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One-plaquette SU(2) Wilson average

Example

Assume ACω. For one square face with free boundary in the finite SU(2) Wilson model at 0<β≤1, take the loop to follow the positively oriented face boundary counterclockwise from its lower-left corner. Let U∂□ be the ordered product of the face's link matrices in traversal order, using inverse matrices on reverse edges, and set W(∂□):=12Tr⁡2(U∂□). Use μ2, the normalized pushforward of round S3 surface area under the matrix-coordinate map in the finite SU(2) plaquette-measure definition, and τ(U)=12Tr⁡2U. The normalized plaquette holonomy has law dνβ(U)=zβ−1eβτ(U)dμ2(U), where zβ=∫eβτdμ2. The Wilson action partition factor is Z□=e−βzβ, and this normalized fundamental loop has expectation ⟨W(∂□)⟩β=λ(β), where λ(β)=∫τeβτdμ2/zβ=β/4+O(β3) as β↓0.

Facts & Assumptions

Given: ACω, the one-face free-boundary SU(2) lattice, 0<β≤1, and W(∂□)=12Tr⁡2(U∂□).

[A1]

The proof is carried out over ACω. The Axiom of Countable Choice (ACω)

[F1]

Finite SU(2) Wilson plaquette measure: Sβ=β(1−τ(Up)), zβ=∫eβτdμ2, and dνβ=zβ−1eβτdμ2.

[F2]

Wilson loop on a finite lattice: the normalized fundamental loop is half the trace of the ordered edge product.

[F3]

Independent plaquette variables on a finite planar rectangle: on a one-face free-boundary rectangle, the plaquette holonomy has law νβ.

[F4]

SU(2) one-plaquette matrix average at strong coupling: μ2 is a normalized bi-invariant Borel probability, ∫τ dνβ=λ(β), and λ(β)=β/4+O(β3).

[F5]

Wilson loop is invariant under vertex gauge transformations: the loop trace in the original link variables equals its value in the tree-gauge representative, while Independent plaquette variables on a finite planar rectangle supplies the law of the representative's plaquette variable.

[F6]

Fubini's theorem for L1 functions on a sigma-finite product: an L1 function on a product of sigma-finite measure spaces has equal product and iterated integrals. In particular this applies successively to bounded measurable functions on a finite product of probability spaces, since their absolute integrals are finite.

Verification

1.1givenA1F3F4F6algebra

Write A,B,C,D∈SU(2) for the eastward bottom, northward right, eastward top, and northward left link matrices. The counterclockwise face word is Up=ABC−1D−1 by [F3]. Multiplication is product-Borel measurable in this finite Lie-group case, as specified in [F3], and inversion is continuous, so this group word is measurable. For any bounded Borel function f:SU(2)→C, its composition with the group word is bounded on the four-link product probability space. By [F6] integrate first in A: for fixed B,C,D, right invariance in [F4] gives ∫f(ABC−1D−1) dμ2(A)=∫f(U) dμ2(U). Integrating the remaining three probability factors therefore proves ∫SU(2)4f(Up) dμ2⊗4=∫SU(2)f(U) dμ2(U). The only inherited choice premise is [A1].

2.1F1step 1.1algebra

On one face the action weight is e−βeβτ(Up) by [F1]. Since ∣τ∣≤1, the function eβτ is bounded. Apply step 1.1 to it to obtain Z□=e−β∫eβτ(U) dμ2(U)=e−βzβ. Applying the same identity to f(U)=1H(U)eβτ(U) for each Borel set H⊆SU(2) and dividing by Z□>0 shows that the original plaquette holonomy has law νβ.

3.1F2F3F4F5step 2.1algebra

In the chosen orientation the loop word is exactly the positively oriented face word, so its ordered holonomy is Up and its Wilson observable is W(∂□)=τ(Up) by [F2]. The equality with the original loop observable follows pointwise from [F5]; integrating against the representative plaquette law in [F3] gives ∫τ dνβ=λ(β) by [F4], also directly from the original holonomy law in step 2.1.

4.1F4step 3.1algebra∎

Substituting the expansion from [F4] yields ⟨W(∂□)⟩β=β/4+O(β3) as β↓0. This is a finite-regulator model expectation.

Depends on

Dependency tree · 2 levels

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