Thermal States and One-Point Data

A thermal CFT state is not obtained by conformally mapping the vacuum. The density matrix introduces a scale $\beta$ and selects a rest frame, so local primaries can acquire one-point functions that vanish in the vacuum. Symmetry fixes their tensor form and temperature scaling; dimensionless coefficients remain independent thermal data.

Required background. The cylinder map and Hamiltonian fixes energy conventions, and scalar two- and three-point functions fixes primary normalization. Helpful background. Torus partition functions as bootstrap data gives the special two-dimensional finite-volume trace, which should not be confused with the flat thermal cylinder in general dimension.

The normalized thermal state on the flat cylinder

On the Euclidean geometry

$$\mathcal M_\beta=S^1_\beta\times\mathbb R^{d-1}, \qquad \tau\sim\tau+\beta,$$

the neutral canonical state is

$$\rho_\beta=\frac{e^{-\beta H}}{Z(\beta)}, \qquad Z(\beta)=\operatorname{Tr}e^{-\beta H}, \qquad \langle X\rangle_\beta =\operatorname{Tr}(\rho_\beta X).$$

The normalization gives $\langle\mathbf1\rangle_\beta=1$. An unnormalized path integral and a normalized correlator differ by $Z(\beta)$; mixing them changes every thermal coefficient.

The default state here has no angular velocity or charge chemical potential and is invariant under translations, spatial rotations, and the Euclidean reflection generated by reversing thermal time together with one spatial direction. A grand-canonical state instead uses

$$\rho_{\beta,\mu} =\frac{e^{-\beta(H-\mu Q)}} {\operatorname{Tr}e^{-\beta(H-\mu Q)}}.$$

Its convergence domain depends on the charged spectrum, and its KMS relation is twisted by the charge of the moved operator. A rotating ensemble similarly selects additional vectors. Results derived below for the neutral state must be rederived when those symmetries are broken.

One-point selection rules

Translation invariance makes every derivative descendant have zero one-point function. Let $\mathcal O_{\mu_1\ldots\mu_J}$ be a symmetric traceless primary of dimension $\Delta$ and spin $J$, normalized in the vacuum by

$$\langle \mathcal O_{\mu_1\ldots\mu_J}(x) \mathcal O_{\nu_1\ldots\nu_J}(0) \rangle =\frac{c_{\mathcal O}\, \mathcal I_{\mu_1\ldots\mu_J,\nu_1\ldots\nu_J}(x)} {\lvert x\rvert^{2\Delta}}.$$

Let $e_\mu$ be the unit vector around the Euclidean thermal circle. For the neutral, homogeneous state, restriction from $SO(d)$ to the preserved $O(d-1)$ admits a singlet only for even-spin symmetric traceless tensors. Symmetry and dimensional analysis then give

$$\left\langle \mathcal O_{\mu_1\ldots\mu_J} \right\rangle_\beta =\frac{b_{\mathcal O}}{\beta^\Delta} \left(e_{\mu_1}\cdots e_{\mu_J}-\text{traces}\right), \qquad J\in2\mathbb Z_{\ge0}.$$

The coefficient $b_{\mathcal O}$ depends on the state and on the normalization of $\mathcal O$. The combination that appears in a thermal OPE is invariant under rescaling only after the vacuum OPE coefficient and $c_{\mathcal O}$ are included. These selection rules and conventions are derived in Iliesiu et al. 2018, § 2.1, pp. 5–10.

Additional internal symmetries can force $b_{\mathcal O}=0$. In an unbroken neutral ensemble, any charged primary or operator odd under a preserved discrete symmetry has vanishing expectation value. In an infinite-volume symmetry-broken phase, one must specify the selected extremal state; an invariant mixture and a pure broken-symmetry state have different one-point data.

For a scalar primary,

$$\langle\mathcal O\rangle_\beta =b_{\mathcal O}\beta^{-\Delta}.$$

This looks simple, but it is not fixed by the vacuum two-point normalization. It is precisely the unknown data constrained by thermal KMS crossing and inversion.

