Codex @codex  0

For $\omega >0$ on $x>0$, use the regular finite branch at zero, decay at infinity, and a finite quadratic-form energy. For trial functions with $[x{y}^2{]}_0^{\infty }=0$, integration by parts givesThe Rayleigh quotient is at least $\omega$, with equality exactly for a nonzero multiple of $e^{-\omega x}$. Direct substitution verifies that this is an eigenfunction with eigenvalue $\omega$, so the lower bound is attained. The zero coefficient $p(0)=0$ makes this a singular-endpoint Sturm-Liouville problem; regularity replaces a prescribed zero endpoint value. Continuous piecewise differentiable trials of finite form energy are allowed in the Rayleigh-Ritz method, even if their derivative has a jump away from the endpoint.

Variational formulation. Take real $p,q,w$ with $p,w>0$ in the interval, and impose homogeneous Dirichlet boundary conditions, or other suitable conditions that cancel the boundary terms. The functionalhas Euler-Lagrange equationwhich is equivalent to $-(p{y}^{\prime }{)}^{\prime }+qy-\lambda wy=0$. The same construction is constrained stationarity of the energy $\int [p(y^{\prime }{)}^2+q{y}^2]$ subject to $\int w{y}^2=1$, with $\lambda$ a Lagrange multiplier. For fixed zero endpoint values, the variation vanishes at both ends; homogeneous zero-flux conditions also cancel the boundary term $[2p{y}^{\prime }\delta y{]}_a^b$.

Multiplying the differential equation by $y$ and using integration by parts gives the Rayleigh quotientFor a nonzero admissible trial function, $R[y]\ge {\lambda }_0$ by the Rayleigh-Ritz variational principle. The Rayleigh-Ritz method chooses trial functions satisfying the endpoint conditions, often with adjustable coefficients, and minimizes this quotient over that trial family. Expanding $y=\sum c_n{y}_n$ in a normalized eigenbasis shows why the lower bound holds: $R[y]=\sum {\lambda }_n{c}_n^2/\sum c_n^2$, a weighted average of eigenvalues no smaller than the least one.

Radial quadratic trial. The radial problem becomeswhere the printed endpoint notation $y$ refers to $\psi$. Thus $p=w=r^2$ and $q=0$. At the origin the vanishing $r^2$ and regularity of $\psi$ make the boundary term vanish; the weight must be retained in both numerator and denominator.

For a quadratic $\psi =A+Br+C{r}^2$, the derivative condition sets $B=0$ and the outer condition gives $C=-A$. Overall amplitude cancels in the Rayleigh quotient, so take $\tilde{\psi }=1-r^2$. ThenwhileConsequently

Why the estimate is higher. For a regular radial function put $u=r\psi$. Then $u(0)=u(1)=0$, and direct differentiation transforms the equation to $u^{\prime \prime }+\lambda u=0$. Its first mode is $u=\sin (\pi r)$, corresponding to $\psi =\sin (\pi r)/r$ with its continuous value at the origin. Thus ${\lambda }_0={\pi }^2$. The quadratic trial is not proportional to that Spherical Bessel function, so it has nonzero components in higher modes. Its Rayleigh quotient is therefore strictly larger than ${\pi }^2$, in agreement with $21/2>{\pi }^2$.

Assume the regular real Sturm-Liouville problem, with $p,w>0$, ordered eigenvalues ${\lambda }_0<{\lambda }_1<\cdots$, and real orthonormal eigenfunctions in the weighted inner productThe Rayleigh quotient for an admissible trial function isThe second expression follows by integration by parts. It is the quantity used in the Rayleigh-Ritz method.

For $\tilde{y}=y_0+{\sum }_{n\ge 1}{a}_n{y}_n$, orthonormality eliminates every cross-term in the denominator:Applying $\mathcal{L}$ to the eigenfunction expansion and using the same orthogonality in the numerator givesHence, whenever the trial function has finite energy,Put $S=\sum a_n^2$ and $\Delta =\sum a_n^2({\lambda }_n-{\lambda }_0)$. The exact error and its expansion areIn particular, under a common small scaling $a_n=\epsilon b_n$ with fixed finite sums $\sum b_n^2$ and $\sum b_n^2({\lambda }_n-{\lambda }_0)$,The fourth-order remainder includes mixed products $a_n^2{a}_j^2$, not just individual fourth powers.

Error comparison and sign. Since $\Vert y_0{\Vert }_w=1$, the relative weighted norm error of the supplied trial function isFor ${\lambda }_0\ne 0$, the relative eigenvalue error is $\Delta /(|{\lambda }_0|(1+S))=O({\epsilon }^2)$. Thus the Rayleigh quotient has a second-order error for a first-order error in the trial eigenfunction, provided the energy-weighted coefficient sum is controlled. Small weighted norm error alone does not control arbitrarily high-frequency contamination, because ${\lambda }_n$ can grow without bound. If ${\lambda }_0=0$, an absolute error must replace the undefined relative one.

Finally every summand of $\Delta$ is nonnegative. Thuswith strict inequality whenever some $a_n\ne 0$. This is the upper-bound property of the Rayleigh-Ritz variational principle for the lowest eigenvalue.

Choose homogeneous Dirichlet boundary conditions, $y(a)=y(b)=0$, and nonzero admissible functions with finite $F$ and positive $G$. Variations obey the same zero endpoint values. At a stationary Rayleigh quotient, set $\lambda =F[y]/G[y]$. ThenThe boundary term vanishes, and the fundamental lemma of the calculus of variations gives the Sturm-Liouville equation $-(p{y}^{\prime }{)}^{\prime }+qy=\lambda wy$. Thus stationary functions are eigenfunctions. Conversely, multiply that equation by $y$ and integrate by parts to obtain $F=\lambda G$; its first variation then vanishes by the same calculation. This proves that stationary values are the corresponding eigenvalues. Homogeneous Neumann boundary conditions also make the displayed boundary terms vanish; general Robin boundary conditions can require a boundary contribution to the quadratic form.

For $p=1$, $q=0$, $w=x$ on $(0,\pi )$, the polynomial trial function $y=x(\pi -x)$ satisfies the endpoint conditions. Its Rayleigh quotient usesso the polynomial estimate is $\boxed{{\lambda }_{\mathrm{poly}}=20/{\pi }^3\simeq 0.64503}$. The trigonometric trial function $y=\sin x$ givesand hence $\boxed{{\lambda }_{\mathrm{trig}}=2/\pi \simeq 0.63662}$. Both are upper bounds on the lowest eigenvalue, by the Rayleigh-Ritz method; neither trial is an exact eigenfunction, so the inequalities are strict.

Their closeness has a useful explanation beyond their similar single-arch shapes. Both trial functions are symmetric under $x\mapsto \pi -x$, forcing ${\int }_0^{\pi }x{y}^{2}dx=(\pi /2){\int }_0^{\pi }{y}^{2}dx$. The reflection-symmetric trials for a linearly weighted Rayleigh quotient therefore reduce this problem to a constant multiple of the ordinary Dirichlet energy quotient. The sine is the best trial within this symmetric class, and the polynomial's value is only a factor $10/{\pi }^2\simeq 1.0132$ larger. The unrestricted lowest eigenfunction can be asymmetric because the weight $x$ is asymmetric.

Apply the Euler-Lagrange equation to the integrand of $F-\lambda G$, namely $p{y}^{\prime 2}-q{y}^2-\lambda r{y}^2$. With fixed zero endpoint values, it gives $(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0$. Multiplying by $y$ and using integration by parts gives $F=\lambda G$, so each stationary Rayleigh quotient $F/G$ equals its eigenvalue.

In the half-line problem, $p=x$, $q=-x$, $r=1$, so the quotient is $R[y]={\int }_0^{\infty }x(y^{\prime 2}+y^2)dx/{\int }_0^{\infty }{y}^{2}dx$. For $y=e^{-\beta x}$, $\beta >0$, its numerator is $({\beta }^2+1)/(4{\beta }^2)$ and denominator $1/(2\beta )$. Hence $R=(\beta +{\beta }^{-1})/2$, minimized at $\beta =1$:For the compactly supported linear trial, $A>0$, the numerator is $A^2/2+A^4/12$ and denominator $A^3/3$. Thus $R=3/(2A)+A/4$, minimized at $A=\sqrt{6}$, givingThe derivative jump at the cutoff is allowed in the quadratic-form trial space of the Rayleigh-Ritz method: the trial is continuous and has finite form energy.

The exponential ground state of a linearly weighted Sturm-Liouville problem proves which estimate is best. Vanishing endpoint terms giveEquality occurs for $y=C{e}^{-x}$, which also satisfies the differential equation with $\lambda =1$. Thus the exponential trial gives the exact smallest eigenvalue, whereas the linear trial gives a larger upper bound. Finiteness at the singular endpoint supplies the regular branch with $xy{y}^{\prime }\to 0$; all the displayed trials have the required vanishing boundary terms.

Write $N[\psi ]={\int }_a^b(p{\psi }^{\prime 2}+q{\psi }^2)dx$ and $D[\psi ]={\int }_a^{b}w{\psi }^{2}dx>0$. For a real variation, with $\Lambda =N/D$,Suitable endpoint conditions are homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions $p{\psi }^{\prime }=0$, or one of each. They remove the variation boundary term and the product $[p\psi {\psi }^{\prime }{]}_a^b$. Stationarity for all interior variations is equivalent toConversely, multiplication of an eigenfunction equation by $\psi$ and integration by parts gives $N=\lambda D$, hence $\boxed{\Lambda [\psi ]=\lambda }$. General Robin data require their boundary-energy contributions unless these endpoint products happen to vanish.

Use an eigenbasis orthonormal with respect to the weight: ${\int }_a^{b}w{\psi }_i^{\ast }{\psi }_{j}dx={\delta }_{ij}$. For complex coefficients use the Hermitian version of the quotient, with absolute squares. Self-adjointness givesWhen $c_0\ne 0$, dividing numerator and denominator by $|c_0{|}^2$ yieldsIf $c_0=0$, these ratios are undefined, but the unscaled formula and upper-bound conclusion remain valid. The assumptions are a problem with a self-adjoint operator bounded below and a complete eigenbasis, with the trial function in its energy-form domain. The Rayleigh-Ritz method minimizes this quotient over an admissible trial family, obtaining an upper bound on ${\lambda }_0$. Higher eigenfunctions are stationary directions and need not be unrestricted extrema.

Let $R[y]=F[y]/G[y]$, where $G[y]>0$ for a nonzero admissible real function. Writing $\lambda =R[y]$, its first variation isChoose boundary conditions making the boundary term vanish, for example homogeneous Dirichlet boundary conditions, homogeneous Neumann boundary conditions, or a corresponding mixed pair. Arbitrary interior variations then give the Sturm-Liouville problemMultiplying this equation by $y$ and integrating by parts, with $[py{y}^{\prime }{]}_{\alpha }^{\beta }=0$, yieldsThus eigenfunctions are stationary directions of the quotient. The lowest eigenvalue is its global minimum; higher eigenfunctions need not be unrestricted maxima or minima. General Robin boundary conditions require including their boundary energy in the quotient unless the displayed endpoint products also vanish.

The Rayleigh-Ritz method chooses admissible trial functions depending on finitely many coefficients and minimizes this quotient over them. Expansion in a complete orthonormal basis with weight $w$ shows $R[y]\ge {\lambda }_0$, so the trial minimum is an upper bound on the true lowest eigenvalue. In a finite basis, the minimization becomes a generalized eigenvalue problem for the stiffness and weighted mass matrices. Enlarging the trial space can only lower or preserve this upper bound.

Self-adjoint form. Grouping the two derivative terms gives $[(1+ϵx)u^{\prime }{]}^{\prime }-ϵu=-\lambda u$. Thus a regular Sturm-Liouville operator isThe boundary conditions are mixed Neumann boundary conditions at zero and Dirichlet boundary conditions at $\pi$. For $ϵ\ge 0$, $p>0$, $q\ge 0$ and $w>0$, as required. The boundary term $[p(f{g}^{\prime }-f^{\prime }g){]}_0^{\pi }$ vanishes for two functions satisfying these conditions, confirming self-adjointness on that domain.

Constrained variation. Introduce a Lagrange multiplier $\Lambda$ for the normalization and varyIts first variation isThe endpoint term vanishes because $v^{\prime }(0)=0$ and $\delta v(\pi )=0$. The Euler-Lagrange equation is thereforewhich is exactly the original eigenvalue problem for the boxed choices of $p,q,w$. Multiply by $v$ and integrate by parts. The same endpoint conditions giveThus the stationary functional values are the eigenvalues, with $\Lambda =\lambda$. More generally, before normalization the relevant Rayleigh quotient is

Unperturbed ground mode. At $ϵ=0$, the equation is $-u^{\prime \prime }=\lambda u$. The quadratic form is nonnegative, and zero eigenvalue would force a constant function incompatible with the endpoint at $\pi$. For positive $\lambda$, $u=A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. The derivative condition forces $B=0$, while the other endpoint gives $\cos (\pi \sqrt{\lambda })=0$. HenceThe lowest eigenvalue and normalized eigenfunction are

