= Solution
A constant electrostatic <energy> offset is removed locally by the <quantum phase> substitution $\psi=e^{-iV_0t/\hbar}\psi_0$, since $(i\hbar\partial_t-V_0)\psi=e^{-iV_0t/\hbar}i\hbar\partial_t\psi_0$. Therefore the two <frequency> branches become
$$
\boxed{\psi=\sum_k\frac1{\sqrt{2\omega_k}}\left[
a_ke^{ikx-i(\omega_k+V_0/\hbar)t}
+b_k^\dagger e^{-ikx+i(\omega_k-V_0/\hbar)t}\right]}.
$$
The normalization uses the kinetic <frequency> $\omega_k$, while the canonical <energy> <eigenvalues> are $E=V_0\pm\hbar\omega_k$. The opposite-frequency branch is not obtained by replacing $\omega_k$ by the same shifted positive <frequency> in both terms. A globally constant offset produces no <electric field>, whereas different offsets on opposite sides of a step cannot be removed by a single smooth <global phase>.
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