State Perturbations and Relative-Entropy Susceptibility

A relative-entropy susceptibility is the Hessian of relative entropy along a normalized state family on one fixed algebra. In a faithful finite regulator it is the Bogoliubov–Kubo–Mori quadratic form. In QFT, the same name is justified only after the tangent, smearing, and regulator-removal limit have been shown to define a finite form.

Required background. The entanglement first law and quadratic correction establish the fixed-reference expansion used here.

Helpful background. Regulated and Araki relative entropy explains how a finite type-I calculation can approximate an intrinsic algebraic comparison.

The chapter overview introduces normalized differentiable families and lists the independent validity gates. Susceptibility requires both.

Tangents and the relative-entropy Hessian

Let $\omega_\lambda$ be a twice-differentiable family of faithful normal states on a fixed von Neumann algebra $\mathcal A$, with faithful reference $\omega_0=\omega$ and

$$\dot\omega_0(\mathbf1)=0.$$

When the derivative exists and is finite, define

$$\chi_\omega(\dot\omega_0) =\left.\frac{d^2}{d\lambda^2} S(\omega_\lambda\Vert\omega)\right|_{\lambda=0}.$$

Araki 1976, §1, pp. 809–810, especially eqs. (1.1), (1.2), and (1.6) supplies the intrinsic relative entropy, its density-matrix reduction, and the subalgebra monotonicity statement under the cases specified there. The displayed derivative is an additional regularity hypothesis; not every normal functional tangent has finite susceptibility.

In a faithful type-I regulator, write $\rho_\lambda=\rho+\lambda X+O(\lambda^2)$ with $\operatorname{Tr}X=0$. Then

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

Equivalently, for $\rho=\sum_n p_n\lvert n\rangle\langle n\rvert$,

$$\chi_\rho(X,X) =\sum_{m,n} \frac{\log p_m-\log p_n}{p_m-p_n} \lvert X_{mn}\rvert^2,$$

where the coincident value is $1/p_n$. This is positive and contractive under a channel when the state and tangent are pushed forward together; Petz 1996, pp. 87–90 gives the monotone-metric classification. Bures or symmetric-logarithmic-derivative susceptibility uses a different noncommutative kernel and must not be substituted silently.

Three domain questions are logically separate:

• Does $X$ define a normalized tangent to states?
• Is $X$ in the quadratic-form domain of the relative modular operator?
• Does a common regulated sequence converge to a finite, scheme-independent value?

Positivity answers none of these existence questions.

Source-deformed Gaussian state

Consider one regulated free-field normal mode,

$$H_0=\omega\left(N+\frac12\right), \qquad Q=\frac{a+a^\dagger}{\sqrt2}, \qquad H_\lambda=H_0-\lambda Q,$$

with $\omega>0$, inverse temperature $\beta>0$, and

$$\rho_\lambda=Z_\lambda^{-1}e^{-\beta H_\lambda}.$$

The source is real, and $Q$ is unbounded, but $H_\lambda$ is a shifted oscillator and $e^{-\beta H_\lambda}$ is trace class for every finite $\lambda$. Completing the square with $b=a-\lambda/(\sqrt2\omega)$ gives

$$H_\lambda =\omega\left(b^\dagger b+\frac12\right) -\frac{\lambda^2}{2\omega}, \qquad \log\frac{Z_\lambda}{Z_0} =\frac{\beta\lambda^2}{2\omega}.$$

It follows that

$$\langle Q\rangle_\lambda=\frac{\lambda}{\omega}$$

and, without a series expansion,

$$\begin{aligned} S(\rho_\lambda\Vert\rho_0) &=\beta\lambda\langle Q\rangle_\lambda -\log\frac{Z_\lambda}{Z_0}\\ &=\frac{\beta\lambda^2}{2\omega}. \end{aligned}$$

Therefore

$$\chi_{\rho_0}=\frac{\beta}{\omega} =\left.\frac{d^2}{d\lambda^2} S(\rho_\lambda\Vert\rho_0)\right|_0.$$

This explicitly verifies agreement between the source susceptibility and the relative-entropy Hessian. For $\beta\omega=\log2$ and $\lambda/\omega=0.1$,

$$S(\rho_\lambda\Vert\rho_0) =\frac{\log2}{200} \simeq3.46574\times10^{-3}.$$

The analytic benchmark has zero truncation and Taylor error. It models a Gaussian QFT mode after a spatial, volume, or mode regulator; it does not by itself establish a continuum local susceptibility.

