Symbol U^dagger

The passage around this formula

so the operator that actually generates |$\psi$'(t)$\rangle$ ’s evolution is $H_c(t)$ := U$H_f$$U^\dagger$ + i$\hbar$$\dot U$ $U^\dagger$ , not the bare conjugate U$H_f$$U^\dagger$ alone. The extra piece, i$\hbar$$\dot U$ $U^\dagger$ , is Hermitian — differentiating U$U^\dagger$ = I gives $\dot U$ $U^\dagger$ = -U$\dot U^\dagger$ , so (i$\hbar$$\dot U$ $U^\dagger$)^$\dagger$ = -i$\hbar$ U$\dot U^\dagger$ = i$\hbar$ $\dot U$ $U^\dagger$ — and it carries units of energy, $\hbar$ times a rate. Call its expectation value the frame’s inertial term, $\Delta_{\rm frame}(t)$ := i$\hbar$$\langle$$\dot U(t)$ U(t)^$\dagger$$\rangle$ .

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