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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.

An instanton describes a finite-action trajectory in imaginary time that crosses between classical vacua. It accounts for quantum tunnelling effects exponentially small in $\hslash$, beyond an ordinary power series around one minimum. A symmetric double well makes the connection between the Euclidean path integral, fluctuation theory and the low-energy spectrum particularly explicit.

For a particle of mass $\mu$, use a Wick rotation $t=-i\tau$ in the kernel. The Euclidean action and its classical equation areThe latter is Newtonian motion in the inverted potential $-V$. Subtract the common minimum energy so $V(\pm a)=0$. A finite-action path approaching $\pm a$ at infinite Euclidean times has conserved Euclidean energy $\mu {\dot{q}}^2/2-V=0$. ThereforeThis gives both the classical trajectory by a quadrature and its tunnelling exponent. The plus solution is the instanton; reversal is the anti-instanton. A single crossing has action $S_I$, not $2S_I$, although a periodic closed path necessarily has equal numbers of crossings in both directions.

An explicit double well. Take $V(q)=\mu {\omega }^2(q^2-a^2{)}^2/(8a^2)$, whose local harmonic frequency at either minimum is $\omega$. Integrating the zero-energy equation gives the quartic double-well instantonThe arbitrary center ${\tau }_0$ is a collective coordinate. The tunnelling regime is $S_I/\hslash \gg 1$, so the crossing width ${\omega }^{-1}$ is short compared with the typical separation between rare crossings.

The exponential alone does not give a dimensionally complete amplitude. Expand $q=q_I+\eta$; the quadratic Euclidean action is $\mu \int \eta L_I\eta d\tau /2$, whereThe derivative ${\dot{q}}_I$ is a normalizable translation instanton translation zero mode. The factorization of this one-dimensional fluctuation operator shows it has no negative modes. In $z=\omega (\tau -{\tau }_0)/2$, its two discrete eigenfunctions are proportional to ${\mathrm{sech}}^{2}z$ and $\mathrm{sech}z\tanh z$, with eigenvalues zero and $3{\omega }^2/4$; the continuum begins at ${\omega }^2$. Treating its zero eigenvalue as an ordinary Gaussian integral would give a divergent determinant. Instead integrate over ${\tau }_0$: since $\mu \int {\dot{q}}_I^{2}d\tau =S_I$, the Jacobian supplies $\sqrt{S_I/(2\pi \hslash )}d{\tau }_0$. The nonzero fluctuations supply the determinant ratio. With common endpoint regularization,This is the instanton fluctuation prefactor, with $K$ and $\kappa$ having frequency units. The prime omits precisely the translation mode.

For this quartic potential the determinant can be evaluated explicitly. Set $z=\omega (\tau -{\tau }_0)/2$ and add a small positive spectral regulator $\epsilon$ to both operators. With $r=2\sqrt{1+\epsilon /{\omega }^2}$, the dimensionless operator is $-{\partial }_z^2+r^2-6{\mathrm{sech}}^{2}z$. Applying successively $-{\partial }_z+\tanh z$ and $-{\partial }_z+2\tanh z$ to the free decaying solution $e^{-rz}$, and normalizing at $z\to +\infty$, gives a growing coefficient at $z\to -\infty$ equal to $(r-1)(r-2)/[(r+1)(r+2)]$. The common Dirichlet functional determinant ratio has this coefficient. Hence the quartic double-well fluctuation determinant isThe frequency factor cannot be discarded when deleting the zero eigenvalue.

Summing rare crossings. Consider Euclidean duration $T$ much longer than a crossing width. In the dilute instanton gas, $n$ ordered centers have integration volume $T^n/n!$, and the amplitude per crossing is $\kappa$. Starting in the left well, directions alternate: an even number returns to the left well, and an odd number ends in the right well. If $E_{\mathrm{well}}=\hslash \omega /2+$ local anharmonic corrections, the leading kernels, with a common endpoint normalization $\mathcal{A}$, areThe even/odd sums must not be replaced by counting all possible crossing directions independently. Comparing their exponential components with the spectral representation givesFor the quartic example, $\Delta E=2\hslash \omega \sqrt{6S_I/(\pi \hslash )}{e}^{-S_I/\hslash }$ to leading semiclassical order. This is the double-well tunneling splitting: the even state is lower because the symmetric wavefunction has no node. The equivalent low-energy Hamiltonian in localized left/right states is $E_{\mathrm{well}}I-\hslash \kappa {\sigma }_x$. An initially localized state tunnels coherently, with probability ${\sin }^2(\Delta Et/(2\hslash ))$ of occupying the other well. Tunnelling does not imply irreversible decay in this closed symmetric system.

