Let us take the electromagnetic potential of the external magnetic field in the axial-symmetric form: $$A^{\mu} = \{ 0, {\bf A} \}, \qquad
    {\bf A} = {{\textstyle}\frac{1}{2}} \{ -H y, H x, 0 \}.$$ It is obvious that $ \left[ p_{z}, H_{\rm D} \right] = 0 $, therefore we shall further consider the problem (\[sol\_HPsi\_EPsi\]) on the subspace with a definite fixed $ p_{z} \equiv p \, $: $$\label{sol_Psi_phi}
    \Psi(x,y,z) =
        {{\textstyle}\frac{1}{\sqrt{2\pi}}} \, e^{i p z} \phi(x,y),$$ so that (\[sol\_HPsi\_EPsi\]) turns into $$\label{sol_Hphi_Ephi}
    H_{\rm D} \phi = E \phi,$$ where in the expression (\[model\_H\]) for the $ H_{\rm D} $ one must take $ P_{3} = p $ so that $ {\bf P} = \{ \hat{P}_{1}, \hat{P}_{2}, p \} $ (we use the “hat” symbol to denote operators in order to distinguish them from $ c $-values where a confusion is possible). Let us introduce now the “mixing angle” $ \delta $ as follows: $$\label{sol_vartheta_def}
    \delta = \arctan \frac{b}{\mu H},$$ so that $$\begin{aligned}
        \mu H &= \tilde{\mu} H \cos \delta, \\
        b     &= \tilde{\mu} H \sin \delta, \\
    \end{aligned}$$ where $$\label{sol_mueff_def}
    \tilde{\mu} H =
        \sqrt{ (\mu H)^{2} + b^{2} }.$$ We shall name the quantity $ \tilde{\mu} $ an effective 