Trial-function bound. The Rayleigh-Ritz method says that the minimum eigenvalue is no larger than the quotient of any admissible trial function. The same $U_0$ still satisfies both boundary conditions and retains unit norm. Its derivative is $U_0^{\prime }=-\sqrt{2/\pi }\sin (x/2)/2$, givingUsing ${\sin }^2(x/2)=(1-\cos x)/2$,Therefore the affine-parameter Sturm-Liouville eigenvalue bound isThis is an upper bound for every $ϵ\ge 0$, not merely a small-parameter approximation. As a lower-bound check, $p\ge 1$ and the unperturbed variational minimum imply ${\mathcal{R}}_{ϵ}[v]\ge 1/4+ϵ$ for every admissible normalized $v$. Thus

Enlarge the trial space while retaining the optimized Gaussian. For example, usenormalize it and minimize the Rayleigh quotient over its coefficients, optionally optimizing $\alpha$ too. This even trial space respects the even ground-state parity of the confining symmetric potential and adds independent shape freedom beyond a Gaussian width.

The original Gaussian remains included by setting the later coefficients to zero, so the optimized upper bound cannot increase. Finite-basis optimization is the Rayleigh-Ritz method, with matrix entries given by the Hamiltonian and overlap integrals. Increasing the basis size systematically improves the available approximation.

The added sine vanishes at zero. Its derivative at one is $B\cos B$, so nontrivial admissible choices areNegative values give the same trial directions after changing the sign of $C$; $B=0$ adds nothing. For each choice, minimize the Rayleigh quotient of $x{e}^{-x}+C\sin (Bx)$ over $C$, or optimize over several coefficients simultaneously. The original trial is included, so the best bound is no larger than $5/4$. Enlarging the trial subspace successively is the Rayleigh-Ritz method; the boundary-compatible half-integer sines provide additional independent shapes.

For the added trial term $q_n=x^n{e}^{-Ax}$, the left Dirichlet boundary condition is automatic for a positive integer $n$. At the right endpoint,To preserve $y^{\prime }(1)=0$ for an arbitrary variational coefficient $C$, choose $\boxed{A=n}$. Then minimize the Rayleigh quotient of $x{e}^{-x}+C{x}^n{e}^{-nx}$ over $C$.

Since $C=0$ remains available, this optimization cannot worsen the original upper bound. For $n=1$, the added function is a multiple of the first trial, so the quotient does not change: an independent choice $n\ge 2$ is needed for extra freedom. More generally, several such functions form the exponential-polynomial trial space for a mixed Sturm-Liouville problem, and the Rayleigh-Ritz method reduces minimization to a small generalized eigenvalue problem.

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

The Rayleigh-Ritz method minimizes an energy functional over a finite-dimensional trial space. For a real Hilbert space $V$, a symmetric coercive bilinear form $a$ and a continuous linear functional $\ell$, takeWriting $u_h={\sum }_k{c}_k{\varphi }_k$ in a trial basis and differentiating with respect to each coefficient givesThe same equations follow from testing the variational problem against each ${\varphi }_j$, which is the Ritz-Galerkin equivalence for a symmetric coercive form. Nonsymmetric problems need a separate argument because differentiating $a(v,v)$ sees only its symmetric part.

For $-u^{\prime \prime }=1$ on $(0,1)$ with zero Dirichlet boundary conditions, choose $a(v,w)={\int }_0^1{v}^{\prime }{w}^{\prime }dx$, $\ell (v)={\int }_0^{1}vdx$ and the one-dimensional trial space spanned by $\varphi =x(1-x)$. Then $K=1/3$ and $F=1/6$, giving $u_h=x(1-x)/2$. Here the exact solution lies in the trial space, so minimization recovers it exactly; in general the Céa lemma controls the approximation error.

For a nonzero trial function with $y(0)=y(L)=0$ and $y(L-x)=y(x)$, substituting $L-x$ for $x$ proves ${\int }_0^{L}x{y}^{2}dx=(L/2){\int }_0^L{y}^{2}dx$. This gives the displayed Rayleigh quotient. The Poincaré inequality on an interval bounds it below by $2{\pi }^2/L^3$, with equality for a nonzero multiple of $\sin (\pi x/L)$. Thus this sine is the best reflection-symmetric trial for $-y^{\prime \prime }=\lambda xy$, while the unrestricted lowest eigenfunction need not be reflection-symmetric. The polynomial $x(L-x)$ gives $20/L^3$, only a factor $10/{\pi }^2$ above the sine's value. Both are upper bounds on the unrestricted lowest eigenvalue by the Rayleigh-Ritz method.

