WKB Recursion and the Even–Odd Split

The Riccati equation turns formal WKB into a triangular calculation: once a sheet $p=\sqrt R$ is fixed, each new coefficient is obtained by one differentiation, one finite convolution, and one division by $2p$. The two branches are not independent. When the normal-form potential is even in $\hbar$, the substitution $\hbar\mapsto-\hbar$ exchanges them and separates the formal momentum into an even phase part and an odd amplitude part.

That simple observation hides a notorious naming trap. This book calls the branch-antisymmetric, $\hbar$-even momentum $P_{\mathrm{even}}$. Much of the exact-WKB literature first divides by $\hbar$ and calls the resulting object the odd Riccati part. This page derives both conventions, rather than asking the reader to guess which parity an author means.

Page question

How does the coefficient recursion determine the phase, amplitude, and formal Wronskian of the two WKB branches without adding any Borel summation or boundary data?

Status ledger

Statement	Status
Coefficient recurrence on a square-root patch	Formal identity
Branch relation under $\hbar\mapsto-\hbar$	Formal identity when $R(z,-\hbar)=R(z,\hbar)$
$P_{\mathrm{amp}}=-(\hbar/2)\partial_z\log P_{\mathrm{even}}$	Formal identity for two branches of the same Riccati equation
Normalized WKB pair	Formal solutions, defined locally up to $z$-independent formal units
Ordered Wronskian $-2/\hbar$	Formal identity for the displayed unit-normalized ordered pair
Convergence, Borel summability, or a preferred exact solution	Not asserted
Vanishing of every closed amplitude integral	False without a divisor and residue check

The coefficient recursion is triangular

Begin with the convention fixed on Page 1,

$$\hbar^2\psi''=R(z)\psi, \qquad P^2+\hbar P'=R,$$

and first assume that $R$ is independent of $\hbar$. Work on a simply connected domain where $R$ is holomorphic and nowhere zero, and choose

$$p(z)=\sqrt{R(z)}.$$

Write the plus branch as

$$P^{(+)}(z,\hbar) = \sum_{n=0}^{\infty} \hbar^n p_n(z), \qquad p_0=p.$$

The coefficient of $\hbar^n$ in ${P^{(+)}}^2+\hbar {P^{(+)}}'-R$ vanishes. For $n\geq1$ this gives

$$\begin{aligned} 2p\,p_n ={}& -p_{n-1}' \\ &- \sum_{j=1}^{n-1} p_jp_{n-j}. \end{aligned}$$

Equivalently,

$$p_n = -\frac1{2p} \left[ p_{n-1}' + \sum_{j=1}^{n-1} p_jp_{n-j} \right].$$

For the general family

$$R(z,\hbar) = \sum_{n=0}^{\infty} \hbar^nR_n(z),$$

expand both branches as

$$P^{(\sigma)} = \sum_{n=0}^{\infty} \hbar^nP_n^{(\sigma)}, \qquad P_0^{(\sigma)} = \sigma\sqrt{R_0}.$$

The same coefficient extraction gives

$$\begin{aligned} 2P_0^{(\sigma)}P_n^{(\sigma)} ={}& R_n - \left( P_{n-1}^{(\sigma)} \right)' \\ &- \sum_{j=1}^{n-1} P_j^{(\sigma)} P_{n-j}^{(\sigma)}, \end{aligned}$$

for $n\geq1$ and $\sigma\in\{+,-\}$. The source-free recurrence above is the special case $R_n=0$ for $n\geq1$. When a plus branch is being discussed, we continue to abbreviate $p_n:=P_n^{(+)}$; in the sourced case these coefficients depend on $R_1,R_2,\ldots$.

The recurrence is triangular in the $\hbar$ order: every term on the right has index smaller than $n$. It is not a two-term recurrence, because the convolution uses all earlier coefficients.

