Instantons and Quantum Tunneling

Put a particle in a symmetric double well and ask a simple question: what is its ground state? Classically the answer is degenerate — the particle sits at the bottom of the left well or the right well, two states of equal energy. Quantum mechanically the degeneracy is broken. The true stationary states are the symmetric and antisymmetric combinations, split by a tiny energy $\Delta E$ , and the particle tunnels back and forth between the wells at the rate $\Delta E/\hbar$ . Ordinary WKB gives that splitting as an exponentially small number $e^{-S_0/\hbar}$ . The path integral gives something better: a picture of what tunnels, and a systematic machine for the prefactor. The picture is the instanton. In real time there is no classical path connecting the two minima — the particle would have to cross a region where its kinetic energy is negative. But rotate to imaginary time, and the Euclidean equation of motion $m\ddot x=+V'(x)$ describes motion in the inverted potential $-V$ , where the two minima become two hilltops with a valley between them. Now a classical trajectory exists: the particle rolls off the left hilltop, coasts through the valley, and climbs the right hilltop, taking infinite imaginary time to complete the journey. That finite-action solution — localized in imaginary time, hence "instant-on" — is the tunnelling event, and its Euclidean action $S_0$ is exactly the WKB under-barrier integral $\int\sqrt{2mV}\,\mathrm dx$ . You arrive here holding the Euclidean path integral (Imaginary-Time and Euclidean Path Integrals), the WKB tunnelling exponent (WKB Tunneling and the Gamow Factor, which explicitly forwards the imaginary-time method to this article), and the harmonic oscillator. Here we solve the double-well kink in closed form $x(\tau)=a\tanh[\tfrac{\omega}{2}(\tau-\tau_0)]$ , prove its action equals the WKB barrier integral, sum a dilute gas of instantons and anti-instantons to get the level splitting $\Delta E=2\hbar K\,e^{-S_0/\hbar}$ , and see how the same construction — with a bounce instead of a kink — computes the decay of a metastable vacuum, the seed of some of the deepest results in quantum field theory.

Key takeaways

A symmetric double well has a tiny ground-state splitting $\Delta E$ ; the particle tunnels between minima at rate $\Delta E/\hbar$ , and the path integral computes both the exponent and the prefactor.
In imaginary time the Euclidean EoM is $m\ddot x=+V'(x)$ — motion in the inverted potential — so a classical path connecting the minima exists: the instanton.
For $V=\tfrac{m\omega^2}{8a^2}(x^2-a^2)^2$ the instanton is the kink $x(\tau)=a\tanh[\tfrac{\omega}{2}(\tau-\tau_0)]$ , solving the Euclidean equation exactly.
Its Euclidean action equals the WKB under-barrier integral: $S_0=\int_{-a}^{a}\sqrt{2mV}\,\mathrm dx=\tfrac23 m\omega a^2$ — two roads to the same exponent.
Summing a dilute gas of well-separated instantons and anti-instantons gives the splitting $\Delta E=2\hbar K\,e^{-S_0/\hbar}$ , with $K$ a one-loop fluctuation prefactor.
Replacing the kink by a bounce in an asymmetric well computes metastable (false-vacuum) decay — the same method, generalized.
Key idea: tunnelling is a classical solution in imaginary time; the instanton's Euclidean action is the WKB exponent, and summing instantons builds the exponentially small level splitting from first principles.

The core insight

Rotating to imaginary time turns the classically forbidden tunnelling event into a finite-action classical trajectory in the inverted potential — the instanton — whose Euclidean action equals the WKB barrier integral, and whose dilute-gas sum yields the double-well level splitting $\Delta E=2\hbar K\,e^{-S_0/\hbar}$ .

Where this lives today

Instanton calculus is a working tool across physics. In quantum field theory, gauge-theory instantons violate baryon-plus-lepton number in the Standard Model and resolve the " $U(1)$ problem" of why the $\eta'$ meson is anomalously heavy; they are central to the strong-CP problem and the axion. In cosmology, the Coleman-De Luccia bounce computes the decay rate of a false vacuum — the calculation behind "is our universe's vacuum stable?", a question that the measured Higgs and top masses place tantalizingly near the metastability boundary. In chemistry, "instanton rate theory" computes low-temperature reaction rates dominated by hydrogen tunnelling, where the reaction path is literally the Euclidean bounce through the barrier. Every one of these is the double-well calculation of this article, generalized to more fields and more dimensions.

