion gives the embedding of $(\mathbb{S},D^{1/2})$.

For states, $d_{1/2}(\rho,\sigma)=D_2(\rho^{1/2},\sigma^{1/2})$ is a function of the quantum affinity $\operatorname{tr}\rho^{1/2}\sigma^{1/2}$ ([11]), so Proposition 3.5 shows that $d_{1/2}$ is of negative type.

The range of $p$

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

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, $\lVert C+B_{\varepsilon}\rVert_{p}^{p}=2+p\varepsilon+O(\varepsilon^{2})$ because $p>2$, so

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 $d\ge2$. $\square$

Proof of Theorem 1.1 for $1<\alpha\le2$

For $\alpha>1$, the proof of Proposition 3.1 with the reverse Hölder inequality gives the equality in (2), with the same minimizing state (14). 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 $X\mapsto\operatorname{tr}X^{p}$ gives $\operatorname{tr}(A_i+A_j)^{p}\ge2^{p}$, with equality only when $A_i=A_j$, so $d_{\alpha}$ separates states. It remains to prove the triangle inequality.

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 $(x+y)/\sqrt{2}$ before averaging.

Proof. Jensen’s inequality gives $r,s\ge\sqrt{2}$. For $a,b\ge0$, put

We first prove the pointwise inequality

For $a\le b$, both $\delta$ and $D$ are nonincreasing in $a$ and nondecreasing in $b$. Thus (4.2) 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

Since $S-xy=x+(1-x)y>0$, the bracket is positive if $xy\le2$. If $xy>2$, it has the sign of

Both factors in the subtracted product are positive and increasing in $y$, since

Thus $D_{xy}$ can change sign only from positive to negative, so $-D_x(x,y)$ has no interior maximum as a function of $y\geq1$. Its limit at infinity is $1/\sqrt{2}$.

At $y=1$, rationalization gives

The fraction is positive and decreasing, so $h>1/\sqrt{2}$, $h'<0$, and $-D_x(x,1)=h(x)-(1-x)h'(x)>1/\sqrt{2}$. Hence $-D_x(x,y)\leq-D_x(x,1)$ for $0<x<1$ and $y\geq1$, and integrating proves (4.2):

To average (4.2), we use concavity of $L$ on $[\sqrt{2},\infty)^2$. 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

where the last inequality follows by weighted Cauchy–Schwarz from $r^2u+s^2v=u^3+v^3+2q\leq q(q^2+2)=qL^2$. Also $L_r=qr/(uL)\geq1$, because $q^2r^2-u^2L^2=2(q^2-u^2)\geq0$, and similarly $L_s\geq1$. Integrating these derivative bounds gives

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

These variables are at least $\sqrt{2}$, and (4.2) is exactly $\widehat T\leq L(\widehat R,\widehat S)$. The mean constraints give $a:=(\mathbb{E}X-1)/\sqrt{2}\geq0$ and $b:=(\mathbb{E}Y-1)/\sqrt{2}\geq0$, so $\mathbb{E}\widehat R=r-a$, $\mathbb{E}\widehat S=s-b$ and $\mathbb{E}\widehat T=t-a-b$. Concavity and (28) now give (4.1):

The endpoint alpha=2

The scalar means in Lemma 4.1 arise from a trace comparison. Commutativity is needed only for the auxiliary matrices in this comparison, not for the states.

Proof. Write $T(X,Y)=\operatorname{tr}\sqrt{X^2+Y^2}$, 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 $\mathcal{D}_d$. A minimizer exists by compactness. If it is an endpoint, there is nothing to prove. Otherwise put

and $t_j=\operatorname{tr}K_j$, so $t_1^2=u^2+2$ and $t_2^2=v^2+2$. The gradient is

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 $-\operatorname{tr}(z^2H)<0$.

