Instantons, quantum periods, and differential equations

Can the coefficient of an instanton correction be recovered by solving an ordinary differential equation? We will match one coefficient in pure $SU(2)$ theory, including its deformation dependence. Then we will change the boundary parameter until the coefficient diverges—and show why the differential equation itself remains well behaved.

Begin with the one-instanton calculation and planning a reproducible research calculation. The new preparation is elementary spectral perturbation theory, developed below. Neither integrable systems nor two-dimensional conformal field theory is an entry requirement.

The parent theory is four-dimensional N=2 $SU(2)$ Yang–Mills, with eight real supercharges and no hypermultiplets. Its generic flat-space Coulomb vacuum has an Abelian rank-one low-energy theory. The equivariant calculation retains selected supersymmetry in a rotational background; it does not preserve the full flat-space Poincaré algebra. No hypermultiplet mass, shifted or physical, occurs in this example.

Two limits and a fixed dictionary

The Omega parameters $\epsilon_1,\epsilon_2$ have mass dimension one. They couple to rotations of the two coordinate planes of Euclidean four-space. Retain the instanton lesson’s conventions:

$$b_1=b=-i\alpha,\qquad b_2=-b,\qquad q=\frac{\Lambda^4}{4},\qquad E=\epsilon_1+\epsilon_2. \tag{D16.1}$$

Here $b$ is the equivariant Coulomb coordinate, $q$ is the dimension-four instanton parameter, and $\Lambda$ is the scale of the site’s first Seiberg–Witten curve. In the undeformed theory $\alpha=a/\sqrt2$, where $a$ is the electric period in the fixed charge basis. This rescaling belongs to the instanton-to-curve dictionary; it does not change the allowed particle or line charges.

Write $Z_{\rm inst}=1+qZ_1+q^2Z_2+\cdots$. The two one-box contributions are

$$Z_1= \frac{1}{\epsilon_1\epsilon_2(2b)(E-2b)} +\frac{1}{\epsilon_1\epsilon_2(-2b)(E+2b)} =\frac{2}{\epsilon_1\epsilon_2(E^2+4\alpha^2)}. \tag{D16.2}$$

The signs and finite scale factor were fixed by the previous contour calculation. In particular, copying the isolated one-instanton sign of Nekrasov’s original contour formula without its parameter translation would change this answer.

There are two distinct operations on the connected quantity $\log Z_{\rm inst}$:

$$\begin{aligned} \mathscr F_{\rm cl,inst} &=\lim_{\epsilon_1,\epsilon_2\to0} \epsilon_1\epsilon_2\log Z_{\rm inst},\\ \mathcal W_{\rm inst}(\alpha,\epsilon_1,q) &=\lim_{\epsilon_2\to0}\epsilon_2\log Z_{\rm inst},\\ \mathscr F_{\rm NS,inst}&=\epsilon_1\mathcal W_{\rm inst}. \end{aligned} \tag{D16.3}$$

We initially take these limits coefficient by coefficient in $q$, away from poles. The first gives the ordinary instanton prepotential in this normalization. The second is the Nekrasov–Shatashvili, or NS, limit: one rotational deformation remains. $\mathcal W$ has dimension one and $\mathscr F_{\rm NS}$ dimension two. Their names do not make them equal to the physical prepotential with its separate $1/(2\pi i)$ normalization. NS (3.6)–(3.8) fixes the logarithm and limit.

Since $\log Z_{\rm inst}=qZ_1+O(q^2)$,

$$\mathscr F_{\rm NS,inst} =\frac{2q}{4\alpha^2+\epsilon_1^2}+O(q^2), \qquad \mathscr F_{\rm cl,inst} =\frac{q}{2\alpha^2}+O(q^2). \tag{D16.4}$$

Keeping $\epsilon_1$ finite and simply renaming the first expression “the Seiberg–Witten answer” loses a physical deformation.

Why an energy involves a derivative of the logarithm

The spectral quantity to compare is a quadratic scalar observable, not the prepotential itself. Its relation to the logarithm is often called a deformed Matone relation. We can see its normalization directly.

