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Solved Gibbs qubit relaxation and finite response coefficients

Example

Let $H = C^{2}$ , $H_{s} = Δ Z / 2$ with $Δ > 0$ joules, $Z = ∣ 0 ⟩ ⟨ 0 ∣ - ∣ 1 ⟩ ⟨ 1 ∣$ , $X = ∣ 0 ⟩ ⟨ 1 ∣ + ∣ 1 ⟩ ⟨ 0 ∣$ , and $Y = - i ∣ 0 ⟩ ⟨ 1 ∣ + i ∣ 1 ⟩ ⟨ 0 ∣$ . Take downward rate $a > 0$ , upward rate $b > 0$ , $b / a = e^{- β Δ}$ , and optional dephasing rate $γ_{ϕ} \geq 0$ . Here $β > 0$ is in joules $^{- 1}$ , $ℏ$ is in joule-seconds, and $a, b, γ_{ϕ}$ are in seconds $^{- 1}$ . Use the finite GKSL generator of Specified finite GNS detailed balance with Hamiltonian $H_{s}$ and exactly the jumps $L_{↓} = \sqrt{a} ∣ 1 ⟩ ⟨ 0 ∣$ , $L_{↑} = \sqrt{b} ∣ 0 ⟩ ⟨ 1 ∣$ , and $L_{ϕ} = \sqrt{γ_{ϕ} / 2} Z$ . Thus the additional dephasing dissipator, on either states or observables, is $D_{ϕ} (A) = (γ_{ϕ} / 2) (Z A Z - A)$ ; when $γ_{ϕ} = 0$ this jump is zero. Let $γ_{1} = a + b$ , $γ_{2} = (a + b) / 2 + γ_{ϕ}$ , $ω = Δ / ℏ$ , and $z_{e q} = - \tanh (β Δ / 2)$ . Set $σ = (I + z_{e q} Z) / 2$ . Define the centered correlations

$C_{X X} (t) : = Tr (σ e^{t G} (X) X), C_{Z Z} (t) : = Tr (σ e^{t G} (Z - z_{e q} I) (Z - z_{e q} I)) .$

Then

$C_{X X} (t) = e^{- γ_{2} t} (\cos (ω t) - i \tanh (β Δ / 2) \sin (ω t)), C_{Z Z} (t) = (1 - z_{e q}^{2}) e^{- γ_{1} t} .$

The real correlation-integral coefficients are

$Γ_{X} = \frac{γ_{2}}{γ_{2}^{2} + ω^{2}}, Γ_{Z} = \frac{1 - z_{e q}^{2}}{γ_{1}} .$

Under the fixed-dissipator perturbation $H_{ϵ} (t) = H_{s} - ϵ h (t) X$ , define $χ_{o p e n} (0) : = \int_{0}^{\infty} R_{X X} (t) d t$ . The open response kernel and its zero-frequency integral are

$R_{X X} (t) = \frac{2}{ℏ} \tanh (β Δ / 2) e^{- γ_{2} t} \sin (ω t), χ_{o p e n} (0) = \frac{2 \tanh (β Δ / 2) ω}{ℏ (γ_{2}^{2} + ω^{2})} .$

For $q \in R$ measured in joules, define $H (q) : = H_{s} - q X$ and $σ_{q} : = e^{- β H (q)} / Tr (e^{- β H (q)})$ . The static Gibbs susceptibility for this family is

$χ_{G i b b s} (0) : = {\frac{d}{d q} Tr (σ_{q} X) ∣}_{q = 0} = \frac{2}{ℏ ω} \tanh (β Δ / 2) .$

Thus $χ_{o p e n} (0)$ and $χ_{G i b b s} (0)$ differ for positive $γ_{2}$ ; the model has a fixed dissipator, so its zero-frequency response is not the derivative of the instantaneous Gibbs preparation.

Facts & Assumptions

Given: $Δ > 0$ , $β > 0$ , $a, b > 0$ , $b / a = e^{- β Δ}$ , $γ_{ϕ} \geq 0$ , Pauli matrices $X, Y, Z$ , and the three explicitly specified GKSL jumps.

[F1]

Connected Gibbs jump generators satisfy detailed balance: the Gibbs density is stationary, the energy-jump dissipator is GNS self-adjoint, and its connected rate graph has scalar kernel.

[F2]

Gap-controlled correlation resolvent and positive coefficients: centered correlations have an integrable resolvent and nonnegative real quadratic coefficients.

[F3]

Fixed-dissipator finite response and integrable kernel: for a perturbation of the Hamiltonian with the dissipator held fixed, the causal response kernel is $\frac{i}{ℏ} Tr (σ [e^{t G} A_{0}, B])$ .

