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For a normalized trial function in the quadratic-form domain of the Schrödinger operator, integration by parts gives $⟨E⟩=⟨\psi ,H\psi ⟩$. Expand in the complete orthonormal energy eigenstates, $\psi ={\sum }_n{c}_n{\psi }_n$ with ${\sum }_n{c}_n^2=1$. ThenMinimizing over a trial family therefore gives the best upper bound within that family, the quantum variational principle.

For $V=|x|$, choose ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$ with $\alpha >0$. The Gaussian integrals giveThus $E(\alpha )=\alpha /2+(\pi \alpha {)}^{-1/2}$, whose derivative vanishes at $\alpha ={\pi }^{-1/3}$. It tends to infinity at both parameter endpoints and has positive second derivative, so this is its minimum. $\boxed{E_0\le \frac{3}{2}{\pi }^{-1/3}\simeq 1.0242}$ is the resulting estimate.

The potential is even; the unique nodeless ground state is even and the first excited state is odd, by the one-dimensional Sturm oscillation theorem. To estimate $E_1$, minimize the same Rayleigh quotient over odd trial functions, for example normalized $x{e}^{-\alpha x^2/2}$, or the compactly supported odd family $C_{a}x(1-|x|/a)$ for $|x|\le a$ and zero otherwise. Oddness gives exact orthogonality to the true ground state, not merely to an approximate trial ground state. The quantum variational principle restricted to the odd sector supplies an upper bound for $E_1$. No further integration is needed.

The ladder operators obey $[a,a^{†}]=1$ and $N=a^{†}a$ has $[N,a]=-a$, $[N,a^{†}]=a^{†}$. Since $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2\ge 0$, its eigenvalues are nonnegative. If $N|\nu ⟩=\nu |\nu ⟩$, lowering gives eigenvalue $\nu -1$ and squared norm $\nu$. Repeated lowering cannot produce a negative nonzero eigenstate; the norm product $\nu (\nu -1)\cdots$ forces $\nu$ to be an integer. The final lowered state has $a|0⟩=0$. With a unique ground state all normalized states are

For time-independent perturbation theory, write $H^{\prime }=H+\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$. The standard first correction is $\lambda ⟨n|W|n⟩$ and the second is ${\lambda }^2{\sum }_{j\ne n}|⟨j|W|n⟩{|}^2/(E_n-E_j)$. The first vanishes. Only $j=n-2,n+2$ occur, with ladder coefficients $\sqrt{n(n-1)}$ and $\sqrt{(n+1)(n+2)}$, respectively. ThereforeThe downward term is zero for $n<2$.

Inverting the given position/momentum expressions givesThus $\boxed{\alpha =1-2\lambda ,\beta =1+2\lambda }$. For $|\lambda |<1/2$ both are positive, and this is a quantum harmonic oscillator of effective mass $m/\alpha$ and frequency $\omega \sqrt{\alpha \beta }$. Its exact energies areExpansion gives $E_n^{\prime }=\hslash \omega (n+1/2)[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with both perturbative corrections.

Hamilton's equations are ${\dot{q}}_j=\partial H/\partial p_j$ and ${\dot{p}}_j=-\partial H/\partial q_j$. The Arnold-Liouville theorem states that $n$ functionally independent first integrals in involution on a $2n$-dimensional symplectic manifold give local integrability; a compact connected regular common level is an $n$-torus. Near such a torus there are action-angle variables $(I,\vartheta )$ with $H=H(I)$, $\dot{I}=0$ and $\dot{\vartheta }=\partial H/\partial I$.

For the displayed diagonal oscillator Hamiltonian set $E_k=(p_k^2+{\omega }_k^2{q}_k^2)/2$. Each is constant, since ${\dot{q}}_k=p_k$ and ${\dot{p}}_k=-{\omega }_k^2{q}_k$. Different $E_k$ depend on disjoint canonical pairs, so their Poisson brackets vanish. Their differentials are independent wherever every $d{E}_k$ is nonzero, an open dense set; this proves complete integrability. For nonzero frequencies put ${\Omega }_k=|{\omega }_k|$. A positive-energy level in each pair is an ellipse with semiaxes $\sqrt{2E_k}$ and $\sqrt{2E_k}/{\Omega }_k$. Its enclosed area is $2\pi E_k/{\Omega }_k$, so the action variables areFor the usual positive oscillator frequencies this is $I_k=E_k/{\omega }_k$. If a printed “constant” frequency is allowed to be zero, that pair is a free particle: $p_k$ is a commuting integral, but its generic level is noncompact and has no closed-cycle oscillator action. Complete integrability persists, while the torus/action formula requires nonzero frequencies.

Watson's lemma gives, under integrability away from zero and the stated endpoint asymptotic expansion,Indeed each power integrates over the positive half-line to $\Gamma (n\beta +1){\lambda }^{-(n\beta +1)}$; replacing a fixed upper endpoint by infinity has exponentially small error. A remainder $o(t^{N\beta })$ near zero gives $o({\lambda }^{-N\beta -1})$ after splitting off a small endpoint neighbourhood and rescaling $u=\lambda t$.

For Laplace's method, assume the phase and amplitude are smooth near the nondegenerate maximum, as required for a full expansion. Localize to a small neighbourhood of $c$: outside it the unique maximum and continuity make the exponential smaller by a factor $e^{-\lambda \epsilon }$. SetNear $c$ this is a smooth increasing coordinate, with ${\zeta }^{\prime }(c)=\sqrt{|{\varphi }^{\prime \prime }(c)|}$. It need not be monotonic on the entire integration interval. In the localized integral put $A(\zeta )=F(t(\zeta ))dt/d\zeta$. Extending its Gaussian integration to both infinite endpoints incurs exponentially small error. Taylor expansion of $A$ and integration of even Gaussian moments givesHere $A(0)=F(c)/\sqrt{|{\varphi }^{\prime \prime }(c)|}$, giving the leading term $\boxed{e^{\lambda \varphi (c)}F(c)\sqrt{2\pi /(\lambda |{\varphi }^{\prime \prime }(c)|)}}$. The asymptotic-equivalence notation for that term presumes $F(c)\ne 0$; if it vanishes the series determines the first nonzero term. The maximum condition alone, without amplitude regularity, would not imply a full expansion.

For the gamma function, set $t=xs$:The phase has unique maximum $-1$ at $s=1$, with second derivative $-1$. To obtain the correction, put $s=1+u/\sqrt{x}$. Expansion givesand hence a Gaussian multiplierThe odd term integrates to zero. The normalized Gaussian fourth and sixth moments are 3 and 15, so the correction is $-3/4+15/18=1/12$. Localization near one, with the exponentially suppressed endpoint tails, justifies termwise expansion. Thus Stirling's formula is