Symbol pi

The passage around this formula

At t=0 this equals -$\hbar$$\omega$/2 , exactly the bath’s own ground-state energy, matching the uncorrelated-bath limit derived above. At the swap time $t^\ast$ = $\pi$/(2g) , the state is |g,e$\rangle$ up to an overall phase — the excitation has moved entirely into the bath — and $\Delta$ $E_\Lambda(t^\ast)$ = +$\hbar$$\omega$/2 , exactly the bath’s excited-state energy with $\langle$ $H_{SB}$$\rangle$ = 0 at that instant, matching the closed form by direct substitution rather than by coincidence. A system-only observer watching only $\langle$ $H_S$$\rangle(t)$ across this half-period would see the tracked energy swing by a full $\hbar$$\omega$ with no term in $H_S$ alone to explain it; the ledger accounts for every joule of…

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