Returning to the $\hbar$-independent case, the first four coefficients make the pattern concrete:

$$\begin{aligned} p_0 &= p, \\ p_1 &= -\frac{p'}{2p}, \\ p_2 &= \frac{p''}{4p^2} - \frac{3(p')^2}{8p^3}, \\ p_3 &= \frac{\dd}{\dd z} \left[ -\frac{p''}{8p^3} + \frac{3(p')^2}{16p^4} \right]. \end{aligned}$$

Page 1 wrote $p_2$ directly in terms of $R$. The $p$-form is more efficient for symbolic generation and makes it visible that $p_3$ is a total derivative.

For a truncation

$$P_{[N]}^{(+)} := \sum_{n=0}^{N} \hbar^np_n,$$

the unambiguous regression test is

$${P_{[N]}^{(+)}}^2 + \hbar {P_{[N]}^{(+)}}' - R = O(\hbar^{N+1}).$$

One should test the Riccati residual, not merely compare a list of printed coefficients: a wrong convolution bound can reproduce the first one or two terms and fail only later.

Sign reversal exchanges the two branches

Suppose more generally that the normal-form potential is even in $\hbar$:

$$R(z,-\hbar)=R(z,\hbar).$$

If $P^{(+)}(z,\hbar)$ solves the formal Riccati equation, then

$$-P^{(+)}(z,-\hbar)$$

solves the same equation and has leading term $-p$. Formal uniqueness on the chosen square-root patch therefore gives

$$P^{(-)}(z,\hbar) = -P^{(+)}(z,-\hbar).$$

For an $\hbar$-independent $R$, the coefficient relation is

$$P_n^{(-)} = (-1)^{n+1}p_n.$$

Thus even-index coefficients change sign between sheets, whereas odd-index coefficients agree. The conclusion remains true when $R=R_0+\hbar^2R_2+\cdots$, although the sourced coefficient recurrence is then the one displayed above. A coordinate-induced Schwarzian starts at $\hbar^2$ and therefore preserves this parity. A term $\hbar R_1$, by contrast, breaks it.

Evenness does not mean $\hbar$-independence. For $R=R_0+\hbar^2R_2+\cdots$, the plus-branch coefficient at order two is

$$p_2 = \frac{R_2}{2p} + \frac{p''}{4p^2} - \frac{3(p')^2}{8p^3}, \qquad p=\sqrt{R_0}.$$

Define the two combinations

$$\begin{aligned} P_{\mathrm{even}} &:= \frac{ P^{(+)}-P^{(-)} }{2}, \\ P_{\mathrm{amp}} &:= \frac{ P^{(+)}+P^{(-)} }{2}. \end{aligned}$$

Under the parity assumption,

$$\begin{aligned} P_{\mathrm{even}} &= p_0+\hbar^2p_2+\hbar^4p_4+\cdots, \\ P_{\mathrm{amp}} &= \hbar p_1+\hbar^3p_3+\hbar^5p_5+\cdots. \end{aligned}$$

The first object is branch-antisymmetric and even in $\hbar$; the second is branch-symmetric and odd in $\hbar$.

Parity ladder for an $\hbar$-independent normal-form potential. The book’s $P_{\mathrm{even}}$ is branch-antisymmetric and even in $\hbar$, while $P_{\mathrm{amp}}$ is branch-symmetric and odd in $\hbar$. Dividing by $\hbar$ produces the traditional exact-WKB $S$-labels.

The source dictionary reverses the parity name

Let

$$S^{(\pm)} := \frac{P^{(\pm)}}{\hbar} = \frac{\widehat\psi_\pm'}{\widehat\psi_\pm}.$$

Here the hats emphasize that the wavefunctions are formal; their explicit normalization is constructed below.

