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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 spectral equation $(S)$ isThis is the KdV Schrodinger spectral problem. Differentiating it with respect to $t$ givesFora direct differentiation, followed by substitution of the two preceding identities, yieldsThereforeComparison with $(\ast )$ givesThis is the reusable KdV Schrodinger spectral problem Wronskian identity.

Now take $u$ to satisfy the Korteweg-De Vries equation and $\varphi ={\phi }_n$, where ${\lambda }_n=-{\kappa }_n^2$. Both ${\phi }_n$ and its derivative decay at infinity, as does $u$, so integration of $(\ast )$ over the real line givesThe normalization was used in the last equality. Hencewhich is the Isospectrality of the KdV discrete spectrum.

With the KdV equation and ${\dot{\lambda }}_n=0$, the Wronskian identity saysDecay at infinity makes the constant zero. Thus $(Q/{\phi }_n{)}_x=0$ between zeros, and continuation across the isolated zeros givesMultiplying by ${\phi }_n$ and integrating,The first integral is zero by differentiating the normalization. Integration by parts turns the last integral into $\int u_x{\phi }_n^{2}dx$, soTo evaluate this, differentiate $(S)$ in $x$ and use it to writeIts integral is $-2\int {\phi }_{n,x}{\phi }_{n,xx}dx=0$ by decay. Hence

As $x\to +\infty$, rapid decay of $u$ and $u_x$, together withreduces $Q=0$ toSince ${\kappa }_n$ is constant,as stated by the Evolution of a KdV discrete norming constant.