Instantons in Quantum Mechanics Preview

An instanton in quantum mechanics is a finite-action solution of the Euclidean classical equations. In a tunneling problem, it connects regions that no real classical trajectory at the relevant energy can connect. Its Euclidean action supplies the leading exponential in a semiclassical tunneling amplitude,

$$\text{instanton contribution} \sim e^{-S_0/\hbar}.$$

This is a path-integral preview of the method. It develops the Euclidean saddle picture and one explicit double-well solution. The quantitative calculation of splittings, multi-instanton sums, collective-coordinate measures, and determinant prefactors belongs to Instantons in Quantum Mechanics and its companion pages.

An instanton is not a particle’s literal trajectory through the barrier in real time. It is a saddle of an imaginary-time integral used to calculate a quantum amplitude.

Euclidean Action

For a particle with

$$H = \frac{p^2}{2m} + V(x),$$

the imaginary-time kernel has the formal path-integral representation

$$K_E(x_f,\tau_f;x_i,\tau_i) = \int_{x(\tau_i)=x_i}^{x(\tau_f)=x_f} \mathcal D x\, \exp\left[ - \frac{1}{\hbar}S_E[x] \right],$$

where

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

The Euclidean and Imaginary-Time Path Integrals page owns the operator and kernel construction. Euclidean Time and Imaginary-Time Action owns the tunneling-oriented signs, background subtraction, boundary data, and fixed-energy exponent. Here the important feature is the real damping weight $e^{-S_E/\hbar}$. When $S_E/\hbar$ is large, a saddle-point expansion can isolate exponentially small sectors that are difficult to see in the oscillatory real-time integral.

Varying the path while holding its endpoints fixed gives

$$\begin{aligned} \delta S_E &= \left[ m\frac{dx}{d\tau}\,\delta x \right]_{\tau_i}^{\tau_f} \\ &\quad+ \int_{\tau_i}^{\tau_f}d\tau\, \left[ - m\frac{d^2x}{d\tau^2} + V'(x) \right]\delta x . \end{aligned}$$

The Euclidean equation of motion is therefore

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

Its force has the opposite sign from the real-time Newton equation. Formally, Euclidean motion in $V(x)$ resembles ordinary motion in the inverted potential $-V(x)$. That analogy is useful for finding solutions, but the parameter $\tau$ remains imaginary time and the solution remains part of an amplitude calculation.

For instantons on an infinite Euclidean-time interval, it is convenient to shift degenerate minima so that $V=0$ there. Finite action then requires

$$\frac{dx}{d\tau}\to0, \qquad V(x)\to0 \qquad \text{as} \qquad \tau\to\pm\infty.$$

Without subtracting a nonzero common vacuum energy, the action contains an uninformative term proportional to the length of the Euclidean-time interval.

Tunneling in Path Integrals

Consider a symmetric double well with low-energy wave packets $\lvert L\rangle$ and $\lvert R\rangle$ localized near opposite minima. The Euclidean transition amplitude is

$$\mathcal K_{RL}(T) = \langle R\rvert e^{-HT/\hbar} \lvert L\rangle.$$

For a high, broad barrier, the two lowest exact states are approximately the even and odd combinations of the localized states. If their energies are $E_0$ and $E_1$, then the two-state approximation gives

$$\mathcal K_{RL}(T) \simeq \frac12 \left( e^{-E_0T/\hbar} - e^{-E_1T/\hbar} \right).$$

Thus the off-diagonal Euclidean kernel contains the tunnel splitting

$$\Delta E = E_1-E_0.$$

In the path integral, histories contributing to $\mathcal K_{RL}$ begin near one minimum and end near the other. The leading semiclassical history in this sector is an instanton. Its weight is $e^{-S_0/\hbar}$, and a controlled calculation leads schematically to

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

where $A$ has dimensions of energy and depends on fluctuations and normalization.

The exponential is nonperturbative. If a perturbative expansion about one minimum is organized in powers of $\hbar$ or a coupling, no finite order can produce a term of the form $e^{-S_0/\hbar}$. The instanton does not replace the local perturbative series; it supplies another saddle sector that the local series alone omits.

The displayed exponential is an amplitude-level contribution, not a probability. A probability or energy splitting follows only after the relevant kernel, boundary conditions, normalization, and saddle sum have been assembled.