Traditional exact-WKB notation often defines

$$\begin{aligned} S_{\mathrm{odd}}^{\mathrm{trad}} &:= \frac{ S^{(+)}-S^{(-)} }{2}, \\ S_{\mathrm{even}}^{\mathrm{trad}} &:= \frac{ S^{(+)}+S^{(-)} }{2}. \end{aligned}$$

Consequently,

$$\begin{aligned} S_{\mathrm{odd}}^{\mathrm{trad}} &= \frac{P_{\mathrm{even}}}{\hbar} = \hbar^{-1}p_0+\hbar p_2+\cdots, \\ S_{\mathrm{even}}^{\mathrm{trad}} &= \frac{P_{\mathrm{amp}}}{\hbar} = p_1+\hbar^2p_3+\cdots. \end{aligned}$$

Both naming schemes are internally consistent. The apparent reversal comes entirely from the factor $\hbar^{-1}$. This book retains $P_{\mathrm{even}}$ because its expansion is literally even in $\hbar$ and because the formal quantum periods on Page 5 integrate $P_{\mathrm{even}}\,\dd z$.

Never import an odd/even label by itself

A source may use $P$ for the unscaled logarithmic derivative that this book calls $S$, and may therefore call the branch difference $P_{\mathrm{odd}}$. Translate the defining formula first. The symbol name alone does not determine either the scaling or the parity.

The amplitude is fixed by the phase momentum

Subtract the two Riccati equations:

$$\begin{aligned} 0 ={}& {P^{(+)}}^2-{P^{(-)}}^2 \\ &+ \hbar \left( {P^{(+)}}'-{P^{(-)}}' \right). \end{aligned}$$

Using

$$P^{(\pm)} = P_{\mathrm{amp}} \pm P_{\mathrm{even}},$$

one obtains

$$2P_{\mathrm{even}}P_{\mathrm{amp}} + \hbar P_{\mathrm{even}}' = 0.$$

Hence

$$P_{\mathrm{amp}} = -\frac{\hbar}{2} \frac{\dd}{\dd z} \log P_{\mathrm{even}}.$$

This identity requires two formal branches of the same Riccati equation, but it does not require $R$ to be even in $\hbar$. The parity interpretation does require that extra hypothesis.

Adding the two Riccati equations supplies the complementary relation

$$P_{\mathrm{even}}^2 + P_{\mathrm{amp}}^2 + \hbar P_{\mathrm{amp}}' = R.$$

Eliminating $P_{\mathrm{amp}}$ gives a single equation for the branch-antisymmetric momentum:

$$\begin{aligned} R ={}& P_{\mathrm{even}}^2 - \frac{\hbar^2}{2} \frac{P_{\mathrm{even}}''}{P_{\mathrm{even}}} \\ &+ \frac{3\hbar^2}{4} \left( \frac{P_{\mathrm{even}}'}{P_{\mathrm{even}}} \right)^2. \end{aligned}$$

This nonlinear equation is another generator for the coefficients of the branch-antisymmetric momentum. Under the parity assumption, those are precisely the even powers. The original triangular recursion is usually simpler for computation, while the reduced equation is better for coordinate-covariance arguments.

Unit normalization fixes the formal Wronskian

Because $P^{(\pm)}=P_{\mathrm{amp}}\pm P_{\mathrm{even}}$, formal integration gives

$$\widehat\psi_\pm(z,\hbar) = C_\pm(\hbar) P_{\mathrm{even}}(z,\hbar)^{-1/2} \exp \left[ \pm\frac1\hbar \int_{z_0}^{z} P_{\mathrm{even}}(\zeta,\hbar)\,\dd\zeta \right].$$

The square root exists formally and locally because the leading term $p$ is nonzero. The $C_\pm$ are arbitrary nonzero formal units that do not depend on $z$.

For the displayed unit choice $C_+=C_-=1$ and the book’s convention

$$\Wr[f,g]=fg'-f'g,$$

direct differentiation gives

$$\Wr[ \widehat\psi_+, \widehat\psi_- ] = -\frac{2}{\hbar}.$$

More generally,

$$\Wr[ \widehat\psi_+, \widehat\psi_- ] = -\frac{2C_+C_-}{\hbar}.$$

This normalization ledger matters. If instead both branches are normalized to equal $1$ at $z_0$, their Wronskian is $-2P_{\mathrm{even}}(z_0,\hbar)/\hbar$. A quoted constant Wronskian is meaningless until the branch order and both scalar normalizations have been stated.