Adversarial test: a UV-singular tangent

Now use a free scalar field in three spatial dimensions with

$$H_\lambda =H_0-\lambda\int d^3x\,f(\mathbf x)\phi(\mathbf x).$$

Adopt

$$\phi(\mathbf x) =\int\frac{d^3k}{(2\pi)^3} e^{i\mathbf k\cdot\mathbf x}\phi_{\mathbf k},$$

and impose a common momentum cutoff $\lvert\mathbf k\rvert\leq\Lambda$. Start with $m>0$ in a finite box, take the infinite-volume limit, and only then take the massless limit below; setting $m=0$ in a finite box would leave an uncontrolled zero mode. Completing the square mode by mode gives the regulated susceptibility

$$\chi_{\Lambda,f} =\beta\int_{\lvert\mathbf k\rvert\leq\Lambda} \frac{d^3k}{(2\pi)^3} \frac{\lvert\widetilde f(\mathbf k)\rvert^2} {\mathbf k^2+m^2}.$$

For the massless theory and normalized Gaussian smearing

$$\widetilde f_\ell(\mathbf k) =e^{-\ell^2\mathbf k^2/2}, \qquad \ell>0,$$

the integral is elementary:

$$\chi_{\Lambda,\ell} =\frac{\beta}{2\pi^2} \int_0^\Lambda dk\,e^{-\ell^2k^2} =\frac{\beta}{4\pi^{3/2}\ell} \operatorname{erf}(\Lambda\ell).$$

At fixed $\ell$, cutoff removal is finite,

$$\chi_{\infty,\ell} =\frac{\beta}{4\pi^{3/2}\ell}.$$

At $\Lambda\ell=4$, the relative cutoff deficit is

$$1-\operatorname{erf}(4) \simeq1.54\times10^{-8}.$$

The adversarial point-source limit fails. If $\ell\to0$ after cutoff removal, $\chi\sim1/\ell$; if $\ell=0$ first, $\chi_{\Lambda,0}=\beta\Lambda/(2\pi^2)$. Smearing therefore restores a finite susceptibility for each fixed test function but does not renormalize the pointlike tangent into an intrinsic finite direction. A subtraction could define a scheme-dependent composite observable, but it would be a new quantity with a stated counterterm, not the original relative-entropy Hessian.

The strongest surviving QFT claim is: smooth finite-width sources define a finite Gaussian susceptibility with controlled cutoff removal. The unsmeared local source is outside this metric domain.

Monotonicity and interpretation

If $\Phi$ is a fixed channel and differentiation can be interchanged with the channel and regulator limit, relative-entropy monotonicity implies

$$\chi_{\Phi(\omega)}(\Phi_*\dot\omega) \leq\chi_\omega(\dot\omega).$$

For restrictions of the same family, a larger observable algebra can therefore distinguish at least as much at quadratic order. This does not imply that a chosen detector or local operator attains the full susceptibility; that requires an allowed measurement class.

Common pitfalls

Calling any second derivative a susceptibility. State the divergence, reference, algebra, tangent normalization, and kernel.

Forgetting the connected subtraction. Differentiating $Z_\lambda^{-1}$ is what enforces $\dot\omega(\mathbf1)=0$.

Removing cutoffs from separate divergent terms. Form the matched relative entropy or quadratic form first, then remove the common regulator.

Exercises

1. Suppose $[H_0,Q]=0$ and $\rho_\lambda\propto e^{-\beta(H_0-\lambda Q)}$. Show that $\chi=\beta^2\operatorname{Var}_{\rho_0}(Q)$.

Solution

Let $F(\lambda)=\log Z_\lambda$. Commutativity gives

$$F'(\lambda)=\beta\langle Q\rangle_\lambda, \qquad F''(0)=\beta^2 \left(\langle Q^2\rangle_0-\langle Q\rangle_0^2\right).$$

Relative entropy is $S(\rho_\lambda\Vert\rho_0)=\lambda F'(\lambda)-F(\lambda)+F(0)$, whose second derivative at zero is $F''(0)$. Hence $\chi=\beta^2\operatorname{Var}(Q)$.

2. Let $\rho=\operatorname{diag}(p,1-p)$ and $\rho_\lambda=e^{-i\lambda G}\rho e^{i\lambda G}$, with $G_{12}=g$. Compute the relative-entropy susceptibility.

Solution

The tangent is $X=-i[G,\rho]$, so

$$\lvert X_{12}\rvert^2 =(2p-1)^2\lvert g\rvert^2,$$

and the conjugate entry has the same modulus. Substituting both ordered pairs into the BKM form gives

$$\chi =2(2p-1)\log\!\left(\frac{p}{1-p}\right) \lvert g\rvert^2\geq0.$$

It vanishes at $p=1/2$, where the state is invariant under every unitary, and also when $g=0$.

3. Starting from $\chi_{\Lambda,f}$, derive the Gaussian-smearing result and compare the two orders of limits $\Lambda\to\infty$ and $\ell\to0$.

Solution

For $m=0$, spherical integration cancels the $k^2$ in the measure against the propagator denominator:

$$\chi_{\Lambda,\ell} =\frac{\beta}{2\pi^2} \int_0^\Lambda e^{-\ell^2k^2}dk.$$

With $u=\ell k$, this is

$$\frac{\beta}{4\pi^{3/2}\ell} \operatorname{erf}(\Lambda\ell).$$

Taking $\Lambda\to\infty$ first gives $\beta/(4\pi^{3/2}\ell)$, which diverges as $\ell\to0$. Taking $\ell=0$ first gives $\beta\Lambda/(2\pi^2)$, which diverges as $\Lambda\to\infty$. Neither order produces a finite point-source susceptibility.

References

• Araki, Huzihiro. “Relative Entropy of States of von Neumann Algebras.” Publications of the Research Institute for Mathematical Sciences 11 (1976): 809–833. DOI.
• Petz, Dénes. “Monotone Metrics on Matrix Spaces.” Linear Algebra and its Applications 244 (1996): 81–96. DOI.