First Law of Entanglement and Quadratic Corrections

The entanglement first law is the linearization of an exact relative-entropy identity. It compares a normalized state family with one fixed reference state on one fixed algebra. The first nonzero relative-entropy term is then a positive quadratic form, not another linear law.

Required background. Relative entropy in QFT supplies the finite continuum comparison, and modular Hamiltonian definitions supply the generator and domain conventions.

The chapter overview first sets up a differentiable family and then separates the independent validity gates. Those checks remain in force below.

The fixed-reference identity

Begin in a finite-dimensional type-I regulator. Let $\sigma>0$ be a normalized reference density operator, let

$$K_\sigma=-\log\sigma,$$

and let $\rho$ be normalized with $\operatorname{supp}\rho\subseteq\operatorname{supp}\sigma$. Direct substitution gives

$$S(\rho\Vert\sigma) =\Delta\langle K_\sigma\rangle-\Delta S,$$

where

$$\Delta\langle K_\sigma\rangle =\operatorname{Tr}[(\rho-\sigma)K_\sigma], \qquad \Delta S=S(\rho)-S(\sigma).$$

This is an exact identity, with the comparison Hamiltonian $K_\sigma$ held fixed. Blanco, Casini, Hung, and Myers 2013, §2, especially eqs. (2.9)–(2.11) use this identity and positivity in QFT. On a local type-III algebra, the intrinsic object is Araki relative entropy; a density matrix, $S(\rho)$, and $\langle K_\sigma\rangle$ may exist only in a common regulator even when their displayed difference has a finite algebraic limit. Araki 1976, §1, pp. 809–810 gives the von Neumann-algebra definition and its density-matrix reduction.

Let

$$\rho_\lambda =\sigma+\lambda X+\frac{\lambda^2}{2}Y+O(\lambda^3), \qquad \operatorname{Tr}X=\operatorname{Tr}Y=0.$$

At a faithful interior point,

$$\left.\frac{d}{d\lambda}S(\rho_\lambda\Vert\sigma)\right|_0=0.$$

Differentiating the exact identity therefore yields

$$\delta S =\delta\langle K_\sigma\rangle =\operatorname{Tr}(XK_\sigma).$$

The normalization condition is essential: in the direct entropy derivative, the otherwise troublesome term is $-\operatorname{Tr}X$.

The positive quadratic term

The Fréchet derivative of the logarithm at a faithful reference is

$$\mathcal T_\sigma(X) =\int_0^\infty dt\, (\sigma+t)^{-1}X(\sigma+t)^{-1}.$$

Consequently,

$$S(\rho_\lambda\Vert\sigma) =\frac{\lambda^2}{2}\, \chi_\sigma(X,X)+O(\lambda^3), \qquad \chi_\sigma(X,X) =\operatorname{Tr}[X\mathcal T_\sigma(X)].$$

The acceleration $Y$ cancels from this Hessian because the first derivative of relative entropy vanishes at coincidence. In an eigenbasis $\sigma=\sum_n p_n\lvert n\rangle\langle n\rvert$,

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

with diagonal value $1/p_n$. Every coefficient is positive. This is the Bogoliubov–Kubo–Mori quadratic form; Petz 1996, pp. 87–90 places it within the monotone-metric classification.

Combining the expansion with the exact identity gives the useful deficit formula

$$\Delta S =\Delta\langle K_\sigma\rangle -\frac{\lambda^2}{2}\chi_\sigma(X,X) +O(\lambda^3).$$

The remainder is not automatic. For the linear family $\rho_\lambda=\sigma+\lambda X$, a sufficient finite-dimensional positivity condition is

$$\lvert\lambda\rvert\, \left\lVert\sigma^{-1/2}X\sigma^{-1/2}\right\rVert<1.$$

A cubic remainder additionally requires three derivatives in a topology in which the logarithm is differentiable. Near a support change, relative entropy can be infinite or nonanalytic. In continuum QFT one should instead state a common algebra, a class of smeared or energy-bounded insertions, and the mode of regulator convergence; an operator-norm Taylor ball usually does not survive the type-III limit.

Gaussian vacuum-mode benchmark

A regulated Gaussian vacuum restricted to a subregion decomposes into mixed Williamson modes. Take one such mode as the exactly solvable benchmark,

$$\sigma=(1-q)q^N, \qquad q=e^{-\varepsilon}, \qquad K_\sigma=-\log(1-q)+\varepsilon N,$$

with $0<q<1$, $N=a^\dagger a$, and $[a,a^\dagger]=1$. This is faithful and trace class. Apply a coherent displacement

$$\rho_\lambda =D(\lambda\alpha)\sigma D(\lambda\alpha)^\dagger, \qquad D(\alpha)=e^{\alpha a^\dagger-\alpha^*a}.$$

Unitary conjugation preserves the spectrum, so $\Delta S=0$. Since $D(\lambda\alpha)^\dagger aD(\lambda\alpha)=a+\lambda\alpha$ and the thermal one-point function of $a$ vanishes,

$$\Delta\langle K_\sigma\rangle =\varepsilon\lambda^2\lvert\alpha\rvert^2.$$

The exact identity now gives

$$S(\rho_\lambda\Vert\sigma) =\varepsilon\lambda^2\lvert\alpha\rvert^2, \qquad \chi_\sigma=2\varepsilon\lvert\alpha\rvert^2.$$

Thus the linear first law reads $0=0$, while the first distinguishable response is exactly quadratic. For the reproducible choice $q=1/2$, $\alpha=1/3$, and $\lambda=0.2$,

$$\Delta S=0, \qquad \Delta\langle K_\sigma\rangle =S(\rho_\lambda\Vert\sigma) =\frac{\log2}{225} \simeq3.08065\times10^{-3}.$$

This analytic single-mode result has zero series and quadrature error. It is a regulator-level QFT benchmark, not a claim that a local continuum algebra factorizes into independent density-matrix modes. A numerical Fock truncation introduces a separate tail error and must be convergence-tested; it is not part of the exact result above.

