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Equation 113 · Part 1 · The Entry a Relabeling Cannot Write

Symbol i

iℏ d∣ψ′⟩dt=(UHfU†+iℏ U˙U†)∣ψ′⟩,i\hbar\,\frac{d|\psi'\rangle}{dt} = \Big(U H_f U^\dagger + i\hbar\,\dot U U^\dagger\Big)|\psi'\rangle,
ii

What this part means

i occurs above the fraction bar. The numerator is divided by the entire denominator below it.

Its job in the formula

i occurs above the fraction bar. The numerator is divided by the entire denominator below it.

The passage around this formula

…be the corresponding time-dependent unitary and |ψ\psi'(t)⟩\rangle := U(t)|ψ(t)\psi(t)⟩\rangle , with |ψ(t)\psi(t)⟩\rangle solving iℏ\hbar\, d|ψ\psi⟩\rangle/dt = HfH_f|ψ\psi⟩\rangle . Differentiating the product and using U†U^\dagger U = I gives iℏ d∣ψ′⟩dt=(UHfU†+iℏ U˙U†)∣ψ′⟩i\hbar\,\frac{d|\psi'\rangle}{dt} = \Big(U H_f U^\dagger + i\hbar\,\dot U U^\dagger\Big)|\psi'\rangle. 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…

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