 Solution

For a spinless particle, minimal coupling gives $H=(\mathbf{p}-q\mathbf{A}{)}^2/(2m)+q\varphi$. Choose the vector potential $\mathbf{A}=(0,\alpha x^2/2,0)$ and $\varphi =0$. Its curl is $\mathbf{B}=\alpha x\hat{\mathbf{z}}$ and $[p_y,A_y]=0$, soThe translation separation in a magnetic field gradient follows from $[H,p_y]=0$: a simultaneous energy eigenstate can be written $\Psi =L_y^{-1/2}{e}^{iky}{\Phi }_k(x)$. The periodic boundary condition requires $e^{ik{L}_y}=1$, hence $k=2\pi n/L_y$, $n\in \mathbb{Z}$. The remaining Schrödinger equation isThis magnetic-gradient quartic confinement has a positive quartic coefficient for $g\ne 0$. If $k/g<0$, it is a single well at zero, with $V_k(0)={\hslash }^2{k}^2/(2m)$. Its harmonic curvature givesThus the two lowest states are respectively even and odd, with one central peak or two opposite-sign lobes, and approximately $E_j=V_k(0)+(j+1/2)\hslash {\omega }_s$. They are not a nearly degenerate pair.

If $k/g>0$, the two minima are $x=\pm x_0$, $x_0=\sqrt{2\hslash k/g}$, with zero potential there and a central barrier ${\hslash }^2{k}^2/(2m)$. For well-separated minima the lowest parity states of a symmetric double well areHere ${\varphi }_L,{\varphi }_R$ are the localized ground-state Gaussians. Their even combination is the lower state; the odd combination has a node at the origin. Weak quantum tunnelling gives an exponentially small double-well tunneling splitting.

To obtain their common leading energy, put $x=x_0+\delta x$. Since $\hslash k=g{x}_0^2/2$,The double-well oscillator scale in a linear magnetic field therefore givesThe asymptotic condition is $x_0\gg \sqrt{\hslash /(m{\omega }_d)}$, so the barrier separates narrow local oscillator states. It is this scale comparison that makes the source's “large” ratio meaningful.

The signed group velocity of a magnetic-gradient doublet follows by differentiating the band energy with respect to the conserved wavevector:For $q\alpha >0$ the double-well branch has $k>0$ and moves along $+y$; reversing both signs reverses its velocity. To this order the two parity states have the same velocity. Corrections include local anharmonicity and the exponentially small splitting. The spinless Hamiltonian specified here has no extra Zeeman effect term.