Stationarity under $\operatorname{tr}z=1$ gives $zH+Hz=2\lambda I$, so $H=\lambda z^{-1}$, that is,

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

Then $\sum_k w_k=1$ and

Since $\rho^2=z^{1/2}(R_1zR_1-z)z^{1/2}$, Lemma 4.2 gives

Applying that lemma once more to $\rho^2+\sigma^2=z^{1/2}(R_1zR_1+R_2zR_2-2z)z^{1/2}$ yields

where the second inequality is Lemma 4.1. Thus $d_2(\rho,\sigma)\leq f(z)$, and minimality gives the triangle inequality for every middle state. $\square$

Transfer to the other orders

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

Proof. Let $S$ have density $ps^{-p-1}$ on $[1,\infty)$, and put $\phi(t)=\mathbb{E}e^{-tS}$. A change of variables gives

where $\ell_p=p\int_0^\infty(1-e^{-v})v^{-p-1}\,dv\in(0,\infty)$.

For independent copies $S_1,\ldots,S_N$, define the positive linear map

on $(\mathbb{C}^d)^{\otimes N}$, where $X^{(r)}$ acts in tensor factor $r$. Writing $\tau_m=\operatorname{tr}/m$, independence and commutativity of the tensor copies of each fixed $X\succeq0$ give

Moreover,

Multiply by $t^{-3/2}$ and use $\sqrt{x}=(2\sqrt{\pi})^{-1}\int_0^\infty(1-e^{-tx})t^{-3/2}\,\mathrm{d}t$. The bound is integrable because $p>1/2$, so dominated convergence gives

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 $\lVert X\rVert_{\mathrm{op}}^{1/2}N^{-1/(2p)}\sum_r\sqrt{S_r}$. Let $s_\ell$ be the nodes and $q_\ell\ge0$ their weights, and set

Their traces are the chosen weighted averages. Linearity gives

Consequently, with $M=\gamma_p d^{-1/(2p)}>0$,

For sufficiently large $N$, normalize $B_i^{(N)}=D_i^{(N)}/m_i^{(N)}$. Operator monotonicity gives

Both bounds have the limit in (31). □

Proof of Theorem 1.1 for $1<\alpha\le2$. Proposition 4.3 proves the triangle inequality at $\alpha=2$. For $1<\alpha<2$, set $p=1/\alpha$ and $A_i=\rho_i^p$. Lemma 4.4 gives

The endpoint triangle inequality passes to this limit. □

AI usage

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 3.4 and Appendix A.2, Remark A.2, Proposition 3.6, and the writing, positioning, and literature search. OpenAI assistants also developed the four-point consequences in Appendix B and assisted with proof simplification and reorganization. The arguments emerged through sustained work involving failed attempts, counterexamples, numerical checks, proof audits, and repairs.

Acknowledgements

The author (SS) thanks the Alexander von Humboldt (AvH) Foundation for their generous support via a Humboldt Professorship in AI.

References

Relation to the cone and Jensen–Shannon metrics

For $0<\alpha<1$, we relate $d_\alpha$ to the two earlier metrics of the introduction: the cone metric of Sra [19], which drops the trace constraint, and the Jensen–Shannon metric, the limit of the rescaled $d_\alpha$ as $\alpha\uparrow1$.

Comparison with the cone minimum

To let the center range over the cone, extend $T_\alpha$ to $X,Z\in\mathbb{P}^d$ by

Minimizing over the cone gives the relative-entropy Jensen–Tsallis divergence of Sra [19]; its commuting case was studied by Österreicher and Vajda [13].

Proof. Write $Z=s\tau$, where $s=\operatorname{tr}Z$ and $\tau\in\mathcal{D}_d$. For fixed $s>0$, Proposition 3.1 gives the unique optimal state $\tau=M_\alpha/h$. Since $\lVert\rho^\alpha+\sigma^\alpha\rVert_{1/\alpha}=2h^\alpha$, the scaled objective reduces to $2\alpha+2(1-\alpha)s-2h^\alpha s^{1-\alpha}$, a strictly convex function of $s\geq0$ minimized at $s=h$, with value $2\alpha(1-h)$. The minimizing matrix is $s\tau=M_\alpha$. □

The limit alpha->1

For the Umegaki relative entropy $\operatorname{tr}\rho(\log\rho-\log\tau)$, the optimal center is $m=(\rho+\sigma)/2$ and the least total is $2\operatorname{QJSD}(\rho,\sigma)$, where

is the quantum Jensen–Shannon divergence.

