Probability measures for field theory

A Euclidean path integral is meant to assign expectations to observables. Before asking whether it has a mass gap, ask a simpler question: what is being averaged, and with respect to which probability law? A formal symbol $D\phi$ does not answer either question.

Prerequisites: ordinary integration and the conventions.
Outcomes: write a normalized expectation; identify the observable and configuration space; distinguish a finite regulated measure from a continuum construction.

Begin with one fluctuating coordinate

Take a real variable $x$ and the dimensionless action $S(x)=x^2/(2\sigma^2)$, with $\sigma>0$. The partition function and expectation are

$$Z=\int_{\mathbb R}e^{-S(x)}dx=\sqrt{2\pi}\sigma,\qquad \mathbb E_\mu[F]=Z^{-1}\int_{\mathbb R}F(x)e^{-S(x)}dx.$$

Here $\mu$ is the probability measure; $F$ is an observable. The identity observable is normalized: $\mathbb E_\mu[1]=1$. Oddness gives $\mathbb E_\mu[x]=0$. Differentiating the Gaussian integral, or integrating by parts with its decaying boundary term, gives $\mathbb E_\mu[x^2]=\sigma^2$.

Normalization matters. Multiplying both numerator and denominator by a constant changes neither the law nor the expectation. Omitting the denominator generally changes every nonzero expectation. An action is dimensionless in units $\hbar=1$; if $x$ has units, $\sigma$ carries the same units.

The small amount of measure language we need

A configuration space $X$ lists possible configurations. Its measurable sets are the events to which probabilities can be assigned. A probability measure $\mu$ has nonnegative, countably additive weights and $\mu(X)=1$. A random variable is a measurable function on $X$; an observable may be real, complex, vector valued, or group valued.

For an integrable complex observable $F$, the expectation is $\int_X F\,d\mu$. Integrable means $\int_X|F|d\mu<\infty$. That last condition is useful: a normalized probability measure does not guarantee that every proposed observable has finite moments. A density is only one way to describe a measure. A delta measure at $x_0$ has expectation $F(x_0)$ without having an ordinary Lebesgue density.

These definitions are developed in Durrett, Probability: Theory and Examples, version 5, January 2019, §§1.1–1.6. The definitions provide language; the calculations here are elementary examples rather than a field construction.

A finite lattice is already a useful measure

For a finite set $E$ of independent oriented links and compact gauge group $G$, take $X=G^{|E|}$. With normalized Haar measure and a real continuous action,

$$d\mu(U)=\frac{e^{-S(U)}}{Z}\prod_{\ell\in E}dU_\ell, \qquad Z=\int_X e^{-S(U)}\prod_{\ell\in E}dU_\ell.$$

Compactness and continuity imply $0<Z<\infty$. A normalized trace of a loop holonomy is a bounded observable. Gauge transformations change link coordinates; Haar invariance and a gauge-invariant action make the expectation independent of that change. The Haar bridge works through the group integral.

This finite model has an actual measure. The continuum question still asks for limits of suitably defined observables, with the relevant covariance, positivity, and physical interpretation. Refining a lattice does not simply produce a larger finite-dimensional Lebesgue integral called $D A$.

Cylinder observables: looking through finitely many windows

For a distribution-valued scalar field $\phi$, one can first inspect finitely many smeared variables $\phi(f_1),\ldots,\phi(f_k)$. An observable of the form

$$F(\phi)=h\bigl(\phi(f_1),\ldots,\phi(f_k)\bigr)$$

is called a cylinder observable. The test functions $f_i$ define the windows; $h$ combines their readings. For example, $h(z_1,z_2)=\exp[i(z_1+z_2)]$ is bounded, whereas $h(z_1,z_2)=z_1^2z_2^2$ asks for moment control.

Compatible finite-dimensional distributions are an important ingredient, but support and regularity on the intended field space need separate arguments. The Gaussian bridge shows a controlled case. The tightness bridge explains why a regulated sequence might fail to converge where we need it.

Exercise: an expectation that does not exist

The Cauchy density is $p(x)=1/[\pi(1+x^2)]$. Is symmetry enough to conclude that $\mathbb E[X]=0$?

Hint: test absolute integrability before canceling positive and negative tails.

Solution. The density is normalized, but $\int|x|p(x)dx$ diverges logarithmically. The symmetric principal value of $\int x p(x)dx$ is zero; this is not a finite probability expectation. A regulator-dependent cancellation cannot substitute for the required definition of an observable.

Return to the construction problem

Continue to the lattice definition or limits and uniformity. The official Jaffe–Witten description, §§4–6 explains why the desired limiting theory requires more than a formal integral. A probability measure by itself supplies neither reflection positivity nor a positive physical spectral gap.