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U(t)=exp⁡(iωrt σz/2)U(t) = \exp(i\omega_r t\,\sigma_z/2)

Why this formula appears here

A single case makes the term concrete and ties it to a result nearly a century old. Let HfH_f = (ℏ\hbarω0\omega_0/2)σz\sigma_z , a spin precessing about a static field at its Larmor frequency, and let U(t) = exp⁡(iωrt σz/2)\exp(i\omega_r t\,\sigma_z/2) describe a frame rotating about the same axis at rate ωr\omega_r . Since U(t) is built entirely from σz\sigma_z , it commutes with HfH_f , so UHfH_fU†U^\dagger = HfH_f exactly and the bare conjugate is unchanged by the relabeling. But U˙\dot U U†U^\dagger = iωr\omega_rσz\sigma_z/2 exactly, giving

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U(t)=exp⁡(iωrt σz/2)U(t) = \exp(i\omega_r t\,\sigma_z/2)

Equation 124 · 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.

A single case makes the term concrete and ties it to a result nearly a century old. Let HfH_f = (ℏ\hbarω0\omega_0/2)σz\sigma_z , a spin precessing about a static field at its Larmor frequency, and let U(t) = exp⁡(iωrt σz/2)\exp(i\omega_r t\,\sigma_z/2) describe a frame rotating about the same axis at rate ωr\omega_r . Since U(t) is built entirely from σz\sigma_z , it commutes with HfH_f , so UHfH_fU†U^\dagger = HfH_f exactly and the bare conjugate is unchanged by the relabeling. But U˙\dot U U†U^\dagger = iωr\omega_rσz\sigma_z/2 exactly, giving

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