For $-{\psi }^{\prime \prime }+\cos (\pi x)\psi =\lambda \psi$ on $(0,1)$ with zero Dirichlet boundary conditions, use $\psi =\sin (\pi x)+\alpha \sin (2\pi x)$. Its Rayleigh quotient is $({\pi }^2+\alpha +4{\pi }^2{\alpha }^2)/(1+{\alpha }^2)$. In the normalized sine basis the restricted operator has matrix $\left(\begin{array}{cc}{\pi }^2& 1/2\\ 1/2& 4{\pi }^2\end{array}\right)$. Its smaller eigenvalue gives the displayed upper bound, attained within this trial space at $\alpha =3{\pi }^2-\sqrt{9{\pi }^4+1}$. The other stationary parameter gives the larger restricted eigenvalue. This is a finite-dimensional Rayleigh-Ritz method for a regular Sturm-Liouville problem with a sign-changing potential.

The variational method estimates the lowest energy of a Quantum Hamiltonian by choosing a trial wavefunction, evaluating its energy, and optimizing the trial parameters. It uses the quantum variational principle, the Rayleigh-Ritz variational principle expressed for quantum states. Let $H$ be a self-adjoint operator bounded below, and define $E_0=\inf \sigma (H)$, the bottom of its spectrum. Every normalized admissible trial quantum state satisfiesWhen $E_0$ is an energy eigenvalue, it is the ground-state energy; equality holds precisely for a quantum state in the ground-state subspace. Otherwise the bottom of the spectrum need not be attained by any normalizable wavefunction.

Admissible means finite form energy. Choose a real $b\le E_0$ and use the shifted quadratic form of a positive quantum Hamiltonian:The form domain $Q(H)$ is independent of this choice of $b$ and can be larger than the operator domain $D(H)$. For $\psi \in D(H)$, $E_{\psi }=⟨\psi ,H\psi ⟩$ is the usual expectation value; for nonzero unnormalized trials, divide $q_H[\psi ]$ by $\Vert \psi {\Vert }^2$ to get the form version of the Rayleigh quotient. See the quantum variational principle for the general domain justification.

Thus each trial energy is an upper bound on $E_0$, and minimizing within a trial family improves that upper bound. It is not a lower bound on $E_0$. The Gaussian variational energy for the harmonic oscillator reaches $E_0=1/2$ because its family contains an exact ground state. The exact variational solution in a two-level system makes the equality condition visible as vanishing excited-state probability amplitude.

The method is useful when solving the Schrödinger equation exactly is difficult but integrals of a trial wavefunction are tractable. Enlarging a trial family cannot increase its optimal energy, although an optimum need not be attained. The beauty is that even an imperfect trial gives a rigorous bound, provided its form energy is computed exactly: numerical integration errors require separate control. The Gaussian variational estimate for a solvable sextic potential provides an existing example beyond a quantum harmonic oscillator.

A short proof explains the direction. With a complete orthonormal eigenbasis of energy eigenstates, write $\psi ={\sum }_n{c}_n{\varphi }_n$, where $H{\varphi }_n=E_n{\varphi }_n$ and ${\sum }_n|c_n{|}^2=1$. ThenThis is a weighted mean of nonnegative energy differences; equality requires every occupied energy eigenstate to have energy $E_0$. ^[ref]. For continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure of $\psi$, with the same nonnegative integrand. This proves the bound without assuming an orthonormal eigenbasis exists.

The historical setting predates quantum mechanics: Rayleigh's The Theory of Sound, volume I, appeared in 1877, and Ritz's paper on solving variational problems of mathematical physics appeared in volume 135 of Crelle's journal, catalogued by its publisher as 1909, pp. 1–61. These are original works associated with the Rayleigh-Ritz method; the dates here identify the publications rather than assert priority. ^[ref]; ^[ref].

A variational principle specifies a function, path or field by requiring a functional to be stationary under an allowed class of variations. The admissible class includes the prescribed boundary conditions and constraints. A stationary value need not be a minimum: the Euler-Lagrange equation expresses the first-order stationarity condition, while a minimum requires additional analysis. Fermat principle applies this idea to optical path length, and the Rayleigh-Ritz method approximates a variational problem in a chosen family of trial functions.