Vacuum Choice and Initial-State Effects

An FLRW metric rarely selects a unique vacuum. Canonical positivity says whether proposed data define a state; ultraviolet regularity says which local observables can be renormalized; symmetry and preparation principles may then select within the admissible family. None of those tests should be silently substituted for another.

Required background. FLRW mode quantization fixes Gaussian mode data; vacuum ambiguity separates state from particle basis; and the Hadamard parametrix fixes the local singularity. Helpful background. Constructing Hadamard states and state-selection failure modes supply general constructions and counterexamples.

Gaussian data on one Cauchy slice

A pure homogeneous isotropic Gaussian state can be specified by one normalized mode $v_k$ for each $k$. A mixed Gaussian state additionally requires occupation and pairing functions $N_k$ and $C_k$. Its equal-time field covariance is

$$\langle\chi_{\mathbf k}\chi_{\mathbf q}\rangle =\delta^3(\mathbf k+\mathbf q) \left[ (2N_k+1)|v_k|^2 +2\operatorname{Re}(C_kv_k^2) \right].$$

The one-mode covariance matrix

$$\Gamma_k= \begin{pmatrix} N_k+1 & C_k\\ C_k^* & N_k \end{pmatrix}$$

must be positive. Hence

$$N_k\ge0, \qquad |C_k|^2\le N_k(N_k+1).$$

The Wronskian fixes the commutator but says nothing about this inequality. Saturation describes a pure squeezed covariance; strict inequality leaves mixed-state entropy.

Spatial homogeneity and isotropy reduce all data to functions of $k$, but they do not determine those functions. Fulling’s analysis of canonical quantization on nonstationary geometries is the foundational warning against inferring a preferred complex structure from the metric alone Fulling 1973.

Finite adiabatic order is not the Hadamard condition

A Hadamard two-point function has the universal microlocal short-distance singularity needed by locally covariant composite observables. Finite-order adiabatic data give only finite Sobolev or large-$k$ control. In the standard spatially flat Robertson–Walker mode construction, Pirk proved that an adiabatic vacuum is Hadamard precisely at infinite order Pirk 1993. Junker and Schrohe formulate finite adiabatic order using Sobolev wavefront sets and place Hadamard states inside the all-orders class Junker and Schrohe 2002, Definition 3.2 and Lemma 3.3.

This distinction matters because Junker’s original 1996 paper stated a stronger Robertson–Walker result that was corrected. Its 2002 erratum explicitly withdraws the claim that the finite-order adiabatic vacua considered there are Hadamard. The correct DOI for the original article is 10.1142/S0129055X9600041X.

“Fourth-order adiabatic” is therefore a useful declared ultraviolet standard for the usual four-dimensional stress mode subtraction, but it is not a synonym for “Hadamard.” When full local QFT control is required, begin with a Hadamard state or prove the corresponding all-orders statement.

Two explicit admissible states

Consider a massive minimally coupled scalar on one smooth FLRW geometry with a stationary asymptotic past. Let $v_k$ be the normalized positive-frequency in-mode, evolved by the exact equation, and assume this reference in-state is Hadamard. This is the reference state.

Construct a second state with

$$\widetilde v_k=A_kv_k+B_kv_k^*, \qquad A_k=\sqrt{1+|B_k|^2},$$

and choose the smooth infrared deformation

$$B_k=b_0e^{i\theta} \begin{cases} \displaystyle \exp\!\left[ -\frac{(k/k_\star)^2}{1-(k/k_\star)^2} \right], & 0\le k<k_\star,\\[8pt] 0, & k\ge k_\star. \end{cases}$$

The function depends smoothly on $k^2$ at the origin and vanishes with all derivatives at $k=k_\star$. For a concrete comparison take

$$b_0=0.15, \qquad \theta=\frac\pi3, \qquad k_\star=2ma(\eta_0),$$

on a common initial slice $\eta_0$ in the stationary past. The identity $A_k^2-|B_k|^2=1$ preserves the Wronskian. Because $B_k$ has compact momentum support, the difference between the two-point functions is smooth whenever the reference state is Hadamard. Thus both states are Hadamard—and consequently better than any fixed finite adiabatic order—while their infrared data differ.

The coefficients $A_k,B_k$ remain constant under the common exact equation, so this is also an explicit propagation prescription rather than a comparison of unrelated slices.

Power and stress differences

The two field spectra differ by

$$\begin{aligned} \Delta\mathcal P_\phi(k,\eta) &=\frac{k^3}{2\pi^2a^2} \left(|\widetilde v_k|^2-|v_k|^2\right)\\ &=\frac{k^3}{\pi^2a^2} \left[ |B_k|^2|v_k|^2 +\operatorname{Re}(A_kB_k^*v_k^2) \right]. \end{aligned}$$

The second term is phase sensitive. Consequently, the same $|B_k|$ can enhance or suppress the spectrum at a particular time; “more particles” does not imply a pointwise larger field spectrum.