In the NS fixed-point description, let $x^0_{u,j}=b_u+(j-1)\epsilon_1$ label the empty columns of the Young diagrams used in instanton counting. A column with many boxes has a finite rescaled length $\lambda_{u,j}$ as $\epsilon_2\to0$, and endpoint $x_{u,j}=x^0_{u,j}+\lambda_{u,j}$. The source input is

$$\begin{aligned} q\partial_q\mathcal W_{\rm inst}&=\sum_{u,j}\lambda_{u,j},\\ \left\langle\operatorname{Tr}_2\Phi_{\rm eq}^{\,2}\right\rangle &=\sum_u b_u^2-\sum_{u,j} \left[(x_{u,j}+\epsilon_1)^2 -(x^0_{u,j}+\epsilon_1)^2-x_{u,j}^2+(x^0_{u,j})^2\right]. \end{aligned} \tag{D16.5}$$

These are the box-counting and scalar-observable formulas in Poghossian (23), (29), (31), with $\mathcal W$ normalized as in (D16.3). The trace is the ordinary fundamental trace. The critical-column construction is a localization input; we do not derive its full saddle here.

Each square bracket is $2\epsilon_1\lambda_{u,j}$. Also $\sum b_u^2=-2\alpha^2$. Define the continued, normalized observable $u_{\rm NS}=-\langle\operatorname{Tr}_2\Phi_{\rm eq}^{\,2}\rangle/2$. Therefore

$$u_{\rm NS}=\alpha^2+q\partial_q\mathscr F_{\rm NS,inst}. \tag{D16.6}$$

The definition fixes the additive convention by $u_{\rm NS}(q=0)=\alpha^2$; no extra $\epsilon_1^2$ constant is inserted. At instanton order $q^k$, the derivative multiplies the connected coefficient by $k$. Thus equality of an energy coefficient and a prepotential coefficient at first order is special. At second order they differ by a factor of two.

Choose the differential equation and its boundary condition

For a self-adjoint circle problem, take $q>0$, $\hbar>0$, and the explicit analytic continuation

$$\epsilon_1=i\hbar,\qquad \alpha\in\mathbb R. \tag{D16.7}$$

$\hbar$ has mass dimension one; it is not a dimensionless copy of the usual laboratory Planck constant. Then the gauge calculation predicts $u_{\rm NS}=\alpha^2+2q/(4\alpha^2-\hbar^2)+O(q^2)$.

The inherited classical curve is $w+q/w=X^2-u$, with differential $X\,dw/w$. On $w=\sqrt q\,e^{i\theta}$ it becomes $X^2=u+2\sqrt q\cos\theta$, and the electric action is $(2\pi)^{-1}\int_0^{2\pi}X\,d\theta$. Quantize $X$ as $-i\hbar\partial_\theta$. Our chosen spectral problem is consequently

$$\left[-\hbar^2\frac{d^2}{d\theta^2} -2\sqrt q\cos\theta\right]\psi=u\psi, \qquad \begin{pmatrix}\psi\\\psi'\end{pmatrix}(\theta+2\pi) =e^{2\pi i\alpha/\hbar} \begin{pmatrix}\psi\\\psi'\end{pmatrix}(\theta). \tag{D16.8}$$

The boundary multiplier is a Floquet multiplier: a wave may return with a phase after one cell. Both conditions are needed to cancel the integration-by-parts boundary term and obtain a self-adjoint operator. The multiplier knows $\alpha$ only modulo $\hbar$; specifying the branch that approaches $u=\alpha^2$ as $q\to0$ also specifies which free mode is being continued.

The NS correspondence motivates this quantization; quantizing a classical curve alone does not prove the correspondence to all instanton orders. NS section 5.1.3 explicitly distinguishes the quasi-periodic Mathieu problem from the square-integrable modified-Mathieu problem on a line. Our circle boundary condition is the former. For comparison with Başar–Dunne (2.1), divide our equation by two and shift $\theta$ by $\pi$: their potential scale squared is $\sqrt q$, their eigenvalue is $u/2$, and their $\hbar$ is ours. No coefficient is compared before this translation.

Reproduce the one-instanton coefficient

Use orthonormal free waves $|n\rangle=(2\pi)^{-1/2}e^{i(\alpha/\hbar+n)\theta}$, $n\in\mathbb Z$. Their energies are $d_n=(\alpha+n\hbar)^2$. The potential connects only neighboring modes, with matrix element $-\sqrt q$. We continue $|0\rangle$.

Write $\psi=|0\rangle+c_1|1\rangle+c_{-1}|-1\rangle+\cdots$. To first order in $\sqrt q$, the equations for the neighboring coefficients give $c_{\pm1}=\sqrt q/(d_{\pm1}-d_0)$. The $n=0$ equation then gives

$$\begin{aligned} u-\alpha^2 &=-\sqrt q(c_1+c_{-1})+O(q^2)\\ &=q\left[\frac{1}{\alpha^2-(\alpha+\hbar)^2} +\frac{1}{\alpha^2-(\alpha-\hbar)^2}\right]+O(q^2)\\ &=\frac{2q}{4\alpha^2-\hbar^2}+O(q^2). \end{aligned} \tag{D16.9}$$

This is the promised match, including sign, denominator and scale. Odd powers of $\sqrt q$ vanish on an isolated branch because changing $\sqrt q$ to $-\sqrt q$ is the unitary translation $\theta\mapsto\theta+\pi$.

For ordinary nondegenerate perturbation theory require $2\alpha/\hbar\notin\mathbb Z$ and a small potential relative to the relevant free-level separations. The first displayed coefficient sees the resonances $2\alpha=\pm\hbar$; higher orders see further mode collisions.

As a numerical example, choose mass units with $\hbar=1$, $\alpha=1/3$, $q=10^{-4}$. The free value is $1/9$ and the first correction is $-3.6\times10^{-4}$. Direct diagonalization of the Fourier operator gives

$$u=0.11075141308533925847\ldots, \qquad \left|u-\left(\frac19-0.00036\right)\right| \simeq3.02\times10^{-7}. \tag{D16.10}$$

The negative shift is expected: this free mode lies below its coupled neighbors. It is not a negative instanton probability; the connected holomorphic coefficient is not a probability distribution.

A separate check integrates the original second-order ODE over one cell. Its fundamental matrix has determinant one because the Wronskian is constant. At the eigenvalue its trace must be $2\cos(2\pi\alpha/\hbar)$. This checks the boundary phase independently of the Fourier matrix. Successive step sizes give the expected fourth-order convergence; numerical precision alone is not an estimate of integration error.

Recover a classical period without changing the question

Set $\psi=\exp[(i/\hbar)\int P\,d\theta]$. Substitution into (D16.8) gives

$$P^2-i\hbar P'=u+2\sqrt q\cos\theta. \tag{D16.11}$$

This is a first-order nonlinear equation for the logarithmic derivative of the wave. A WKB expansion means solving it in powers of $\hbar$. Its leading term is $P_0=\sqrt{u+2\sqrt q\cos\theta}$, on a chosen branch. The next term is $P_1=iP_0'/(2P_0)$; its integral vanishes on a rotation contour with no zero or winding of $P_0$.

For $u>2\sqrt q$, expand the leading period:

$$\alpha_{\rm cl} =\frac1{2\pi}\int_0^{2\pi}P_0\,d\theta =\sqrt u\left(1-\frac{q}{4u^2} -\frac{15q^2}{64u^4}+\cdots\right). \tag{D16.12}$$

The linear cosine averages to zero; its square averages to $1/2$. Inversion begins with $u=\alpha_{\rm cl}^2+q/(2\alpha_{\rm cl}^2)+\cdots$, agreeing with (D16.9) as $\hbar\to0$ coefficient by coefficient.

At finite $\hbar$, the continued $\alpha$ is a quantum action/Floquet coordinate. It must not simply be substituted for the undeformed physical period in a four-dimensional BPS mass formula. Nor does the coefficientwise limit prove uniform convergence of a chosen eigenvalue branch: the conditions $2\alpha/\hbar\in\mathbb Z$ accumulate as $\hbar\to0$ at fixed $\alpha$.

