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JT disk spectral preparation and ensemble boundaries

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

PostulateStatement: Literature-sourcedProof: Not applicable‡ sources checked 2026-10-22Open item page →

Adopted JT disk spectral model

Physical scope

An adopted continuous genus-zero disk density for the specified two-dimensional Jackiw–Teitelboim (JT) model in natural units. It asserts no particular individual exact energy spectrum, higher-topology contribution, or empirical validity.

Postulate

Use natural units c=ℏ=kB=1. Thus energy E has units s−1, inverse temperature β and the positive model parameter C have units s, and S0∈R is dimensionless. In this item, JT disk model means only the disk-sector spectral model with the density specified below; the label refers to the cited two-dimensional Jackiw–Teitelboim calculations and adopts no additional field equation or path-integral construction. Genus-zero disk contribution means the single connected disk term, with one boundary and no handles, in the formal expansion by spacetime topology. A nonperturbative completion means additional exact model data that select an exact theory beyond the terms fixed by that formal expansion.

A continuous spectral density here is a nonnegative function ρ0:(0,∞)→[0,∞) whose measure ρ0(E) dE assigns model spectral weight to energy intervals. Adopt

ρ0(E):=eS0C2π2sinh⁡ ⁣(2π2CE),E>0.

The density has units s, so ρ0(E) dE is dimensionless. This postulate specifies a continuous disk-level input only. It does not specify an individual discrete energy spectrum, sum higher-topology terms, choose a nonperturbative completion, or assert that the model describes nature.

Remarks

The prior restricted semiclassical AdS entropy dictionary Restricted semiclassical entropy dictionary concerns bulk dimension d+1 with d≥2, and is background context only for this two-dimensional disk model: none of its entropy-dictionary assumptions is adopted here, and it does not derive this disk density. The disk expression remains a separate adopted input; any exact boundary quantum system, higher-genus expansion, or nonperturbative completion is outside this postulate.

Physics theoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Exact disk-model Laplace integral and heat capacity

Physical scope

A conditional exact calculation for the adopted continuous genus-zero JT disk density, at fixed C and S0 in c=ℏ=kB=1 units. It does not assert a discrete microscopic spectrum, a higher-topology result, a nonperturbative completion, or empirical validity.

Statement

Let C>0 and β>0 have units of seconds, let S0∈R be dimensionless, and use c=ℏ=kB=1 units as in the adopted JT disk spectral model. Then E>0 has units s−1 and the density ρ0(E) from that model has units seconds. Define the disk partition function as the Laplace integral

Z0(β):=∫0∞ρ0(E)e−βE dE.

When Z0(β) is finite and positive, define the normalized canonical energy distribution by dPβ(E):=Z0(β)−1ρ0(E)e−βE dE and its mean energy by Eˉ(β):=∫0∞E dPβ(E). Put T:=1/β and, at fixed C,S0, write CV/kB:=dEˉ/dT for the dimensionless canonical heat-capacity coefficient in these units. Then the integral is finite and

Z0(β)=eS0C2C2π β3/2exp⁡ ⁣(2π2Cβ),Eˉ(β)=32β+2π2Cβ2,CVkB=32+4π2Cβ>0.

These are deductions within the adopted continuous disk model, not claims about an individual microscopic spectrum or nature.

Facts & Assumptions

Given: C>0, S0∈R, β>0, the units in the Statement, and the definitions of Z0, Pβ, Eˉ, and CV/kB there.

[A1]

Adopted JT disk spectral model: in these units, ρ0(E)=eS0Csinh⁡(2π2CE)/(2π2) for E>0, as a continuous disk-level model density; no discrete Hamiltonian or completion is supplied.

[F1]

The Gaussian integral ∫−∞∞e−x2 dx=π: ∫−∞∞e−x2 dx=π; its complete proof squares the integral, evaluates the plane Gaussian in polar coordinates, and selects the positive square root.

[F2]

Differentiation under an improper multiple integral under an integrable derivative bound: for open D⊆Rn and open interval I, if f and ∂tf are continuous, one slice is absolutely improperly integrable, and on every compact parameter interval J⊂I the derivative is bounded by a nonnegative improperly integrable gJ, then every slice is absolutely improperly integrable and ddt∫Df(x,t) dx=∫D∂tf(x,t) dx.

