Ambiguity Cancellation and Transseries Consistency

For a real observable, an imaginary ambiguity created by choosing a lateral Borel contour must cancel against an equal and opposite ambiguity in another sector. This cancellation tests the normalization, Stokes data, and sector content at the stated exponential order. It does not prove that every sector has been found, that the resummation converges numerically, or that the full QFT exists.

Required background. Instantons and large-order relations supplies the neighboring-sector normalization and the symmetric-double-well action.

Helpful background. Complex saddles, Lefschetz thimbles, and integration cycles explains why a contour can fix the physically admissible combination.

The cancellation condition

Let a perturbative sector have lateral sums

$$\mathcal S_{0^\pm}\Phi_0(g) =P(g)\pm i\,C(g)e^{-A/g}, \qquad g>0,$$

where $P$ and $C$ are real in the stated regime. Suppose an independently derived nonperturbative sector has matched boundary values

$$\mathcal N_\pm(g) =N(g)\mp i\,C(g)e^{-A/g}.$$

Then

$$\mathcal O_\pm(g) \equiv\mathcal S_{0^\pm}\Phi_0(g)+\mathcal N_\pm(g) =P(g)+N(g),$$

and

$$\mathcal O_+(g)-\mathcal O_-(g)=0$$

at that exponential order. The signs are not assigned by taste: the same analytic continuation must define both terms. If one reverses the discontinuity convention, both signs reverse.

In a transseries

$$\mathcal O(g;\sigma) =\mathcal S\Phi_0(g) +\sigma e^{-A/g}\mathcal S\Phi_1(g)+\cdots,$$

the Stokes jump of $\Phi_0$ is canceled by a compensating jump of $\sigma$. Reality may select a median resummation. For a single conjugate pair at leading order, its perturbative piece resembles the average

$$\frac12\left( \mathcal S_{0^+}\Phi_0+\mathcal S_{0^-}\Phi_0 \right),$$

but the general median operation includes half of the full Stokes automorphism and acts on every coupled sector. It is therefore more than a universal principal-value rule.

The symmetric double well at pair order

Use the fixed action

$$S_E[x]=\frac1g\int\mathrm d\tau \left[\frac12\dot x^2+\frac12(x^2-1)^2\right], \qquad \mathcal S_I=\frac43.$$

The one-instanton sector has weight $e^{-4/(3g)}$ and splits the even and odd energies. The perturbative series common to both wells returns to its original endpoint only after an instanton–anti-instanton pair, whose weight is

$$e^{-2\mathcal S_I/g}=e^{-8/(3g)}.$$

The pair separation $R$ is a quasi-zero mode. At large $R$, the attractive interaction makes the schematic integral

$$\int_0^\infty \mathrm dR\, \exp\!\left[\frac{c}{g}e^{-2R}+\cdots\right], \qquad c>0,$$

ill-defined if expanded naively at positive $g$. The Bogomolny–Zinn-Justin prescription first evaluates the repulsive continuation $g\to-g$, subtracts the uncorrelated large-$R$ contribution, and analytically continues back above or below the positive axis. The logarithm generated by the separation integral then has two boundary values,

$$\log(-g)=\log g\pm i\pi.$$

For the dimensionless spectral quantity $\mathcal E=E_{\mathrm{can}}/g$ fixed on the preceding page, translating the standard double-well result gives the leading magnitudes

$$\begin{aligned} \operatorname{Im}\mathcal E_{\mathrm{pert},\pm}(g) &=\mp\frac{16}{g}e^{-8/(3g)} \left[1+O(g)\right],\\ \operatorname{Im}\mathcal E_{I\bar I,\pm}(g) &=\pm\frac{16}{g}e^{-8/(3g)} \left[1+O(g)\right]. \end{aligned}$$

Their sum vanishes. The coefficient also reproduces the late perturbative growth

$$a_n\sim-\frac6\pi \frac{\Gamma(n+1)}{(8/3)^n},$$

so one calculation checks the action, sign, power of $g$, and normalization. Bogomolny introduced the correlated-pair continuation in Bogomolny 1980, pp. 431–435; Zinn-Justin developed the multi-instanton sectors and their logarithms in Zinn-Justin 1981, pp. 125–140. The convention translation is detailed in Mariño 2015, §3.5, pp. 102–104.

