Modular Dynamics and Equilibrium Representations

A faithful KMS state does more than implement time translations: in its GNS representation it makes physical equilibrium dynamics a rescaled modular flow. The statement is exact only after the represented von Neumann algebra, support of the state, sign convention, and inverse temperature are fixed. It does not say that the modular flow of every region or every state is laboratory time.

Required background. Modular automorphisms, conjugations, and standard forms supplies $J$, $\Delta$, and $\sigma_s$; C*-dynamical systems and the KMS condition supplies the strip boundary identity. Helpful background. Modular KMS relations and correlators develops the information-theoretic use, and thermal density operators and the KMS condition provides the finite-dimensional model.

KMS dynamics in the GNS representation

Let $(\mathcal A,\alpha)$ be a C*-dynamical system and let $\omega$ be a $\beta$-KMS state with GNS triple $(\mathcal H_\omega,\pi_\omega,\Omega_\omega)$. Invariance of $\omega$ gives a strongly continuous unitary group

$$U_\omega(t)\pi_\omega(A)\Omega_\omega =\pi_\omega(\alpha_t(A))\Omega_\omega, \qquad U_\omega(t)\Omega_\omega=\Omega_\omega.$$

Write $\mathcal M_\omega=\pi_\omega(\mathcal A)''$. If the vector state is faithful on $\mathcal M_\omega$, then $\Omega_\omega$ is separating as well as cyclic, so its modular operator $\Delta_\omega$ is defined. With

$$\sigma_s^\omega(M)=\Delta_\omega^{is}M\Delta_\omega^{-is},$$

the KMS uniqueness theorem identifies the two flows:

$$\boxed{\ \sigma_s^\omega\!\left(\pi_\omega(A)\right) =\pi_\omega\!\left(\alpha_{-\beta s}(A)\right)\ }.$$

The minus sign follows from the convention $\alpha_t(A)=e^{itH}Ae^{-itH}$ together with $\sigma_s(M)=\Delta^{is}M\Delta^{-is}$. The finite and infinite equilibrium representations in Haag, Hugenholtz, and Winnink 1967, §§ 2–4, pp. 221–236 exhibit precisely this commutant and modular structure.

If $\omega$ is not faithful, one first passes to its support projection in $\mathcal M_\omega$ or to the faithful quotient determined by the GNS kernel. Omitting that reduction can make the Tomita map ill-defined. If the KMS state is not factorial, the theorem still acts on the supported von Neumann algebra, but its central decomposition may contain distinct equilibrium phases; no single density matrix is thereby produced.

Proof mechanism and converse

The essential point is uniqueness, not a formal comparison of generators. The KMS strip identity extends from $\pi_\omega(\mathcal A)$ to a $\sigma$-weakly dense analytic subalgebra of $\mathcal M_\omega$. Tomita–Takesaki theory says that the vector state has the corresponding unit-height boundary property for its modular group. After the reparametrization $t=-\beta s$, both automorphism groups have the same boundary values on the same faithful normal state. The uniqueness theorem for modular automorphisms makes them equal.

Conversely, suppose a faithful normal vector state on $\mathcal M$ has modular group $\sigma_s$ and a strongly continuous dynamics $\widetilde\alpha_t$ satisfies

$$\widetilde\alpha_t=\sigma_{-t/\beta}.$$

Then the modular boundary relation gives the $\beta$-KMS condition for $\widetilde\alpha$. This converse is representation-level. To descend it to an abstract C*-algebra one must also know that $\widetilde\alpha_t$ preserves the represented copy of that algebra and is compatible with its kernel.

The generator relation is equally precise. If $U_\omega(t)=e^{itL_\omega}$ is the standard implementation, then

$$\Delta_\omega=e^{-\beta L_\omega}$$

on the supported standard representation. Since a modular operator generally has spectrum on both sides of $1$, $L_\omega$ generally has two-sided spectrum. Calling it a Hamiltonian bounded below would erase the doubled equilibrium representation.

