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A finite bath revives coherence

Physical scope

A conditional two-qubit finite-environment unitary model with fixed product preparation. The coherence revival is an exact model prediction, not a reported experiment or a general claim about finite reservoirs.

Example

For $H_{S} = H_{E} = C^{2}$ , $Z = diag (1, - 1)$ , $g > 0$ in $s^{- 1}$ , and $ℏ > 0$ in joule-seconds, choose $H_{t o t} = ℏ g Z \otimes Z$ , $U_{t} = e^{- i t H_{t o t} / ℏ}$ , and the fixed environment state $∣ + ⟩ ⟨ + ∣$ with $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ in the displayed environment $Z$ eigenbasis. Here coherence means an off-diagonal matrix entry in the displayed system basis. The reduced map leaves the system populations unchanged and multiplies each coherence by $\cos (2 g t)$ . At a positive time $s$ where the multiplier is zero, it erases the distinction between the two $X$ eigenstates; at $2 s$ it restores them with their signs exchanged. Thus this finite reduced family is neither a semigroup nor CP-divisible.

Example references

Product-state reduced evolution is a channel

Finite channel families, semigroups, and CP-divisibility

Facts & Assumptions

Given: The two-qubit model, $g > 0$ , $ℏ > 0$ , and the definitions of channel semigroup and CP-divisibility.

[F1]

The two-qubit calculation in Product-state reduced evolution is a channel uses this exact Hamiltonian and environment state; it defines the matrix exponential by its power series, derives $U_{t} = \sum_{b = \pm 1} V_{b} (t) \otimes ∣ b ⟩ ⟨ b ∣$ with $V_{b} (t) = diag (e^{- i b g t}, e^{i b g t})$ , and obtains the reduced channel's coherence multiplier $κ (t) = \frac{1}{2} (e^{- 2 i g t} + e^{2 i g t})$ .

[F2]

Finite channel families, semigroups, and CP-divisibility: the semigroup law is $Φ_{t + s} = Φ_{t} \circ Φ_{s}$ with $Φ_{0} = id$ , and CP-divisibility requires a channel $V_{t, s}$ for every $s \leq t$ with $Φ_{t} = V_{t, s} \circ Φ_{s}$ .

[F3]

$t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle: some $θ \in [0, 2 π)$ satisfies $(\cos θ, \sin θ) = (0, 1)$ .

[F4]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : $e^{i θ} = \cos θ + i \sin θ$ for real $θ$ .

[F5]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential: $e^{z + w} = e^{z} e^{w}$ for complex $z, w$ .

[F6]

The complex exponential by its power series: $e^{0} = 1$ and $e^{z}$ is the scalar exponential power series.

[F7]

Finite matrix spectral and state calculus: every finite Hermitian matrix has an orthonormal eigenbasis with real eigenvalues; for a real-valued function on its finite spectrum, the corresponding spectral operator is basis-independent.

Verification

technique · compute the conditional qubit rotations and compare two exact times

1.1givenF1F4F7algebra

The Hermitian matrix $Z \otimes Z$ has the orthonormal eigenbasis of product $Z$ eigenvectors and eigenvalues $\pm 1$ by direct multiplication, consistent with [F7]. For this exact matrix and environment preparation, [F1] defines the matrix exponential by its power series and derives $U_{t} = \sum_{b = \pm 1} V_{b} (t) \otimes ∣ b ⟩ ⟨ b ∣$ , where $V_{b} (t) = diag (e^{- i b g t}, e^{i b g t})$ . By [F4], each diagonal phase has modulus one, so these blocks and $U_{t}$ are unitary. The environment state has equal weight on $b = \pm 1$ , so the reduced map is $Φ_{t} (X) = \frac{1}{2} (V_{+ 1} X V_{+ 1}^{*} + V_{- 1} X V_{- 1}^{*})$ ; multiplying the two diagonal matrices shows that the off-diagonal entries are multiplied by $κ (t) = \frac{1}{2} (e^{- 2 i g t} + e^{2 i g t}) = \cos (2 g t)$ , while the diagonal entries are unchanged.

1.2F3F4F5F6algebra

By [F3] choose $θ \in [0, 2 π)$ with $\cos θ = 0$ and $\sin θ = 1$ , and put $s = θ / (2 g) > 0$ . Then [F4] gives $e^{i θ} = i$ , while [F5] and [F6] with $e^{i θ} e^{- i θ} = e^{0} = 1$ give $e^{- i θ} = - i$ ; another application of [F5] yields $κ (s) = 0$ and $κ (2 s) = \frac{1}{2} (e^{- 2 i θ} + e^{2 i θ}) = - 1$ .

2.1F1F2step 1.1step 1.2algebra∎

Let $X_{P} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ and $ρ_{\pm} = (I \pm X_{P}) / 2$ . Since $X_{P} = X_{P}^{*}$ , $X_{P}^{2} = I$ , and $Tr X_{P} = 0$ , each $ρ_{\pm}$ is a positive trace-one matrix. Steps 1.1 and 1.2 give $Φ_{s} (ρ_{+}) = Φ_{s} (ρ_{-}) = I / 2$ , but $Φ_{2 s} (ρ_{+}) = ρ_{-}$ and $Φ_{2 s} (ρ_{-}) = ρ_{+}$ . Thus $Φ_{2 s} (ρ_{+}) \neq Φ_{s} (Φ_{s} (ρ_{+}))$ , so the semigroup law in [F2] fails. If a channel $V_{2 s, s}$ satisfied $Φ_{2 s} = V_{2 s, s} \circ Φ_{s}$ , it would map the equal intermediate states to equal outputs, contradicting their distinct final states; by [F1], every $Φ_{t}$ is a channel, so the family is not CP-divisible by [F2].

A finite bath revives coherence

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources