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Harmonic BAOAB thermostat: exact position variance and kinetic bias

Example

Take a one-dimensional harmonic oscillator with mass m=2, stiffness k=8, inverse temperature β=1, friction γ=1, and BAOAB step h=1/2. The stability threshold is h<2m/k=1. The example concerns the exact stationary covariance of the discrete BAOAB chain, not Monte Carlo error from a finite trajectory.

Facts & Assumptions

Given: m=2, k=8, β=1, γ=1, h=1/2, and the harmonic BAOAB theorem.

[F1]

Exact invariant covariance and stability threshold for scalar harmonic BAOAB: when 0<h<2m/k, Var⁡(Q)=1/(βk) and Var⁡(P)=(m/β)(1−h2k/(4m)).

[A1]

The Axiom of Choice: AC is inherited through [F1]; the parameter substitution and arithmetic use no additional choice.

Verification

Given: m=2, k=8, β=1, γ=1, h=1/2, and the harmonic BAOAB theorem.

[F1] Exact invariant covariance and stability threshold for scalar harmonic BAOAB: when 0<h<2m/k, Var⁡(Q)=1/(βk) and Var⁡(P)=(m/β)(1−h2k/(4m)).

[A1] The Axiom of Choice: AC is inherited through [F1]; the parameter substitution and arithmetic use no additional choice.

Given: The parameter values in the Example.

1.1F1givenalgebra

Here 2m/k=22/8=1, so h=1/2 is in the strict stable range of [F1]. The dimensionless quantity is r=h2k/(4m)=(1/4)8/8=1/4.

2.1A1F1step 1.1algebra∎

The discrete stationary position variance is 1/(βk)=1/8, exactly the canonical Gibbs position variance. The momentum variance is (2/1)(1−1/4)=3/2, whereas the continuous canonical value is m/β=2, so the momentum bias is −1/2=−h2k/(4β).

Depends on

Used by

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