Magnetic-gradient quartic confinement ($V_k(x)=[\hslash k-q\alpha x^2/2{]}^2/(2m)$)

A spinless charged particle in $B_z=\alpha x$ admits vector potential $A_y=\alpha x^2/2$. Separating a plane wave along y produces a one-dimensional quartic potential energy. For nonzero $q\alpha$, $k/(q\alpha )>0$ gives two minima at $\pm \sqrt{2\hslash k/(q\alpha )}$; the opposite sign gives one minimum at zero. The field gradient creates confinement even though there is no scalar potential. The quartic potential is even, so its nondegenerate eigenstates have definite spatial parity.

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.

In a gauge with potentials independent of y, the canonical momentum $p_y$ commutes with the Hamiltonian. A simultaneous eigenbasis can be chosen as $e^{iky}{\Phi }_k(x)$, with $p_y=\hslash k$. Periodic length $L_y$ gives $k=2\pi n/L_y$. The kinetic y-momentum is instead $\hslash k-q{A}_y(x)$ and is not spatially constant. Confusing these two momenta loses the magnetic contribution to the velocity.