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$\langle H_f \rangle$

Why this formula appears here

the standard two-level exchange solution, exact for all t under this Hamiltonian. From it, $\langle$ $H_S$ $\rangle(t)$ = ( $\hbar$ $\omega$ /2)[ $\cos^2(gt)$ - $\sin^2(gt)$ ] = ( $\hbar$ $\omega$ /2) $\cos(2gt)$ , while $\langle$ $H_f$ $\rangle$ 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| $H_f$ |e,g $\rangle$ = $\hbar$ $\omega$ /2 - $\hbar$ $\omega$ /2 + 0 = 0 , since $H_{SB}$ is purely off-diagonal in this basis. The ledger reading follows immediately:

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H_f

Symbol H_f

$H_f$ is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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Research cited beside this formula

[12] Circuit Quantum Electrodynamics ↗

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\langle H_f \rangle

Equation 89 · Evolutionary Physics

The Entry a Relabeling Cannot Write

This mathematical expression combines the displayed quantities; its precise role follows from the surrounding article text.

the standard two-level exchange solution, exact for all t under this Hamiltonian. From it, $\langle$ $H_S$ $\rangle(t)$ = ( $\hbar$ $\omega$ /2)[ $\cos^2(gt)$ - $\sin^2(gt)$ ] = ( $\hbar$ $\omega$ /2) $\cos(2gt)$ , while $\langle$ $H_f$ $\rangle$ 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| $H_f$ |e,g $\rangle$ = $\hbar$ $\omega$ /2 - $\hbar$ $\omega$ /2 + 0 = 0 , since $H_{SB}$ is purely off-diagonal in this basis. The ledger reading follows immediately:

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⟨Hf⟩\langle H_f \rangle⟨Hf​⟩