[F3]

Change of variable in an improper integral: if ϕ:I→J is a monotone differentiable surjection between intervals, its derivative is locally Riemann integrable, the proper change-of-variable hypotheses hold on compact truncations, and f is locally Riemann integrable on J, then ∫Jf(x) dx and ∫If(ϕ(t))∣ϕ′(t)∣ dt converge simultaneously and are equal when convergent.

Proof

1.1givenF1F3algebra

Fix β>0 and set a:=2π2C. Completing the square gives −βx2+ax=−β(x−a/(2β))2+a2/(4β). The affine map u=β(x−a/(2β)) is an increasing C1 bijection R→R; its proper change-of-variable hypotheses hold on every compact truncation. Thus [F3] and [F1] give I(a):=∫−∞∞e−βx2+ax dx=ea2/(4β)β−1/2∫−∞∞e−u2 du=ea2/(4β)π/β. The same calculation gives I(−a)=I(a). Since e−βx2cosh⁡(ax) is even and cosh⁡(ax)=(eax+e−ax)/2, its full-line integral is I(a) and F(a):=∫0∞e−βx2cosh⁡(ax) dx=I(a)/2=π2βea2/(4β).

2.1givenF1F2F3step 1.1algebra

For each compact J=[−A,A]⊂R, f(x,t):=e−βx2cosh⁡(tx) and ∂tf=xe−βx2sinh⁡(tx) are continuous on (0,∞)×R. The inequality Ax≤βx2/2+A2/(2β) bounds ∣∂tf∣ on J by eA2/(2β)xe−βx2/2, whose integral is eA2/(2β)/β<∞; the base slice f(x,0)=e−βx2 is absolutely integrable by [F1] and [F3] with the increasing C1 bijection u=βx:(0,∞)→(0,∞), whose derivative is the positive constant β. Thus [F2] applies and F′(a)=∫0∞xe−βx2sinh⁡(ax) dx. Differentiating the closed form in step 1.1 yields F′(a)=aπ ea2/(4β)/(4β3/2).

3.1givenA1F3step 2.1algebra

Let h(E):=ρ0(E)e−βE and ϕ(x):=x2 on (0,∞). The map ϕ is an increasing C1 bijection with ϕ′(x)=2x>0; on each compact truncation its proper change-of-variable hypotheses hold, and h is locally continuous. By [A1], the transformed integrand is 2xh(x2)=eS0Cπ−2xe−βx2sinh⁡(ax), whose improper integral is finite by step 2.1. Thus [F3] gives Z0(β)=∫0∞h(E) dE=∫0∞h(x2)2x dx=eS0Cπ−2F′(a), where a=2π2C. Substitution of step 2.1 gives the stated formula for Z0. It is finite and strictly positive because C,β>0 and the displayed factors are finite and positive.

4.1givenA1F2step 3.1algebra

For f(E,β):=ρ0(E)e−βE on (0,∞)×(0,∞), both f and ∂βf=−Eρ0(E)e−βE are continuous. On a compact J⊂(0,∞) let β0:=min⁡J>0. Since sinh⁡(aE)≤eaE/2 and aE≤β0E/2+a2/(2β0), the derivative magnitude is bounded on J by a constant times Ee−β0E/2, which has finite integral; the same estimate without the factor E gives an integrable base slice. Apply [F2] on (0,∞) to differentiate Z0: Z0′(β)=−∫0∞Eρ0(E)e−βE dE. Hence Eˉ=−∂βlog⁡Z0=3/(2β)+2π2C/β2.

5.1givenstep 4.1algebra∎

At fixed C,S0, T=1/β gives dβ/dT=−β2, so CV/kB=dEˉ/dT=−β2∂βEˉ=3/2+4π2C/β. Since C>0 and β>0, this coefficient is strictly positive.

Remarks

The proof uses the published Gaussian integral supplier, the improper-substitution theorem for the affine, rescaling, and square substitutions, and the dominated-differentiation theorem for both parameter-dependent improper integrals. The JT density itself remains the adopted premise [A1]. No path-integral or ensemble assumption is used in the calculation.

5 · Examples, experiments and counterexamples

None yet.

Sources