text:
\section*{Preamble}

The term \textit{duality}, in the meaning we use it here, was introduced in \cite{Dual} in the studies of the Lam\'e system. There,  potentials with different values of the elliptic parameter $\,\nu\,$ were named dual if their spectra (i.e., the band-gap edges) interchange under the switch $\,E\,\to\,-\,E\,$. In the self-dual case (the one corresponding to $\,\nu=1/2\,$), the spectrum of the potential maps onto itself under the said switch. One way to understand this phenomenon lies through  algebraisation of the Lam\'e potential, i.e., through writing the Hamiltonian in a matrix form.

The property of the spectrum to transform into itself under the change $\,E \to -E\,$ was originally established in the studies of a sextic quasi-exactly solvable (QES) potential. This property was termed as the $\,${\it{energy reflection (ER) symmetry}}$\,$. $\,$In our article, we shall refer to it as the $\,${\it{energy spectrum reflection (ESR) property}}$\,$. $\,$In contrast to the Lam\'e model, wherein \textit{all} the band-gap edges can be found by the matrix means, the QES problems normally permit only for partial algebraisation of the, generally infinite, spectrum. Consequently, the ESR symmetry refers only to its algebraic part. In this, restricted sense, the sextic potential can be regarded self-dual. Therefore, it is natural to hypothesise that there also exist \textit{dual} sextic potentials whose algebraic parts of the spectra interchange under ${\,E\,\to\,-\,E\,}$. Such potentials were found in \cite{MyNew}, and their duality was proven both by matrix methods and by the quantum-mechanical means (with the Schr\"odinger equation employed). Here we study in a great detail the latter of these two approaches.

In~\cite{Alvarez}, it was spelled out for the first time, that the symmetry of the quantum energy levels of a QES system stems from a symmetry of the corresponding classical system. In {\it{Ibid.}}, the symmetry of the classical system was discussed in the canonical language of phase space. In the present work, we shall show that classical and quantum symmetries can be uniformly described in the language of the action, for which, as we shall demonstrate, the term \textit{duality} can be applied as well. This formalism allows one to explore at a deeper level the connection between the symmetries of the classical and quantum systems. Specifically, it turns out that the symmetries of both kinds of systems can be understood from studying the global properties of the Riemannian surfaces of classical and quantum momenta, correspondingly.

\newpage
\section{Duality, self-duality, and action reflection symmetry for classical systems}

This section addresses the ESR symmetry of two classical systems governed by sextic potentials. To explore the behaviour of such systems, we shall employ some tools from the complex analysis, with elements of topology involved. Aside from theoretical yield for classical mechanics, this study will lay the basis for the subsequent research on symmetry between the quantised counterparts of the classical systems considered.

\subsection{The sextic potential}

We begin with exploring one-dimensional motion in the potential
\begin{equation}\begin{gathered}
    V_{a,\,c}\,(x) = a + b\,x^2 + c\,x^4 + d \,x^6\quad,\\
    a,\,b,\,c,\in \mathbb{R}\quad,\qquad d\in\mathbb{R}^+\quad.
    \label{Vcl}
\end{gathered}\end{equation}
Two sextic potentials of the type (\ref{Vcl}), which share the values of the parameters $\,b\,$ and $\,d\,$, and have opposite values of the parameters $\,a\,$ and $\,c\,$, possess the following duality property:
\begin{equation}
    V_{-a,\,-c}\,(x) = - V_{a,\,c}\,({i\mkern1mu}\,x) \quad.
\end{equation}
In the special case of $\,a=c=0\,$, this becomes self-duality:
\begin{equation}
    V_{0,\,0}\,(x)

    = - V_{0,\,0}\,({i\mkern1mu}\,x) \quad.
\end{equation}
We now introduce the canonical momentum as
\begin{equation}
    p\,(x, E) = \sqrt{2\,(E - V\,(x))}\quad,\label{p}
\end{equation}
with the mass set to be $\,m=1\,$, and with the positive sign in front of the square root chosen.

For the two said potentials, the duality property of the corresponding momenta  reads as:
\begin{equation}
    p_{a,\,c}\,(x,\, E) = {i\mkern1mu}\,p_{-a,\,-c}\,({i\mkern1mu}\,x,-E)\quad. \label{pdual0}
\end{equation}
For an arbitrary initial energy, the motion in each of these potentials is periodic, as both potentials become infinite at $\,|x| \to \infty\,$.

The abbreviated classical action is defined as an integral of the momentum over a period:
\footnote{~The so-defined action is but a Legendre transform of the standard action~$\int^t L(q, \dot{q})\,\d t\,$, with $L$ being the Lagrangian.}
\begin{equation}
    S\,(E) = \oint \limits_{\mathcal{C}_R} p\,(x, E) \, \d x \quad,
    \label{18}
\end{equation}
where the contour $\mathcal{C}_R$ encloses the two (real) turning points $x_1$ and $x_2$ in the complex $x$-plane, between which oscillations are taking place.$\,$\footnote{~The contour $\mathcal{C}_R$ should be close enough to the real axis, in order to avoid enclosing the singularities other than those located on the branch cut of the momentum between the two real turning points.}$\,$
These two turning points are such two real solutions of the equation
\begin{equation}
V\,(x) = E \quad \quad,
\label{turn}
\end{equation}
that
\begin{equation}
    V\,(x) < E \qquad \text{if} \qquad x_1 < x < x_2 \quad
\end{equation}
and no other real solutions of~\eqref{turn} lies between them. \text
In~\eqref{18}, the integral along the closed contour $\mathcal{C}_R$ is nonzero due to the presence of a branch cut of the momentum inside it, between the turning points.$\,$\footnote{~Generally speaking, one has freedom in defining the branch cuts. For the real turning points, it is usually convenient to make the cut go along the finite interval of the real axis between the turning points.}
This singularity makes the contour non-contractible or, in the parlance of topology, homotopically non-trivial.

The momentum is set positive along the bottom of the cut, in order for the action to stay positive when the direction of the contour in~\eqref{gammas} is chosen  counter-clockwise.

For various values of the parameters entering the potential, and for various values of energy, the equation~\eqref{turn} may have up to six real roots, see Figure~\ref{F1}.
We shall be interested in the following special cases:
\begin{figure}
\centering
\subfloat[][
The potential $V_{a,c}\,(x)$ for $b<0$.
]{
\includegraphics[width=.45\textwidth]{aug11}
\label{aug11}
}\qquad
\subfloat[][
The potential $V_{-a,-c}\,(x)$ for $b<0$.
]{
\centering
\includegraphics[width=.45\textwidth]{aug12}
\label{aug12}
}
\\
\subfloat[][
The potential $V_{a,c}\,(x)$ for $b>0$.
]{
\includegraphics[width=.45\textwidth]{aug21}
}\qquad
\subfloat[][
The potential $V_{-a,-c}\,(x)$ for $b>0$.
]{
\includegraphics[width=.45\textwidth]{aug22}
}
\caption{}
\label{F1}
\end{figure}
\begin{enumerate}
\newcounter{potcases}
\item In the case of
\begin{equation}
\addtocounter{equation}{1}
    b < 0\quad,
    \addtocounter{potcases}{1}
    \tag{\theequation\alph{potcases}}
    \label{case1}
\end{equation}
the potential has a double-well shape, with two absolute minima and one local maximum.
\item In the case
\begin{equation}
    0 < b < \dfrac{c^2}{4\,d}\quad,
    \addtocounter{potcases}{1}
    \tag{\theequation\alph{potcases}}
    \label{case2}
\end{equation}
the following two subcases will be considered:
\begin{itemize}
\item If
\begin{equation}
    c > 0 \quad,
    \addtocounter{potcases}{1}
    \tag{\theequation\alph{potcases}}
    \label{case21}
\end{equation}
the potential is of a single-well type, i.e., has a single absolute minimum and no maxima;
\item If
\begin{equation}

    c<0\quad,
    \addtocounter{potcases}{1}
    \tag{\theequation\alph{potcases}}
    \label{case22}
\end{equation}
the potential has three wells, and has one of the local minima located at $\,x=0\,$.
\end{itemize}
\end{enumerate}
The case of $\,b>\dfrac{ c^2}{ 4\,d}\;$ is of no interest to us and will not be considered.

In both cases,~\eqref{case1} and~\eqref{case2}, there exists a range of energies for which one of the equations
\begin{subequations}
\begin{alignat}{3}
    V_{a,c}\,(x)\, &=\,&-&\,E \quad,\label{9a}\\
    V_{-a,-c} \,(x)\, &=\,& &\,E \quad,\label{9b}
\end{alignat}
\end{subequations}
has four real solutions, while another equation has only two. Below we assume the energy $E$ to stay in the range where~\eqref{9b} has two solutions, while~\eqref{9a} has four.
\footnote{~See Section~\ref{shapes} for a detailed discussion in application to a particular form of the potential~\eqref{Vcl}.}

Let us write down the action for (a) the potential $\,V_{a,c}\,(x)\,$ and energy $(-E)$, and (b) the potential $\,V_{-a,-c}\,(x)\,$ and energy $E$. We define the integration cycles $\,\gamma_{n}\,(E)\,$ and $\,\gamma_{n}^{\,\prime}\,(E)\,$ as
\begin{equation}
\begin{array}{cc}
{\displaystyle
\gamma_{n}\,(E)=~\oint \limits_{\mathcal{C}_n} p_{a,c}\,(x,\,-E)\,\d x} \\
~\\
{\displaystyle
\gamma_{n}^{\,\prime}\,(E)=~\oint \limits_{\mathcal{C}_n'} p_{-a,-c}\,(x,\,E)\,\d x}
\end{array}
\qquad\qquad,\qquad
n=1,2,3 \quad
\label{gammas}
\end{equation}
with $\,{\mathcal{C}_n}\,$ and $\,{\mathcal{C}^{\,\prime}_n}\,$ being contours in the complex $\,x$-planes of $\,p_{a,c}\,(x,\,-E)\,$ and $\,p_{-a,-c}\,(x,\,E)\,$.
\footnote{~Be mindful that Figure~\ref{roots6} represents only one of the two sheets of the Riemannian surface of the momentum. The second sheet corresponds to the negative sign in front of the square root in~\eqref{p}. However, it turns out that all the calculations needed in this section can be performed on one sheet only.}

Now we shall express the action of the potential $V_{-a,-c}\,(x)$ in terms of the integration cycles of the potential $V_{a,c}\,(x)$.
Since Figures~\ref{roots6} and~\ref{roots62} are identical up to a rotation by the angle $\pi/2$, identical are
the integrals over the contours $\,{\mathcal{C}_n}\,$ and $\,{\mathcal{C}_n^{\,\prime}}\,$. Indeed,
\begin{eqnarray}
\begin{gathered}
    \gamma_1 \,(E) = \gamma_2 \,(E) =
    \oint \limits_{\mathcal{C}_1} p_{a,c} \, (x, \,-E) \d x =
    2  \int  \limits_A^B p_{a,c} \, (x, \,-E) \d x =
    2  \int  \limits_A^B p_{-a,-c} \, ({i\mkern1mu}\,x, \,E) \d ({i\mkern1mu}\,x) \\=
    2  \int  \limits_{A'}^{B'} p_{-a,-c} \, (x, \,E) \d x =
    \oint \limits_{\mathcal{C}_1'} p_{-a,-c} \, (x, \,E) \d x =
    \gamma_1^{\,\prime} \,(E) = \gamma_2^{\,\prime} \,(E) \quad.\label{rotate}
\end{gathered}
\end{eqnarray}
An analogous derivation works for $\gamma_3 \,(E)$. So one obtains:
\begin{subequations}
\begin{equation}
    \gamma_1\,(E) = \gamma_2\,(E) =
    \gamma_1^{\,\prime}\,(E)= \gamma_2^{\,\prime}\,(E) \quad,
\end{equation}
\begin{equation}
    \gamma_3\,(E) = \gamma_3^{\,\prime}\,(E) \quad.
\end{equation}\label{primeq}
\end{subequations}
These integration cycles are linked to the actions as
\begin{subequations}
\begin{alignat}{9}
    S_{a,c} \,(-E)
    &\equiv \gamma_{1,2} \,(E) &&= \gamma_{1,2}^{\,\prime} \,(E)\quad&&,\\
    S_{-a,-c} \,(E)
    &\equiv \gamma_{3}^{\,\prime} \,(E) &&= \gamma_{3} \,(E)\quad&&.

 \end{alignat}
 \end{subequations}
Hereafter primes will be omitted, because the equalities~\eqref{primeq} make them redundant.
\begin{figure}
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1\textwidth]{roots6}
\caption{The integration cycles for $p_{a,c}\,(x,\,-E)$.}\label{roots6}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1\textwidth]{roots62}
\caption{The integration cycles for $p_{-a,-c}\,(x,\,E)$.}\label{roots62}
\end{minipage}
\end{figure}

It will be helpful to introduce the cycle
\begin{equation}
    \gamma_{\infty}\,(E) \equiv
    \oint \limits_{\mathcal{C}_{\infty}} p_{a,c}\,(x,\,-E)\,\d x =
    \gamma_{1}\,(E) + \gamma_{2}\,(E) + \gamma_{3}\,(E) \quad.
    \label{gammas6}
\end{equation}
This is but an integral over a contour $\mathcal{C_\infty}$ enclosing all branching points. So evaluation of this integral can be carried out with the aid of the residue at infinity:
\begin{equation}
    \gamma_\infty \, (E) =
     2 \pi {i\mkern1mu} \Res \{ p_{a,c}\,(x, -E), \infty \}
    = \dfrac{\pi  \left(c^2-4\,b\,d\right)}{(2\,d)^{3/2}}\quad.
\end{equation}
The key property of this result is that $\gamma_{\infty}$ does $\,${\it{not}}$\,$ depend on the energy. In other words, $\gamma_1\,(E)$ and $\gamma_3\,(E)$, as functions of energy, turn out to be linearly dependent. A similar result can be obtained for $V_{a,c} \,(E)$ and $V_{-a,-c} \,(E)$. Thus we arrive at
\begin{gather}
    2\,S_{a,c}\,(-E) + S_{-a,-c} \,(E)
    = 2\,S_{-a,-c}\,(-E) + S_{a,c} \,(E)
    = \dfrac{\pi  \left(c^2-4\,b\,d\right)}{(2\,d)^{3/2}}\quad, \label{S}
\end{gather}
an equality to be referred to as \textit{duality of the action}.
For the sum $~2\,S_{\pm a,\pm c}\,(-E) + S_{\mp a,\mp c} \,(E)~$
to come out energy-independent, not only the duality property was essential but also a simple analytic structure of the potential. Had we started with potentials of a higher order, we would have encountered more branching points in the complex plane. So the sum of the cycles responsible for the classical actions would no longer be proportional to the residue at infinity, because other nontrivial cycles
(enclosing additional branching points)
would have emerged:
\begin{subequations}
\begin{equation}
    \gamma_\infty
    = \gamma_1\,(E) + \gamma_2\,(E) + \gamma_3\,(E) + \gamma_4\,(E) +\ldots~~.
 \label{26a}
\end{equation}
The latter is equivalent to
\begin{equation}
\gamma_1\,(E) + \gamma_2\,(E) + \gamma_3\,(E)\,=\,\gamma_\infty\,-\,\gamma_4\,(E)\,-\,\ldots\,\;,
\label{26b}
\end{equation}
\label{26}
\end{subequations}
with the right-hand side (and, therefore, also the left-hand side) being energy-dependent.

\newpage
\subsection{Elliptic potential}
This section was largely motivated by \cite{Dual}. \vspace{1.5mm}

It turns out that the method developed above for a polynomial potential can be easily applied to a periodic one. Before demonstrating this, let us recall the major differences between polynomial potentials and periodic potentials (the latter assumed continuous along the real direction).

First of all, if a potential (as a function of a complex variable) has a single period, then, instead of the entire complex plane, it can be defined on a strip. In this case, one of the coordinates can be compactified, and the function can be treated as if it were defined on a cylinder. If the potential is doubly-periodic, the situation is even better, in that the function can be naturally defined on a parallelogram compactifiable into a torus deprived of, at least, one point. This point corresponds to the pole which the periodic potential does necessarily have. Indeed, while the polynomial potential has its only pole at infinity, the doubly-periodic function, in order not to be constant, must have poles inside the fundamental parallelogram,  according to the Liouville theorem. The Riemannian surface of the corresponding momentum function can, generally, be of a different genus.

Another fundamental difference between the polynomial and periodic potentials comes from the fact that, while the former only allow for the bounded motion, the latter permit for both bounded and unbounded motion, see Table~\ref{tab:motcla}. Unbounded motion corresponds to the energy set above the absolute maximum of the potential. In this case, there are no turning points, and the momentum behaves quite differently for the energies below and above the absolute maximum of the potential. For the bounded motion, the integration contour in the action integral is non-contractible due to the presence of the branch cut inside it. For the unbounded motion, it is homotopically non-trivial due to the  global properties of the Riemannian surface.

\begin{table}
    \centering
\begin{tabular}{|c|c|c|}
\hline
&
Non-periodic potentials &
Periodic potentials \\
\hline
Below absolute maximum &
\specialcell{Bounded: periodic,\\with turning points} &
\specialcell{Bounded: periodic,\\with turning points} \\
\hline
Above absolute maximum &
\specialcell{Unbounded: non-periodic} &
\specialcell{Unbounded: periodic,\\with no turning points} \\
\hline
\end{tabular}
\caption{Types of motion in classical mechanics}
\label{tab:motcla}
\end{table}

We begin with introducing the potential
\begin{equation}\begin{gathered}
V\,(x\,|\,\nu) = a\,\nu \,\snt (x,\, \nu) + b \quad,\\
a \in \mathbb{R}^+\quad,\qquad b \in \mathbb{R} \quad,\qquad\nu \in [0,1]\quad,
\label{Ve}
\end{gathered}\end{equation}
where $\sn (x, \nu)$ is the Jacobi elliptic function \cite{NIST}. The potential~\eqref{Ve} is a doubly-periodic meromorphic function. Its real and imaginary periods are $2\,K \equiv 2\,K(\nu)$ and $2\,{i\mkern1mu}\, K' \equiv 2\,{i\mkern1mu}\, K(1-\nu)$, where
\begin{equation}
    K(\nu) = \int \limits_0^{\pi/2}
    \dfrac{\d\,\theta}{\sqrt{1-\nu\,\sin^2 \theta}}
\end{equation}
is the complete elliptic integral of the first kind. After gluing the opposite sides of the fundamental parallelogram (Figure~\ref{vxnuriem}), one obtains a torus (deprived of a point corresponding to the pole) that represents the Riemannian surface of the potential.

The potential $V\,(x\,|\,\nu)$ has three critical points:
\begin{equation}
    \dfrac{d}{dx}\, V\,(x\,|\,\nu) = 0
    \qquad \Longrightarrow \qquad
    x = \{ 0,\,K,\,K+{i\mkern1mu}\,K' \}\quad.
\end{equation}
The point $x=0$ is a zero of the second order. Hence the existence of two turning points
\begin{equation}
    V\,(x\,|\,\nu) = E \qquad\Longrightarrow\qquad x_1,\,x_2\quad
\end{equation}
that are real iff the energy stays between the minimum and maximum of the potential.

The potential $V\,(x\,|\,1-\nu)$ is dual to the one defined in~\eqref{Ve}, in the sense that
\begin{equation}
V\,({i\mkern1mu}\,x + K + {i\mkern1mu}\,K'\,|\,\nu) + V\,(x\,|\,1 - \nu) = a+2\,b \quad.
\end{equation}
For the maxima of these potentials, it will be convenient to introduce the notations
\begin{equation}\begin{aligned}
    V_{\nu,\,\text{max}} = V\,(K\,|\,\nu) = a\,\nu + b \quad,\qquad
    V_{1-\nu,\,\text{max}} = V\,(K'\,|\,1-\nu) = a\,(1-\nu) + b \quad.
\end{aligned}\end{equation}

With the mass value set to be $\,m=1/2\,$, the momentum takes form
\begin{equation}
p\,(x,\,E\,|\,\nu) = \sqrt{E  - a\,\nu\,\snt(x,\,\nu)-b}\quad.
\end{equation}
The duality property for the momentum looks as
\begin{equation}\begin{aligned}
p\,({i\mkern1mu}\,x+K+{i\mkern1mu}\,K',\,V_{\nu,\,\text{max}}-\epsilon\,|\,\nu) =
{i\mkern1mu}\,p\,(x,\,V_{1-\nu,\,\text{max}}+\epsilon\,|\,1-\nu) \quad.
\end{aligned}
\label{plame}
\end{equation}

The momentum $p\,(x,\,E|\,\nu)$ is a doubly-periodic meromorphic function. Its real and imaginary periods are $4\,K$ and $2\,{i\mkern1mu}\,K'$. The two halves of the fundamental parallelogram in Figure~\ref{pxnuriem} correspond to the two sheets of the square root. On each sheet, the function has a cut and a pole of the first order. One can travel between the sheets either by moving parallel to the real axis or by crossing the branch cuts. After gluing the opposite sides of the parallelogram and the cuts, one ends up with a genus-$2$ surface (deprived of two points).
\begin{figure}
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1\textwidth]{vxnu}
\caption{The potential $V\,(x\,|\,\nu)$ for $a=1$ and $b=0$.}\label{vxnu}
\end{minipage}\hfill
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics[width=1\textwidth]{vxnuriem}
\caption{The Riemannian surface of the potential  $V\,(x\,|\,\nu)$.}\label{vxnuriem}
\end{minipage}
\end{figure}

The potentials considered in this section are continuous and, therefore, bounded along the real direction. This indicates that we have two types of motion: (a) bounded, with energies lower than the absolute maximum of the potential; and (b) unbounded, with energies above the absolute maximum. For the action of the bounded motion, we can employ the definition~\eqref{18}. For the unbounded motion, the definition of the action has to be slightly altered. As the energy reaches the absolute maximum of the potential, the period of the motion experiences a jump and becomes twice smaller.
\footnote{~When a particle moving in the periodic potential reaches one end of the interval, it immediately appears on the opposite end, not bounces back.
}

We introduce the integration contours for the momentum $p\,(x,\,V_{\nu,\,\text{max}}-\epsilon\,|\,\nu)$ as shown in Figure~\ref{pxnuriem}:
\begin{equation}\begin{alignedat}{4}
    \mathcal{C}_c \equiv& &\mathcal{C}_1 + \mathcal{C}_{2(c)} + \mathcal{C}_3 + \mathcal{C}_{4(c)} \quad,\qquad
    \mathcal{C}_{2} &=&  \mathcal{C}_{2(c)} +  \mathcal{C}_{2(p)} \quad, \\
    \mathcal{C}_p \equiv& -&\mathcal{C}_3 + \mathcal{C}_{2(p)} - \mathcal{C}_1 + \mathcal{C}_{4(p)} \quad,\qquad
    \mathcal{C}_{4} &=&  \mathcal{C}_{4(c)} +  \mathcal{C}_{4(p)} \quad.
    \label{Ccp}
\end{alignedat}\end{equation}
The contours $\mathcal{C}_c$, $\mathcal{C}_p$, $\mathcal{C}_2$ and $\mathcal{C}_4$ are closed and homotopically non-trivial. What precludes us from writing $(- \mathcal{C}_2)$ instead of $\mathcal{C}_4\;$ (the way we did it with $\mathcal{C}_1$ in the definition of $\mathcal{C}_p$ in Figure~\ref{pxnuriem} ) is the presence of the second sheet of the Riemannian surface with singularities on it. In other words, we cannot continuously deform $\mathcal{C}_4$ into $(-\mathcal{C}_2)$.

\begin{figure}
\centering
\includegraphics[width=1\textwidth]{pxnuriem}
\caption{The integration contours on the Riemannian surface of the $p\,(x,\,V_{\nu,\,\text{max}}-\epsilon\,|\,\nu)$.}
\label{pxnuriem}
\end{figure}

After summing up the two lines of~\eqref{Ccp}, one arrives at
\begin{equation}

    \mathcal{C}_c + \mathcal{C}_p

    = \mathcal{C}_2 + \mathcal{C}_4 \quad.
\end{equation}
Let us now define the following integrals along these contours:
\begin{equation}\begin{alignedat}{2}
    \gamma_c\,(\epsilon) &= \phantom{-}\oint \limits_{\mathcal{C}_c} p\,(x,\,V_{\nu,\,\text{max}}-\epsilon\,|\,\nu) \,\d\,x \quad,\\
    \gamma_n\,(\epsilon) &= - \oint \limits_{\mathcal{C}_n} p\,(x,\,V_{\nu,\,\text{max}}-\epsilon\,|\,\nu) \,\d\,x \quad,\qquad
    n=p,\,2,\,4\quad.
    \label{ellcont}
\end{alignedat}\end{equation}
The signs are chosen for the integration cycles to stay positive when the direction along the contours is set counter-clockwise and the  momentum is set positive along the bottom of the cut.
These integration cycles obey an equality similar to~\eqref{gammas6}:
\begin{equation}
    \gamma_{c}\,(\epsilon) + \gamma_2\,(\epsilon) + \gamma_4\,(\epsilon) = \gamma_{p}\,(\epsilon)\quad.
    \label{29}
\end{equation}
The integral along the contour enclosing the pole can be evaluated with the aid of the residue:
\begin{equation}
    \gamma_p\,(\epsilon)
    = -2\,\pi\,{i\mkern1mu}\,\Res \{ p\,(x,\,V_{\nu,\,\text{max}}-\epsilon\,|\,\nu),\, {i\mkern1mu}\,K' \}
    = -2\,\pi\,{i\mkern1mu} ({i\mkern1mu}\,\sqrt{a})
    = 2\,\pi\, \sqrt{a}\quad.
    \label{Ssres}
\end{equation}
It is now the right time
to identify the integration cycles with the appropriate actions:
\begin{subequations}\begin{alignat}{4}
    &S\,(V_{\nu,\,\text{max}}-\epsilon\,|\,\nu)
    &\,\equiv \,&\gamma_c \,(\epsilon)& \quad,\\
    &S\,(V_{1-\nu,\,\text{max}} + \epsilon\,|\,1-\nu)
    &\,\equiv &\gamma_2 \,(\epsilon)\,&
    =\gamma_4 \,(\epsilon) \quad.
\end{alignat}
\label{31}
\end{subequations}
Together, the formul\ae~\eqref{29} and~\eqref{31} entail a relation between the actions of the two  potentials:
\begin{equation}
    S\,(V_{\nu,\,\text{max}}-\epsilon\,|\,\nu) + 2\,S\,(V_{1-\nu,\,\text{max}}+\epsilon\,|\,1-\nu)
    = 2\,S\,(V_{\nu,\,\text{max}}+\epsilon\,|\,\nu) + S\,(V_{\nu,\,\text{max}}-\epsilon\,|\,1-\nu)
    = 2\,\pi\,\sqrt{a}\quad,
    \label{Ss}
\end{equation}
which is a close analogue to the relation~\eqref{S}.

It would be worth noting that the origin of the factor of two in the equation~\eqref{S} is slightly different from that of the one emerging in~\eqref{Ss}. In the former case, one has two symmetric wells in the potential. In the latter, we have two options for the particle~---~either to move from left to right or from right to left. In both cases, the action is an integral along the non-contractible loop.

\newpage

\section{A symmetry of the quantum energy levels}
\subsection{Prolegomena \label{prolego}}
The process of quantisation of a classical system can be thought of as replacing the classical position and momentum by operators obeying the canonical commutation relations:
\begin{equation}
    [\,\hat{x}\,,\,\hat{p}\, ] = {i\mkern1mu} \,\hbar \quad,
\end{equation}
and acting on the wave vectors ascribed to the states of a quantum system. Oftentimes, the so-called position representation is used:
\begin{equation}
    x \to \hat{x} = x \quad,\qquad p \to \hat{p} = - {i\mkern1mu}\,\hbar\,\dfrac{d}{dx}\quad,
\end{equation}
with the wave vectors implemented as functions of $x$. After the quantisation, the Hamiltonian
\begin{equation}
    H = \dfrac{p^2}{2\,m} + V\,(x)
\end{equation}
becomes an operator, and the latter energy-conservation law
\footnote{~We limit ourselves with the consideration of the time-independent Hamiltonians.}
\begin{equation}
    H\,(x,\,p) = E
\end{equation}
turns into a differential equation named after Erwin Schr\"odinger:
\begin{equation}
    \widehat{H} \, \psi\,(x) = \left[ -\dfrac{\hbar^2}{2\,m} \dfrac{d^2}{dx^2} + V(x) \right] \, \psi \,(x) = E \, \psi \,(x) \quad.\label{Sch}
\end{equation}
To know the permissible energies, we need to find the spectrum of the differential operator $\hat{H}$.

The spectra of periodic and non-periodic potentials have to be discussed separately, see Tables~\ref{tab:motcla} and~\ref{tab:motquant}.

For bound states in a non-periodic potential, the spectrum turns out to be discrete. These states are quantum counterparts to periodic bounded motion in a corresponding classical system. This is the only type of motion available for potentials that become infinite at $\,|x|\to\infty\,$. In this case, the process of solving the Schr\"odinger equation gets reduced to diagonalisation of a matrix of a countable (but, generally, infinite) size. There are few well-known \textit{exactly solvable} problems, for which the entire matrix can be diagonalised. In general, though, this kind of problem cannot be solved analytically. However, there exist the so-called \textit{quasi-exactly solvable} (QES) problems, the ones permitting partial algebraisation.$\,$\footnote{~For a detailed introduction into the topic, see~\cite{Shifman}.} In these problems, the matrix of the Hamiltonian permits to single out a block of a finite size~---~which reduces the search of a limited number of energy levels to diagonalisation of a finite-sized matrix.

For periodic potentials, the situation is quite different. Typically, their spectrum is continuous, with gaps in it.
The search for band-edges in periodic potentials is analogous, in a certain sense, to the search for the bound states in non-periodic potentials. In particular, for the QES periodic potential considered in Section~\ref{ellqes}, it is possible to reduce the problem of finding the locations of the band-edges
to diagonalisation of a finite-sized matrix.

\begin{table}
    \centering
\begin{tabular}{|c|c|c|}
\hline
&
Non-periodic potentials &
Periodic potentials \\
\hline
Below absolute maximum &
\specialcell{Bound states:\\discrete spectrum} &
\multirow{4}{*}{\specialcell{Continuous spectrum\\with gaps\\~}}
 \\
\cline{1-2}
Above absolute maximum &
\specialcell{Unbounded states:\\continuous spectrum}
&\\
\hline
\end{tabular}
\caption{Types of motion in quantum mechanics}
\label{tab:motquant}
\end{table}

To solve the Schr\"odinger equation~\eqref{Sch}, it is often convenient to change variables:
\begin{equation}
    \psi\,(x) = \exp\left( {i\mkern1mu}\,\sigma\,(x, E) / \hbar \right) \quad.\label{ch}
\end{equation}
Then the function $\sigma \,(x)$ satisfies the Ricatti equation:
\begin{equation}
    (\sigma'\,(x, E))^2 + \dfrac{\hbar}{{i\mkern1mu}} \, \sigma''\,(x, E) = 2\,m\,(E - V(x)) \quad.\label{ric1}
\end{equation}
In the limit $\hbar \to 0$, the function $\sigma'\,(x)$ becomes the classical momentum~\eqref{p}. So we shall call it the quantum momentum function (QMF) and introduce the notation:
\begin{equation}
    \qm\,(x, E) =  \sigma'\,(x, E)\quad,
    \label{290}
\end{equation}
so the equation~\eqref{ric1} takes the form:
\begin{equation}
    \qm^2\,(x, E) + \dfrac{\hbar}{{i\mkern1mu}} \, \qm'\,(x, E) = 2\,m\,(E - V(x)) \quad.\label{ric2}
\end{equation}

The quantisation condition, whence the $n$-th energy level is determined, is normally obtained from the requirement of single-valuedness of the function $\psi\,(x)$. However, in \cite{HamJa} a more interesting option was proposed. It was based on the fact that the wave function corresponding to the $n$-th energy level has $n$ zeros on the real axis, between the classical turning points (the latter points being the zeros of the classical momentum) \cite{LL}.

In these zeroes, the QMF  has poles. Indeed, it trivially follows from (\ref{ch}) and (\ref{290}) that
\begin{equation}
    \qm\,(x, E) = \dfrac{\hbar}{{i\mkern1mu}} \dfrac{1}{\psi\,(x)} \dfrac{d \psi(x)}{dx}\quad.
\end{equation}
For analytic potentials, the pole of the function $\qm\,(x)$ is of the first order, and the residue at this pole is $(-{i\mkern1mu} \hbar)$. Therefore, the integral of the QMF along the contour $\mathcal{C}_R$ enclosing classical turning points is:
\begin{equation}
    B\,(E) = \oint \limits_{\mathcal{C}_R} \qm\,(x, E) \,\d x = 2 \, \pi \,n\,\hbar\quad,\label{GBS}
\end{equation}
where the contour $\mathcal{C}_R$ should be close enough to the real axis, in order to avoid containing the poles and branch cuts of $\qm\,(x)$, that are off the real axis. In the classical limit ($\hbar \to 0$), the series of poles inside $\mathcal{C}_R$ coalesces into a branch cut of the classical momentum \cite{LP}.

The function $B\,(E)$ is sometimes referred to as the quantum action function (QAF) \cite{LP,LP2}, and the equality~\eqref{GBS} itself~---~as the generalised Bohr-Sommereld quantisation condition (GBS). The GBS is often employed as a starting point in studies of the spectra of quantum systems.
The two common approaches to extract information from the Ricatti equation are:
\begin{enumerate}
\item To employ the expansion of the QMF in the powers of $\hbar$ (the WKB method):
\begin{equation}
    \qm(x) = \sum \limits_{k=0}^{\infty} \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^k \qm_k(x) \quad.
    \label{L8}
\end{equation}
A strong side of this approach lies in its clear physical meaning, in that the first two terms give the ordinary Bohr-Sommerfeld quantisation condition.
A disadvantage of this method is that the form of~\eqref{L8} implies the neglect of exponentially small non-perturbative terms. Hence all the results are valid only when tunneling is unimportant (i.e., at a distance from the relative maxima of the potential). This drawback can, however, be overcome, if one begins looking for the quantum action (and, thereby, the energy) in a form of trans-series in $\hbar$ rather then of a polynomial. Moreover, until recently, the usage of the GBS was the only systematic way of finding the multi-instanton corrections to the energy in quantum mechanics \cite{Mult1}.
\footnote{~A recently discovered way of constructing the resurgent expansions with the aid of the so-called Uniform WKB method provides a more efficient way of calculating the exponentially small corrections to the energy, for certain QM potentials. This method also reveals some other interesting properties of such potentials, see \cite{UWKB}.}

Substituting~\eqref{L8} into the Ricatti equation~\eqref{ric2} gives a recursive relation
\begin{gather}
    \sum \limits_{l=0}^{k} \qm_l\,(x)\,\qm_{k-l}\,(x) + \qm_{k-1}'\,(x) = 0 \quad,
\end{gather}
which allows to express all the higher terms~\eqref{L8} through the classical momentum:
\begin{gather}\begin{gathered}
    \qm_k\,(x) = - \dfrac{1}{2\,\qm_0\,(x)} \left(\qm'_{k-1}\,(x)+\sum \limits_{l=1}^{k-1} \qm_l\,(x) \,\qm_{k-l}\,(x) \right)\quad,\\
    \qm_0 \,(x) = p\,(x)\quad.
    \label{qmkp}
\end{gathered}\end{gather}

Next, we substitute the expansion~\eqref{L8} into~\eqref{GBS} and get:
\footnote{~As we have already mentioned, the form of the equation above implies the neglect of the tunneling effects. We will discuss this issue in Section~\ref{nonpert}.}
\begin{equation}
    \oint \limits_{\mathcal{C}_R} \sum \limits_{k=0}^{\infty} \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^k \qm_k\,(x) \,\d \,x
    =2\,\pi\,n\,\hbar \quad.
    \label{GBSser}
\end{equation}

The zeroth and first terms in~\eqref{GBSser} render:
\begin{equation}
    \oint \limits_{\mathcal{C}_R} p(x) \d\,x + \dfrac{\hbar}{{i\mkern1mu}} \oint \limits_{\mathcal{C}_R} \qm_1\,(x) \d\,x =
    S\,(E) + \dfrac{\hbar}{{i\mkern1mu}}\,2\,\pi\,{i\mkern1mu}\,\left(-\dfrac{1}{2}\right)
    = S\,(E) - \pi\,\hbar
    = 2\,\pi\,n\,\hbar \quad.
    \label{derBS}
\end{equation}

The constant arising from the first term is often referred to as Maslov index and can be calculated in various ways \cite{LL}. Importantly, it does not depend on the form of the potential well. After moving it to the RHS of~\eqref{derBS}, we arrive to the famous Bohr-Sommerfeld quantisation condition:
\begin{equation}
    S\,(E) = 2\,\pi\,\hbar \,\left(n+\dfrac{1}{2}\right)\quad.
    \label{BS}
\end{equation}

One may also proceed with calculating the higher-order terms in the LHS of~\eqref{GBSser}. This will, for example, provide a way to generate the perturbative expansion in the cases where it exists. To this end, one will have to solve~\eqref{GBSser} for the energy, inverting the series term by term.
When taking the integrals, one should take into account then all the odd terms in the expansion of the QMF starting from $k=3$ are total derivatives and, so, the corresponding integrals in~\eqref{GBSser} vanish.

\item To employ the Laurent expansion of the QMF:
\begin{equation}
    \qm(x) =  \sum \limits_{k=-\infty}^{\infty} a_k\, x^k\label{laurent}
\end{equation}
This approach has not been studied in depth so far. In \cite{Nan2} and \cite{Nan3} an approximate method of finding the energies was suggested based on this idea.
In this paper, we use the Laurent expansion to find the residue of the QMF at infinity, in order to prove the ESR property in the language of the QAF.
These results rely on those obtained in \cite{HamJa}. We also correct some overlooks made in \textit{Ibid}.
\end{enumerate}

\subsection{The sextic QES potentials \label{sexticqes}}

In this section we introduce the sextic QES potentials and discuss how the energy spectrum reflection property follows from the matrix form of the Hamiltonians. Then we learn how the ESR can be formulated and proved on the language of the quantum action. Lastly, we discuss in greater detail connection between the duality of classical and quantum actions.

Through Section~\ref{sexticqes}, we shall employ the units in which $\,\dfrac{\hbar^2}{2\,m} = \dfrac{1}{2}\,$.

\subsubsection{An algebraic approach \label{algappr}}

The most well-known examples of QES problems are the even and odd sixth-order polynomial potentials:~\footnote{~These potentials are discussed in~\cite{Shifman}. Be mindful, though, that the constant parts of these potentials are omitted there.}
\begin{subequations}
\begin{gather}
    V_{\text{even}}\,(\nu,\mu,j,x) = \dfrac{\nu^2}{8}\,x^6 + \dfrac{\mu\,\nu}{4} \, x^4 + \left[  \dfrac{\mu^2}{8} - 2 \, \nu\, \left( j + \dfrac{3}{8} \right)\right]\,x^2 - \mu\left(j+\dfrac{1}{4}\right) \quad,\vspace{2mm}
    \label{even}\\
    V_{\text{odd}}\,(\nu,\mu,j,x) = \dfrac{\nu^2}{8}\,x^6 + \dfrac{\mu\,\nu}{4} \, x^4 + \left[  \dfrac{\mu^2}{8} - 2 \, \nu\, \left( j + \dfrac{5}{8} \right)\right]\,x^2  - \mu\left(j+\dfrac{3}{4}\right)\quad,\vspace{4mm}
    \label{odd}\\
    \nonumber
    \nu \in \mathbb{R}^+,\quad,\qquad \mu \in \mathbb{R} \quad,\qquad
    j = 1/2,\, 1,\, 3/2,\, 2,\,\ldots
\end{gather}\label{V}
\end{subequations}
These potentials are even and odd in the following sense. For them, the first $J = 2\,j + 1$ lowest even (odd) energy levels can be obtained through diagonalisation of the following tridiagonal matrices:
\begin{equation}
\begin{gathered}
\mathcal{H}_{\text{``G'',\,even}}:\\
\begin{alignedat}{4}
    &\mathcal{H}_{k, k+1} & \ = \ & - k \, (2\,k - 1) \\
    &\mathcal{H}_{k, k}   & \ = \ & \mu \,(k - j - 1)\\
    &\mathcal{H}_{k, k-1} & \ = \ & (k-2)\,\nu - 2 \,j \,\nu
\end{alignedat}
\end{gathered}
\qquad\qquad
\begin{gathered}
\mathcal{H}_{\text{``G'',\,odd}}:\\
\begin{alignedat}{4}
    &\mathcal{H}_{k, k+1} & \ = \ & - k \, (2\,k + 1) \\
    &\mathcal{H}_{k, k}   & \ = \ & \mu \,(k - j - 1)\\
    &\mathcal{H}_{k, k-1} & \ = \ & (k-2)\,\nu - 2 \,j \,\nu
\end{alignedat}
\end{gathered}
\qquad
\begin{aligned}
k = 1,2,\ldots,J\quad.
\end{aligned}\label{HH}
\end{equation}
This is owing to the fact that, after an eigenvalues-preserving quasi-gauge transformation,$\,$\footnote{~Quasi-gauge is a transformation of the form $\;\psi\,(x) \to \psi_{\text{``G''}}(x)\,\mathrm{e}^{-a\,(x)}\,,\;\,H\to \mathrm{e}^{-a\,(x)}\,H_{\text{``G''}}\,\mathrm{e}^{a\,(x)}\,$. The function $a(x)$ is set real, wherefore the transformation is \textit{non-unitary}, and the norm of the wave function is not preserved. The physical meaning of such a transformation is straightforward: solutions to a one-dimensional Schr\"odinger equation are normally sought in the form of  $~\left(\,\sum_n a_n\, x^n\,\right)\,\exp(.\,.\,.)\,$ where the exponent defines the asymptotic behaviour. When the exponential is found somehow, it remains to determine the prefactor $\,\sum_n a_n\, x^n\,$. In special cases, this series truncates to a finite sum.
\par ~A key property of quasi-gauge transformations is that they preserve the eigenvalues of the operator. See~\cite{Shifman} for details.}
the Hamiltonians corresponding to potentials~\eqref{V} can be written as:
\begin{equation}
\begin{alignedat}{3}
    &H_{\text{``G'',\,even}} & \ = \ & -2\,T^0\,T^- - (2j+1)\,T^- - \nu\,T^+ + \mu\,T^0\quad&,\\
    &H_{\text{``G'',\,odd}}  & \ = \ & -2\,T^0\,T^- - (2j+1)\,T^- - \nu\,T^+ + \mu\,T^0 -2\,T^-\quad&,
\end{alignedat}
\end{equation}
where the operators
\begin{equation}\begin{gathered}
    T^+ = 2 \,j\,\xi - \xi^2 \,\dfrac{d}{d\xi} \quad,\qquad
    T^0 = - j + \xi \dfrac{d}{d\xi}\quad,\qquad
    T^- = \dfrac{d}{d\xi}\quad,\\ \xi=x^2\quad
\end{gathered}\end{equation}
obey the commutation relation:
\begin{equation}
    [T^+,T^-] = 2T^0\quad,\qquad [T^+,T^0] = -T^+\quad,\qquad[T^-,T^0] = T^-\quad
\end{equation}
and therefore can be interpreted as the generators of a representation of the $SU(2)$ group, with $j$ being the spin and $J=2j+1$ the representation dimension.

Since the subsequent discussion applies to both potentials shown in (\ref{V}), we shall concentrate on the even one, Eqn. (\ref{even}). The latter can be obtained from the classical potential~\eqref{Vcl} by setting
\begin{gather}
    a = -\mu\left(j+\dfrac{1}{4}\right) \quad,\qquad
    b = \left[  \dfrac{\mu^2}{8} - 2 \, \nu\, \left( j + \dfrac{3}{8} \right)\right] \quad,\qquad
    c = \dfrac{\mu\,\nu}{4} \quad,\qquad
    d = \dfrac{\nu^2}{8} \quad.\label{abc}
\end{gather}
The change of sign $\mu \to - \mu$ leads to $\,\{a \to -a,\,b \to \,b,\,c \to -c,\,d \to d\}\,$, whence the potentials with opposite values of $\mu$ are dual.
In the notation
\begin{subequations}
\begin{alignat}{2}
    V_{\{ \mu \}} \,(x) &= V_{\text{even}}\,&(\nu,\mu,j,x) \quad,\\
    \widetilde{V}_{\{ \mu \}} \,(x) &= V_{\text{odd}}\,&(\nu,\mu,j,x) \quad.
\end{alignat}
\end{subequations}
the duality and self-duality conditions assume the form of
\begin{subequations}
\begin{alignat}{4}
    V_{\{ \mu \}} \,(x)& = - V_{\{ - \mu \}} \,({i\mkern1mu} \,x) \quad&,\qquad
    \widetilde{V}_{\{ \mu \}} \,(x)& = - \widetilde{V}_{\{ - \mu \}} \,({i\mkern1mu} \,x) \quad&,
    \label{selfd1}
    \\
    V_{\{ 0 \}} \,(x)& = - V_{\{ 0 \}} \,({i\mkern1mu} \,x)\quad&,\qquad
    \widetilde{V}_{\{ 0 \}} \,(x)& = - \widetilde{V}_{\{ 0 \}} \,({i\mkern1mu} \,x)
    \quad&.
    \label{selfd2}
\end{alignat}
\end{subequations}

In \cite{MyNew}, a theorem was proven stating that the eigenvalues of a tridiagonal matrix change their signs, when one changes the signs of the diagonal elements. The matrix~\eqref{HH} is exactly the case. Thus, in accordance with the said theorem, the energies from the algebraic parts of the spectra of the even and odd potentials change their signs under the switch $\mu \to -\mu$:
\begin{subequations}
\begin{alignat}{9}
    E_{\{\mu\},\,n} &= - E_{\{-\mu\},\,N-n} \quad&&,\qquad
    n=0,2,4\ldots,N
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{even}}\,$)}
    \quad&&,\\
    \widetilde{E}_{\{\mu\},\,\widetilde{n}} &= - \widetilde{E}_{\{-\mu\},\,\widetilde{N} + 1 -\widetilde{n}} \quad&&,\qquad
    \widetilde{n}=1,3,5\ldots,\widetilde{N}
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{odd}}\,$)}
    \quad&&.
\end{alignat}
\label{symmy}
\end{subequations}
Here $N$ and $\widetilde{N}$ denote the highest energy levels in the algebraic sectors of the potentials:
\begin{subequations}
\begin{alignat}{6}
    N &= 2\,J-2 &&= 4\,j \quad&&,\\
    \widetilde{N} &= 2\,J-1 &&= 4\,j + 1 \quad&&.
\end{alignat}
\end{subequations}

\subsubsection{The possible shapes of the sextic QES potentials \label{shapes}}

For different values of the parameters, both QES sextic potentials fall either into the category~\eqref{case1} or into~\eqref{case2}. This means that the corresponding classical systems are dual to each other. Let us discuss these two cases using our new notations. Here we shall limit our discussion to the even potential.
\begin{enumerate}
\item In the case
\begin{equation}
    \dfrac{\mu^2}{8} < 2\,\nu\,\left(j+\dfrac{3}{8}\right)\quad,
    \label{qcond1}
\end{equation}
both potentials, $V_{\{\mu\}}\,(x)$ and $V_{\{-\mu\}}\,(x)$ are of the double-well shape, see Figure~\ref{F1}, (a) and (b). The duality takes place in the following range of energies:
\begin{equation}
\begin{gathered}
\text{Two wells} \hspace{45mm} \text{Single well} \hspace{45mm} \\
\begin{alignedat}{14}

    &V_{\{\mu\},\,\text{min}} &<& \ E < V_{\{\mu\}}\,(0) \qquad \qquad &V_{\{\mu\}}\,(0) &<& \ E  &<& -&V_{\{-\mu\},\,\text{min}}
    &\quad\quad\qquad &\text{(for $V_{\{\mu\}}$)} \quad&,\\
    &V_{\{-\mu\},\,\text{min}} &<& \ E < V_{\{-\mu\}}\,(0) \qquad \qquad &V_{\{-\mu\}}\,(0) &<& \ E  &<& -&V_{\{\mu\},\,\text{min}}
    &\quad\quad\qquad &\text{(for $V_{\{-\mu\}}$)} \quad&.
\end{alignedat}
\end{gathered}
\label{zho1}
\end{equation}
\item In the case
\begin{equation}
    \dfrac{\mu^2}{8} > 2\,\nu\,\left(j+\dfrac{3}{8}\right)\quad,
    \label{qcond2}
\end{equation}
one of the potentials has a single minimum, while another one has three of them, see Figure~\ref{F1}, (c) and (d). For definiteness, assume $\mu < 0$. Then the energy ranges wherein the duality occurs look as follows:
\begin{equation}
\begin{gathered}
\text{Two wells} \hspace{45mm} \text{Single well} \hspace{45mm} \\
\begin{alignedat}{14}

    &V_{\{\mu\},\,\text{min}} &<& \ E < V_{\{\mu\}}\,(0) \qquad \qquad
    & & & & & &
    &\quad\quad\qquad &\text{(for $V_{\{\mu\}}$)} \quad&,\\
    & & &
    \qquad \qquad &V_{\{-\mu\}}\,(0) &<& \ E  &<& -&V_{\{\mu\},\,\text{min}}
    &\quad\quad\qquad &\text{(for $V_{\{-\mu\}}$)} \quad&,
\end{alignedat}
\end{gathered}
\label{zho2}
\end{equation}
\end{enumerate}
In both cases, $V_{\{\mu\},\,\text{min}}$ and $V_{\{-\mu\},\,\text{min}}$ denote the values of the corresponding potentials at the minima, and the following equality takes place:
\begin{equation}
    V_{\{\mu\}}\,(0) = - V_{\{-\mu\}}\,(0) \quad.
\end{equation}
From the formul\ae~\eqref{zho1} and~\eqref{zho2} it ensues that, without loss of generality, we can assume that the equation
\begin{equation}
     V_{\{\mu\}}\,(x) = -E
     \label{Veps}
\end{equation}
has four solutions. Consequently, the equation
\begin{equation}
     V_{\{-\mu\}}\,(x) = E
\end{equation}
will then have two real solutions.

For the odd potential, one has to replace $3/8$ with $5/8$ in both~\eqref{qcond1} and~\eqref{qcond2}.

\subsubsection{The quantum action \label{QAsex}}

A different way to prove the ESR property~\eqref{symmy} of the energy levels of  dual potentials follows from relating their quantum actions. This implies derivation of a quantum analogue to~\eqref{S}. Despite the fact that we do not know the expression for the QMF (obtaining whereof would be equivalent to solving the Schr\"odinger equation), we can calculate its residue at infinity by substituting the Laurent expansion into the Ricatti equation~\eqref{ric2}, as was proposed in \cite{LP}.
\footnote{~For the algebraic part of the spectrum of QES potentials, the solutions of the Schr\"odinger equation (and, thereby, the QMF) can, in principle, be obtained by means of the matrix algebra. Technically, though, this can be performed only if the size of the matrix, $J$, is small. Here we discuss a general method applicable to arbitrary potentials.}

For the time being, we shall perform our calculations for the even potential. Thereafter, we shall provide the results for the odd potential, and shall briefly discuss the difference between these two situations.

Our starting point will be the equation~\eqref{GBS}. As was agreed above, we assume that~\eqref{Veps} has four solutions. Then~\eqref{GBS} reads as:
\begin{equation}\begin{gathered}
    \oint \limits_{\mathcal{C}_1} \qm_{\{\mu\}}\,(x, \,-E) \,\d x=
    \oint \limits_{\mathcal{C}_2} \qm_{\{\mu\}}\,(x, \,-E) \,\d x
    =2 \, \pi\,\hbar \,m\quad,\\
    V_{\{\mu\},\,\text{min}} < -E < V_{\{\mu\}}\,(0) \quad.
    \label{GBSb}
\end{gathered}\end{equation}
where, as before, the contours are defined as in Figure~\ref{roots6}.

From the equation~\eqref{ric2} and the form of the potential~\eqref{even}, it follows that the duality property for the QMF has exactly the same form as that for the classical momenta:
\begin{equation}
    \qm_{\{-\mu\}}\,(x\,, E) = {i\mkern1mu}\,\qm_{\{\mu\}}\,({i\mkern1mu}\,x,-E)\quad. \label{pdual}
\end{equation}
It allows us to calculate the LHS of the~\eqref{GBS} for the positive energy levels of the potential $V_{\{-\mu\}}\,(x)$, as an integral of $\qm_{\{\mu\}}\,(x,\,-E)$ along the imaginary axis:
\begin{equation}
    \oint \limits_{\mathcal{C}'_3} \qm_{\{-\mu\}}\,(x,\, E) \,\d x =
    \oint \limits_{\mathcal{C}_3} \qm_{\{\mu\}}\,(x,\, -E) \,\d x \quad.
\end{equation}
The sum of the integrals of $\,\qm_{\{\mu\}}\,(x,\,-E)\,$ over the contours $\,\mathcal{C}_1\,$, $\,\mathcal{C}_2\,$ and $\,\mathcal{C}_3\,$ is equal to an integral over the large contour $\,\mathcal{C}_\infty\,$ enclosing \textit{all} the poles of $\,\qm_{\{\mu\}}\,(x,\,-E)\,$.
The latter integral can be evaluated with the aid of the residue of $\qm_{\{\mu\}}\,(x,\,-E)$ at infinity:
\begin{equation}
    2 \oint \limits_{\mathcal{C}_1} \qm_{\{\mu\}}\,(x, \,-E) \,\d x +
    \oint \limits_{\mathcal{C}_3} \qm_{\{\mu\}}\,(x, \,-E) \,\d x =
    \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}}\,(x, \,-E) \,\d x =
    2\,\pi\,{i\mkern1mu}\,\Res\{\qm_{\{\mu\}}\,(x, \,-E), \infty\}\quad.
    \label{36}
\end{equation}
Here we made use of the fact that the function $\,\qm_{\{\mu\}}\,(x,\,-E)\,$ has no poles other than those inside the contours $\,\mathcal{C}_1\,$, $\,\mathcal{C}_2\,$ or $\,\mathcal{C}_3\,$. Indeed, as was demonstrated in \cite{LP}, the poles of the QMF emerge only on the branch lines of the classical momentum, which in our case are the segments of the real and imaginary axes, enclosed by $\,\mathcal{C}_1\,$, $\,\mathcal{C}_2\,$ and $\,\mathcal{C}_3\,$.
This remarkable property allows us to employ the same integration contours for the QMF, as those employed for the classical momentum. We also assume the point $x=\infty$ to be an isolated singularity, which is indeed the case for the considered QES potential, as  was demonstrated in \cite{HamJa}.

The equality (\ref{36}) establishes a relation between the quantisation condition for a system described by the potential $V_{\{\mu\}}\,(x)$, with the energy $-E$, and the quantisation condition for a system described by $V_{\{-\mu\}}\,(x)$, with the opposite energy $E$. In other words, starting from~\eqref{GBSb}, we can \textit{derive} the quantisation condition in the form~\eqref{GBS} for $V_{\{-\mu\}}\,(x)$.

To calculate the residue at infinity, we introduce the variable $y$:
\begin{subequations}
\begin{gather}
    y \equiv \dfrac{1}{x} \quad,\\
    \widetilde{\qm}_{\{\mu\}}\,(y, \,-E) \equiv \qm_{\{\mu\}}\,(1 / y, \, -E)\quad,
\end{gather}
\end{subequations}
in terms whereof the Ricatti equation~\eqref{ric2} assumes the shape of
\footnote{~Recall that in this section we are using the units in which $\,m=\hbar^2\,$.}
\begin{equation}
    \widetilde{\qm}^{\ 2}_{\{\mu\}}\,(y, \,-E) + {i\mkern1mu}\,\hbar\,y^2 \, \widetilde{\qm}^{\ \prime}_{\{\mu\}} \,(y,\,-E) = 2\,\hbar^2\,(E - V_{\{\mu\}}(1/y)) \quad.\label{ric3}
\end{equation}
The Laurent expansion for $\widetilde{\qm}_{\{\mu\}}\,(y,\,-E)$ reads as
\begin{equation}
    \widetilde{\qm}_{\{\mu\}}\,(y, \,-E) = \sum \limits_{k=-3}^{\infty} a_k\, y^k\quad.\label{laurent0}
\end{equation}
The lowest power of $y$ is determined by the highest power of $x$ in $V\,(x)$ in~\eqref{ric2}.
Insertion of the above expansion into the integral (\ref{36}) gives:
\begin{equation}
    \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}}\,(x, \,-E) \,\d x =
    \oint \limits_{\mathcal{C}''_0} \widetilde{\qm}_{\{\mu\}}\,(y, \,-E) \,\dfrac{1}{y^2}\,\d y = 2\,\pi\,{i\mkern1mu}\,a_1 \quad,
\end{equation}
where the contour $\mathcal{C}''_0$ encloses the origin in the complex $y$-plane.
To find $a_0\,$, we insert the Laurent series~\eqref{laurent} in the Ricatti equation~\eqref{ric3} and match the coefficients of the five lowest powers of $y$:
\begin{equation}\begin{aligned}
k &= -6:\qquad& \dfrac{\hbar^2\,\nu^2}{4} + a_{-3}^2 &= 0 \\
k &= -5:\qquad& 2\, a_{-3}\,a_{-2} &= 0 \\
k &= -4:\qquad& \dfrac{\hbar^2\,\nu\,\mu}{2} + a_{-2}^2 + 2\,a_{-3}\,a_{-1} &= 0 \\
k &= -3:\qquad& 2\, (a_{-3}\,a_0 + a_{-2}\,a_{-1}) &= 0 \\
k &= -2:\qquad& \dfrac{\hbar^2\,\mu ^2}{4}-\dfrac{1}{2}\, (8 j+3)\,\hbar^2 \,\nu +a_{-1}^2+2 a_{-2}\, a_{0}+a_{-3} (2 a_{1}-3 {i\mkern1mu})&=0\qquad.
\end{aligned}
\end{equation}
The two sets of solutions are:
\begin{equation}
\begin{cases}
a_{-3}= -\dfrac{{i\mkern1mu}\,\hbar\,\nu}{2} \\
a_{-2}= 0 \\
a_{-1}= -\dfrac{{i\mkern1mu}\,\hbar\,\mu}{2} \\
a_{0}=0\\
a_{1}= {i\mkern1mu} \,\hbar\,(4\, j+3)
\end{cases} \quad,\qquad
\begin{cases}
a_{-3}= \dfrac{{i\mkern1mu}\,\hbar\,\nu}{2} \\
a_{-2}= 0 \\
a_{-1}= \dfrac{{i\mkern1mu}\,\hbar\,\mu}{2}\\
a_{0}=0\\
a_{1}=-4 \,{i\mkern1mu} \,\hbar\,j
\end{cases} \quad.
\end{equation}
Following \cite{HamJa}, we employ the condition of square integrability of the wave function
\begin{equation}
\psi\,(x) = \exp \left[ {i\mkern1mu} \int \limits^x \qm_{\{\mu\}}\,(\tilde{x}, \,-E) \, \d \,\tilde{x} \right]\quad.
\end{equation}
For large $x$ (and small $y$) one gets:
\begin{equation}
\psi\,(x) \approx \exp \left[ {i\mkern1mu}\; \dfrac{a_{-3}\,\;x^4}{4}\, \right] \quad,
\end{equation}
and one has to choose the second set of solution with $a_{-3} = {i\mkern1mu}\,\hbar\,\nu/2$ and $a_1 = - 4 \,{i\mkern1mu}\,\hbar\,j$. This yields:
\begin{equation}
    2 \oint \limits_{\mathcal{C}_1} \qm_{\{\mu\}}\,(x, \,-E) \,\d x +
    \oint \limits_{\mathcal{C}'_3} \qm_{\{-\mu\}}\,(x, \,E) \,\d x =
    \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}}\,(x, \,-E) \,\d x =
    2\,\pi\,{i\mkern1mu}\,(- 4\, {i\mkern1mu}\, j) =
    8 \, \pi \,j\,\hbar \quad
    \label{qduali}
\end{equation}
or, equivalently:
\begin{equation}
    \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}}\,(x, \,-E) \,\d x =
    2\,\pi\,N\,\hbar
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{even}}\,$)}
    \quad,\label{ALLeven}
\end{equation}
where
\begin{equation}
N = 2\,J-2 = 4\,j
\end{equation}
is the number of the highest energy level in the algebraic sector.
Moving the first integral in~\eqref{qduali}, with an opposite sign, from the LHS to the RHS, and then using the equality~\eqref{GBSb}, one arrives at
\begin{equation}
    \oint \limits_{\mathcal{C}'_3} \qm_{\{-\mu\}}\,(x, \,E) \,\d x =
    \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}}\,(x, \,-E) \,\d x
    - 2 \oint \limits_{\mathcal{C}_1} \qm_{\{\mu\}}\,(x, \,-E) \,\d x =
    2\,\pi\,(N - 2\,m)\,\hbar
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{even}}\,$)}
    \quad,\label{GBSa}
\end{equation}
while~\eqref{qduali} can be written down as
\begin{equation}
    2\,B_{\{\mu\}}\,(-E) + B_{\{-\mu\}}\,(E) = 2\,\pi\,N\,\hbar\quad.
    \label{ququdual}
\end{equation}

From looking at the formul\ae~\eqref{GBSb} and~\eqref{GBSa}, one may misconclude that the $m$-th energy level of the potential $V_{\{\mu\}}\,(x)$ is symmetric to the $\,(N-2\,m)$-th energy level of $V_{\{-\mu\}}\,(x)$.
This, however, would be a mistake.
There is a fundamental difference between the cases of two and four classical turning points for a given energy.
In the former case, the equation~\eqref{GBS} has a single solution for the energy, while in the latter case it, usually, possesses two solutions (an exception to this rule to be discussed shortly):
\begin{equation}
        \oint \limits_{\mathcal{C}_1} \qm_{\{\mu\}}\,(x, E) \,\d x = 2 \, \pi \,m\,\hbar \qquad\rightarrow\quad
    \begin{cases}
        E_{\{\mu\},\,m} \qquad&\text{if} \qquad E_{\{\mu\},\,m\phantom{\,(+)}} > V_{\{\mu\}}\,(0) \\
        E_{\{\mu\},\,m\,(-)},\:E_{\{\mu\},\,m\,(+)} \qquad&\text{if}\qquad E_{\{\mu\},\,m\,(-)} < V_{\{\mu\}}\,(0)
    \end{cases}\quad.
    \label{solving}
\end{equation}

We intend to comply with a  standard convention wherein
(a) the notation $\,E_{\{\mu\},\,n}\,$ stands for the $n$-th energy level from the bottom, and (b) the bottom level corresponds to $\,n=0\,$. To that end, the energy levels below $V_{\{\mu\}\,(0)}$ should be enumerated as:
\begin{equation}
\begin{cases}
E_{\{\mu\},\,2m} &= E_{\{\mu\},\,m\,(-)} \\
E_{\{\mu\},\,2m+1} &= E_{\{\mu\},\,m\,(+)}
\end{cases}
\qquad \text{if}\quad
E_{\{\mu\},\,m\,(-)} < V_{\{\mu\}}\,(0)\quad.\label{numbering}
\end{equation}

Strictly speaking,~\eqref{solving} does not cover all the possibilities. A situation may emerge, wherein the highest level below $V_{\{\mu\}\,(0)}$ is $E_{\{\mu\},\,m^*\,(-)}$, and it lacks a counterpart $E_{\{\mu\},\,m^*\,(+)}$. Nonetheless, in this case the proper numbering is $E_{\{\mu\},\,2m^*} = E_{\{\mu\},\,m^*\,(-)}$, as is~\eqref{numbering}. See also the discussion in Section~\ref{exampleee}.

Now it becomes evident that, on the right-hand side of the equation~\eqref{GBSa}, the quantity $2\,m$  is the number of the even energy level, $E_{\{\mu\},\,2m}$.
In other words, for an even $n$, the $n$-th energy level is symmetric to the $(N-n)$-th energy level:
\begin{equation}
    E_{\{\mu\},\,n} = - E_{\{-\mu\},\,N-n} \quad,\qquad
    n=0,2,4\ldots,N
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{even}}\,$)}

    \quad.\label{EReven}
\end{equation}

It is worth noting that the energies $\,E_{\{\mu\},\,m\,(-)}\,$ and $\,E_{\{\mu\},\,m\,(+)}\,$ share a perturbative part and differ only due to the non-perturbative splitting called into being by the tunneling effects.
These effects get smaller near the bottom of the wells, and become strong closer to the maximum. At the perturbative level (in neglect of the splitting), one can say that the level $\,E_{\{-\mu\},\,N-n}\,$ is symmetric to two levels: $\,E_{\{\mu\},\,n}\,$ and $\,E_{\{\mu\},\,n+1}\,$.

For the odd potential~\eqref{odd},  integrating $\widetilde{\qm}_{\{\mu\}}\,(x, \,-E)$ over the contour $\mathcal{C}_\infty$ furnishes:
\begin{equation}
    \oint \limits_{\mathcal{C}_\infty} \widetilde{\qm}_{\{\mu\}}\,(x, \,-E) \,\d x
    = 2\,\pi\,(\widetilde{N}+1)\,\hbar
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{odd}}\,$)}
    \quad,
    \label{ALLodd}
\end{equation}
where the number of the highest energy level in the algebraic sector is now given by:
\begin{equation}
\widetilde{N} = 2\,J-1 = 4\,j + 1 \quad.
\end{equation}
Then, starting again with~\eqref{GBSb}, we arrive at:
\begin{equation}
    \oint \limits_{\mathcal{C}'_3} \widetilde{\qm}_{\{-\mu\}}\,(x, \,E) \,\d x =
    2\,\pi\,(\widetilde{N}  - 2\,m)\,\hbar =
    2\,\pi\,\left(\widetilde{N}  + 1 - (2\,m + 1)\right)\hbar
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{odd}}\,$)}
    \quad.\label{GBSaodd}
\end{equation}
Be mindful that in the case of the odd potential, one chooses $\widetilde{E}_{\{\mu\},\,m\,(+)} = \widetilde{E}_{\{\mu\},\,2m+1}$ when solving~\eqref{solving}. Therefrom, the energy spectrum reflection reads:
\begin{equation}
    \widetilde{E}_{\{\mu\},\,\widetilde{n}} = - \widetilde{E}_{\{-\mu\},\,\widetilde{N} + 1 -\widetilde{n}} \quad,\qquad
    \widetilde{n}=1,3,5\ldots,\widetilde{N}
    \quad\quad\qquad \text{($\,$for~$\,V_{\text{odd}}\,$)}
    \quad.\label{ERodd}
\end{equation}

\subsubsection{The action in the potentials with multiple minima \label{quaction}}

It would be worth pointing out that the equation~\eqref{solving} reveals a peculiarity inherent in all potentials with degenerate minima.
For such potentials, the set of integer numbers in the RHS of~\eqref{GBS}, for which GBS has solutions, typically has gaps in it. In the case of the double-well potential, the set looks as:
\begin{equation}
    n \in \{0,\,1,\,2,\,\ldots,\,m^*-1,\,m^*,\,k^*,\,k^*+1,\,k^*+2,\,\ldots\} \quad.
\end{equation}
Here $m^*$ denotes the largest value of $n$ for which~\eqref{GBS} has solutions (or a solution) below its maximum, while $k^*$ is the smallest one for which the solution above the maximum exists.
This, however, should not surprise us.
Indeed, since the LHS of~\eqref{GBS} has the meaning of action, it should experience a jump, as the energy crosses the level of the maximum of the potential. For symmetric double-well potentials, we can expect
\begin{equation}
    2\,m^* \approx k^* \quad.
    \label{app}
\end{equation}
In quantum mechanics, the above equality is not, generally, exact~---~though it becomes asymptotically exact for highly excited states (i.e., for $\,m^* \to \infty$).
In the next section, this will be illustrated with examples.

\subsubsection{Examples \label{exampleee}}

In this section, we consider the even potential~\eqref{even} with $j=1$ (i.e., $N = 4$) and $\nu = 1$, and with different values of $\mu$. Let $m_{\{\mu\},\,n}$ be such a value of the integer $\,m\,$ in
\begin{equation}
    B_{\{\mu\}}\,(E) = \oint \limits_{\mathcal{C}_R} \qm_{\{\mu\}}\,(x, E) \,\d x = 2 \, \pi \,m\,\hbar\,\quad,\label{GBSm}
\end{equation}
that solution of this equation gives the energy level $E_{\{\mu\},\,n}$.
As we discussed above, in the case of a single well it is but the number of the energy level, $m_{\{\mu\},\,n} = n$. In the case of two wells, one either has two energy levels with numbers $n=2\,m_{\{\mu\},\,n}$ and $n=2\,m_{\{\mu\},\,n}+1$ or, possibly, the level $n=2\,m_{\{\mu\},\,n}$ solely. Due to the ESR property, these numbers for the dual potentials are linked via
\begin{equation}
    \left[
    \begin{alignedat}{4}
    &m_{\{\mu\},\,n} + 2\,m_{\{-\mu\},\,N-n} & \ = \ & N\\
    &2 \,m_{\{\mu\},\,n} + m_{\{-\mu\},\,N-n} & \ = \ & N
    \end{alignedat}
    \right.
    \quad,\qquad
    n=0,2,4\ldots,N
    \quad.
\end{equation}
In other words, if one of two symmetric levels lies below the maximum of the potential, another one has to reside above the maximum.

\begin{enumerate}
\item $\mu = 5$

In this case, the potential $V_{\{\mu\}}\,(x)$ is of a single-well shape, while $V_{\{-\mu\}}\,(x)$ is of a triple-well shape. The numbering of the levels is
\begin{equation}
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{$\mu = 5$} \\
\hline
    $n$ & $m_{\{\mu\},\,n}$ & $m_{\{-\mu\},\,n}$ \\
\hline
    5 & 5 & 2 \\
    4 & 4 & 2 \\
    3 & 3 & 1 \\
    2 & 2 & 1 \\
    1 & 1 & 0 \\
    0 & 0 & 0 \\
\hline
\end{tabular}
\label{tabu5}
\end{equation}
In the right column, one can clearly observe the pattern usual for  energy splitting.
\begin{subequations}
\begin{alignat}{6}
m_{\{\mu\},\,0} + 2\,m_{\{-\mu\},\,4} = 0 + 2 \cdot 2 = 4 \quad,\label{sums51}\\
m_{\{\mu\},\,2} + 2\,m_{\{-\mu\},\,2} = 2 + 2 \cdot 1 = 4 \quad,\\
m_{\{\mu\},\,4} + 2\,m_{\{-\mu\},\,0} = 4 + 2 \cdot 0 = 4 \quad.
\end{alignat}
\label{sums5}
\end{subequations}

\item $\mu = 4$

In this case, both potentials are of a double-well shape. However, the ground state energy in $V_{\{\mu\}}\,(x)$ is above the maximum of the potential:
\begin{equation}
    E_{\{\mu\},\,0} > V_{\{\mu\}}\,(0) \quad,
\end{equation}
and the numbering of the levels is the same as in~\eqref{tabu5}.

\item $\mu = 3$

In this case, the ground state energy in $V_{\{\mu\}}\,(x)$ is still above the maximum of the potential. However, the numbering of the levels changes slightly (though this change does not touch the levels possessing the ESR property).
\begin{equation}
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{$\mu = 3$} \\
\hline
    $n$ & $m_{\{\mu\},\,n}$ & $m_{\{-\mu\},\,n}$ \\
\hline
    5 & 5 & 5 \\
    4 & 4 & 2 \\
    3 & 3 & 1 \\
    2 & 2 & 1 \\
    1 & 1 & 0 \\
    0 & 0 & 0 \\
\hline
\end{tabular}
\label{tabu3}
\end{equation}
The fifth energy level in the potential $V_{\{-\mu\}}\,(x)$ is no longer below the maximum of the potential. We can summarise the difference as follows:
\begin{equation}
\begin{alignedat}{6}
    &\mu = 4\text{:} \qquad m_{\{-\mu\},\,5} &= 2 \quad,\\
    &\mu = 3\text{:} \qquad m_{\{-\mu\},\,5} &= 5 \quad.
\end{alignedat}
\end{equation}
In other words, for the potential $V_{\{-4\}}\,(x)$, the fifth energy level can be found from the equation
\begin{equation}
    \mu = 4\text{:} \qquad
    \oint \limits_{\mathcal{C}_R} \qm_{\{-\mu\}}\,(x, E) \,\d x = 2\,\pi\,(2)\,\hbar \qquad \to \qquad E_{\{-\mu\},\,5} \quad,
\end{equation}
while for the potential $V_{\{-3\}}\,(x)$, the fifth energy level is obtained from the equation
\begin{equation}
    \mu = 3\text{:} \qquad
    \oint \limits_{\mathcal{C}_R} \qm_{\{-\mu\}}\,(x, E) \,\d x = 2\,\pi\,(5)\,\hbar \qquad \to \qquad E_{\{-\mu\},\,5} \quad.
\end{equation}

\item $\mu = 2.5$

In this case, the ground state energy in the potential $V_{\{\mu\}}\,(x)$ is below its maximum, while the first excited state is above the maximum.
This is sufficient for us, to be able to state that the fourth state of the potential $V_{\{-\mu\}}\,(x)$ is above its maximum.
\begin{equation}
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{$\mu = 2.5$} \\
\hline
    $n$ & $m_{\{\mu\},\,n}$ & $m_{\{-\mu\},\,n}$ \\
\hline
    5 & 5 & 5 \\
    4 & 4 & 4 \\
    3 & 3 & 1 \\
    2 & 2 & 1 \\
    1 & 1 & 0 \\
    0 & 0 & 0 \\
\hline
\end{tabular}
\label{tabu25}
\end{equation}
\begin{subequations}
\begin{alignat}{6}
&2\,m_{\{\mu\},\,0} + m_{\{-\mu\},\,4} & \ = \ & 2 \cdot 0 + 4 & \ = \ & 4 \quad,\label{sums251}\\
&m_{\{\mu\},\,2} + 2\,m_{\{-\mu\},\,2} & \ = \ & 2 + 2 \cdot 1 & \ = \ & 4 \quad,\\
&m_{\{\mu\},\,4} + 2\,m_{\{-\mu\},\,0} & \ = \ & 4 + 2 \cdot 0 & \ = \ & 4 \quad.
\end{alignat}
\label{sums25}
\end{subequations}

Mind the difference between~\eqref{sums51} and~\eqref{sums251}. We had to add the factor of two to $m_{\{\mu\},\,0}$ due to the fact that the ground state is below $V_{\{\mu\}}\,(0)$ for $\mu = 2.5$.

One may have noticed from~\eqref{tabu25} that, as we go from the third to the fourth energy level in $V_{\{-\mu\}}\,(x)$, the quantum action becomes four times larger. This is, of course, a purely quantum effect. The general rule for our potential is:
\begin{equation}
    B\,(E_{\{\mu,\,k^*\}}) - 2\, B\,(E_{\{\mu,\,m^*\}}) \leq 4 \, \pi \, \hbar \quad,
\end{equation}
with $k^*$ and $m^*$ introduced in Section~\ref{quaction}.

\item $\mu = 2$

In this case, for $V_{\{\mu\}}\,(x)$ one has two levels below its maximum, and has four levels below the maximum for $V_{\{-\mu\}}\,(x)$.
\begin{equation}
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{$\mu = 2$} \\
\hline
    $n$ & $m_{\{\mu\},\,n}$ & $m_{\{-\mu\},\,n}$ \\
\hline
    5 & 5 & 5 \\
    4 & 4 & 4 \\
    3 & 3 & 1 \\
    2 & 2 & 1 \\
    1 & 0 & 0 \\
    0 & 0 & 0 \\
\hline
\end{tabular}
\label{tabu2}
\end{equation}
The enumeration of the energy levels possessing the ESR property has not changed, and stays same as in~\eqref{sums25}.

\item $\mu = 0$

In this case, the self-duality takes place, see Eq.~\eqref{selfd2}.
\begin{equation}
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{$\mu = 0$} \\
\hline
    $n$ & $m_{\{\mu\},\,n}$ & $m_{\{-\mu\},\,n}$ \\
\hline
    5 & 5 & 5 \\
    4 & 4 & 4 \\
    3 & 3 & 3 \\
    2 & 1 & 1 \\
    1 & 0 & 0 \\
    0 & 0 & 0 \\
\hline
\end{tabular}
\label{tabu0}
\end{equation}
The energy of the second excited state is equal precisely to the value of the potential at the maximum, which is a general rule for the integer values of $j$.
\begin{equation}
    \begin{cases}
    \mu = 0 \\
    j \in \mathbb{N}
    \end{cases}
    \qquad \to \qquad
    E_{\{\mu\},\,N/2} = V_{\{\mu\}}\,(0) = 0 \quad.
\end{equation}
This energy level can be treated both as (a)~a level below the maximum, with zero non-perturbative splitting; or (b)~a level above the maximum, with a zero non-perturbative shift. The summation reads as:
\begin{subequations}
\begin{align}
    & \hspace{4.2mm}         2\, m_{\{\mu\},\,2} + 2\,m_{\{-\mu\},\,2} =  2 \cdot 1           +  2 \cdot 1 = 4 \quad,\\
    &\left[
    \begin{aligned}
    & 2\,m_{\{\mu\},\,2} + 2\,m_{\{-\mu\},\,2} = 2 \cdot 1  +  2 \cdot 1 =  4 \\
    & \phantom{2\,}m_{\{\mu\},\,2}  +  2\,m_{\{-\mu\},\,2} = \hspace{5.2mm} 2  +  2 \cdot 1  =  4  \\
    & 2\,m_{\{\mu\},\,2}  + \phantom{2\,}m_{\{-\mu\},\,2} =  2 \cdot 1  +  \hspace{5.2mm} 2 =  4 \\
    & \phantom{2\,}m_{\{\mu\},\,2}  +  \phantom{2\,}m_{\{-\mu\},\,2} = \hspace{5.2mm} 2  +  \hspace{5.2mm} 2  = 4
    \end{aligned}
    \right.\quad,\\
    &\hspace{4.2mm} \phantom{2\,}m_{\{\mu\},\,4} + 2\,m_{\{-\mu\},\,0} =  \hspace{5.2mm} 4 + 2  \cdot 0 = 4 \quad.
\end{align}
\end{subequations}
\end{enumerate}

\subsubsection{A discussion of results by Geojo, Ranjani and Kapoor}

As we have already mentioned, our studies of the QES systems in the language of the QAF formalism  were largely inspired by the results obtained in \cite{HamJa} by Geojo, Ranjani and Kapoor. We agree with {\it{Ibid.}} on the following items:
\begin{itemize}
    \item For the sextic potential, the assumption of the QMF to have an isolated pole at infinity leads to the condition of quasi-exact solvability for this potential.
    \item The residue of the QMF can be calculated explicitly by substituting the Laurent expansion of the QMF into the Ricatti equation.
\end{itemize}
However, in \cite{HamJa} an oversight was made when the authors implicitly assumed the contour enclosing the real poles of the QFM to enclose \textit{all} the poles of the QMF. In our notation (see Figure~\ref{roots6}), this oversight reads as:
\begin{equation}
    \xcancel{ \oint \limits_{\mathcal{C}_1} \qm_{\{\mu\}}\,(x,\,E) \d\,x = \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}}\,(x,\,E) \d\,x} \quad.
\end{equation}
In reality, however, the singularities which are off the real axis should also be taken into account, in order to obtain the correct expression~\eqref{36}.

Due to the said omission, the authors got a wrong result for the residue of the QMF at infinity, which in their case turned out to be dependent on the number of the energy level.$\,$\footnote{~Specifically, the meaning of $n$ from the equation (6) in \cite{HamJa} had to be different from the meaning of $n$ from the equation (20) in {\it{Ibid}}. In the former equation, $\,n\,$ is related to the number of the energy level, while in the latter it is the number of the highest energy level from the algebraic sector ($N$ in our notation).}

\subsubsection{The WKB approach \label{wkbappr}}

Now we would like to trace the connection between the duality property of the classical action~\eqref{S} and its quantum analogue~\eqref{ququdual}. To do so, we shall employ the WKB method and show that the equations~\eqref{EReven} and~\eqref{ERodd} can be obtained in the Bohr-Sommerfeld approximation (i.e., by taking into account the leading-order and next-to-leading-order terms in the expansion of the QMF in powers of $\,\hbar\,$). This however is not a general rule. In the next section, we shall perform an analogous calculation for a periodic potential. As we shall see, in that case it will be necessary to take into account \textit{all} the perturbative orders of the WKB expansion, in order to reconstruct the exact ESR property.

We begin with writing down the classical duality condition~\eqref{S} in the notation introduced in~\eqref{V}. For the even potential, we arrive at:
\begin{equation}\begin{gathered}
    \oint \limits_{\mathcal{C}_\infty} p_{\{\mu\}}\,(x,\,-E)\,\d\,x =
    2\,\pi\left(4\,j + \dfrac{3}{2}\right)\hbar =
    2\,\pi\left(N + \dfrac{3}{2}\right)\hbar
    \quad,\label{symBS}
\end{gathered}\end{equation}
which virtually coincides with~\eqref{ALLeven}. It is only the extra term $\,\dfrac{3}{2}\,$ in the brackets, that makes~\eqref{symBS} different from~\eqref{ALLeven}.
This term arises owing to the order-$\hbar$ correction to the classical action, a sum of the Maslov indices. Indeed,
insertion of the series expansion~\eqref{L8} into the RHS of~\eqref{36} furnishes us:
\footnote{~In Section~\ref{prolego}, we mentioned that, beginning from $k=2$, it is only the even terms that appear in the LHS of~\eqref{sumcancel1}.}
\begin{equation}
\begin{gathered}
    2\,B_{\{\mu\}}\,(-E) + B_{\{-\mu\}}\,(E) =
    2\,\pi\,{i\mkern1mu}\,\Res \{ \qm_{\{\mu\}}\,(x,\,-E),\,\infty \} =
    \oint \limits_{\mathcal{C}_\infty} \sum \limits_{k=0}^{\infty} \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^k \qm_{\{\mu\}\,k}\,(x,\,-E) \,\d \,x \\=
    \oint \limits_{\mathcal{C}_\infty} p_{\{\mu\}}\,(x,\,-E)\,\d\,x
    + \dfrac{\hbar}{{i\mkern1mu}} \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}\,1}\,(x,\,-E)\,\d\,x
    + \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^2 \oint \limits_{\mathcal{C}_\infty} \qm_{\{\mu\}\,2}\,(x,\,-E)\,\d\,x +
    \ldots
    \\= 2\,\pi\,{i\mkern1mu}\,\Bigg(
    \underbrace{ \vphantom{\left(\dfrac{\hbar}{{i\mkern1mu}}\right)^2}
    \Res\left\{ p_{\{\mu\}}\,(x,\,-E),\,\infty \right\}
    }_{\text{The classical result, eq.~\eqref{symBS}}} +
    \underbrace{ \dfrac{\hbar}{{i\mkern1mu}}\, \vphantom{\left(\dfrac{\hbar}{{i\mkern1mu}}\right)^2}
    \Res\left\{ \qm_{\{\mu\}\,1}\,(x,\,-E),\,\infty \right\}
    }_{\text{The sum of three Maslov indices}}
    +
    \underbrace{ \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^2
    \Res\left\{ \qm_{\{\mu\}\,2}\,(x,\,-E),\,\infty \right\} + \ldots
    }_{\text{Vanishes for the sextic potential}}

    \Bigg) \\
    = 2\,\pi\,\left(N + \dfrac{3}{2}\right)\hbar - 3\,\pi\,\hbar
    = 2\,\pi\,N\,\hbar \quad.
\end{gathered}
\label{sumcancel1}
\end{equation}
Thus, due to the unique feature of the considered QES potential,
\begin{equation}
     \Res\{ \qm_{\{\mu\}\,k}\,(x,\,-E), \infty \} = 0 \qquad\text{for}\qquad k \geq 2 \quad,\label{res0}
\end{equation}
the quantum ESR symmetry can be explained in the Bohr-Sommerfeld approximation.

Had we carried out our proof starting from the usual Bohr-Sommerfeld quantisation condition~\eqref{BS} instead of~\eqref{GBS}, we would have been able to prove the ESR symmetry at the perturbative level only. In other words, we would have proved that the two levels below in the double-well energy region of one of two dual potentials are symmetric to some level in the single-well region of another one.
\subsubsection{Non-perturbative effects \label{nonpert}}

We now offer a couple of comments on  non-perturbative corrections which never showed up in our developments.
As one calculates the energy in potentials with degenerate minima, such corrections typically arise:
(a) below the absolute maximum of the potentials, due to the tunneling between the two wells; (b) above the absolute maximum (in the single-well region), owing to the over-the-barrier reflection. In the case of
deep wells, these corrections become exponentially small for the energy levels far from the maximum of the potential. In the double-well (single-well) region of the sextic potential, the corrections are negative (positive) for even levels, and are positive (negative) for the odd ones.

It would be important to undescore the following two circumstances:

\begin{enumerate}
\item
Up to the equation
\eqref{sumcancel1}, no non-perturbative corrections have been omitted, since we had made no assumption on the specific form of the QMF. For this reason, the equations~\eqref{EReven} and~\eqref{ERodd} are \textit{exact}~---~a fact already known from the algebraic considerations, see Section~\ref{algappr}).

\item As we already stated, the form of the series in~\eqref{sumcancel1} implies the neglect of the non-perturbative terms. The more remarkable it is that we have nevertheless arrived at the exact result~\eqref{sumcancel1}.
The reason for this is easy to guess. The instanton action (a cycle enclosing $BC$ in Figure~\ref{roots6}) and the action emerging due to the over-the-barrier reflection (a cycle enclosing $B'C'$ in Figure~\ref{roots62}) have opposite signs and cancel in~\eqref{sumcancel1}.
Accordingly, the corrections to the symmetric energy levels have opposite signs. The exponentially small leading-order corrections are easy to obtain, and they indeed do match. From the duality property of the potential, it ensues that
\begin{equation}\begin{aligned}
    \Delta_{-\epsilon} \propto - \exp\left(- \dfrac{2}{\hbar} \int \limits_B^C p_{\{\mu\}}(x,\,-E) \right) \quad,\qquad
    \Delta_{\epsilon} \propto \exp\left(- \dfrac{2}{\hbar} \int \limits_{B'}^{C'} p_{\{-\mu\}}({i\mkern1mu} \,x,\,E) \right)  \sim -\Delta_{-\epsilon} \quad.\\
\end{aligned}\end{equation}
\end{enumerate}
\subsection{The elliptic potential \label{ellqes}}

In this section, we consider a QES periodic potential and perform  calculations analogous to those done in Section~\ref{sexticqes} for the sextic potentials. First, we find an exact expression interrelating the QMF's of dual potentials. Then we study its WKB expansion.

Through Section~\ref{ellqes}, we shall be using the units in which $\,\dfrac{\hbar^2}{2\,m} = 1\,$.

\subsubsection[Algebraic approach]{Algebraic approach\footnote{~The material of this section is largely based on \cite{Dual}.}}

The potential~\eqref{Ve} turns out to be quasi-exactly solvable if one sets
\begin{equation}
a = J\,(J+1) \quad,\qquad J =1,\,2,\,3,\,\ldots
\end{equation}
For convenience, we also set
\begin{equation}
b = -\dfrac{1}{2} J\,(J+1) \quad.
\end{equation}
Thus we arrive at the potential of the Lam\'e model:
\begin{equation}
V^{\text{L}}\,(x\,|\,\nu) = J\,(J+1)
\left( \nu \snt(x\,|\,\nu) - \dfrac{1}{2} \right)\quad.
\label{LAME}
\end{equation}
The corresponding Shr\"odinger equation
\begin{equation}
    \left[-\dfrac{d^2}{dx^2} + V^{\text{L}}\,(x\,|\,\nu)\right] \psi\,(x) = E \psi\,(x)
\end{equation}
has a continuous spectrum consisting of $J$ finite bands and an upper band extending to infinity. For the potential $V\,(x\,|\,\nu)$, we label the band edges as $E_l\,[\nu]$, with ${l=1,\,2,\,\ldots,\,(2\,J+1)}$. In this notation, the band-gap structure of the potential reads as
\begin{equation}\begin{alignedat}{12}
        &\text{Bands:}\qquad &E_{2k-1} &\leq &E &\leq E_{2k}  \quad&\text{and}\quad &E\geq E_{2k+1}\\
        &\text{Gaps:} \qquad &E_{2k}   &\leq &E &\leq E_{2k+1}\quad&\text{and}\quad &E\leq E_{1}
\end{alignedat}
\quad,\qquad k=1,\,2\,\ldots J \quad.
\end{equation}
The band-edge energies for this potential can be found as eigenvalues of the operator
\begin{equation}
H\,[\nu] = J^2_x + \nu\,J^2_y - \dfrac{1}{2}\,J\,(J+1)\,\unit \quad,
\label{lamop}
\end{equation}
where $\,\unit\,$ is a unity matrix,
while $\,J_k\,$ are  generators of the $\,SU(2)\,$ group in the spin $J$ representation, obeying the usual commutation relation $\,{[J_x,\,J_y] = {i\mkern1mu}\,J_z}\,$.

The duality property
\begin{equation}
    E_l\,[\nu] = - E_{2J+2-l}\,[1-\nu]
    \label{ERlamee}
\end{equation}
immediately ensues from
\begin{equation}
H\,[\nu] = J^2_x + \nu\,J^2_y - \dfrac{1}{2}\,J\,(J+1)\,\unit =
- \left[ \left(J^2_z + (1-\nu)\,J^2_y\right) - \dfrac{1}{2}\,J\,(J+1)\,\unit\right]\quad,
\label{lamopdual}
\end{equation}
because the operator $\,\left[ J^2_z +(1-\nu)\,J^2_y \right]\,$ has the same eigenvalues as $\,\left[ J^2_z +(1-\nu)\,J^2_y \right]\,$.

In the next section, we see study the duality property in the language of the quantum action.

\subsubsection{The quantum action}

An equation defining the QMF for the Lam\'e potential comes up through insertion of~\eqref{LAME} into the Ricatti equation~\eqref{ric2}. The duality property has the same shape as~\eqref{plame}:
\begin{equation}\begin{aligned}
\qm\,({i\mkern1mu}\,x+K+{i\mkern1mu}\,K',\,V^{\text{L}}_{\nu,\,\text{max}}-\epsilon\,|\,\nu) =
{i\mkern1mu}\,\qm\,(x,\,V^{\text{L}}_{1-\nu,\,\text{max}}+\epsilon\,|\,1-\nu) \quad.
\end{aligned}
\end{equation}
Now, taking into account that
\begin{equation}
    V^{\text{L}}_{\nu,\,\text{max}} = \dfrac{1}{2}\,J\,(J+1)\,(-1+2\,\nu) = - V^{\text{L}}_{1-\nu,\,\text{max}} \quad,
\end{equation}
and changing
\begin{equation}
    \epsilon \to \epsilon + V^{\text{L}}_{\nu,\,\text{max}} \quad,
\end{equation}
we finally arrive at:
\begin{equation}\begin{aligned}
\qm\,({i\mkern1mu}\,x+K+{i\mkern1mu}\,K',\,-\epsilon\,|\,\nu) =
{i\mkern1mu}\,\qm\,(x,\,\epsilon\,|\,1-\nu) \quad.
\end{aligned}
\label{qplame}
\end{equation}
The latter equality looks very much like the equation~\eqref{pdual} for the sextic potential, and leads us to analogous results.

In terms of the quantum actions, the ESR property~\eqref{ERlamee} reads as:
\begin{subequations}
\begin{alignat}{5}
    &&\oint\limits_{\mathcal{C}_c} &\qm\,(x,\,-\epsilon\,|\,\nu) &&= 2\,\pi\,(J-n)\,\hbar \quad,\label{dudu1}\\
    &&\oint\limits_{-\mathcal{C}_2} &\qm\,(x,\,\epsilon\,|\,1-\nu) &&= \pi\,n\,\hbar \quad.\label{dudu2}
\end{alignat}
\label{dudu}
\end{subequations}
Starting from~\eqref{dudu2}, we shall demonstrate that condition~\eqref{dudu1} has to be fulfilled.

The integration contours for the QMF stay the same as in the classical case, see Figure~\ref{pxnuriem}.
The LHS of~\eqref{dudu1} can be expressed as:
\begin{equation}\begin{alignedat}{2}
    \oint \limits_{\mathcal{C}_c} \qm\,(x,\,-\epsilon\,|\,\nu) =
    \oint \limits_{-\mathcal{C}_p} \qm\,(x,\,-\epsilon\,|\,\nu)
    & -2\, \oint \limits_{-\mathcal{C}_2} \qm\,(x,\,\epsilon\,|\,1-\nu) \\
    = - 2\,\pi\,{i\mkern1mu} \, \Res\{\qm\,(x,\,-\epsilon\,|\,\nu),\,{i\mkern1mu}\,K'\}
    &- 2\, \oint \limits_{-\mathcal{C}_2} \qm\,(x,\,\epsilon\,|\,1-\nu) \quad.
    \label{snres}
\end{alignedat}\end{equation}
The residue of the QMF at its pole is calculated by expanding it into the Laurent series and plugging the result into the Ricatti equation:
\begin{alignat}{2}
    B\,(-\epsilon\,|\,\nu) + 2\,B\,(\epsilon\,|\,1-\nu)=
    - 2\,\pi\,{i\mkern1mu}\,\Res\{\qm\,(x,\,-\epsilon\,|\,\nu),\,{i\mkern1mu}\,K'\} =

    2\,\pi\,J\,\hbar\quad.\label{snfullres}
\end{alignat}
Insertion of the latter equation into the former furnishes:
\begin{equation}
        \oint \limits_{\mathcal{C}_c} \qm\,(x,\,-\epsilon\,|\,\nu) =
        2\,\pi\,J\,\hbar - 2\,(\pi\,n\,\hbar) =
        2\,\pi\,(J - n)\,\hbar\quad.
        \label{snansw}
\end{equation}
To understand the physical meaning of the index $\,n\,$, recall that the equation~\eqref{dudu2} renders us the band-edges $\,E_{2n+1}\,$ and $\,E_{2n+2}\,$. In the situations where the non-perturbative terms can be neglected, this equation defines the \textit{locations of bands and gaps}.

\subsubsection{The WKB approach}

Now let us employ the expansion~\eqref{L8} for the QMF, and derive the equation~\eqref{snansw} order by order in $\hbar$. The classical part of~\eqref{snfullres} is given by~\eqref{Ssres}:
\begin{equation}
    -2\,\pi\,{i\mkern1mu}\,\Res \{ p\,(x,-\epsilon\,|\,\nu),\,{i\mkern1mu}\,K' \} =
    \oint \limits_{-\mathcal{C}_p} p^{\text{L}}(x,\,-\epsilon\,|\,\nu) = 2\,\pi\,\sqrt{J\,(J+1)}\,\hbar
    =2\,\pi\left( J + \dfrac{1}{2} -\dfrac{1}{8\,J} + \ldots \right)\hbar
    \quad,
\end{equation}
which in the limit $J\to \infty$ differs from~\eqref{snfullres} by Maslov index $\dfrac{1}{2}$. For convenience, we introduce the notation.
\begin{equation}
    \kappa = \sqrt{J\,(J+1)} \quad.
\end{equation}
Taking into consideration the higher-order WKB terms in~\eqref{snfullres} renders:
\begin{equation}
\begin{gathered}
    B\,(-\epsilon\,|\,\nu) + 2\,B\,(\epsilon\,|\,1-\nu) =
    -2\,\pi\,{i\mkern1mu}\,\Res \{ \qm\,(x,-\epsilon\,|\,\nu),\,{i\mkern1mu}\,K' \} =
    \oint \limits_{-\mathcal{C}_p} \sum \limits_{k=0}^{\infty} \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^k \qm_k\,(x,\,-\epsilon\,|\,\nu) \,\d \,x \\
    \\
    =

    \oint \limits_{-\mathcal{C}_p} p^{\text{L}}(x,\,-\epsilon)\,\d\,x
    + \dfrac{\hbar}{{i\mkern1mu}} \oint \limits_{-\mathcal{C}_p} \qm_{1}\,(x,\,-\epsilon\,|\,\nu)\,\d\,x
    + \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^2 \oint \limits_{-\mathcal{C}_p} \qm_{2}\,(x,\,-\epsilon\,|\,\nu)\,\d\,x +

    \ldots
    \\
    \\
    =

    -2\,\pi\,{i\mkern1mu}
    \left[
    ({i\mkern1mu}\, \kappa\,\hbar) +
    \dfrac{\hbar}{{i\mkern1mu}} \left(\dfrac{1}{2}\right) +
    \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^2 \left( \dfrac{1}{8\,{i\mkern1mu}\,\hbar\,\kappa} \right) +
    \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^3 ( 0 ) +
    \left(\dfrac{\hbar}{{i\mkern1mu}}\right)^4 \left( \dfrac{1}{128\,{i\mkern1mu}\,\hbar^3\,\kappa^3} \right) + \ldots
    \right]
    \\
    \\=

    2\,\pi\,\hbar
    \left[
    \kappa - \dfrac{1}{2}  + \dfrac{1}{8\,\kappa} - \dfrac{1}{128\,\kappa^3} + \ldots
    \right]
    = 2\,\pi\,\hbar\, \dfrac{\sqrt{4\,\kappa^2+1}-1}{2}
    = 2\,\pi\,J\,\hbar
    \quad.
\end{gathered}
\label{sumcancel2}
\end{equation}
We see that summation of all (perturbative) orders of the QAF reproduces the exact equality~\eqref{snfullres}.
The explicit form of the higer-order terms in the WKB expansion of the QMF can be used to perform the calculation of the residues, see the equation~\eqref{qmkp}. We remind that all the odd terms starting from $k=3$ vanish
(and the corresponding residues have to be equal to zero).

In other words, in the Bohr-Sommerfeld approximation, having started from~\eqref{dudu2}, we would end up with $\;2\,\pi\,( \sqrt{J\,(J+1)} - 1/2 - n)\,\hbar\;$ in the RHS of~\eqref{dudu1}.

In the latter calculation, we managed to obtain the correct result, having completely ignored the non-perturbative effects. The reason for this is the same as was in the case of the double-well potential. The integration cycles corresponding to the tunneling and over-the-barrier reflection have opposite signs and cancel each other in~\eqref{sumcancel2}.

\section{Summary and discussion}

In our paper, we have introduced the notion of reflection symmetry of the action in classical systems, see the equations~\eqref{S} and~\eqref{Ss}. Generally, we expect equalities of this kind to hold for the classical counterparts of all quantum systems having the ESR symmetry of the quantum energy levels.

We also have discussed in detail a general method of constructing the Picard-Fuchs equation for a function defined as an integral, and we have illustrated this method with two examples.

Lastly, we have discussed the energy spectrum reflection property for dual quasi-exactly solvable potentials. In the language of the quantum action, this property can be written as
\begin{subequations}
\begin{alignat}{10}
    &B_{\{\mu\}}\,(E) + 2\,B_{\{-\mu\}}\,(-E) &&= 2\,B_{\{\mu\}}\,(-E) &&+ B_{\{-\mu\}}\,(E) & \ = \ & 2\,\pi\,N\,\hbar
    \quad\quad &&\text{($\,$for~$\,V_{\text{even}}\,$)}    \quad&,\label{resu1}\\
    &\widetilde{B}_{\{\mu\}}\,(E) + 2\,\widetilde{B}_{\{-\mu\}}\,(-E) &&= 2\,\widetilde{B}_{\{\mu\}}\,(-E) &&+ \widetilde{B}_{\{-\mu\}}\,(E) & \ = \ & 2\,\pi\,(\widetilde{N}+1)\,\hbar
    \quad\quad &&\text{($\,$for~$\,V_{\text{odd}}\,$)}     \quad&,\label{resu2}\\
    &B\,(-\epsilon\,|\,\nu) + 2\,B\,(\epsilon\,|\,1-\nu) &&= 2\,B\,(-\epsilon\,|\,\nu) &&+ B\,(\epsilon\,|\,1-\nu) & \ = \ & 2\,\pi\,J\,\hbar
    \quad\quad &&\text{($\,$for~$\,V^L\,$)}     \quad&,\label{resu3}
\end{alignat}
\label{resu}
\end{subequations}
where $N$ and $\widetilde{N}$ stand for the numbers of the highest energy levels in the algebraic sector for the even and odd potentials, correspondingly;
while $J$ stands for the number of bands for the elliptic potential.

For a particular form of the sextic potential, we discussed in great detail how the numbers of the energy levels are related to the integer in the RHS of the generalised Bohr-Sommerfeld quantisation condition.

We have also considered the WKB expansions of the formul\ae~\eqref{resu} and  investigated how these equalities work out in each consecutive order in $\,\hbar\,$.

In our calculations, the non-perturbative corrections to the action never appeared explicitly. They are hidden inside the quantum action, but turn out to be ignored in the WKB approximation thereof.

redpajama_set_name: RedPajamaArXiv


On Jun. 16-17, SPHARC held the 2015 Peer-to-Peer Exchange on "Considering Culture in Autism Screening and Systems of Care." Hosted by the Massachusetts HRSA Autism Planning Grant team and Act Early State team, the meeting was held at the Boston Children's Museum and brought together seven state teams (CT, IA, MA, ME, MN, NH, and VA) to brainstorm and problem solve educating parents from culturally and linguistically diverse backgrounds about healthy developmental milestones in young children and reducing early identification barriers for autism and other developmental disorders. The meeting included a training on the Considering Culture in Autism Screening curriculum, sharing strategies and models of intervention, and creating action plans inspired by the wisdom shared from the states' efforts.
Video: "Top 10 things every child with Autism would like you to know"


text:
The World (film)
The World (radio program)
Agallah
05:22:09:12 Off
Jump (Flo Rida song)
Jump (Madonna song)
This page contains text from Wikipedia, the Free Encyclopedia - https://wn.com/World
The World (Chinese: 世界; pinyin: Shìjiè) is a 2004 Chinese film written and directed by Jia Zhangke. Starring Jia's muse, Zhao Tao, as well as Chen Taisheng, The World was filmed on and around an actual theme park located in Beijing, Beijing World Park, which recreates world landmarks at reduced scales for Chinese tourists. The World was Jia's first to gain official approval from the Chinese government. Additionally, it was the first of his films to take place outside of his home province of Shanxi.
The World was a joint-production by Jia Zhangke's own Xstream Pictures, Japan's Office Kitano, and France's Lumen Films. It received additional financial support from the Shanghai Film Studio and several Japanese corporations including Bandai Visual and Tokyo FM, among others.
The film premiered in competition at the 2004 Venice Film Festival on September 4, 2004, but failed to win the coveted Golden Lion, the festival's top award, which ultimately went to Mike Leigh's drama Vera Drake, but which Jia would win two years later with Still Life. The World also premiered in 2004 at the New York Film Festival and would go on to receive a limited release in New York City the following year on July 1, 2005.
This page contains text from Wikipedia, the Free Encyclopedia - https://wn.com/The_World_(film)
PRI's The World is a global news radio, audio and multi-platform program created by Public Radio International based on the program's congruence with PRI's mission, and in order to fulfill the critical need for more original global news created for and provided to Americans. This was partly a response to declining investments by commercial media in international news. The program is co-produced by the BBC World Service of the United Kingdom, and Public Radio International and WGBH of the United States. The program's goal is to bring international journalism/news that illuminates the world specifically for US listeners.
The World was PRI's first co-production—the company now has a number of productions. It was also the first news co-production of the BBC World Service. For the BBC, The World was conceived as stand-alone program and also as a template for future co-productions which might serve to expand the reach of BBC World Service resources. At its launch, it was the first program dedicated to focusing on providing global news and making the global-local connection for Americans on a daily basis.
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Agallah (also known as 8-Off the Assassin, Don Bishop, Swagallah, or Brad Piff), born Angel Aguilar, May 8, 1974, is an American rapper who was a member of The Diplomats-affiliated group Purple City and the group Propain campaign. He was formerly known as 8-Off Agallah, but changed the name because it was controversial; 8-Off is a variation of the name Adolf.
He has ghostwritten songs for many well-known artists. In addition to producing his own music, he also produced tracks for Busta Rhymes, Remy Ma, The Diplomats, Guru, Rockin' Squat, Sean Price, Saigon, Game Theory, Mobb Deep and Big Pun. In 1996, he finished his debut album, Wrap Your Lips Around This, but it was never properly released because the album was shelved. In 2012, Wrap Your Lips Around This became available for digital download from Amazon.com and iTunes after being unreleased for many years.
A native of Brooklyn's Brownsville Projects, Agallah started out as a producer, then turned to rhyming and released a major-label single on East West America/Elektra Records in 1995. In 1997, he featured on R.A. the Rugged Man's track "Till My Heart Stops" on the Rawkus Records Soundbombing compilation an underground classic . A collaboration with Mr. Cheeks of the Lost Boyz followed, although the album the single was to appear on, Wrap Your Lips Around This, was never released. Additionally, there were production stints for the likes of EPMD, Das EFX, Onyx, and Group Home. In 2000, Agallah recorded the Sesame Street-themed single "Crookie Monster," which became an underground hit, and featured production by The Alchemist.
This page contains text from Wikipedia, the Free Encyclopedia - https://wn.com/Agallah
05:22:09:12 Off is an album by Industrial/EBM group Front 242. It was released by Sony on 2 November 1993 (see 1993 in music). The album's title is a simple substitution cipher for the word "evil"; where each letter is represented by its equivalent numerical position in the alphabet.
The album was unusual in that it primarily features a female voice, that of Christine "99" Kowalski, on vocals, with Jean-Luc De Meyer only appearing on a few tracks.
All songs written and composed by Front 242.
Track 14 & 15 on CD version only.
Daniel Bressanutti – Producer, Mixing
Patrick Codenys – Producer, Mixing
John Dubs – Producer
Bob Ludwig – Mastering
Eran Westwood – Producer, Mixing
Jean-Luc De Meyer – vocals
Christine "99 Kowalski" – vocals
Daniel Bressanutti – keyboards, programming, live mixing
Patrick Codenys – keyboards, programming, samplers
Richard Jonckheere, often credited as "Richard 23" – percussion, vocals
05:22:09:12 Off at MusicBrainz (list of releases)
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"Jump" is the fourth official single taken from Flo Rida's second album R.O.O.T.S. Nelly Furtado makes an appearance on the song with an upbeat hook. Flo-Rida said: "I used a whole different delivery on this one. We're talking about different situations to get people hype in the club. 'Jump!' Whether you're an athlete running in the stadium or you're in the club. Get hype!"
The song was used to promote the Disney film G-Force, and an edited version was added to Radio Disney's playlist.
It's Flo Rida's eighth consecutive Hot 100 entry, and one of his only singles that didn't make it to the Top 30 of the Hot 100 chart.
"Jump" made a brief visit to the Canadian Hot 100 Chart for 3 weeks and The Billboard Hot 100 chart for 1 week. Since the song has not been released as a single or for airplay, the chart performance relied on only digital sales. The single has re-entered at number 76 on Billboard Hot 100, due to G-Force taking the #1 spot its opening week. So far the single has peaked at #54, making this the first single from Flo Rida not to crack Top 20 in the US. The song made appearance on the American Top 40 for one week. It debuted at #40, and it fell outside the top 40 on the next week. The song has been most successful in Australia where it has so far peaked at #18, making it his 5th song to reach the top 20 there.
This page contains text from Wikipedia, the Free Encyclopedia - https://wn.com/Jump_(Flo_Rida_song)
"Jump" is a song by American singer Madonna from her tenth studio album Confessions on a Dance Floor (2005). Written by Madonna, Stuart Price and Joe Henry, the song was supposed to be released as the third single of the album. However, since "Get Together" was decided as the third single, "Jump" was released as the fourth and final single from the album, on October 31, 2006 by Warner Bros. Records. The song incorporates techno music with tributes to Pet Shop Boys. Madonna sings in her lower register in the song. Its lyrics talk about self-empowerment and sufficiency while looking for the prospects of a new relationship.
Contemporary critics complimented the song and its empowerment theme. They compared it to Madonna's '80s music and praised the club-anthem like quality of the song. The song peaked inside the top ten of the charts of some European countries, while peaking the charts in Italy and Hungary. In the United States, "Jump" placed in several Billboard dance charts and became one of the most popular dance hits of the decade.
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changetheworlds.com
offtherunway.org
intheunderworlds.com
worldsbrightestwomen.com
worldpress.com
theunderworlds.com
thewaroftheworlds.org
thebeatoff.com
leadinghotelsoftheworld.net
portsoftheworld.org
worldscientist.com
thechroniclesoffunemployment.com
atomicmarinesupply.com
menoftheworld.net
thepriceoffracking.com
Jump Off The World, School Of Fish
Jump Off The World
3000 miles underneath the pavement
holds you to something you don't understand
staring for hours at the spot on the ceiling
it needs to be painted but it's out of your hands
and i don't wanna be soul free
and that we can't even get on
everything i should have never thought have has finally come
so don't paint my candles red again
don't crawl into my head again
stuck in a hit in a staircase around you
waiting for all of the trust in the sky
write up on pictures you should have avoided
you gladly trade your mind for a write but
for many lines of action
will follow you forever and ever
don't jump off the world
My Pillow CEO Mike Lindell arrived at the White House on Friday for an apparent meeting with President Donald Trump, raising alarms after a press photographer shared a close-up shot of the visitor's notes ... @MyPillowUSA CEO Michael Lindell shows off his notes before going into the West Wing at the White House on Friday,… https.//t.co/zSFxpnRVbI ... In light of everything that has happened, however, the meeting is highly alarming....
Coronavirus : Les lieux culturels sont-ils majoritairement fermés en Europe, comme l'affirme Bachelot ?
Les Européens, interdits de se rendre dans leur ciné ou salle de spectacle préférés ? C'est ce qu'a avancé Roselyne Bachelot, jeudi, pour justifier la prolongation de la fermeture des lieux culturels en France ... Cinémas, salles de spectacles et musées encore fermés pour cause de crise sanitaire ... » ... 20 Minutes a vérifié ces déclarations. FAKE OFF ... Signaler une information qui vous paraît fausse à l'équipe « Fake Off » de « 20 Minutes ». ....
Du "Loft" aux "Anges", 20 ans de clashs
"Onze célibataires coupés du monde, filmés dans un loft de 225 m2, 24 h/24 par 26 caméras et 50 micros… Dans neuf semaines, ils ne seront plus que deux". Le 26 avril 2001, pour la première fois de l'histoire de la télévision en France, une voix off énonçait le synopsis de "Loft Story", émission d'un genre nouveau qui allait bouleverser le paysage audiovisuel ... Point commun entre tous ces programmes ... Buzz sur buzz ... ....
Toulouse relance la course à la montée
C'est devenu un classique de la Ligue 2. Un duel souvent acharné et serré. La rencontre d'hier soir n'a pas fait mentir l'historique récent entre Angers et le Toulouse Métropole Basket. Surtout, elle pouvait prendre des allures de premier virage de la saison ... Après avoir encaissé un 10-2 (47-42, 26e), les Toulousaines cessaient de souffler le chaud et le froid ... Là, c'était comme un premier tour de play-off ... ANGERS 67 – TOULOUSE 72 ... ....
Trèbes. Cazilhac : "On va voir pour faire des séances le week-end"
Alors que le Premier ministre vient d'annoncer l'abaissement du couvre-feu à 18 heures, nous avons sollicité les réactions des coaches seniors du FC Briolet. Ca y est, le couvre-feu est avancé à 18 heures . comment allez-vous vous organiser pour les entraînements ? ... J. B. ... R. Z. et H. K. ... R ... K ... ... J'espère vraiment qu'on pourra reprendre la compétition en mars, ce qui permettrait de finir les matchs aller et de faire des play-off aux beaux jours....
Faisons le point sur un potin quelque peu tiré par les cheveux. Le prince Harry a perdu une placepar rtl.fr ... "Il habite assez près de chez moi et depuis qu'ils se sont installés, ils donnent l'impression de vivre reclus", a commencé par expliquer le comédien vu dans "Wayne's World" avant de poursuivre en révélant qu'il guettait le fils de Lady Diana depuis quelque temps avant de finalement réussir à l'apercevoir ... ....
Carcassonne. Micka Simon trace sa route
Après onze saisons de Super League – neuf aux Dragons Catalans et deux à Wakefield – Mickaël Simon a décidé de poursuivre sa carrière en Élite, à Carcassonne XIII. Il s'y est imposé comme un des leaders du pack "jaune et noir". Demain après-midi, l'ancien limouxin tentera de mener les siens vers la victoire contre son ancien club ... Nous devons continuer à travailler pour être à cent pour cent lors des plays-offs ... Actuellement, non ... Oui ... ....
L'Œil de l'INA : Comment Georges Pernoud a vogué vers le succès
C'était en un temps que les journalistes de télévision de moins de 20 ans ne peuvent pas connaître . celui où l'on partait en reportage accompagné d'un cameraman ... De retour sur la terre ferme, il lance à Monique, sa femme ... Le record absolu de longévité dans l'histoire du petit écran ! En voix off, pendant les cinq premières saisons, Pernoud commence à apparaître à l'écran lorsque le rendez-vous devient hebdomadaire ... ....
Des centres de vaccination de proximité, une coopération renforcée entre le pouvoir fédéral et les Etats, plus de campagnes de prévention ... Joe Biden, 78 ans, cherche à éradiquer la racine de la crise économique et sanitaire qui mine les Etats-Unis ... En proportion, le pays fait toutefois mieux que de nombreux autres pays, dont la France, mais moins bien qu'Israël ou le Royaume-Uni, selon le site d'analyse des données Our world in data....

redpajama_set_name: RedPajamaCommonCrawl