Double-Well Potential Preview

Take the quartic double well

$$V(x) = \lambda \left( x^2-a^2 \right)^2, \qquad \lambda\gt0.$$

The degenerate minima are at $x=\pm a$, and the small-oscillation frequency in either well is

$$\omega = \sqrt{ \frac{V''(\pm a)}{m} } = \sqrt{ \frac{8\lambda a^2}{m} }.$$

Because the Euclidean Lagrangian has no explicit $\tau$ dependence, the quantity

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

is conserved. An instanton approaching the two minima with zero velocity has $E_E=0$, so

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

For an increasing path from $-a$ to $a$, this first-order equation becomes

$$\frac{dx}{d\tau} = \sqrt{\frac{2\lambda}{m}} \left( a^2-x^2 \right).$$

Its solution is

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

The center $\tau_0$ is arbitrary because the Euclidean action is invariant under translations of $\tau$. Reversing the orientation gives the anti-instanton,

$$x_{\mathrm{anti}}(\tau) = - x_{\mathrm{inst}}(\tau)$$

after a suitable choice of center.

Using the zero-Euclidean-energy relation, the one-instanton action can be reduced to an ordinary integral:

$$\begin{aligned} S_0 &= \int_{-\infty}^{\infty}d\tau\, \left[ \frac{m}{2} \left( \frac{dx}{d\tau} \right)^2 + V(x) \right] \\ &= \int_{-a}^{a}dx\, \sqrt{2mV(x)} \\ &= \frac{4}{3} a^3 \sqrt{2m\lambda}. \end{aligned}$$

This leading exponent is closely related to the under-barrier WKB action. The canonical Double-Well Tunneling page compares the parity splitting, WKB estimate, and instanton estimate without identifying their prefactors by analogy.

Instanton as a Saddle Point

The boundary conditions select a distinct sector of path space:

$$x(-\infty)=-a, \qquad x(+\infty)=a.$$

Within that sector, the instanton is a stationary configuration of $S_E$. The term saddle point is used broadly because the semiclassical expansion is an expansion around a stationary configuration in an infinite-dimensional integral. A double-well instanton should not be confused with a bounce describing decay from a metastable state. A bounce typically has a negative fluctuation mode; the translationally invariant double-well instanton instead has a zero mode.

Three distinctions matter:

• The instanton is a Euclidean classical solution, not a real-time classical trajectory.
• Its existence depends on boundary conditions and finite action, not merely on solving the differential equation locally.
• A single instanton gives one saddle sector; a physical kernel may also require anti-instantons, repeated events, or the trivial vacuum sector.

For a long Euclidean interval, widely separated instantons and anti-instantons can often be treated approximately as a dilute collection. Summing those sectors is what turns individual transition events into a corrected spectrum. The accuracy of that dilute-gas approximation requires $S_0/\hbar\gg1$ and separations large compared with the instanton width, of order $\omega^{-1}$ in the quartic model.

Fluctuations and Determinants Preview

The action $S_0$ gives only the leading exponential. To obtain the prefactor, write

$$x(\tau) = x_{\mathrm{inst}}(\tau) + \eta(\tau).$$

The quadratic expansion is

$$S_E[x] = S_0 + \frac12 \int d\tau\, \eta(\tau) \mathcal M_{\mathrm{inst}} \eta(\tau) + O(\eta^3),$$

with fluctuation operator

$$\mathcal M_{\mathrm{inst}} = - m\frac{d^2}{d\tau^2} + V''\left( x_{\mathrm{inst}}(\tau) \right).$$

Formally, Gaussian integration produces a determinant factor. A physical quantity normally involves a regulated ratio such as

$$\left[ \frac{ \det\mathcal M_0 }{ \det{}'\mathcal M_{\mathrm{inst}} } \right]^{1/2},$$

where $\mathcal M_0$ is the operator around a reference minimum. The prime is essential. Translating $\tau_0$ changes the instanton’s center without changing its action, so

$$\eta_0(\tau) \propto \frac{dx_{\mathrm{inst}}}{d\tau}$$

is a zero mode of $\mathcal M_{\mathrm{inst}}$. It must be removed from the ordinary determinant and replaced by integration over the collective coordinate $\tau_0$.

Boundary conditions, determinant regularization, zero-mode normalization, and multi-instanton combinatorics all enter $A$. That machinery is developed in Fluctuation Determinants Preview. Quoting $e^{-S_0/\hbar}$ alone is appropriate for a leading exponential estimate, but not for a normalized splitting or rate.

QFT Instanton Bridge

Quantum mechanics is a field theory in $0+1$ dimensions, so its instanton logic provides a clean prototype. In a scalar field theory, paths $x(\tau)$ are replaced by fields $\phi(\tau,\mathbf x)$ and the saddle equation becomes

$$\frac{\delta S_E[\phi]}{\delta\phi} = 0.$$

The structural pattern survives:

• finite-action Euclidean configurations define nontrivial saddle sectors;
• their actions produce exponential weights;
• quadratic fluctuations produce determinant prefactors;
• continuous symmetries produce zero modes and collective coordinates.

The detailed calculation changes substantially. Spatial dependence introduces partial differential equations, gauge theories require gauge fixing, finite-action boundary conditions can carry topological information, fermions can introduce additional zero modes, and ultraviolet fluctuations require renormalization. Not every nonperturbative effect in field theory is instanton-dominated.

