Instantons, Fermion Zero Modes, and Tunneling

Instanton calculus begins with a finite-action Euclidean saddle and ends, when it succeeds, with a contribution to a specified amplitude or correlation function. Between those points lie boundary conditions, topology, gauge fixing, collective coordinates, determinants, fermion saturation, renormalization, and integration-range tests. This chapter develops that complete chain and marks the places where a correct saddle does not imply a controlled physical conclusion.

Helpful background. Wick rotation and analytic continuation supplies the Euclidean continuation, while theta terms, periodicity, and vacuum sectors supplies the sector phase and global periodicity.

Enter the chapter

Choose the route by the unresolved step in your calculation.

If your question is…	Read…	Result you should be able to use
Which Euclidean boundary conditions compute tunneling, a trace, or decay?	Euclidean Tunneling Saddles and Boundary Conditions	Distinguish heteroclinic instantons, periodic saddles, gauge instantons, and false-vacuum bounces
How does one-dimensional tunneling split a nearly degenerate pair?	Quantum-Mechanical Instantons and Tunnel Splitting	Derive the normalized instanton, determinant prefactor, dilute sum, and $\Delta E$
How does a four-dimensional gauge field realize $Q=1$?	Gauge Instantons, Topological Charge, and Moduli	Verify BPST self-duality, charge, action, boundary behavior, and moduli
How does a classical family become a one-instanton density?	Instanton Measures, Zero Modes, and Determinants	Assemble the gauge quotient, Jacobians, determinants, running factor, and normalization checks
When does a dilute gas produce a cosine in theta?	Dilute Instanton Ensembles and Theta Dependence	Exponentiate independent events and derive the susceptibility and failure conditions
Which fermion correlators can an instanton contribute to?	Fermion Zero Modes, Index Data, and Selection Rules	Turn an index into normalizable modes, Grassmann saturation, and a flavor vertex
Does the size integral remain in weak coupling?	The Instanton Size Modulus and Infrared Limitations	Derive every $\rho$ power, test both endpoints, and identify loss of control
When can a charge-one caloron have fractional constituents?	Fractional Events, Calorons, and Monopole Constituents	Reconstruct charge, action, and magnetic neutrality from declared circle holonomy

The main dependency chain is

$$\begin{gathered} \text{observable and boundary data} \longrightarrow \text{finite-action saddle} \longrightarrow \text{topology and moduli},\\ \text{gauge quotient} \longrightarrow \text{zero-mode Jacobians and determinants} \longrightarrow \text{renormalized one-event measure},\\ \text{zero-mode saturation and integration-range tests} \longrightarrow \text{bounded contribution to an observable}. \end{gathered}$$

The arrows do not reverse. A unit topological charge does not guarantee a finite size integral; an allowed fermion vertex does not prove a condensate; and periodic theta dependence does not prove that a dilute gas describes a strongly coupled vacuum.

Shared conventions

The gauge-instanton pages use oriented Euclidean four-space with $\epsilon_{1234}=+1$, Hermitian generators normalized by

$$\operatorname{tr}(T^aT^b)=\frac12\delta^{ab},$$

and a mathematical connection $\mathcal A=gA$, so the coupling sits outside the action:

$$S_E=\frac{1}{4g^2}\int d^4x\, \mathcal F_{\mu\nu}^a\mathcal F_{\mu\nu}^a, \qquad Q=\frac{1}{32\pi^2}\int d^4x\, \mathcal F_{\mu\nu}^a\widetilde{\mathcal F}_{\mu\nu}^a.$$

A self-dual $Q=+1$ BPST instanton therefore has

$$S_I=\frac{8\pi^2}{g^2}.$$

For fermions,

$$\gamma_5\psi_L=-\psi_L, \qquad \operatorname{ind}\mathcal D_R =n_L-n_R=2T(R)Q.$$

Thus a self-dual $SU(2)$ instanton has one left-handed fundamental zero mode or four left-handed adjoint zero modes. Anti-instantons reverse the chirality.

The quantum-mechanical pages use the fixed dimensionless double well

$$S_E[x] =\frac1g\int d\tau \left[\frac12\dot x^2+\frac12(x^2-1)^2\right],$$

for which

$$x_I=\tanh(\tau-\tau_0), \qquad S_I=\frac{4}{3g}, \qquad \Delta E_{\rm 1-loop} =8\sqrt{\frac{2}{\pi g}}e^{-4/(3g)}.$$

Keeping these normalizations fixed makes the conceptual comparison meaningful without conflating the one-dimensional and gauge-theory measures.

Three distinct control questions

An instanton calculation should answer three independent questions.

Is the local saddle expansion controlled? The action must be large in the expansion parameter, and the gauge-fixed nonzero-mode operator must be well defined. Zero and negative modes require their own treatments.

