Semigroups and the two meanings of time

Two different evolutions appear in the Yang–Mills literature. One evolves a physical quantum state through Euclidean time. The other evolves a stochastic algorithm or stochastic quantization process through an auxiliary time. Both can be described by semigroups, but their generators and gaps are not automatically the same.

Composing evolution

Prerequisites: exponentials of matrices, basic probability, and the spectral theorem.
Outcomes: distinguish the two clocks; compute a relaxation scale; identify why an unchanged invariant measure can have different stochastic relaxation rates.

A semigroup is a family of operators $T_t$ for $t\geq0$ satisfying

$$T_0=I,\qquad T_{t+s}=T_tT_s.$$

For a nonnegative self-adjoint Hamiltonian $H$, spectral calculus defines

$$T_t=e^{-tH}.$$

The restriction to nonnegative time makes decay natural. Unlike real-time evolution $e^{-itH}$, Euclidean evolution is a contraction rather than a unitary group.

In a chosen sector with a unique normalized vacuum $\Omega$, for $\psi$ orthogonal to $\Omega$ a physical gap $\Delta$ gives

$$\|e^{-tH}\psi\|\leq e^{-t\Delta}\|\psi\|.$$

This is a statement about the quantum Hamiltonian on its physical Hilbert space.

A Markov example

First compare two finite matrices. A two-level quantum Hamiltonian $H=\operatorname{diag}(0,\Delta)$ has a physical energy gap $\Delta$. A symmetric two-state Markov chain with transition rate $r$ has generator

$$\mathcal L=\begin{pmatrix}-r&r\\r&-r\end{pmatrix}.$$

Its constant vector has eigenvalue zero and its centered vector $(1,-1)$ has eigenvalue $-2r$. The stochastic relaxation gap is $2r$, while its stationary probabilities are $(1/2,1/2)$ for every $r>0$. The two matrices do different jobs: one acts on physical quantum states, the other evolves functions of a sampled state. Similar-looking spectra do not identify their clocks.

Now consider the Ornstein–Uhlenbeck process

$$dX_s=-\kappa X_s\,ds+\sqrt{2D}\,dB_s, \qquad \kappa,D>0.$$

The parameter $s$ is stochastic time, and $B_s$ is standard real Brownian motion: its independent increments over a time interval of length $h$ have mean zero and variance $h$. Its Markov semigroup acts on functions:

$$P_sf(x)=\mathbb E[f(X_s)\mid X_0=x].$$

Its generator is

$$\mathcal Lf=Df''-\kappa x f'.$$

The stationary distribution is Gaussian with variance $D/\kappa$. In the corresponding $L^2$ space, $-\mathcal L$ has gap $\kappa$, and centered observables relax exponentially.

This gap measures relaxation toward equilibrium. It is not a particle mass.

For the linear observable this follows directly from the solution

$$X_s=e^{-\kappa s}X_0+\sqrt{2D}\int_0^s e^{-\kappa(s-r)}\,dB_r.$$

For a deterministic integrand $h(r)$, the Brownian integral has mean zero and variance $\int_0^s h(r)^2\,dr$. The whole noise term above therefore has variance $2D\int_0^s e^{-2\kappa(s-r)}\,dr=(D/\kappa)(1-e^{-2\kappa s})$. Taking $X_0$ independent of future Brownian increments and distributed according to the stationary Gaussian law gives $\mathbb E[X_sX_0]=(D/\kappa)e^{-\kappa s}$. The full Gaussian $L^2$ generator has Hermite-polynomial modes with eigenvalues $0,\kappa,2\kappa,\ldots$ for $-\mathcal L$; the stated gap refers to that space and normalization, not an arbitrary observable or physical clock.

A decisive test

Multiply both $D$ and $\kappa$ by a positive constant $c$. The stationary measure is unchanged because $D/\kappa$ is unchanged. But the stochastic relaxation rate becomes $c\kappa$.

Therefore the numerical value of the Markov gap cannot be identified directly with an observable determined solely by the equilibrium field theory. The evolution’s clock can be rescaled while preserving its invariant measure.

This does not make stochastic analysis irrelevant. Locality and functional inequalities can help derive spatial correlation estimates for that measure. Those estimates may then enter a separate reconstruction and physical-gap argument. The connecting steps must be supplied.

Poincaré inequalities

For a reversible Markov process with invariant measure $\mu$, a Poincaré inequality has the form

$$\operatorname{Var}_\mu(f)\leq \frac1\lambda\,\mathcal E(f,f), \qquad \mathcal E(f,f)=-\langle f,\mathcal Lf\rangle_\mu.$$

It controls the generator on centered functions. Under the standard domain assumptions it implies an $L^2$ relaxation estimate involving $\lambda$. Its constants may depend on coupling, volume, cutoff, and the normalization of stochastic time.

The inequality is powerful precisely because it is explicit. Retaining its hypotheses is essential when translating it into another kind of statement.

Exercise. For $f(x)=x$ in the Ornstein–Uhlenbeck example, compute $\operatorname{Var}_\mu(f)$ and $\mathcal E(f,f)$.

Solution. The variance is $D/\kappa$. Since $\mathcal Lx=-\kappa x$, the Dirichlet form is $\kappa\mathbb E[X^2]=D$. The Poincaré bound is saturated with $\lambda=\kappa$.

Continue

The three-dimensional model makes another distinction: local existence of a stochastic process does not by itself establish global existence or an invariant field measure. Operator domains and forms explains the domain assumptions behind an energy estimate.

Read the spectral-theorem bridge and uniform estimates. Shen, Zhu, and Zhu’s stochastic analysis of lattice Yang–Mills is a concrete research example in which functional inequalities and locality lead to further correlation results under stated strong-coupling assumptions.