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UU†=IUU^\dagger = I

Why this formula appears here

so the operator that actually generates |ψ\psi'(t)⟩\rangle ’s evolution is Hc(t)H_c(t) := UHfH_fU†U^\dagger + iℏ\hbarU˙\dot U U†U^\dagger , not the bare conjugate UHfH_fU†U^\dagger alone. The extra piece, iℏ\hbarU˙\dot U U†U^\dagger , is Hermitian — differentiating UU†U^\dagger = I gives U˙\dot U U†U^\dagger = -UU˙†\dot U^\dagger , so (iℏ\hbarU˙\dot U U†U^\dagger)^†\dagger = -iℏ\hbar UU˙†\dot U^\dagger = iℏ\hbar U˙\dot U U†U^\dagger — and it carries units of energy, ℏ\hbar times a rate. Call its expectation value the frame’s inertial term, Δframe(t)\Delta_{\rm frame}(t) := iℏ\hbar⟨\langleU˙(t)\dot U(t) U(t)^†\dagger⟩\rangle .

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U†U^\dagger

Symbol U^dagger

UdU^dagger is part of the quantity the equation computes from the expression on the right.

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UU†=IUU^\dagger = I

Equation 118 · 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.

so the operator that actually generates |ψ\psi'(t)⟩\rangle ’s evolution is Hc(t)H_c(t) := UHfH_fU†U^\dagger + iℏ\hbarU˙\dot U U†U^\dagger , not the bare conjugate UHfH_fU†U^\dagger alone. The extra piece, iℏ\hbarU˙\dot U U†U^\dagger , is Hermitian — differentiating UU†U^\dagger = I gives U˙\dot U U†U^\dagger = -UU˙†\dot U^\dagger , so (iℏ\hbarU˙\dot U U†U^\dagger)^†\dagger = -iℏ\hbar UU˙†\dot U^\dagger = iℏ\hbar U˙\dot U U†U^\dagger — and it carries units of energy, ℏ\hbar times a rate. Call its expectation value the frame’s inertial term, Δframe(t)\Delta_{\rm frame}(t) := iℏ\hbar⟨\langleU˙(t)\dot U(t) U(t)^†\dagger⟩\rangle .

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