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Exactly computed two-level Kubo response

Example

Let $H = C^{2}$ , $Δ > 0$ in J, and $ℏ > 0$ in J s; put $ω = Δ / ℏ$ . Set $X = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}), Y = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array}), Z = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}), H = \frac{Δ}{2} Z .$ Take $A = B = X$ (dimensionless), $β > 0$ in J $^{- 1}$ , and a finite interval $[0, T]$ with $T > 0$ in s. Use dimensionless $ϵ \in R$ and a constant real control $h (t) = h_{0}$ such that $h_{0} X$ has unit J. Write $σ = e^{- β H} / Tr (e^{- β H})$ and $x = β Δ / 2$ ; the initial Gibbs matrix is fixed while the control acts.

The horizontal lines denote the two eigenvalues of $H$ ; the arrows mark the off-diagonal matrix elements of $X$ and their Bohr frequency $ω = Δ / ℏ$ . The drawing encodes the finite transition structure only; all response claims below follow from the explicit matrix calculation.

Facts & Assumptions

Given: The matrices, $Δ > 0$ , $ℏ > 0$ , $β > 0$ , and real constant $h_{0}$ .

[F1]

Finite-time Kubo formula with norm remainder: the fixed-Gibbs finite-time response uses $R_{X X} (t) = \frac{i}{ℏ} Tr (σ [α_{t} (X), X])$ and has the stated quadratic remainder.

[F2]

Finite spectral balance and distributional fluctuation response: the equilibrium correlation has a positive atomic spectral measure, detailed balance, and the stated zero-frequency qualification.

[F3]

Duhamel derivative and finite Kubo–Mori response: differentiating a finite noncommuting Gibbs family gives the Kubo–Mori covariance; it equals $β$ times the ordinary variance only in the commuting case.

Verification

technique · multiply the Pauli matrices, then apply the finite response and static-derivative suppliers

Given: [F1]–[F3] and the two-level data above.

1.1givenconstructalgebra

Direct multiplication gives $X^{2} = Y^{2} = Z^{2} = I$ , $Y X = - i Z$ , and $[Z, X] = 2 i Y$ . The Gibbs matrix is $σ = (I - \tanh (x) Z) / 2$ , so $Tr (σ Z) = - \tanh (x)$ and $Tr (σ X) = 0$ .

2.1givenF1F2step 1.1algebra

Conjugation by $e^{- i t H / ℏ}$ yields $α_{t} (X) = X \cos (ω t) - Y \sin (ω t)$ . Hence $C_{X X} (t) = Tr (σ α_{t} (X) X) = \cos (ω t) - i \tanh (x) \sin (ω t)$ and $R_{X X} (t) = 2 \tanh (x) \sin (ω t) / ℏ$ . The spectral measure has its two transition atoms at $\pm ω$ and satisfies $S_{X X} (- ω) = e^{- β Δ} S_{X X} (ω)$ , in agreement with [F2].

2.2givenF3step 1.1algebra

For comparison, the static Gibbs family for $H (h) = H - h X$ has $Z (h) = 2 \cosh (β \sqrt{(Δ / 2)^{2} + h^{2}})$ . By [F3], $\partial_{h} \log Z (h) = β Tr (σ_{h} X)$ , so $Tr (σ_{h} X) = \frac{h}{\sqrt{(Δ / 2)^{2} + h^{2}}} \tanh ⁣ (β \sqrt{(Δ / 2)^{2} + h^{2}})$ and ${\partial_{h} Tr (σ_{h} X) ∣}_{h = 0} = \frac{2}{Δ} \tanh (x)$ . Since ${Var}_{σ} (X) = 1$ and $\tanh (x) < x$ for $x > 0$ (the derivative of $x - \tanh x$ is $\tanh^{2} x > 0$ ), this static susceptibility is strictly below $β {Var}_{σ} (X) = β$ . Thus the noncommuting Gibbs derivative is not replaced by $β$ times ordinary variance.

3.1F1step 2.1algebra

Integrating the kernel for constant $h_{0}$ gives $δ ⟨ X (t) ⟩ = \frac{2 ϵ h_{0}}{Δ} \tanh (x) (1 - \cos (ω t)) + r_{ϵ} (t)$ , with $∣ r_{ϵ} (t) ∣ \leq (∣ ϵ ∣ C_{T})^{2} e^{∣ ϵ ∣ C_{T}} / 2$ and $C_{T} = 2 ∣ h_{0} ∣ T / ℏ$ , because $∥ X ∥ = 1$ .

4.1givenstep 3.1step 2.2∎

When $ϵ h_{0} \neq 0$ , the leading response oscillates without a long-time limit; the finite-time theorem does not imply relaxation to the static Gibbs derivative. This is an exact model calculation, not an observation.

Exactly computed two-level Kubo response

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources