††institutetext: Department of Physics, The University of Tokyo, 7-3-1 Hongo,
Bunkyo-ku, Tokyo 113-0033, Japan

# Exact WKB analysis for ${\cal PT}$ symmetric quantum mechanics:
Study of the Ai-Bender-Sarkar conjecture

Syo Kamata<EMAIL_ADDRESS>
###### Abstract

We consider exact WKB analysis to a ${\cal PT}$ symmetric quantum mechanics
defined by the potential, $V(x)=\omega^{2}x^{2}+gx^{2}(ix)^{\varepsilon=2}$
with $\omega\in{\mathbb{R}}_{\geq 0}$, $g\in{\mathbb{R}}_{>0}$. We in
particular aim to verify a conjecture proposed by Ai-Bender-Sarkar (ABS), that
pertains to a relation between $D$-dimensional ${\cal PT}$-symmetric theories
and analytic continuation (AC) of Hermitian theories concerning the energy
spectrum or Euclidean partition function. For the purpose, we construct energy
quantization conditions by exact WKB analysis and write down their transseries
solution by solving the conditions. By performing alien calculus to the energy
solutions, we verify validity of the ABS conjecture and seek a possibility of
its alternative form by Borel resummation theory if it is violated. Our
results claim that the validity of the ABS conjecture drastically changes
depending on whether $\omega>0$ or $\omega=0$: If ${\omega}>0$, then the ABS
conjecture is violated when exceeding the semi-classical level, but its
alternative form is constructable by Borel resummation theory. The ${\cal PT}$
and the AC energies are related to each other by a one-parameter Stokes
automorphism, and a median resummed form, which corresponds to a formal exact
solution, of the AC energy (resp. ${\cal PT}$ energy) is directly obtained by
acting Borel resummation to a transseries solution of the ${\cal PT}$ energy
(resp. AC energy). If $\omega=0$, then, with respect to the inverse energy
level-expansion, not only perturbative/non-perturbative structures of the
${\cal PT}$ and the AC energies but also their perturbative parts do not match
with each other. These energies are independent solutions, and no alternative
form of the ABS conjecture can be reformulated by Borel resummation theory.

##  1 Introduction

Non-Hermitian theory is an exciting topic in a broad area of theoretical and
experimental physics. Although ${\cal CPT}$ invariance is a theoretically
fundamental concept for field theories, non-Hermitian theories show rich
physics depending on relaxed symmetry. Non-Hermitian effects were
experimentally observed, such as in open systems and non-equilibrium systems,
and such a non-Hermitian physics is expected to offer further insights into
these interesting phenomena, providing us with a deeper understanding
el2018non ; Ashida:2020dkc ; Okuma:2022bnb .

It is known that ${\cal PT}$ symmetric quantum mechanics have a special
property in their energy spectrum Bender:1998gh . When we consider the ${\cal
PT}$ symmetric potential defined by

$\displaystyle V_{\cal
PT}(x)=\omega^{2}x^{2}+gx^{2}(ix)^{\varepsilon},\qquad(\omega\in{\mathbb{R}}_{\geq
0},\ g,\varepsilon\in{\mathbb{R}}_{>0},\ x\in{\mathbb{C}})$ (1)

despite lack of ${\cal CPT}$ invariance, the ${\cal PT}$ symmetric Hamiltonian
gives a real and bounded energy spectrum due to ${\cal PT}$ invariance under
the ${\cal P}$ and ${\cal T}$ transforms,

$\displaystyle{\cal P}:\ x\rightarrow-x,\qquad{\cal T}:\ x\rightarrow\bar{x},\
i\rightarrow-i,$ (2)

where $\bar{x}$ is complex conjugate of $x$ Bender:1998ke ; Dorey:2001uw ;
Jones:2006qs ; Bender:2019 . Generalizations to $D$-dimensional ${\cal PT}$
symmetric field theories were also considered in Refs.Bender:2018pbv ;
Felski:2021evi . In such a generalization to field theories, it was suggested
that Hermitian theories given by a quartic potential,

$\displaystyle V_{\cal H}(x)=\omega^{2}x^{2}+\lambda
x^{4},\qquad(\omega\in{\mathbb{R}}_{\geq 0},\lambda\in{\mathbb{R}})$ (3)

with a negative coupling $\lambda$ relates to the ${\cal PT}$ symmetric
theories with $\varepsilon=2$ in Eq.(1) Bender:2021fxa ; Mavromatos:2021hpe ;
Grunwald:2022kts ; Ai:2022csx ; Lawrence:2023woz . In particular, Ai-Bender-
Sarkar (ABS) made a conjectured relation of Euclidean partition functions for
$D\geq 1$, that takes the form that

$\displaystyle\log Z_{\cal PT}(g)$ $\displaystyle=$ $\displaystyle{\rm Re}\log
Z_{\cal H}(\lambda=-g+i0_{\pm}),\qquad g\in{\mathbb{R}}_{>0},$ (4)

from a semi-classical analysis Ai:2022csx . However, a contradiction was
reported in Ref. Lawrence:2023woz , and the authors claimed violation of the
ABS conjecture for the pure quartic potential by the study of quantum
mechanics and the $D=0$ $N$-component scalar model. By revealing validity and
a limitation of the ABS conjecture, it is expected to provide us physical
meaning of theories with an unbounded potential from the relation to ${\cal
PT}$ symmetric theories. Symmetry is quite crucial for any physical phenomena,
and thus the relation would be possibly helpful to characterize their physical
features. Especially, applications to four-dimensional theories are important
issues, such as their IR physics. See Refs. Romatschke:2022jqg ;
Romatschke:2022llf ; Grable:2023paf ; Weller:2023jhc ; Romatschke:2023sce ,
for example. In order to theoretically investigate the ABS conjecture in a
more rigorous way, however, a method beyond-semi-classical analysis must be
necessary.

Since a Euclidean partition function can be written down by an energy
spectrum, analysis of an energy spectrum is direct investigation of the ABS
conjecture expressed by partition functions. In quantum mechanics ($D=1$),
exact WKB analysis (EWKB) is quite powerful for such a problem related to an
energy spectrum Berry ; BPV ; Voros1983 ; Silverstone ; DDP2 ; DP1 ; Takei1 ;
Takei2 ; Takei3 ; Kawai1 ; AKT1 ; Schafke1 ; Iwaki1 ; Alvarez1 ; Zinn-
Justin:2004vcw ; Zinn-Justin:2004qzw ; Dunne:2013ada ; Dunne:2014bca . EWKB is
not only a generalization of semi-classical analysis treating all order of
$\hbar$ but also some sort of Borel resummation theory and resurgence theory,
specialized in a one-dimensional Schrödinger equation. EWKB offers an energy
quantization condition (QC) by performing analytic continuation on the complex
$x$-plane, and then its energy spectrum can be found by solving it. When a
perturbative part of the energy solution is divergent series and Borel non-
summable, one can access to non-perturbative informations from the
perturbative expansion through Borel resummation. Such a relation among
perturbative and non-perturbative parts is called as resurgence or resurgent
relation Ec1 ; delabaere1994introduction ; Sauzin:1405 ; Marino:2012zq ;
Dorigoni2019 ; Aniceto:2018bis ; delabaere2016divergent ; Costin2008 . Non-
perturbative effects crucially affect many physical phenomena, and thus EWKB
and related approaches were employed in many physical contexts of non-
perturbative physics Mironov:2009uv ; Gaiotto:2009hg ; ashok2016exact ;
taya2021exact ; enomoto2022exact ; kashani2015pure ; Basar:2017hpr ;
Cavusoglu:2023bai ; suzuki2023exact ; Sueishi:2020rug ; Sueishi:2021xti ;
Kamata:2021jrs ; Bucciotti:2023trp ; Grassi:2014uua . It is also an
interesting topic in mathematical physics, such as a relation to integrable
systems takei2004toward ; dunne2017wkb ; hollands2020exact ; imaizumi2020exact
; Emery:2020qqu ; Ito:2018eon ; ito2023exact ; vanSpaendonck:2022kit ;
Dorey:2007zx ; Franco:2015rnr ; Hatsuda:2015qzx , and analyses of ${\cal PT}$
symmetric quantum mechanics were considered in Ref. Emery:2019znd ;
Emery:2020qqu . In addition, ${\cal PT}$ symmetric eigenvalue problems with
numerical approaches were studied in, e.g. Ref. Noble2013 ; Noble2017 ;
Khan:2022uyz .

In this paper, we verify the ABS conjecture for $D=1$ in Eq.(4) by EWKB which
is a beyond-semi-classical analysis. We construct QCs of the theories by EWKB
and then obtain formal transseries solutions of the energy spectra by solving
them. From those results, we consider validity of the ABS conjecture and seek
a possibility of its alternative form by Borel resummation theory if it is
violated.

##  2 Summary of our results and structure of this paper

As described in the end of Sec. 1, our main purpose of this paper is
verification of the ABS conjecture for $D=1$ by using EWKB and reformulation
by Borel resummation theory if it is violated. Since a Euclidean partition
function can consist of an energy spectrum, the ABS conjecture expressed by
partition functions can be known by investigation of the relation of the
energies. This study is a benchmark for generalization to field theories.

We summarize our results below:

  * •

If $\omega>0$, then the ABS conjecture is violated when exceeding the semi-
classical level. While the ${\cal PT}$ energy is purely perturbative and Borel
non-summable, the AC energy contains non-perturbative contributions in
addition to the same perturbative part to the ${\cal PT}$ energy. The ABS
conjecture is violated in the second non-perturbative sector and higher, which
contains both real and imaginary parts in the AC energy, but an alternative
form is constructable by a one-parameter Stokes automorphism and Borel
resummation. The ${\cal PT}$ and the AC energies are related to each other by
the one-parameter Stokes automorphism, and a median resummed form, which
corresponds to a formal exact solution, of the AC energy (resp. the ${\cal
PT}$ energy) is directly obtained by acting Borel resummation to a transseries
solution of the ${\cal PT}$ energy (resp. the AC energy). The relations of the
energy solutions for the formal transseries and the Borel resummed forms are
represented in Eqs.(144) and (147), respectively. The overall view is
schematically shown in Fig. 1. The Euclidean partition functions also exhibit
the same relations, and those are expressed by Eqs.(151) and (153).

  * •

If $\omega=0$, then the ABS conjecture is not satisfied. With respect to the
inverse energy level-expansion, the ${\cal PT}$ energy is purely perturbative
and Borel non-summable, and the AC energy has non-perturbative parts. In
addition to the difference of the perturbative/non-perturbative structures
between the ${\cal PT}$ and the AC energies, even their perturbative parts do
not match with each other unlike the case of $\omega=0$. Since their QCs can
not take the same form by a one-parameter Stokes automorphism, those are
independent solutions. No alternative form of the ABS conjecture can be
reformulated by Stokes automorphism and Borel resummation.

In this paper, we mainly address the case that $\varepsilon=2$, but this
result is extendable to $\varepsilon\in 4{\mathbb{N}}_{0}+2$.

This paper is organized as follows: In Sec. 3, we review Borel resummation
theory and EWKB. We firstly explain main concepts of Borel resummation and
Stokes automorphism in a slightly generic point of view in Sec. 3.1, and then
introduce EWKB to provide QCs in Sec. 3.2. After the review part, we consider
a quartic potential with a quadratic term in Sec. 4. As warm-up, we apply EWKB
to the Hermitian potential with a positive coupling $\lambda$ in Sec. 4.1.
Then, in Sec. 4.2, we derive the energy solutions of the ${\cal PT}$ symmetric
potential. From the results, we modify the ABS conjecture for $\omega>0$ based
on Borel resummation theory in Sec. 4.3. In Sec. 5, we consider a quartic
potential without a quadratic term. In Sec. 5.1, we begin with the Hermitian
potential with a positive coupling $\lambda$ and obtain the energy solution by
using the inverse energy level-expansion. In Sec. 5.2, we consider the ${\cal
PT}$ symmetric potential. In Sec. 5.3, we show the fact that no alternative
form of the ABS conjecture can be reformulated by Stokes automorphism and
Borel resummation. In Sec. 6, we make some remarks about generalizations to
even $\varepsilon$ and spectral reality from the viewpoint of EWKB. Technical
computations, such as derivation of non-perturbative parts and alien calculus
for the energy spectrum, are summarized in Appendices A and B, respectively.

For QCs, denoted as ${\mathfrak{D}}=0$, we frequently use the symbol that
${\mathfrak{D}}_{1}\propto{\mathfrak{D}}_{2}$ through this paper, which means
that ${\mathfrak{D}}_{1}$ and ${\mathfrak{D}}_{2}$ are equivalent to each
other except an overall factor not to affect their energy solutions.

Figure 1:  Schematic figure of our modified ABS conjecture for $\omega>0$.
Borel resummation ${\cal S}_{0_{\pm}}$ and a one-parameter Stokes automorphism
${\mathfrak{S}}^{\nu\in{\mathbb{R}}}_{0}$ are defined around $\arg(g)=0$.
Formal transseries of the ${\cal PT}$ and the AC energies (resp. QCs), denoted
by $E_{{\cal PT}/{\rm AC}}$ (resp. ${\mathfrak{D}}_{{\cal PT}/{\rm AC}}$), are
connected to each other by the one-parameter Stokes automorphism with $\nu=\pm
1/2$. The superscript, $\arg(\lambda)=\pm\pi$, corresponds to
$\lambda=-g+i0_{\pm}$ in Eq.(4). Acting the median resummation ${\cal S}_{{\rm
med}}$ to the formal transseries gives their median resummed forms,
$\widehat{E}_{{\cal PT}/{\rm AC}}$ (resp. $\widehat{{\mathfrak{D}}}_{{\cal
PT}/{\rm AC}}$), that reproduce the original transseries by taking
$\hbar\rightarrow 0_{+}$. By introducing another one-parameter Stokes
automorphism acting to Borel resummed forms,
$\widehat{\mathfrak{S}}^{\nu\in{\mathbb{R}}}$, the similar relations among the
Borel resummed forms of the ${\cal PT}$ and the AC energies (resp. QCs) hold
formally. Borel resummation ${\cal S}_{0_{\pm}}$ makes a direct connection
from $E_{{\cal PT}/{\rm AC}}$ (resp. ${\mathfrak{D}}_{{\cal PT}/{\rm AC}}$) to
$\widehat{E}_{{\rm AC}/{\cal PT}}$ (resp. $\widehat{\mathfrak{D}}_{{\rm
AC}/{\cal PT}}$).

##  3 Preliminary: Borel resummation theory and exact WKB analysis

In this section, we introduce Borel resummation and EWKB. In Sec. 3.1, we
review Borel resummation and Stokes automorphism. Then, in Sec. 3.2, we
explain some basics of EWKB.

###  3.1 Borel resummation and Stokes automorphism

We introduce concepts of Borel resummation and Stokes automorphism. There are
many textbooks and reviews of Borel resummation theory and resurgence theory.
See Refs. Ec1 ; delabaere1994introduction ; Sauzin:1405 ; Marino:2012zq ;
Dorigoni2019 ; Aniceto:2018bis ; delabaere2016divergent ; Costin2008 , for
example.

Let us begin with the following formal power series expanded by $\hbar$:

$\displaystyle
f(\hbar)\sim\sum_{n\in{\mathbb{N}}}c_{n}\hbar^{n}\quad\mbox{as}\quad\hbar\rightarrow
0_{+}.\qquad(c_{n}\in{\mathbb{C}})$ (5)

For simplicity, we assumed that $f$ does not have non-perturbative
contributions and oscillations such as $e^{-S/\hbar}$ with ${\rm Re}(S)\geq
0$. We also assume that $f$ is a divergent series and Gevrey-1 class, i.e.,

$\displaystyle c_{n}\sim AS^{-n}n!\quad\mbox{as}\quad
n\rightarrow\infty,\qquad\lim_{n\rightarrow\infty}\left|\frac{c_{n}}{c_{n+1}}\right|=0.\qquad(A,S\in{\mathbb{C}})$
(6)

Borel resummation ${\cal S}_{\theta}$ is defined as an operator acting to a
formal power series and being composition of Borel transform ${\cal B}$ and
Laplace integral ${\cal L}_{\theta}$, i.e., ${\cal S}_{\theta}={\cal
L}_{\theta}\circ{\cal B}$. Those are defined as

$\displaystyle{\cal
B}[f](\xi):=\sum_{n\in{\mathbb{N}}}\frac{c_{n}}{\Gamma(n)}\xi^{n-1}=f_{B}(\xi),\qquad{\cal
L}_{\theta}[f_{B}](\hbar):=\int_{0}^{\infty
e^{i\theta}}d\xi\,e^{-\frac{\xi}{\hbar}}f_{B}(\xi).$ (7)

The Borel transformed expansion, $f_{B}$, is an analytic function in a
neighborhood of $\xi=0$, but it is not on the entire complex $\xi$-plane when
$f$ is a divergent series. If no singular point lays on the integration ray
with $\arg(\xi)=\theta$, then the Laplace integral is formally performable. In
such a case, we say $f$ is Borel (re)summable along $\arg(\xi)=\theta$.
Otherwise, $f$ is Borel non-(re)summable. Borel resummation is a homomorphism:

$\displaystyle{\cal S}_{\theta}[f_{1}+f_{2}]={\cal S}_{\theta}[f_{1}]+{\cal
S}_{\theta}[f_{2}],\qquad{\cal S}_{\theta}[f_{1}f_{2}]={\cal
S}_{\theta}[f_{1}]\cdot{\cal S}_{\theta}[f_{2}].$ (8)

Suppose the case that $f$ is Borel non-summable along $\theta=0$. When $f$ is
Borel non-summable, one usually adds a infinitesimal phase to the integration
ray to make $f$ Borel summable, as $\theta=0_{\pm}$. However, because of a
discontinuity from the singular point, the Borel resummed form depends on the
sign of the phase, namely,

$\displaystyle{\cal S}_{0_{+}}[f]\neq{\cal S}_{0_{-}}[f].$ (9)

This implies that the resulting form of ${\cal S}_{0_{\pm}}[f]$ does not match
with $f$ in the asymptotic limit, i.e.,111 In this expression, we used a
symbol “$0_{\mp}$” instead of “$0_{\pm}$” in $f^{0_{\mp}}$ to adjust to the
notation in Eq.(22).

$\displaystyle{\cal S}_{0_{\pm}}[f]\overset{\hbar\rightarrow
0_{+}}{\sim}f^{0_{\mp}}\neq f,$ (10)

where $\overset{\hbar\rightarrow 0_{+}}{\sim}$ denotes reduction of a Borel
resummed form to a formal transseries by taking a small $\hbar$. Although our
input in Borel resummation is a formal power series as Eq.(6), $f^{0_{\mp}}$
in Eq.(10) generally becomes a formal transseries taking the form, for
example, that Costin2008 ; Aniceto:2018bis 222 This form is constrained by
underlying mathematical structures such as a differential equation. Eq.(11) is
just an example, and a resurgent structure of our problem slightly differs
from Eq.(11).

$\displaystyle
f^{0_{\mp}}(\hbar)\sim\sum_{n\in{\mathbb{N}}}c^{(0)}_{n}\hbar^{n}+\sum_{\ell\in{\mathbb{N}}}\sum_{n\in{\mathbb{N}}_{0}}c^{(\ell)}_{n}\left(\pm\frac{{\cal
A}}{2}\right)^{\ell}e^{-\frac{\ell
S}{\hbar}}\hbar^{n+\beta\ell},\qquad(c^{(\ell)}_{n}\in{\mathbb{C}},{\cal A}\in
i{\mathbb{R}})$ (11)

with a non-zero ${\cal A}$. The deviation from $f$ can be roughly estimated by
a radius of convergence of $f_{B}$, and it is evaluated as $r_{*}=S$. Hence,

$\displaystyle{\cal S}_{0_{+}}[f]-{\cal S}_{0_{-}}[f]\overset{\hbar\rightarrow
0_{+}}{\sim}ie^{-\frac{S}{\hbar}}\left[c+O(\hbar)\right],$ (12)

with a constant $c$. Therefore, in general, Borel summability does not
guarantee to give a Borel resummed form perfectly reproducing the original
formal expansion.

The question is how to construct a Borel resummed form to reproduce the
original formal power series, $f$, in the small $\hbar$ limit. It is possible
by introducing Stokes automorphism, ${\mathfrak{S}}_{\theta}$, which has a key
role through this paper, defined as

$\displaystyle{\cal S}_{\theta+0_{+}}={\cal
S}_{\theta+0_{-}}\circ{\mathfrak{S}}_{\theta}.$ (13)

Roughly speaking, it is a mapping from a transseries to a transseries to
compensate a difference due to the discontinuity along the integration ray
with $\arg(\xi)=\theta$. Stokes automorphism is a homomorphism:

$\displaystyle{\mathfrak{S}}_{\theta}[f_{1}+f_{2}]={\mathfrak{S}}_{\theta}[f_{1}]+{\mathfrak{S}}_{\theta}[f_{2}],\qquad{\mathfrak{S}}_{\theta}[f_{1}f_{2}]={\mathfrak{S}}_{\theta}[f_{1}]\cdot{\mathfrak{S}}_{\theta}[f_{2}].$
(14)

If $f$ is Borel summable along $\arg(\xi)=\theta$, then
${\mathfrak{S}}_{\theta}$ is an identity mapping. Stokes automorphism defined
in Eq.(13) is just an automorphism determined by action of ${\cal
S}_{\theta+0_{\pm}}$ to $f$, but it is extendable to a one-parameter group
generated by a generator, alien derivative
$\overset{\underset{\bullet}{}}{\Delta}_{\theta}$, as

$\displaystyle{\mathfrak{S}}^{\nu}_{\theta}=\exp\left[\nu\overset{\underset{\bullet}{}}{\Delta}_{\theta}\right],\qquad\overset{\underset{\bullet}{}}{\Delta}_{\theta}=\sum_{w\in{\Gamma(\theta)}}\overset{\underset{\bullet}{}}{\Delta}_{w},\qquad(\nu\in{\mathbb{R}})$
(15)

where $\Gamma(\theta)$ denotes a set of singular points along the integration
ray with $\theta$. It satisfies

$\displaystyle{\mathfrak{S}}^{\nu=0}_{\theta}=1,\qquad{\mathfrak{S}}^{\nu_{1}}_{\theta}\circ{\mathfrak{S}}_{\theta}^{\nu_{2}}={\mathfrak{S}}_{\theta}^{\nu_{2}}\circ{\mathfrak{S}}_{\theta}^{\nu_{1}}={\mathfrak{S}}_{\theta}^{\nu_{1}+\nu_{2}},\qquad(\nu_{1},\nu_{2}\in{\mathbb{R}})$
(16)

and taking $\nu=1$ leads to ${\mathfrak{S}}_{\theta}$ in Eq.(13). Intuitively,
action of $\overset{\underset{\bullet}{}}{\Delta}_{w\in\Gamma(\theta)}$ to a
formal power series means pre-extracting a non-perturbative contribution
originated by taking a Hankel contour clockwisely going around a singular
point located at $\xi=w$ in the Laplace integral. An alien derivative is
additive and satisfies Leibniz rule:

$\displaystyle\overset{\underset{\bullet}{}}{\Delta}_{w}[f_{1}+f_{2}]=\overset{\underset{\bullet}{}}{\Delta}_{w}[f_{1}]+\overset{\underset{\bullet}{}}{\Delta}_{w}[f_{2}],\qquad\overset{\underset{\bullet}{}}{\Delta}_{w}[f_{1}f_{2}]=\overset{\underset{\bullet}{}}{\Delta}_{w}[f_{1}]\cdot
f_{2}+f_{1}\cdot\overset{\underset{\bullet}{}}{\Delta}_{w}[f_{2}].$ (17)

Let us come back to our question. Again, we take $\theta=0$. We suppose that
there exists formal transseries $f^{0_{\pm}}$ and a Borel resummed form
$\widehat{f}$ such that ${\cal S}_{0_{+}}[f^{0_{+}}]={\cal
S}_{0_{-}}[f^{0_{-}}]=\widehat{f}\overset{\hbar\rightarrow 0_{+}}{\sim}f$.
Thus, knowing a relation between $f$ and $f^{0_{\pm}}$ is the solution. This
problem can be solved by introducing median resummation defined as333 These
exponents of Stokes automorphism in this definition, $\mp 1/2$, are determined
by the conditions that  $\displaystyle{\cal C}\circ{\cal S}_{\theta}={\cal
S}_{-\theta}\circ{\cal C},\qquad{\cal C}\circ{\cal S}_{{\rm
med},{\theta}}={\cal S}_{{\rm med},{-\theta}}\circ{\cal C},$ (18) where ${\cal
C}$ is complex conjugate. The first condition in Eq.(18) can be expressed by
Stokes automorphism and alien derivative as  $\displaystyle{\cal
C}\circ{\mathfrak{S}}^{\nu}_{\theta}={\mathfrak{S}}^{-\nu}_{-\theta}\circ{\cal
C},\qquad{\cal
C}\circ\overset{\underset{\bullet}{}}{\Delta}_{\theta}=-\overset{\underset{\bullet}{}}{\Delta}_{-\theta}\circ{\cal
C},$ (19) and requiring the second condition in Eq.(18) gives the exponent,
$\mp 1/2$, using Eq.(19).

$\displaystyle{\cal S}_{{\rm med},0}[f]:={\cal
S}_{0_{+}}\circ{\mathfrak{S}}^{-1/2}_{0}[f]={\cal
S}_{0_{-}}\circ{\mathfrak{S}}^{+1/2}_{0}[f],$ (20)

where $f^{0_{\pm}}$ are related to $f$ as

$\displaystyle f^{0_{\pm}}={\mathfrak{S}}^{\mp 1/2}_{0}[f],\qquad
f={\mathfrak{S}}^{\pm 1/2}_{0}[f^{0_{\pm}}],$ (21)

and thus, ${\cal S}_{{\rm med},0}[f]={\cal S}_{0_{\pm}}[f^{0_{\pm}}]$444 In
the aspect of Lefschetz thimble decomposition, the Stokes automorphism in
Eq.(13) has a role of reconstruction of an integration-path to keep the same
homology class against a discontinuity induced by a Stokes phenomenon. In
addition, ${\mathfrak{S}}_{0}^{\mp 1/2}$ and ${\cal S}_{0_{\pm}}$ correspond
to appropriately adding non-trivial saddles to reproduce its original
integration-path and performing the integration along the thimbles,
respectively. . The median resummation directly provides a Borel resummed form
which is continuous at $\arg(\hbar)=0$ from $f$. Action of the alien
derivative to a formal power expansion in Gevrey-1 class is well-defined, so
that the median resummed form naturally returns ${\cal S}_{{\rm
med},0}[f]\overset{\hbar\rightarrow 0_{+}}{\sim}f$. Conversely, one can
express the discontinuity in ${\cal S}_{0_{\pm}}[f]$ using the median
resummation as

$\displaystyle{\cal S}_{0_{\pm}}[f]$ $\displaystyle=$ $\displaystyle{\cal
S}_{{\rm med},0}\circ{\mathfrak{S}}_{0}^{\pm 1/2}[f]={\cal S}_{{\rm
med},0}[f^{0_{\mp}}]\overset{\hbar\rightarrow 0_{+}}{\sim}f^{0_{\mp}}.$ (22)

In addition, the procedure (21) can be interpreted to enable to construct $f$
from $f^{0_{+}}$ and/or $f^{0_{-}}$ by the Stokes automorphism,
${\mathfrak{S}}_{0}^{\pm 1/2}$, without passing through Borel resummation.

It is useful to introduce another Stokes automorphism,
$\widehat{\mathfrak{S}}_{\theta}$, that acts to a Borel resummed form and
satisfies

$\displaystyle{\cal S}_{{\rm
med},\theta}\circ{\mathfrak{S}}^{\nu}_{\theta}=\widehat{\mathfrak{S}}_{\theta}^{\nu}\circ{\cal
S}_{{\rm med},\theta}.$ (23)

Choosing $\nu=\pm 1/2$ in the l.h.s. is equivalent to ${\cal
S}_{\theta+0_{\pm}}$. In this sense, this generalized Stokes automorphism,
$\widehat{\mathfrak{S}}^{\nu}_{\theta}$, can be regarded to be a continuous
transform to Borel resummed forms, and the Borel resummation, ${\cal
S}_{\theta+0_{\pm}}$, is a special case of Eq.(23).

Depending on problems, one obtains $f^{0_{\pm}}$ prior to knowing a specific
form of $f$. EWKB is the case. In such a case, although taking Borel
resummation ${\cal S}_{0_{\pm}}$ to $f^{0_{\pm}}$ formally solves the problem,
it is in practice too tough to find their exact resummed forms except in
special cases. Instead of that, we take another strategy for consideration of
the ABS conjecture, that is, constructing a formal transseries $f$ from
$f^{0_{\pm}}$ by Stokes automorphism using Eq.(21).

Through this paper, we usually deal with formal transseries except where
specifically noted. The hat symbol, $\widehat{f}$, denotes the median resummed
form of $f$, i.e., $\widehat{f}={\cal S}_{{\rm med}}[f]$.

####  3.1.1 Example 1: $c_{n}=AS^{-n}n!$

We consider $c_{n}=AS^{-n}n!$ with $S\in{\mathbb{R}}_{>0}$ in Eq.(5). Acting
Borel transform defined in Eq.(7) to the formal power series gives

$\displaystyle f_{B}:={\cal B}[f]=\frac{AS}{(S-\xi)^{2}},$ (24)

and one can immediately see that $f$ is Borel non-summable along $\theta=0$.
Taking Laplace integral with $\theta=0_{\pm}$ leads to

$\displaystyle{\cal S}_{0_{\pm}}[f]$ $\displaystyle=$ $\displaystyle{\cal
L}_{0_{\pm}}[f_{B}]=\frac{ASe^{-\frac{S}{\hbar}}}{\hbar}\left[\text{Ei}\left(\frac{S}{\hbar}\right)\pm\pi
i\right]-A$ (25) $\displaystyle\overset{\hbar\rightarrow 0_{+}}{\sim}$
$\displaystyle\sum_{n\in{\mathbb{N}}}AS^{-n}n!\hbar^{n}\pm\pi
i\frac{ASe^{-\frac{S}{\hbar}}}{\hbar},$

where ${\rm Ei}(x)$ is the exponential integral. Apparently, this result has a
discontinuity at $\arg(\xi)=0$, that is $\pm\pi
i\frac{ASe^{-\frac{S}{\hbar}}}{\hbar}$. From Eq.(24), one can see that a set
of singular points consists of only the double pole at $\xi=S$, i.e.
$\Gamma(\theta=0)=\\{S\\}$, and thus the alien derivative is easily obtained
by a residue integration around $\xi=S$ clockwisely, as

$\displaystyle\overset{\underset{\bullet}{}}{\Delta}_{S}[f]=-\oint_{\xi=S}d\xi\,e^{-\frac{\xi}{\hbar}}f_{B}(\xi)=2\pi
i\frac{ASe^{-\frac{S}{\hbar}}}{\hbar},\qquad(\overset{\underset{\bullet}{}}{\Delta}_{S})^{n>1}[f]=0.$
(26)

From this result and Eq.(15), one can obtain action of the Stokes automorphism
to $f$ as

$\displaystyle{\mathfrak{S}}^{\nu}_{0}[f]=f+2\pi\nu
i\frac{ASe^{-\frac{S}{\hbar}}}{\hbar},$ (27)

and Eq.(21) gives $f^{0_{\pm}}$ as

$\displaystyle f^{0_{\pm}}={\mathfrak{S}}^{\mp 1/2}_{0}[f]=f\mp\pi
i\frac{ASe^{-\frac{S}{\hbar}}}{\hbar}.$ (28)

Taking Borel resummation to $f^{0_{\pm}}$ yields

$\displaystyle\widehat{f}$ $\displaystyle=$ $\displaystyle{\cal
S}_{0_{\pm}}[f^{0_{\pm}}]=\frac{ASe^{-\frac{S}{\hbar}}}{\hbar}\text{Ei}\left(\frac{S}{\hbar}\right)-A\overset{\hbar\rightarrow
0_{+}}{\sim}\sum_{n\in{\mathbb{N}}}AS^{-n}n!\hbar^{n}.$ (29)

This result could correctly reproduce $c_{n}=AS^{-n}n!$ and remove the
discontinuity, $\pm\pi i\frac{ASe^{-\frac{S}{\hbar}}}{\hbar}$, appeared in
Eq.(25). In addition, acting $\widehat{\mathfrak{S}}_{0}^{\nu}$ to
$\widehat{f}$ is written as

$\displaystyle\widehat{\mathfrak{S}}_{0}^{\nu}[\widehat{f}]=\widehat{f}+2\pi\nu
i\frac{ASe^{-\frac{S}{\hbar}}}{\hbar},$ (30)

and taking $\nu=\pm 1/2$ corresponds to ${\cal S}_{0_{\pm}}[f]$ in Eq.(25).

####  3.1.2 Example 2: Zero-dimensional ${\cal PT}$ symmetric toy model

Another example is the zero-dimensional model considered in Ref. Ai:2022csx ,
which is given by

$\displaystyle Z_{\cal PT}=\int_{\gamma_{\cal
PT}}dx\,\exp\left[-x^{2}+gx^{4}\right],\qquad g\in{\mathbb{R}}_{>0},$ (31)
$\displaystyle\gamma_{\cal
PT}:=se^{+\frac{\pi}{4}i}\theta(-s)+se^{-\frac{\pi}{4}i}\theta(+s),\qquad
s\in{\mathbb{R}},$ (32)

where $\theta(s)$ is the step function. The integration in Eq.(31) is
analytically performable, and one can obtain the exact solution as

$\displaystyle\widehat{Z}_{\cal PT}=\frac{\pi
e^{-\frac{1}{8g}}}{4\sqrt{g}}\left[I_{-\frac{1}{4}}\left(\frac{1}{8g}\right)+I_{+\frac{1}{4}}\left(\frac{1}{8g}\right)\right],$
(33)

where $I_{\nu}(x)$ is the modified Bessel function of the first kind. We apply
the above methods to a formal transseries of the exact solution expanded by
$g$. This example might be helpful to understand the whole story in this paper
and Fig. 1.

Expanding Eq.(33) gives

$\displaystyle\widehat{Z}_{\cal PT}\sim Z_{\cal
PT}=\sqrt{\pi}\sum_{n\in{\mathbb{N}}_{0}}\frac{\left(\frac{1}{4}\right)_{n}\left(\frac{3}{4}\right)_{n}}{n!}(4g)^{n}\quad\mbox{as}\quad
g\rightarrow 0_{+},$ (34)

where $(a)_{n}:=\Gamma(a+n)/\Gamma(a)$ is the Pochhammer symbol. Here, we
define ${\cal J}_{0}:=gZ_{\cal PT}$ and deal with ${\cal J}_{0}$. The
coefficients of ${\cal J}_{0}$ are given by

$\displaystyle{\cal J}_{0}=\sum_{n\in{\mathbb{N}}}c_{n}g^{n},\qquad
c_{n}=\sqrt{\pi}\frac{\left(\frac{1}{4}\right)_{n-1}\left(\frac{3}{4}\right)_{n-1}}{(n-1)!}4^{n-1},$
(35)

and Borel transform (7) gives

$\displaystyle{\cal J}_{0,B}:={\cal B}[{\cal
J}_{0}]=\frac{2K\left(\frac{4\sqrt{\xi}}{2\sqrt{\xi}+1}\right)}{\sqrt{\pi}\sqrt{2\sqrt{\xi}+1}},$
(36)

where $K(x)$ is the elliptic integral. This has a singular point at
$\xi=\frac{1}{4}$, i.e. $\Gamma(\theta=0)=\\{\frac{1}{4}\\}$. Taking Borel
resummation yields

$\displaystyle{\cal S}_{0_{\pm}}[{\cal J}_{0}]$ $\displaystyle=$
$\displaystyle\frac{\pi
e^{-\frac{1}{8g}}\sqrt{g}}{4}\left[I_{-\frac{1}{4}}\left(\frac{1}{8g}\right)+I_{+\frac{1}{4}}\left(\frac{1}{8g}\right)\pm
i\frac{\sqrt{2}}{\pi}K_{\frac{1}{4}}\left(\frac{1}{8g}\right)\right]$ (37)
$\displaystyle=$ $\displaystyle\widehat{\cal J}_{0}+\widehat{\cal J}_{\pm},$

where $\widehat{\cal J}_{0}:=g\widehat{Z}_{\cal PT}$, and $\widehat{\cal
J}_{\pm}:=\pm
i\frac{e^{-\frac{1}{8g}}\sqrt{2g}}{4}K_{\frac{1}{4}}\left(\frac{1}{8g}\right)$
with the modified Bessel function of the second kind, $K_{\nu}(x)$. Formal
transseries of $\widehat{\cal J}_{\pm}$, denoted by ${\cal J}_{\pm}$, can be
obtained by the asymptotic expansion of
$K_{\frac{1}{4}}\left(\frac{1}{8g}\right)$. Notice that ${\cal J}_{+}+{\cal
J}_{-}=0$ and ${\cal J}_{+}-{\cal J}_{-}=\pm 2{\cal J}_{\pm}$. From the
results, one can find actions of alien derivatives and Stokes automorphism to
${\cal J}_{0}$ and ${\cal J}_{\pm}$ as

$\displaystyle(\overset{\underset{\bullet}{}}{\Delta}_{\frac{1}{4}})[{\cal
J}_{0}]=\int_{+\infty+i0_{-}}^{+\infty+i0_{+}}d\xi\,e^{-\frac{\xi}{g}}{\cal
J}_{0,B}(\xi)={\cal J}_{+}-{\cal
J}_{-},\qquad(\overset{\underset{\bullet}{}}{\Delta}_{\frac{1}{4}})^{n>1}[{\cal
J}_{0}]=0,$ (38)
$\displaystyle(\overset{\underset{\bullet}{}}{\Delta}_{\theta=0})^{n\in{\mathbb{N}}}[{\cal
J}_{\pm}]=0,$ (39)

where $\int_{+\infty+i0_{-}}^{+\infty+i0_{+}}d\xi$ denotes integration along a
Hankel contour going around the singular point at $\xi=\frac{1}{4}$
clockwisely, and

$\displaystyle{\mathfrak{S}}_{0}^{\nu}[{\cal J}_{0}]={\cal J}_{0}+\nu({\cal
J}_{+}-{\cal J}_{-}),\qquad{\mathfrak{S}}_{0}^{\nu}[{\cal J}_{\pm}]={\cal
J}_{\pm},$ (40) $\displaystyle\widehat{\mathfrak{S}}_{0}^{\nu}[\widehat{\cal
J}_{0}]=\widehat{\cal J}_{0}+\nu(\widehat{\cal J}_{+}-\widehat{\cal
J}_{-}),\qquad\widehat{\mathfrak{S}}_{0}^{\nu}[\widehat{\cal
J}_{\pm}]=\widehat{\cal J}_{\pm}.$ (41)

Hence, ${\cal J}_{0}^{0_{\pm}}$ which satisfy ${\cal S}_{0_{+}}[{\cal
J}_{0}^{0_{+}}]={\cal S}_{0_{-}}[{\cal J}_{0}^{0_{-}}]$ are obtained as

$\displaystyle{\cal J}_{0}^{0_{\pm}}={\mathfrak{S}}_{0}^{\mp 1/2}[{\cal
J}_{0}]={\cal J}_{0}\mp\frac{1}{2}({\cal J}_{+}-{\cal J}_{-})={\cal
J}_{0}-{\cal J}_{\pm}={\cal J}_{0}+{\cal J}_{\mp},$ (42) $\displaystyle{\cal
S}_{0_{\pm}}[{\cal J}_{0}^{0_{\pm}}]={\cal S}_{{\rm med},0}[{\cal
J}_{0}]=\widehat{\cal J}_{0}.$ (43)

Similarly, we consider analytic continuation of the Hermitian model given by

$\displaystyle\widehat{Z}_{\cal
H}:=\int_{-\infty}^{+\infty}dx\,\exp\left[-x^{2}-\lambda
x^{4}\right]=\frac{e^{\frac{1}{8\lambda}}}{2\sqrt{\lambda}}K_{\frac{1}{4}}\left(\frac{1}{8\lambda}\right),\qquad\lambda\in{\mathbb{R}}_{>0},$
(44)

and define ${\cal J}_{\rm AC}^{\arg(\lambda)=\pm\pi}:=gZ_{\cal
H}^{\arg(\lambda)=\pm\pi}$. By taking $\lambda=e^{\pm\pi i}g$ with
$g\in{\mathbb{R}}_{>0}$, those can be expressed by

$\displaystyle{\cal J}_{\rm AC}^{\arg(\lambda)=\pm\pi}$ $\displaystyle=$
$\displaystyle{\cal J}_{0}+{\cal J}_{\mp}={\cal J}_{0}\mp\frac{1}{2}({\cal
J}_{+}-{\cal J}_{-}),$ (45) $\displaystyle\widehat{\cal J}_{\rm
AC}^{\arg(\lambda)=\pm\pi}$ $\displaystyle=$ $\displaystyle\frac{\pi
e^{-\frac{1}{8g}}\sqrt{g}}{4}\left[I_{\frac{1}{4}}\left(\frac{1}{8g}\right)+I_{-\frac{1}{4}}\left(\frac{1}{8g}\right)\mp
i\frac{\sqrt{2}}{\pi}K_{\frac{1}{4}}\left(\frac{1}{8g}\right)\right]$ (46)
$\displaystyle=$ $\displaystyle\widehat{\cal J}_{0}+\widehat{\cal
J}_{\mp}=\widehat{\cal J}_{0}\mp\frac{1}{2}(\widehat{\cal J}_{+}-\widehat{\cal
J}_{-}).$

From Eqs.(40)(41), one can find that

$\displaystyle{\mathfrak{S}}_{0}^{\pm 1/2}[{\cal J}_{0}]={\cal J}_{\rm
AC}^{\arg(\lambda)=\mp\pi},\qquad{\mathfrak{S}}_{0}^{\pm 1/2}[{\cal J}_{\rm
AC}^{\arg(\lambda)=\pm\pi}]={\cal J}_{0},$
$\displaystyle{\mathfrak{S}}_{0}^{\pm 1}[{\cal J}_{\rm
AC}^{\arg(\lambda)=\pm\pi}]={\cal J}_{\rm AC}^{\arg(\lambda)=\mp\pi},$ (47)
$\displaystyle\widehat{\mathfrak{S}}_{0}^{\pm 1/2}[\widehat{\cal
J}_{0}]=\widehat{\cal J}_{\rm
AC}^{\arg(\lambda)=\mp\pi},\qquad\widehat{\mathfrak{S}}_{0}^{\pm
1/2}[\widehat{\cal J}_{\rm AC}^{\arg(\lambda)=\pm\pi}]=\widehat{\cal J}_{0},$
$\displaystyle\widehat{\mathfrak{S}}_{0}^{\pm 1}[\widehat{\cal J}_{\rm
AC}^{\arg(\lambda)=\pm\pi}]=\widehat{\cal J}_{\rm AC}^{\arg(\lambda)=\mp\pi},$
(48) $\displaystyle{\cal S}_{0_{\pm}}[{\cal J}_{0}]=\widehat{\cal J}_{\rm
AC}^{\arg(\lambda)=\mp\pi},\qquad\ \ \ \,{\cal S}_{0_{\pm}}[{\cal J}_{\rm
AC}^{\arg(\lambda)=\pm\pi}]=\widehat{\cal J}_{0}.$ (49)

Therefore, the zero-dimensional ${\cal PT}$ symmetric model satisfies the
relations in Fig. 1.

It is remarkable to mention that this picture directly corresponds to
Lefschetz thimble decomposition. ${\cal S}_{0_{\pm}}[{\cal J}_{0}]$ and ${\cal
S}_{0_{\pm}}[{\cal J}_{\pm}]$ (divided by $g$) are nothing but thimble-
integrations from a perturbative saddle at $x=0$ and non-perturbative saddles
at $x=\pm\frac{1}{\sqrt{2g}}$, respectively. See Ref. Ai:2022csx  in detail.

###  3.2 Exact WKB analysis

We review EWKB and explain a procedure to obtain energy quantization
conditions (QCs). See, for example, Refs. Voros1983 ; Silverstone ; Schafke1 ;
DDP2 ; DP1 ; Kawai1 ; AKT1 ; Iwaki1  in detail.

We consider the following Schrödinger equation given by a potential $V(x)$:

$\displaystyle{\cal L}=-\hbar^{2}\partial_{x}^{2}+V(x)-E,\qquad{\cal
L}\psi(x)=0,\qquad(x\in{\mathbb{C}},\ \ E,\hbar\in{\mathbb{R}}_{>0})$ (50)

where $E$ is an energy and $\psi$ is a wavefunction. The variable, $x$, is
normally taken as a real value, but we extend it to be complex-valued for
analytic continuation of the wavefunction. For simplicity, we assume that the
potential $V$ is independent on $\hbar$ and a polynomial of $x$ bounded in the
asymptotic limit, $x\rightarrow\pm\infty$, i.e., the wavefunction $\psi$ is
normalizable along the real axis. A main difference from the standard WKB is
to take into account of its all orders with respect to $\hbar$. We construct
the wavefunction by preparing an ansatz expanded by $\hbar$, which is given by

$\displaystyle\psi_{a}(x,\hbar)=\sigma(\hbar)\exp\left[\int_{a}^{x}dx^{\prime}\,S(x^{\prime},\hbar)\right],\qquad(x\in{\mathbb{C}})$
(51) $\displaystyle
S(x,\hbar)=\sum_{n\in{\mathbb{N}}_{0}}S_{n-1}(x)\hbar^{n-1}\quad\mbox{as}\quad\hbar\rightarrow
0_{+},$ (52)

where $\sigma(\hbar)$ is an integration constant generally depending on
$\hbar$, and $a$ is a normalization point. Substituting $\psi$ into the
Schrödinger equation (50) leads to Riccati equation in terms of $S(x,\hbar)$
as

$\displaystyle S(x,\hbar)^{2}+\partial_{x}S(x,\hbar)=\hbar^{-2}Q(x),\qquad
Q(x):=V(x)-E.$ (53)

From Riccati equation (53), $S(x,\hbar)$ is recursively obtained order by
order, as

$\displaystyle S_{-1}(x)=\pm\sqrt{Q(x)},\qquad
S_{0}(x)=-\frac{\partial_{x}\log Q(x)}{4},$ $\displaystyle
S_{+1}(x)=\pm\frac{1}{8\sqrt{Q(x)}}\left[\partial_{x}^{2}\log
Q(x)-\frac{(\partial_{x}\log Q(x))^{2}}{4}\right],\qquad\cdots.$ (54)

Since Riccati equation is quadratic, we have two independent solutions which
enable to be distinguished by the sign, $\pm$, in $S(x,\hbar)$. One can also
find from Eq.(53) that $S(x,\hbar)$ can be decomposed into two parts; the one
part consists of terms with the sign $\pm$ determined by choice of
$S_{-1}(x)=\pm\sqrt{Q(x)}$ and the other part contains terms independent on
the sign of $S_{-1}(x)$. By denoting $S_{\rm od}(x,\hbar)$ and $S_{\rm
ev}(x,\hbar)$ for the former and latter parts, respectively, one can express
the total $S(x,\hbar)$ as $S(x,\hbar)=\pm S_{\rm od}(x,\hbar)+S_{\rm
ev}(x,\hbar)$, and $S_{\rm ev}(x,\hbar)$ can be written down by $S_{\rm
od}(x,\hbar)$ as555 If $Q(x)$ includes $\hbar^{\alpha}$ with
$\alpha\in{\mathbb{N}}$, then $S_{\rm odd}(x,\hbar)$ contains both odd and
even powers of the $\hbar$-expansion. The degenerate Weber-type Stokes graph
in Appendix A is the case.

$\displaystyle S_{\rm
od}(x,\hbar)=\sum_{n\in{\mathbb{N}}_{0}}S_{2n-1}(x)\hbar^{2n-1},\qquad
S_{-1}(x)=\sqrt{Q(x)},$ (55) $\displaystyle S_{\rm
ev}(x,\hbar)=\sum_{n\in{\mathbb{N}}_{0}}S_{2n}(x)\hbar^{2n}=-\frac{1}{2}\partial_{x}\log
S_{\rm od}(x,\hbar).$ (56)

Therefore, the wavefunction (51) is given by

$\displaystyle\psi_{a\pm}(x,\hbar)=\frac{\sigma_{\pm}(\hbar)}{\sqrt{S_{\rm
od}(x,\hbar)}}\exp\left[\pm\int_{a}^{x}dx^{\prime}\,S_{\rm
od}(x^{\prime},\hbar)\right].$ (57)

After formulating the wavefunction, we then draw Stokes graph, that is defined
by $\int dx\,S_{{\rm od},-1}(x)$ of $x\in{\mathbb{C}}$ and provides a
structure of Borel summability of the wavefunction. A Stokes graph normally
consists of three kinds of objects: turning points, Stokes lines, and branch-
cuts. Turning points are defined from $Q(x)$ as

$\displaystyle{\rm TP}:=\\{x\in{\mathbb{C}}\,|\,Q(x)=0\\}.$ (58)

These are used as a normalization point, denoted by $a$ in Eq.(57), in the
wavefunction. A Stokes line is defined as a line on the complex $x$-plane
emerging from a turning point and satisfying the condition that

$\displaystyle{\rm
Im}\left[\frac{1}{\hbar}\int_{a_{j}}^{x}dx^{\prime}\,S_{{\rm
od},-1}(x^{\prime})\right]=0,\qquad a_{j}\in{\rm TP}.$ (59)

Along the lines, ${\rm Re}\left[\int_{a_{j}}^{x}dx^{\prime}\,S_{{\rm
od},-1}(x^{\prime})\right]$ is monotonically increasing (resp. decreasing), so
that we add labels, $+$ (resp. $-$), to an asymptotic domain where ${\rm
Re}\left[\int_{a_{j}}^{x}dx^{\prime}\,S_{{\rm
od},-1}(x^{\prime})\right]\rightarrow+\infty$ (resp. $-\infty$). In addition,
without some exceptions, branch-cuts normally have a role of swapping the
independent solutions, $\psi_{\pm}$. Fig. 2 is an example of Stokes graph in
the case of a double-well potential, which would be also helpful to follow the
below explanations.

(a) $\arg(\hbar)=0_{+}$ (b) $\arg(\hbar)=0_{-}$
---
Figure 2: Example of Stokes graph, a double-well potential. Turning points are
denoted by blue dots, $a_{1,\cdots,4}$. The black solid and red wave lines
mean Stokes lines and branch-cuts, respectively. The label, $+$ (resp. $-$),
denotes an asymptotic domain where ${\rm
Re}\left[\int_{a_{j}}^{x}dx^{\prime}\,S_{{\rm
od},-1}(x^{\prime})\right]\rightarrow+\infty$ (resp. $-\infty$). We perform
analytic continuation along the blue line slightly below the real axis from
$x=-\infty$ to $x=+\infty$. In these cycles, only the $B$-cycle is non-
perturbative, and ${\rm C}_{\rm NP,\theta=0}=\\{B\\}$. The other cycles, $A$
and $C$, intersect with $B$ once, so that action of the Stokes automorphism to
them is non-trivial and related to $B$. In this case, the intersection number
is given by $\langle A,B\rangle=-\langle C,B\rangle=-1$.

When the wavefunction is Borel non-summable, there exist two possibilities for
the reason: movable singularity and fixed singularity. A movable singularity
means a singularity on the Borel plane expressed by Stokes lines, and the
wavefunction becomes Borel non-summable when it stays exactly on the lines.
Because of that, the wavefunctions defined on each the domain surrounded by
the Stokes lines are generally discontinuous to each other. A fixed
singularity arises when a Stokes line reaches another turning point by taking
certain values of $\hbar\in{\mathbb{C}}$, and then two Stokes lines normally
degenerate. This situation is called Stokes phenomenon. In such a case, the
wavefunction is Borel non-summable on the complex $x$-plane entirely, but it
can be resolved by adding an infinitesimal complex phase to $\hbar$ or
parameters in $Q(x)$. A example is drawn in Fig. 3. Below, we define the angle
of the Laplace integral by $\hbar$, as $\theta=\arg(\hbar)$.

Figure 3:  Example of degeneracy of two Stokes lines. When $\arg(\hbar)=0$,
two Stokes lines emerging from turning points (blue dots) on the one side go
into the other side and degenerate. This degeneracy can be resolved by adding
an infinitesimal complex phase to $\hbar$ as $\arg(\hbar)=0_{\pm}$.  Figure 4:
Airy-type Stokes graph defined by a simple turning point, $a$ (blue dot). The
colored arrows crossing the Stokes lines and the branch-cut are expressed by
$M_{+}$ (green), $M_{-}$ (red), and $T$ (blue) in Eq.(63), respectively.

Our main task in EWKB is to obtain a QC by performing analytic continuation
along a certain path on the complex $x$-plane. For the computation, one has to
connect the wavefunctions on each the domain separated by a Stokes line. Here,
we express the wavefunction by a vector form using two independent solutions
of Riccati equation (53), $\psi_{\pm}$, as

$\displaystyle\psi_{a}=\begin{pmatrix}\psi_{a+}\\\ \psi_{a-}\end{pmatrix}.$
(60)

We suppose that $\psi^{\rm I}$ and $\psi^{\rm II}$ are wavefunctions on two
domains separated by a Stokes line, ${\rm I}$ and ${\rm II}$, respectively. A
connection matrix crossing the Stokes line at $x=x_{*}$ is defined such that
their Borel resummed wavefunctions become continuous at $x=x_{*}$, i.e.,

$\displaystyle{\cal S}_{\theta}[\psi^{\rm I}(x_{*}+0_{-})]={\cal
S}_{\theta}[\psi^{\rm I}(x_{*}+0_{+})],$ (61) $\displaystyle\psi^{\rm
I}(x_{*}+0_{+}):=M^{{\rm I}\rightarrow{\rm II}}\psi^{\rm II}(x_{*}+0_{+}),$
(62)

where $M^{\rm I\rightarrow II}$ is the connection matrix, and $x_{*}+0_{-}$
and $x_{*}+0_{+}$ belong to the domain ${\rm I}$ and ${\rm II}$, respectively.
In this paper, we mainly deal with Airy-type Stokes graph, shown in Fig. 4.
Anti-clockwisely crossing Stokes lines and branch-cut emerging from a turning
point, $a$, can be expressed by the connection matrices $M_{\pm}$ and the
branch-cut matrix $T$ given by

$\displaystyle M_{+}=\begin{pmatrix}1&&i\\\ 0&&1\end{pmatrix},\qquad
M_{-}=\begin{pmatrix}1&&0\\\ i&&1\end{pmatrix},\qquad T=\begin{pmatrix}0&-i\\\
-i&0\end{pmatrix},$ (63)

where the subscript of $M_{\pm}$ indicates the label of asymptotic behavior
attached to the Stokes line we are crossing. In addition, their inverse
matrices correspond to crossing them clockwisely. Such a connection formula is
locally formulated around a turning point, so that one has to change the
normalization point depending on which turning point the Stokes line emerges
from. The normalization matrix changing a normalization point from $a_{k}$ to
$a_{j}$ is given by

$\displaystyle
N_{a_{j},a_{k}}:=\begin{pmatrix}e^{+\int_{a_{k}}^{a_{j}}dx\,S_{{\rm
od}}(x,\hbar)}&\\\ &e^{-\int_{a_{k}}^{a_{j}}dx\,S_{{\rm
od}}(x,\hbar)}\end{pmatrix}=N_{a_{k},a_{j}}^{-1},\qquad a_{j},a_{k}\in{\rm
TP}.$ (64)

Acting those matrices (63)(64) to the wavefunction one by one along a path of
analytic continuation yields a monodromy matrix. When degeneracies are induced
at $\theta=0$ like Fig. 3, one has to resolve the degeneracies by taking
$\theta=0_{\pm}$, and thus the resulting monodromy matrix depends on the
complex phase. For example, for a double-well potential shown in Fig. 2, those
are given by

$\displaystyle{\cal S}_{0_{\pm}}[\psi_{a_{1}}^{0_{\pm}{\rm I}}]={\cal
S}_{0_{\pm}}[{\cal M}^{0_{\pm}}\psi_{a_{1}}^{0_{\pm}{\rm VI}}],\qquad$
$\displaystyle{\cal
M}^{0_{+}}=M_{+}N_{a_{1},a_{2}}M_{+}N_{a_{2},a_{3}}M_{+}M_{-}N_{a_{3},a_{4}}M_{-}N_{a_{4},a_{1}},$
(65) $\displaystyle{\cal
M}^{0_{-}}=M_{+}N_{a_{1},a_{2}}M_{+}M_{-}N_{a_{2},a_{3}}M_{-}N_{a_{3},a_{4}}M_{-}N_{a_{4},a_{1}},$
(66)

where ${\cal S}_{0_{+}}[\psi_{a}^{0_{+}\bullet}]={\cal
S}_{0_{-}}[\psi_{a}^{0_{-}\bullet}]$666 The wavefunctions,
$\psi_{a}^{0_{\pm}}$, also have a discontinuity caused by a fixed singularity
at $\arg(\hbar)=0$ and can be constructed from $\psi_{a}$ in Eq.(60). Since
this effect is consequently irrelevant to QCs and their solutions, we do not
argue $\psi^{0_{\pm}}_{a}$. See Ref. Takei2  for the construction. .
Normalizability to the wavefunction in the asymptotic limit,
$x\rightarrow\pm\infty$, requires being zero to a corresponding component of
the monodromy matrix, that gives a QC denoted by ${\mathfrak{D}}^{0_{\pm}}$:

$\displaystyle{\cal M}^{0_{\pm}}_{jk}={\mathfrak{D}}^{0_{\pm}}=0,$ (67)

where $(j,k)$ is determined by normalizability of the wavefunction as

$\displaystyle(j,k)=\begin{cases}(1,1)&\mbox{if}\ \ \psi_{-}^{0_{\pm}{\rm
I}}(-\infty)=\psi_{+}^{0_{\pm}{\rm I}}(+\infty)=0\\\ (1,2)&\mbox{if}\ \
\psi_{-}^{0_{\pm}{\rm I}}(-\infty)=\psi_{-}^{0_{\pm}{\rm I}}(+\infty)=0\\\
(2,1)&\mbox{if}\ \ \psi_{+}^{0_{\pm}{\rm I}}(-\infty)=\psi_{+}^{0_{\pm}{\rm
I}}(+\infty)=0\\\ (2,2)&\mbox{if}\ \ \psi_{+}^{0_{\pm}{\rm
I}}(-\infty)=\psi_{-}^{0_{\pm}{\rm I}}(+\infty)=0\end{cases}.$ (68)

The QCs are functions of the energy $E$, so that solving
${\mathfrak{D}}^{0_{\pm}}(E)=0$ gives a energy spectrum, but containing a
discontinuity at $\theta=0$.

Even if the resulting QCs are Borel non-summable at $\theta=0$ such that
${\mathfrak{D}}^{0_{+}}\neq{\mathfrak{D}}^{0_{-}}$, their Borel resummed forms
are equivalent to each other, i.e., ${\cal
S}_{0_{+}}[{\mathfrak{D}}^{0_{+}}]\propto{\cal
S}_{0_{-}}[{\mathfrak{D}}^{0_{-}}]$. This fact implies existence of Stokes
automorphism that offers a relation between ${\mathfrak{D}}^{0_{+}}$ and
${\mathfrak{D}}^{0_{-}}$. This Stokes automorphism is known as Delabaere-
Dillinger-Pham (DDP) formula DDP2 ; DP1  in the context of EWKB. Generally,
QCs consist of Voros symbols (cycles) defined as a contour integration of
$S_{{\rm od}}$ going around two turning points such that $A=e^{a_{ij}}$, where

$\displaystyle a_{ij}:=\oint_{a_{i}}^{a_{j}}dx\,S_{\rm
od}(x,\hbar)=-a_{ji},\qquad
a_{ij}+a_{jk}=a_{ik},\qquad(a_{i},a_{j},a_{k}\in{\rm TP})$ (69)

and DDP formula tells us perturbative/non-perturbative relations, i.e.,
resurgent relations, among cycles. For cycles constructed only by simple
turning points, e.g. in the case of Fig. 2, DDP formula can be expressed by

$\displaystyle{\cal S}_{\theta+0_{+}}[A_{j}]={\cal
S}_{\theta+0_{-}}[A_{j}]\prod_{B_{k}\in{\rm C}_{{\rm NP},\theta}}(1+{\cal
S}_{\theta+0_{-}}[B_{k}])^{\langle A_{j},B_{k}\rangle},$ (70)
$\displaystyle{\cal S}_{\theta+0_{+}}[B_{k}]={\cal
S}_{\theta+0_{-}}[B_{k}],\qquad B_{k}\in{\rm C}_{{\rm NP},\theta},$ (71)

or identically,

$\displaystyle{\mathfrak{S}}_{\theta}^{\nu=1}[A_{j}]=A_{j}\prod_{B_{k}\in{\rm
C}_{{\rm NP},\theta}}(1+B_{k})^{\langle A_{j},B_{k}\rangle},$ (72)
$\displaystyle{\mathfrak{S}}_{\theta}^{\nu=1}[B_{k}]=B_{k},\qquad B_{k}\in{\rm
C}_{{\rm NP},\theta},$ (73)

where ${\rm C}_{{\rm NP},\theta}$ is a set of purely non-perturbative cycles
that correspond to a degenerated Stokes line induced by a Stokes phenomenon at
$\arg(\hbar)=\theta$, and orientation of the cycles in ${\rm C}_{{\rm
NP},\theta}$ is fixed such that $\lim_{\hbar\rightarrow 0_{+}}B_{k}(\hbar)=0$.
The other cycles labeled by $j\in{\mathbb{N}}$, $A_{j}$, can be arbitrarily
taken as far as being transversal to $B_{k}\in{\rm C}_{{\rm NP},\theta}$, but
those are usually given as a set of cycles going around two turning points
that enable to be connected by a branch-cut. In addition, $\langle A,B\rangle$
denotes the intersection number between two cycles, $A$ and $B$, determined by
their orientation as

$\displaystyle\langle\rightarrow,\uparrow\rangle=\langle\leftarrow,\downarrow\rangle=+1,\qquad\langle\rightarrow,\downarrow\rangle=\langle\leftarrow,\uparrow\rangle=-1.$
(74)

See also Ref. Iwaki1  for the derivation. Notice that ${\rm C}_{{\rm
NP},\theta}$ depends on the value of $\theta$, and thus ${\rm C}_{{\rm
NP},\theta}=\emptyset$ when no Stokes phenomenon happens on the Stokes graph
at $\theta$. The generalization to a one-parameter Stokes automorphism with
$\nu\in{\mathbb{R}}$, denoted by ${\mathfrak{S}}^{\nu}_{\theta}$, is available
by replacing the intersection number in Eq.(72) as $\langle
A_{j},B_{k}\rangle\rightarrow\langle A_{j},B_{k}\rangle\times\nu$. Eq.(73)
does not change for any $\nu$. From Eqs.(72)(73), the median resummed forms
are obtained as

$\displaystyle\widehat{A}_{j}={\cal
S}_{\theta+0_{\pm}}[A_{j}^{\theta+0_{\pm}}],\qquad
A^{\theta+0_{\pm}}_{j}:={\mathfrak{S}}^{\mp
1/2}_{\theta}[A_{j}]=A_{j}\prod_{B_{k}\in{\rm C}_{{\rm
NP},\theta}}(1+B_{k})^{\mp\langle A_{j},B_{k}\rangle/2},$ (75)
$\displaystyle\widehat{B}_{k}={\cal
S}_{\theta+0_{\pm}}[B_{k}^{\theta+0_{\pm}}],\qquad
B_{k}^{\theta+0_{\pm}}=B_{k},$ (76)

where $\widehat{f}={\cal S}_{\rm med,\theta}[f]$. From the properties of Borel
resummation and Stoke automorphism in Eqs.(8)(14)(16), the QCs satisfy

$\displaystyle{\mathfrak{S}}^{+1/2}_{\theta}[{\mathfrak{D}}^{\theta+0_{+}}(A,B)]\propto{\mathfrak{S}}^{-1/2}_{\theta}[{\mathfrak{D}}^{\theta+0_{-}}(A,B)]\
\ =:{\mathfrak{D}}^{\theta}(A,B),$ (77)
$\displaystyle{\mathfrak{D}}^{\theta}(A^{\theta+0_{\pm}},B^{\theta+0_{\pm}})\propto{\mathfrak{D}}^{\theta+0_{\pm}}(A,B).$
(78)

Apparently, ${\cal
S}_{\theta+0_{+}}[{\mathfrak{D}}^{\theta}(A^{\theta+0_{+}},B^{\theta+0_{+}})]\propto{\cal
S}_{\theta+0_{-}}[{\mathfrak{D}}^{\theta}(A^{\theta+0_{-}},B^{\theta+0_{-}})]={\mathfrak{D}}^{\theta}(\widehat{A},\widehat{B})$.
This is a simple way to make sure of the relation that

$\displaystyle{\cal S}_{{\rm
med},\theta}[{\mathfrak{D}}^{\theta}(A,B)]\overset{\hbar\rightarrow
0_{+}}{\sim}{\mathfrak{D}}^{\theta}(A,B).$ (79)

It is quite crucial to remind that the energy, $E$, is a free-parameter in the
DDP formula in Eqs.(70)-(73) and determined by solving the QC,
${\mathfrak{D}}^{\theta}=0$ (or $\widehat{\mathfrak{D}}^{\theta}=0$). In this
sense, the DDP formula of cycles is not relevant to the energy solution
directly. However, one can formulate a one-parameter Stokes automorphism for a
formal transseries of the energy solution by combining the DDP formula for
cycles with the QC. This issue would be discussed in Sec. 4.2 and Appendix B.

The complex phase $\theta$ to resolve a degeneracy of Stokes lines is normally
introduced by $\arg(\hbar)$, but the other parameters in a given theory can be
used if it induces the similar effect. In our analysis, we use a coupling
constant, $\lambda$ (resp. $g$), as $\theta=\arg(\lambda)$ (resp.
$\theta=\arg(g)$).

##  4 Quartic potential with a quadratic term

In this section, we apply EWKB to a quartic potential with $\omega>0$. In Sec.
4.1, we firstly analyze the Hermitian potential as a warm-up. In Sec. 4.2, we
then consider the ${\cal PT}$ symmetric potential. In Sec. 4.3, from
transseries solutions of the ${\cal PT}$ and the AC QCs obtained in Sec. 4.2,
we reformulate the ABS conjecture by Borel resummation theory.

###  4.1 Hermitian potential: $V=\omega^{2}x^{2}+\lambda x^{4}$

We consider the Hermitian potential. In Sec. 4.1.1, we demonstrate the
procedure that we repeatedly perform in the later analyses. In Sec. 4.1.2, we
also discuss analytic continuation of the QCs which will be helpful for
analysis of the ${\cal PT}$ symmetric potential and a consideration in Sec.
5.3.

####  4.1.1 Quantization condition and energy spectrum

(a) $\arg(\lambda)=0_{+}$ (b) $\arg(\lambda)=0_{-}$
---
Figure 5: Stokes graph of the Hermitian potential with $\omega>0$. We take the
analytic continuation along the path denoted the blue line to obtain the
monodromy matrix.

We apply EWKB to a Schrödinger equation of the Hermitian potential defined as

$\displaystyle{\cal L}=-\hbar^{2}\partial_{x}^{2}+\omega^{2}x^{2}+\lambda
x^{4}-E.\qquad{\cal L}\psi=0,$ (80)

where $\omega,\lambda,E\in{\mathbb{R}}_{>0}$, and we take $E$ sufficiently
small. From Eq.(58), turning points are given by

$\displaystyle{\rm TP}$ $\displaystyle=$
$\displaystyle\left\\{a_{1}=-\sqrt{\frac{\sqrt{4E\lambda+\omega^{4}}-\omega^{2}}{2\lambda}},\
a_{2}=-i\sqrt{\frac{\sqrt{4E\lambda+\omega^{4}}+\omega^{2}}{2\lambda}},\right.$
(81) $\displaystyle\left.\ \
a_{3}=+i\sqrt{\frac{\sqrt{4E\lambda+\omega^{4}}+\omega^{2}}{2\lambda}},\
a_{4}=+\sqrt{\frac{\sqrt{4E\lambda+\omega^{4}}-\omega^{2}}{2\lambda}}\right\\}.$

Fig. 5 shows the Stokes graph drawn by Eq.(80). Because of ${\cal P}$ symmetry
in the potential, $V(-x)=V(x)$, one has two independent cycles, $A_{1,2}$,
defined as

$\displaystyle A_{1}:=e^{a_{12}}=e^{a_{34}},\qquad
A_{2}:=e^{a_{13}}=e^{a_{24}},$ (82)

where $a_{jk}$ is defined by Eq.(69). Performing analytic continuation along a
line slightly below the real axis generates a QC by connection matrices, as
explained in Sec. 3.2. Since a Stokes phenomenon happens at $\arg(\lambda)=0$,
the monodromy matrix depends on the complex phase $\arg(\lambda)=0_{\pm}$ and
is obtained as

$\displaystyle{\cal
M}^{0_{+}}=M_{+}N_{a_{1},a_{2}}M_{-}^{-1}N_{a_{2},a_{3}}M_{+}N_{a_{3},a_{4}}M_{+}N_{a_{4},a_{1}},$
(83) $\displaystyle{\cal
M}^{0_{-}}=M_{+}N_{a_{1},a_{3}}M_{+}N_{a_{3},a_{2}}M_{-}^{-1}N_{a_{2},a_{4}}M_{+}N_{a_{4},a_{1}}.$
(84)

Normalizability of the wavefunction determines the boundary condition and
requires ${\cal M}^{0_{\pm}}_{12}=0$ by Eq.(68). As a result, one can write
down the QCs represented by cycles as

$\displaystyle{\mathfrak{D}}^{0_{+}}_{\cal H}\propto
1+\frac{(2+A_{1})A_{2}}{1+A_{1}^{-1}A_{2}}=1+\frac{(2+A_{1})A_{1}D}{1+D},$
(85) $\displaystyle{\mathfrak{D}}^{0_{-}}_{\cal
H}\propto(1+A_{2})^{2}+A_{1}A_{2}=(1+A_{1}D)^{2}+A_{1}^{2}D,$ (86)

where we defined $D=A_{1}^{-1}A_{2}$ exponentially damping without oscillation
as a non-perturbative cycle, ${\rm C}_{{\rm NP},\arg(\lambda)=0}=\\{D\\}$.

From Eqs.(72)-(73), the DDP formula for $\arg(\lambda)=0$ is given by

$\displaystyle{\mathfrak{S}}_{0}^{\nu}[A_{j=1,2}]=A_{j}(1+D)^{\nu},\qquad{\mathfrak{S}}_{0}^{\nu}[D]=D.\qquad(\nu\in{\mathbb{R}})$
(87)

The QC removed the discontinuity, ${\mathfrak{D}}^{0}_{\cal H}$, is available
by Eq.(77) as ${\mathfrak{S}}_{0}^{+1/2}[{\mathfrak{D}}^{0_{+}}_{\cal
H}]\propto{\mathfrak{S}}_{0}^{-1/2}[{\mathfrak{D}}^{0_{-}}_{\cal
H}]=:{\mathfrak{D}}^{0}_{\cal H}$, where

$\displaystyle{\mathfrak{D}}^{0}_{\cal H}\propto
1+A_{2}\left(A_{1}+\frac{2}{\sqrt{1+A_{1}^{-1}A_{2}}}\right)=1+A_{1}D\left(A_{1}+\frac{2}{\sqrt{1+D}}\right).$
(88)

It is worth to see that the median resummed form of ${\mathfrak{D}}^{0}_{\cal
H}$ has the same to Eq.(88). The discontinuous QCs in Eqs.(85)(86) can be
reexpressed as

$\displaystyle{\mathfrak{D}}^{0_{\pm}}_{\cal H}\propto
1+A_{2}^{0_{\pm}}\left(A_{1}^{0_{\pm}}+\frac{2}{\sqrt{1+(A_{1}^{0_{\pm}})^{-1}A_{2}^{0_{\pm}}}}\right)=1+A_{1}^{0_{\pm}}D^{0_{\pm}}\left(A_{1}^{0_{\pm}}+\frac{2}{\sqrt{1+D^{0_{\pm}}}}\right),$
$\displaystyle A_{j=1,2}^{0_{\pm}}={\mathfrak{S}}^{\mp
1/2}_{0}[A_{j}]=A_{j}(1+D)^{\mp 1/2},\qquad D^{0_{\pm}}=D.$ (89)

Since $\widehat{A}_{j}={\cal S}_{{\rm med},0}[A_{j}]={\cal
S}_{0_{\pm}}[A_{j}^{0_{\pm}}]$ and $\widehat{D}={\cal S}_{{\rm
med},0}[D]={\cal S}_{0_{\pm}}[D^{0_{\pm}}]$, one can immediately find that
$\widehat{\mathfrak{D}}^{0}_{\cal H}={\cal S}_{{\rm
med},0}[{\mathfrak{D}}^{0}_{\cal H}]\propto{\cal
S}_{0_{\pm}}[{\mathfrak{D}}^{0_{\pm}}_{\cal H}]$.

(a) $\arg(\lambda)=0_{+}$ (b) $\arg(\lambda)=0_{-}$
---
Figure 6: Stokes graph of the Hermitian potential with a quadratic term.
Varying $E_{0}\rightarrow 0_{+}$, the two simple turning points, $a_{1}$ and
$a_{4}$, collide to each other and then become a double turning point (green
dot) at $E_{0}=0$, where $E_{0}$ is the classical part of the energy.

One can formally find the energy spectrum by solving the median resummed QC,
but it is technically almost impossible to directly see details from it. So
that, our strategy is to find a transseries solution from
${\mathfrak{D}}^{0}_{\cal H}=0$ in Eq.(88). However, QCs formulated by the
Airy-type Stokes graph usually have a problem such that the ordering of a
double-expansion in terms of $E$ and $\hbar$ is generally non-commutative with
each other. Thus, the resulting transseries from the QC does not always
correspond to a correct energy solution. The main reason is that the energy
solution is $O(\hbar)$ and exceeds the applicable limit or the assumption of
the Airy-type connection formula. In such a case, one needs to modify the
Schrödinger equation as

$\displaystyle{\cal L}=-\hbar^{2}\partial_{x}^{2}+\omega^{2}x^{2}+\lambda
x^{4}-\widetilde{E}\hbar,\qquad{\cal L}\psi=0,$ (90)

where $E=E_{0}+\widetilde{E}\hbar>0$ with $E_{0}=0$ and
$\widetilde{E}=O(\hbar^{0})$. $E_{0}$ has an important role for the Stokes
graph and is indeed a control parameter of it. Taking $E_{0}=0$ causes a
bifurcation such that two simple turning points corresponding to a harmonic
oscillator collide to each other and become a double turning point. In
consequence, the set of turning points in Eq.(81) changes to

$\displaystyle{\rm TP}$ $\displaystyle=$ $\displaystyle\left\\{a_{1}=a_{4}=0,\
a_{2}=-i\frac{\omega}{\sqrt{\lambda}},\
a_{3}=+i\frac{\omega}{\sqrt{\lambda}}\right\\}.$ (91)

The Stokes graph is shown in Fig. 6. The local Stokes graph defined by a
double turning point is called as degenerate Weber (DW)-type Stokes graph,
that gives another form of connection matrices Takei3 ; DDP2 . See also
Appendix A. In spite of the difference between the Airy-type and the DW-type
connection formulas, the QCs and the DDP formula expressed by the Airy-type
cycles can be directly translated into the DW-type DP1 ; Zinn-Justin:2004qzw ;
Sueishi:2021xti . For a simplified notation, by scaling dimensions in Eq.(90)
given by

$\displaystyle[x]=\frac{1}{2},\qquad[\hbar]=1,\qquad[\omega]=0,\qquad[\lambda]=-1,\qquad[\widetilde{E}]=0,$
(92)

we rescale the parameters as

$\displaystyle
x\rightarrow\frac{\omega}{\sqrt{\lambda}}x,\qquad\widetilde{E}\rightarrow\omega\widetilde{E},\qquad\hbar\rightarrow\frac{\omega^{3}}{\lambda}\hbar.$
(93)

Then, the Schrödinger equation (90) becomes dimensionless as

$\displaystyle\frac{\lambda}{\omega^{4}}{\cal L}\rightarrow{\cal
L}=-\hbar^{2}\partial_{x}^{2}+x^{2}+x^{4}-\widetilde{E}\hbar.$ (94)

The QCs of the DW-type can be directly obtained by replacing $A_{1,2}$ as

$\displaystyle A_{1}\rightarrow{\mathfrak{A}}^{1/2}{\mathfrak{B}},\qquad
A_{2}\rightarrow{\mathfrak{A}}^{1/2}{\mathfrak{B}}^{-1},$ (95)
$\displaystyle{\mathfrak{A}}=e^{-2\pi
iF},\qquad{\mathfrak{B}}=\frac{\sqrt{2\pi}e^{-G}}{\Gamma(1/2+F)}\left(\frac{\hbar}{2}\right)^{-F},$
(96)

where $F$ and $G$ are formal expansions of $\hbar$ and $\widetilde{E}$. $F$
can be computed by a residue integration rotating clockwisely around the
double turning point, but we do not argue $G$ in this subsection. By this
procedure, one can obtain the modified QC from Eq.(88) as777 Of course, one
can obtain QCs by beginning with the DW connection formula. In this case, the
QCs are given by  $\displaystyle{\mathfrak{D}}^{0_{+}}_{\cal
H}\propto{\mathfrak{D}}^{0_{-}}_{\cal H}\propto{\mathfrak{B}},$ (97) but the
resulting QCs do not have dependence on $\arg(\lambda)=0_{\pm}$. As we will
see, this observation is not a contradiction to Eqs.(88)(98).

$\displaystyle{\mathfrak{D}}^{0}_{\cal H}\propto
1+{\mathfrak{A}}\left(1+2\frac{{\mathfrak{A}}^{-1/2}{\mathfrak{B}}}{\sqrt{1+{\mathfrak{B}}^{2}}}\right).$
(98)

The next step is to solve the QC (98). The perturbative part is given by
${\mathfrak{D}}^{0}_{\cal H}\sim 1+{\mathfrak{A}}$ = 0, i.e.,

$\displaystyle F(\widetilde{E},\hbar)=\frac{q}{2},\qquad q\in
2{\mathbb{Z}}+1,$ (99)

where $q$ is an energy level. As described above, $F$ is obtained by the
residue integration around the double turning point, as

$\displaystyle F(\widetilde{E};\hbar)=-{\rm Res}_{x=0}S_{\rm
od}(x,\widetilde{E};\hbar),$ (100)

and the formal power expansion of $F$ is given by

$\displaystyle F(\widetilde{E},\hbar)$ $\displaystyle=$
$\displaystyle-\frac{\widetilde{E}}{2}+\frac{3\left(1+\widetilde{E}^{2}\right)}{16}\hbar-\frac{5\left(17\widetilde{E}+7\widetilde{E}^{3}\right)}{128}\hbar^{2}$
(101)
$\displaystyle+\frac{105\left(19+50\widetilde{E}^{2}+11\widetilde{E}^{4}\right)}{2048}\hbar^{3}+O(\hbar^{4}).$

By setting the asymptotic form to $\widetilde{E}$ as $\widetilde{E}_{\cal
H}(\hbar)\sim\sum_{n\in{\mathbb{N}}_{0}}e_{n}\hbar^{n}$, the coefficients
$e_{n}$ can be obtained by solving the QC order by order. Specifically, the
first four coefficients are given by

$\displaystyle\widetilde{E}_{\cal H}(\hbar)$ $\displaystyle=$ $\displaystyle
q+\frac{3\left(q^{2}+1\right)}{8}\hbar-\frac{q\left(17q^{2}+67\right)}{64}\hbar^{2}$
(102)
$\displaystyle+\frac{3\left(125q^{4}+1138q^{2}+513\right)}{1024}\hbar^{3}+O(\hbar^{4}),$

where the energy level is constrained as $q\in 2{\mathbb{N}}_{0}+1$ to give a
positive energy. This also means that $F$ is a negative half-integer,
$F\in-{\mathbb{N}}+\frac{1}{2}$, and thus the non-perturbative effects do not
appear because ${\mathfrak{B}}=0$ due to the gamma function in Eq.(96).
Therefore, the energy solution is Borel summable even though the DDP formula
in Eq.(87) looked non-trivial Bender:1969si ; DDP2 ; Bucciotti:2023trp .

It might be meaningful to turn on the parameters by the scaling law (93). From
$E_{\cal H}=\widetilde{E}_{\cal H}\hbar$, the energy solution can be written
down as the following form:

$\displaystyle\frac{E_{\cal H}}{\omega\hbar}$ $\displaystyle=$ $\displaystyle
q+\frac{3\left(q^{2}+1\right)}{8}\left(\frac{\lambda\hbar}{\omega^{3}}\right)-\frac{q\left(17q^{2}+67\right)}{64}\left(\frac{\lambda\hbar}{\omega^{3}}\right)^{2}$
(103)
$\displaystyle+\frac{3\left(125q^{4}+1138q^{2}+513\right)}{1024}\left(\frac{\lambda\hbar}{\omega^{3}}\right)^{3}+O(\hbar^{4}).$

One can see that taking $\omega\rightarrow 0$ induces divergence. Therefore,
the formal power series is ill-defined at $\omega=0$.

####  4.1.2 Analytic continuation of the quantization condition

(a) $\arg(\lambda)=0$ (b) $\arg(\lambda)=\theta_{*}$ (c) $\arg(\lambda)=+\pi$
---
Figure 7:  Stokes graphs for $\arg(\lambda)=0,\theta_{*}=\frac{3\pi
E|\lambda|}{4\omega^{4}}+O(E^{3/2})$, and $+\pi$. The similar phenomena occur
at $-\theta_{*}$ and $-\pi$ by varying $\arg(\lambda)$ from $0$ to $-\pi$.

Before moving on the ${\cal PT}$ symmetric case, it is worth to consider
analytic continuation of the QCs by varying $\theta=\arg(\lambda)$ for a
complex $\lambda$. When QCs are expressed by a formal transseries, one
generally needs to take into account of each Stokes phenomenon appearing
during the analytic continuation. Below, we consider the Airy-type cycles, but
the similar consideration works for the DW-type thanks to the correspondence
of cycle representations.

Varying $\arg(\lambda)$ from $0$ to $+\pi$ (resp. $-\pi$) rotates the two
turning points, $a_{2}$ and $a_{3}$, (resp. anti-)clockwisely, and they reach
the real axis when $E_{0}>0$ is small enough. During the process, after
resolving the degeneracy at $\arg(\lambda)=0$, Stokes phenomena happen twice
at $\arg(\lambda)=\pm\frac{3\pi
E|\lambda|}{4\omega^{4}}+O(E^{3/2})=:\pm\theta_{*}$ and
$\arg(\lambda)=\pm\pi$. Fig. 7 shows the Stokes graphs in which a Stokes
phenomenon appears during the analytic continuation. Each the phenomenon has
its own DDP formula, and those can be expressed by using $A_{1,2}$ in Eq.(82)
as

$\displaystyle{\mathfrak{S}}_{+\theta_{*}}^{\nu}[A_{1}^{-1}]=A^{-1}_{1},\qquad\qquad\quad\
\ {\mathfrak{S}}_{+\theta_{*}}^{\nu}[A_{2}]=A_{2}(1+A_{1}^{-1})^{2\nu},$
$\displaystyle{\mathfrak{S}}_{+\pi}^{\nu}[{\cal A}]={\cal A}(1+{\cal
B})^{-2\nu},\quad\quad\quad\,{\mathfrak{S}}_{+\pi}^{\nu}[{\cal B}]={\cal
B},\qquad({\cal A}=A_{1}A_{2},\ {\cal B}=A_{1})$ (104)
$\displaystyle{\mathfrak{S}}_{-\theta_{*}}^{\nu}[A_{1}]=A_{1}(1+A_{2})^{2\nu},\qquad\,{\mathfrak{S}}_{-\theta_{*}}^{\nu}[A_{2}]=A_{2},$
$\displaystyle{\mathfrak{S}}_{-\pi}^{\nu}[{\cal A}]={\cal A}(1+{\cal
B})^{-2\nu},\quad\quad\ \ \ \,{\mathfrak{S}}_{-\pi}^{\nu}[{\cal B}]={\cal
B},\qquad({\cal A}=A_{1}A_{2},\ {\cal B}=A_{2}^{-1})$ (105)

where ${\rm C}_{{\rm NP},+\theta_{*}}=\\{A_{1}^{-1}\\}$, ${\rm C}_{{\rm
NP},-\theta_{*}}=\\{A_{2}\\}$, and ${\rm C}_{{\rm NP},\pm\pi}=\\{{\cal B}\\}$.
These give us analytic continuation of the QCs from $\arg(\lambda)=0_{+}$
(resp. $0_{-}$) to $\arg(\lambda)=+\pi+0_{-}$ (resp. $-\pi+0_{+}$) as888 In
these equations, the symbol “$0_{\pm}$” is a different sense from the same
symbol in the ABS conjecture. $\lambda=-g+i0_{\pm}$ in the ABS conjecture
corresponds to $\lambda=e^{\pm\pi i}g$ in our notation. Our “$0_{+}$” (resp.
“$0_{-}$”) means the limit from the right (resp. left) to the discontinuity on
the Borel plane.

$\displaystyle{\mathfrak{D}}^{+\theta_{*}+0_{+}}_{\cal
H}={\mathfrak{S}}_{\theta_{*}}^{-1}[{\mathfrak{D}}^{+\theta_{*}+0_{-}}_{\cal
H}],\quad\ \ \qquad{\mathfrak{D}}^{+\pi+0_{+}}_{\cal
H}={\mathfrak{S}}_{+\pi}^{-1}[{\mathfrak{D}}^{+\pi+0_{-}}_{\cal H}],$ (106)
$\displaystyle{\mathfrak{D}}^{-\theta_{*}+0_{-}}_{\cal
H}={\mathfrak{S}}_{-\theta_{*}}^{+1}[{\mathfrak{D}}^{-\theta_{*}+0_{+}}_{\cal
H}],\qquad\quad\ {\mathfrak{D}}^{-\pi+0_{-}}_{\cal
H}={\mathfrak{S}}_{-\pi}^{+1}[{\mathfrak{D}}^{-\pi+0_{+}}_{\cal H}].$ (107)

Notice that the cycles and the QCs are analytic under the change of
$\arg(\lambda)$ until encountering the next Stokes phenomenon, i.e.,

$\displaystyle{\mathfrak{D}}^{0_{+}}_{\cal H}\overset{\underset{{\rm
sym}.}{}}{=}{\mathfrak{D}}^{+\theta_{*}+0_{-}}_{\cal H}\quad\quad\ \
\mbox{for}\quad 0<\arg(\lambda)<\theta_{*},$
$\displaystyle{\mathfrak{D}}^{+\theta_{*}+0_{+}}_{\cal
H}\overset{\underset{{\rm sym}.}{}}{=}{\mathfrak{D}}^{+\pi+0_{-}}_{\cal
H}\quad\mbox{for}\quad\theta_{*}<\arg(\lambda)<\pi,$ (108)
$\displaystyle{\mathfrak{D}}^{0_{-}}_{\cal H}\overset{\underset{{\rm
sym}.}{}}{=}{\mathfrak{D}}^{-\theta_{*}+0_{+}}_{\cal H}\quad\quad\ \
\mbox{for}\quad-\theta_{*}<\arg(\lambda)<0,$
$\displaystyle{\mathfrak{D}}^{-\theta_{*}+0_{-}}_{\cal
H}\overset{\underset{{\rm sym}.}{}}{=}{\mathfrak{D}}^{-\pi+0_{+}}_{\cal
H}\quad\mbox{for}\quad-\pi<\arg(\lambda)<-\theta_{*},$ (109)

where $\overset{\underset{{\rm sym}.}{}}{=}$ means that those cycle
representations are analytic and keep the same symbolic forms in the domain.
Combining the above relations yields QCs at $\arg(\lambda)=+\pi+0_{-}$ and
$\arg(\lambda)=-\pi+0_{+}$ as

$\displaystyle{\mathfrak{D}}^{+\pi+0_{-}}_{\cal H}\overset{\underset{{\rm
sym}.}{}}{=}{\mathfrak{S}}_{+\theta_{*}}^{-1}[{\mathfrak{D}}^{0_{+}}_{\cal
H}]\propto 1+{\cal A},$ (110) $\displaystyle{\mathfrak{D}}^{-\pi+0_{+}}_{\cal
H}\overset{\underset{{\rm
sym}.}{}}{=}{\mathfrak{S}}_{-\theta_{*}}^{+1}[{\mathfrak{D}}^{0_{-}}_{\cal
H}]\propto 1+{\cal A}.$ (111)

As a result, by removing the discontinuity at $\arg(\lambda)=\pm\pi$ from
${\mathfrak{D}}^{\pm\pi+0_{\mp}}_{\cal H}$, one finds

$\displaystyle{\mathfrak{D}}^{+\pi}_{\cal
H}={\mathfrak{S}}_{+\pi}^{-1/2}[{\mathfrak{D}}^{+\pi+0_{-}}_{\cal H}]\propto
1+{\cal A}(1+{\cal B}),$ (112) $\displaystyle{\mathfrak{D}}^{-\pi}_{\cal
H}={\mathfrak{S}}_{-\pi}^{+1/2}[{\mathfrak{D}}^{-\pi+0_{+}}_{\cal H}]\propto
1+\frac{{\cal A}}{1+{\cal B}}.$ (113)

As one can readily see, the difference of the QCs by $\arg(\lambda)=\pm\pi$
appears in their non-perturbative part corresponding to ${\cal B}$ and is
expected to eventually propagates to their energy solution. It is notable
that, as long as ${\cal B}\neq 0$, the energy solutions should be complex
values because ${\mathfrak{D}}^{\pm\pi}_{\cal H}$ is not invariant under
complex conjugate. By denoting ${\cal C}$ as complex conjugate, since
$\log{\cal A}\in i{\mathbb{R}}$ and $\log{\cal B}\in{\mathbb{R}}$, one can
readily find that

$\displaystyle{\cal C}[{\mathfrak{D}}^{\pm\pi}_{\cal
H}]\propto{\mathfrak{D}}^{\mp\pi}_{\cal H},\qquad{\cal C}[{\cal A}]={\cal
A}^{-1},\qquad{\cal C}[{\cal B}]={\cal B}.$ (114)

Complex conjugate is nothing but ${\cal T}$-transform, so that Eq.(114) can be
regarded that ${\cal T}$-transform swaps the QCs, ${\mathfrak{D}}^{+\pi}_{\cal
H}\leftrightarrow{\mathfrak{D}}^{-\pi}_{\cal H}$.

In Sec. 4.2, we would construct the ${\cal PT}$ and the AC QCs from the ${\cal
PT}$ symmetric potential. As we will see, one can actually obtain the same
forms to Eqs.(112)(113).

###  4.2 ${\cal PT}$ symmetric potential: $V=\omega^{2}x^{2}-gx^{4}$

(a) $\arg(g)=0_{+}$ (b) $\arg(g)=0_{-}$
---
Figure 8:  Stokes graph of the ${\cal PT}$ symmetric potential with a
quadratic term. The paths for the analytic continuation are denoted by colored
lines, $\gamma_{3\rightarrow 1}$ (red), $\gamma_{2\rightarrow 4}$ (green), and
$\gamma_{3\rightarrow 4}$ (blue).

We consider the ${\cal PT}$ symmetric potential. The procedure of EWKB is
almost parallel to Sec. 4.1.1.

We begin with the following Schrödinger equation:

$\displaystyle{\cal L}=-\hbar^{2}\partial_{x}^{2}-gx^{4}-E,\qquad{\cal
L}\psi(x)=0,$ (115)

where $\omega,g,E,\in{\mathbb{R}}_{>0}$, and we take $E$ sufficient small.
From Eq.(58), turning points are given by

$\displaystyle{\rm TP}$ $\displaystyle=$
$\displaystyle\left\\{a_{2}=-\sqrt{\frac{\omega^{2}+\sqrt{\omega^{4}-4Eg}}{2g}},\
a_{2}=-\sqrt{\frac{\omega^{2}-\sqrt{\omega^{4}-4Eg}}{2g}},\right.$ (116)
$\displaystyle\left.\ \
a_{4}=+\sqrt{\frac{\omega^{2}-\sqrt{\omega^{4}-Eg}}{2g}},\
a_{4}=+\sqrt{\frac{\omega^{2}+\sqrt{\omega^{4}-4Eg}}{2g}}\right\\},$

and we define cycles as

$\displaystyle A=e^{a_{23}},\qquad B=e^{a_{21}}=e^{a_{34}}.$ (117)

Similar to the Hermitian case, a real positive-valued $g$ induces a Stokes
phenomenon, so that we introduce an infinitesimal phase to $g$. The Stokes
graph is shown in Fig. 8. In this case, the wavefunction at $x=\pm\infty$ is
not normalizable because of oscillation, but there are six asymptotic domains
to give a convergent (or divergent) wavefunction on the complex $x$-plane.
Hence, we take the following paths for analytic continuation:

$\displaystyle\gamma_{3\rightarrow 1}:-e^{\frac{\pi}{4}i}\infty\ \rightarrow\
e^{\frac{\pi}{4}i}\infty,\qquad\gamma_{2\rightarrow
4}:-e^{-\frac{\pi}{4}i}\infty\ \rightarrow\ e^{-\frac{\pi}{4}i}\infty,$ (118)

for the AC energy, and

$\displaystyle\gamma_{3\rightarrow 4}$ $\displaystyle:$
$\displaystyle-e^{\frac{\pi}{4}i}\infty\ \rightarrow\
e^{-\frac{\pi}{4}i}\infty,$ (119)

for the ${\cal PT}$ energy. By taking the above paths, one can obtain the
monodromy matrices for $\arg(g)=0_{\pm}$ as

$\displaystyle{\cal M}_{{\rm AC}(3\rightarrow
1)}^{0_{+}}=N_{a_{1},a_{2}}M_{+}N_{a_{2},a_{3}}M_{+}M_{-}N_{a_{3},a_{1}},$
(120) $\displaystyle{\cal M}_{{\rm AC}(3\rightarrow
1)}^{0_{-}}=M_{+}N_{a_{1},a_{2}}M_{+}N_{a_{2},a_{3}}M_{+}M_{-}N_{a_{3},a_{4}}M_{+}^{-1}N_{a_{4},a_{1}},$
(121) $\displaystyle{\cal M}_{{\rm AC}(2\rightarrow
4)}^{0_{+}}=M_{+}^{-1}N_{a_{1},a_{2}}M_{-}M_{+}N_{a_{2},a_{3}}M_{+}N_{a_{3},a_{4}}M_{+}N_{a_{4},a_{1}},$
(122) $\displaystyle{\cal M}_{{\rm AC}(2\rightarrow
4)}^{0_{-}}=N_{a_{1},a_{2}}M_{-}M_{+}N_{a_{2},a_{3}}M_{+}N_{a_{3},a_{1}},$
(123) $\displaystyle{\cal M}_{\cal
PT}^{0_{+}}=N_{a_{1},a_{2}}M_{+}N_{a_{2},a_{3}}M_{+}N_{a_{3},a_{4}}M_{+}N_{a_{4},a_{1}},$
(124) $\displaystyle{\cal M}_{\cal
PT}^{0_{-}}=M_{+}N_{a_{1},a_{2}}M_{+}N_{a_{2},a_{3}}M_{+}N_{a_{3},a_{1}}.$
(125)

By using Eq.(68), normalizability of the wavefunction determines the boundary
condition as ${\cal M}^{0_{\pm}}_{12}=0$ for both the ${\cal PT}$ and the AC
matrices, and one can consequently find the QCs for $\arg(g)=0_{\pm}$ as

$\displaystyle{\mathfrak{D}}_{{\rm AC}(3\rightarrow 1)}^{0_{+}}\propto
1+A,\qquad\qquad\quad\ \ \,{\mathfrak{D}}_{{\rm AC}(3\rightarrow
1)}^{0_{-}}\propto 1+\frac{A}{(1+B)^{2}},$ (126)
$\displaystyle{\mathfrak{D}}_{{\rm AC}(2\rightarrow 4)}^{0_{+}}\propto
1+A(1+B)^{2},\qquad{\mathfrak{D}}_{{\rm AC}(2\rightarrow 4)}^{0_{-}}\propto
1+A,$ (127) $\displaystyle{\mathfrak{D}}_{\cal PT}^{0_{+}}\propto
1+A(1+B),\qquad\qquad\ \,{\mathfrak{D}}_{\cal PT}^{0_{-}}\propto
1+\frac{A}{1+B},$ (128)

where ${\rm C}_{{\rm NP},\arg(g)=0}=\\{B\\}$. From Eqs.(72)-(73), DDP formula
is also obtained by counting intersection numbers between the cycles as

$\displaystyle{\mathfrak{S}}_{0}^{\nu}[A]=A(1+B)^{-2\nu},\qquad{\mathfrak{S}}_{0}^{\nu}[B]=B.$
(129)

Thus, the QCs removed the discontinuity are obtained from Eq.(77) as

$\displaystyle{\mathfrak{D}}^{0}_{{\rm AC}(3\rightarrow 1)}\propto
1+\frac{A}{1+B},\qquad{\mathfrak{D}}^{0}_{{\rm AC}(2\rightarrow 4)}\propto
1+A(1+B),$ (130) $\displaystyle{\mathfrak{D}}^{0}_{\cal PT}\propto 1+A.$ (131)

It is remarkable that we could reproduce the QCs in Eqs.(112)-(113), that
correspond to ${\mathfrak{D}}^{\arg(\lambda)=+\pi}_{\cal
H}\propto{\mathfrak{D}}_{{\rm AC}(2\rightarrow 4)}^{\arg(g)=0}$ and
${\mathfrak{D}}^{\arg(\lambda)=-\pi}_{\cal H}\propto{\mathfrak{D}}_{{\rm
AC}(3\rightarrow 1)}^{\arg(g)=0}$. It is also notable that the path-dependence
of the analytic continuation, given by Eq.(118), can be seen as $A(1+B)^{\mp
1}$ in Eq.(130). The difference consequently should appear in non-perturbative
parts of the energy solution, as we pointed out in Sec. 4.1.2. In contrast,
the ${\cal PT}$ QC does not have such a non-perturbative effect, and the
energy solution should contain only a perturbative part. However, it does not
mean that the ${\cal PT}$ energy solution should be Borel summable. It is
indeed Borel non-summable unlike the Hermitian case, as we can see later.

(a) $\arg(g)=0_{+}$ (b) $\arg(g)=0_{-}$
---
Figure 9:  Stokes graph with $E_{0}=0$ corresponding to Fig. 8, where $E_{0}$
is the classical part of the energy. By taking $E_{0}\rightarrow 0_{+}$, two
simple turning points, $a_{2}$ and $a_{3}$, collide to each other and then
become a double turning point (green dot) at $E_{0}=0$.

Let us solve the QCs (130)(131) and obtain their formal transseries solutions.
As we have done in Sec. 4.1.1, it is useful to use the DW-type by taking
$E_{0}\rightarrow 0_{+}$, where $E_{0}$ is a classical part of the energy. The
Stokes graph is shown in Fig. 9. For a simplified notation, we rescale
variables as

$\displaystyle
x\rightarrow\frac{\omega}{\sqrt{g}}x,\qquad\widetilde{E}\rightarrow\omega\widetilde{E},\qquad\hbar\rightarrow\frac{\omega^{3}}{g}\hbar,$
(132)

where $E=E_{0}+\widetilde{E}\hbar>0$, and the Schrödinger equation (115)
becomes dimensionless as

$\displaystyle\frac{g}{\omega^{4}}{\cal L}\rightarrow{\cal
L}=-\hbar^{2}\partial_{x}^{2}+x^{2}-x^{4}-\widetilde{E}\hbar.$ (133)

By this procedure, the $A$\- and $B$-cycles are replaced with

$\displaystyle A\rightarrow{\mathfrak{A}}=e^{-2\pi iF},\qquad
B\rightarrow{\mathfrak{B}}=\frac{\sqrt{2\pi}e^{-G}}{\Gamma(1/2-F)}\left(\frac{\hbar}{2}\right)^{F},$
(134)

where $F$ and $G$ are formal power series of $\hbar$ and $\widetilde{E}$. One
can easily obtain $F$ by Eq.(100), but computation of $G$ is more technical.
We summarize the derivation of $G$ in Appendix A. Notice that the overall sign
to $F$ in ${\mathfrak{B}}$ is flipped to Eq.(96) because this sign is related
to the intersection number $\langle{\mathfrak{A}},{\mathfrak{B}}\rangle$ and
takes the opposite to the Hermitian case. By this procedure, the QCs
(130)(131) lead to

$\displaystyle{\rm AC}:\ \log\frac{{\mathfrak{A}}^{\pm
1}}{1+{\mathfrak{B}}}=\pi iq,\qquad{\cal PT}:\ \log{\mathfrak{A}}=\pi
iq,\qquad q\in 2{\mathbb{Z}}+1,$ (135)

where $q$ is an energy level, and $\pm 1$ in ${\rm AC}$ corresponds to
${\mathfrak{D}}_{{\rm AC}(3\rightarrow 1)}$ and ${\mathfrak{D}}_{{\rm
AC}(2\rightarrow 4)}$, respectively. According to Eq.(100) and Appendix A, $F$
and $G$ can be specifically written down as

$\displaystyle
F=-\frac{\widetilde{E}}{2}-\frac{3(1+\widetilde{E}^{2})}{16}\hbar-\frac{5\widetilde{E}(17+7\widetilde{E}^{2})}{128}\hbar^{2}-\frac{105(19+50\widetilde{E}^{2}+11\widetilde{E}^{4})}{2048}\hbar^{3}+O(\hbar^{4}),$
$\displaystyle
G=\frac{2}{3\hbar}+\frac{\widetilde{E}^{2}}{8}\hbar+\frac{\widetilde{E}(55+23\widetilde{E}^{2})}{128}\hbar^{2}+\frac{441+5388\widetilde{E}^{2}+1091\widetilde{E}^{4}}{3072}\hbar^{3}+O(\hbar^{4}).$
(136)

Perturbative parts of the ${\cal PT}$ and the AC energy solutions are obtained
from the same condition, $\log{\mathfrak{A}}=\pi iq$. Substituting the ansatz,
$\widetilde{E}^{(0)}=\sum_{n\in{\mathbb{N}}_{0}}e^{(0)}_{n}\hbar^{n}$, into
$F$ in the condition leads to

$\displaystyle\widetilde{E}^{(0)}$ $\displaystyle=$ $\displaystyle
q-\frac{3\left(q^{2}+1\right)}{8}\hbar-\frac{q\left(17q^{2}+67\right)}{64}\hbar^{2}-\frac{3\left(125q^{4}+1138q^{2}+513\right)}{1024}\hbar^{3}+O(\hbar^{4}),$

with $q\in 2{\mathbb{N}}_{0}+1$. Since ${\mathfrak{D}}_{\cal PT}$ does not
contain non-perturbative contributions, Eq.(4.2) is the transseries solution
of the ${\cal PT}$ energy. Hence, the perturbative part of the AC energy
equals to the ${\cal PT}$ energy:

$\displaystyle\widetilde{E}_{\cal PT}=\widetilde{E}^{(0)}_{{\rm
AC}(3\rightarrow 1)}=\widetilde{E}^{(0)}_{{\rm AC}(2\rightarrow
4)}\in{\mathbb{R}}_{>0}.$ (138)

In contrast, ${\mathfrak{D}}_{\rm AC}$ has non-perturbative effects, and those
can be computed by expanding the energy $\widetilde{E}$ around its
perturbative part $\widetilde{E}^{(0)}$ in Eq.(135). Precisely, by shifting
the energy as $\widetilde{E}=\widetilde{E}^{(0)}+\delta\widetilde{E}$, the AC
QCs in Eq.(135) can be written as

$\displaystyle\sum_{t\in{\mathbb{N}}}\left.\frac{\partial^{s}F}{\partial\widetilde{E}^{s}}\right|_{\widetilde{E}=\widetilde{E}^{(0)}}\frac{(\delta\widetilde{E})^{s}}{s!}$
$\displaystyle=$ $\displaystyle\mp\frac{1}{2\pi
i}\sum_{s\in{\mathbb{N}}_{0}}\left.\frac{\partial^{s}\log(1+{\mathfrak{B}})}{\partial\widetilde{E}^{s}}\right|_{\widetilde{E}=\widetilde{E}^{(0)}}\frac{(\delta\widetilde{E})^{s}}{s!},$
(139)

where $\delta\widetilde{E}=\sum_{\ell\in{\mathbb{N}}}\widetilde{E}^{(\ell)}$,
and
$\widetilde{E}^{(\ell)}=e^{-\frac{2\ell}{3\hbar}}\times\left(\cdots\right)$ is
a non-perturbative part originated by $\ell$ bions (instanton-antiinstanton
pairs). Solving Eq.(139) yields transseries solutions of
$\widetilde{E}^{(\ell)}_{{\rm AC}(3\rightarrow 1)}$, and the first two non-
perturbative parts $(\ell=1,2)$ can be written down as (cf. Bender:1973rz )

$\displaystyle\widetilde{E}_{{\rm AC}(3\rightarrow 1)}^{(1)}$ $\displaystyle=$
$\displaystyle-i\sigma\left[1-\frac{q(q+6)}{8}\hbar+\frac{q^{4}+q^{3}-102q^{2}-43q-134}{128}\hbar^{2}\right.$
(140)
$\displaystyle\left.-\frac{q\left(q^{5}-15q^{4}-184q^{3}+4371q^{2}+2400q+20484\right)}{3072}\hbar^{3}+O(\hbar^{4})\right],$
$\displaystyle\widetilde{E}_{{\rm AC}(3\rightarrow 1)}^{(2)}$ $\displaystyle=$
$\displaystyle\sigma^{2}\left[\frac{\zeta_{+}}{2}+\frac{2q+3}{8}\hbar-\frac{q(q+3)}{8}\zeta_{+}\hbar\right.$
(141)
$\displaystyle\left.-\frac{8q^{3}+3q^{2}-102q-43}{128}\hbar^{2}+\frac{2q^{4}+q^{3}-51q^{2}-43q-67}{128}\zeta_{+}\hbar^{2}\right.$
$\displaystyle\left.+\frac{12q^{5}-75q^{4}-368q^{3}+2988q^{2}+2400q+5121}{1536}\hbar^{3}\right.$
$\displaystyle\left.-\frac{q\left(2q^{5}-15q^{4}-92q^{3}+996q^{2}+1200q+5121\right)}{1536}\zeta_{+}\hbar^{3}+O(\hbar^{4})\right],$

where $\sigma$ and $\zeta_{\pm}$ are defined as

$\displaystyle\sigma:=\sqrt{\frac{2}{\pi}}\frac{e^{-\frac{2}{3\hbar}}}{\Gamma(\frac{q+1}{2})}\left(\frac{\hbar}{2}\right)^{-\frac{q}{2}},\qquad\zeta_{\pm}:=\psi^{(0)}\left(\frac{q+1}{2}\right)+\log\left(\frac{\hbar}{2}\right)\pm\pi
i.$ (142)

One can directly obtain $\widetilde{E}_{{\rm AC}(2\rightarrow 4)}$ by complex
conjugate of $\widetilde{E}_{{\rm AC}(3\rightarrow 1)}$ as ${\cal
C}[\widetilde{E}_{{\rm AC}(3\rightarrow 1)}]=\widetilde{E}_{{\rm
AC}(2\rightarrow 4)}$, i.e. $i\rightarrow-i$ and
$\zeta_{+}\rightarrow\zeta_{-}$ in Eqs.(140)(141).

The important fact is that the non-perturbative parts of the AC energy,
$\widetilde{E}^{(\ell\in{\mathbb{N}})}_{\rm AC}$, are complex values, and the
signs of their imaginary parts correspond to the path of analytic
continuation. In particular, $\widetilde{E}^{(1)}_{\rm AC}$ is pure imaginary,
but $\widetilde{E}^{(\ell>1)}_{\rm AC}$ contains both real and imaginary
parts. This point is extremely crucial for the ABS conjecture; when exceeding
the semi-classical level, the ABS conjecture (4) is violated.

###  4.3 Reformulation of the ABS conjecture

We reformulate the ABS conjecture by Borel resummation theory using the above
results. Although $\widetilde{E}_{\rm AC}$ and $\widetilde{E}_{\cal PT}$ are
not naively related to each other, one can formulate a direct relation between
$\widetilde{E}_{\rm AC}$ and $\widetilde{E}_{\cal PT}$ using the DDP formula
(129) thanks to the quadratic term.

Let us see this fact. From the DDP formula (129) and the QCs (130)(131), one
finds that

$\displaystyle{\mathfrak{D}}_{\cal
PT}={\mathfrak{S}}^{-1/2}_{0}[{\mathfrak{D}}_{{\rm AC}(3\rightarrow
1)}]={\mathfrak{S}}^{+1/2}_{0}[{\mathfrak{D}}_{{\rm AC}(2\rightarrow 4)}].$
(143)

Since the QCs are directly related to the energy solutions, Eq.(143) implies
existence of Stokes automorphism for the energy solutions such that999 Remind
that $E=\widetilde{E}\hbar$, and this redefinition does not change the
transformation law of Stokes automorphism.

$\displaystyle E_{\cal PT}={\mathfrak{S}}_{0}^{-1/2}[E_{{\rm AC}(3\rightarrow
1)}]={\mathfrak{S}}^{+1/2}_{0}[E_{{\rm AC}(2\rightarrow 4)}].$ (144)

Notice that $E_{{\rm AC}(3\rightarrow 1)}$ and $E_{{\rm AC}(2\rightarrow 4)}$
also have to be related to each other as

$\displaystyle E_{{\rm AC}(3\rightarrow 1)}={\mathfrak{S}}^{+1}_{0}[E_{{\rm
AC}(2\rightarrow 4)}].$ (145)

Thus, our task is specifically formulating the Stokes automorphism for the
energy solutions. It is actually performable by computation of alien calculus,
and we summarize the derivation in Appendix B. By the results in Eqs.(B)(B),
one can reproduce the non-perturbative parts of $\widetilde{E}_{\rm AC}$ in
Eqs.(140)(141) from its perturbative part in Eq.(4.2), i.e.
$\widetilde{E}_{\cal PT}$, by101010 The subscript, “0”, in the alien
derivative denotes $\arg(g)=0$, not locations of the singular points. The
alien derivative contains non-perturbative effects from all the singular
points along the real axis on the Borel plane. See Appendix B.

$\displaystyle{\mathfrak{S}}_{0}^{\nu}[\widetilde{E}_{\cal PT}]$
$\displaystyle=$
$\displaystyle\left(1+\nu\overset{\underset{\bullet}{}}{\Delta}_{0}+\frac{(\nu\overset{\underset{\bullet}{}}{\Delta}_{0})^{2}}{2}+O(\nu^{3})\right)[\widetilde{E}_{\cal
PT}]$ (146) $\displaystyle=$ $\displaystyle\begin{cases}\widetilde{E}_{{\rm
AC}(3\rightarrow 1)}&\quad\mbox{for}\quad\nu=+\frac{1}{2}\\\
\widetilde{E}_{{\rm AC}(2\rightarrow
4)}&\quad\mbox{for}\quad\nu=-\frac{1}{2}\end{cases}.$

This result insists that Stokes automorphism for the energy solutions is
actually constructable by alien calculus without any contradictions to
Eqs.(4.2)(140)(141). Notice that showing Eq.(145) is trivial by using the
property of Stokes automorphism in Eq.(16). We should remind that analytic
continuation of the Hermitian QC with $\arg(\lambda)=\pm\pi$ and the AC QCs
with $\arg(g)=0$ are equivalent as ${\mathfrak{D}}^{\arg(\lambda)=+\pi}_{\cal
H}\propto{\mathfrak{D}}_{{\rm AC}(2\rightarrow 4)}^{\arg(g)=0}$ and
${\mathfrak{D}}^{\arg(\lambda)=-\pi}_{\cal H}\propto{\mathfrak{D}}_{{\rm
AC}(3\rightarrow 1)}^{\arg(g)=0}$. Hence, the energy solutions should hold the
same equivalence, i.e. $E^{\arg(\lambda)=+\pi}_{\cal H}=E_{{\rm
AC}(2\rightarrow 4)}^{\arg(g)=0}$ and $E^{\arg(\lambda)=-\pi}_{\cal H}=E_{{\rm
AC}(3\rightarrow 1)}^{\arg(g)=0}$. One can also provide the same results for
the Borel resummed forms using Eq.(23). From Eqs.(144)(145), those are given
by

$\displaystyle\widehat{E}_{\cal PT}={\cal S}_{0_{-}}[E_{{\rm AC}(3\rightarrow
1)}]=\widehat{{\mathfrak{S}}}_{0}^{-1/2}[\widehat{E}_{{\rm AC}(3\rightarrow
1)}]$ $\displaystyle\qquad\,\,={\cal S}_{0_{+}}[E_{{\rm AC}(2\rightarrow
4)}]=\widehat{{\mathfrak{S}}}_{0}^{+1/2}[\widehat{E}_{{\rm AC}(2\rightarrow
4)}],$ (147) $\displaystyle\widehat{E}_{{\rm AC}(3\rightarrow 1)}={\cal
S}_{0_{+}}[E_{\cal PT}]=\widehat{\mathfrak{S}}^{+1/2}_{0}[\widehat{E}_{\cal
PT}],$ $\displaystyle\qquad\qquad\,\,={\cal
S}_{0_{+}}\circ{\mathfrak{S}}^{+1/2}_{0}[E_{{\rm AC}(2\rightarrow
4)}]=\widehat{\mathfrak{S}}^{+1}_{0}[\widehat{E}_{{\rm AC}(2\rightarrow 4)}],$
(148) $\displaystyle\widehat{E}_{{\rm AC}(2\rightarrow 4)}={\cal
S}_{0_{-}}[E_{\cal PT}]=\widehat{\mathfrak{S}}^{-1/2}_{0}[\widehat{E}_{\cal
PT}]$ $\displaystyle\qquad\qquad\,\,={\cal
S}_{0_{-}}\circ{\mathfrak{S}}^{-1/2}_{0}[E_{{\rm AC}(3\rightarrow
1)}]=\widehat{\mathfrak{S}}^{-1}_{0}[\widehat{E}_{{\rm AC}(3\rightarrow 1)}],$
(149)

where $\widehat{E}={\cal S}_{{\rm med},0}[E]$. Therefore, the modified ABS
conjecture for the energy solutions are reformulated by Stokes automorphism
and Borel resummation, as Eqs.(144)(147).

One can directly map the results of the energies to Euclidean partition
functions by using the definition,

$\displaystyle Z=\sum_{q\in 2{\mathbb{N}}_{0}+1}\exp\left[-\beta
E_{q}\right],$ (150)

where $q$ denotes the energy level. Formal transseries of the ${\cal PT}$ and
the AC partition functions are related as

$\displaystyle Z_{\cal PT}={\mathfrak{S}}^{-1/2}_{0}[Z_{{\rm AC}(3\rightarrow
1)}]={\mathfrak{S}}^{+1/2}_{0}[Z_{{\rm AC}(2\rightarrow 4)}],$ (151)

where

$\displaystyle{\mathfrak{S}}_{0}^{\nu}[Z]=\sum_{q\in
2{\mathbb{N}}_{0}+1}\exp\left[-\beta{\mathfrak{S}}_{0}^{\nu}[E_{q}]\right].$
(152)

The Borel resummed forms are expressed by

$\displaystyle\widehat{Z}_{\cal PT}={\cal S}_{0_{-}}[Z_{{\rm AC}(3\rightarrow
1)}]=\widehat{{\mathfrak{S}}}_{0}^{-1/2}[\widehat{Z}_{{\rm AC}(3\rightarrow
1)}]$ $\displaystyle\qquad\,\,={\cal S}_{0_{+}}[Z_{{\rm AC}(2\rightarrow
4)}]=\widehat{{\mathfrak{S}}}_{0}^{+1/2}[\widehat{Z}_{{\rm AC}(2\rightarrow
4)}].$ (153)

Finally, we briefly see the energy solutions from the viewpoint of the
continuous deformation by Stokes automorphism. The QCs in Eqs.(130)(131) can
be unifiedly expressed by a real parameter, $\nu$, as

$\displaystyle{\mathfrak{D}}^{\nu\in{\mathbb{R}}}:=1+A(1+B)^{-2\nu}\propto\begin{cases}{\mathfrak{D}}_{{\cal
PT}}&\quad\mbox{if}\quad\nu=0\\\ {\mathfrak{D}}_{{\rm AC}(3\rightarrow
1)}&\quad\mbox{if}\quad\nu=+\frac{1}{2}\\\ {\mathfrak{D}}_{{\rm
AC}(2\rightarrow 4)}&\quad\mbox{if}\quad\nu=-\frac{1}{2}\end{cases},$ (154)
$\displaystyle{\mathfrak{S}}^{\nu\in{\mathbb{R}}}_{0}[{\mathfrak{D}}^{\nu_{0}}]={\mathfrak{D}}^{\nu_{0}+\nu}.$
(155)

Denoting $\widetilde{E}(\nu_{0})$ as the energy solution of
${\mathfrak{D}}^{\nu_{0}}$, one can calculate the $\nu$-evolution from
$\widetilde{E}(\nu_{0})$ by modifying Eq.(139) as

$\displaystyle\left.F\right|_{\widetilde{E}=\widetilde{E}(\nu_{0})}=-\frac{q}{2}-\frac{\nu_{0}}{\pi
i}\left.\log(1+{\mathfrak{B}})\right|_{\widetilde{E}=\widetilde{E}(\nu_{0})},$
(156)
$\displaystyle\sum_{s\in{\mathbb{N}}}\left.\frac{\partial^{s}}{\partial\widetilde{E}^{s}}\left(F+\frac{\nu_{0}}{\pi
i}\log(1+{\mathfrak{B}})\right)\right|_{\widetilde{E}=\widetilde{E}(\nu_{0})}\frac{(\delta\widetilde{E})^{s}}{s!}$
$\displaystyle\qquad=-\frac{\nu}{\pi
i}\sum_{s\in{\mathbb{N}}_{0}}\left.\frac{\partial^{s}\log(1+{\mathfrak{B}})}{\partial\widetilde{E}^{s}}\right|_{\widetilde{E}=\widetilde{E}(\nu_{0})}\frac{(\delta\widetilde{E})^{s}}{s!}.$
(157)

One can express $\delta\widetilde{E}$ by the alien derivatives as

$\displaystyle\delta\widetilde{E}=\sum_{n\in{\mathbb{N}}}\frac{\nu^{n}}{n!}(\overset{\underset{\bullet}{}}{\Delta}_{0})^{n}[\widetilde{E}(\nu_{0})],$
(158)

and substituting Eq.(158) into Eq.(157) determines
$(\overset{\underset{\bullet}{}}{\Delta}_{0})^{n}[\widetilde{E}(\nu_{0})]$
recursively. Taking $\nu_{0}=0$ and $\nu=\pm 1/2$ reproduces Eq.(146).
Therefore, the ${\cal PT}$ and the AC energy solutions are parameterized by
$\nu$ and continuously connected by the one-parameter Stokes automorphism
${\mathfrak{S}}^{\nu}_{0}$, as
${\mathfrak{S}}^{\nu}_{0}[E(\nu_{0})]=E(\nu_{0}+\nu)$. In addition, these
Borel resummed forms obey the following initial value problem:

$\displaystyle\frac{d\widehat{E}(\nu)}{d\nu}$ $\displaystyle=$
$\displaystyle\frac{d\widehat{\mathfrak{S}}_{0}^{\nu}[\widehat{E}(0)]}{d\nu}={\cal
S}_{\rm med,0}\circ\overset{\underset{\bullet}{}}{\Delta}_{0}[E(\nu)]$
$\displaystyle=$ $\displaystyle-\frac{\hbar}{\pi
i}\left.\frac{\log(1+\widehat{\mathfrak{B}})}{\frac{\partial}{\partial\widetilde{E}}\left(\widehat{F}+\frac{\nu}{\pi
i}\log(1+\widehat{\mathfrak{B}})\right)}\right|_{\widetilde{E}=\widehat{E}(\nu)/\hbar},$
$\displaystyle\widehat{E}(\nu=0)$ $\displaystyle=$
$\displaystyle\widehat{E}_{\cal PT},$ (159)

where
$\widehat{\mathfrak{S}}^{\nu}_{0}[\widehat{E}(\nu_{0})]=\widehat{E}(\nu_{0}+\nu)$.

We have now completed to make all the relations in Fig. 1.

##  5 Pure quartic potential

In this section, we consider the Schrödinger equation given by the pure
quartic potential, i.e. $\omega=0$. In this case, since no quadratic local
minimum exists in the potential, one might imagine that EWKB expanding $\hbar$
is meaningless. Indeed, the scaling law in the Hamiltonian gives a monomial
with a rational power of $\hbar$ such that

$\displaystyle E=c(k)(\lambda\hbar^{4})^{1/3},\qquad
c(k)\in{\mathbb{R}}_{>0},$ (160)

with an overall function, $c(k)$, depending on the energy level, $k$. However,
although the $\hbar$-expansion is meaningless for the potential, there is
another parameter for a transseries solution of the energy, that is the
(inverse) energy level, $k$ BPV ; Bucciotti:2023trp . Hence, our main strategy
for the pure quartic potential is to derive a formal transseries of $c(k)$ in
Eq.(160).

We firstly consider EWKB of the Hermitian potential, $V(x)=\lambda x^{4}$, in
Sec. 5.1, and then analyze the ${\cal PT}$ symmetric potential,
$V(x)=-gx^{4}$, in Sec. 5.2. From these results, we finally discuss
(im)possibility of an alternative form of the ABS conjecture in Sec. 5.3.

###  5.1 Hermitian potential: $V=\lambda x^{4}$

The Schrödinger equation with the Hermitian potential, $V(x)=\lambda x^{4}$,
is defined as

$\displaystyle{\cal L}=-\hbar^{2}\partial_{x}^{2}+\lambda x^{4}-E,\qquad{\cal
L}\psi=0,$ (161)

where $\lambda,E\in{\mathbb{R}}_{>0}$. When naively applying EWKB to Eq.(161),
a remarkable feature of the pure quartic potential can be readily seen from
$\int dx\,S_{\rm od}$, which takes the form that

$\displaystyle\int dx\,S_{\rm
od}(x,\hbar)=\Phi_{-1}(x)\eta^{-1}+\Phi_{+1}(x)\eta^{+1}+\cdots,\qquad\eta:=\frac{\lambda^{1/4}\hbar}{E^{3/4}},$
(162)

where the coefficients, $\Phi_{n\in 2{\mathbb{N}}_{0}-1}(x)\in{\mathbb{C}}$,
are independent on $\lambda$ and $E$ for all $n$. This suggests that taking
$\eta$ should be a more reasonable choice of the expansion parameter. The
Schrödinger equation of $\eta$ is easily obtained by scaling dimensions of the
parameters in Eq.(161), that are

$\displaystyle[x]=\frac{1}{4},\qquad[\hbar]=\frac{3}{4},\qquad[\lambda]=0,\qquad[E]=1.$
(163)

Rescaling $x\rightarrow\left(\frac{E}{\lambda}\right)^{1/4}x$ and the relation
of $\eta$ with $\hbar$ in Eq.(162) lead to the dimensionless operator ${\cal
L}$ as

$\displaystyle\frac{1}{E}{\cal L}\rightarrow{\cal
L}=-\eta^{2}\partial_{x}^{2}+Q,\qquad Q=x^{4}-1.$ (164)

Let us perform EWKB for the dimensionless equation (164). The Stokes graph has
the same topology as Fig. 5 consisting of turning points with different values
given by

$\displaystyle{\rm TP}=\left\\{a_{1}=-1,a_{2}=-i,a_{3}=+i,a_{4}=+1\right\\}.$
(165)

Similarly, the Stokes phenomenon happens at $\arg(\lambda)=0$. Thanks to the
same topology of the Stokes graph, the monodromy matrix, the QC, and the DDP
formula have the same cycle representations to Eqs.(83)-(89). The difference
is only specific forms of $A_{j=1,2}$, and those are given by

$\displaystyle A_{1}=e^{\phi(e^{-\pi i/4}\eta)},\qquad A_{2}=e^{-\phi(e^{+\pi
i/4}\eta)},$ (166)

where $\phi=\phi(\eta)$ is a formal power expansion of $\eta$ as

$\displaystyle\phi(\eta):=\sum_{n\in{\mathbb{N}}_{0}}v_{2n-1}\eta^{2n-1},\qquad
v_{2n-1}\in{\mathbb{R}}\ \ \mbox{for all}\ \ n\in{\mathbb{N}}_{0}.$ (167)

Those coefficients, $v_{2n-1}$, are evaluated as

$\displaystyle v_{-1}=\frac{4\sqrt{2}K(-1)}{3},\quad
v_{1}=-\frac{4\sqrt{2\pi}\Gamma(7/4)}{9\Gamma(-3/4)},\quad
v_{3}=-\frac{11\sqrt{2\pi}\Gamma(5/4)}{384\Gamma(3/4)},\quad\cdots$ (168)

where $K(x)$ denotes the incomplete elliptic integral of the first kind. For
solving the QC, it is technically useful to decompose the exponents of
$A_{j=1,2}$ into the real and the imaginary parts. By defining

$\displaystyle\phi(e^{-\pi i/4}\eta)=\phi_{\rm R}(\eta)+i\phi_{\rm
I}(\eta),\qquad\phi_{\rm R,I}(\eta)\in{\mathbb{R}},$ (169)

the cycles are expressed by $\log A_{1}=\phi_{\rm R}+i\phi_{\rm I}$ and $\log
A_{2}=-\phi_{\rm R}+i\phi_{\rm I}$, and thus one can write down the QC, which
has the same cycle representation to Eq.(88), as

$\displaystyle{\mathfrak{D}}_{\cal H}^{\arg(\eta)=0}\propto\cos\phi_{\rm
I}+\frac{e^{-\phi_{\rm R}}}{\sqrt{1+e^{-2\phi_{\rm R}}}}.$ (170)

As we described above, $E$ appears only in $\eta$ as coupling to $\hbar$. This
fact implies that any ansatz expanded by $\hbar$ does not work for the energy
solution. The coupling constant, $\lambda$, is also the case. Although any
extra parameters explicitly do not exist in the potential, solving QCs
generally gains another parameter to distinguish the energy state, that is an
energy level. For this reason, we use the (inverse) energy level for our
ansatz. Some properties of formal transseries with the $\kappa^{-1}$-expansion
for $V=\lambda x^{4}$ are discussed in Ref. Bucciotti:2023trp . Here, we
prepare the following ansatz for $\eta$111111 The transmonomial, $\sigma$, is
available after determining the leading order, $\phi_{\rm R}=\phi_{\rm
I}=\kappa$. By taking $\phi_{\rm R,I}=\kappa+\delta\phi_{\rm R,I}$, and the
linearized Eq.(170) in terms of $\delta\phi_{\rm R,I}$ leads to
$\displaystyle-(-1)^{k}\delta\phi_{\rm
I}+e^{-\kappa}+O(\delta\phi^{2},e^{-\kappa}\delta\phi)=0\quad\Rightarrow\quad\delta\phi_{\rm
I}=(-1)^{k}e^{-\kappa}.$ (171) :

$\displaystyle\eta^{-1}=\frac{E^{3/4}}{\lambda^{1/4}\hbar}\sim\sum_{n\in{\mathbb{N}}_{0}}e^{(0)}_{2n-1}\kappa^{{1-2n}}+\sum_{\ell\in{\mathbb{N}}}\sum_{n\in{\mathbb{N}}_{0}}e^{(\ell)}_{n}\sigma^{\ell}\kappa^{-n}\quad\mbox{as}\quad\kappa\rightarrow+\infty,$
(172)
$\displaystyle\kappa=\kappa(k)=\pi\left(k+\frac{1}{2}\right),\qquad\sigma:=e^{-\kappa},\qquad
k\in{\mathbb{N}}_{0},\qquad e^{(\ell)}_{n}\in{\mathbb{R}}.$ (173)

Notice that we use the $\kappa^{-1}$-expansion for technical simplicity, but
the $k^{-1}$-expansion is always available from it. The coefficients in the
perturbative part, $e^{(0)}_{2n-1}$, can be determined by the first term in
Eq.(170), i.e. $\cos\phi_{\rm I}(\eta)=0$, and are obtained as

$\displaystyle e^{(0)}_{-1}$ $\displaystyle=$
$\displaystyle\frac{3}{4K(-1)},\qquad\qquad\ \
e^{(0)}_{1}=-\frac{4\sqrt{\pi}\Gamma\left(7/4\right)}{9\Gamma\left(-3/4\right)},$
$\displaystyle e^{(0)}_{3}$ $\displaystyle=$
$\displaystyle-\frac{11\Gamma\left(5/4\right)[\Gamma\left(1/4\right)]^{4}}{6912\sqrt{\pi}\Gamma\left(3/4\right)}-\frac{8\sqrt{2\pi}[\Gamma\left(11/4\right)]^{2}}{1323},\quad\cdots.$
(174)

The second term in Eq.(170) generates the non-perturbative parts, and the
coefficients for the first two bion contributions, $e^{(\ell=1,2)}_{n}$, can
be written down as

$\displaystyle e^{(1)}_{0}=(-1)^{k}\frac{3}{4K(-1)},\qquad\qquad\qquad
e^{(1)}_{1}=(-1)^{k}\frac{8\sqrt{\pi}\Gamma\left(7/4\right)}{9\Gamma\left(-3/4\right)},$
$\displaystyle
e^{(1)}_{2}=-(-1)^{k}\frac{2\sqrt{2}(6-\pi)\pi^{3/2}}{27[\Gamma\left(-3/4\right)]^{2}},\qquad
e^{(1)}_{3}=(-1)^{k}\frac{\sqrt{2}(36-\pi)\pi^{5/2}}{243[\Gamma\left(-3/4\right)]^{2}},\quad\cdots,$
(175) $\displaystyle e^{(2)}_{0}=-\frac{3}{4K(-1)},\qquad\qquad\qquad\qquad
e^{(2)}_{1}=-\frac{16\sqrt{\pi}\Gamma\left(7/4\right)}{9\Gamma\left(-3/4\right)},$
$\displaystyle
e^{(2)}_{2}=\frac{4\sqrt{2}(9-2\pi)\pi^{3/2}}{27[\Gamma\left(-3/4\right)]^{2}},\qquad\qquad\quad
e^{(2)}_{3}=-\frac{4\sqrt{2}(2\pi(18-\pi)-27)\pi^{3/2}}{243[\Gamma\left(-3/4\right)]^{2}},\quad\cdots,$
(176)

Taking $E_{\cal H}=\eta^{4/3}(\lambda\hbar^{4})^{1/3}$ yields the energy
solution, and $E^{(\ell=0,1,2)}_{\cal H}$ are obtained as (cf. BPV )

$\displaystyle\frac{E^{(0)}_{\cal H}}{(\lambda\hbar^{4})^{1/3}}$
$\displaystyle=$
$\displaystyle\kappa^{4/3}\left[\frac{(3/4)^{4/3}}{[K(-1)]^{4/3}}-\frac{2^{10/3}\sqrt{\pi}\Gamma\left(7/4\right)}{3^{8/3}[K(-1)]^{1/3}\Gamma\left(-3/4\right)}\kappa^{-2}+O(\kappa^{-4})\right],$
(177) $\displaystyle\frac{E^{(1)}_{\cal H}}{(\lambda\hbar^{4})^{1/3}}$
$\displaystyle=$
$\displaystyle(-1)^{k}\sigma\kappa^{1/3}\left[\frac{(3/4)^{1/3}}{[K(-1)]^{4/3}}+\frac{2^{13/3}\sqrt{\pi}\Gamma\left(7/4\right)}{3^{8/3}[K(-1)]^{1/3}\Gamma\left(-3/4\right)}\kappa^{-1}\right.$
(178)
$\displaystyle\left.-\frac{4\sqrt{\pi}\left(2^{5/6}\pi(6-\pi)+2^{4/3}\Gamma\left(7/4\right)\Gamma\left(-3/4\right)\right)}{3^{11/3}[K(-1)]^{1/3}[\Gamma\left(-3/4\right)]^{2}}\kappa^{-2}+O(\kappa^{-3})\right],$
$\displaystyle\frac{E^{(2)}_{\cal H}}{(\lambda\hbar^{4})^{1/3}}$
$\displaystyle=$
$\displaystyle\sigma^{2}\kappa^{1/3}\left[-\frac{(3/4)^{1/3}}{[K(-1)]^{4/3}}-\frac{128\sqrt{\pi}[K(-1)]\Gamma\left(7/4\right)-9\Gamma\left(-3/4\right)}{3\cdot
6^{5/3}[K(-1)]^{4/3}\Gamma\left(-3/4\right)}\kappa^{-1}\right.$ (179)
$\displaystyle\left.+\frac{2^{10/3}\sqrt{\pi}\left(\sqrt{2}\pi(9-2\pi)+3\Gamma\left(7/4\right)\Gamma\left(-3/4\right)\right)}{3^{11/3}[K(-1)]^{1/3}[\Gamma\left(-3/4\right)]^{2}}\kappa^{-2}+O(\kappa^{-3})\right].$

In the above, we naively applied the ansatz (172) for solving the QC, but need
to clarify what we actually did from the aspect of EWKB of the
$\kappa^{-1}$-expansion. The point is that the dimensionless operator (161)
can be interpreted to be given by another potential, $V(x)=x^{4}-1$, and a
fixed energy already determined as zero121212 This “zero-energy” is not
directly relevant to $E$ in Eq.(161) and just an analogy with a usual
Schrödinger equation. . Here, let us replace $\eta$ with the formal expansion,
$\eta^{-1}\sim\sum_{n\in{\mathbb{N}}_{0}}e^{(0)}_{2n-1}\kappa^{1-2n}$ as
$\kappa\rightarrow\infty$, in Eq.(164)131313 We omitted non-perturbative
sectors, $\sigma^{\ell}=e^{-\ell\kappa}$, for simplicity, but the similar
discussion works for the case including them. . Then, we define a modified
potential
$\widetilde{Q}(\kappa)\sim\sum_{n\in{\mathbb{N}_{0}}}\widetilde{Q}_{n}\kappa^{-2n}$
as

$\displaystyle\eta^{-2}{\cal L}=-\partial_{x}^{2}+\eta^{-2}Q,$
$\displaystyle\eta^{-2}Q\sim\kappa^{2}\left[\widetilde{Q}_{0}+\widetilde{Q}_{1}\kappa^{-2}+\widetilde{Q}_{2}\kappa^{-4}+O(\kappa^{-6})\right],$
(180)

where

$\displaystyle\widetilde{Q}_{0}=(e^{(0)}_{-1})^{2}Q,\qquad\widetilde{Q}_{1}=2e^{(0)}_{-1}e^{(0)}_{1}Q,\qquad\widetilde{Q}_{2}=(2e^{(0)}_{-1}e^{(0)}_{3}+(e^{(0)}_{1})^{2})Q.$
(181)

By analogy with a standard $\hbar$-expansion, this potential
$\widetilde{Q}(\kappa)$ can be regarded as a quantum deformed potential with
respect to the $\kappa^{-1}$-expansion. Assuming $e^{(0)}_{-1}\neq 0$, one can
in principle draw a Stokes graph and construct a QC by using the Airy-type
connection formula without details of the coefficients, $c_{n}$, because the
Stokes graph depends only on $\arg(\kappa)$ and $\widetilde{Q}_{0}$. After
that, the coefficients are determined by solving the resulting QC. In other
words, EWKB of the $\kappa^{-1}$-expansion can be interpreted as an inverse
problem; we determined the specific potential form of $\widetilde{Q}(\kappa)$
by solving the QC constrained only by $\widetilde{Q}_{0}$ and topology of its
Stokes graph.

Unlike the case that $\omega>0$ discussed in Sec. 4, one can directly obtain
the AC energy solution from Eqs.(177)-(179) by taking $\lambda=e^{\pm\pi i}g$
and fixing $\arg(\eta)=0$. By this procedure, one finds

$\displaystyle\frac{E_{\rm
AC}(g)}{(g\hbar^{4})^{1/3}}=e^{\pm\frac{\pi}{3}i}\frac{E_{\cal
H}(\lambda)}{(\lambda\hbar^{4})^{1/3}},$ (182)

where $E_{\cal H}$ is given by Eqs.(177)-(179). Notice that $E_{\rm AC}$
contains non-perturbative parts with respect to $\kappa^{-1}$. We would
further discuss the resulting AC energy in Sec. 5.2.

###  5.2 ${\cal PT}$ symmetric potential: $V=-gx^{4}$

(a) $\arg(g)=0_{+}$ (b) $\arg(g)=0_{-}$
---
Figure 10: Stokes graph of the ${\cal PT}$ symmetric potential without a
quadratic term. The paths for analytic continuation are denoted by colored
lines, $\gamma_{3\rightarrow 1}$ (red), $\gamma_{2\rightarrow 4}$ (green), and
$\gamma_{3\rightarrow 4}$ (blue).

We consider the ${\cal PT}$ symmetric potential with $\omega=0$. The
Schrödinger equation is given by

$\displaystyle{\cal L}=-\hbar^{2}\partial_{x}^{2}-gx^{4}-E,\qquad{\cal
L}\psi=0,$ (183)

where $g,E\in{\mathbb{R}}_{>0}$. By using the scaling law (163) and rescaling
$x\rightarrow\left(\frac{E}{g}\right)^{1/4}x$, one obtains the dimensionless
equation as

$\displaystyle\frac{1}{E}{\cal L}\rightarrow{\cal
L}=-\widetilde{\eta}^{2}\partial_{x}^{2}+Q,\qquad
Q=-x^{4}-1,\qquad\widetilde{\eta}:=\frac{g^{1/4}\hbar}{E^{3/4}}.$ (184)

From Eq.(58), turning points are given by

$\displaystyle{\rm
TP}=\left\\{a_{1}=e^{-\frac{3\pi}{4}i},a_{2}=e^{-\frac{\pi}{4}i},a_{3}=e^{\frac{3\pi}{4}i},a_{4}=e^{\frac{\pi}{4}i}\right\\},$
(185)

and the Stokes graph is shown in Fig. 10. Performing analytic continuation
taking the paths in Eqs.(118)(119) yields the monodromy matrices, ${\cal
M}^{\arg(\widetilde{\eta})=0_{\pm}}$, as

$\displaystyle{\cal M}_{{\rm AC}(3\rightarrow
1)}^{0_{+}}=M_{-}^{-1}M_{+}^{-1}N_{a_{1},a_{3}}M_{-}N_{a_{3},a_{2}}M_{-}^{-1}N_{a_{2},a_{4}}M_{-}M_{+}N_{a_{4},a_{1}},$
(186) $\displaystyle{\cal M}_{{\rm AC}(3\rightarrow
1)}^{0_{-}}=M_{-}^{-1}M_{+}^{-1}N_{a_{1},a_{4}}M_{-}M_{+}N_{a_{4},a_{1}},$
(187) $\displaystyle{\cal M}_{{\rm AC}(2\rightarrow
4)}^{0_{+}}=N_{a_{1},a_{3}}M_{+}M_{-}N_{a_{3},a_{2}}M_{+}^{-1}M_{-}^{-1}N_{a_{2},a_{1}},$
(188) $\displaystyle{\cal M}_{{\rm AC}(2\rightarrow
4)}^{0_{-}}=N_{a_{1},a_{3}}M_{+}M_{-}N_{a_{3},a_{1}}M_{+}^{-1}N_{a_{1},a_{4}}M_{-}N_{a_{4},a_{2}}M_{+}^{-1}M_{-}^{-1}N_{a_{2},a_{1}},$
(189) $\displaystyle{\cal M}_{\cal
PT}^{0_{+}}=M_{+}N_{a_{1},a_{3}}M_{+}N_{a_{3},a_{2}}M_{+}N_{a_{2},a_{1}},$
(190) $\displaystyle{\cal M}_{\cal
PT}^{0_{-}}=M_{+}N_{a_{1},a_{4}}M_{+}N_{a_{4},a_{2}}M_{+}N_{a_{2},a_{1}}.$
(191)

From Eq.(68), normalizability of the wavefunctions determines a matrix
component to be ${\cal M}^{\arg(\widetilde{\eta})=0_{\pm}}_{ij}=0$. For each
the QC, those are given as ${\mathfrak{D}}_{{\rm AC}(3\rightarrow
1)}^{\arg(\widetilde{\eta})=0_{\pm}}={\cal
M}^{\arg(\widetilde{\eta})=0_{\pm}}_{{\rm AC}(3\rightarrow 1),11}$,
${\mathfrak{D}}_{{\rm AC}(2\rightarrow
4)}^{\arg(\widetilde{\eta})=0_{\pm}}={\cal
M}^{\arg(\widetilde{\eta})=0_{\pm}}_{{\rm AC}(2\rightarrow 4),22}$, and
${\mathfrak{D}}_{\cal PT}^{\arg(\widetilde{\eta})=0_{\pm}}={\cal
M}^{\arg(\widetilde{\eta})=0_{\pm}}_{{\cal PT},12}$. Specifically,

$\displaystyle{\mathfrak{D}}_{{\rm AC}(3\rightarrow 1)}^{0_{+}}\propto
1+\frac{AB}{(1+B)^{2}},\qquad{\mathfrak{D}}_{{\rm AC}(3\rightarrow
1)}^{0_{-}}\propto 1+AB,$ (192) $\displaystyle{\mathfrak{D}}_{{\rm
AC}(2\rightarrow 4)}^{0_{+}}\propto
1+A^{-1}B,\qquad\quad\,\,{\mathfrak{D}}_{{\rm AC}(2\rightarrow
4)}^{0_{-}}\propto 1+\frac{A^{-1}B}{(1+B)^{2}},$ (193)
$\displaystyle{\mathfrak{D}}_{\cal PT}^{0_{+}}\propto
1+\frac{A}{1+B},\qquad\qquad\quad\,\,{\mathfrak{D}}_{\cal PT}^{0_{-}}\propto
1+A(1+B),$ (194)

where

$\displaystyle A=e^{a_{12}}=e^{a_{43}},\qquad B=e^{a_{31}}=e^{a_{42}},$ (195)

and ${\rm C}_{{\rm NP},\arg(\widetilde{\eta})=0}=\\{B\\}$. From Eqs.(72)-(73),
one can obtain the DDP formula for $\arg(\widetilde{\eta})=0_{\pm}$ as

$\displaystyle{\mathfrak{S}}^{\nu}_{0}[A]=A(1+B)^{2\nu},\qquad{\mathfrak{S}}^{\nu}_{0}[B]=B,\qquad(\nu\in{\mathbb{R}})$
(196)

and thus the QCs removed the discontinuity are derived by Eq.(77) as

$\displaystyle{\mathfrak{D}}^{0}_{{\rm AC}(3\rightarrow 1)}\propto
1+\frac{AB}{1+B},\qquad{\mathfrak{D}}^{0}_{{\rm AC}(2\rightarrow 4)}\propto
1+\frac{A^{-1}B}{1+B},$ (197) $\displaystyle{\mathfrak{D}}^{0}_{\cal
PT}\propto 1+A.$ (198)

Then, we solve the QCs in Eqs.(197)(198). The cycles can be written as

$\displaystyle A=e^{i\widetilde{\phi}(\widetilde{\eta})},\qquad
B=e^{-\phi\left(\widetilde{\eta}\right)},$ (199)

where $\phi(\widetilde{\eta})\in{\mathbb{R}}$ is given by Eq.(167), and
$\widetilde{\phi}(\widetilde{\eta})$ is defined from $\phi$ as

$\displaystyle\widetilde{\phi}(\widetilde{\eta}):=-i\phi(\widetilde{\eta}e^{-\frac{\pi
i}{2}})=\sum_{n\in{\mathbb{N}}_{0}}v_{2n-1}(-1)^{n}\widetilde{\eta}^{2n-1}\quad\in{\mathbb{R}}.$
(200)

Thus, the QCs (197)(198) can be rewritten as

$\displaystyle{\rm AC}:\ \pm
i\widetilde{\phi}-\phi-\log(1+e^{-\phi})=2i\kappa,\qquad{\cal PT}:\
\widetilde{\phi}=2\kappa,$ (201)

where the sign, $\pm$ , in ${\rm AC}$ corresponds to ${\mathfrak{D}}^{0}_{{\rm
AC}(3\rightarrow 1)}$ and ${\mathfrak{D}}^{0}_{{\rm AC}(2\rightarrow 4)}$,
respectively, and $\kappa$ is defined by Eq.(173). In the similar way to the
analysis in Sec. 5.1, we set the following ansatz to $\widetilde{\eta}$:

$\displaystyle\widetilde{\eta}^{-1}=\frac{E^{3/4}}{g^{1/4}\hbar}\sim\sum_{n\in{\mathbb{N}}_{0}}\widetilde{e}^{(0)}_{2n-1}\kappa^{{1-2n}}+\sum_{\ell\in{\mathbb{N}}}\sum_{n\in{\mathbb{N}}_{0}}\widetilde{e}^{(\ell)}_{n}\sigma^{\ell}\kappa^{-n}\quad\mbox{as}\quad\kappa\rightarrow+\infty,$
(202)

where $\sigma=e^{-\kappa}$. One can easily solve ${\mathfrak{D}}_{\cal
PT}^{0}=0$ and obtains

$\displaystyle\widetilde{e}^{(0)}_{-1}=\frac{3}{2\sqrt{2}K(-1)},\qquad\widetilde{e}^{(0)}_{1}-\frac{2\sqrt{2\pi}\Gamma\left(7/4\right)}{9\Gamma\left(-3/4\right)},$
$\displaystyle\widetilde{e}^{(0)}_{3}=\frac{3\sqrt{2}\pi^{3}\Gamma\left(7/4\right)-704[\Gamma\left(5/4\right)]^{7}}{324\pi^{3/2}\Gamma\left(-3/4\right)},\qquad\cdots$
(203)

and $\widetilde{e}^{(\ell\in{\mathbb{N}})}_{n}=0$ for all
$n\in{\mathbb{N}}_{0}$. Taking $E_{\cal
PT}=\widetilde{\eta}(g\hbar^{4})^{1/3}$ yields

$\displaystyle\frac{E_{\cal PT}}{(g\hbar^{4})^{1/3}}$ $\displaystyle=$
$\displaystyle\kappa^{4/3}\left[\frac{3^{4/3}}{4[K(-1)]^{4/3}}-\frac{8\sqrt{\pi}\Gamma\left(7/4\right)}{3^{8/3}[K(-1)]^{1/3}\Gamma\left(-3/4\right)}\kappa^{-2}+O(\kappa^{-4})\right].$
(204)

Notice that the solution is real and Borel non-summable. In Fig. 11, we show
the truncated ${\cal PT}$ energy solutions defined as

$\displaystyle E_{{\cal PT}}^{\rm
trn}=(g\hbar^{4})^{1/3}\kappa^{4/3}\sum_{n=0}^{n_{\rm
max}-1}c_{n}\kappa^{-2n},$ (205)

where $c_{n\in{\mathbb{N}}_{0}}$ are coefficients obtained by our EWKB. These
plots are quite close to the exact values in $n_{\rm max}\lesssim 5$. The
lowest energy starts to deviate from the exact value around $n_{\rm
max}\approx 6$ and then becomes divergent. Since the energy spectrum with any
energy level is a divergent series of $\kappa^{-1}$, all of them should be
eventually divergent by taking $n_{\rm max}\rightarrow+\infty$. Hence, one can
interpret these plots as follows: only the lowest energy goes beyond the
limitation of optimal truncation in $n_{\rm max}\gtrsim 6$, and the others are
still stable as keeping almost the exact value up to, at least, $n_{\rm
max}=12$ boyd1999devil . Therefore, Fig. 11 insists that EWKB of the
$\kappa^{-1}$-expansion works well.

Figure 11: Truncated ${\cal PT}$ energy spectrum for first five energy levels,
$k=0,\cdots,4$, evaluated by Eq.(205) with $g\hbar^{4}=1$. $n_{\rm max}$ in
the horizontal axis denotes the truncation order of the truncated ${\cal PT}$
energy solution. The black dashed lines are exact values from Ref. Bender:2019
.  (a) $\arg[\widetilde{Q}_{0}\kappa^{2}]=0$ (b)
$\arg[\widetilde{Q}_{0}\kappa^{2}]=-\frac{\pi}{2}$ (c)
$\arg[\widetilde{Q}_{0}\kappa^{2}]=+\frac{\pi}{2}$
---
Figure 12:  Stokes graphs for
$\arg[\widetilde{Q}_{0}\kappa^{2}]=0,\mp\frac{\pi}{2}$. The QCs in
Eqs.(197)(198) are constructed under the assumption that
$\arg[\widetilde{Q}_{0}\kappa^{2}]=0$ shown in (a). The solution of
${\mathfrak{D}}_{\rm AC}^{\arg(\widetilde{\eta})=0}=0$ violates the assumption
and reproduces either (b) or (c), where another Stokes phenomenon occurs,
depending on the paths of analytic continuation, $\gamma_{3\rightarrow 1}$ or
$\gamma_{2\rightarrow 4}$.

We show the fact that no appropriate solution of
${\mathfrak{D}}^{\arg(\widetilde{\eta})=0}_{\rm AC}$ exists. It can be seen by
constructing the modified potential $\widetilde{Q}(\kappa)$ in Eq.(180). Since
the AC QCs (197) have a special form such that the perturbative part is given
by ${\mathfrak{D}}_{\rm AC}^{\arg(\widetilde{\eta})=0}\sim 1+A^{\pm 1}B$, the
leading order is a complex value, that is
$\widetilde{e}_{-1}^{(0)}=\frac{3e^{\pm\pi i/4}}{4K(-1)}$. This coefficient
makes the leading order of $\widetilde{Q}$ complex-valued as
$\widetilde{Q}_{0}(x)\kappa^{2}=Q(x)\left(\frac{3}{4K(-1)}\right)^{2}(e^{\pm\pi
i/4}\kappa)^{2}$. Due to the phase in $\widetilde{Q}_{0}$, i.e. $e^{\pm\pi
i/2}$, it does not give the same Stokes graph to Fig. 12(a), but (b) or (c).
Since DDP formulas are generally defined from each Stokes phenomenon, the QCs
(197) formulated by the DDP formula corresponding to Fig. 12(a) violates the
assumption that its solution must reproduce the same Stokes graph. Therefore,
the solution of ${\mathfrak{D}}_{\rm AC}^{\arg(\widetilde{\eta})=0}=0$ in
Eq.(197) should be rejected141414 We should also remind that $\kappa$ is now a
real value. . Instead of Fig. 12(a), one can begin with Fig. 12(b)(c) by
taking $\arg(\widetilde{\eta})=\pm\frac{\pi}{4}$ and perform EWKB to obtain
the AC energy. By this procedure, appropriate solutions are available from
either (b) or (c) depending on the path of analytic continuation, as (b) for
$\gamma_{3\rightarrow 1}$ and (c) for $\gamma_{2\rightarrow 4}$. The Stokes
graph and the details of the calculations are parallel to the analysis in Sec.
5.1, and one can consequently find the same the Hermitian energy in
Eqs.(177)-(179) except an overall phase due to the phase rotation of the
coupling constant, $\lambda=ge^{\pm\pi i}$:

$\displaystyle E^{\arg(\widetilde{\eta})=+\pi/4}_{{\rm AC}(3\rightarrow
1)}(g)={\cal C}\left[E^{\arg(\widetilde{\eta})=-\pi/4}_{{\rm AC}(2\rightarrow
4)}(g)\right]=E_{\cal H}(\lambda=ge^{+\pi i}),\qquad g\in{\mathbb{R}}_{>0},$
(206)

where ${\cal C}$ is complex conjugate. This is the same conclusion to
Eq.(182), and we should accept this solution as $E_{\rm AC}$.

###  5.3 Impossibility of an alternative form of the ABS conjecture

Finally, we point out that the ${\cal PT}$ and the AC energies for the pure
quartic potential are independent solutions of each other, meaning that those
can not be related by Stokes automorphism and Borel resummation.

Let us see this fact. In order to make such a relation, if there exists, those
QCs need to be equivalent by DDP formula at a certain $\arg(\lambda)$ and/or
$\arg(\kappa)$151515 These two complex phases are distinguished for the ansatz
in Eq.(172). The former is an overall phase, and the latter affects the
coefficients. . The discussion in Sec. 4.1.2 is helpful for the consideration.
Since no Stokes phenomenon occurs in $0<|\arg(\eta)|<\pi/4$, the QC for
$\arg(\eta)=0_{\pm}$ in Eqs.(85)(86) can be directly analytic-continued to the
AC QCs for $\arg(\widetilde{\eta})=0_{\mp}$ in Eqs.(192)(193) by replacing
symbols $(A_{1},A_{2})\rightarrow(A,B)$ as

$\displaystyle{\mathfrak{D}}_{{\rm AC}(3\rightarrow
1)}^{\arg(\widetilde{\eta})=0_{+}}\propto\left.{\mathfrak{D}}_{\cal
H}^{\arg(\eta)=0_{-}}\right|_{(A_{1},A_{2})\rightarrow(A,B)},\qquad{\mathfrak{D}}_{{\rm
AC}(2\rightarrow
4)}^{\arg(\widetilde{\eta})=0_{-}}\propto\left.{\mathfrak{D}}_{\cal
H}^{\arg(\eta)=0_{+}}\right|_{(A_{1},A_{2})\rightarrow(A,B)}.$

This procedure is the same to rotating complex phases of $\lambda$ and
$\kappa$ at once. Thus, we now have the direct relation between
${\mathfrak{D}}_{\cal H}^{\arg(\eta)=0_{\pm}}$ and ${\mathfrak{D}}_{\rm
AC}^{\arg(\widetilde{\eta})=0_{\pm}}$ given by Eq.(5.3). However, no
connection from ${\mathfrak{D}}_{\rm AC}^{\arg(\widetilde{\eta})=0_{\pm}}$ to
${\mathfrak{D}}_{\cal PT}^{\arg(\widetilde{\eta})=0}$ is constructable by the
DDP formula in Eq.(196). Especially, eliminating the non-perturbative part,
$B$, from ${\mathfrak{D}}_{\rm AC}^{\arg(\widetilde{\eta})=0_{\pm}}$ is
extremely tough, which implies that the perturbative/non-perturbative
structures between the ${\cal PT}$ and the AC energies are not related to each
other. Notice that $E_{\cal PT}$ in Eq.(204) is purely perturbative with
respect to the $\kappa^{-1}$-expansion, but $E_{\rm AC}$ in Eq.(182) is not.
Furthermore, even the coefficients in their perturbative part do not match
with each other. Therefore, the ABS conjecture is not satisfied, and those
must be independent solutions. No alternative form of the ABS conjecture can
be reformulated by Stokes automorphism and Borel resummation. In summary, the
above consideration can be schematically expressed by

$\displaystyle{\mathfrak{D}}_{\cal
H}^{\arg(\eta)=0_{-}}\xrightarrow{\lambda=ge^{+\pi i}}{\mathfrak{D}}_{{\rm
AC}(3\rightarrow
1)}^{\arg(\widetilde{\eta})=+\frac{\pi}{4}+0_{-}}\xrightarrow{\arg(\kappa)=0\rightarrow+\frac{\pi}{4}+0_{-}}{\mathfrak{D}}_{{\rm
AC}(3\rightarrow
1)}^{\arg(\widetilde{\eta})=0_{+}}\mathrel{\vphantom{\xrightarrow[]{{\mathfrak{S}}^{\forall\nu\in{\mathbb{R}}}_{\arg(\widetilde{\eta})=0}}}\ooalign{$\vbox{\hbox{\rotatebox[origin={c}]{-45.0}{\scalebox{1.5}[1.0]{$\scriptscriptstyle|$}}}}$\cr$\xrightarrow[]{{\mathfrak{S}}^{\forall\nu\in{\mathbb{R}}}_{\arg(\widetilde{\eta})=0}}$\cr}}{\mathfrak{D}}^{\arg(\widetilde{\eta})=0}_{\cal
PT},$ (210) $\displaystyle{\mathfrak{D}}_{\cal
H}^{\arg(\eta)=0_{+}}\xrightarrow{\lambda=ge^{-\pi i}}{\mathfrak{D}}_{{\rm
AC}(2\rightarrow
4)}^{\arg(\widetilde{\eta})=-\frac{\pi}{4}+0_{+}}\xrightarrow{\arg(\kappa)=0\rightarrow-\frac{\pi}{4}+0_{+}}{\mathfrak{D}}_{{\rm
AC}(2\rightarrow
4)}^{\arg(\widetilde{\eta})=0_{-}}\mathrel{\vphantom{\xrightarrow[]{{\mathfrak{S}}^{\forall\nu\in{\mathbb{R}}}_{\arg(\widetilde{\eta})=0}}}\ooalign{$\vbox{\hbox{\rotatebox[origin={c}]{-45.0}{\scalebox{1.5}[1.0]{$\scriptscriptstyle|$}}}}$\cr$\xrightarrow[]{{\mathfrak{S}}^{\forall\nu\in{\mathbb{R}}}_{\arg(\widetilde{\eta})=0}}$\cr}}{\mathfrak{D}}^{\arg(\widetilde{\eta})=0}_{\cal
PT}.$ (213)

Notice that the violation of the ABS conjecture for $\omega=0$ is essentially
irrelevant to the values of $\beta$, $g$, and $\hbar$. It is because our EWKB
of the $\kappa^{-1}$-expansion does not depend on them, and their dependence
appears as an overall factor of the energy only. Our result is consistent with
the observations in Ref. Lawrence:2023woz .

##  6 Additional remarks

###  6.1 Generalization to even $\varepsilon$

We would make comments on generalizations of the modified ABS conjecture to
$\varepsilon\in 2{\mathbb{N}}$ in Eq.(1). The cases of $\varepsilon\in
4{\mathbb{N}}$ are trivial because the interaction term is non-negative in the
${\cal PT}$ symmetric potential and is equivalent to the Hermitian potential.
For this reason, we address the cases of $\varepsilon\in 4{\mathbb{N}}_{0}+2$.
In such a case, the paths of analytic continuation in Eqs.(118)(119) are
modified as

$\displaystyle\gamma_{3\rightarrow 1}:-e^{\alpha\pi i}\infty\ \rightarrow\
e^{\alpha\pi i}\infty,\qquad\gamma_{2\rightarrow 4}:-e^{-\alpha\pi i}\infty\
\rightarrow\ e^{-\alpha\pi i}\infty,$ (215)

for the AC energy, and

$\displaystyle\gamma_{3\rightarrow 4}$ $\displaystyle:$
$\displaystyle-e^{\alpha\pi i}\infty\ \rightarrow\ e^{-\alpha\pi i}\infty,$
(216)

for the ${\cal PT}$ energy, where $\alpha=\frac{1}{\varepsilon+2}$. Along
these paths, one can directly apply the same analyses in Secs. 4 and 5, and
the similar results can be eventually found.

For $\omega>0$, similar to the case that $\varepsilon=2$, the ${\cal PT}$
energy is purely perturbative (but Borel non-summable), and the AC energy
includes non-perturbative contributions in addition to the same perturbative
${\cal PT}$ energy. The QCs and the DDP formula also take the same forms. As a
result, Eq.(144) is satisfied, and our modified ABS conjecture is unchanged
for any $\varepsilon\in 4{\mathbb{N}}_{0}+2$.

For $\omega=0$, the generalization of $\varepsilon$ to the energy in Eq.(160)
gives

$\displaystyle E=c(k)(\lambda\hbar^{2+\varepsilon})^{2/(4+\varepsilon)},\qquad
c(k)\in{\mathbb{R}}_{>0},$ (217)

and the scaling law in Eq.(163) is modified as

$\displaystyle[x]=\frac{1}{2+\varepsilon},\qquad[\hbar]=\frac{4+\varepsilon}{4+2\varepsilon},\qquad[\lambda]=0,\qquad[E]=1.$
(218)

The ansatz in Eqs.(172)(202) based on the $\kappa^{-1}$-expansion works by a
slight modification of the dimensionless parameters, $\eta$ and
$\widetilde{\eta}$, as

$\displaystyle\eta=\frac{\lambda^{1/(2+\varepsilon)}\hbar}{E^{(4+\varepsilon)/(4+2\varepsilon)}},\qquad\widetilde{\eta}=\frac{g^{1/(2+\varepsilon)}\hbar}{E^{(4+\varepsilon)/(4+2\varepsilon)}}.$
(219)

The ${\cal PT}$ energy is purely perturbative with respect to the
$\kappa^{-1}$ expansion, and the AC energy contains non-perturbative parts not
only the perturbative part like the case of $\varepsilon=2$. The relation
between the Hermitian and the AC energies in Eq.(182) is modified as

$\displaystyle\frac{E_{\rm
AC}(g)}{(g\hbar^{2+\varepsilon})^{2/(4+\varepsilon)}}=e^{\pm\frac{2\pi}{4+\varepsilon}i}\frac{E_{\cal
H}(\lambda)}{(\lambda\hbar^{2+\varepsilon})^{2/(4+\varepsilon)}}.$ (220)

No relation between $E_{\rm AC}(g)$ and $E_{\cal PT}(g)$ is constructable by
Stokes automorphism and Borel resummation.

###  6.2 Spectral reality

Observations from QCs and DDP formula are quite useful to know properties of a
energy spectrum of a given theory. In EWKB, spectral reality can be determined
solely by the corresponding QC without specifically solving them, and showing
spectral reality is equivalent to obtaining a real QC by analytic
continuation. For example, by denoting $a_{A}(E,\hbar)=-i\log
A(E,\hbar)\in{\mathbb{R}}$, Eqs.(130)(131) are reexpressed by

$\displaystyle{\mathfrak{D}}^{0}_{{\rm AC}(3\rightarrow 1)/(2\rightarrow
4)}\propto\cos\left[\frac{a_{A}}{2}\pm\frac{i}{2}\log(1+B)\right],\qquad{\mathfrak{D}}^{0}_{\cal
PT}\propto\cos\frac{a_{A}}{2}.$ (221)

Since ${\mathfrak{D}}^{0}_{\cal PT}$ is real, the ${\cal PT}$ energy solution
should be real. In addition, one can immediately see that the spectral reality
of the above AC energy solutions are violated by the non-perturbative effect,
$B$, because of the complex ${\mathfrak{D}}^{0}_{\rm AC}$.

The methods used in our analysis based on the Airy-type and the DW-type
connection formulas should work for ${\cal PT}$ symmetric polynomial
potentials. By taking appropriate paths of analytic continuation, one can make
sure that Eq.(1) with $\omega\in{\mathbb{R}}_{>0}$ and
$\varepsilon\in{\mathbb{N}}$ provides the same cycle representation to
Eq.(221) for ${\mathfrak{D}}^{0}_{\cal PT}$. For more complicated potentials
such as $V_{\cal PT}(x)=g(ix)^{\alpha}\log(ix)$, one has to formulate a
connection formula beyond the Airy- and DW-type. This would be an interesting
problem as a future work.

###### Acknowledgements.

We would thank Toshiaki Fujimori, Tatsuhiro Misumi, Muneto Nitta, Norisuke
Sakai, Naohisa Sueishi, and Hidetoshi Taya for helpful discussion about exact
WKB analysis. We would thank Marco Serone for helpful discussion about
connection formula of the degenerate Weber-type Stokes graph. We would thank
Mithat Ünsal for comments on a draft of this paper. S. K is supported by JSPS
KAKENHI Grant No. 22H05118.

##  Appendix A Derivation of $G$

Figure 13: Stokes graph for the degenerate Weber equation. The green dot is a
double turning point. The black solid and red wave lines denote Stokes lines
and a branch-cut, respectively.

We show the derivation of $G$ in Eq.(134) using a coordinate transform from
the degenerate Weber (DW)-type Stokes graph to our potential form Takei3 ;
DDP2 ; Sueishi:2021xti ; Bucciotti:2023trp . There is another method using
Mellin transform, see Refs. DP1 ; Zinn-Justin:2004qzw ; Kamata:2021jrs .

The DW-type Stokes graph offers a connection formula around a double turning
point. In general, any connection formulas are firstly defined in a local
coordinate and then mapped to a coordinate used in a given problem. The local
DW-type connection formula is defined by the following Schrödinger equation:

$\displaystyle\widehat{\cal
L}:=-\hbar^{2}\partial_{y}^{2}+\frac{y^{2}}{4}-\kappa\hbar,\qquad\widehat{\cal
L}\widehat{\psi}(y,\hbar)=0,$ (222)

where $\kappa\in{\mathbb{R}}$. The Stokes graph is shown in Fig. 13, and the
connection matrices, $\widehat{\cal M}^{\bullet\rightarrow\bullet}$, are
expressed by DDP2 ; Kawai1 ; Takei3 ; Bucciotti:2023trp

$\displaystyle\widehat{\cal M}^{{\rm IV}\rightarrow{\rm I}}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&0\\\ i\frac{\sqrt{2\pi}e^{+\pi
i\kappa}\hbar^{+\kappa}}{\Gamma(1/2-\kappa)}&1\end{pmatrix},\qquad\widehat{\cal
M}^{{\rm I}\rightarrow{\rm
II}}=\begin{pmatrix}1&i\frac{\sqrt{2\pi}\hbar^{-\kappa}}{\Gamma(1/2+\kappa)}\\\
0&1\end{pmatrix},$ $\displaystyle\widehat{\cal M}^{{\rm II}\rightarrow{\rm
III}}$ $\displaystyle=$ $\displaystyle\begin{pmatrix}1&0\\\
i\frac{\sqrt{2\pi}e^{-\pi
i\kappa}\hbar^{+\kappa}}{\Gamma(1/2-\kappa)}&1\end{pmatrix},\qquad\widehat{\cal
M}^{{\rm III}\rightarrow{\rm IV}}=\begin{pmatrix}1&i\frac{\sqrt{2\pi}e^{-2\pi
i\kappa}\hbar^{-\kappa}}{\Gamma(1/2+\kappa)}\\\ 0&1\end{pmatrix}.$ (223)

In addition, the branch-cut matrix $\widehat{T}$ is given by

$\displaystyle\widehat{T}=\begin{pmatrix}-e^{+2\pi i\kappa}&0\\\ 0&-e^{-2\pi
i\kappa}\end{pmatrix}.$ (224)

By a coordinate transform $y=y(x)$, the connection matrices in the
$x$-coordinate, ${\cal M}^{\bullet\rightarrow\bullet}$, take the form that

$\displaystyle{\cal M}^{{\rm IV}\rightarrow{\rm I}}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&0\\\
i\frac{C_{-}}{C_{+}}\frac{\sqrt{2\pi}e^{+\pi
iF}\hbar^{+F}}{\Gamma(1/2-F)}&1\end{pmatrix},\qquad{\cal M}^{{\rm
I}\rightarrow{\rm
II}}=\begin{pmatrix}1&i\frac{C_{+}}{C_{-}}\frac{\sqrt{2\pi}\hbar^{-F}}{\Gamma(1/2+F)}\\\
0&1\end{pmatrix},$ $\displaystyle{\cal M}^{{\rm II}\rightarrow{\rm III}}$
$\displaystyle=$ $\displaystyle\begin{pmatrix}1&0\\\
i\frac{C_{-}}{C_{+}}\frac{\sqrt{2\pi}e^{-\pi
iF}\hbar^{+F}}{\Gamma(1/2-F)}&1\end{pmatrix},\qquad{\cal M}^{{\rm
III}\rightarrow{\rm
IV}}=\begin{pmatrix}1&i\frac{C_{+}}{C_{-}}\frac{\sqrt{2\pi}e^{-2\pi
iF}\hbar^{-F}}{\Gamma(1/2+F)}\\\ 0&1\end{pmatrix}.$ (225)

and the branch-cut matrix $T$ is given by

$\displaystyle T=\begin{pmatrix}-e^{+2\pi iF}&0\\\ 0&-e^{-2\pi
iF}\end{pmatrix}.$ (226)

The non-perturbative ${\mathfrak{B}}$ in Eq.(134) is given by the above
connection formula and takes the form as

$\displaystyle{\mathfrak{B}}=\frac{C_{-}}{C_{+}}\frac{\sqrt{2\pi}{\mathfrak{B}}_{0}e^{+\pi
iF}\hbar^{+F}}{\Gamma(1/2-F)},\qquad{\mathfrak{B}}_{0}:=e^{-\frac{S_{\mathfrak{B}}}{\hbar}},$
(227)

where $S_{\mathfrak{B}}\in{\mathbb{R}}_{>0}$ is determined by distance of two
turning points defining the ${\mathfrak{B}}$-cycle and $S_{\mathfrak{B}}=2/3$
in our case. Hence, in order to know the specific form of $G$ in Eq.(134), our
task is to compute $C_{\pm}$ by the coordinate transform.

Those are available by constructing the wavefunctions of Eqs.(133)(222)
specifically. Here, $Q(x)$ and $\widehat{Q}(y)$ are defined as
$Q(x)=x^{2}-x^{4}-\widetilde{E}\hbar$ and
$\widehat{Q}(y)=\frac{y^{2}}{4}-\kappa\hbar$, respectively. One should be
careful that $S_{\rm od}(\hbar)$ given by them contains not only odd powers of
the $\hbar$-expansion but also even powers due to the energy terms
proportional to $\hbar$. The coordinate transform can be obtained from the
relation of $Q(x)$ and $\widehat{Q}(y)$ given by

$\displaystyle
Q(x,\hbar)=\left[\partial_{x}y(x,\hbar)\right]^{2}\widehat{Q}(y(x,\hbar),\hbar)-\frac{\hbar^{2}}{2}\\{y(x,\hbar);x\\},$
(228)

with the Schwarzian derivative defined as

$\displaystyle\\{y(x,\hbar),x\\}:=\frac{\partial_{x}^{3}y(x,\hbar)}{\partial_{x}y(x,\hbar)}-\frac{3}{2}\left(\frac{\partial_{x}^{2}y(x,\hbar)}{\partial_{x}y(x,\hbar)}\right)^{2}.$
(229)

By denoting $S_{\rm od}(x)$ and $\widehat{S}_{\rm od}(y)$ as solutions of
Riccati equation defined by $Q(x)$ and $\widehat{Q}(y)$, respectively, the
wavefunctions constructed from $S_{\rm od}(x)$ and $\widehat{S}_{\rm od}(y)$
are also related to each other as

$\displaystyle\psi_{\pm}(x,\hbar)=C_{\pm}(\hbar)[\partial_{x}y(x,\hbar)]^{-1/2}\widehat{\psi}_{\pm}(y(x,\hbar),\hbar),$
(230)

where the sign, “$\pm$”, in the wavefunctions is a label to distinguish the
upper and lower components. From Eq.(230), the coefficients, $C_{\pm}(\hbar)$,
can be obtained as

$\displaystyle C_{\pm}(\hbar)=\lim_{x\rightarrow
0}[\partial_{x}y(x,\hbar)]^{1/2}\frac{\psi_{\pm}(x,\hbar)}{\widehat{\psi}_{\pm}(y(x,\hbar),\hbar)}.$
(231)

In our analysis, we define the upper and lower components of the wavefunctions
such that $\psi_{\pm}(x,\hbar)=\frac{\exp[\mp\int dxS_{\rm
od}(x,\hbar)]}{\sqrt{S_{\rm od}(x,\hbar)}}$ and
$\widehat{\psi}_{\pm}(y,\hbar)=\frac{\exp[\pm\int dy\widehat{S}_{\rm
od}(y,\hbar)]}{\sqrt{\widehat{S}_{\rm od}(y,\hbar)}}$ to adjust to the
asymptotic behavior in Fig. 13. One can easily check that $\kappa=-{\rm
Res}_{y=0}\widehat{S}_{\rm od}(y,\hbar)=-{\rm Res}_{x=0}S_{\rm od}(x,\hbar)$.
From Eq.(228), one can obtain $y(x,\hbar)$ and $\kappa(\hbar)$ as

$\displaystyle y(x,\hbar)$ $\displaystyle=$
$\displaystyle\frac{i}{\sqrt{2}}\left[2x-\frac{\widetilde{E}}{4}x\hbar-\frac{49+16\widetilde{E}^{2}}{64}x\hbar^{2}\right.$
(232)
$\displaystyle\qquad\left.+\frac{\widetilde{E}(11+\widetilde{E}^{2})}{32x^{3}}\hbar^{3}-\frac{\widetilde{E}(1379+226\widetilde{E}^{2})}{512}x\hbar^{3}+O(\hbar^{4})\right]+O(x^{2}),$
$\displaystyle\kappa(\hbar)$ $\displaystyle=$
$\displaystyle-\frac{\widetilde{E}}{2}-\frac{3(1+\widetilde{E}^{2})}{16}\hbar-\frac{5\widetilde{E}(17+7\widetilde{E}^{2})}{128}\hbar^{2}-\frac{105(19+50\widetilde{E}^{2}+11\widetilde{E}^{4})}{2048}\hbar^{3}+O(\hbar^{4}).$

Thus, Eq.(231) gives

$\displaystyle C_{+}(\hbar)$ (234) $\displaystyle=$
$\displaystyle\exp\left[\frac{\widetilde{E}^{2}}{16}\hbar+\frac{\widetilde{E}(55+23\widetilde{E}^{2})}{256}\hbar^{2}+\frac{441+5388\widetilde{E}^{2}+1091\widetilde{E}^{4}}{6144}\hbar^{3}+O(\hbar^{4})\right]2^{\frac{F(\hbar)}{2}}e^{\frac{\pi}{2}iF(\hbar)}$
$\displaystyle=$ $\displaystyle\frac{1}{C_{-}(\hbar)},$

where we used the fact that $F(\hbar)=\kappa(\hbar)$ by Eqs.(136)(A).
Substituting Eq.(234) into Eq.(227) and comparing with Eq.(134) give the form
of $G$ in Eq.(136).

##  Appendix B Alien calculus for energy spectra

We formulate Stokes automorphism (or DDP formula) for the transseries solution
of the energies in Sec. 4.2 by performing alien calculus. We omit the
subscript “$0$”, e.g.
${\mathfrak{S}}_{0}^{\nu}\rightarrow{\mathfrak{S}}^{\nu}$ and only address the
case that $\arg(g)=0$. Since expanding the Stokes automorphism in Eq.(15)
around $\nu=0$ gives

$\displaystyle{\mathfrak{S}}^{\nu}=1+\nu\overset{\underset{\bullet}{}}{\Delta}+\frac{\nu^{2}}{2}(\overset{\underset{\bullet}{}}{\Delta})^{2}+O(\nu^{3}),$
(235)

our main task is to calculate action of
$(\overset{\underset{\bullet}{}}{\Delta})^{n\in{\mathbb{N}}}$ to the
transseries solutions order by order.

Before discussion about our case, we suppose a function $f(E(\hbar),\hbar)$
with an indirect dependence of $\hbar$ in the variable $E=E(\hbar)$ and
consider action of the alien derivative to $f(E(\hbar),\hbar)$. In this case,
the alien derivative can be decomposed into two parts, i.e. actions to
$E(\hbar)$ and to $f(E,\hbar)$ with a fixed $E$. Using the chain rule, it can
be expressed by161616 This procedure is the same methodology to split a total
derivative into partial derivatives for a function with indirect independences
Bucciotti:2023trp .

$\displaystyle\overset{\underset{\bullet}{}}{\Delta}[f](E(\hbar),\hbar)=\left.\frac{\partial
f(E,\hbar)}{\partial