Instantons in Quantum Mechanics

In quantum mechanics, an instanton is a finite-action classical solution of the Euclidean equations of motion. It gives a semiclassical estimate of tunneling effects that are exponentially small and invisible to ordinary perturbation theory around a single classical minimum.

The Euclidean action for one coordinate is

$$S_E[x] = \int d\tau \left[ \frac{m}{2} \left(\frac{dx}{d\tau}\right)^2 + V(x) \right].$$

Euclidean path integrals are weighted schematically by

$$e^{-S_E[x]/\hbar},$$

so finite-action saddle points can dominate exponentially small effects.

Double-Well Splitting evaluates the quartic trajectory and $S_0=4/3$ in a convention designed for direct comparison with numerical spectral gaps. The reusable saddle construction remains here.

Euclidean Equation of Motion

Stationarity of $S_E$ gives

$$m\frac{d^2x}{d\tau^2} = \frac{dV}{dx}.$$

This differs in sign from the real-time Newton equation. Equivalently, Euclidean motion in $V(x)$ resembles real-time motion in the inverted potential $-V(x)$.

For finite action, the path must approach minima of $V$ as $\tau\to\pm\infty$, so that both the kinetic and potential contributions remain integrable.

Boundary Conditions

For a degenerate double well with minima at $x=-a$ and $x=a$, an instanton satisfies

$$x(\tau)\to -a \quad \text{as} \quad \tau\to-\infty,$$

and

$$x(\tau)\to a \quad \text{as} \quad \tau\to+\infty.$$

The anti-instanton has the opposite orientation. Multi-instanton configurations describe repeated tunneling events in Euclidean time.

First Integral

For a double well whose minima have $V=0$, the Euclidean energy

$$\frac{m}{2} \left(\frac{dx}{d\tau}\right)^2 - V(x)$$

is conserved. The finite-action boundary conditions set this conserved value to zero, so

$$\frac{m}{2} \left(\frac{dx}{d\tau}\right)^2 = V(x).$$

Therefore the instanton action can be written as

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

This is the same under-barrier action that appears in WKB estimates of tunneling exponents.

Quartic Double-Well Instanton

For

$$V(x)=\lambda(x^2-a^2)^2,$$

the minima are at $x=\pm a$. The small-oscillation frequency near a minimum is

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

The instanton solution can be written

$$x_{\mathrm{inst}}(\tau) = a \tanh \left[ \frac{\omega}{2} (\tau-\tau_0) \right].$$

The parameter $\tau_0$ is the instanton center. Translating the instanton in Euclidean time costs no action, so $\tau_0$ is a collective coordinate.

The action is

$$S_0 = \int_{-a}^{a} dx\, \sqrt{2m\lambda}(a^2-x^2) = \frac{4}{3} a^3 \sqrt{2m\lambda}.$$

Tunneling Splitting

For the symmetric double well, instantons produce a splitting between even and odd low-energy states:

$$\Delta E \sim A e^{-S_0/\hbar}.$$

The exponent $S_0$ comes from the classical Euclidean saddle. The prefactor $A$ comes from quadratic fluctuations around the saddle, normalization of the zero mode, and multi-instanton combinatorics.

For many purposes, the exponent is the main physical information. Precise prefactors require more careful analysis than the leading saddle picture.

Tunneling Splittings derives how the dilute instanton sum becomes an even–odd spectral splitting and matches that result to Herring and WKB formulas.

Double-Well Instanton Numerical Check tests the normalized one-loop formula against converged parity gaps while keeping exponent accuracy, prefactor accuracy, and numerical resolution separate.

A metastable well poses a different boundary-value problem. Bounce Solutions develops the Euclidean round trip, its decay exponent, and the additional negative fluctuation mode.

Collective Coordinate Preview

Because the instanton can be centered at any $\tau_0$, the quadratic fluctuation operator around the instanton has a zero mode proportional to

$$\frac{dx_{\mathrm{inst}}}{d\tau}.$$

This zero mode must not be treated like an ordinary Gaussian fluctuation. Instead, one integrates over the collective coordinate $\tau_0$. This is the simplest example of a technical issue that becomes central in field-theory instanton calculations.

Relation to Perturbation Theory

Perturbation theory around one minimum produces a series in local fluctuations. It does not connect the two classical minima at any finite order. The instanton contribution has the form

$$e^{-S_0/\hbar},$$

which is nonanalytic at $\hbar=0$. This is why tunneling splittings are called nonperturbative.

QFT Bridge

Quantum-mechanical instantons are instantons in $0+1$ dimensions: one time coordinate and no spatial field profile. In quantum field theory, instantons are finite-action Euclidean field configurations. The same saddle-point logic appears, but new issues enter:

• spatial dependence and topology;
• gauge redundancy;
• zero modes from symmetries;
• fluctuation determinants over fields;
• renormalization and scale dependence.

The quantum-mechanical case is the cleanest place to learn the mechanism before those complications enter.

Common Mistakes

• Treating Wick rotation as a purely formal substitution without checking boundary conditions.
• Forgetting that instantons are Euclidean, not real-time classical, solutions.
• Confusing the instanton exponent with the full prefactor.
• Ignoring zero modes when discussing fluctuation determinants.
• Presenting instanton methods as replacing WKB rather than matching the same semiclassical exponent in one-dimensional tunneling.

Exercises

1. Derive the Euclidean equation of motion from $S_E$.

Solution

Vary

$$S_E = \int d\tau \left[ \frac{m}{2}\dot x^2+V(x) \right].$$

The variation is

$$\delta S_E = \int d\tau \left[ m\dot x\,\delta\dot x + V'(x)\delta x \right].$$

Integrating the first term by parts and taking endpoint variations to vanish gives

$$\delta S_E = \int d\tau \left[ -m\ddot x+V'(x) \right]\delta x.$$

Stationarity for arbitrary $\delta x$ gives

$$m\ddot x=V'(x).$$

2. Show that the instanton action can be written as an integral over $x$ when the Euclidean energy is zero.

Solution

Zero Euclidean energy gives

$$\frac{m}{2}\dot x^2=V(x).$$

Then

$$S_E = \int d\tau \left[ \frac{m}{2}\dot x^2+V(x) \right] = \int d\tau\,2V(x).$$

Using

$$d\tau = \frac{dx}{\dot x} = dx \sqrt{\frac{m}{2V(x)}},$$

one obtains

$$S_0 = \int dx\,\sqrt{2mV(x)}.$$

3. Why does the instanton center $\tau_0$ lead to a zero mode?

Solution

The Euclidean action is invariant under translations of $\tau$. If $x_{\mathrm{inst}}(\tau)$ is a solution, then $x_{\mathrm{inst}}(\tau-\tau_0)$ is also a solution with the same action. Differentiating this family with respect to $\tau_0$ gives a fluctuation proportional to $dx_{\mathrm{inst}}/d\tau$, and the action does not change to quadratic order in that direction. Hence it is a zero mode.

References

• S. Coleman, Aspects of Symmetry, Cambridge University Press, 1985.
• R. Rajaraman, Solitons and Instantons, North-Holland, 1982.
• J. Zinn-Justin, Path Integrals in Quantum Mechanics, Oxford University Press, 2005.
• H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 5th ed., World Scientific, 2009.