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Consider hard walls at $x=0,3L$, zero potential energy on $0<x<L$ and $2L<x<3L$, and a central rectangular barrier whose height equals the ground state energy $E$. The time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=0$ on the central interval. Reflection symmetry and the nodeless theorem for a one-dimensional ground state require an even, positive state about $3L/2$, hence a constant central segment. Wavefunction matching at a finite potential step makes the derivatives vanish at $L,2L$. The side segments are proportional to $\sin (\pi x/(2L))$ and $\sin (\pi (3L-x)/(2L))$. Their common plateau value is $1/\sqrt{2L}$ after wavefunction normalization, and $E={\pi }^2{\hslash }^2/(8m{L}^2)$. Equality of barrier height and energy is essential: a higher barrier gives positive curvature of a positive wavefunction in the middle, while a lower barrier gives negative curvature.

A nodeless real eigenfunction has no zero in the interior of the domain. For a one-dimensional confining Schrödinger operator, the nodeless theorem for a one-dimensional ground state identifies the ground state as the eigenfunction that can be chosen strictly positive. Normalizable excited states have nodes.

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 any nonzero state, the Rayleigh quotient isExpand $|\psi ⟩={\sum }_n{c}_n|{\psi }_n⟩$ in normalized energy eigenstates. ThenBecause the ground state is unique, equality holds exactly when all coefficients except $c_0$ vanish. Thus the Rayleigh-Ritz variational principle gives its minimum at the ray of $|{\psi }_0⟩$.

For the proposed exact state, write $\tilde{\psi }=e^{-\beta x^n}$. Its logarithmic derivatives giveThe stationary Schrödinger equation, after multiplication by $2m/{\hslash }^2$, isMatching the highest power requires $2n-2=6$, so $n=4$, and cancellation of $x^6$ requires $16{\beta }^2=1$. Normalizability selects $\beta =1/4$. Then $12\beta =3$ also cancels the quadratic term, leaving $\epsilon =2$. ThereforeThis is the exact ground state of a solvable sextic potential candidate $e^{-x^4/4}$.

For the Gaussian trial state ${\psi }_{\alpha }=e^{-\alpha x^2/2}$, normalization cancels from the quotient. With respect to the probability density proportional to $e^{-\alpha x^2}$,HenceThe stationary equation isWriting $z={\alpha }^2$, the quadratic has exactly one positive root,Since the quotient tends to infinity as $\alpha \downarrow 0$ or $\alpha \to \infty$, this is the unique global minimizer. ThusUsing the stationary equation to replace $15/(8{\alpha }_{\ast }^3)$ by ${\alpha }_{\ast }/6+1/(2{\alpha }_{\ast })$ gives the best estimateThis is the Gaussian variational estimate for a solvable sextic potential.

The exact eigenfunction $e^{-x^4/4}$ is positive and has no nodes. The nodeless theorem for a one-dimensional ground state therefore identifies it as the true ground state, with dimensionless energy $2$. The variational value is consistent: since ${\alpha }_{\ast }^2>3/2$,so $E_0^{\ast }>{\hslash }^2/m=E_0$, as every trial-state upper bound must satisfy.

The original diagram places hard walls at $0$ and $3L$, and the central plateau on $L<x<2L$ is exactly at the indicated ground-state energy. This equality determines the middle of the requested sketch. In the time-independent Schrödinger equation,the central interval has $V=E$, hence ${\psi }^{\prime \prime }=0$. The ground state can be chosen real and positive inside the walls by the nodeless theorem for a one-dimensional ground state. Reflection symmetry makes it even about $x=3L/2$, so its linear central segment must be constant.

The hard-wall boundary conditions are $\psi (0)=\psi (3L)=0$. On both outer intervals $V=0<E$, so the positive wavefunction has negative curvature and is sinusoidal. Wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to remain continuous at $L,2L$; the derivative therefore vanishes where each sinusoidal flank meets the flat middle. A normalized representative isEach side contributes $L/2$ and the plateau contributes $L$ to the squared norm before multiplying by $1/(2L)$. The corresponding consistency condition is $E={\pi }^2{\hslash }^2/(8m{L}^2)$; a numerical energy is not required for the sketch.

Thus the ground state at the top of a central rectangular barrier has no internal nodes, continuous value and slope, and a flat middle, rather than an exponential dip through a classically forbidden region. Multiplication by an overall constant phase changes none of these physical features.