The physical picture

The whole subject turns on one inversion. In the real double well the region between the two minima is a barrier: to pass through it a classical particle would need negative kinetic energy, which is why there is no real-time classical tunnelling path. Flip to imaginary time and the sign of the potential in the equation of motion flips — the Euclidean particle moves in $-V$ . The barrier becomes a valley, the two minima become two equal hilltops, and now a perfectly ordinary classical trajectory connects them: release the particle at rest on the left hilltop, let it roll down and up to the right hilltop, arriving at rest after infinite imaginary time. This finite-action "rolling" solution is the instanton.

Figure

Real barrier vs inverted-potential valley: the instanton is classical in imaginary time

The payoff picture is the kink profile itself and the splitting it produces. Plotted against imaginary time, the instanton is a smooth step $a\tanh[\tfrac{\omega}{2}(\tau-\tau_0)]$ : the particle sits near $-a$ for a long time, transits rapidly near $\tau_0$ over a width $\sim1/\omega$ , then sits near $+a$ . Because the transition is fast and localized, well-separated instantons and anti-instantons behave like a dilute gas of independent events, and summing over any number of them exponentiates into the splitting of the two lowest levels — the symmetric ground state below, the antisymmetric first excited state above, separated by $\Delta E$ .

Figure

The kink profile and the level splitting it produces

The core equations

The starting point is the Euclidean equation of motion, obtained by varying the Euclidean action $S_E=\int(\tfrac12 m\dot x^2+V)\,\mathrm d\tau$ inherited from Imaginary-Time and Euclidean Path Integrals. The plus sign on $V$ is the whole story: it flips the force.

m\,\ddot x=+\,V'(x),\qquad \dot x=\frac{\mathrm dx}{\mathrm d\tau} \qquad (1)

where: the sign of the force term is $+V'$ , not $-V'$ — motion in the inverted potential $-V$ . Reading it physically: the minima of $V$ (where the real particle sits) become maxima of $-V$ (hilltops the Euclidean particle can sit on and roll off of), so a finite-action trajectory linking two degenerate minima exists. The conserved "Euclidean energy" is $\tfrac12 m\dot x^2-V$ ; for a trajectory that starts and ends at rest at $V=0$ minima, this energy is zero, giving the first integral $\tfrac12 m\dot x^2=V$ . For the standard double well we pin the potential by its small-oscillation frequency $\omega$ at each minimum, which keeps every formula clean.

V(x)=\frac{m\omega^2}{8a^2}\,(x^2-a^2)^2,\qquad V''(\pm a)=m\omega^2 \qquad (2)

where: $a$ — the location of each minimum; $\omega$ — the frequency of small oscillations in either well. Reading it physically: near $x=+a$ the potential looks like $\tfrac12 m\omega^2(x-a)^2$ , an ordinary oscillator, so $\omega$ sets both the harmonic spectrum in each well and the width $1/\omega$ of the instanton transition. The instanton is the zero-energy solution of (1) in this potential, and it is an elementary hyperbolic tangent.

x_{\rm inst}(\tau)=a\tanh\!\Big[\frac{\omega}{2}(\tau-\tau_0)\Big],\qquad S_0=\int_{-a}^{a}\!\sqrt{2mV}\,\mathrm dx=\frac{2}{3}m\omega a^2 \qquad (3)

where: $\tau_0$ — the (arbitrary) imaginary-time center of the transition, a zero mode; $S_0$ — the instanton action. Reading it physically: the action is identical to the WKB under-barrier integral $\int\sqrt{2mV}\,\mathrm dx$ — the two calculations, path integral and WKB, compute the same exponent by different routes. The freedom in $\tau_0$ (the instanton can happen at any imaginary time) is the collective coordinate that, integrated over, supplies a factor of the total imaginary-time interval $T$ and is the reason the dilute gas exponentiates. Summing the dilute gas of instantons and anti-instantons gives the level splitting.

