Double-well oscillator scale in a linear magnetic field (${\omega }_d=\sqrt{2\hslash q\alpha k}/m$)

For $q\alpha k>0$, let $x_0=\sqrt{2\hslash k/(q\alpha )}$. Expansion of the magnetic-gradient quartic confinement at either minimum gives $V(x_0+\delta x)=(q\alpha {)}^2{x}_0^2(\delta x{)}^2/(2m)+\cdots$. The local oscillator frequency is therefore $|q\alpha |x_0/m$. Both lowest parity states have leading energy $\hslash {\omega }_d/2$ if $x_0$ greatly exceeds the oscillator width. Local anharmonicity changes their common energy, while quantum tunnelling gives their much smaller splitting.

Differentiating $E_k\simeq \sqrt{{\hslash }^{3}q\alpha k/2}/m$ gives the displayed group velocity when $q\alpha k>0$. Its sign is the sign of $q\alpha$, equivalently of k on that branch. Writing only a positive square root of $q\alpha /k$ would lose this sign. Exactly, the Hellmann–Feynman theorem gives $v_y=⟨\hslash k-q\alpha x^2/2⟩/m$. Evaluating this small drift from a localized wavefunction requires enough anharmonic accuracy: a leading centred Gaussian alone need not reproduce it.