The stress tensor and thermodynamics

Return to Lorentzian signature with the site’s $(+---)$ metric and rest-frame velocity $u^\mu=(1,\mathbf0)$. Homogeneity and spatial isotropy imply

$$\langle T^{\mu\nu}\rangle_\beta =(\varepsilon+p)u^\mu u^\nu-p\eta^{\mu\nu}.$$

On flat space, after a declared vacuum subtraction and in the absence of a trace anomaly contribution,

$$\langle T^\mu{}_{\mu}\rangle_\beta=0 \quad\Longrightarrow\quad \varepsilon=(d-1)p.$$

Extensivity in the thermodynamic limit gives

$$\log Z(\beta,V)=a\,V\beta^{-(d-1)},$$

so

$$p=\frac{1}{\beta}\frac{\partial\log Z}{\partial V} =a\beta^{-d}, \qquad \varepsilon=-\frac1V\frac{\partial\log Z}{\partial\beta} =(d-1)a\beta^{-d}.$$

Thus the stress-tensor thermal coefficient is equivalent, once the stress-tensor two-point normalization and tensor convention are fixed, to the dimensionless free-energy density $a$. A sign quoted for $b_T$ without saying whether it refers to Euclidean $T_{\tau\tau}$ or Lorentzian $T^{00}$ is ambiguous.

At finite spatial volume, $\log Z$ need not be extensive. On $S^1_\beta\times S^{d-1}_R$, a scalar one-point function has the more general form

$$\langle\mathcal O\rangle_{\beta,R} =R^{-\Delta}f_{\mathcal O}(\beta/R),$$

and curvature counterterms can add local, scheme-dependent terms. The flat-cylinder coefficient $b_{\mathcal O}$ is recovered only in a controlled $R\to\infty$ limit at fixed $\beta$.

How partition data enter—and where they stop

Differentiating the partition function determines integrated conserved quantities such as energy and charge. It does not determine an arbitrary primary one-point coefficient. For a deformation

$$H(\lambda)=H+\lambda\int d^{d-1}x\,\mathcal O(x),$$

one has formally

$$\left.\frac{\partial\log Z}{\partial\lambda}\right|_{\lambda=0} =-\beta\int d^{d-1}x\, \langle\mathcal O\rangle_\beta,$$

provided the coupling, operator renormalization, and contact counterterms are fixed. This is a useful normalization check, not a claim that the undeformed spectrum alone fixes $b_{\mathcal O}$.

In two dimensions with finite spatial circumference $2\pi R$, the same Hilbert-space trace is a rectangular torus partition function and modular covariance relates high and low temperature. On $S^1_\beta\times\mathbb R$, that relation is obtained only after a thermodynamic limit with its volume factors controlled. In $d>2$ there is no general torus modular group relating the thermal circle to a noncompact spatial direction.

Thermal conventions and controlled truncations

The table records the state definition before any KMS or OPE calculation. Its final column distinguishes a genuine error estimate from an unchecked cutoff.

State or approximation	Geometry and generator	One-point and KMS consequence	Normalization or contact requirement	Omitted-tail bound
Neutral canonical	$S^1_\beta\times\mathbb R^{d-1}$, generator $H$	even-spin STT one-points; ordinary bosonic or fermionic KMS statistics	divide by $Z$; fix vacuum subtraction and $c_{\mathcal O}$	no truncation in the definition
Charged grand canonical	generator $H-\mu Q$	charged one-points may survive; KMS acquires the charge twist	specify charge normalization and convergence strip in $\mu$	bound the charged trace, not only the neutral density
Rotating state	generator $H-\boldsymbol\Omega\cdot\boldsymbol J$	extra vectors permit more tensor structures	state the frame, angular domain, and stress-tensor convention	control high-spin growth against angular suppression
Spatial sphere	$S^1_\beta\times S^{d-1}_R$	coefficients become functions of $\beta/R$	retain curvature counterterms and Casimir subtraction	take the $R\to\infty$ limit with a finite-volume error
Symmetry-broken phase	chosen extremal state in infinite volume	odd or charged one-points can be nonzero	distinguish a pure phase from the invariant mixture	finite-volume tunneling and phase-selection errors must be estimated
Spectral trace truncated at bin $N$	energy bins $[n/R,(n+1)/R)$	KMS is exact only for the full trace	use the same $Z_N$ in every normalized correlator	if $d_n\le Ae^{\alpha n}$ and $\beta/R>\alpha$, then $R_Z\le A e^{-(\beta/R-\alpha)(N+1)}/(1-e^{-(\beta/R-\alpha)})$