\Delta E=E_-\!-E_+ = 2\hbar K\,e^{-S_0/\hbar},\qquad K=\omega\sqrt{\frac{S_0}{2\pi\hbar}} \qquad (4)

where: $E_\pm$ — antisymmetric/symmetric levels; $K$ — the one-loop prefactor from Gaussian fluctuations about the instanton (with the zero mode handled by the $\tau_0$ integration). Reading it physically: the exponent $e^{-S_0/\hbar}$ is the WKB tunnelling suppression; the path integral's contribution is the prefactor $K$ , computed systematically as a fluctuation determinant. An equivalent WKB-connection-formula estimate gives $\Delta E\approx(\hbar\omega/\pi)e^{-S_0/\hbar}$ — same exponent, an $O(1)$ -different prefactor.

Derivation AThe kink solves the Euclidean equation of motion

Our goal is to verify that $x(\tau)=a\tanh[\tfrac{\omega}{2}(\tau-\tau_0)]$ is an exact solution of $m\ddot x=+V'(x)$ for the double well (2), and to extract its action. The clean route is the first integral: any zero-Euclidean-energy trajectory obeys $\tfrac12 m\dot x^2=V$ , a first-order equation, and we check the tanh satisfies it, then integrate the action along the path. Watch that $V$ is a perfect square, which is exactly what makes the first integral separable.

Step 1. First integral. Zero Euclidean energy means $\tfrac12 m\dot x^2=V=\tfrac{m\omega^2}{8a^2}(x^2-a^2)^2$ , so $\dot x=\tfrac{\omega}{2a}(a^2-x^2)$ (taking the root with $\dot x>0$ for the left-to-right kink). Why: the potential being a perfect square lets us take a clean square root without branch trouble on the interval $-a<x<a$ ; this separable ODE is the BPS-type equation whose solution is the instanton.

Step 2. Solve by separation: $\displaystyle\int\frac{\mathrm dx}{a^2-x^2}=\frac{\omega}{2a}\int\mathrm d\tau\ \Rightarrow\ \frac{1}{a}\,\mathrm{artanh}\frac{x}{a}=\frac{\omega}{2a}(\tau-\tau_0)$ , giving $x=a\tanh[\tfrac{\omega}{2}(\tau-\tau_0)]$ . Why: the artanh integral is exactly the inverse of the tanh; the integration constant is the free center $\tau_0$ , the zero mode of the solution.

Step 3. Check the full second-order equation. Differentiating twice and substituting into $m\ddot x-V'(x)$ gives identically zero — the tanh solves not just the first integral but the equation of motion (1). Why: any solution of the zero-energy first integral automatically solves the second-order equation, but the explicit substitution certifies it with no hidden assumption.

Step 4. The action. Along the trajectory $\tfrac12 m\dot x^2=V$ , so $S_0=\int(\tfrac12 m\dot x^2+V)\,\mathrm d\tau=\int 2V\,\mathrm d\tau=\int m\dot x\,\mathrm dx=\int_{-a}^{a}\sqrt{2mV}\,\mathrm dx$ . Evaluating: $\int_{-a}^{a}\tfrac{m\omega}{2a}(a^2-x^2)\,\mathrm dx=\tfrac23 m\omega a^2$ . Why: on a zero-energy path the two terms of $S_E$ are equal, converting the imaginary-time integral into a spatial one — and that spatial integral is precisely the WKB barrier integral $\int\sqrt{2mV}\,\mathrm dx$ . This is the exact statement that instanton and WKB compute the same exponent. Verified: `verify_aq-qm-instantons-tunneling.py` confirms the residual $m\ddot x-V'(x)\equiv0$ for the tanh, the first integral $(m\dot x)^2=2mV$ , and $S_0=\tfrac23 m\omega a^2$ symbolically.

Go deeper · Derivation B · 1 minFrom one instanton to the level splitting

Derivation BFrom one instanton to the level splitting

Here we sketch how summing instantons produces the splitting $\Delta E=2\hbar K e^{-S_0/\hbar}$ . The strategy is the dilute-gas approximation: because a single instanton has action $S_0$ and is localized in imaginary time, a configuration of $n$ well-separated instantons and anti-instantons contributes $e^{-nS_0/\hbar}$ times a combinatorial factor, and summing over all $n$ exponentiates. We keep this at the level of the counting that fixes the structure of (4), not the full fluctuation determinant.

Step 1. Amplitude with $n$ kinks. The Euclidean transition amplitude $\langle\pm a|e^{-\hat H T/\hbar}|\mp a\rangle$ receives contributions from any odd (for $-a\to+a$ ) or even (for $-a\to-a$ ) number of alternating instantons/anti-instantons; each carries $e^{-S_0/\hbar}$ and a prefactor. Why: a single kink flips the particle between wells, so the parity of the kink number is fixed by the endpoints; this is what separates the symmetric and antisymmetric channels.

Step 2. Collective-coordinate factor. Each instanton can sit at any center $\tau_i$ in $[0,T]$ , contributing $\int_0^T K\,\mathrm d\tau_i$ ; for $n$ ordered kinks this gives $K^n T^n/n!$ . Why: the $\tau_0$ zero mode integrates to the interval length $T$ ; the $1/n!$ is the ordering of identical events — the hallmark of a dilute gas.

Step 3. Sum the series. $\sum_n \dfrac{(KT e^{-S_0/\hbar})^n}{n!}$ over even/odd $n$ gives $\cosh$ and $\sinh$ of $KT e^{-S_0/\hbar}$ , so the two amplitudes are $e^{-\omega T/2}\big[\cosh\mp\sinh\big]=e^{-\omega T/2}e^{\mp KT e^{-S_0/\hbar}}$ . Why: the harmonic prefactor $e^{-\omega T/2}$ is the zero-point energy $\tfrac12\hbar\omega$ common to both wells; the instanton sum shifts it by $\mp\hbar K e^{-S_0/\hbar}$ .

Step 4. Read the levels. Matching $e^{-E_\pm T/\hbar}$ gives $E_\pm=\tfrac12\hbar\omega\mp\hbar K e^{-S_0/\hbar}$ , so the splitting is $\Delta E=E_--E_+=2\hbar K e^{-S_0/\hbar}$ . Why: the symmetric state (even kink number resummed as $\cosh$ ) is lowered, the antisymmetric raised; their gap is the tunnelling splitting, exponentially small in $S_0/\hbar$ . Verified: the verifier confirms $\Delta E=2\hbar K e^{-S_0/\hbar}>0$ numerically for $m=\omega=\hbar=1,a=3$ ( $S_0=6$ , suppression $e^{-6}=0.00248$ ), and that the WKB-form prefactor $(\hbar\omega/\pi)$ shares the identical exponent.

Limiting cases

case	result	meaning
single well ( $a\to0$ )	$S_0\to0$ , $\Delta E\to$ finite	no barrier, no tunnelling suppression
deep well ( $S_0/\hbar\gg1$ )	$\Delta E\sim e^{-S_0/\hbar}\to0$	strongly suppressed tunnelling; classical localization
$\hbar\to0$	$\Delta E\to0$	quantum tunnelling vanishes in the classical limit
asymmetric well	kink → bounce	metastable (false-vacuum) decay rather than degenerate splitting
WKB comparison	same $S_0$ , prefactor $\hbar\omega/\pi$ vs $2\hbar K$	identical exponent, $O(1)$ -different prefactor

Common trapThe instanton is not a real-time trajectory — and S₀ is a Euclidean, not a classical, action

Two linked misconceptions. First, the instanton does not describe the particle "really moving" from one well to the other in ordinary time — there is no such real-time classical path, which is the whole reason tunnelling is nonclassical. The instanton lives in imaginary time and is a saddle point of the Euclidean action. Second, $S_0=\int\sqrt{2mV}\,\mathrm dx$ is not the ordinary action $\int L\,\mathrm dt=\int(T-V)\,\mathrm dt$ ; it is the Euclidean action $\int(T+V)\,\mathrm d\tau$ , which on the zero-energy path equals $\int\sqrt{2mV}\,\mathrm dx$ . Confusing the two produces a wrong sign and a growing (rather than decaying) tunnelling amplitude.

Worked examples

Example 1 — the instanton action and its suppression. For the double well with $m=1$ , $\omega=1$ , $a=3$ (natural units $\hbar=1$ ), compute $S_0$ and the tunnelling suppression $e^{-S_0/\hbar}$ .