A controlled variation: resolve the first resonance

Let $\alpha=\hbar/2+\delta$, with $|\delta|\ll\hbar$. The free modes $0$ and $-1$ are now close. Keeping their mixing exactly gives

$$H_{\rm pair}= \begin{pmatrix} (\hbar/2+\delta)^2&-\sqrt q\\ -\sqrt q&(-\hbar/2+\delta)^2 \end{pmatrix}, \qquad u_\pm^{\rm pair} =\frac{\hbar^2}{4}+\delta^2 \pm\sqrt{\hbar^2\delta^2+q}. \tag{D16.13}$$

The other free modes are separated by order $\hbar^2$. For $\sqrt q\ll\hbar^2$ their leading effect is of order $q/\hbar^2$. At exact resonance the leading gap is $2\sqrt q$, and both eigenvalues remain finite. The expansion in integer powers of $q$ has failed because it divided by the splitting of two states that had to be diagonalized together.

For $\hbar=1$, $q=10^{-4}$, the full Fourier problem gives a gap $0.0199995000153\ldots$, close to $0.02$. The midpoint is $0.2499499995834\ldots$, rather than exactly $1/4$: remote modes also contribute. Halving $\sqrt q$ tests both the leading gap and the correction scaling.

This is a bounded research-training extension, not a newly discovered spectrum. A useful deliverable is a table over the dimensionless ratio $\hbar\delta/\sqrt q$ at fixed $\hbar$ comparing (i) ordinary one-instanton order, (ii) the two-state approximation, and (iii) a converged ODE or Fourier calculation. Record the free gap, the cutoff residual and the numerical step error separately. Agreement only away from $\delta=0$ diagnoses nonuniform perturbation theory, rather than a failure of the differential equation or a new singularity of the undeformed Coulomb branch.

Exercises

1. Continue one more order without claiming two instantons

For the same circle problem, normalize the Fourier coefficient $c_0=1$ and use $(d_n-u)c_n=\sqrt q(c_{n-1}+c_{n+1})$. Find the coefficient of $q^2$ in $u$. Determine the connected NS coefficient that a future two-instanton calculation would have to reproduce, and check the classical limit.

Hint 1

The $n=\pm2$ modes first enter at order $q$. Retain the order-$q$ shift of $u$ in the denominators for $c_{\pm1}$.

Hint 2

Let $D_n=d_n-d_0$ and $A=2/(4\alpha^2-\hbar^2)$. The coefficient is $-\sum_{\eta=\pm1}[A/D_\eta^2+1/(D_\eta^2D_{2\eta})]$.

Solution

Writing $g=\sqrt q$, the recurrence gives $c_\eta=g/D_\eta+g^3[A/D_\eta^2+1/(D_\eta^2D_{2\eta})]+O(g^5)$. Here the second term comes from $c_{2\eta}=g^2/(D_\eta D_{2\eta})+O(g^4)$ and from expanding $D_\eta-(u-d_0)$. Use $u-d_0=-g(c_{-1}+c_1)$ to obtain

$$u=\alpha^2+\frac{2q}{4\alpha^2-\hbar^2} +\frac{2(20\alpha^2+7\hbar^2)q^2} {(4\alpha^2-\hbar^2)^3(4\alpha^2-4\hbar^2)} +O(q^3).$$

The required $q^2$ coefficient of $\mathscr F_{\rm NS,inst}$ is half the displayed coefficient, by (D16.6). We have derived an ODE prediction, not performed the two-instanton fixed-point sum. As $\hbar\to0$, the energy coefficient becomes $5/(32\alpha^6)$. Squaring (D16.12) gives $\alpha_{\rm cl}^2=u-q/(2u)-13q^2/(32u^3)+\cdots$; inversion reproduces the same $5/32$. The poles at $\alpha=\pm\hbar$ mark new free-mode degeneracies relevant at this order.

