transformations produce physically acceptable results. Therefore, gauge transformations are not unitary transformations. This conclusion is supported by the basic structure of transition amplitudes.

Transition amplitudes without resort to perturbation theory are best expressed by S matrices. These are of two (equivalent) types. The *direct-time* or *post* amplitude is$$\left(  S-1\right)  _{fi}=-\frac{i}{\hslash}\int_{-\infty}^{\infty}dt\left(
\Phi_{f},H_{I}\Psi_{i}\right)  , \label{aq}$$ and the *time-reversed* or *prior* amplitude is$$\left(  S-1\right)  _{fi}=-\frac{i}{\hslash}\int_{-\infty}^{\infty}dt\left(
\Psi_{f},H_{I}\Phi_{i}\right)  . \label{ar}$$ The indices $f$ and $i$ label the final and initial states. The $\Phi$ states are non-interacting states and the $\Psi$ states are fully interacting states satisfying, respectively, the Schrödinger equations$$\begin{aligned}
i\hslash\partial_{t}\Phi &  =H_{0}\Phi,\label{as}\\
i\hslash\partial_{t}\Psi &  =\left(  H_{0}+H_{I}\right)  \Psi, \label{at}$$ where $H_{I}$ is the interaction Hamiltonian.

In a gauge transformation, the matrix elements within the time integrations in Eqs. (\[aq\]) and (\[ar\]) transform as$$\begin{aligned}
\left(  \Phi_{f},H_{I}\Psi_{i}\right)   &  \rightarrow\left(  \Phi
_{f},\widetilde{H}_{I}\widetilde{\Psi}_{i}\right)  ,\label{au}\\
\left(  \Psi_{f},H_{I}\Phi_{i}\right)   &  \rightarrow\left(  \widetilde{\Psi
}_{f},\widetilde{H}_{I}\Phi_{i}\right)  . \label{av}$$ Because the noninteracting states are unaltered in a gauge transformation, there is no necessary ** equivalence between the two sides of the expressions in Eqs. (\[au\]) and (\[av\]).

Those authors that endorse the favored status of the length gauge [@yang; @kobesmirl; @beckerss; @lss; @jbauer] solve this problem by attaching a unitary operator to all states, including non-interacting 