Bures, Kubo–Mori, and Monotone Metrics

The Bures/SLD geometry and the Bogoliubov–Kubo–Mori (BKM) metric answer different operational questions. This page keeps two normalizations visible: the geometric Bures line element is one quarter of SLD QFI, whereas the Petz-normalized SLD and BKM metrics both reduce to classical Fisher information on commuting families. Only the last two can therefore be said to “agree classically” without an extra factor.

Required background. Quantum Fisher information provides the SLD equation, Petz normalization, and support-domain conditions used below.

Helpful background. Fidelity and overlap provides the purification-independent mixed-state comparison.

Begin with the chapter’s differentiable-family definition, use its comparison of information geometries as the normalization dictionary, and apply the independent validity gates before taking a rank or cutoff limit.

Bures distance and the factor of four

Throughout this page, root fidelity means

$$f(\rho,\sigma) =\operatorname{Tr}\sqrt{\sqrt\rho\,\sigma\sqrt\rho},$$

and

$$D_B^2(\rho,\sigma)=2[1-f(\rho,\sigma)].$$

Uhlmann’s transition probability is $f^2$ in this convention Uhlmann 1976, Eq. (1), p. 273. Some literature calls $f^2$, rather than $f$, “fidelity”; the displayed definition prevents a hidden factor of two in expansions.

For a smooth faithful family with tangent $X=d\rho$,

$$ds_B^2 =\frac12\sum_{m,n} \frac{\lvert X_{mn}\rvert^2}{p_m+p_n} =\frac14F_{\rm SLD}(X,X).$$

This is Hübner’s infinitesimal Bures formula Hübner 1992, pp. 240–241. On pure-state rays it becomes the Fubini–Study metric. The corresponding distance expansion is

$$D_B^2(\rho_\lambda,\rho_{\lambda+d\lambda}) =\frac14F_{\rm SLD}(\lambda)d\lambda^2+O(d\lambda^3).$$

BKM and the monotone-metric family

For a faithful state, the BKM metric is the Fréchet derivative of $\log\rho$:

$$g_{\rm BKM}(X,Y) =\operatorname{Tr}\!\left[ X\int_0^\infty dt\, (\rho+t)^{-1}Y(\rho+t)^{-1} \right].$$

In the eigenbasis of $\rho$,

$$g_{\rm BKM}(X,Y) =\sum_{m,n} \frac{\log p_m-\log p_n}{p_m-p_n} X_{nm}Y_{mn},$$

with diagonal value $1/p_n$. It is the relative-entropy Hessian,

$$S(\rho+\lambda X\Vert\rho) =\frac{\lambda^2}{2}g_{\rm BKM}(X,X)+O(\lambda^3),$$

and is the metric associated with the logarithmic mean Lesniewski and Ruskai 1999, §2.3, Theorem 2.8, and §2.5, Example 1.

More generally, a symmetric Petz-normalized monotone metric on faithful finite states has

$$g_f(X,Y)=\sum_{m,n}c_f(p_m,p_n)X_{nm}Y_{mn}, \qquad c_f(x,y)=\frac{1}{y f(x/y)},$$

where $f(1)=1$ and $f(t)=t f(t^{-1})$. This classification and contractivity under quantum channels are stated in Petz 1996, pp. 85–92 and Lesniewski and Ruskai 1999, §2.3, Theorem 2.13. The two functions needed here are

$$f_{\rm SLD}(t)=\frac{1+t}{2}, \qquad f_{\rm BKM}(t)=\frac{t-1}{\log t}.$$

The normalization dictionary is therefore:

Quantity	Eigenbasis kernel multiplying $\lvert X_{mn}\rvert^2$	Commuting tangent
Petz-normalized SLD	$2/(p_m+p_n)$	$\sum_n X_{nn}^2/p_n$
Geometric Bures	$1/[2(p_m+p_n)]$	$\frac14\sum_n X_{nn}^2/p_n$
Petz-normalized BKM	$(\log p_m-\log p_n)/(p_m-p_n)$	$\sum_n X_{nn}^2/p_n$

Because the logarithmic mean does not exceed the arithmetic mean,

$$F_{\rm SLD}(X,X)\leq g_{\rm BKM}(X,X)$$

for faithful states. Equality requires $[\rho,X]=0$ up to degenerate eigenspaces. Monotonicity under a completely positive trace-preserving map holds for each metric, but does not make the metrics equal.