Proof. Put $Y_\alpha=\rho^\alpha+\sigma^\alpha$, $g(\alpha)=\operatorname{tr}Y_\alpha^{1/\alpha}$ and $f(\alpha)=g(\alpha)^\alpha=\lVert Y_\alpha\rVert_{1/\alpha}$. Then $f(1)=2$, so $d_\alpha(\rho,\sigma)^2=f(1)-f(\alpha)$ by Proposition 3.1, and the limit is $f'(1)/2$. A brief calculation gives

so $(\log f)'(1)=\log g(1)+g'(1)/g(1)=\operatorname{QJSD}(\rho,\sigma)$ and $f'(1)=2\operatorname{QJSD}(\rho,\sigma)$. Since $d_\alpha/(2(1-\alpha))^{1/2}$ is a metric for $1/2\leq\alpha<1$ by Theorem 1.1, its pointwise limit $\operatorname{QJSD}^{1/2}$ satisfies the triangle inequality, and strict convexity of $X\mapsto\operatorname{tr}X\log X$ makes it separate states. □

Appendix B. Quantum Jensen–Tsallis and Jensen–Rényi divergences

We apply Theorem 1.2 to entropy-based Jensen divergences. For $\rho,\sigma\in\mathcal{D}_d$ and $m=(\rho+\sigma)/2$, let

and put $\operatorname{QJSD}_1=\operatorname{QJSD}$. The quantum Jensen–Rényi divergence is

with $\operatorname{QJRD}_1=\operatorname{QJSD}$. These divergences differ from both minima in Proposition A.1.

The square-root metric properties were proved in Sra [19]. 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 Sra [19]. For commuting states, $\operatorname{QJSD}_q$ is of negative type on arbitrary finite sets [2, 12].

Proof. For $0<q<2$,

This follows from $c_q\int_0^\infty t^{q-1}(x+t)^{-2}\,\mathrm{d}t=qx^{q-2}$ 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 $O(1+\lvert\log t\rvert)$ at zero and $O(t^{-2})$ at infinity. Since $c_q>0$, Theorem 1.2 proves four-point negative type for $\operatorname{QJSD}_q$. At $q=2$ the same conclusion follows from $\operatorname{QJSD}_2(\rho,\sigma)=\frac{1}{4}\lVert\rho-\sigma\rVert_2^2$.

For the Rényi case, we apply the logarithmic argument of Sra [19] to four states. Fix $0<q<1$ and $\rho_0,\ldots,\rho_3\in\mathcal{D}_d$, and put $h_{ij}=\operatorname{tr}((\rho_i+\rho_j)/2)^q$, so that $h_{ij}=(1-q)\operatorname{QJSD}_q(\rho_i,\rho_j)+(h_{ii}+h_{jj})/2$. By the first part and Schoenberg’s theorem, $[e^{-s(1-q)\operatorname{QJSD}_q(\rho_i,\rho_j)}]\succeq0$ for every $s>0$, and rescaling rows and columns by $e^{-sh_{ii}/2}$ gives $[e^{-sh_{ij}}]\succeq0$. For any $\xi\in\mathbb{R}^4$ with $\sum_i\xi_i=0$, the identity $\log r=\int_0^\infty(e^{-s}-e^{-sr})\,\mathrm{d}s/s$ therefore gives

Since $(1-q)\operatorname{QJRD}_q(\rho_i,\rho_j)=\log h_{ij}-\frac{1}{2}\log h_{ii}-\frac{1}{2}\log h_{jj}$ and $\sum_i\xi_i=0$, this proves four-point negative type for $\operatorname{QJRD}_q$.

For distinct states, $\delta_S^2(\rho+tI,\sigma+tI)>0$, so (B.3) gives $\operatorname{QJSD}_q(\rho,\sigma)>0$. For $0<q<1$, this also gives $h_{ij}>(h_{ii}+h_{jj})/2\geq\sqrt{h_{ii}h_{jj}}$, hence $\operatorname{QJRD}_q(\rho_i,\rho_j)>0$ when $\rho_i\ne\rho_j$. Schoenberg’s criterion now gives the claimed Euclidean embeddings, so the square roots are metrics. □