Step 1. Use $S_0=\tfrac23 m\omega a^2$ : $S_0=\tfrac23(1)(1)(3^2)=\tfrac23\cdot9=\mathbf{6}$ .

Step 2. Cross-check by the integral $S_0=\int_{-a}^{a}\tfrac{m\omega}{2a}(a^2-x^2)\,\mathrm dx=\tfrac{m\omega}{2a}\big[a^2x-\tfrac{x^3}{3}\big]_{-a}^{a}=\tfrac{m\omega}{2a}\cdot\tfrac{4a^3}{3}=\tfrac23 m\omega a^2=6$ . Same value.

Step 3. Suppression: $e^{-S_0/\hbar}=e^{-6}=\mathbf{0.00248}$ . A tunnelling amplitude suppressed by a factor of ~400 — the exponentially small splitting is entirely controlled by the barrier size $S_0$ .

Example 2 — the level splitting and its prefactor. With the same parameters, estimate $\Delta E=2\hbar K e^{-S_0/\hbar}$ and compare the two prefactor conventions.

Step 1. Coleman prefactor $K=\omega\sqrt{S_0/2\pi\hbar}=1\cdot\sqrt{6/2\pi}=\sqrt{0.9549}=\mathbf{0.977}$ .

Step 2. Splitting: $\Delta E=2\hbar K e^{-S_0/\hbar}=2(1)(0.977)(0.00248)=\mathbf{0.00485}$ (natural units).

Step 3. WKB-form comparison: $(\hbar\omega/\pi)e^{-S_0/\hbar}=(1/\pi)(0.00248)=0.000789$ . The two differ by the $O(1)$ prefactor ratio $2K/( \omega/\pi)=2(0.977)\pi\approx6.1$ , but share the identical exponent $e^{-6}$ — the robust, physical part of the answer. The prefactor is where the path integral's one-loop machinery earns its keep over the crudest WKB estimate.

Origins · 1 minsidney coleman and how this was discovered

sidney coleman

1937–2007 · American theoretical physicist at Harvard; his 1977 Erice lectures "The Uses of Instantons" made the Euclidean saddle-point method a standard tool

Coleman's 1977 Erice lectures, "The Uses of Instantons," systematized the double-well splitting, the dilute-instanton-gas sum, and — with Curtis Callan — the "bounce" solution that computes false-vacuum decay (Phys. Rev. D 16, 1762). The name "instanton," for a solution localized like a particle but in time as well as space (an "instant"), dates to this era and 't Hooft's parallel gauge-theory work. Coleman was famous at Harvard for lectures of unusual clarity; the instanton calculus he laid out is now indispensable in quantum field theory, cosmology (vacuum decay), and chemical-reaction rate theory.

History

1928

— Gamow, Gurney and Condon explain alpha decay by barrier tunnelling — the WKB exponent this method reproduces

1967

— Langer introduces the Euclidean "bounce" for metastable-state decay rates

1975

— Belavin, Polyakov, Schwartz and Tyupkin find the self-dual Yang-Mills instanton

1976

— 't Hooft computes instanton effects in gauge theory and resolves the U(1) problem

1977

— Coleman's "Uses of Instantons" and the Callan-Coleman bounce systematize the method

Going further

This article cashes the forward reference from WKB Tunneling and the Gamow Factor, which explicitly names the imaginary-time/instanton method as the field-theory generalization of the Gamow exponent, and it builds directly on Imaginary-Time and Euclidean Path Integrals (the Wick rotation and the inverted-potential equation of motion) and the small-oscillation analysis of the harmonic oscillator. The exponent it computes is the same $\int\sqrt{2mV}\,\mathrm dx$ derived by connection formulae in The WKB Approximation. Beyond quantum mechanics, the bounce solution flagged in the limiting-cases table opens onto false-vacuum decay and Yang-Mills instantons — the frontier this method was built for.

Provenance. Base article QUANT.RQM.INST.1097 (AllQFT "quantum-mechanics" track, Coleman ch. 7). Upgraded to v3 canon; the kink solution, the action $S_0=\tfrac23 m\omega a^2$ , its equality with the WKB barrier integral, and the splitting structure are verified in `verify_aq-qm-instantons-tunneling.py`. The base's action formula $S_0=\tfrac23 m\omega a^2$ was independently re-derived and confirmed.

