anomalous magnetic moment. Using the angle $ \delta $, we go over to the effective mass and momentum: $$\label{sol_mpeff_def}
    \begin{pmatrix}
        \tilde{m} \\
        \tilde{p} \\
    \end{pmatrix}
    =
    \begin{pmatrix}
        \cos \delta  & \sin \delta \\
        -\sin \delta & \cos \delta \\
    \end{pmatrix}
    \begin{pmatrix}
        m \\
        p \\
    \end{pmatrix}.$$ It should be noted that the effective mass $ \tilde{m} $ may take negative values, when $ p < - m
\cot \delta $. It is easy to see that with the help of the unitary transformation $$U^{-1} H_{\rm D} U = \tilde{H}_{\rm D}, \qquad$$ where $$\label{sol_U_def}
    U
    = \exp({{\textstyle}-\frac{\delta}{2} \gamma^{3}})
    = \cos{{\textstyle}\frac{\delta}{2}} - \gamma^{3} \sin{{\textstyle}\frac{\delta}{2}},$$ the hamiltonian (\[model\_H\]) can be brought to the following form: $$\label{sol_tilde_H}
    \tilde{H}_{\rm D} =
        {\boldsymbol}\alpha \tilde{\bf P} + \gamma^{0} \tilde{m}
        + \tilde{\mu} H \gamma^{0} \Sigma_{3},$$ where $ \tilde{\bf P} = \{ \hat{P}_{1}, \hat{P}_{2}, \tilde{p} \} $ ($ \hat{P}_{1} $, $ \hat{P}_{2} $ are the same as in the hamiltonian (\[model\_H\])). Thus the problem (\[sol\_Hphi\_Ephi\]) is formally equivalent to the problem $$\label{sol_tilde_Hphi_Ephi}
    \tilde{H}_{\rm D} \tilde{\phi} = E \tilde{\phi},$$ since the operators $ \tilde{H}_{\rm D} $ and $ H_{\rm D} $ have identical eigenvalues, and their eigenvectors are related by the transformation (\[sol\_U\_def\]): $ \phi = U \tilde{\phi} $. The 