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ΔEΛ(t)=⟨Hf⟩−⟨HS⟩(t)=−ℏω2cos⁡(2gt)\Delta E_\Lambda(t) = \langle H_f\rangle - \langle H_S\rangle(t) = -\frac{\hbar\omega}{2}\cos(2gt)

Why this formula appears here

the standard two-level exchange solution, exact for all t under this Hamiltonian. From it, ⟨\langle HSH_S ⟩(t)\rangle(t) = (ℏ\hbarω\omega/2)[cos⁡2(gt)\cos^2(gt) - sin⁡2(gt)\sin^2(gt)] = (ℏ\hbarω\omega/2)cos⁡(2gt)\cos(2gt) , while ⟨\langle HfH_f ⟩\rangle is exactly time-independent — a general fact for any state evolving under its own generator, not special to this one — and equal here to ⟨\langle e,g|HfH_f|e,g⟩\rangle = ℏ\hbarω\omega/2 - ℏ\hbarω\omega/2 + 0 = 0 , since HSBH_{SB} is purely off-diagonal in this basis. The ledger reading follows immediately: ΔEΛ(t)=⟨Hf⟩−⟨HS⟩(t)=−ℏω2cos⁡(2gt)\Delta E_\Lambda(t) = \langle H_f\rangle - \langle H_S\rangle(t) = -\frac{\hbar\omega}{2}\cos(2gt). At t=0 this equals -ℏ\hbarω\omega/2 , exactly the bath’s own ground-state energy, matching the uncorrelated-bath limit derived above. At the swap time t∗t^\ast = π\pi/(2g) ,…

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ΔEΛ\Delta E_\Lambda

Symbol Δ E_Lambda

Δ ELE_Lambda is part of the quantity the equation computes from the expression on the right.

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Published contexts (1)

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ΔEΛ(t)=⟨Hf⟩−⟨HS⟩(t)=−ℏω2cos⁡(2gt).\Delta E_\Lambda(t) = \langle H_f\rangle - \langle H_S\rangle(t) = -\frac{\hbar\omega}{2}\cos(2gt).

Equation 92 · Evolutionary Physics

The Entry a Relabeling Cannot Write

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

the standard two-level exchange solution, exact for all t under this Hamiltonian. From it, ⟨\langle HSH_S ⟩(t)\rangle(t) = (ℏ\hbarω\omega/2)[cos⁡2(gt)\cos^2(gt) - sin⁡2(gt)\sin^2(gt)] = (ℏ\hbarω\omega/2)cos⁡(2gt)\cos(2gt) , while ⟨\langle HfH_f ⟩\rangle is exactly time-independent — a general fact for any state evolving under its own generator, not special to this one — and equal here to ⟨\langle e,g|HfH_f|e,g⟩\rangle = ℏ\hbarω\omega/2 - ℏ\hbarω\omega/2 + 0 = 0 , since HSBH_{SB} is purely off-diagonal in this basis. The ledger reading follows immediately: ΔEΛ(t)=⟨Hf⟩−⟨HS⟩(t)=−ℏω2cos⁡(2gt)\Delta E_\Lambda(t) = \langle H_f\rangle - \langle H_S\rangle(t) = -\frac{\hbar\omega}{2}\cos(2gt). At t=0 this equals -ℏ\hbarω\omega/2 , exactly the bath’s own ground-state energy, matching the uncorrelated-bath limit derived above. At the swap time t∗t^\ast = π\pi/(2g) ,…

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