2. Estimate the contribution of the remote modes

At $\alpha=\hbar/2$, derive the leading common shift of the two resonant levels from the modes $n=1,-2$. Explain why this correction changes the midpoint before it changes the leading gap. Compare its value with the numerical example.

Hint 1

The pair eigenvectors are $(|0\rangle\pm|-1\rangle)/\sqrt2$. Each remote mode couples to only one component.

Hint 2

To leading order in $q$, both energy denominators are $\hbar^2/4-9\hbar^2/4=-2\hbar^2$.

Solution

Each coupling has squared modulus $q/2$. Summing over the two remote modes gives $2(q/2)/(-2\hbar^2)=-q/(2\hbar^2)$ for either pair state. Thus

$$u_\pm=\frac{\hbar^2}{4}\pm\sqrt q-\frac{q}{2\hbar^2} +O(q^{3/2}/\hbar^4).$$

The common shift cancels in the gap, leaving $2\sqrt q$ at leading order. For $\hbar=1$, $q=10^{-4}$ the predicted midpoint is $0.24995$, in agreement with $0.2499499995834\ldots$. The remaining gap correction is smaller by a relative factor of order $q/\hbar^4$. This calculation requires $\sqrt q/\hbar^2\ll1$; it does not authorize using the two-state matrix at arbitrary potential strength.

3. A change of phase and a change of spectral problem

Replace $\cos\theta$ by $\cos(\theta+\varphi)$ for real $\varphi$. Determine whether the eigenvalues and Floquet multiplier change. Then explain why replacing the circle by a real line with a $\cosh x$ potential is a different problem, even though the potentials are related by analytic continuation.

Hint 1

Translate the coordinate, or multiply Fourier coefficient $c_n$ by a phase. Check the boundary condition as well as the differential expression.

Hint 2

For the line problem, ask what replaces the return phase after one period.

Solution

The translation operator $\psi(\theta)\mapsto\psi(\theta+\varphi)$ is unitary on the same quasi-periodic domain. It preserves the derivative seam and multiplier. Equivalently the two neighboring Fourier entries gain conjugate factors $e^{\pm i\varphi}$, removable by a diagonal unitary matrix. Every eigenvalue is unchanged.

Analytically continuing $\theta$ to an imaginary coordinate changes the real contour, kinetic sign and admissible wavefunctions. For a confining positive $\cosh x$ problem on $\mathbb R$, square integrability requires decay at both ends; there is no circle multiplier to hold fixed. An analytic continuation of a differential expression alone does not establish a unitary equivalence of these domains. One must translate scales and specify a connection or quantization condition. This is the type A/type B distinction in NS, rather than a new choice of four-dimensional electric charge unit.

Literature and a bounded next calculation

The literature check for this dossier is dated 20 October 2026. Nekrasov supplies the instanton construction; NS identifies the retained-deformation limit and distinguishes its spectral problems. Poghossian supplies the scalar-observable relation used above. Başar and Dunne analyze why perturbative periods and nonperturbative band or gap information must be distinguished. Our one-coefficient and two-state calculations reproduce established structures; neither is a novelty claim.

The later François–Grassi study, published June 2026, constructs entire off-shell eigenfunctions in a different spectral/topological-string setting and studies modified-Mathieu limits. “Off-shell” there concerns an eigenvalue parameter before spectral quantization. It does not turn our circle calculation into a proof of that construction.

There is also a separate conformal-block branch: AGT section 4 connects the four-flavor $SU(2)$ problem to Liouville four-point data, distinguishing the instanton block from the full correlator. That continuation needs a mass/momentum map, the Abelian prefactor and a dimensionless coupling; our dimension-four $q$ cannot be substituted for its cross-ratio.

The next achievable task is to calculate the actual two-instanton connected term and compare it with Exercise 1, at generic $\alpha,\hbar$ before approaching resonance. A discrepancy by two suggests confusion between the logarithm and the quadratic observable; a wrong denominator sign suggests a lost $\epsilon_1=i\hbar$ continuation. Agreement establishes a further local coefficient. It does not establish all-order equality, a unique nonperturbative completion, or a spectrum without boundary conditions.