Araki–Woods realization of a free Bose field

Let $\mathfrak h$ be the one-particle Hilbert space, $h\geq m>0$ its energy, and

$$\rho_\beta=(e^{\beta h}-1)^{-1}.$$

The thermal representation of the Weyl algebra acts on the symmetric Fock space over $\mathfrak h\oplus\overline{\mathfrak h}$. In a standard convention,

$$\pi_\beta(W(f)) =W_{\!F}\!\left((1+\rho_\beta)^{1/2}f \oplus\rho_\beta^{1/2}\overline f\right),$$

and the Fock vacuum $\Omega_\beta$ represents the thermal state. The physical dynamics is implemented by

$$L_\beta=d\Gamma(h\oplus(-\overline h)), \qquad U_\beta(t)=e^{itL_\beta}.$$

Thus particles in the first copy and the commutant copy carry opposite Liouvillean signs. The modular data are

$$\Delta_\beta=e^{-\beta L_\beta}, \qquad \Delta_\beta^{is}=U_\beta(-\beta s),$$

while the modular conjugation exchanges the two copies, with the conjugations dictated by the chosen one-particle realization. Detailed CCR constructions and their standard representations appear in Dereziński 2006, §§ 9 and 11, pp. 39–54 (PDF) and originate in Araki and Woods 1963, §§ 3–5, pp. 644–657.

This is the exact first application: the modular operator of the free Bose thermal vector recovers the doubled Liouvillean evolution. Its physical infinite-volume interpretation belongs with infinite-volume KMS states, passivity, and phase multiplicity.

Adversarial test: a bounded-region vacuum

Take the vacuum vector and the algebra of a generic bounded double cone. Reeh–Schlieder may make the pair standard, hence it has a modular group. But standardness alone supplies no spacetime formula for that group and no physical thermal dynamics selected in advance. Declaring the modular parameter to be laboratory time therefore fails at the missing covariance theorem. Wedges in a Poincaré-covariant vacuum theory are special because a Bisognano–Wichmann theorem can identify their modular flow with boosts; generic bounded regions do not inherit that conclusion.

Common pitfalls

Dropping the inverse-temperature rescaling. Modular parameter $s$ is dimensionless. Physical time $t$ enters through $t=-\beta s$ for the conventions used here.

Treating the Liouvillean as a positive Hamiltonian. Its job is to implement equilibrium automorphisms in a standard representation. Its negative copy is essential, not an instability.

Exercises

1. Let $\rho=Z^{-1}e^{-\beta H}>0$ act on a finite-dimensional Hilbert space. Compute the modular action in Hilbert–Schmidt standard form.

Solution

Left multiplication represents $B(\mathcal H)$ and $\Omega_\rho=\rho^{1/2}$. The modular operator acts as $\Delta(X)=\rho X\rho^{-1}$. Hence

$$\sigma_s(A)=\rho^{is}A\rho^{-is} =e^{-i\beta sH}Ae^{i\beta sH} =\alpha_{-\beta s}(A).$$

2. Why does $L_\beta\Omega_\beta=0$ not imply $L_\beta\geq0$?

Solution

The vacuum of the doubled Fock space is fixed by both second-quantized terms, so its eigenvalue is zero. But a one-particle excitation in the conjugate copy has energy $-\lambda$ when $h$ has energy $\lambda>0$. The spectrum is therefore generally symmetric enough to extend in both directions.

References

• Araki, Huzihiro, and E. J. Woods. “Representations of the Canonical Commutation Relations Describing a Nonrelativistic Infinite Free Bose Gas.” Journal of Mathematical Physics 4 (1963): 637–662. DOI.
• Dereziński, Jan. “Introduction to Representations of the Canonical Commutation and Anticommutation Relations.” In Large Coulomb Systems, Lecture Notes in Physics 695 (2006): 63–143. DOI. Open PDF.
• Haag, Rudolf, Nicolaas M. Hugenholtz, and Marinus Winnink. “On the Equilibrium States in Quantum Statistical Mechanics.” Communications in Mathematical Physics 5 (1967): 215–236. DOI.