Another common source normalization uses $\left(S_{\mathrm{odd}}^{\mathrm{trad}}\right)^{-1/2} =\sqrt{\hbar}\,P_{\mathrm{even}}^{-1/2}$. It multiplies both columns by $\sqrt{\hbar}$ and therefore prints the ordered Wronskian as $-2$.

The hats are also essential: these are formal solutions. An actual analytic pair with these asymptotics requires a sector, a summation direction, and normalization data from Chapter 9.

A logarithmic derivative is locally exact, not globally disposable

The amplitude one-form satisfies

$$P_{\mathrm{amp}}\,\dd z = -\frac{\hbar}{2} \dd\log P_{\mathrm{even}}.$$

Locally it integrates to the prefactor $P_{\mathrm{even}}^{-1/2}$. For the $\hbar$-independent potential used in the coefficient formulas above, expanding the logarithm gives

$$\begin{aligned} P_{\mathrm{amp}}\,\dd z ={}& -\frac{\hbar}{2} \dd\log p \\ &- \frac{\hbar^3}{2} \dd \left( \frac{p_2}{p} \right) + O(\hbar^5). \end{aligned}$$

The coefficients from order $\hbar^3$ onward are derivatives of meromorphic functions on the spectral cover. The leading $-(\hbar/2)\dd\log p$ can nevertheless have residues at the divisor of $p$. Even on a closed path, its integral can record a nonzero winding number. On an open path it contributes endpoint normalization.

Thus “the odd-power part is a total derivative” is a local algebraic statement, not permission to erase residues, endpoint terms, or half-density monodromy. Page 5 makes the required subtraction and residue ledger explicit.

The phase one-form is covariant and the amplitude is a connection

Let $z=z(w)$ be an $\hbar$-independent local biholomorphism and write

$$s(w)=\frac{\dd z}{\dd w}.$$

Thus $s(w)\neq0$ on the coordinate patch.

Page 1 derived the affine branch transformation

$$\widetilde P^{(\pm)} = sP^{(\pm)} - \frac{\hbar}{2} \frac{s'}{s}.$$

Taking the difference and the sum gives

$$\begin{aligned} \widetilde P_{\mathrm{even}} &= sP_{\mathrm{even}}, \\ \widetilde P_{\mathrm{amp}} &= sP_{\mathrm{amp}} - \frac{\hbar}{2} \frac{s'}s. \end{aligned}$$

Therefore

$$\widetilde P_{\mathrm{even}}\,\dd w = P_{\mathrm{even}}\,\dd z,$$

while

$$\widetilde P_{\mathrm{amp}}\,\dd w = P_{\mathrm{amp}}\,\dd z - \frac{\hbar}{2} \dd\log s.$$

The first is the canonical WKB one-form. The second is the connection term that makes the normalized formal wavefunction transform as

$$\widetilde{\widehat\psi}_\pm = s^{-1/2} \widehat\psi_\pm.$$

The same inverse-half-density factor preserves the ordered Wronskian:

$$\Wr_w[ s^{-1/2}f(z(w)), s^{-1/2}g(z(w)) ] = \Wr_z[f,g].$$

This covariance uses the Schwarzian correction to the transformed normal-form potential. Treating $R$ as a scalar would invalidate the branch equations from which these formulas follow.

