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Exact centered qubit-record displacement variance

Example

Assume full AC and initialize the selected energy-resolved record with the Gibbs law π. Use the two energy states ∣0⟩,∣1⟩ of Connected Gibbs jump generators satisfy detailed balance with energies E0=Δ/2 and E1=−Δ/2, where Δ>0 is in joules and β>0 is in joules−1. Let a>0 be the rate from 0→1 and b>0 the rate from 1→0, both in seconds−1, with b/a=e−βΔ. The stationary mean is zeq:=(b−a)/(a+b). Write Z=diag⁡(1,−1) and let u be a speed scale in metres per second; assign centered pointer velocities va=u(Zaa−zeq). The monitored record displacement YT=∫0Tva(s)ds has

Cv(s)=u2(1−zeq2)e−(a+b)s,

and exact variance

Var⁡(YT)=2u2(1−zeq2)[Ta+b−1−e−(a+b)T(a+b)2].

Consequently

lim⁡T→∞Var⁡(YT)2T=u2(1−zeq2)a+b.

This is a variance of the selected classical record with units of squared metres. The record is not asserted to be a continuous quantum position or an experimental velocity.

Facts & Assumptions

Given: Full AC, a two-state connected Gibbs jump model, its ideal energy-resolved record instrument, and u∈R in metres per second.

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function, retained by the finite joint-path supplier.

[F1]

Stationary monitored-chain transport variance coefficient: for the stationary selected record and any centered real velocity labels, the covariance integral and variance formulas hold.

[F2]

Connected Gibbs jump generators satisfy detailed balance: for energies E0=Δ/2,E1=−Δ/2, the Gibbs weights obey detailed balance with the stated rates, and diagonal populations follow the two-state generator.

Verification

1.1givenF2algebra

The Gibbs weights are π0=b/(a+b) and π1=a/(a+b): their ratio is b/a=e−βΔ=e−β(E0−E1), and they sum to one. Thus π0a=π1b and π(v)=0. The two-state generator has centered eigenvector Z−zeqI with eigenvalue −(a+b); therefore esWv=e−(a+b)sv and ⟨v,v⟩π=u2(1−zeq2).

2.1A1F1step 1.1algebra∎

Substituting this covariance into [F1] gives

Var⁡(YT)=2u2(1−zeq2)∫0T(T−s)e−(a+b)s ds.

Evaluating the integral gives the displayed finite-time formula. Dividing by 2T and letting T→∞ gives u2(1−zeq2)/(a+b).

Depends on

Dependency tree · 2 levels

Sources