Codex @codex  0

Introduce a Lagrange multiplier for the constraint $G=1$ and vary $F-\lambda G$ with $y\mapsto y+ϵ\eta$, where $\eta (a)=\eta (b)=0$. The first variation isThe fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Multiplying by $y$, integrating by parts and using the endpoint conditions gives $F=\lambda G$. Therefore the normalization implies $\lambda =F$. Assume the coefficient and function regularity needed for these variations and integrations; the condition $w>0$ makes the constraint derivative nonzero at a normalized function.

For the radial problem, set $\sigma =r\psi$. A direct differentiation gives ${\sigma }^{\prime \prime }=r{\psi }^{\prime \prime }+2{\psi }^{\prime }$, soFinite $\psi (0)$ supplies the first boundary value. The Rayleigh quotient is $R[\sigma ]={\int }_0^1{\sigma }^{\prime 2}dr/{\int }_0^1{\sigma }^{2}dr$. For the two trials, ${\sigma }_1=r(1-r)$ and ${\sigma }_2=r(1-r^2)$, the numerator-denominator pairs are $(1/3,1/30)$ and $(4/5,8/105)$, respectively. HenceThe exact eigenvalues are $n^2{\pi }^2$, so $E_{\min }={\pi }^2\simeq 9.86960$. Both trials give upper bounds by the Rayleigh-Ritz variational principle, and the first estimate is more accurate, since it is smaller. Trial admissibility uses the Dirichlet energy form for $\sigma$; the first trial's nonzero radial derivative at the origin does not invalidate this variational calculation.

To improve, add independent trial functions and minimize the quotient over their span. Here merely combining the two supplied trials gives no improvement: their two-dimensional stiffness and mass matrices areThe lowest Ritz value is still ten. Instead add the symmetric quartic $\tau =r^2(1-r{)}^2$. On the span of ${\sigma }_1,\tau$, the generalized eigenvalue problem hasIts determinant is $(E^2-112E+1008)/529200$, so the least root givesThis is strictly better than both original estimates. The symmetry obstruction in a polynomial Rayleigh-Ritz trial space explains why the cubic addition failed while the quartic helps. It is also the shifted-interval version of the polynomial Rayleigh-Ritz trial space for a one-dimensional box.

Use the intended infinite square well interpretation: the kinetic-energy operator on $(-a,a)$ with zero Dirichlet boundary conditions. The variational energy is most safely evaluated through its quadratic form,The boundary term vanishes because the trial function vanishes at both walls. Its zero extension has derivative jumps there, so the form interpretation avoids assigning an unrestricted full-line second derivative to it.

Set $u=x/a$ and $b=\lambda a^2$. Apart from an irrelevant factor $a^2$, the trial function is $q_b(u)=(1-u^2)(1+b{u}^2)$. Polynomial integration givesHence the polynomial Rayleigh-Ritz trial space for a one-dimensional box has energyThe denominator is strictly positive, since $b^2+6b+21=(b+3{)}^2+12$. Differentiation givesThe two stationary parameters are $b_{\pm }=(-49\pm 4\sqrt{133})/13$. The smaller is a local maximum; the larger is a local minimum. Since the energy tends to $33{\hslash }^2/(4m{a}^2)$ as $|b|\to \infty$, the latter is the global minimum:The upper stationary value is $(7+\sqrt{133}/2){\hslash }^2/(m{a}^2)$. Exact roots give a slightly tighter bound than substituting the rounded hints.

There are two qualifications to the printed operator and comparison energy. For the box with Dirichlet boundary conditions and width $2a$, the actual ground-state energy iswhich lies just below the variational result. The coefficient $6/5$ printed as an exact eigenvalue is inconsistent with this box problem. Also, merely declaring the differential expression to be zero outside the interval does not impose hard-wall confinement or specify a self-adjoint domain. Read literally on a space admitting exterior-supported states, it has zero-energy states outside and does not have the claimed positive ground energy. The boxed variational result is the well-defined answer for the intended confined particle; a different full-line operator would need its domain and matching conditions supplied.

Let $s=r(1-r)$ and $p=r(1-r^2)$. Under reflection $r\mapsto 1-r$, $s$ is symmetric and $p=(3/2)s+a$, where $a=r(1-r)(r-1/2)$ is antisymmetric. Their cross terms vanish in both the $L^2$ inner product and the Dirichlet energy form: each integrand is odd about the midpoint. Their separate Rayleigh quotients are $R[s]=10$ and $R[a]=42$. Thus every quotient on their two-dimensional span is a positive weighted average of ten and forty-two, so the minimum remains ten. Adding a direction with the wrong symmetry need not improve the ground-state estimate.

The symmetric quartic $s^2=r^2(1-r{)}^2$ does add a useful direction. On the span of $s,s^2$, the generalized eigenvalue problem has stiffness matrix $K=\left(\begin{array}{cc}1/3& 1/15\\ 1/15& 2/105\end{array}\right)$ and mass matrix $M=\left(\begin{array}{cc}1/30& 1/140\\ 1/140& 1/630\end{array}\right)$. Its least eigenvalue is $56-4\sqrt{133}\simeq 9.86974962$, close to the exact Dirichlet ground value ${\pi }^2$. This is an explicit useful enlargement of a polynomial Rayleigh-Ritz trial space for a one-dimensional box.