Asymptotic Series and Nonperturbative Corrections

A perturbative calculation can be accurate even when its infinite series diverges. The apparent paradox disappears once two limits are kept separate. An asymptotic expansion controls the error as a parameter $g$ tends to zero at each fixed truncation order. Convergence would instead control the limit of the partial sums as the truncation order tends to infinity at fixed nonzero $g$.

In many quantum problems, perturbative coefficients eventually grow like $n!$. The terms first decrease, reach a least term near an order of size $1/g$, and then increase. The most accurate raw perturbative approximation is usually obtained by stopping near that least term. The error of this optimally truncated raw series is often exponentially small,

$$\text{least-term scale} \sim e^{-A/g},$$

which is also the characteristic size of tunneling and saddle contributions.

Required background. Asymptotic analysis defines the limiting notation, higher-order perturbation theory supplies the coefficients, and instantons explains exponential saddle contributions. Resurgence Preview develops Borel transforms and relations between sectors.

The plotted points are $\log_{10}(n!g^n)$ at $g=0.1$. The two equal least terms occur at $n=9,10$ because the consecutive-term ratio is $(n+1)g$. This identifies a sensible truncation region; an actual error bound requires information about the function represented by the series.

Convergent and Asymptotic Expansions

Suppose a quantity has a formal expansion

$$F(g) \sim \sum_{n=0}^{\infty} a_n g^n, \qquad g\to0^+.$$

The asymptotic symbol means that, for every fixed nonnegative integer $N$,

$$F(g) - \sum_{n=0}^{N} a_n g^n = o(g^N) \qquad \text{as }g\to0^+.$$

For an expansion available to every order, applying the definition at order $N+1$ also implies

$$F(g) - \sum_{n=0}^{N} a_n g^n = O(g^{N+1}),$$

but the constant hidden in the $O$ symbol may grow rapidly with $N$. Nothing in either statement requires

$$\lim_{N\to\infty} \sum_{n=0}^{N} a_n g^n$$

to exist at fixed $g\ne0$.

A convergent Taylor series and an asymptotic series therefore answer different questions.

Property	Convergent power series	Asymptotic power series
Limit being controlled	$N\to\infty$ at fixed $g$	$g\to0$ at fixed $N$
Effect of adding terms	The error tends to zero inside the convergence disk, though not necessarily monotonically	Decreasing terms often improve the sum; the actual error still needs a remainder estimate
Remainder	Tends to zero as $N\to\infty$	Often smallest near a finite, $g$-dependent order
Uniqueness	Determines its sum in the convergence disk	Does not determine exponentially small additions by itself

The last row is the gateway to nonperturbative physics. Two functions that differ by $e^{-A/g}$ have the same power-series asymptotic expansion as $g\to0^+$. Local power-series data alone cannot decide which function a physical boundary-value problem selects.

An Exactly Solvable Model

Consider, for $g\gt0$,

$$I(g) = \int_0^\infty \frac{e^{-x}}{1+gx} \,dx.$$

This integral is finite and unambiguous. The finite geometric identity

$$\frac{1}{1+gx} = \sum_{n=0}^{N} (-gx)^n + \frac{(-gx)^{N+1}}{1+gx}$$

gives an exact decomposition after integration:

$$I(g) = \sum_{n=0}^{N} (-1)^n n!g^n + R_N(g),$$

where

$$R_N(g) = (-g)^{N+1} \int_0^\infty \frac{e^{-x}x^{N+1}}{1+gx} \,dx.$$

For each fixed $N$,

$$\lvert R_N(g)\rvert \lt (N+1)!g^{N+1},$$

so

$$I(g) \sim \sum_{n=0}^{\infty} (-1)^n n!g^n.$$

The expansion is asymptotic. It cannot converge for any nonzero $g$, because the ratio of consecutive term magnitudes is

$$\frac{(n+1)!g^{n+1}}{n!g^n} = (n+1)g,$$

which eventually exceeds one no matter how small $g$ is.

The exact answer can also be written

$$I(g) = \frac{e^{1/g}}{g} E_1\left(\frac{1}{g}\right),$$

where

$$E_1(z) = \int_z^\infty \frac{e^{-t}}{t} \,dt.$$

Thus a perfectly smooth, exactly defined function can possess a divergent weak-coupling expansion. Divergence is not evidence that the function or the original problem is ill-defined.

At $g=0.1$, the term magnitudes are

$$1,\; 0.1,\; 0.02,\; 0.006,\; 0.0024,\; \ldots$$

and reach their minimum near $n=9$ or $10$. The first several partial sums improve rapidly. Summing indefinitely would eventually destroy that accuracy.

What the model does and does not prove

Because the model series alternates and the exact remainder has a definite sign, its first omitted term supplies a useful bound. A general quantum perturbation series need not have either property. Its coefficients may have several competing large-order contributions, complex phases, logarithms, or singularities on different rays of the coupling plane.

The durable lesson is narrower:

1. factorial coefficient growth is compatible with an accurate low-order approximation;
2. the useful truncation order depends on the numerical value of the small parameter;
3. the smallest attainable raw perturbative error can be exponentially small.

Factorial Growth and Optimal Truncation

Assume the large-order coefficients behave as

$$a_n \sim C A^{-n} \Gamma(n+\beta), \qquad A\gt0.$$

The $n$th term is $T_n(g)=a_ng^n$. At large $n$,

$$\left| \frac{T_{n+1}(g)} {T_n(g)} \right| \sim \frac{g}{A} (n+\beta).$$

The terms stop decreasing when this ratio reaches one, so the optimal order is approximately

$$N_* \simeq \frac{A}{g} -\beta.$$

Since $N_*$ must be an integer, one compares the neighboring terms and truncates at the smaller one. The formula is an asymptotic estimate, not a command to retain a fractional number of terms.

To estimate the least term, set $z=N_*+\beta\simeq A/g$ and use Stirling’s formula:

$$\Gamma(z) \sim \sqrt{2\pi}\, z^{z-\frac12} e^{-z}.$$

Then

$$\begin{aligned} \lvert T_{N_*}(g)\rvert &\sim \lvert C\rvert A^{-N_*} \Gamma(N_*+\beta) g^{N_*} \\ &\sim \lvert C\rvert \sqrt{2\pi} \left( \frac{A}{g} \right)^{\beta-\frac12} e^{-A/g}. \end{aligned}$$

The power multiplying the exponential depends on the large-order index $\beta$, but the decisive scale is $e^{-A/g}$.

Why the best order moves

As $g$ decreases, $N_*$ increases. More perturbative information becomes useful because the terms continue decreasing for longer. At a larger coupling, the asymptotic tail arrives earlier. A statement such as “fourth order is enough” has no invariant meaning unless the parameter range and desired accuracy are specified.

This also explains an apparently surprising numerical pattern. At fixed $g$, the sequence of errors often falls, flattens near the least term, and rises. At fixed $N$, however, the error still improves as $g\to0$. The first pattern diagnoses asymptotic divergence in truncation order; the second is exactly what an asymptotic expansion promises.

The least term as an error estimate

For many one-saddle, sign-regular problems, the least term gives the right order of magnitude for the optimally truncated remainder. It is not a universal theorem. A reliable use of this estimate requires checking:

• whether the calculated coefficients have entered their large-order regime;
• whether cancellation makes the last retained term atypically small;
• whether another saddle has a smaller action;
• whether numerical, basis-truncation, or roundoff errors are already larger;
• whether the physical observable contains a separate nonperturbative sector of comparable size.

The least term is a scale diagnosis. It does not by itself determine the coefficient, sign, or phase of the missing contribution.

Beyond Every Power

For fixed $A\gt0$, define

$$f(g) = \begin{cases} e^{-A/g}, & g\gt0,\\ 0, & g=0. \end{cases}$$

Repeated differentiation gives

$$f^{(k)}(g) = P_k\left(\frac{1}{g}\right) e^{-A/g},$$

where $P_k$ is a polynomial. Because the exponential defeats every inverse power,

$$\lim_{g\to0^+} g^{-m}e^{-A/g} = 0$$

for every fixed $m$, all right derivatives of $f$ at the origin vanish. Its Taylor series is identically zero, although $f(g)$ is nonzero for every $g\gt0$.

The fixed positive $A$ is essential: if the action tends to zero together with $g$, this beyond-all-orders conclusion need not hold. In semiclassical notation, $e^{-S/\hbar}$ is smaller than every power of $\hbar$ as $\hbar\to0^+$ at fixed $S>0$. A finite power expansion in that limit cannot recover it, although a different approximation can. “Nonperturbative” therefore identifies a limitation of the chosen expansion, not an absence of controlled methods.

Consequently,

$$F_1(g) \sim \sum_{n=0}^{\infty}a_ng^n$$

does not distinguish $F_1$ from

$$F_2(g) = F_1(g) + Cg^\alpha e^{-A/g}.$$

They share the same expansion in integer powers of $g$ to every algebraic order. The boundary conditions, integration contour, global potential, or spectral problem must supply the missing information.

The coincidence between the optimally truncated error scale and an exponential saddle scale is profound but should be stated carefully. It does not mean that every remainder is literally an instanton, or that inspecting a few coefficients identifies a unique saddle. It means that factorial divergence makes ordinary perturbation theory sensitive, at large order, to information of the same exponential size as other saddle sectors. The precise relation belongs to Borel and resurgent analysis.

Why Local Perturbation Misses Tunneling

Consider a symmetric double well in a semiclassical family with small dimensionless parameter $g$. Perturbation theory about the left minimum constructs an intrawell energy

$$E_{\mathrm{loc}}(g) \sim \sum_{n=0}^{\infty} a_n g^n.$$

Reflection symmetry gives the same local series about the right minimum. Exact low-energy eigenstates, however, have definite parity. In the deep-well regime their energies have the schematic form

$$\begin{aligned} E_{\mathrm{even}}(g) &\sim E_{\mathrm{loc}}(g) - C(g)e^{-S_I/g}, \\ E_{\mathrm{odd}}(g) &\sim E_{\mathrm{loc}}(g) + C(g)e^{-S_I/g}, \end{aligned}$$

where $S_I$ is the dimensionless instanton action and $C(g)$ contains a power of $g$ and a fluctuation expansion.

Both parity levels have the same ordinary power-series expansion. Subtracting them gives

$$\Delta E = E_{\mathrm{odd}} - E_{\mathrm{even}} \sim 2C(g)e^{-S_I/g},$$

while every coefficient of the formal power series for $\Delta E$ is zero. No finite perturbative order around one minimum can create communication with the other minimum.

This is a global statement, not a failure of algebra. The local recursion knows the potential and wavefunction near one classical configuration. The splitting depends on propagation through the entire forbidden region and on the parity boundary condition connecting both wells. Tunneling Splittings develops the quantitative relation among this exponential, the two-state coupling, WKB flux, instanton sums, and exact spectral doublets.

Mean energy and splitting are different observables

Define the center of the doublet,

$$\bar E = \frac12 \left( E_{\mathrm{even}} + E_{\mathrm{odd}} \right).$$

The leading one-instanton shifts cancel in $\bar E$, so local perturbation theory can approximate the mean energy well while predicting no splitting at all. A numerically excellent answer for $\bar E$ is therefore not evidence that the same method resolves $\Delta E$.

This distinction recurs throughout semiclassical physics. A nonperturbative effect may be tiny compared with an absolute energy but be the leading contribution to a carefully chosen difference, transition amplitude, or decay width.

Large Order Encodes Global Instability

The connection between factorial coefficients and nonperturbative saddles can be previewed without developing the full Borel formalism. Suppose an energy $E(g)$ is analytic in the complex $g$ plane except for a cut along negative coupling. A dispersion relation can express its perturbative coefficients through the discontinuity across that cut. For the decaying resonance continued below the negative-coupling cut, write $E_{\mathrm{dec}}(-s)=E(-s-i0)$, with negative imaginary part. The upper-edge value is its complex conjugate for a real Hamiltonian family. With this convention,

$$a_n \sim \frac{(-1)^{n+1}}{\pi} \int_0^\infty \frac{ -\operatorname{Im}E(-s-i0) }{s^{n+1}} \,ds.$$

If the analytically continued problem is metastable and

$$\begin{aligned} -\operatorname{Im}E(-s-i0) &\sim D s^{-\beta}e^{-A/s}, \\ s&\to0^+. \end{aligned}$$

then changing variables to $u=A/s$ gives

$$\begin{aligned} \lvert a_n\rvert &\sim \frac{D}{\pi} \int_0^\infty s^{-n-\beta-1} e^{-A/s} \,ds \\ &= \frac{D}{\pi} A^{-n-\beta} \Gamma(n+\beta). \end{aligned}$$

A decay saddle at negative coupling has produced factorial growth at positive coupling. The alternating sign follows from the negative-axis cut in this convention.

This calculation is schematic. Subtractions may be required in a full dispersion relation, the normalization of $A$, $D$, and $\beta$ depends on the Hamiltonian, and several saddles can contribute. Its conceptual content is robust: late perturbative coefficients can remember global analytic structure far from the stable expansion point.

The Quartic Oscillator

In units $\hbar=m=\omega=1$, consider

$$H(g) = \frac{p^2}{2} + \frac{x^2}{2} + gx^4, \qquad g\gt0.$$

The ground-state weak-coupling expansion begins

$$E_0(g) \sim{} \frac12 + \frac34g - \frac{21}{8}g^2 + \frac{333}{16}g^3 - \frac{30885}{128}g^4 + \cdots.$$

For positive $g$, the potential is confining and the exact spectrum is real and discrete. Nevertheless, the Rayleigh–Schrödinger series diverges factorially. At weak coupling its early alternating terms give excellent approximations, and rigorous results establish an appropriate Borel summability of the stable oscillator’s energy levels. Divergence and physical ambiguity are therefore not synonyms.

Why does the stable problem have large coefficients? Continue to $g=-s\lt0$:

$$V(x) = \frac{x^2}{2} - sx^4.$$

The origin becomes a metastable well and the potential falls without bound at large $\lvert x\rvert$. An outgoing resonance acquires an imaginary part controlled semiclassically by barrier escape. Through the dispersion mechanism above, that exponentially small imaginary part determines the factorial growth of the coefficients in the stable positive-$g$ expansion.

This example sharpens three separate claims:

1. Stable positive coupling: the Hamiltonian is well-defined and the divergent series is summable by suitable methods.
2. Negative-coupling continuation: the local state is metastable and its imaginary part contains nonperturbative information.
3. Large-order relation: the instability of the continued problem controls the late coefficients of the stable problem.

The Anharmonic Oscillator compares low-order perturbative, variational, and numerical calculations. Here the model illustrates a different question: it explains how a physical saddle outside the stable real-coupling problem leaves a measurable signature in perturbative large order.

Exercises

1. Exact upper and lower estimates

For the integral $I(g)$ above, use the sign of $R_N$ to obtain upper and lower bounds from successive partial sums. Does this prove the same bounds for an arbitrary factorial series?

Solution

Writing $S_N=\sum_{n=0}^N(-1)^n n!g^n$, the integral remainder has sign $(-1)^{N+1}$. Thus

$$S_{2k+1}<I(g)<S_{2k}.$$

Their separation is $(2k+1)!g^{2k+1}$, which decreases only up to the least-term region. The rigorous bound comes from the exact positive remainder integral, not from factorial growth alone. Other asymptotic series can have additional exponentially small contributions and lack this bracketing property.

2. The quartic escape action

For $V(x)=x^2/2-sx^4$, $s>0$, find the positive zero-energy escape turning point and the round-trip bounce action. Infer the exponential constant controlling large order.

Solution

The turning point is $x_t=(2s)^{-1/2}$. In the stated units,

$$B=2\int_0^{x_t}x\sqrt{1-2sx^2}\,dx=\frac{1}{3s}.$$

Therefore the decay exponential is $e^{-1/(3s)}$ and the large-order action constant is $A=1/3$ in this coupling convention. Two symmetric escape directions affect the prefactor, not this action. This result predicts the scale $N_*\sim1/(3g)$ for the stable quartic series, rather than the $1/g$ scale of the model integral.

3. A numerical stopping rule

For the model series

$$\sum_{n=0}^{\infty} (-1)^n n!g^n,$$

take $g=0.08$. Determine the least-term order and estimate its magnitude.

Solution

The ratio of neighboring magnitudes is

$$\frac{\lvert T_{n+1}\rvert} {\lvert T_n\rvert} = (n+1)g.$$

At $g=0.08$, this ratio crosses one when $n+1\simeq12.5$. Direct comparison therefore places the least term at $n=12$:

$$\lvert T_{12}\rvert = 12!(0.08)^{12} \simeq 3.29\times10^{-5}.$$

The Stirling estimate gives

$$\sqrt{\frac{2\pi}{g}} e^{-1/g} = \sqrt{\frac{2\pi}{0.08}} \times e^{-12.5} \simeq 3.30\times10^{-5}.$$

in excellent agreement. The raw series should be truncated near this order, not summed through arbitrarily large $n$.

References

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