Turning points predict coefficient growth

For an $\hbar$-independent potential, every $p_n$ produced by the recurrence is holomorphic at an ordinary point where $p$ is holomorphic and nonzero. Near a generic simple turning point, write

$$R(z)=a(z-z_t)+O((z-z_t)^2), \qquad a\neq0.$$

Set $t=z-z_t$ and choose $c^2=a$. The recurrence gives, for $n\geq1$,

$$p_n(z) \sim A_n c^{1-n}t^{-(3n-1)/2},$$

where

$$\begin{aligned} A_1 &= -\frac14, \\ A_n &= \frac{3n-4}{4}A_{n-1} - \frac12 \sum_{j=1}^{n-1} A_jA_{n-j}. \end{aligned}$$

Induction shows $A_n<0$: if the earlier coefficients are negative, both terms in the second line are negative. Thus no leading cancellation occurs. In the cover coordinate $t=\xi^2$, $p_n$ has a meromorphic pole of order $3n-1$. The rapidly increasing pole order explains why an outer WKB series is nonuniform near a turning point. It is also compatible with, but does not by itself prove, the factorial large-order growth found under the hypotheses of Chapter 9.

The pole estimate is local bookkeeping, not yet a large-order theorem. Page 3 constructs the normalized cover and classifies zeros and poles; Chapter 9 supplies the Gevrey and Borel statements.

Reproducible coefficient audit

The formal recursion check generates the coefficients through a declared finite order, verifies the Riccati residual on both branches, tests the parity split and logarithmic amplitude identity, and audits the normalized Wronskian. Run

python3 public/code/advanced-ode/wkb-formal-recursion-check.py

The script uses exact SymPy algebra, prints its interpreter and dependency versions, and raises explicit errors rather than relying on Python assertions.

Common pitfalls

Calling $P_{\mathrm{even}}$ a Riccati solution. It is a linear combination of two Riccati solutions, and the Riccati equation is nonlinear. It instead satisfies the reduced equation displayed above.

Using parity without checking $R(z,-\hbar)=R(z,\hbar)$. The branch relation survives an $\hbar^2$ Schwarzian term but generally fails when $R_1\neq0$. The definitions by branch sum and difference still make sense, but their series no longer contain only one parity.

Forgetting the scaling in a source. The traditional odd part of $S=\psi'/\psi$ is $P_{\mathrm{even}}/\hbar$. It is not a contradiction that this book calls the numerator even.

Dropping every amplitude integral. A logarithmic derivative is locally exact. Around zeros and poles it can carry residues and winding, and along an open path it changes endpoint normalization.

Quoting $-2/\hbar$ without a normalization. That value belongs to the ordered pair $(\widehat\psi_+,\widehat\psi_-)$ with unit formal constants. Reversing the order or rescaling either column changes it.

Exercises

1. Derive the triangular recurrence

Insert $P^{(+)}=\sum_{n\geq0}\hbar^np_n$ into $P^2+\hbar P'=R$ and derive the recurrence for $p_n$.

Solution

The coefficient of $\hbar^n$ is

$$\sum_{j=0}^{n} p_jp_{n-j} + p_{n-1}',$$

where $p_{-1}:=0$. At order zero, $p_0^2=R$. For $n\geq1$, isolate the $j=0$ and $j=n$ terms:

$$2p_0p_n + \sum_{j=1}^{n-1} p_jp_{n-j} + p_{n-1}' = 0.$$

Putting $p_0=p$ and dividing by $2p$ gives the displayed formula.

2. Prove the branch-parity relation

Assume $R(z,-\hbar)=R(z,\hbar)$. Show that $-P^{(+)}(z,-\hbar)$ solves the same Riccati equation and identify its leading term.

Solution

Set

$$\widetilde P(z,\hbar) := -P^{(+)}(z,-\hbar).$$

Then

$$\begin{aligned} \widetilde P^2+\hbar\widetilde P' &= {P^{(+)}(z,-\hbar)}^2 - \hbar {P^{(+)}}'(z,-\hbar) \\ &= R(z,-\hbar) = R(z,\hbar). \end{aligned}$$

Its leading term is $-p$, so formal uniqueness identifies it with $P^{(-)}$. Coefficient comparison yields $P_n^{(-)}=(-1)^{n+1}P_n^{(+)}$.

3. Recover the amplitude and Wronskian

Subtract the two Riccati equations, derive $P_{\mathrm{amp}}=-(\hbar/2)\partial_z\log P_{\mathrm{even}}$, and compute $\Wr[\widehat\psi_+,\widehat\psi_-]$ for unit formal constants.

