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For a compact connected regular level of $n$ independent Poisson-commuting functions, the periods of their joint flow form a full-rank discrete subgroup $\Lambda \subset {\mathbb{R}}^n$. The level is ${\mathbb{R}}^n/\Lambda$. A local smooth basis of these periods determines action integrals and the normalized periodic fields used to construct action-angle variables.

For a harmonic oscillator with Hamiltonian $H=p^2/(2m)+m{\omega }^2{q}^2/2$, where $m,\omega >0$, defineThese action-angle variables are defined for $I>0$, with a local angle $\theta$ modulo $2\pi$. A proof of canonicity is ${\theta }_q=p/(2I)$ and ${\theta }_p=-q/(2I)$, which give $\{\theta ,I\}=1$ using the Poisson bracket. Substitution gives $H=\omega I$. Hamilton's equations then prove $I(t)=I(0)$ and $\theta (t)=\theta (0)+\omega t$. The action integral around the corresponding phase space ellipse is $\oint pdq=2\pi I$, so $I$ is the usual normalized action variable. The angle is undefined at the zero-energy origin.

In the central classically forbidden region, define the positive decay momentum $\chi (x)=\sqrt{2m[V(x)-E]}$ and the dimensionless action integral $K(x)={\hslash }^{-1}{\int }_x^{x_1}\chi (y)dy$. The general leading WKB approximation isBoth coefficients are allowed: this interval has finite length, so the exponential that grows towards the barrier centre cannot be discarded by a condition at infinity. The WKB validity condition holds away from the classical turning point $x_1$; a local Airy function solution supplies its continuation there.

Use the amplitudes and positive action integrals of part (a), and define the right-well phaseMatching at the inner classical turning point gives $C=A$, $D=2B$. Matching the exterior decaying wavefunction at the outer classical turning point givesThe trigonometric identitytherefore yieldsAll three pieces on $x>0$ are now fixed up to $E$ and $F$:The Airy function descriptions replace the apparent singularities at the classical turning points. For the permitted energy eigenvalues, part (d) extends this right-hand wavefunction to the left by even or odd parity; $F$ is then determined by wavefunction normalization. At arbitrary $E$, the decaying right-hand solution instead has a unique continuation by the Schrödinger equation, which need not decay at negative infinity.

For $V(-x)=V(x)$, the parity operator $P\psi (x)=\psi (-x)$ commutes with the Hamiltonian operator. Also $P^2=I$, so its eigenvalues are $\pm 1$ and the energy eigenspaces admit a basis of definite parity. More strongly, nondegeneracy of one-dimensional bound states says that each bound state energy here is nondegenerate: the Wronskian of two solutions at the same energy is constant by the Schrödinger equation, and decay at infinity makes it zero, forcing linear dependence. Thus each bound state itself is even or odd.

An even function obeys the Neumann boundary condition ${\psi }^{\prime }(0)=0$; an odd function obeys the Dirichlet boundary condition $\psi (0)=0$. WriteSmooth reflection symmetry gives ${\chi }^{\prime }(0)=0$, while $K^{\prime }(0)=-\chi (0)/\hslash$. Hence, for the central WKB approximation from part (c),The two boundary conditions consequently give the WKB quantization condition for a symmetric double well:The full wavefunction is ${\psi }_{+}(x)={\psi }_R(|x|)$ in the even case and ${\psi }_{-}(x)=\mathrm{sgn}(x){\psi }_R(|x|)$ in the odd case, with the value at zero fixed continuously. The relevant boundary condition makes these extensions differentiable at zero. Both action integrals $\theta ,\varphi$ depend on $E$ through the integrands and the classical turning points.

A tall, broad central potential barrier has $\varphi \gg 1$. Let $\sigma =+1$ denote even parity and $\sigma =-1$ odd parity, and write $\theta =(n+\frac{1}{2})\pi +{\delta }_{\sigma }$, where $n=0,1,\dots$. Since $\tan [(n+\frac{1}{2})\pi +\delta ]=-\cot \delta$, the condition in part (d) givesThus the leading correction to the isolated-well Bohr-Sommerfeld quantization condition isThis is an implicit condition, with all action integrals evaluated at the corrected energy. To convert it into an explicit leading energy correction, let $E_n^{(0)}$ solve $\theta (E_n^{(0)})=(n+\frac{1}{2})\pi$ and define the full classical oscillation period in one well by $\mathcal{T}(E)=2m{\int }_{x_1}^{x_2}dx/p(x)$. Differentiation of the action integral has no endpoint terms because $p(x_1)=p(x_2)=0$, and givesConsequently, to leading exponentially small order within the WKB approximation,The even member lies below its odd partner. For an approximately harmonic oscillator well, $\mathcal{T}\simeq 2\pi /\omega$, giving the familiar double-well tunneling splitting $\Delta E\simeq (\hslash \omega /\pi )e^{-\varphi }$. The explicit energy formula retains only its leading exponential correction; varying $\varphi$ and ${\theta }^{\prime }$ with the shifted energy also contributes at higher orders.

The ground state has an even, everywhere same-sign wavefunction with a peak in each well and a small nonzero value beneath the central potential barrier. Its derivative vanishes at zero. The first excited state is odd, has opposite-sign lobes in the two wells, and has exactly one interior node of a wavefunction, at zero. Both wavefunctions decay outside the wells. In terms of localized normalized wavefunctions ${\phi }_L,{\phi }_R$, with overlap $s=⟨{\phi }_L|{\phi }_R⟩$, their qualitative shapes are the lowest parity states of a symmetric double well:The overall sign of the odd wavefunction is arbitrary. The sketch uses localized Gaussian profiles to show the parity and the central node of a wavefunction, rather than claiming exact energy eigenfunctions for an unspecified potential energy.

For a regular periodic energy orbit, differentiate the action integral with respect to its energy. On a branch $H(p,q)=E$, implicit differentiation gives $\partial p/\partial E=1/H_p=1/\dot{q}$. ThusThe integral follows the entire orbit, with its orientation; at turning points this formula is interpreted by the limiting integral from each branch. It is valid for a smooth family of nonseparatrix periodic orbits.

An adiabatic invariant means that the action variable remains constant to leading order when the external parameter changes on a timescale much longer than the orbital period. Denote that parameter by $\lambda$. With $H(0,\lambda )=0$, the displayed identity integrates toIf the parameter change decreases $T(J,\lambda )$ for every intervening action $J$, this integrand increases pointwise. At the adiabatically fixed positive $I$, the energy increases. This uses the phrase “with any $I$” as a statement about the family of orbits, not merely a period decrease at a single selected action; that weaker premise alone would not determine the integral.

The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincaré lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

For the half-line square-root potential well, the dimensionless action integral isHence $E_n/E_{\ast }=[15\pi (n+3/4)/8{]}^{2/5}$. The ground-state estimate is $1.811694E_{\ast }$. This is a semiclassical estimate, not a variational upper bound.