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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.

A tall, broad central potential barrier has $\varphi \gg 1$. Let $\sigma =+1$ denote even parity and $\sigma =-1$ odd parity, and write $\theta =(n+\frac{1}{2})\pi +{\delta }_{\sigma }$, where $n=0,1,\dots$. Since $\tan [(n+\frac{1}{2})\pi +\delta ]=-\cot \delta$, the condition in part (d) givesThus the leading correction to the isolated-well Bohr-Sommerfeld quantization condition isThis is an implicit condition, with all action integrals evaluated at the corrected energy. To convert it into an explicit leading energy correction, let $E_n^{(0)}$ solve $\theta (E_n^{(0)})=(n+\frac{1}{2})\pi$ and define the full classical oscillation period in one well by $\mathcal{T}(E)=2m{\int }_{x_1}^{x_2}dx/p(x)$. Differentiation of the action integral has no endpoint terms because $p(x_1)=p(x_2)=0$, and givesConsequently, to leading exponentially small order within the WKB approximation,The even member lies below its odd partner. For an approximately harmonic oscillator well, $\mathcal{T}\simeq 2\pi /\omega$, giving the familiar double-well tunneling splitting $\Delta E\simeq (\hslash \omega /\pi )e^{-\varphi }$. The explicit energy formula retains only its leading exponential correction; varying $\varphi$ and ${\theta }^{\prime }$ with the shifted energy also contributes at higher orders.

The ground state has an even, everywhere same-sign wavefunction with a peak in each well and a small nonzero value beneath the central potential barrier. Its derivative vanishes at zero. The first excited state is odd, has opposite-sign lobes in the two wells, and has exactly one interior node of a wavefunction, at zero. Both wavefunctions decay outside the wells. In terms of localized normalized wavefunctions ${\phi }_L,{\phi }_R$, with overlap $s=⟨{\phi }_L|{\phi }_R⟩$, their qualitative shapes are the lowest parity states of a symmetric double well:The overall sign of the odd wavefunction is arbitrary. The sketch uses localized Gaussian profiles to show the parity and the central node of a wavefunction, rather than claiming exact energy eigenfunctions for an unspecified potential energy.