Solution

Factoring the difference gives

$$\left( P^{(+)}-P^{(-)} \right) \left( P^{(+)}+P^{(-)} \right) + \hbar \left( P^{(+)}-P^{(-)} \right)' = 0.$$

Division by $2(P^{(+)}-P^{(-)})$ gives the amplitude identity. For the unit-normalized pair,

$$\frac{\widehat\psi_\pm'}{\widehat\psi_\pm} = -\frac12 \frac{P_{\mathrm{even}}'}{P_{\mathrm{even}}} \pm \frac{P_{\mathrm{even}}}{\hbar}.$$

Therefore

$$\begin{aligned} \Wr[ \widehat\psi_+, \widehat\psi_- ] &= \widehat\psi_+\widehat\psi_- \left( -\frac{2P_{\mathrm{even}}}{\hbar} \right) \\ &= -\frac2\hbar, \end{aligned}$$

because $\widehat\psi_+\widehat\psi_-=P_{\mathrm{even}}^{-1}$.

4. Transform phase and amplitude separately

Starting from

$$\widetilde P^{(\pm)} = sP^{(\pm)} - \frac{\hbar}{2} \frac{s'}s,$$

derive the transformation laws for $P_{\mathrm{even}}$ and $P_{\mathrm{amp}}$.

Solution

The affine term is the same on both branches, so it cancels in the difference and doubles in the sum:

$$\widetilde P_{\mathrm{even}} = sP_{\mathrm{even}},$$ $$\widetilde P_{\mathrm{amp}} = sP_{\mathrm{amp}} - \frac{\hbar}{2} \frac{s'}s.$$

Multiplication by $\dd w$ and use of $\dd z=s\,\dd w$ gives the two one-form laws on this page.

5. See how an odd source breaks parity

Let

$$R(z,\hbar) = R_0(z)+\hbar R_1(z)+O(\hbar^2).$$

Compute $P_1^{(\pm)}$ and show that it is no longer branch-symmetric when $R_1\neq0$.

Solution

At order $\hbar$,

$$2P_0P_1+P_0'=R_1.$$

With $P_0^{(\pm)}=\pm\sqrt{R_0}$,

$$P_1^{(\pm)} = -\frac{R_0'}{4R_0} \pm \frac{R_1}{2\sqrt{R_0}}.$$

The second term changes sign between branches, so already

$$P_{\mathrm{even}} = \sqrt{R_0} + \hbar \frac{R_1}{2\sqrt{R_0}} + O(\hbar^2).$$

The branch difference is no longer an even series. This is precisely the obstruction excluded by $R(z,-\hbar)=R(z,\hbar)$.

References

• K. Iwaki, “Les Houches Lectures on Exact WKB Analysis and Painlevé Equations”, arXiv:2512.17599v2, May 2026, §1.1.1. Derives the Riccati recurrence, branch decomposition, and normalized formal solutions, from which the constant Wronskian follows in an unscaled logarithmic-derivative convention.
• K. Iwaki and T. Nakanishi, “Exact WKB Analysis and Cluster Algebras”, Journal of Physics A: Mathematical and Theoretical 47 (2014), 474009, §§2.1 and 2.3–2.4, especially equations (2.4)–(2.18) and (2.27). These sections fix widely used odd/even exact-WKB terminology, the associated WKB solutions, and their coordinate behavior.
• T. Kawai and Y. Takei, Algebraic Analysis of Singular Perturbation Theory, Translations of Mathematical Monographs 227, American Mathematical Society, 2005. Gives the systematic formal and exact WKB normalization framework.
• J. L. Dunham, “The Wentzel–Brillouin–Kramers Method of Solving the Wave Equation”, Physical Review 41 (1932), 713–720. Develops the all-orders one-dimensional WKB expansion and the closed-contour quantization series that motivates later quantum-period methods.