 Solution

A constant electrostatic energy offset is removed locally by the quantum phase substitution $\psi =e^{-i{V}_{0}t/\hslash }{\psi }_0$, since $(i\hslash {\partial }_t-V_0)\psi =e^{-i{V}_{0}t/\hslash }i\hslash {\partial }_t{\psi }_0$. Therefore the two frequency branches becomeThe normalization uses the kinetic frequency ${\omega }_k$, while the canonical energy eigenvalues are $E=V_0\pm \hslash {\omega }_k$. 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.