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A two-component truncated RG flow with linear mixing

Example

Take a supplied fixed-weight quotient $V = Q_{6}$ of complex dimension two, with ordered basis representatives $O_{1}, O_{2}$ of engineering weight six, a cutoff $Λ > 0$ , and a scheme whose renormalization-scale interval is $J = (0, \infty)$ . In natural units use the dimensionless coefficient vector $C (t) = (C_{1} (t), C_{2} (t))^{T}$ and the formal quotient expression $L_{t r u n c} (t) = \frac{C_{1} (t) O_{1} + C_{2} (t) O_{2}}{Λ^{2}}, t = \log (μ / μ_{0}) .$ Supply the constant complex-linear coefficient-flow matrix $A = (\begin{array}{cc} - 1 & 1 \\ 0 & - 2 \end{array})$ and initial vector $C (0) = (1, 1)^{T}$ . In real coordinates the supplied vector field is $β (t, ρ_{2} (C)) = ρ_{2} (A C)$ on $R \times R^{4}$ . The resulting truncated system is $C^{'} = A C$ , and its solution on all $t \in R$ is $C (t) = (\begin{array}{c} 2 e^{- t} - e^{- 2 t} \\ e^{- 2 t} \end{array}) .$ The matrix is supplied as part of this mathematical example; it is not calculated from a QFT and gives no estimate for omitted terms.

Example references

Finite renormalization schemes and Wilson coefficient vectors

Local well-posedness of a supplied truncated coupled RG system

Facts & Assumptions

Given: The supplied two-dimensional complex coefficient quotient and basis, the RG interval $J = (0, \infty)$ , matrix $A$ , and initial value $C (0) = (1, 1)^{T}$ in the Example.

[F1]

Finite renormalization schemes and Wilson coefficient vectors: the coefficient space is $C^{N}$ , its realification $ρ_{N}$ is an isometry to $R^{2 N}$ , and $t = \log (μ / μ_{0})$ is dimensionless.

[F2]

Local well-posedness of a supplied truncated coupled RG system: a supplied continuous vector field locally Lipschitz in the coefficient state has unique local solutions, unique agreement on common intervals, and the stated local continuous-dependence estimate.

Verification

technique · direct differentiation, followed by the local well-posedness theorem

Given: [F1], [F2], and the supplied data in the Example.

1.1givenalgebra

At $t = 0$ , the displayed vector has components $2 - 1 = 1$ and $e^{0} = 1$ , so it satisfies the stated initial condition.

2.1givenstep 1.1algebra

Differentiating gives $C_{1}^{'} (t) = - 2 e^{- t} + 2 e^{- 2 t}$ and $C_{2}^{'} (t) = - 2 e^{- 2 t}$ . The first component of $A C$ is $- C_{1} + C_{2} = - 2 e^{- t} + 2 e^{- 2 t}$ , and the second is $- 2 C_{2} = - 2 e^{- 2 t}$ . Thus the displayed curve solves $C^{'} = A C$ for every real $t$ .

3.1F1F2step 1.1step 2.1algebra∎

For $Δ = C - D \in C^{2}$ , the triangle inequality and $(a - b)^{2} \geq 0$ for $a, b \geq 0$ give $∣ - Δ_{1} + Δ_{2} ∣^{2} \leq (∣ Δ_{1} ∣ + ∣ Δ_{2} ∣)^{2} \leq 2 (∣ Δ_{1} ∣^{2} + ∣ Δ_{2} ∣^{2})$ . Also $∣ - 2 Δ_{2} ∣^{2} = 4 ∣ Δ_{2} ∣^{2}$ . Adding these bounds gives $∥ A Δ ∥_{2}^{2} \leq 2 ∣ Δ_{1} ∣^{2} + 6 ∣ Δ_{2} ∣^{2} \leq 6 ∥ Δ ∥_{2}^{2}$ . By the isometry $ρ_{2}$ , the real vector field is globally Lipschitz with constant $\sqrt{6}$ and is continuous. Theorem [F2] gives uniqueness on every common interval. The displayed solution exists on all $R$ , so it is the unique global solution of this supplied system.

A two-component truncated RG flow with linear mixing

Example

Example references

Facts & Assumptions

Verification

Depends on

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