From reflection positivity to a physical Hilbert space

Prerequisites: the reflection-positivity lesson and the spectral theorem.

You will learn to calculate a reflected norm, explain why null vectors must be quotiented out, and distinguish a positive Hamiltonian from a gapped one.

Reflect across a time slice

Ordinary covariance positivity says that the variance of a linear combination of random variables is nonnegative. Reflection positivity tests a different form: observables on positive Euclidean time are paired with their reflected, complex-conjugated partners on negative time.

For positive-time observables $F,G$, write

$$(F,G)_{\mathrm{OS}}=\langle(\Theta F)G\rangle.$$

Here $\Theta$ reflects time and conjugates coefficients. The measure, observable algebra and reflection operation are part of the definition. A positive probability density alone does not ensure this form is positive.

If it is positive semidefinite, the Cauchy–Schwarz inequality for that form shows that every null vector is orthogonal to every vector. Consequently the quotient

$$\mathcal H_{\mathrm{pre}}=\mathcal A_+/\mathcal N,\qquad \mathcal N=\{F:(F,F)_{\mathrm{OS}}=0\},$$

has an actual inner product. Completing it produces a Hilbert space. Here $\mathcal A_+$ denotes the chosen positive-time observable space, not automatically every formal polynomial in point fields.

Worked example: a massive Gaussian time process

Take a centered Gaussian process with covariance

$$K(t-s)=\frac{e^{-m|t-s|}}{2m},\qquad m>0.$$

The parameter $m$ has inverse-time units. For $F=\sum_i c_i\phi(t_i)$ with $t_i>0$, the reflected form is

$$(F,F)_{\mathrm{OS}} =\frac1{2m}\sum_{i,j}\overline{c_i}c_j e^{-m(t_i+t_j)} =\frac1{2m}\left|\sum_i c_i e^{-mt_i}\right|^2\geq0.$$

The last equality supplies an explicit reason for positivity; no finite numerical scan is needed for this linear sector.

Many different functions have the same quotient vector. For $m=1$ in chosen units,

$$F=\phi(1)-e\,\phi(2)$$

has zero reflected norm. It is not the zero random variable: its ordinary variance is $(e^2-1)/2>0$. The quotient removes redundancy relative to the reflected form, not randomness in the original probability space.

Where time evolution enters

Shift every insertion by $s\geq0$. In the linear example, its quotient coordinate $\sum_i c_i e^{-mt_i}$ is multiplied by $e^{-ms}$. Constants give a vacuum sector unchanged by the shift. Thus on the constants-plus-linear quotient the time-shift operator has eigenvalues $1$ and $e^{-ms}$, and its generator has energies $0$ and $m$.

This small calculation illustrates the mechanism; it does not prove a general reconstruction theorem. In a field theory one must establish that shifts respect the null space, satisfy the required boundedness and continuity, and fit the covariance and regularity conditions. Locality and analytic continuation involve further structure. The semigroup bridge explains how $e^{-sH}$ relates to a self-adjoint generator.

A useful theorem reference is Osterwalder–Schrader, Axioms for Euclidean Green’s functions II, CMP 42, 281–305 (1975), which extends and corrects their 1973 formulation. The example above is a direct calculation, not a claim to have reproduced those full hypotheses or their proof.

Positivity does not force a gap

Consider instead the positive mixture

$$K(t)=\int_0^1 e^{-m|t|}\,dm.$$

Its reflected linear form is an integral of squared moduli, hence nonnegative. But it has spectral weight arbitrarily close to $m=0$ and $K(t)=(1-e^{-|t|})/|t|$ for $t\ne0$. Its long-time decay is not controlled by any strictly positive exponential rate.

Thus positivity and positive energy are compatible with a gapless spectrum. A separate uniform estimate on physical spectral support is needed. The gap-and-decay page states a conditional converse with the necessary density assumption.

Exercise: test the wrong positivity

The kernel $K(t-s)=\cos(\omega(t-s))$ is ordinarily positive because it is the Gram kernel of $(\cos\omega t,\sin\omega t)$. Test its reflected form for the single insertion $t=\pi/(2\omega)$.

Solution. Reflection produces $K(2t)=\cos\pi=-1$. One negative value is enough to disprove reflection positivity. Ordinary covariance positivity survived because it evaluated time differences, not reflected sums. The positivity lab lets you compare these exact identities with finite tests.

Return to Yang–Mills

A gauge-theory construction must supply a suitable physical observable algebra, deal with gauge constraints, and preserve these structures through regulator and volume limits. Continue to the transfer matrix for a regulated time-evolution interpretation, or return to lesson 10.