[F4]

Duhamel derivative and finite Kubo–Mori response: for finite Hermitian $H (q)$ , differentiation of $Tr e^{- β H (q)}$ gives the exact Kubo–Mori static susceptibility; no commutation of $H$ and $X$ is assumed.

[F5]

Specified finite GNS detailed balance: $G$ is the finite trace-dual observable generator with the stated Hamiltonian, jump matrices, and $σ$ -weighted inner product.

Verification

1.1givenF1F2F3F5algebra

Since $Z$ commutes with $σ$ and $Z^{2} = I$ , the map $A \mapsto Z A Z$ is self-adjoint in the $σ$ -weighted inner product and is an involution. Hence $D_{ϕ}$ is self-adjoint and nonpositive, annihilates diagonal matrices, and sends $X, Y$ to $- γ_{ϕ} X, - γ_{ϕ} Y$ . Adding it to the Gibbs-jump dissipator preserves stationarity and the scalar kernel: the sum of its two nonpositive quadratic forms can vanish only in the scalar kernel of the Gibbs-jump part. Thus [F2] and [F3] apply to this full generator. The two-state population equation is $\dot{z} = - γ_{1} (z - z_{e q})$ , while the off-diagonal equation is ${\dot{ρ}}_{01} = - (γ_{2} + i ω) ρ_{01}$ , by direct multiplication of the two jump matrices and the dephaser. The dual observables satisfy $G (X) = - γ_{2} X - ω Y$ and $G (Y) = ω X - γ_{2} Y$ , so $e^{t G} X = e^{- γ_{2} t} (X \cos ω t - Y \sin ω t)$ ; also $e^{t G} (Z - z_{e q} I) = e^{- γ_{1} t} (Z - z_{e q} I)$ .

1.2F1F3algebra

Applying [F3] with $A = B = X$ , the commutator identity $[Y, X] = - 2 i Z$ and $z_{e q} = - \tanh (β Δ / 2)$ give $R_{X X} (t) = 2 \tanh (β Δ / 2) e^{- γ_{2} t} \sin (ω t) / ℏ$ . Since $\int_{0}^{\infty} e^{- γ_{2} t} \sin (ω t) d t = ω / (γ_{2}^{2} + ω^{2})$ , the displayed $χ_{o p e n} (0)$ follows.

1.3F4algebra

For $q \in R$ , $H (q) = Δ Z / 2 - q X$ satisfies $H (q)^{2} = ((Δ / 2)^{2} + q^{2}) I$ . Put $λ (q) = \sqrt{(Δ / 2)^{2} + q^{2}}$ . The finite spectral calculus gives $e^{- β H (q)} = \cosh (β λ (q)) I - \frac{\sinh (β λ (q))}{λ (q)} H (q)$ , hence the partition function $Z (q) : = Tr (e^{- β H (q)}) = 2 \cosh (β λ (q))$ and $Tr (e^{- β H (q)} X) / Z (q) = q \tanh (β λ (q)) / λ (q)$ . Differentiating at $q = 0$ gives $2 \tanh (β Δ / 2) / Δ = 2 \tanh (β Δ / 2) / (ℏ ω)$ ; [F4] identifies this derivative as the static Gibbs susceptibility.

2.1F1F2step 1.1algebra

Using $X^{2} = Y^{2} = Z^{2} = I$ , $Y X = - i Z$ , $Tr (σ Z) = z_{e q}$ , and $Tr (σ X) = Tr (σ Y) = 0$ in step 1.1 gives the stated $C_{X X}$ and $C_{Z Z}$ . Integrating their real parts yields $Γ_{X} = \int_{0}^{\infty} e^{- γ_{2} t} \cos (ω t) d t = γ_{2} / (γ_{2}^{2} + ω^{2})$ and $Γ_{Z} = (1 - z_{e q}^{2}) \int_{0}^{\infty} e^{- γ_{1} t} d t = (1 - z_{e q}^{2}) / γ_{1}$ , consistent with [F2].

3.1F3F4step 1.2step 1.3algebra∎

The two susceptibilities differ because their denominators are respectively $γ_{2}^{2} + ω^{2}$ and $ω^{2}$ , with $γ_{2} > 0$ . The dynamic formula used [F3]'s fixed-dissipator perturbation, whereas the static formula differentiates a Gibbs state whose Hamiltonian changes with $q$ . Therefore no equilibrium identity equating them is asserted for this model.

Solved Gibbs qubit relaxation and finite response coefficients

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

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