Double-Well Tunneling

A symmetric double well is the canonical model of tunneling splitting. Classically, a low-energy particle trapped in one well cannot cross the barrier. Quantum mechanically, the two localized well states mix through the barrier, producing even and odd energy eigenstates separated by an exponentially small splitting.

Assume a regular one-dimensional scalar Schrödinger Hamiltonian on the connected line, with a smooth confining symmetric potential and two minima:

$$V(x)=V(-x), \qquad V(\pm a)=0, \qquad V(0)>0.$$

A standard example is

$$V(x)=\lambda(x^2-a^2)^2, \qquad \lambda>0.$$

Required background. Barrier penetration supplies the forbidden-region action; degenerate perturbation theory supplies the two-state diagonalization; parity distinguishes the two eigenstates.

Tunneling Splittings derives quantitative WKB and instanton prefactors. Double-Well Splitting compares the quartic model with numerical even–odd gaps.

Localized Well States

Near either minimum, the potential is approximately harmonic. For the quartic double well,

$$V''(a)=8\lambda a^2,$$

so the local oscillator frequency is

$$\omega = \sqrt{\frac{8\lambda a^2}{m}}.$$

If the wells were completely isolated, there would be a left-localized ground state $\lvert L\rangle$ and a right-localized ground state $\lvert R\rangle$ with nearly equal local energy

$$E_{\mathrm{loc}} \approx \frac12\hbar\omega.$$

The barrier is finite, so these states are not exact eigenstates.

Even and Odd Eigenstates

Because the Hamiltonian has parity symmetry, exact energy eigenstates can be chosen with definite parity. The two lowest states are approximately

$$\lvert +\rangle = \frac{1}{\sqrt2} \left( \lvert L\rangle+\lvert R\rangle \right),$$

and

$$\lvert -\rangle = \frac{1}{\sqrt2} \left( \lvert L\rangle-\lvert R\rangle \right).$$

For the regular scalar problem specified above, the one-dimensional node theorem makes the ground state even and the first excited state odd. This ordering is strict, not a statistical tendency.

Two-State Effective Hamiltonian

The displayed superpositions require orthonormal $\lvert L\rangle,\lvert R\rangle$. One precise construction starts from the exact lowest even and odd states and defines their sum and difference as localized states. Then $E_{\mathrm{loc}}=(E_0+E_1)/2$ and $K=(E_1-E_0)/2$ give an exact representation within this doublet. Localization becomes sharp in the deep-well limit. Overlapping isolated-well trial functions must first be orthonormalized, or treated with a generalized eigenvalue problem.

In this orthonormal localized basis,

$$H_{\mathrm{eff}} = \begin{pmatrix} E_{\mathrm{loc}} & -K \\ -K & E_{\mathrm{loc}} \end{pmatrix}, \qquad K>0.$$

Its eigenvalues are

$$E_\pm = E_{\mathrm{loc}}\mp K.$$

The splitting is

$$\Delta E = E_- - E_+ = 2K.$$

The off-diagonal matrix element $K$ is exponentially small in the deep-well regime. Starting in $\lvert L\rangle$, the projected unitary evolution gives

$$P_{L\to R}(t)=\sin^2(Kt/\hbar).$$

The first complete transfer occurs at $t=\pi\hbar/\Delta E$. This is coherent oscillation in a closed doublet, not exponential decay from a metastable well.

WKB Estimate

Let $E_{\mathrm{loc}}$ lie below the barrier, and let $x_1$ and $x_2$ be the turning points that bound the forbidden barrier region between the wells. Define

$$\kappa(x) = \sqrt{2m(V(x)-E_{\mathrm{loc}})}.$$

The splitting has the leading exponential form

$$\Delta E \sim A \exp\left[ - \frac{1}{\hbar} \int_{x_1}^{x_2} \kappa(x)\,dx \right].$$

The prefactor $A$ depends on the well frequency, matching conventions, and corrections near the turning points. The exponent is the most robust part of the estimate.

The important lesson is not the exact prefactor but the nonanalytic dependence:

$$\Delta E \propto e^{-S/\hbar}.$$

No finite power series in $\hbar$ around one isolated well can reproduce this splitting.

Instanton Estimate

In Euclidean time, tunneling is described semiclassically by a finite-action path connecting one minimum to the other. For a one-dimensional double well with minima normalized to $V=0$, the instanton action is

$$S_0 = \int_{-a}^{a} dx\, \sqrt{2mV(x)}.$$

For the quartic example,

$$S_0 = \frac{4}{3} a^3 \sqrt{2m\lambda}.$$

The splitting has the instanton form

$$\Delta E \sim A_{\mathrm{inst}} e^{-S_0/\hbar},$$

with a prefactor determined by fluctuations around the instanton, including its translation zero mode. Here $S_0$ uses the potential minimum energy, whereas the WKB integral uses $E_{\mathrm{loc}}\simeq\hbar\omega/2$. Their common leading exponential scale does not make the two actions identical: the energy-dependent logarithmic corrections contribute to the power and numerical coefficient of the prefactor. Garg’s matching analysis in the references explains this distinction.

Numerical Comparison

A numerical one-dimensional diagonalization gives a direct check:

1. choose a grid or basis that resolves both wells and the barrier;
2. diagonalize the Hamiltonian;
3. identify the two lowest parity eigenstates;
4. compute $\Delta E=E_1-E_0$;
5. compare the logarithm of the splitting to the WKB or instanton action.

The logarithm isolates the dominant dependence only after the gap itself is numerically resolved. Converge each parity energy tightly enough that its absolute error is much smaller than the gap; subtraction and roundoff can otherwise manufacture a plausible-looking logarithm. Prefactors demand still tighter control.

Double-Well Instanton Numerical Check carries out this comparison with parity-resolved basis diagonalization, an independent finite-difference discretization, retained WKB quadratures, and explicit basis and roundoff diagnostics.

Exercises

1. Add a small bias between the wells, $H=\bar E I+(\epsilon/2)\sigma_z-K\sigma_x$ in the ordered basis $(L,R)$. Find the gap and the largest possible transfer probability from $L$ to $R$.

Solution

The gap is $\Delta=\sqrt{\epsilon^2+4K^2}$. Exponentiating the Pauli-matrix part gives

$$P_{L\to R}(t)=\frac{4K^2}{\epsilon^2+4K^2} \sin^2\!\left(\frac{\Delta t}{2\hbar}\right).$$

Thus the maximum transfer is $4K^2/(\epsilon^2+4K^2)$. A bias much larger than the exponentially small coupling suppresses transfer even when it is tiny relative to the intrawell excitation energy.

2. Why is the splitting called nonperturbative in $\hbar$?

Solution

The leading dependence is of the form

$$\Delta E\propto e^{-S/\hbar}.$$

As $\hbar\to0^+$, this is smaller than any finite power of $\hbar$. A power series in $\hbar$ around one well cannot produce such an exponential term at finite order.

References

• S. Coleman, Aspects of Symmetry, Cambridge University Press, 1985.
• A. Garg, “Tunnel splittings for one-dimensional potential wells revisited”, American Journal of Physics 68, 430–437 (2000); author preprint.
• H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 5th ed., World Scientific, 2009.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Butterworth-Heinemann, 1981.