Adversarial test: move the reference

Now replace the fixed comparison Hamiltonian by

$$K_\lambda=-\log\rho_\lambda =D(\lambda\alpha)K_\sigma D(\lambda\alpha)^\dagger.$$

Then $\langle K_\lambda\rangle_{\rho_\lambda}=S(\rho_\lambda)=S(\sigma)$, so comparing the moving quantities gives the tautology $0=0$ and hides the positive fixed-reference relative entropy. In general,

$$\left.\frac{d^2}{d\lambda^2} \langle K_\lambda\rangle_{\rho_\lambda}\right|_0 =\operatorname{Tr}(YK_\sigma) +2\operatorname{Tr}(X\dot K_0) +\operatorname{Tr}(\sigma\ddot K_0).$$

Only the first term is the second variation of the fixed-reference modular energy. For the displaced mode, the last two terms sum to $-2\varepsilon\lvert\alpha\rvert^2$ and cancel its fixed-reference curvature. Without a prescription for $\dot K_0$ and $\ddot K_0$, the moving-reference comparison has an uncontrolled contribution.

The strongest surviving statement is therefore precise: normalization gives $\delta S=\delta\langle K_\sigma\rangle$ on a fixed algebra about a fixed faithful reference. It does not license replacing $K_\sigma$ by the modular Hamiltonian of the perturbed state, nor does it control support-changing or moving-region families.

Common pitfalls

Treating the first law as dynamics. It follows from normalization and differentiability. Equations of motion or gravitational field equations require additional input.

Subtracting separately regulated quantities at different cutoffs. Entropy and modular energy must use the same algebra and regulator before their finite relative-entropy combination is formed.

Writing $O(\lambda^3)$ at a support boundary. The logarithm is singular at zero eigenvalues. Restrict to the supported face or use an algebraic statement whose differentiability domain is established.

Exercises

1. Let $\rho_\lambda=\sigma+\lambda X+\lambda^2Y/2+O(\lambda^3)$ be normalized. Show directly that $Y$ does not enter the quadratic coefficient of $S(\rho_\lambda\Vert\sigma)$.

Solution

Write $F(\rho)=S(\rho\Vert\sigma)$. At $\rho=\sigma$, the first derivative on normalized tangents is zero. The chain rule gives

$$\left.\frac{d^2}{d\lambda^2}F(\rho_\lambda)\right|_0 =D^2F_\sigma[X,X]+DF_\sigma[Y].$$

The second term vanishes because $\operatorname{Tr}Y=0$. The Hessian is $D^2F_\sigma[X,X]=\operatorname{Tr}[X\mathcal T_\sigma(X)]$, so only $X$ remains.

2. For $\sigma=\operatorname{diag}(p,1-p)$ and

$$X=\begin{pmatrix}x&z\\ z^*&-x\end{pmatrix},$$

compute $\chi_\sigma(X,X)$ and take the limit $p\to1/2$.

Solution

The two diagonal entries contribute $x^2(1/p+1/(1-p))$. The two ordered off-diagonal entries contribute

$$2\frac{\log p-\log(1-p)}{2p-1}\lvert z\rvert^2.$$

Hence

$$\chi_\sigma =\frac{x^2}{p(1-p)} +2\frac{\log[p/(1-p)]}{2p-1}\lvert z\rvert^2.$$

Both divided differences tend to $2$ at $p=1/2$, giving $\chi_\sigma=4(x^2+\lvert z\rvert^2)$.

3. For the coherent Gaussian family above, verify explicitly that the moving modular Hamiltonian cancels the fixed-reference quadratic response.

Solution

The fixed Hamiltonian gives

$$\langle K_\sigma\rangle_{\rho_\lambda} =\langle K_\sigma\rangle_\sigma +\varepsilon\lambda^2\lvert\alpha\rvert^2,$$

so its second derivative is $2\varepsilon\lvert\alpha\rvert^2$. By unitary covariance,

$$\langle K_\lambda\rangle_{\rho_\lambda} =\operatorname{Tr} \left[D\sigma D^\dagger D K_\sigma D^\dagger\right] =\operatorname{Tr}(\sigma K_\sigma),$$

whose second derivative is zero. Therefore the terms containing $\dot K_0$ and $\ddot K_0$ in the product rule sum to $-2\varepsilon\lvert\alpha\rvert^2$. They remove, rather than compute, the fixed-reference relative-entropy curvature.

References

• Araki, Huzihiro. “Relative Entropy of States of von Neumann Algebras.” Publications of the Research Institute for Mathematical Sciences 11 (1976): 809–833. DOI.
• Blanco, David D., Horacio Casini, Ling-Yan Hung, and Robert C. Myers. “Relative Entropy and Holography.” Journal of High Energy Physics 2013, no. 8 (2013): 060. DOI; arXiv.
• Petz, Dénes. “Monotone Metrics on Matrix Spaces.” Linear Algebra and its Applications 244 (1996): 81–96. DOI.