Symbol U^dagger

The passage around this formula

A single case makes the term concrete and ties it to a result nearly a century old. Let $H_f$ = ($\hbar$$\omega_0$/2)$\sigma_z$ , a spin precessing about a static field at its Larmor frequency, and let U(t) = $\exp(i\omega_r t\,\sigma_z/2)$ describe a frame rotating about the same axis at rate $\omega_r$ . Since U(t) is built entirely from $\sigma_z$ , it commutes with $H_f$ , so U$H_f$$U^\dagger$ = $H_f$ exactly and the bare conjugate is unchanged by the relabeling. But $\dot U$ $U^\dagger$ = i$\omega_r$$\sigma_z$/2 exactly, giving

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