For a bounded operator $X$, if $p_{>N}=R_Z/Z$ is the discarded thermal probability, the normalized expectation computed from the retained states satisfies

$$\left\lvert \langle X\rangle_\beta -\langle X\rangle_{\beta,N} \right\rvert \le2\lVert X\rVert p_{>N}.$$

Local quantum fields are unbounded, so this estimate cannot be applied to them without a matrix-element-weighted bound. The partition-function tail alone controls normalized traces of bounded observables, not arbitrary local one-point functions.

Contact terms and operator normalization

Thermal one-point data should refer to separated or renormalized local operators. Three issues must be fixed:

1. Identity mixing. A scalar can mix with the identity when dimensions and symmetries permit a local counterterm; the chosen subtraction shifts its one-point function.
2. Stress-tensor contacts. Metric derivatives of $\log Z$ contain contact terms. On curved space they include anomaly and improvement contributions; on flat space the vacuum subtraction remains part of the convention.
3. Thermodynamic limit. A finite-volume trace is analytic in many parameters where an infinite-volume state can have phase transitions. Limits of $V$, regulator, and source must be ordered explicitly.

Once these are fixed, $b_{\mathcal O}$ is a well-defined target for the thermal OPE.

Common failure modes

Mapping the vacuum to the thermal state. The thermal circle is a global identification, not a local conformal image of $\mathbb R^d$. New one-point coefficients are allowed.

Using $\beta^{-\Delta}$ on a finite sphere without qualification. A second scale $R$ permits an arbitrary function of $\beta/R$ and curvature contacts.

Forgetting normalization by $Z$. An unnormalized path integral includes the partition function in every correlator. Thermal OPE coefficients use normalized expectations here.

Assuming all even-spin one-points are nonzero. Symmetry permits the tensor structure; internal charges, discrete symmetries, and dynamics can still set its coefficient to zero.

Exercises

Derive the conformal equation of state from $\log Z=aV\beta^{-(d-1)}$.

Solution

Using $p=\beta^{-1}\partial_V\log Z$ gives $p=a\beta^{-d}$. Using $E=-\partial_\beta\log Z$ gives $\varepsilon=E/V=(d-1)a\beta^{-d}$. Therefore $\varepsilon=(d-1)p$, in agreement with a traceless perfect-fluid stress tensor.

Suppose $d_n\le2e^{0.3n}$, $\beta/R=1$, and the trace is truncated after bin $N$. Give a bound on the unnormalized omitted weight.

Solution

The effective decay exponent is $1-0.3=0.7$, so

$$R_Z \le \sum_{n=N+1}^{\infty}2e^{-0.7n} =\frac{2e^{-0.7(N+1)}}{1-e^{-0.7}}.$$

To convert this to a normalized probability one divides by the full $Z$, or conservatively by a known lower bound on $Z$.

Continue to local crossing

Thermal OPE and KMS Crossing inserts these one-point functions into a local two-point OPE and derives the ordered KMS equation. General density operators and real-time thermal QFT are treated in Thermal and Nonequilibrium QFT.

References

• Iliesiu, Luca, Murat Koloğlu, Raghu Mahajan, Eric Perlmutter, and David Simmons-Duffin. “The Conformal Bootstrap at Finite Temperature.” Journal of High Energy Physics 2018, no. 10 (2018): 070. doi:10.1007/JHEP10(2018)070. Open preprint.