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ΔEΛ(t∗)=+ℏω/2\Delta E_\Lambda(t^\ast) = +\hbar\omega/2

Why this formula appears here

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) , the state is |g,e⟩\rangle up to an overall phase — the excitation has moved entirely into the bath — and Δ\Delta EΛ(t∗)E_\Lambda(t^\ast) = +ℏ\hbarω\omega/2 , exactly the bath’s excited-state energy with ⟨\langle HSBH_{SB}⟩\rangle = 0 at that instant, matching the closed form by direct substitution rather than by coincidence. A system-only observer watching only ⟨\langle HSH_S⟩(t)\rangle(t) across this half-period would see the tracked energy swing by a full ℏ\hbarω\omega with no term in HSH_S alone to explain it; the ledger accounts for every joule of…

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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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t∗t^\ast

Symbol t^ast

tat^ast is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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Published contexts (1)

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ΔEΛ(t∗)=+ℏω/2\Delta E_\Lambda(t^\ast) = +\hbar\omega/2

Equation 97 · 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.

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) , the state is |g,e⟩\rangle up to an overall phase — the excitation has moved entirely into the bath — and Δ\Delta EΛ(t∗)E_\Lambda(t^\ast) = +ℏ\hbarω\omega/2 , exactly the bath’s excited-state energy with ⟨\langle HSBH_{SB}⟩\rangle = 0 at that instant, matching the closed form by direct substitution rather than by coincidence. A system-only observer watching only ⟨\langle HSH_S⟩(t)\rangle(t) across this half-period would see the tracked energy swing by a full ℏ\hbarω\omega with no term in HSH_S alone to explain it; the ledger accounts for every joule of…

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