Use exactly the same local subtraction for both states. For the massive minimal example, set $D_k=v_k'-\mathcal Hv_k$ and $\widetilde D_k=\widetilde v_k'-\mathcal H\widetilde v_k$. The subtraction terms cancel in the difference, leaving

$$\Delta\rho(\eta) =\frac1{4\pi^2a^4} \int_0^{k_\star}dk\,k^2 \left[ |\widetilde D_k|^2-|D_k|^2 +\omega_k^2 \left(|\widetilde v_k|^2-|v_k|^2\right) \right].$$

The integral is finite because the deformation is smooth and compactly supported. Pressure follows from the independently computed spatial-stress integrand on the adiabatic-subtraction page. Evolving $v_k$ and $\widetilde v_k$ with the same solver then gives two different, finite, conserved stress tensors. Local ultraviolet regularity has licensed both calculations; it has not made the answers identical.

This example cleanly separates three statements:

Test	What it establishes	What it does not establish
Wronskian and covariance positivity	valid canonical Gaussian data	ultraviolet regularity
Hadamard or sufficient adiabatic control	admissibility for declared local composites	a unique vacuum
Asymptotic, Euclidean, or preparation rule	a preferred member when its hypotheses hold	preference outside that domain

Selection criteria and their boundaries

An asymptotic positive-frequency prescription requires the asymptotic region to exist. Euclidean continuation requires an appropriate analytic continuation and separate treatment of zero modes. Minimizing an instantaneous Hamiltonian depends on the canonical variable and time coordinate. Minimizing a smeared energy is better behaved, but the sampling function is physical input rather than a theorem that the geometry has one vacuum.

These criteria can agree in special spacetimes. For example, de Sitter symmetry plus Euclidean analyticity selects the usual Euclidean state for suitable free fields. That conclusion cannot be transferred unchanged to a finite inflationary patch, a field with a problematic zero mode, or an arbitrary initial hypersurface.

The adversarial test is the pair constructed above. Both states share the same ultraviolet singularity, yet their low-$k$ spectra and finite stresses differ. Therefore any argument that uses Hadamard regularity alone to identify them has silently added a selection principle.

Common pitfalls

Using regularity as uniqueness. The Hadamard condition defines an admissible class. Infrared occupation and pairing can vary smoothly within that class.

Changing the subtraction with the state. State differences are computed with one fixed local prescription. Re-fitting counterterms to remove an infrared difference erases physics and violates state independence.

Exercises

1. Show that the deformation preserves ultraviolet regularity

Why does the compactly supported $B_k$ leave the Hadamard singularity of the reference state unchanged?

Solution

The difference of two-point functions is built from terms proportional to $B_k$, $|B_k|^2$, and their conjugates. Every such momentum integral is over the compact interval $0\le k<k_\star$, and the integrand is smooth at both endpoints. Arbitrarily many spacetime derivatives can therefore be passed under the finite integral without producing a high-$k$ divergence. The difference is smooth, so it cannot alter the wavefront set or the universal short-distance singularity. If the reference state is Hadamard, the deformed state is Hadamard as well.

2. Derive the pairing bound

Use positivity of $\Gamma_k$ to derive $|C_k|^2\le N_k(N_k+1)$. What does equality mean?

Solution

A Hermitian $2\times2$ matrix is positive only if its diagonal entries and determinant are nonnegative. Here $N_k\ge0$ and

$$\det\Gamma_k=N_k(N_k+1)-|C_k|^2\ge0.$$

This is the stated bound. Equality makes the covariance determinant minimal, so the uncertainty relation is saturated and the one-mode Gaussian covariance is pure. Strict inequality adds noise and hence mixed-state entropy.

References

• Fulling, Stephen A. “Nonuniqueness of Canonical Field Quantization in Riemannian Space-Time.” Physical Review D 7 (1973): 2850–2862. DOI.
• Junker, Wolfgang. “Hadamard States, Adiabatic Vacua and the Construction of Physical States for Scalar Quantum Fields on Curved Spacetime.” Reviews in Mathematical Physics 8 (1996): 1091–1159. DOI.
• Junker, Wolfgang. “Erratum: Hadamard States, Adiabatic Vacua and the Construction of Physical States for Scalar Quantum Fields on Curved Spacetime.” Reviews in Mathematical Physics 14 (2002): 511–517. DOI.
• Junker, Wolfgang, and Elmar Schrohe. “Adiabatic Vacuum States on General Spacetime Manifolds: Definition, Construction, and Physical Properties.” Annales Henri Poincaré 3 (2002): 1113–1181. DOI. Open PDF.
• Pirk, K.-T. “Hadamard States and Adiabatic Vacua.” Physical Review D 48 (1993): 3779–3783. DOI.