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A fixed source breaks the matching symmetry

Physical scope

Exact symmetry check in a finite four-dimensional periodic complex-scalar lattice model with an explicit source; no anomaly phase, continuum limit, or empirical conclusion is asserted.

Example

Let $Λ_{a} = (a Z / N a Z)^{4}$ be the periodic four-dimensional lattice with $N \geq 2$ sites in each direction and spacing $a > 0$ ; each coordinate is taken modulo $N a$ . At each site $x$ take a complex scalar $ϕ_{x}$ of mass dimension one. In natural units $c = ℏ = 1$ , choose a mass $m_{0} > 0$ and a dimensionless coupling $λ > 0$ , and define the finite Euclidean action

S_{j} [ϕ] = a^{4} \sum_{x \in Λ_{a}} (\sum_{ν = 1}^{4} {∣ \frac{ϕ_{x + a \hat{ν}} - ϕ_{x}}{a} ∣}^{2} + m_{0}^{2} ∣ ϕ_{x} ∣^{2} + λ ∣ ϕ_{x} ∣^{4} + j (ϕ_{x} + \overline{ϕ_{x}})),

where the fixed real source $j \neq 0$ has mass dimension three. The finite-dimensional Lebesgue integral converges because the positive quartic term dominates the linear source. The source-free action has a global $U (1)$ transformation $ϕ_{x} \mapsto e^{i α} ϕ_{x}$ , but the fixed-source theory does not. Consequently this endpoint fails the exact-symmetry hypothesis of the anomaly-matching theorem; the example asserts no change, absence, or value of an anomaly.

Example references

Harvey, TASI 2003 Lectures on Anomalies, §2.3

Tong, Gauge Theory, §3.5

Verification

Given: The finite periodic lattice, the action and units in the Example, $m_{0} > 0$ , $λ > 0$ , and a fixed $j \neq 0$ .

[F1] Symmetry-preserving RG anomaly matching: the matching conclusion assumes the same exact global symmetry is unbroken along the RG comparison.

[A1] The source $j$ is a fixed parameter and is not transformed as a spurion.

1.1givenA1algebra

Choose $R > 0$ with $R^{3} \geq 4 ∣ j ∣ / λ$ . For $r \geq R$ , $λ r^{4} - 2 ∣ j ∣ r \geq (λ / 2) r^{4}$ ; on $0 \leq r < R$ , the continuous function $(λ / 2) r^{4} - λ r^{4} + 2 ∣ j ∣ r$ has a finite maximum $K$ , so $λ r^{4} - 2 ∣ j ∣ r \geq (λ / 2) r^{4} - K$ for every $r \geq 0$ . Since the kinetic and mass terms are nonnegative, $S_{j} [ϕ] \geq a^{4} (λ / 2) \sum_{x} ∣ ϕ_{x} ∣^{4} - a^{4} K N^{4}$ . In polar coordinates, each one-site factor $\int_{C} e^{- a^{4} (λ / 2) ∣ z ∣^{4}} d^{2} z = 2 π \int_{0}^{\infty} r e^{- a^{4} (λ / 2) r^{4}} d r$ is finite; the finite product over $N^{4}$ sites proves convergence of the full Boltzmann integral. Under the global rotation, the source summand changes from $j (ϕ_{x} + \overline{ϕ_{x}})$ to $j (e^{i α} ϕ_{x} + e^{- i α} \overline{ϕ_{x}})$ , whose derivative at $α = 0$ is $i j (ϕ_{x} - \overline{ϕ_{x}})$ .

2.1F1A1step 1.1algebra∎

Choose the allowed configuration $ϕ_{x} = i r$ at every site with real $r \neq 0$ . The derivative of the full source term is $- 2 a^{4} j r N^{4} \neq 0$ . Therefore the fixed-source action is not invariant under $U (1)$ , so this model cannot be the endpoint of an exact $U (1)$ -preserving flow as required by [F1]. No anomaly-matching conclusion follows for this explicitly broken model.

Depends on

Symmetry-preserving RG anomaly matching

A fixed source breaks the matching symmetry

Example

Example references

Verification

Depends on

Dependency tree · 2 levels

Sources