Prerequisites. WKB Tunneling and the Gamow Factor · The Feynman Path Integral Formulation · Imaginary-Time and Euclidean Path Integrals · Gaussian functional integrals and determinants. Leads to. False vacuum decay (the bounce) · Theta vacua and Yang-Mills instantons · lattice field theory.

Why is the tunnelling "path" a solution of the equation of motion at all — isn't tunnelling nonclassical?

It is nonclassical in real time: there is no real-time classical trajectory through a barrier. But the exact statement of the WKB/saddle-point method is that the tunnelling amplitude is dominated by a saddle point of the action, and after Wick rotation that saddle is a genuine solution of the Euclidean equation of motion — the instanton. So "tunnelling is nonclassical" and "the instanton is a classical (Euclidean) solution" are both true and not in conflict: the classicality is in imaginary time, where the forbidden region has become allowed.

What exactly is the zero mode, and why does it matter?

The instanton can be centred at any imaginary time $\tau_0$ — shifting $\tau_0$ gives another solution of the same action. This one-parameter family is a "zero mode" of the fluctuation operator: a direction in which the action does not change, so its Gaussian integral would naively diverge. The fix is to trade the divergent Gaussian for an integral over the collective coordinate $\tau_0$ , which supplies a factor of the total imaginary-time interval $T$ (and a Jacobian $\sqrt{S_0/2\pi\hbar}$ ). That $T$ is exactly what makes the dilute-gas sum exponentiate into an energy shift. Handling the zero mode correctly is the technically delicate heart of the prefactor $K$ .

Does the instanton method give the same answer as WKB connection formulae?

For the exponent, always — both give $e^{-S_0/\hbar}$ with $S_0=\int\sqrt{2mV}\,\mathrm dx$ . For the prefactor they can differ by an $O(1)$ factor depending on how carefully each is done. The crude WKB estimate gives $\Delta E\approx(\hbar\omega/\pi)e^{-S_0/\hbar}$ ; the one-loop instanton calculation gives $\Delta E=2\hbar\omega\sqrt{S_0/2\pi\hbar}\,e^{-S_0/\hbar}$ . A careful WKB treatment with the correct connection-formula matching reproduces the instanton prefactor. The instanton method's advantage is that it generalizes systematically to field theory and higher loops, where connection formulae have no analogue.

What is the difference between an instanton and a bounce?

An instanton connects two degenerate minima (like the symmetric double well): it produces a level splitting. A bounce lives in a potential with a metastable minimum and a lower-lying (true) vacuum: the Euclidean solution rolls out of the false minimum, bounces off the far turning point, and returns — it produces a decay rate, not a splitting, and its fluctuation operator has a negative mode signalling instability. Coleman's double-well instanton and the Callan-Coleman bounce are the two faces of the same Euclidean saddle-point method; which one you get depends on whether the wells are degenerate.

Why does the dilute-gas approximation require the instantons to be "well separated"?

Each instanton has a width $\sim1/\omega$ in imaginary time. The dilute-gas sum treats a configuration of $n$ instantons as $n$ independent events, which is valid only if their typical separation $T/n$ is much larger than $1/\omega$ — i.e. if the gas is dilute. Because each instanton is weighted by $e^{-S_0/\hbar}\ll1$ , configurations with many instantons are exponentially rare, so the gas is dilute whenever the barrier is high ( $S_0/\hbar\gg1$ ), and the approximation is self-consistent exactly in the regime where tunnelling is a small effect. Instanton-anti-instanton pairs at short separation are a known correction (the "instanton-anti-instanton interaction").

How can an exponentially small ΔE ever be measured?

In several ways. Ammonia (NH₃) is the textbook realization: the nitrogen tunnels through the plane of the three hydrogens between two equivalent pyramidal configurations, and the tunnelling splitting is the famous 24 GHz inversion line used in the first maser. Molecular double wells, hydrogen-bonded systems, and superconducting flux qubits (where the two wells are two circulating-current states) all show measurable tunnelling splittings. The exponential smallness is exactly why these splittings are so sensitive to barrier height — a superb probe of the potential — and why they can be engineered over many orders of magnitude in qubit design.