Are the collective integrals controlled? The size, position, orientation, and quasi-zero-mode ranges must stay within the regime in which the saddle and determinants were derived. For four-dimensional asymptotically free Yang–Mills theory, the BPST size integral commonly reaches $\rho\Lambda\sim1$ and loses weak-coupling control.

Is the multi-event ensemble controlled? A dilute sum requires small overlap or packing fraction. A large total event number in a large volume is compatible with diluteness; large local overlap is not.

Passing one test does not imply the other two. Self-duality controls the classical action in fixed $Q$, but not the size integral. A convergent Gaussian determinant does not saturate fermion zero modes. A calculable one-event density does not make its gas dilute.

One instanton, several outputs

The same formal weight can enter different observables only after the boundary and zero-mode data are changed appropriately:

• In a symmetric double well, a heteroclinic event changes wells. Alternating events and parity projection produce a real level splitting.
• A false-vacuum bounce begins and ends at the metastable vacuum. Its one physical negative mode produces an imaginary part and a decay rate.
• A BPST instanton contributes to a fixed topological sector or to theta-weighted correlation functions. Fermion insertions or masses must saturate every chiral zero mode.
• On $\mathbb R^3\times S^1$, nontrivial holonomy can resolve a caloron into monopole constituents, as constructed by Kraan and van Baal 1998, pp. 627–659. Their fractional charges belong to that compactified boundary-value problem.

These distinctions trace back to the original gauge solution of Belavin, Polyakov, Schwartz, and Tyupkin 1975, pp. 85–87, the determinant and fermion analysis of ‘t Hooft 1976, pp. 3432–3450, and the metastable-decay construction of Callan and Coleman 1977, pp. 1762–1768.

Review the chapter

Why does self-duality give $S_I=8\pi^2/g^2$ for the displayed normalization?

Solution

For a self-dual field, $\widetilde{\mathcal F}=\mathcal F$, so

$$Q =\frac{1}{32\pi^2} \int d^4x\,\mathcal F_{\mu\nu}^a\mathcal F_{\mu\nu}^a.$$

With $Q=1$, the integral is $32\pi^2$. Multiplying by the action coefficient $1/(4g^2)$ gives $8\pi^2/g^2$.

Why is a translation zero mode omitted from a determinant while a bounce negative mode is not simply omitted?

Solution

The zero mode is tangent to an exact family of equal-action solutions. Its Gaussian amplitude is replaced by integration over the collective coordinate with a Jacobian. A negative mode is a direction along which the real quadratic form decreases; it signals that the original real contour is not a convergent steepest-descent contour. Its treatment requires analytic continuation or contour deformation and produces the phase associated with decay.

For pure $SU(3)$, derive the explicit power of $\rho$ in the one-loop vacuum density and interpret it.

Solution

The coefficient is $b_0=11$, so

$$\rho^{-5}(\mu\rho)^{b_0} \propto\rho^6.$$

The integral converges at $\rho=0$ but grows toward large $\rho$. It becomes sensitive to $\rho\Lambda\sim1$, where the one-loop instanton measure is not controlled; the growth is a limitation, not a prediction of an infrared phase.

For $N_f$ massless fundamental Dirac fermions in a $Q=+1$ instanton, why does the vacuum amplitude vanish while a $2N_f$-fermion correlator need not?

Solution

Each flavor supplies one $\psi_L$ and one $\bar\psi_R$ Grassmann zero-mode coefficient. Integrating the vacuum integrand leaves those variables unsaturated and gives zero. A correlator with the flavor-determinant combination contains every coefficient once, so its Grassmann integral can be nonzero. This selection rule does not fix the remaining size integral or prove a condensate.

An $SU(N)$ caloron has center-symmetric holonomy. Verify charge and action recombination.

Solution

Center symmetry gives $N$ gaps $\nu_i=1/N$. Each constituent has

$$Q_i=\frac1N, \qquad S_i=\frac{8\pi^2}{g^2N}.$$

Summing over all $N$ constituents gives $Q=1$ and $S=8\pi^2/g^2$. Their simple and affine magnetic co-roots sum to zero.

References

• Belavin, Alexander A., Alexander M. Polyakov, Albert S. Schwartz, and Yuri S. Tyupkin. “Pseudoparticle Solutions of the Yang–Mills Equations.” Physics Letters B 59 (1975): 85–87. DOI.
• Callan, Curtis G., Jr., and Sidney Coleman. “Fate of the False Vacuum. II. First Quantum Corrections.” Physical Review D 16 (1977): 1762–1768. DOI.
• Kraan, Thomas C., and Pierre van Baal. “Periodic Instantons with Non-Trivial Holonomy.” Nuclear Physics B 533 (1998): 627–659. DOI.
• ‘t Hooft, Gerard. “Computation of the Quantum Effects Due to a Four-Dimensional Pseudoparticle.” Physical Review D 14 (1976): 3432–3450; erratum 18 (1978): 2199. DOI.