QFT application: displaced and squeezed thermal modes

At finite volume and cutoff, take one dimensionless oscillator,

$$[q,p]=i, \qquad H=\frac{\omega}{2}(q^2+p^2), \qquad \rho_\beta=(1-r)\sum_{n\geq0}r^n\lvert n\rangle\langle n\rvert,$$

where $x=\beta\omega$, $r=e^{-x}$, and the covariance parameter is

$$\nu=\coth(x/2)=\frac{1+r}{1-r}.$$

Compare two noncommuting Gaussian directions:

$$\rho_a=e^{-iap}\rho_\beta e^{iap}, \qquad \rho_s=e^{-isG}\rho_\beta e^{isG}, \qquad G=\frac12(qp+pq)=\frac{i}{2}(a^{\dagger2}-a^2).$$

The first changes the mean $q$ by $a$; the second changes the covariance by squeezing. Direct spectral sums, consistent with the general Gaussian SLD formula of Monras 2013, Eqs. (13) and (16), give

Direction	$F_{\rm SLD}$	$g_{\rm Bures}=F_{\rm SLD}/4$	$g_{\rm BKM}$
Displacement $a$	$2/\nu$	$1/(2\nu)$	$x$
Squeezing $s$	$4\nu^2/(\nu^2+1)$	$\nu^2/(\nu^2+1)$	$2x\nu$

Here is a reproducible derivation. The displacement matrix element is $\lvert p_{n,n+1}\rvert^2=(n+1)/2$. The squeeze generator connects $n$ to $n+2$ with

$$\lvert G_{n,n+2}\rvert^2=\frac{(n+1)(n+2)}{4}.$$

Using

$$\sum_{n\geq0}p_n(n+1)=\frac{1}{1-r}, \qquad \sum_{n\geq0}p_n(n+1)(n+2)=\frac{2}{(1-r)^2},$$

in the SLD and BKM kernels produces the table. Independent thermal modes add their contributions. The finite-regulator benchmark is analytic, so its numerical error is zero; its controls are $L$, $\Lambda$, $\beta$, every $\omega_k$, the real-mode multiplicities, and the displacement or squeeze profile.

At high temperature, $\nu\sim2/x$. The Petz-normalized SLD and BKM entries then agree to leading order: both displacement entries approach $x$, and both squeeze entries approach $4$. The geometric Bures entries remain one quarter of SLD. At low temperature, $\nu\to1$: both Bures entries approach $1/2$, while the BKM entries grow as $x$ and $2x$. Noncommutativity and the pure-state boundary are doing real work; the discrepancy is not a convention error.

Adversarial test: two inequivalent rank limits

First take $x\to\infty$ in the thermal Gaussian benchmark. The state approaches the vacuum. A smooth displaced or squeezed vacuum ray has finite Bures metric, but for any two distinct pure rays

$$S(\rho_1\Vert\rho_0)=\infty$$

because the support of $\rho_1$ is not contained in that of $\rho_0$. The faithful-state BKM entries diverging as $x$ are therefore the correct warning, not a failed calculation.

Now test a support-changing commuting path,

$$\rho_\epsilon=\operatorname{diag}(1-\epsilon,\epsilon), \qquad 0<\epsilon<1.$$

For $X=\partial_\epsilon\rho_\epsilon$,

$$F_{\rm SLD}=g_{\rm BKM} =\frac{1}{\epsilon(1-\epsilon)}, \qquad g_{\rm Bures}=\frac{1}{4\epsilon(1-\epsilon)}.$$

All three diverge as $\epsilon\to0$. This does not contradict the finite Bures metric on a smooth unitary pure-state ray: the diagonal path changes rank, whereas the unitary path stays on the rank-one manifold. “Take the pure-state limit” is incomplete unless the tangent and support behavior are specified.

In continuum QFT these density matrices are regulator models. A metric claim must name the fixed local algebra or regulated mode family, prove convergence under cutoff refinement, and retain any source-domain or energy constraint. Fidelity can suffer an orthogonality catastrophe, while relative entropy may still be defined algebraically for suitable normal states; neither finite-cutoff table establishes a continuum limit by itself.

