Skip to content
Yang–MillsA QFT.ORG PROJECT

The transfer matrix and physical time

Prerequisites: the lattice definition, semigroups and reflection positivity.

You will learn to convert transfer eigenvalues into energies with correct units, identify the physical gauge-invariant sector, and distinguish time evolution from Monte Carlo updates.

A Euclidean lattice can be sliced into spatial configurations. Integrating over the fields between adjacent slices suggests an operator TT that advances by a time spacing ata_t. Repeating the step gives TnT^n, and inserting observables on slices produces correlation functions separated by t=natt=na_t.

In the pure SU(N)SU(N) Wilson-action setting, this construction must account for the gauge transformations on each spatial slice. The physical space consists of gauge-invariant states; temporal-link integration implements the corresponding constraints. A transfer operator on an unconstrained space and a physical transfer operator are not interchangeable.

The Wilson convention used here is

SW=β∑p(1−1NRe⁡tr⁡Up),β=2Ng02S_W=\beta\sum_p\left(1-\frac1N\operatorname{Re}\operatorname{tr}U_p\right), \qquad \beta=\frac{2N}{g_0^2}

for isotropic spacing, with each plaquette counted once. If temporal and spatial spacings differ, the couplings for their plaquettes must be specified separately. We do not take a Hamiltonian anisotropic limit by silently replacing ata_t with the spatial spacing.

For TT self-adjoint and positive, normalize its largest vacuum eigenvalue λ0\lambda_0 to one. On a suitable subspace where the logarithm is defined,

H=−1atlog⁡(T/λ0).H=-\frac1{a_t}\log(T/\lambda_0).

An eigenvalue 0<λn≤λ00<\lambda_n\leq\lambda_0 then corresponds to

En−E0=−1atlog⁡λnλ0.E_n-E_0=-\frac1{a_t}\log\frac{\lambda_n}{\lambda_0}.

The logarithm is defined by spectral calculus, not by taking logarithms of matrix entries. Zero spectral values require attention to its domain; “positive semidefinite” and “strictly positive with a bounded inverse” are different conditions.

A kernel with positive entries need not be a positive operator. The matrix

(1221)\begin{pmatrix}1&2\\2&1\end{pmatrix}

has positive entries but eigenvalues three and minus one. An ordinary positive Boltzmann weight is therefore not an adequate substitute for a transfer-positivity argument. Reflection positivity supplies the underlying kind of structure.

Worked conversion: hold the physical mass fixed

Section titled “Worked conversion: hold the physical mass fixed”

Take the exact two-state model

T(at)=(100e−atm),m>0.T(a_t)=\begin{pmatrix}1&0\\0&e^{-a_t m}\end{pmatrix},\qquad m>0.

At at=0.1/ma_t=0.1/m, the excited transfer eigenvalue is e−0.1e^{-0.1}; at at=0.05/ma_t=0.05/m, it is e−0.05e^{-0.05}. Both give energy mm. The eigenvalue approaches the vacuum eigenvalue as the lattice is refined, although the physical gap remains fixed.

A claim that a transfer-eigenvalue separation must remain a fixed positive number as at→0a_t\to0 would exclude this elementary massive scaling example. The dimensionless energy is atma_t m, not mm itself.

Monte Carlo time can change without changing the theory

Section titled “Monte Carlo time can change without changing the theory”

For a two-state uniform probability measure, consider a reversible sampling matrix

Pϵ=(1−ϵϵϵ1−ϵ),0<ϵ≤12.P_\epsilon= \begin{pmatrix}1-\epsilon&\epsilon\\ \epsilon&1-\epsilon\end{pmatrix}, \qquad0<\epsilon\leq\tfrac12.

Its stationary distribution is always (1/2,1/2)(1/2,1/2), while its nonconstant eigenvalue is 1−2ϵ1-2\epsilon. The relaxation rate depends on ϵ\epsilon. Altering the algorithm changes the rate without changing the equilibrium measure.

This exact toy calculation shows why a sampling gap needs a separate argument before it becomes a statement about physical energies. The auxiliary dynamics in the lattice area-law dossier are used to derive a specified static estimate; they are not simply identified with the physical Hamiltonian.

Exercise: do the units survive a refinement?

Section titled “Exercise: do the units survive a refinement?”

A transfer eigenvalue ratio is e−0.3e^{-0.3} at time spacing at=0.1a_t=0.1 in chosen length units. What energy does it give? If that same ratio is retained while ata_t is halved, has the physical energy stayed fixed?

Solution. The first energy is 0.3/0.1=30.3/0.1=3 in inverse-length units. Keeping the ratio fixed at at=0.05a_t=0.05 gives six. To retain energy three, the ratio must instead become e−0.15e^{-0.15}. “Same dimensionless eigenvalue” is not “same physics.”

Creutz, Physical Review D 15, 1128 (1977) treats gauge fixing and the lattice transfer matrix. Kogut–Susskind, Physical Review D 11, 395 (1975) develops the Hamiltonian formulation. For the explicit transfer-positivity assumptions used in spectral extraction, see Blossier et al., arXiv:0902.1265v2, §2.1.

The calculations here are conditional operator identities and exact toy examples, not a reconstruction proof for the continuum target. Continue to spectral extraction or continuum scaling.