The calculation does not say that the real part of the pair amplitude is determined by cancellation alone. It comes from the prescribed quasi-zero-mode integral and higher fluctuations.

A reusable consistency test

For any proposed cancellation, record:

1. the observable and its reality or boundary-value condition;
2. the coefficient, Borel, and lateral conventions;
3. the exponential action $A$ and algebraic power of $g$;
4. the sector that carries the compensating ambiguity;
5. the contour or analytic continuation defining that sector;
6. the order through which coefficients have been computed; and
7. an independent check, such as late coefficients or exact spectral data.

Then compare the complete ambiguous terms, not only their exponentials:

$$\operatorname{Disc}\mathcal S\Phi_0 +\operatorname{Disc}\mathcal N =0.$$

A mismatch in the power of $g$ usually signals a missing zero mode or determinant factor. A sign mismatch often signals inconsistent contour orientations. A residue mismatch can reveal an omitted sector or a normalization error.

The Borel and transseries map shows where this test sits in the larger chain. The exact and rigorous status comparison prevents internal cancellation from being labeled a QFT construction.

What cancellation proves

Successful cancellation establishes that the included sectors can represent a prescription-independent answer through the checked order. Together with large-order data, it can strongly test the proposed resurgent relation.

It does not establish:

• that omitted exponentials or logarithmic sectors are absent;
• that the real parts are accurate;
• that the transseries sums to the intended global solution;
• that the infinite-volume or continuum limit exists; or
• that the same mechanism applies to a different observable.

A residual ambiguity may therefore be a diagnostic, but a vanishing residual is not a completeness theorem.

Common pitfalls

Canceling magnitudes without conventions. Equal absolute values are insufficient. The upper/lower orientation, sector normalization, and analytic continuation must be the same on both sides.

Calling the arithmetic average universally physical. A median prescription is justified by the full Stokes action and boundary data. In a metastable problem, a selected lateral imaginary part may instead be physical.

Using cancellation to infer the real nonperturbative coefficient. Cancellation fixes the ambiguous part. The real part requires an independent sector calculation or global condition.

Exercises

1. Suppose

$$\mathcal S_\pm\Phi_0=P\pm iC g^{-2}e^{-A/g}, \qquad \mathcal N_\pm=N\mp iD g^{-2}e^{-A/g}.$$

What does prescription independence require?

Solution

The total discontinuity is $2i(C-D)g^{-2}e^{-A/g}$, so cancellation requires $C=D$ in the same normalization. Matching only $A$ and the power $g^{-2}$ is not enough.

2. Translate a pair ambiguity of magnitude $g_M^{-1}e^{-1/(3g_M)}$ under $g_M=g/8$ and $E=2E_M$.

Solution

Multiplying the energy by two and substituting $g_M=g/8$ gives

$$2\frac1{g_M}e^{-1/(3g_M)} =\frac{16}{g}e^{-8/(3g)}.$$

The two lateral signs are opposite, and the perturbative ambiguity has the negative of the pair ambiguity.

References

• Bogomolny, E. B. “Calculation of Instanton–Anti-Instanton Contributions in Quantum Mechanics.” Physics Letters B 91 (1980): 431–435. doi:10.1016/0370-2693(80)91014-X.
• Mariño, Marcos. Instantons and Large N: An Introduction to Non-Perturbative Methods in Quantum Field Theory. Cambridge University Press, 2015. doi:10.1017/CBO9781107705968.
• Zinn-Justin, Jean. “Multi-Instanton Contributions in Quantum Mechanics.” Nuclear Physics B 192 (1981): 125–140. doi:10.1016/0550-3213(81)90197-8.