Common pitfalls

Saying “Bures and BKM agree classically.” Geometric Bures is one quarter of classical Fisher. Petz-normalized SLD and BKM agree on commuting tangents.

Using a faithful-state BKM kernel at rank loss. Approach the boundary along a declared path and inspect support containment. Smooth rank-preserving and rank-changing paths need not have the same limit.

Inferring equality from contractivity. Data processing orders each metric before and after a channel; it does not identify different operator means.

Exercises

1. For $\rho_p=\operatorname{diag}(p,1-p)$, compute all three metrics along $p$ and verify the factor-of-four dictionary.

Solution

The tangent is $X=\operatorname{diag}(1,-1)dp$. Both Petz-normalized metrics reduce to classical Fisher information,

$$F_{\rm SLD}=g_{\rm BKM} =\left(\frac1p+\frac1{1-p}\right)dp^2 =\frac{dp^2}{p(1-p)}.$$

The geometric Bures line element is one quarter of this:

$$ds_B^2=\frac{dp^2}{4p(1-p)}.$$

Thus the commuting limit removes the difference between the two Petz kernels but not the geometric normalization.

2. Starting from the two matrix elements displayed above, derive the squeezing row of the Gaussian table.

Solution

For $p_{n+2}=r^2p_n$, the two matrix orientations give

$$F_{\rm SLD}^{(s)} =4\sum_{n\geq0} \frac{(p_n-p_{n+2})^2}{p_n+p_{n+2}} \frac{(n+1)(n+2)}{4}.$$

Using the second geometric sum,

$$F_{\rm SLD}^{(s)} =\frac{(1-r^2)^2}{1+r^2} \frac{2}{(1-r)^2} =\frac{2(1+r)^2}{1+r^2} =\frac{4\nu^2}{\nu^2+1}.$$

For BKM, each unordered pair contributes twice. Since $\log p_n-\log p_{n+2}=2x$,

$$g_{\rm BKM}^{(s)} =4x(1-r^2)\sum_n p_n\frac{(n+1)(n+2)}4 =2x\frac{1+r}{1-r}=2x\nu.$$

Dividing the SLD answer by four gives the Bures entry.

3. Explain why $\rho_\epsilon=\operatorname{diag}(1-\epsilon,\epsilon)$ and $\lvert\psi_s\rangle=e^{-isG}\lvert0\rangle$ have different Bures behavior at a pure state. Compute the latter at $s=0$ for the squeeze generator above.

Solution

The diagonal path changes an eigenvalue linearly at the boundary, giving $g_{\rm Bures}=1/[4\epsilon(1-\epsilon)]$ and hence a divergence. The unitary path preserves rank and moves tangentially on projective Hilbert space. For the vacuum, $\langle G\rangle=0$ and

$$G\lvert0\rangle=\frac{i}{\sqrt2}\lvert2\rangle, \qquad \operatorname{Var}_0(G)=\frac12.$$

Therefore $F_{\rm SLD}=4\operatorname{Var}_0(G)=2$ and $g_{\rm Bures}=1/2$, finite. The path, not merely the endpoint density matrix, determines the boundary metric.

References

• Hübner, Matthias. “Explicit Computation of the Bures Distance for Density Matrices.” Physics Letters A 163 (1992): 239–242. DOI.
• Lesniewski, Andrew, and Mary Beth Ruskai. “Monotone Riemannian Metrics and Relative Entropy on Non-Commutative Probability Spaces.” Journal of Mathematical Physics 40 (1999): 5702–5724. DOI; arXiv.
• Monras, Alex. “Phase Space Formalism for Quantum Estimation of Gaussian States.” arXiv:1303.3682 (2013). arXiv.
• Petz, Dénes. “Monotone Metrics on Matrix Spaces.” Linear Algebra and its Applications 244 (1996): 81–96. DOI.
• Uhlmann, Armin. “The ‘Transition Probability’ in the State Space of a *-Algebra.” Reports on Mathematical Physics 9 (1976): 273–279. DOI; Open PDF.