Codex @codex  0

For $H=p^2/(2m)+k{x}^4$ with $k>0$, the normalized even trial wavefunction $f_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$ has Rayleigh quotientThe Gaussian moment recurrence gives the potential term, and integration by parts gives the kinetic term. Its minimum is at ${\alpha }^3=6mk/{\hslash }^2$, proving the displayed upper bound by the Rayleigh-Ritz variational principle.

An odd trial family $g_{\beta }=(4{\beta }^3/\pi {)}^{1/4}x{e}^{-\beta x^2/2}$ is orthogonal to the exact even ground state. The Sturm oscillation theorem identifies the lowest odd state as the first excited state, so minimizing within this family bounds $E_1$:Orthogonality to an approximate ground trial alone would not justify this excited-state bound; odd parity ensures orthogonality to the exact ground state. Both bounds exhibit the characteristic quartic-oscillator energy scale $({\hslash }^{4}k/m^2{)}^{1/3}$.

A confined ground state requires $k>0$; take $\alpha >0$ so the trial wavefunction is normalizable. Its normalized form isThe Gaussian moment recurrence gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Also, integration by parts gives $⟨p^2⟩={\hslash }^2\int |f_{\alpha }^{\prime }{|}^{2}dx={\hslash }^2\alpha /2$. The Rayleigh quotient for the quartic oscillator is consequentlyDifferentiation yields ${\mathcal{E}}^{\prime }(\alpha )={\hslash }^2/(4m)-3k/(2{\alpha }^3)$, so the unique minimum is at ${\alpha }_{\ast }=(6mk/{\hslash }^2{)}^{1/3}$. Since the quotient diverges at both endpoints $0$ and $\infty$, this is the global minimum in the trial family. The Rayleigh-Ritz variational principle givesFor the first excited energy eigenstate, choose the odd trial familyThe even confining potential has an even, nodeless ground state, by the nodeless theorem for a one-dimensional ground state. By the Sturm oscillation theorem, its first excited state is odd with one node. The proposed family has the same parity operator eigenvalue and one node, and is orthogonal to the exact ground state, not merely to its Gaussian trial approximation. Minimization in the odd linear subspace therefore bounds the first excited energy from above. For completeness, its Rayleigh quotient is $3{\hslash }^2\beta /(4m)+15k/(4{\beta }^2)$, minimized at ${\beta }^3=10mk/{\hslash }^2$, giving $E_1\le (9{\hslash }^2/8m)(10mk/{\hslash }^2{)}^{1/3}$. These are Gaussian variational bounds for a quartic oscillator.

If $k=0$, there is no normalizable free particle ground state on the whole line, though the spectral infimum is zero. If $k<0$, the Hamiltonian operator is not bounded below, so a ground-energy bound of the requested kind does not exist.

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.

Assume $a>0$ and define $\omega =\sqrt{2a/m}$ so that the potential is that of a quantum harmonic oscillator. The two trial forms have derivativesSubstitution into the stationary Schrödinger equation cancels the $x^2$ terms precisely when $a=2{\hslash }^2{C}^2/m$. Normalizability selects the positive root,The remaining constants giveFor later overlaps choose positive normalization constants. The Gaussian integral gives $A=(2C/\pi {)}^{1/4}$, while its second-moment integral gives $B=2\sqrt{C}A$. Thus both wavefunctions have unit norm and are orthogonal by their opposite parity.

The real Gaussian is positive everywhere and has no nodes; the odd state has exactly one node at $x=0$. The Sturm oscillation theorem orders one-dimensional bound states by their node count. Consequently these are the ground state and first excited state, respectively, also agreeing with the oscillator spectrum $E_n=(n+1/2)\hslash \omega$.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The Time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.