The instanton approach thus yields the same leading forbidden-region action as the WKB approximation, while identifying the zero-mode measure, fluctuation normalization and multiple tunnelling events systematically. Perturbation theory about either separate minimum cannot generate $e^{-S_I/\hslash }$. Multi-instanton interactions and higher-loop fluctuations correct the leading dilute-gas result; at finite temperature one instead uses periodic Euclidean boundary conditions, whose saddles may differ when the period is comparable to the crossing width. A false-vacuum bounce returning to the same metastable minimum has a negative fluctuation mode and describes a decay rate. It should not be confused with the stable-vacuum connecting instanton above, which has a translation zero mode but no negative mode.

For a trajectory connecting the degenerate minima in infinite Euclidean time, the first integral gives $m{\dot{q}}^2/2=V(q)$. The quartic double-well instanton and its reverse areTheir action isThe instanton fluctuation prefactor $K$ has dimensions of inverse time. It incorporates nonzero Gaussian fluctuation eigenvalues and the translation zero-mode Jacobian; a conventional determinant expression iswith common endpoint regularization and the prime removing the translation mode. Only its positive quantum tunnelling rate $\kappa =K{e}^{-S_0/\hslash }$ is needed here.

In the dilute instanton gas, well-separated crossings alternate in direction. Integrating $n$ ordered centers gives ${\tau }^n/n!$. A path returning to the same well has $n$ an even number, and one connecting opposite wells has $n$ an odd number. The common leading harmonic endpoint factor follows from the supplied harmonic oscillator transition kernel:Expanding these functions gives the sums over even numbers and odd numbers, with the position-kernel normalization ${\mathcal{A}}_0$ that its abbreviated formula suppresses. Validity requires $S_0/\hslash \gg 1$, $\omega \tau \gg 1$ and $\kappa \ll \omega$, so typical crossing separation greatly exceeds an instanton width. Local anharmonic corrections replace the harmonic well energy by its perturbatively corrected value; the leading formula does not claim uniformly negligible relative error at arbitrarily large $\tau$.

The two exponents identify the double-well tunneling splitting:The superposition with an even spatial wavefunction is the lower state. Coherent quantum tunnelling removes the classical degeneracy, and the exponentially small splitting sets the long quantum tunnelling timescale.

The quartic double-well instanton has $L_I=-{\partial }_{\tau }^2+{\omega }^2[1-3{\mathrm{sech}}^2(\omega \tau /2)/2]$. For a positive regulator $\epsilon$ and $r=2\sqrt{1+\epsilon /{\omega }^2}$, the common infinite-line determinant ratio is $(r-1)(r-2)/[(r+1)(r+2)]$. Its zero-mode factor is $\epsilon /(12{\omega }^2)$ to leading order, giving the displayed primed ratio. Therefore the instanton fluctuation prefactor is $K=\omega \sqrt{6S_I/(\pi \hslash )}$ and the double-well tunneling splitting is $2\hslash K{e}^{-S_I/\hslash }$.

For a smooth symmetric confining potential energy with four separated simple classical turning points $-x_2<-x_1<x_1<x_2$, defineThe Airy turning-point connection formula applied to a decaying exterior wavefunction gives a central WKB approximation proportional to ${\chi }^{-1/2}[2\cos \theta e^K+\sin \theta e^{-K}]$, where $\chi =\sqrt{2m(V-E)}$ and $K(x)={\hslash }^{-1}{\int }_x^{x_1}\chi dx$. Since ${\chi }^{\prime }(0)=0$ and $2K(0)=\varphi$, the even Neumann boundary condition gives $\tan \theta =2e^{\varphi }$, while the odd Dirichlet boundary condition gives $\tan \theta =-2e^{\varphi }$. For $\varphi \gg 1$, $\theta =(n+\frac{1}{2})\pi \mp \frac{1}{2}{e}^{-\varphi }+O(e^{-3\varphi })$, with the minus correction for even parity. Since ${\theta }^{\prime }=\mathcal{T}/(2\hslash )$ for the single-well classical oscillation period $\mathcal{T}$, the leading double-well tunneling splitting is $2\hslash e^{-\varphi }/\mathcal{T}$.