Bridge to QFT Instantons develops that translation while keeping the full field-theory instanton calculus outside the scope of this preview.

Common Mistakes

• Treating the instanton as the particle’s hidden real-time route through a barrier.
• Forgetting to normalize degenerate minima to the same vacuum energy before taking an infinite-time action.
• Calling every Euclidean solution an instanton without checking its boundary conditions and action.
• Interpreting $e^{-S_0/\hbar}$ as a tunneling probability or a complete energy-splitting formula.
• Including the translational zero eigenvalue in an ordinary determinant.
• Confusing a double-well instanton with a metastable bounce and importing the latter’s negative-mode interpretation.
• Assuming a one-instanton contribution is sufficient when the observable requires a sum over several saddle sectors.
• Transferring a quantum-mechanical prefactor directly to field theory without rebuilding the measure, gauge treatment, and renormalization.

References

• S. Coleman, Aspects of Symmetry, Cambridge University Press, 1985, chapter 7.
• R. Rajaraman, Solitons and Instantons, North-Holland, 1982.
• J. Zinn-Justin, Path Integrals in Quantum Mechanics, Oxford University Press, 2005.
• L. S. Schulman, Techniques and Applications of Path Integration, Dover, 2005.
• H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 5th ed., World Scientific, 2009.

Exercises

1. Derive the Euclidean equation of motion for a particle.

Solution

Start from

$$S_E[x] = \int_{\tau_i}^{\tau_f}d\tau\, \left[ \frac{m}{2}(x')^2 + V(x) \right].$$

Under $x\mapsto x+\delta x$,

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

Integrating the first term by parts and imposing $\delta x(\tau_i)=\delta x(\tau_f)=0$ gives

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

Stationarity for arbitrary interior variations requires

$$mx'' = V'(x).$$

2. Show that a finite-action path between degenerate minima has zero Euclidean energy.

Solution

Multiply the Euclidean equation $mx''=V'(x)$ by $x'$:

$$mx''x' = V'(x)x'.$$

Both sides are total derivatives, so

$$\frac{d}{d\tau} \left[ \frac{m}{2}(x')^2 - V(x) \right] = 0.$$

The bracket is the conserved Euclidean energy $E_E$. At either asymptotic minimum, finite action requires $x'\to0$ and the chosen normalization gives $V\to0$. Hence $E_E=0$ throughout the solution, and

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

3. Verify the quartic instanton without substituting into the second-order equation.

Solution

Let

$$u = \frac{\omega}{2} \left( \tau-\tau_0 \right), \qquad x = a\tanh u.$$

Then

$$x' = \frac{a\omega}{2} \operatorname{sech}^2u$$

and

$$V(x) = \lambda a^4 \operatorname{sech}^4u.$$

Using $\omega^2=8\lambda a^2/m$,

$$\frac{m}{2}(x')^2 = \frac{ma^2\omega^2}{8} \operatorname{sech}^4u = \lambda a^4 \operatorname{sech}^4u = V(x).$$

The trajectory therefore satisfies the zero-energy first-order equation. Differentiating that equation where $x'\neq0$ gives the Euclidean equation of motion, and continuity extends it through the asymptotic regions. Since $\tanh u\to\pm1$, the required boundary conditions also hold.

4. Compute the action of the quartic instanton.

Solution

For the increasing solution, $m(x')^2/2=V(x)$ and

$$x' = \sqrt{\frac{2V(x)}{m}}.$$

Therefore

$$\begin{aligned} S_0 &= \int d\tau\,2V(x) \\ &= \int_{-a}^{a}dx\, \sqrt{2mV(x)} \\ &= \sqrt{2m\lambda} \int_{-a}^{a}dx\, \left( a^2-x^2 \right). \end{aligned}$$

The remaining integral is

$$\left[ a^2x-\frac{x^3}{3} \right]_{-a}^{a} = \frac{4a^3}{3},$$

so

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

5. Prove that translating the instanton produces a zero mode.

Solution

The instanton equation can be written

$$- mx_{\mathrm{inst}}'' + V'(x_{\mathrm{inst}}) = 0.$$

Differentiate it with respect to $\tau$:

$$\left[ - m\frac{d^2}{d\tau^2} + V''(x_{\mathrm{inst}}) \right] x_{\mathrm{inst}}' = 0.$$

The operator in brackets is $\mathcal M_{\mathrm{inst}}$, hence

$$\mathcal M_{\mathrm{inst}} x_{\mathrm{inst}}' = 0.$$

The mode $x_{\mathrm{inst}}'$ is tangent to the family $x_{\mathrm{inst}}(\tau-\tau_0)$ and represents shifting $\tau_0$. It must be replaced by collective-coordinate integration rather than included in the Gaussian determinant.